idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
|---|---|---|---|---|---|---|
39,901 | Regression as mutual information minimization | You're proposing to minimize the mutual information between the residuals and the inputs. Minimal information between the residuals and inputs is indeed a desirable property. But, it's not generally sufficient for regression because it doesn't force the residuals to be small. For example, note that any constant value c... | Regression as mutual information minimization | You're proposing to minimize the mutual information between the residuals and the inputs. Minimal information between the residuals and inputs is indeed a desirable property. But, it's not generally s | Regression as mutual information minimization
You're proposing to minimize the mutual information between the residuals and the inputs. Minimal information between the residuals and inputs is indeed a desirable property. But, it's not generally sufficient for regression because it doesn't force the residuals to be smal... | Regression as mutual information minimization
You're proposing to minimize the mutual information between the residuals and the inputs. Minimal information between the residuals and inputs is indeed a desirable property. But, it's not generally s |
39,902 | Reproducing t-test in R gives different result than built-in function | You are making mistake in calculating your degree of freedom.
Here is the code, that exactly reproduce the R t.test results.
a <- c(5.36, 16.57, 0.62, 1.41, 0.64, 7.26)
b <- c(19.12, 3.52, 3.38, 2.5, 3.6, 1.74)
v1 <- var(a)
v2 <- var(b)
n1 <- length(a)
n2 <- length(b)
se <- sqrt(v1/n1 + v2/n2)
nu <- se^4 / ((v1^2... | Reproducing t-test in R gives different result than built-in function | You are making mistake in calculating your degree of freedom.
Here is the code, that exactly reproduce the R t.test results.
a <- c(5.36, 16.57, 0.62, 1.41, 0.64, 7.26)
b <- c(19.12, 3.52, 3.38, 2.5 | Reproducing t-test in R gives different result than built-in function
You are making mistake in calculating your degree of freedom.
Here is the code, that exactly reproduce the R t.test results.
a <- c(5.36, 16.57, 0.62, 1.41, 0.64, 7.26)
b <- c(19.12, 3.52, 3.38, 2.5, 3.6, 1.74)
v1 <- var(a)
v2 <- var(b)
n1 <- len... | Reproducing t-test in R gives different result than built-in function
You are making mistake in calculating your degree of freedom.
Here is the code, that exactly reproduce the R t.test results.
a <- c(5.36, 16.57, 0.62, 1.41, 0.64, 7.26)
b <- c(19.12, 3.52, 3.38, 2.5 |
39,903 | Difference between using cv=5 or cv=KFold(n_splits=5) in cross_val_score()? [closed] | When an integer is passed to the cv parameter of cross_val_score():
StratifiedKFold is used if the estimator is a classifier and y is either binary or multiclass.
In all other cases, KFold is used.
from sklearn import datasets
from sklearn.tree import DecisionTreeClassifier
from sklearn.model_selection import cross_... | Difference between using cv=5 or cv=KFold(n_splits=5) in cross_val_score()? [closed] | When an integer is passed to the cv parameter of cross_val_score():
StratifiedKFold is used if the estimator is a classifier and y is either binary or multiclass.
In all other cases, KFold is used.
| Difference between using cv=5 or cv=KFold(n_splits=5) in cross_val_score()? [closed]
When an integer is passed to the cv parameter of cross_val_score():
StratifiedKFold is used if the estimator is a classifier and y is either binary or multiclass.
In all other cases, KFold is used.
from sklearn import datasets
from ... | Difference between using cv=5 or cv=KFold(n_splits=5) in cross_val_score()? [closed]
When an integer is passed to the cv parameter of cross_val_score():
StratifiedKFold is used if the estimator is a classifier and y is either binary or multiclass.
In all other cases, KFold is used.
|
39,904 | Confusion on hinge loss and SVM | Searching for the quoted text, it seems the book is Data Science for Business (Provost and Fawcett), and they're describing the soft-margin SVM. Their description of the hinge loss is wrong. The problem is that it doesn't penalize misclassified points that lie within the margin, as you mentioned.
In SVMs, smaller weigh... | Confusion on hinge loss and SVM | Searching for the quoted text, it seems the book is Data Science for Business (Provost and Fawcett), and they're describing the soft-margin SVM. Their description of the hinge loss is wrong. The probl | Confusion on hinge loss and SVM
Searching for the quoted text, it seems the book is Data Science for Business (Provost and Fawcett), and they're describing the soft-margin SVM. Their description of the hinge loss is wrong. The problem is that it doesn't penalize misclassified points that lie within the margin, as you m... | Confusion on hinge loss and SVM
Searching for the quoted text, it seems the book is Data Science for Business (Provost and Fawcett), and they're describing the soft-margin SVM. Their description of the hinge loss is wrong. The probl |
39,905 | Confusion on hinge loss and SVM | The (A) hinge function can be expressed as
$$y_{i} = \gamma \max{\left(x_{i}-\theta, 0\right)} + \varepsilon_{i},$$
where:
$\gamma$ is the change in slope after the hinge. In your example, this amounts to the slope following the hinge, since your hinge-only model (see below) assumes zero effect of $x$ on $y$ until th... | Confusion on hinge loss and SVM | The (A) hinge function can be expressed as
$$y_{i} = \gamma \max{\left(x_{i}-\theta, 0\right)} + \varepsilon_{i},$$
where:
$\gamma$ is the change in slope after the hinge. In your example, this amou | Confusion on hinge loss and SVM
The (A) hinge function can be expressed as
$$y_{i} = \gamma \max{\left(x_{i}-\theta, 0\right)} + \varepsilon_{i},$$
where:
$\gamma$ is the change in slope after the hinge. In your example, this amounts to the slope following the hinge, since your hinge-only model (see below) assumes ze... | Confusion on hinge loss and SVM
The (A) hinge function can be expressed as
$$y_{i} = \gamma \max{\left(x_{i}-\theta, 0\right)} + \varepsilon_{i},$$
where:
$\gamma$ is the change in slope after the hinge. In your example, this amou |
39,906 | MLE: Marginal vs Full Likelihood | The usual way of doing likelihood inference on a parameter of interest in the presence of nuisance parameters consists of using the Profile Likelihood function (see this link). In your context, the profile likelihood is:
$$\mathcal{L}_P(\theta_1;\mathbf{x}) = \max_{\theta_2,\dots,\theta_n} \mathcal{L(\boldsymbol{\theta... | MLE: Marginal vs Full Likelihood | The usual way of doing likelihood inference on a parameter of interest in the presence of nuisance parameters consists of using the Profile Likelihood function (see this link). In your context, the pr | MLE: Marginal vs Full Likelihood
The usual way of doing likelihood inference on a parameter of interest in the presence of nuisance parameters consists of using the Profile Likelihood function (see this link). In your context, the profile likelihood is:
$$\mathcal{L}_P(\theta_1;\mathbf{x}) = \max_{\theta_2,\dots,\theta... | MLE: Marginal vs Full Likelihood
The usual way of doing likelihood inference on a parameter of interest in the presence of nuisance parameters consists of using the Profile Likelihood function (see this link). In your context, the pr |
39,907 | MLE: Marginal vs Full Likelihood | A general answer is tough and +1 to @Shifer, but if you're looking for one particular example when integrating outperforms profiling then you might find the Neyman-Scott "paradox" interesting.
The problem: suppose we have $\{(x_1,y_1),\dots,(x_n,y_n)\}$ where each pair $(x_i, y_i)$ is independent and
$$
{x_i \choose y_... | MLE: Marginal vs Full Likelihood | A general answer is tough and +1 to @Shifer, but if you're looking for one particular example when integrating outperforms profiling then you might find the Neyman-Scott "paradox" interesting.
The pro | MLE: Marginal vs Full Likelihood
A general answer is tough and +1 to @Shifer, but if you're looking for one particular example when integrating outperforms profiling then you might find the Neyman-Scott "paradox" interesting.
The problem: suppose we have $\{(x_1,y_1),\dots,(x_n,y_n)\}$ where each pair $(x_i, y_i)$ is i... | MLE: Marginal vs Full Likelihood
A general answer is tough and +1 to @Shifer, but if you're looking for one particular example when integrating outperforms profiling then you might find the Neyman-Scott "paradox" interesting.
The pro |
39,908 | What is the CDF of the Shapiro-Wilk $W$ statistic? | There are many tests of normality. Some of them are based on Q-Q plots (normal probability plots) of the data, measuring according to various criteria how "nearly linear" the Q-Q plot is.
Intuitive view of Shapiro-Wilk test. In visual inspection of a Q-Q plot, the extreme values of a truly normal sample can appear to ... | What is the CDF of the Shapiro-Wilk $W$ statistic? | There are many tests of normality. Some of them are based on Q-Q plots (normal probability plots) of the data, measuring according to various criteria how "nearly linear" the Q-Q plot is.
Intuitive v | What is the CDF of the Shapiro-Wilk $W$ statistic?
There are many tests of normality. Some of them are based on Q-Q plots (normal probability plots) of the data, measuring according to various criteria how "nearly linear" the Q-Q plot is.
Intuitive view of Shapiro-Wilk test. In visual inspection of a Q-Q plot, the ext... | What is the CDF of the Shapiro-Wilk $W$ statistic?
There are many tests of normality. Some of them are based on Q-Q plots (normal probability plots) of the data, measuring according to various criteria how "nearly linear" the Q-Q plot is.
Intuitive v |
39,909 | What is the CDF of the Shapiro-Wilk $W$ statistic? | tl;dr: There is no known distribution. i.e. it doesn't have a name. You use simulations instead.
I found this to give some insight:
Basically, Shapiro and Wilk calculated the distribution of their statistic only for $n=3$, to be a truncated $Beta(\frac{1}{2}, \frac{1}{2})$ distribution for $\frac{3}{4} \le w \le 1$ and... | What is the CDF of the Shapiro-Wilk $W$ statistic? | tl;dr: There is no known distribution. i.e. it doesn't have a name. You use simulations instead.
I found this to give some insight:
Basically, Shapiro and Wilk calculated the distribution of their sta | What is the CDF of the Shapiro-Wilk $W$ statistic?
tl;dr: There is no known distribution. i.e. it doesn't have a name. You use simulations instead.
I found this to give some insight:
Basically, Shapiro and Wilk calculated the distribution of their statistic only for $n=3$, to be a truncated $Beta(\frac{1}{2}, \frac{1}{... | What is the CDF of the Shapiro-Wilk $W$ statistic?
tl;dr: There is no known distribution. i.e. it doesn't have a name. You use simulations instead.
I found this to give some insight:
Basically, Shapiro and Wilk calculated the distribution of their sta |
39,910 | ANOVA vs pairwise t-tests with multiple test correction | The difference becomes clear if you understand the null/alternative hypothesis of each test.
ANOVA's null hypothesis is that the group means are the same, while the alternative is that at least one group mean is different from the others. This analysis does not tell you which group mean is different, or which differen... | ANOVA vs pairwise t-tests with multiple test correction | The difference becomes clear if you understand the null/alternative hypothesis of each test.
ANOVA's null hypothesis is that the group means are the same, while the alternative is that at least one gr | ANOVA vs pairwise t-tests with multiple test correction
The difference becomes clear if you understand the null/alternative hypothesis of each test.
ANOVA's null hypothesis is that the group means are the same, while the alternative is that at least one group mean is different from the others. This analysis does not t... | ANOVA vs pairwise t-tests with multiple test correction
The difference becomes clear if you understand the null/alternative hypothesis of each test.
ANOVA's null hypothesis is that the group means are the same, while the alternative is that at least one gr |
39,911 | Pairwise comparisons with emmeans for a mixed three-way interaction in a linear mixed-effects model | It shouldn't be necessary to fit a separate model just to do the post-hoc comparisons you want. You had tried:
emms <- emmeans(fit1b, ~ AB*C)
contrast(emms, interaction = "pairwise")
but you can get the same results from the original model using by variables judiciously:
emms1 <- emmeans(fit1, ~ A*B | C)
con1 <- cont... | Pairwise comparisons with emmeans for a mixed three-way interaction in a linear mixed-effects model | It shouldn't be necessary to fit a separate model just to do the post-hoc comparisons you want. You had tried:
emms <- emmeans(fit1b, ~ AB*C)
contrast(emms, interaction = "pairwise")
but you can get | Pairwise comparisons with emmeans for a mixed three-way interaction in a linear mixed-effects model
It shouldn't be necessary to fit a separate model just to do the post-hoc comparisons you want. You had tried:
emms <- emmeans(fit1b, ~ AB*C)
contrast(emms, interaction = "pairwise")
but you can get the same results fr... | Pairwise comparisons with emmeans for a mixed three-way interaction in a linear mixed-effects model
It shouldn't be necessary to fit a separate model just to do the post-hoc comparisons you want. You had tried:
emms <- emmeans(fit1b, ~ AB*C)
contrast(emms, interaction = "pairwise")
but you can get |
39,912 | Does a density forecast add value beyond a point forecast when the loss function is given? | I can think of one-and-a-half more or less realistic situations where a full density is better than a point forecast, even if the loss function is known.
The nitpicky situation is the one where the user's loss function depends not only on the point forecast, but on a two-sided prediction-interval, or even the entire d... | Does a density forecast add value beyond a point forecast when the loss function is given? | I can think of one-and-a-half more or less realistic situations where a full density is better than a point forecast, even if the loss function is known.
The nitpicky situation is the one where the u | Does a density forecast add value beyond a point forecast when the loss function is given?
I can think of one-and-a-half more or less realistic situations where a full density is better than a point forecast, even if the loss function is known.
The nitpicky situation is the one where the user's loss function depends n... | Does a density forecast add value beyond a point forecast when the loss function is given?
I can think of one-and-a-half more or less realistic situations where a full density is better than a point forecast, even if the loss function is known.
The nitpicky situation is the one where the u |
39,913 | Does a density forecast add value beyond a point forecast when the loss function is given? | Background (may be skipped)
I will be thinking in decision-theoretic terms as follows. A user must choose an action $a$ among a set of possibilities $A$. The action will bring him/her some "utility" (a notion commonly used in economics) $u(a;s)$ depending on the state of nature $s$ that will be realized in the future, ... | Does a density forecast add value beyond a point forecast when the loss function is given? | Background (may be skipped)
I will be thinking in decision-theoretic terms as follows. A user must choose an action $a$ among a set of possibilities $A$. The action will bring him/her some "utility" ( | Does a density forecast add value beyond a point forecast when the loss function is given?
Background (may be skipped)
I will be thinking in decision-theoretic terms as follows. A user must choose an action $a$ among a set of possibilities $A$. The action will bring him/her some "utility" (a notion commonly used in eco... | Does a density forecast add value beyond a point forecast when the loss function is given?
Background (may be skipped)
I will be thinking in decision-theoretic terms as follows. A user must choose an action $a$ among a set of possibilities $A$. The action will bring him/her some "utility" ( |
39,914 | Intuition behind Correlated Terms in Mixed Effect Models | Suppose, for concreteness, that we have a model
y ~ 1 + x1 + x2 + (1|g)
where x1 and x2 are (for simplicity) continuous predictor variables (g is a categorical grouping variable). This model states that the expected value of y changes linearly with changes in x1 and x2 and that there are differences in the intercept ... | Intuition behind Correlated Terms in Mixed Effect Models | Suppose, for concreteness, that we have a model
y ~ 1 + x1 + x2 + (1|g)
where x1 and x2 are (for simplicity) continuous predictor variables (g is a categorical grouping variable). This model states | Intuition behind Correlated Terms in Mixed Effect Models
Suppose, for concreteness, that we have a model
y ~ 1 + x1 + x2 + (1|g)
where x1 and x2 are (for simplicity) continuous predictor variables (g is a categorical grouping variable). This model states that the expected value of y changes linearly with changes in x... | Intuition behind Correlated Terms in Mixed Effect Models
Suppose, for concreteness, that we have a model
y ~ 1 + x1 + x2 + (1|g)
where x1 and x2 are (for simplicity) continuous predictor variables (g is a categorical grouping variable). This model states |
39,915 | Prove that the mean value of a convolution is the sum of the mean values of its individual parts | Like many demonstrations involving convolutions, it comes down to applying Fubini's Theorem.
Let's establish notation and assumptions.
Let $f$ and $g$ be integrable real-valued functions defined on $\mathbb{R}^n$ having unit integrals (with respect to Lebesgue measure): that is,
$$1=\int_{\mathbb{R}^n} f(x) dx = \int_{... | Prove that the mean value of a convolution is the sum of the mean values of its individual parts | Like many demonstrations involving convolutions, it comes down to applying Fubini's Theorem.
Let's establish notation and assumptions.
Let $f$ and $g$ be integrable real-valued functions defined on $\ | Prove that the mean value of a convolution is the sum of the mean values of its individual parts
Like many demonstrations involving convolutions, it comes down to applying Fubini's Theorem.
Let's establish notation and assumptions.
Let $f$ and $g$ be integrable real-valued functions defined on $\mathbb{R}^n$ having uni... | Prove that the mean value of a convolution is the sum of the mean values of its individual parts
Like many demonstrations involving convolutions, it comes down to applying Fubini's Theorem.
Let's establish notation and assumptions.
Let $f$ and $g$ be integrable real-valued functions defined on $\ |
39,916 | Prove that the mean value of a convolution is the sum of the mean values of its individual parts | Convolving distributions corresponds to adding independent random variables. Given PDFs $f_X$ and $f_Y$, let $f_Z = f_X * f_Y$ denote their convolution. $f_Z$ is the PDF of a random variable $Z = X + Y$, where $X \sim f_X$, $Y \sim f_Y$, and $X$ and $Y$ are independent. By linearity of expectation, $E[Z] = E[X+Y] = E[X... | Prove that the mean value of a convolution is the sum of the mean values of its individual parts | Convolving distributions corresponds to adding independent random variables. Given PDFs $f_X$ and $f_Y$, let $f_Z = f_X * f_Y$ denote their convolution. $f_Z$ is the PDF of a random variable $Z = X + | Prove that the mean value of a convolution is the sum of the mean values of its individual parts
Convolving distributions corresponds to adding independent random variables. Given PDFs $f_X$ and $f_Y$, let $f_Z = f_X * f_Y$ denote their convolution. $f_Z$ is the PDF of a random variable $Z = X + Y$, where $X \sim f_X$,... | Prove that the mean value of a convolution is the sum of the mean values of its individual parts
Convolving distributions corresponds to adding independent random variables. Given PDFs $f_X$ and $f_Y$, let $f_Z = f_X * f_Y$ denote their convolution. $f_Z$ is the PDF of a random variable $Z = X + |
39,917 | Is PCG random number generator as good as claimed? | Other people have looked at the statistical qualities of the PCG generators and found them to be good, see for example https://lemire.me/blog/2017/08/22/testing-non-cryptographic-random-number-generators-my-results/. On that page you will also find references to other RNGs, that are all faster than MT. If being publish... | Is PCG random number generator as good as claimed? | Other people have looked at the statistical qualities of the PCG generators and found them to be good, see for example https://lemire.me/blog/2017/08/22/testing-non-cryptographic-random-number-generat | Is PCG random number generator as good as claimed?
Other people have looked at the statistical qualities of the PCG generators and found them to be good, see for example https://lemire.me/blog/2017/08/22/testing-non-cryptographic-random-number-generators-my-results/. On that page you will also find references to other ... | Is PCG random number generator as good as claimed?
Other people have looked at the statistical qualities of the PCG generators and found them to be good, see for example https://lemire.me/blog/2017/08/22/testing-non-cryptographic-random-number-generat |
39,918 | Is PCG random number generator as good as claimed? | PCG generators are more or less fine if you use them single-stream in a single application (albeit they will be very slow if you use a state larger than 64 bits). In all other situations, they should be avoided. You can find a very detailed discussion here: http://prng.di.unimi.it/pcg.php .
As for benchmarking: a PCG32... | Is PCG random number generator as good as claimed? | PCG generators are more or less fine if you use them single-stream in a single application (albeit they will be very slow if you use a state larger than 64 bits). In all other situations, they should | Is PCG random number generator as good as claimed?
PCG generators are more or less fine if you use them single-stream in a single application (albeit they will be very slow if you use a state larger than 64 bits). In all other situations, they should be avoided. You can find a very detailed discussion here: http://prng... | Is PCG random number generator as good as claimed?
PCG generators are more or less fine if you use them single-stream in a single application (albeit they will be very slow if you use a state larger than 64 bits). In all other situations, they should |
39,919 | Compute standard error from beta, p-value, sample size, and the number of regression parameters | In a linear regression, the $p$-value is calculated from a $t$-value, which is the coefficient divided by its standard error ($t=\hat{\beta}/\mathrm{SE}_{\hat{\beta}}$). The degrees of freedom used in the $t$-distribution for calculating the $p$-value are the residual degrees of freedom ($\mathrm{SE}_{\hat{\beta}}=\hat... | Compute standard error from beta, p-value, sample size, and the number of regression parameters | In a linear regression, the $p$-value is calculated from a $t$-value, which is the coefficient divided by its standard error ($t=\hat{\beta}/\mathrm{SE}_{\hat{\beta}}$). The degrees of freedom used in | Compute standard error from beta, p-value, sample size, and the number of regression parameters
In a linear regression, the $p$-value is calculated from a $t$-value, which is the coefficient divided by its standard error ($t=\hat{\beta}/\mathrm{SE}_{\hat{\beta}}$). The degrees of freedom used in the $t$-distribution fo... | Compute standard error from beta, p-value, sample size, and the number of regression parameters
In a linear regression, the $p$-value is calculated from a $t$-value, which is the coefficient divided by its standard error ($t=\hat{\beta}/\mathrm{SE}_{\hat{\beta}}$). The degrees of freedom used in |
39,920 | Geometric distribution with random, varying success probability | Your algorithm independently draws a sequence of probabilities from random variables $P_i$ with distributions $F_i.$ Since these distributions must be bounded between $0$ and $1$ they all have expectations $p_i.$ It then observes a parallel sequence of independent Bernoulli$(P_i)$ variates $X_i,$ stopping the first t... | Geometric distribution with random, varying success probability | Your algorithm independently draws a sequence of probabilities from random variables $P_i$ with distributions $F_i.$ Since these distributions must be bounded between $0$ and $1$ they all have expect | Geometric distribution with random, varying success probability
Your algorithm independently draws a sequence of probabilities from random variables $P_i$ with distributions $F_i.$ Since these distributions must be bounded between $0$ and $1$ they all have expectations $p_i.$ It then observes a parallel sequence of i... | Geometric distribution with random, varying success probability
Your algorithm independently draws a sequence of probabilities from random variables $P_i$ with distributions $F_i.$ Since these distributions must be bounded between $0$ and $1$ they all have expect |
39,921 | Geometric distribution with random, varying success probability | My main question is: Is X still distributed Geometric?
No.
Indeed, the mean of the resulting distribution is not the same as the mean of a geometric with parameter equal to $E(p)$; this is because the mean of the geometric is not linear in $p$. The mean of the resulting distribution would correspond instead to the ha... | Geometric distribution with random, varying success probability | My main question is: Is X still distributed Geometric?
No.
Indeed, the mean of the resulting distribution is not the same as the mean of a geometric with parameter equal to $E(p)$; this is because t | Geometric distribution with random, varying success probability
My main question is: Is X still distributed Geometric?
No.
Indeed, the mean of the resulting distribution is not the same as the mean of a geometric with parameter equal to $E(p)$; this is because the mean of the geometric is not linear in $p$. The mean ... | Geometric distribution with random, varying success probability
My main question is: Is X still distributed Geometric?
No.
Indeed, the mean of the resulting distribution is not the same as the mean of a geometric with parameter equal to $E(p)$; this is because t |
39,922 | Geometric distribution with random, varying success probability | The most common continuous distribution with the random variable with support between 0 and 1 is the beta distribution. When the "p" in the geometric distribution is a beta variable, you will end up with a geometric mixture distribution called "beta-geometric" distribution whose parameters can easily be derived.
Howeve... | Geometric distribution with random, varying success probability | The most common continuous distribution with the random variable with support between 0 and 1 is the beta distribution. When the "p" in the geometric distribution is a beta variable, you will end up w | Geometric distribution with random, varying success probability
The most common continuous distribution with the random variable with support between 0 and 1 is the beta distribution. When the "p" in the geometric distribution is a beta variable, you will end up with a geometric mixture distribution called "beta-geomet... | Geometric distribution with random, varying success probability
The most common continuous distribution with the random variable with support between 0 and 1 is the beta distribution. When the "p" in the geometric distribution is a beta variable, you will end up w |
39,923 | How to extract specific information from text using Machine learning? | The problem you pose here is called named entity recognition (NER), or named entity extraction.
There are multiple technologies (not necessary neural networks) that can be used for this problem, and some of them are quite mature. See e.g. this repo for an easy-to-plug-in solution, or try to apply the ne_chunk_sents fu... | How to extract specific information from text using Machine learning? | The problem you pose here is called named entity recognition (NER), or named entity extraction.
There are multiple technologies (not necessary neural networks) that can be used for this problem, and | How to extract specific information from text using Machine learning?
The problem you pose here is called named entity recognition (NER), or named entity extraction.
There are multiple technologies (not necessary neural networks) that can be used for this problem, and some of them are quite mature. See e.g. this repo ... | How to extract specific information from text using Machine learning?
The problem you pose here is called named entity recognition (NER), or named entity extraction.
There are multiple technologies (not necessary neural networks) that can be used for this problem, and |
39,924 | How to extract specific information from text using Machine learning? | I think you could look into dependency parsing. Your fact tuples could be extracted from edges in depencency graph.
PS1 If you want to do something on NLP you should check cs224n and not cs231n. I also recall cs224 contains a section on DL for dependency parsing.
PS2 The dependency tree is taken from Stanford Neural N... | How to extract specific information from text using Machine learning? | I think you could look into dependency parsing. Your fact tuples could be extracted from edges in depencency graph.
PS1 If you want to do something on NLP you should check cs224n and not cs231n. I al | How to extract specific information from text using Machine learning?
I think you could look into dependency parsing. Your fact tuples could be extracted from edges in depencency graph.
PS1 If you want to do something on NLP you should check cs224n and not cs231n. I also recall cs224 contains a section on DL for depen... | How to extract specific information from text using Machine learning?
I think you could look into dependency parsing. Your fact tuples could be extracted from edges in depencency graph.
PS1 If you want to do something on NLP you should check cs224n and not cs231n. I al |
39,925 | Beta regression and regression diagnostics. Do we need to check for normality and other diagnostics? | Raw residuals will not necessarily be normally distributed. Here are some simulated data, following the Ferrari & Cribari-Neto (2004) reparameterization of the beta distribution:
> set.seed(1839)
> library(betareg)
> inv_logit <- function(logit) exp(logit) / (1 + exp(logit))
> n <- 500
> x <- rnorm(n)
> mu <- inv_logit... | Beta regression and regression diagnostics. Do we need to check for normality and other diagnostics? | Raw residuals will not necessarily be normally distributed. Here are some simulated data, following the Ferrari & Cribari-Neto (2004) reparameterization of the beta distribution:
> set.seed(1839)
> li | Beta regression and regression diagnostics. Do we need to check for normality and other diagnostics?
Raw residuals will not necessarily be normally distributed. Here are some simulated data, following the Ferrari & Cribari-Neto (2004) reparameterization of the beta distribution:
> set.seed(1839)
> library(betareg)
> in... | Beta regression and regression diagnostics. Do we need to check for normality and other diagnostics?
Raw residuals will not necessarily be normally distributed. Here are some simulated data, following the Ferrari & Cribari-Neto (2004) reparameterization of the beta distribution:
> set.seed(1839)
> li |
39,926 | Linear regression, independent variable stationarity | What you assume in a linear regression model is that the error term is a white noise process and, therefore, it must be stationary. There is no assumption that either the independent or dependant variables are stationary.
However, consider the following simple linear regression model for time series data:
$$Y_t = a + b... | Linear regression, independent variable stationarity | What you assume in a linear regression model is that the error term is a white noise process and, therefore, it must be stationary. There is no assumption that either the independent or dependant vari | Linear regression, independent variable stationarity
What you assume in a linear regression model is that the error term is a white noise process and, therefore, it must be stationary. There is no assumption that either the independent or dependant variables are stationary.
However, consider the following simple linear... | Linear regression, independent variable stationarity
What you assume in a linear regression model is that the error term is a white noise process and, therefore, it must be stationary. There is no assumption that either the independent or dependant vari |
39,927 | What if Markov chain does not converge in a reasonable amount of time? | To answer your original question
I am not a huge fan of using Gelman-Rubin mainly because it is somewhat handwavy for my taste. However, if you still want to use it, maybe try a multivariate Gelman-Rubin since it is possible the joint posterior of the weights have a complicated dependence structure that the univariat... | What if Markov chain does not converge in a reasonable amount of time? | To answer your original question
I am not a huge fan of using Gelman-Rubin mainly because it is somewhat handwavy for my taste. However, if you still want to use it, maybe try a multivariate Gelman- | What if Markov chain does not converge in a reasonable amount of time?
To answer your original question
I am not a huge fan of using Gelman-Rubin mainly because it is somewhat handwavy for my taste. However, if you still want to use it, maybe try a multivariate Gelman-Rubin since it is possible the joint posterior of... | What if Markov chain does not converge in a reasonable amount of time?
To answer your original question
I am not a huge fan of using Gelman-Rubin mainly because it is somewhat handwavy for my taste. However, if you still want to use it, maybe try a multivariate Gelman- |
39,928 | What if Markov chain does not converge in a reasonable amount of time? | This was touched on by @Greenparker but the easiest way to speed a slowly converging model is to pick the right starting values. This is counterintuitive, as if you knew the right parameter values in the first place why would you fit the model. But, if you can fit a similar frequentist model quickly, use those paramete... | What if Markov chain does not converge in a reasonable amount of time? | This was touched on by @Greenparker but the easiest way to speed a slowly converging model is to pick the right starting values. This is counterintuitive, as if you knew the right parameter values in | What if Markov chain does not converge in a reasonable amount of time?
This was touched on by @Greenparker but the easiest way to speed a slowly converging model is to pick the right starting values. This is counterintuitive, as if you knew the right parameter values in the first place why would you fit the model. But,... | What if Markov chain does not converge in a reasonable amount of time?
This was touched on by @Greenparker but the easiest way to speed a slowly converging model is to pick the right starting values. This is counterintuitive, as if you knew the right parameter values in |
39,929 | When does deep learning fail? | One can generally think of two types of hardness results in machine learning: Information theoretic hardness in the context of statistical learning (namely, giving a lower bound to the minimal number of examples required to learn) and algorithmic hardness (i.e, a bad algorithmic choice means that the optimization beco... | When does deep learning fail? | One can generally think of two types of hardness results in machine learning: Information theoretic hardness in the context of statistical learning (namely, giving a lower bound to the minimal number | When does deep learning fail?
One can generally think of two types of hardness results in machine learning: Information theoretic hardness in the context of statistical learning (namely, giving a lower bound to the minimal number of examples required to learn) and algorithmic hardness (i.e, a bad algorithmic choice me... | When does deep learning fail?
One can generally think of two types of hardness results in machine learning: Information theoretic hardness in the context of statistical learning (namely, giving a lower bound to the minimal number |
39,930 | When does deep learning fail? | It fails when you don't impose the right structure on the problem. Convolutional neural networks work because they assume pixels which are close to each other are related, so it makes sense to apply spatial convolutions to your features. And in so doing, you've reduced the hypothesis search space dramatically, which me... | When does deep learning fail? | It fails when you don't impose the right structure on the problem. Convolutional neural networks work because they assume pixels which are close to each other are related, so it makes sense to apply s | When does deep learning fail?
It fails when you don't impose the right structure on the problem. Convolutional neural networks work because they assume pixels which are close to each other are related, so it makes sense to apply spatial convolutions to your features. And in so doing, you've reduced the hypothesis searc... | When does deep learning fail?
It fails when you don't impose the right structure on the problem. Convolutional neural networks work because they assume pixels which are close to each other are related, so it makes sense to apply s |
39,931 | Cook's distance vs. hat values | The cook's distance is given by the formula:
$D_{i} = \frac{\sum_{j = 1}^{n} (\hat Y_j - \hat Y_{j(i)})^2}{pMSE}$
Where:
$\hat Y_j$ is the fitted value for the j observation;
$ \hat Y_{j(i)}$ is the fitted value for the j observation without including the i-th observation in the data that will generate the model;
p... | Cook's distance vs. hat values | The cook's distance is given by the formula:
$D_{i} = \frac{\sum_{j = 1}^{n} (\hat Y_j - \hat Y_{j(i)})^2}{pMSE}$
Where:
$\hat Y_j$ is the fitted value for the j observation;
$ \hat Y_{j(i)}$ is th | Cook's distance vs. hat values
The cook's distance is given by the formula:
$D_{i} = \frac{\sum_{j = 1}^{n} (\hat Y_j - \hat Y_{j(i)})^2}{pMSE}$
Where:
$\hat Y_j$ is the fitted value for the j observation;
$ \hat Y_{j(i)}$ is the fitted value for the j observation without including the i-th observation in the data t... | Cook's distance vs. hat values
The cook's distance is given by the formula:
$D_{i} = \frac{\sum_{j = 1}^{n} (\hat Y_j - \hat Y_{j(i)})^2}{pMSE}$
Where:
$\hat Y_j$ is the fitted value for the j observation;
$ \hat Y_{j(i)}$ is th |
39,932 | Cook's distance vs. hat values | Cooks distance shows how much the whole regression model would change if $(x_i, y_i)$ is removed. I am not quite clear what you mean by "hat value". Do you mean $e_i = y_i - \hat{y}_i$, or $h_{ii}$ in the hat matrix $H$ (i.e. leverage)? Either way they are different from cooks distance.
Note that cooks distance takes t... | Cook's distance vs. hat values | Cooks distance shows how much the whole regression model would change if $(x_i, y_i)$ is removed. I am not quite clear what you mean by "hat value". Do you mean $e_i = y_i - \hat{y}_i$, or $h_{ii}$ in | Cook's distance vs. hat values
Cooks distance shows how much the whole regression model would change if $(x_i, y_i)$ is removed. I am not quite clear what you mean by "hat value". Do you mean $e_i = y_i - \hat{y}_i$, or $h_{ii}$ in the hat matrix $H$ (i.e. leverage)? Either way they are different from cooks distance.
N... | Cook's distance vs. hat values
Cooks distance shows how much the whole regression model would change if $(x_i, y_i)$ is removed. I am not quite clear what you mean by "hat value". Do you mean $e_i = y_i - \hat{y}_i$, or $h_{ii}$ in |
39,933 | Classification algorithm based on average distances from a test point to the points in each class | It is a nice idea, but has one major flaw - it is too sensitive to the spread of the data.
To clarify the question, given $k$ disjoint clusters $ C_1, \ldots, C_k $, you ask whether it makes sense to classify a new sample $ x^* $ according to the rule $$ \arg\min_{i\in \left[k\right]} \frac{1}{\left| C_i \right|} \sum... | Classification algorithm based on average distances from a test point to the points in each class | It is a nice idea, but has one major flaw - it is too sensitive to the spread of the data.
To clarify the question, given $k$ disjoint clusters $ C_1, \ldots, C_k $, you ask whether it makes sense to | Classification algorithm based on average distances from a test point to the points in each class
It is a nice idea, but has one major flaw - it is too sensitive to the spread of the data.
To clarify the question, given $k$ disjoint clusters $ C_1, \ldots, C_k $, you ask whether it makes sense to classify a new sample ... | Classification algorithm based on average distances from a test point to the points in each class
It is a nice idea, but has one major flaw - it is too sensitive to the spread of the data.
To clarify the question, given $k$ disjoint clusters $ C_1, \ldots, C_k $, you ask whether it makes sense to |
39,934 | Can you calculate $R^2$ from correlation coefficents in multiple linear regression? | The coefficient-of-determination can be determined from the correlations: Consider a multiple linear regression with $m$ explanatory vectors and an intercept term. First we define the correlation values for all the variables in the problem $r_i = \mathbb{Corr}(\mathbf{y},\mathbf{x}_i)$ and $r_{i,j} = \mathbb{Corr}(\ma... | Can you calculate $R^2$ from correlation coefficents in multiple linear regression? | The coefficient-of-determination can be determined from the correlations: Consider a multiple linear regression with $m$ explanatory vectors and an intercept term. First we define the correlation val | Can you calculate $R^2$ from correlation coefficents in multiple linear regression?
The coefficient-of-determination can be determined from the correlations: Consider a multiple linear regression with $m$ explanatory vectors and an intercept term. First we define the correlation values for all the variables in the pro... | Can you calculate $R^2$ from correlation coefficents in multiple linear regression?
The coefficient-of-determination can be determined from the correlations: Consider a multiple linear regression with $m$ explanatory vectors and an intercept term. First we define the correlation val |
39,935 | Can you calculate $R^2$ from correlation coefficents in multiple linear regression? | If you have both the absolute errors from the regression and the dependent variable values, you can calculate $R^2$ as:
$$1 - ({\rm absolute\ error\ variance} / {\rm dependent\ variable\ variance})$$
Using the above, the number of independent variables or their correlations is not needed, which is quite handy for mult... | Can you calculate $R^2$ from correlation coefficents in multiple linear regression? | If you have both the absolute errors from the regression and the dependent variable values, you can calculate $R^2$ as:
$$1 - ({\rm absolute\ error\ variance} / {\rm dependent\ variable\ variance})$$ | Can you calculate $R^2$ from correlation coefficents in multiple linear regression?
If you have both the absolute errors from the regression and the dependent variable values, you can calculate $R^2$ as:
$$1 - ({\rm absolute\ error\ variance} / {\rm dependent\ variable\ variance})$$
Using the above, the number of inde... | Can you calculate $R^2$ from correlation coefficents in multiple linear regression?
If you have both the absolute errors from the regression and the dependent variable values, you can calculate $R^2$ as:
$$1 - ({\rm absolute\ error\ variance} / {\rm dependent\ variable\ variance})$$ |
39,936 | How do I perform a statistical test for a Difference-in-Differences analysis? | The difference in differences is what is called an interaction in statistics (as Dimitriy Masterov has already pointed out). You want to test whether the time effect is different when you intervene compared with when you don't.
Your data is most naturally modelled as binomial, i.e., the number of top scores out of tota... | How do I perform a statistical test for a Difference-in-Differences analysis? | The difference in differences is what is called an interaction in statistics (as Dimitriy Masterov has already pointed out). You want to test whether the time effect is different when you intervene co | How do I perform a statistical test for a Difference-in-Differences analysis?
The difference in differences is what is called an interaction in statistics (as Dimitriy Masterov has already pointed out). You want to test whether the time effect is different when you intervene compared with when you don't.
Your data is m... | How do I perform a statistical test for a Difference-in-Differences analysis?
The difference in differences is what is called an interaction in statistics (as Dimitriy Masterov has already pointed out). You want to test whether the time effect is different when you intervene co |
39,937 | How do I perform a statistical test for a Difference-in-Differences analysis? | The easiest solution is to use the linear regression formulation of DID:
Regress the binary customer ratings on a constant, a post dummy, an area A dummy, and the interaction of last two. It may be appropriate to add other regressors measuring characteristics that are time-invariant at the individual level, but whose ... | How do I perform a statistical test for a Difference-in-Differences analysis? | The easiest solution is to use the linear regression formulation of DID:
Regress the binary customer ratings on a constant, a post dummy, an area A dummy, and the interaction of last two. It may be a | How do I perform a statistical test for a Difference-in-Differences analysis?
The easiest solution is to use the linear regression formulation of DID:
Regress the binary customer ratings on a constant, a post dummy, an area A dummy, and the interaction of last two. It may be appropriate to add other regressors measuri... | How do I perform a statistical test for a Difference-in-Differences analysis?
The easiest solution is to use the linear regression formulation of DID:
Regress the binary customer ratings on a constant, a post dummy, an area A dummy, and the interaction of last two. It may be a |
39,938 | How do I perform a statistical test for a Difference-in-Differences analysis? | Thanks for the question and response.
I have the exact same issue, just that instead of 4 variables i have 8 variables as below:
N_for_KPI <- c(683,538,2225,1458,294,307,922,781)
N <- c(1951,1564,5683,4507,819,862,2479,2511)
Wave <- factor(c("A","A","B","B","C","C"))
Brand <- factor(c(0,1,0,1,0,1))
data = data.frame(N_... | How do I perform a statistical test for a Difference-in-Differences analysis? | Thanks for the question and response.
I have the exact same issue, just that instead of 4 variables i have 8 variables as below:
N_for_KPI <- c(683,538,2225,1458,294,307,922,781)
N <- c(1951,1564,5683 | How do I perform a statistical test for a Difference-in-Differences analysis?
Thanks for the question and response.
I have the exact same issue, just that instead of 4 variables i have 8 variables as below:
N_for_KPI <- c(683,538,2225,1458,294,307,922,781)
N <- c(1951,1564,5683,4507,819,862,2479,2511)
Wave <- factor(c(... | How do I perform a statistical test for a Difference-in-Differences analysis?
Thanks for the question and response.
I have the exact same issue, just that instead of 4 variables i have 8 variables as below:
N_for_KPI <- c(683,538,2225,1458,294,307,922,781)
N <- c(1951,1564,5683 |
39,939 | Support vector machine margin term, why norm squared? | As far as I know, the square is introduced in the formulation for convenience. The norm will reach the optimum at the same point, and we get rid of an ugly square root.
With respect to the hinge loss term, the square just makes no difference either, because of the presence of $\lambda$. Both $f(x)=\|x\|$ and $g(x)=\|x\... | Support vector machine margin term, why norm squared? | As far as I know, the square is introduced in the formulation for convenience. The norm will reach the optimum at the same point, and we get rid of an ugly square root.
With respect to the hinge loss | Support vector machine margin term, why norm squared?
As far as I know, the square is introduced in the formulation for convenience. The norm will reach the optimum at the same point, and we get rid of an ugly square root.
With respect to the hinge loss term, the square just makes no difference either, because of the p... | Support vector machine margin term, why norm squared?
As far as I know, the square is introduced in the formulation for convenience. The norm will reach the optimum at the same point, and we get rid of an ugly square root.
With respect to the hinge loss |
39,940 | Why does keras binary_crossentropy loss function return wrong values? [closed] | A mistake in your code:
$$-\frac{1}{N}\sum_{i=1}^N [\color{red}{\hat{y}_i} \log(\hat{y}_i)+(1-y_i) \log(1-\hat{y}_i)]$$
It should be
$$-\frac{1}{N}\sum_{i=1}^N [\color{blue}{y_i} \log(\hat{y}_i)+(1-y_i) \log(1-\hat{y}_i)]$$
Your code:
result.append([y_pred[i][j] * math.log(y_pred[i][j]) + (1 - y_true[i][j]) * math.lo... | Why does keras binary_crossentropy loss function return wrong values? [closed] | A mistake in your code:
$$-\frac{1}{N}\sum_{i=1}^N [\color{red}{\hat{y}_i} \log(\hat{y}_i)+(1-y_i) \log(1-\hat{y}_i)]$$
It should be
$$-\frac{1}{N}\sum_{i=1}^N [\color{blue}{y_i} \log(\hat{y}_i)+(1- | Why does keras binary_crossentropy loss function return wrong values? [closed]
A mistake in your code:
$$-\frac{1}{N}\sum_{i=1}^N [\color{red}{\hat{y}_i} \log(\hat{y}_i)+(1-y_i) \log(1-\hat{y}_i)]$$
It should be
$$-\frac{1}{N}\sum_{i=1}^N [\color{blue}{y_i} \log(\hat{y}_i)+(1-y_i) \log(1-\hat{y}_i)]$$
Your code:
resu... | Why does keras binary_crossentropy loss function return wrong values? [closed]
A mistake in your code:
$$-\frac{1}{N}\sum_{i=1}^N [\color{red}{\hat{y}_i} \log(\hat{y}_i)+(1-y_i) \log(1-\hat{y}_i)]$$
It should be
$$-\frac{1}{N}\sum_{i=1}^N [\color{blue}{y_i} \log(\hat{y}_i)+(1- |
39,941 | Multivariate binary responses - advice on regression strategy | You are making a strong assumption that all the childhood events have equal weight in predicting adult outcomes. But given that, there are several possible ways to proceed. Here are three main approaches, one of which you've already mentioned.
Turn the problem backwards to predict the number of childhood events give... | Multivariate binary responses - advice on regression strategy | You are making a strong assumption that all the childhood events have equal weight in predicting adult outcomes. But given that, there are several possible ways to proceed. Here are three main appro | Multivariate binary responses - advice on regression strategy
You are making a strong assumption that all the childhood events have equal weight in predicting adult outcomes. But given that, there are several possible ways to proceed. Here are three main approaches, one of which you've already mentioned.
Turn the pr... | Multivariate binary responses - advice on regression strategy
You are making a strong assumption that all the childhood events have equal weight in predicting adult outcomes. But given that, there are several possible ways to proceed. Here are three main appro |
39,942 | Multivariate binary responses - advice on regression strategy | The multivariate probit model might be considered, as described in the Greene book mentioned by Frank Harrell. See also (Lesaffre and Mohlenberghs, 1991 Stat. Med 10, 1391-1403). The idea is to think of a multivariate Normal (4 dimensions) distribution of propensity or tolerance towards each event. You model the multiv... | Multivariate binary responses - advice on regression strategy | The multivariate probit model might be considered, as described in the Greene book mentioned by Frank Harrell. See also (Lesaffre and Mohlenberghs, 1991 Stat. Med 10, 1391-1403). The idea is to think | Multivariate binary responses - advice on regression strategy
The multivariate probit model might be considered, as described in the Greene book mentioned by Frank Harrell. See also (Lesaffre and Mohlenberghs, 1991 Stat. Med 10, 1391-1403). The idea is to think of a multivariate Normal (4 dimensions) distribution of pr... | Multivariate binary responses - advice on regression strategy
The multivariate probit model might be considered, as described in the Greene book mentioned by Frank Harrell. See also (Lesaffre and Mohlenberghs, 1991 Stat. Med 10, 1391-1403). The idea is to think |
39,943 | Difference between smoothing splines and splines in R | Smoothing splines have all the knots (knots at each point), but then regularizes (shrinks the coefficients/smooths the fit) by adding a roughness penalty term (integrated squared second derivative times a smoothing parameter/tuning parameter) to the least squares criterion.
In one way, it's sort of analogous to a kind ... | Difference between smoothing splines and splines in R | Smoothing splines have all the knots (knots at each point), but then regularizes (shrinks the coefficients/smooths the fit) by adding a roughness penalty term (integrated squared second derivative tim | Difference between smoothing splines and splines in R
Smoothing splines have all the knots (knots at each point), but then regularizes (shrinks the coefficients/smooths the fit) by adding a roughness penalty term (integrated squared second derivative times a smoothing parameter/tuning parameter) to the least squares cr... | Difference between smoothing splines and splines in R
Smoothing splines have all the knots (knots at each point), but then regularizes (shrinks the coefficients/smooths the fit) by adding a roughness penalty term (integrated squared second derivative tim |
39,944 | Is variance homogeneity check necessary before t-test? | No, it is not necessary. Given that there is a test that accounts for heterogeneous variances (Welch's t-test), you can simply conduct it. For one, the tests for homogeneity of variance (HOV) are problematic in a number of ways. Some lack power, they - like other statistical tests - are too powerful with large sample s... | Is variance homogeneity check necessary before t-test? | No, it is not necessary. Given that there is a test that accounts for heterogeneous variances (Welch's t-test), you can simply conduct it. For one, the tests for homogeneity of variance (HOV) are prob | Is variance homogeneity check necessary before t-test?
No, it is not necessary. Given that there is a test that accounts for heterogeneous variances (Welch's t-test), you can simply conduct it. For one, the tests for homogeneity of variance (HOV) are problematic in a number of ways. Some lack power, they - like other s... | Is variance homogeneity check necessary before t-test?
No, it is not necessary. Given that there is a test that accounts for heterogeneous variances (Welch's t-test), you can simply conduct it. For one, the tests for homogeneity of variance (HOV) are prob |
39,945 | Is variance homogeneity check necessary before t-test? | According to statistics textbooks, t-tests require the dependent
variable to be normally distributed and the variance to be homogenous
across conditions
This is misleading. Generally, introductory statistics textbooks teach 2 (maybe 3 if you count paired stuff) two sample t-tests. Both tests assume that each of th... | Is variance homogeneity check necessary before t-test? | According to statistics textbooks, t-tests require the dependent
variable to be normally distributed and the variance to be homogenous
across conditions
This is misleading. Generally, introductor | Is variance homogeneity check necessary before t-test?
According to statistics textbooks, t-tests require the dependent
variable to be normally distributed and the variance to be homogenous
across conditions
This is misleading. Generally, introductory statistics textbooks teach 2 (maybe 3 if you count paired stuff... | Is variance homogeneity check necessary before t-test?
According to statistics textbooks, t-tests require the dependent
variable to be normally distributed and the variance to be homogenous
across conditions
This is misleading. Generally, introductor |
39,946 | Is variance homogeneity check necessary before t-test? | Not only is it not necessary, see user162986's answer, it can also imperil the interpretability of your test. | Is variance homogeneity check necessary before t-test? | Not only is it not necessary, see user162986's answer, it can also imperil the interpretability of your test. | Is variance homogeneity check necessary before t-test?
Not only is it not necessary, see user162986's answer, it can also imperil the interpretability of your test. | Is variance homogeneity check necessary before t-test?
Not only is it not necessary, see user162986's answer, it can also imperil the interpretability of your test. |
39,947 | Calculating error of mean of time series | Clearly you have good statistical intuition, because you are exactly right! Because of correlations between the individual terms, the standard error of the mean of the observations is not an accurate estimate of the error bar on the population mean from time series data.
The actual variance of the sample mean $m$ is
$$... | Calculating error of mean of time series | Clearly you have good statistical intuition, because you are exactly right! Because of correlations between the individual terms, the standard error of the mean of the observations is not an accurate | Calculating error of mean of time series
Clearly you have good statistical intuition, because you are exactly right! Because of correlations between the individual terms, the standard error of the mean of the observations is not an accurate estimate of the error bar on the population mean from time series data.
The act... | Calculating error of mean of time series
Clearly you have good statistical intuition, because you are exactly right! Because of correlations between the individual terms, the standard error of the mean of the observations is not an accurate |
39,948 | Calculating error of mean of time series | I'm answering my own question for the reference of future people who might find it helpful. I've accepted David Wright's answer, though, because it contains the actual solution and he did all the work.
In this case, I used ARIMA, specifically the auto.arima function in the R package forecast. I wanted to be able to imp... | Calculating error of mean of time series | I'm answering my own question for the reference of future people who might find it helpful. I've accepted David Wright's answer, though, because it contains the actual solution and he did all the work | Calculating error of mean of time series
I'm answering my own question for the reference of future people who might find it helpful. I've accepted David Wright's answer, though, because it contains the actual solution and he did all the work.
In this case, I used ARIMA, specifically the auto.arima function in the R pac... | Calculating error of mean of time series
I'm answering my own question for the reference of future people who might find it helpful. I've accepted David Wright's answer, though, because it contains the actual solution and he did all the work |
39,949 | Calculating error of mean of time series | Maybe "the blocking method" proposed by H. Flyvbjerg and H. G. Petersen in "Error Estimates on Averages of Correlated Data", J. Chem. Phys. 91, 461 (1989) would be useful. | Calculating error of mean of time series | Maybe "the blocking method" proposed by H. Flyvbjerg and H. G. Petersen in "Error Estimates on Averages of Correlated Data", J. Chem. Phys. 91, 461 (1989) would be useful. | Calculating error of mean of time series
Maybe "the blocking method" proposed by H. Flyvbjerg and H. G. Petersen in "Error Estimates on Averages of Correlated Data", J. Chem. Phys. 91, 461 (1989) would be useful. | Calculating error of mean of time series
Maybe "the blocking method" proposed by H. Flyvbjerg and H. G. Petersen in "Error Estimates on Averages of Correlated Data", J. Chem. Phys. 91, 461 (1989) would be useful. |
39,950 | How can the confidence interval for standard deviation not include the sample standard deviation? | [Aside: We're using a chi-squared distribution to obtain the confidence interval because this interval is obtained assuming we're sampling from a normal distribution.]
While the interval for $\sigma$ not including the observed sample value of $s$ might at first glance seem surprising, it occurs for the simple reason th... | How can the confidence interval for standard deviation not include the sample standard deviation? | [Aside: We're using a chi-squared distribution to obtain the confidence interval because this interval is obtained assuming we're sampling from a normal distribution.]
While the interval for $\sigma$ | How can the confidence interval for standard deviation not include the sample standard deviation?
[Aside: We're using a chi-squared distribution to obtain the confidence interval because this interval is obtained assuming we're sampling from a normal distribution.]
While the interval for $\sigma$ not including the obse... | How can the confidence interval for standard deviation not include the sample standard deviation?
[Aside: We're using a chi-squared distribution to obtain the confidence interval because this interval is obtained assuming we're sampling from a normal distribution.]
While the interval for $\sigma$ |
39,951 | Multiple regression avPlots vs termplot | termplot and crPlot: Partial-residual plots
These functions display partial residuals on the y-axis and the focal variable on the x-axis together with the corresponding regression line. The slope of the regression line will be identical with the coefficient of the focal variable in the full model. Such type of graphs a... | Multiple regression avPlots vs termplot | termplot and crPlot: Partial-residual plots
These functions display partial residuals on the y-axis and the focal variable on the x-axis together with the corresponding regression line. The slope of t | Multiple regression avPlots vs termplot
termplot and crPlot: Partial-residual plots
These functions display partial residuals on the y-axis and the focal variable on the x-axis together with the corresponding regression line. The slope of the regression line will be identical with the coefficient of the focal variable ... | Multiple regression avPlots vs termplot
termplot and crPlot: Partial-residual plots
These functions display partial residuals on the y-axis and the focal variable on the x-axis together with the corresponding regression line. The slope of t |
39,952 | Adding a white noise process to an AR(P) process | You have
$$y_t = x_t + v_t \tag{1}
$$
and
$$
\phi(B)x_t = e_t.
$$
Applying $\phi(B)$ to both sides of (1) yields
\begin{align}
\phi(B)y_t &= \phi(B)x_t + \phi(B) v_t
\\ &= e_t + \phi(B) v_t. \tag{2}
\end{align}
Consider the right hand side of (2). This is clearly a covariance stationary process. By the Wold decompos... | Adding a white noise process to an AR(P) process | You have
$$y_t = x_t + v_t \tag{1}
$$
and
$$
\phi(B)x_t = e_t.
$$
Applying $\phi(B)$ to both sides of (1) yields
\begin{align}
\phi(B)y_t &= \phi(B)x_t + \phi(B) v_t
\\ &= e_t + \phi(B) v_t. \tag{2}
| Adding a white noise process to an AR(P) process
You have
$$y_t = x_t + v_t \tag{1}
$$
and
$$
\phi(B)x_t = e_t.
$$
Applying $\phi(B)$ to both sides of (1) yields
\begin{align}
\phi(B)y_t &= \phi(B)x_t + \phi(B) v_t
\\ &= e_t + \phi(B) v_t. \tag{2}
\end{align}
Consider the right hand side of (2). This is clearly a cov... | Adding a white noise process to an AR(P) process
You have
$$y_t = x_t + v_t \tag{1}
$$
and
$$
\phi(B)x_t = e_t.
$$
Applying $\phi(B)$ to both sides of (1) yields
\begin{align}
\phi(B)y_t &= \phi(B)x_t + \phi(B) v_t
\\ &= e_t + \phi(B) v_t. \tag{2}
|
39,953 | What is the meaning of || (double vertical bar) in this KL divergence equation? | My understanding is that the double bar emphasises that the order of the arguments matters. The reminder is perhaps helpful because KL is used much like a distance, but it's not symmetric, so it's not a distance. The double bars don't actually mean something special over and above, say, a comma. | What is the meaning of || (double vertical bar) in this KL divergence equation? | My understanding is that the double bar emphasises that the order of the arguments matters. The reminder is perhaps helpful because KL is used much like a distance, but it's not symmetric, so it's not | What is the meaning of || (double vertical bar) in this KL divergence equation?
My understanding is that the double bar emphasises that the order of the arguments matters. The reminder is perhaps helpful because KL is used much like a distance, but it's not symmetric, so it's not a distance. The double bars don't actua... | What is the meaning of || (double vertical bar) in this KL divergence equation?
My understanding is that the double bar emphasises that the order of the arguments matters. The reminder is perhaps helpful because KL is used much like a distance, but it's not symmetric, so it's not |
39,954 | What is implied by i.i.d.? | We often assume that random variables of our interest are independent and identically distributed (i.i.d.). You may also be interested in more broader term exchangability (see also here).
What this means is that they are independent, so loosely speaking, knowing one value tells us nothing new about other value, and mor... | What is implied by i.i.d.? | We often assume that random variables of our interest are independent and identically distributed (i.i.d.). You may also be interested in more broader term exchangability (see also here).
What this me | What is implied by i.i.d.?
We often assume that random variables of our interest are independent and identically distributed (i.i.d.). You may also be interested in more broader term exchangability (see also here).
What this means is that they are independent, so loosely speaking, knowing one value tells us nothing new... | What is implied by i.i.d.?
We often assume that random variables of our interest are independent and identically distributed (i.i.d.). You may also be interested in more broader term exchangability (see also here).
What this me |
39,955 | Why are all the permutations of i.i.d. samples from a continuous distribution equally likely? | Although this is intuitively obvious, there is merit in providing a rigorous proof because it helps demonstrate that our definitions really do capture what we think ought to be true.
The basic idea
The independence of two variables $X$ and $Y$ is defined in terms of the probabilities of rectangles: $X$ and $Y$ are ind... | Why are all the permutations of i.i.d. samples from a continuous distribution equally likely? | Although this is intuitively obvious, there is merit in providing a rigorous proof because it helps demonstrate that our definitions really do capture what we think ought to be true.
The basic idea
T | Why are all the permutations of i.i.d. samples from a continuous distribution equally likely?
Although this is intuitively obvious, there is merit in providing a rigorous proof because it helps demonstrate that our definitions really do capture what we think ought to be true.
The basic idea
The independence of two var... | Why are all the permutations of i.i.d. samples from a continuous distribution equally likely?
Although this is intuitively obvious, there is merit in providing a rigorous proof because it helps demonstrate that our definitions really do capture what we think ought to be true.
The basic idea
T |
39,956 | What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent Normal(0,1) variables U and V? | $\newcommand{\E}{\mathrm{E}}$ $\newcommand{\Var}{\mathrm{Var}}$ $\newcommand{\cov}{\mathrm{Cov}}$ $\newcommand{\Expect}{{\rm I\kern-.3em E}}$
As a direct consequence of the definition of covariance, $\cov (X,Y)= \E(XY)-\E(X)\E(Y)$.
Fact 1:
$U, V \overset{i.i.d.}{\sim} \mathcal{N}(0,1)$
$\Rightarrow U - V \sim \mathcal... | What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent Normal(0,1) variables U and V? | $\newcommand{\E}{\mathrm{E}}$ $\newcommand{\Var}{\mathrm{Var}}$ $\newcommand{\cov}{\mathrm{Cov}}$ $\newcommand{\Expect}{{\rm I\kern-.3em E}}$
As a direct consequence of the definition of covariance, $ | What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent Normal(0,1) variables U and V?
$\newcommand{\E}{\mathrm{E}}$ $\newcommand{\Var}{\mathrm{Var}}$ $\newcommand{\cov}{\mathrm{Cov}}$ $\newcommand{\Expect}{{\rm I\kern-.3em E}}$
As a direct consequence of the definition of covariance, $\cov (X,Y)= \E(XY)-\E(X... | What is cov(X,Y), where X=min(U,V) and Y=max(U,V) for independent Normal(0,1) variables U and V?
$\newcommand{\E}{\mathrm{E}}$ $\newcommand{\Var}{\mathrm{Var}}$ $\newcommand{\cov}{\mathrm{Cov}}$ $\newcommand{\Expect}{{\rm I\kern-.3em E}}$
As a direct consequence of the definition of covariance, $ |
39,957 | How to interpret Bootstrap? | The intuitive idea behind the bootstrap is this: if your original dataset was a random draw from the full population, then if you take subsample from the sample (with replacement), then that too represents a draw from the full population. You can then estimate your model on all of those bootstrapped datasets. This give... | How to interpret Bootstrap? | The intuitive idea behind the bootstrap is this: if your original dataset was a random draw from the full population, then if you take subsample from the sample (with replacement), then that too repre | How to interpret Bootstrap?
The intuitive idea behind the bootstrap is this: if your original dataset was a random draw from the full population, then if you take subsample from the sample (with replacement), then that too represents a draw from the full population. You can then estimate your model on all of those boot... | How to interpret Bootstrap?
The intuitive idea behind the bootstrap is this: if your original dataset was a random draw from the full population, then if you take subsample from the sample (with replacement), then that too repre |
39,958 | How to interpret Bootstrap? | As @Superpronker mentioned it really depends on what SPSS is doing with the bootstrap. Including your code and the output would help a great deal. Also the bootstrap is a subject with a vast amount of literature. You could see this by simply looking at the bibliography in my 2007 edition of Bootstrap Methods publish... | How to interpret Bootstrap? | As @Superpronker mentioned it really depends on what SPSS is doing with the bootstrap. Including your code and the output would help a great deal. Also the bootstrap is a subject with a vast amount | How to interpret Bootstrap?
As @Superpronker mentioned it really depends on what SPSS is doing with the bootstrap. Including your code and the output would help a great deal. Also the bootstrap is a subject with a vast amount of literature. You could see this by simply looking at the bibliography in my 2007 edition ... | How to interpret Bootstrap?
As @Superpronker mentioned it really depends on what SPSS is doing with the bootstrap. Including your code and the output would help a great deal. Also the bootstrap is a subject with a vast amount |
39,959 | How to interpret Bootstrap? | As a quick summary, the general bootstrap in SPSS Statistics is described thusly in the help.
The Simple method is case resampling with replacement from the original dataset. The Stratified method is case resampling with replacement from the original dataset, within the strata defined by the cross-classification of str... | How to interpret Bootstrap? | As a quick summary, the general bootstrap in SPSS Statistics is described thusly in the help.
The Simple method is case resampling with replacement from the original dataset. The Stratified method is | How to interpret Bootstrap?
As a quick summary, the general bootstrap in SPSS Statistics is described thusly in the help.
The Simple method is case resampling with replacement from the original dataset. The Stratified method is case resampling with replacement from the original dataset, within the strata defined by the... | How to interpret Bootstrap?
As a quick summary, the general bootstrap in SPSS Statistics is described thusly in the help.
The Simple method is case resampling with replacement from the original dataset. The Stratified method is |
39,960 | How to simulate random effects models? | Just write down an (algebraic) formula for the model, and simulate from that description. I will give a very simple example, a model with multiple observations of the same subjects, with an exchangeable covariance structure. Such a structure can be represented with a random intercept for each subject. Also an subject-l... | How to simulate random effects models? | Just write down an (algebraic) formula for the model, and simulate from that description. I will give a very simple example, a model with multiple observations of the same subjects, with an exchangeab | How to simulate random effects models?
Just write down an (algebraic) formula for the model, and simulate from that description. I will give a very simple example, a model with multiple observations of the same subjects, with an exchangeable covariance structure. Such a structure can be represented with a random interc... | How to simulate random effects models?
Just write down an (algebraic) formula for the model, and simulate from that description. I will give a very simple example, a model with multiple observations of the same subjects, with an exchangeab |
39,961 | How to simulate random effects models? | Here is how I simulate random effects. I'll demonstrate for linear regression, but extending it to a different GLM should be straight forward.
Let's start with a random intercept model. The model is usually written as
$$ y = XB + Z\gamma $$
Where $Z$ is an indicator for the group and $\gamma_i$ is normally distribute... | How to simulate random effects models? | Here is how I simulate random effects. I'll demonstrate for linear regression, but extending it to a different GLM should be straight forward.
Let's start with a random intercept model. The model is | How to simulate random effects models?
Here is how I simulate random effects. I'll demonstrate for linear regression, but extending it to a different GLM should be straight forward.
Let's start with a random intercept model. The model is usually written as
$$ y = XB + Z\gamma $$
Where $Z$ is an indicator for the grou... | How to simulate random effects models?
Here is how I simulate random effects. I'll demonstrate for linear regression, but extending it to a different GLM should be straight forward.
Let's start with a random intercept model. The model is |
39,962 | CDF of $\frac{X_1X_2}{X_1+X_2+a}$, where $X_1$ and $X_2$ have exp. distributions | Thanks to the indications that @Did gave, I was able to derive the CDF of $Y=\frac{X_1X_2}{X_1+X_2+a}$ (note that in the question I use $X$ instead of $Y$) as follows:
Based on the identity of events
\begin{align}
\left[ Y <y \right] = \left[ X_1 < y \right] \cup \left[ X_1 \ge y, X_2 < y (X_1+a)(X_1-y)^{-1} \right]... | CDF of $\frac{X_1X_2}{X_1+X_2+a}$, where $X_1$ and $X_2$ have exp. distributions | Thanks to the indications that @Did gave, I was able to derive the CDF of $Y=\frac{X_1X_2}{X_1+X_2+a}$ (note that in the question I use $X$ instead of $Y$) as follows:
Based on the identity of event | CDF of $\frac{X_1X_2}{X_1+X_2+a}$, where $X_1$ and $X_2$ have exp. distributions
Thanks to the indications that @Did gave, I was able to derive the CDF of $Y=\frac{X_1X_2}{X_1+X_2+a}$ (note that in the question I use $X$ instead of $Y$) as follows:
Based on the identity of events
\begin{align}
\left[ Y <y \right] = ... | CDF of $\frac{X_1X_2}{X_1+X_2+a}$, where $X_1$ and $X_2$ have exp. distributions
Thanks to the indications that @Did gave, I was able to derive the CDF of $Y=\frac{X_1X_2}{X_1+X_2+a}$ (note that in the question I use $X$ instead of $Y$) as follows:
Based on the identity of event |
39,963 | CDF of $\frac{X_1X_2}{X_1+X_2+a}$, where $X_1$ and $X_2$ have exp. distributions | Let $X$ and $Y$ denote two independent exponential random variables and suppose that $V = \frac{XY}{X+Y+a}$ where $a > 0$. What is the CDF of $V$?
First, note that $V > 0$. Let $v$ denote a positive constant, and let us try to determine the complementary CDF $P\{V > v\}$ by integrating the joint density of $X$ and $Y$... | CDF of $\frac{X_1X_2}{X_1+X_2+a}$, where $X_1$ and $X_2$ have exp. distributions | Let $X$ and $Y$ denote two independent exponential random variables and suppose that $V = \frac{XY}{X+Y+a}$ where $a > 0$. What is the CDF of $V$?
First, note that $V > 0$. Let $v$ denote a positive | CDF of $\frac{X_1X_2}{X_1+X_2+a}$, where $X_1$ and $X_2$ have exp. distributions
Let $X$ and $Y$ denote two independent exponential random variables and suppose that $V = \frac{XY}{X+Y+a}$ where $a > 0$. What is the CDF of $V$?
First, note that $V > 0$. Let $v$ denote a positive constant, and let us try to determine t... | CDF of $\frac{X_1X_2}{X_1+X_2+a}$, where $X_1$ and $X_2$ have exp. distributions
Let $X$ and $Y$ denote two independent exponential random variables and suppose that $V = \frac{XY}{X+Y+a}$ where $a > 0$. What is the CDF of $V$?
First, note that $V > 0$. Let $v$ denote a positive |
39,964 | Why sum of squared errors for logistic regression not used and instead maximum likelihood estimation is used to fit the model? [duplicate] | Firstly, least squares (or sum of squared errors) is a possible loss function to use to fit your coefficients. There's nothing technically wrong about it.
However there are number of reasons why MLE is a more attractive option. In addition to those in the comments, here are two more:
Computational efficiency
Because th... | Why sum of squared errors for logistic regression not used and instead maximum likelihood estimation | Firstly, least squares (or sum of squared errors) is a possible loss function to use to fit your coefficients. There's nothing technically wrong about it.
However there are number of reasons why MLE i | Why sum of squared errors for logistic regression not used and instead maximum likelihood estimation is used to fit the model? [duplicate]
Firstly, least squares (or sum of squared errors) is a possible loss function to use to fit your coefficients. There's nothing technically wrong about it.
However there are number o... | Why sum of squared errors for logistic regression not used and instead maximum likelihood estimation
Firstly, least squares (or sum of squared errors) is a possible loss function to use to fit your coefficients. There's nothing technically wrong about it.
However there are number of reasons why MLE i |
39,965 | Why sum of squared errors for logistic regression not used and instead maximum likelihood estimation is used to fit the model? [duplicate] | Maybe I'm not getting the point of ilanman's answer as well as of some of the comments here, but afaiks, the answer is simply that
OLS = log L(Gaussian)
i.e. the OLS corresponds to the log likelihood of a regression with a normal / Gaussian distribution. You can see this by logging the formula for the Gaussian - the $... | Why sum of squared errors for logistic regression not used and instead maximum likelihood estimation | Maybe I'm not getting the point of ilanman's answer as well as of some of the comments here, but afaiks, the answer is simply that
OLS = log L(Gaussian)
i.e. the OLS corresponds to the log likelihood | Why sum of squared errors for logistic regression not used and instead maximum likelihood estimation is used to fit the model? [duplicate]
Maybe I'm not getting the point of ilanman's answer as well as of some of the comments here, but afaiks, the answer is simply that
OLS = log L(Gaussian)
i.e. the OLS corresponds to... | Why sum of squared errors for logistic regression not used and instead maximum likelihood estimation
Maybe I'm not getting the point of ilanman's answer as well as of some of the comments here, but afaiks, the answer is simply that
OLS = log L(Gaussian)
i.e. the OLS corresponds to the log likelihood |
39,966 | Prove that a distribution is symmetric using moments | It is a fact that the function $$\phi_Y: n \to \mathbb{E}(|Y|^n)^{1/n},\ n \gt 0$$
(known as the $L_n$ norm of $Y$) is nondecreasing. The demonstration at https://stats.stackexchange.com/a/244221 uses Jensen's Inequality (as applied to a strictly convex function). That inequality is a strict inequality whenever $|Y|$... | Prove that a distribution is symmetric using moments | It is a fact that the function $$\phi_Y: n \to \mathbb{E}(|Y|^n)^{1/n},\ n \gt 0$$
(known as the $L_n$ norm of $Y$) is nondecreasing. The demonstration at https://stats.stackexchange.com/a/244221 use | Prove that a distribution is symmetric using moments
It is a fact that the function $$\phi_Y: n \to \mathbb{E}(|Y|^n)^{1/n},\ n \gt 0$$
(known as the $L_n$ norm of $Y$) is nondecreasing. The demonstration at https://stats.stackexchange.com/a/244221 uses Jensen's Inequality (as applied to a strictly convex function). ... | Prove that a distribution is symmetric using moments
It is a fact that the function $$\phi_Y: n \to \mathbb{E}(|Y|^n)^{1/n},\ n \gt 0$$
(known as the $L_n$ norm of $Y$) is nondecreasing. The demonstration at https://stats.stackexchange.com/a/244221 use |
39,967 | Prove that a distribution is symmetric using moments | Here's one approach. This answers $a$ , probably $b$, and hopefully $c$.
Summarizing what we know: $E[X]=0$, $E[X^2]=\mbox{Var}(X)=2$ and $E[X^4]=4$. Let $m_i:=E[X^i]$. Moments of any probability distribution must satisfy positive-definiteness, in the sense that any proper $n\times n$ sub matrix of the Hankel Moment M... | Prove that a distribution is symmetric using moments | Here's one approach. This answers $a$ , probably $b$, and hopefully $c$.
Summarizing what we know: $E[X]=0$, $E[X^2]=\mbox{Var}(X)=2$ and $E[X^4]=4$. Let $m_i:=E[X^i]$. Moments of any probability dis | Prove that a distribution is symmetric using moments
Here's one approach. This answers $a$ , probably $b$, and hopefully $c$.
Summarizing what we know: $E[X]=0$, $E[X^2]=\mbox{Var}(X)=2$ and $E[X^4]=4$. Let $m_i:=E[X^i]$. Moments of any probability distribution must satisfy positive-definiteness, in the sense that any... | Prove that a distribution is symmetric using moments
Here's one approach. This answers $a$ , probably $b$, and hopefully $c$.
Summarizing what we know: $E[X]=0$, $E[X^2]=\mbox{Var}(X)=2$ and $E[X^4]=4$. Let $m_i:=E[X^i]$. Moments of any probability dis |
39,968 | Asymptotically Normally Distributed | But when we say "an estimator is asymptotically normally distributed", what does it mean?
Using similar language to your first sentence, when we say an estimator is asymptotically normally distributed, we mean something like as the sample size increases, the sampling distribution of a suitably standardized version of ... | Asymptotically Normally Distributed | But when we say "an estimator is asymptotically normally distributed", what does it mean?
Using similar language to your first sentence, when we say an estimator is asymptotically normally distribute | Asymptotically Normally Distributed
But when we say "an estimator is asymptotically normally distributed", what does it mean?
Using similar language to your first sentence, when we say an estimator is asymptotically normally distributed, we mean something like as the sample size increases, the sampling distribution of... | Asymptotically Normally Distributed
But when we say "an estimator is asymptotically normally distributed", what does it mean?
Using similar language to your first sentence, when we say an estimator is asymptotically normally distribute |
39,969 | Can a neural network tell if it has seen an image before? | I suppose you are not interested in only recognizing exactly the same image [1]. Instead, you want to know if an image is extremely similar to one you have already seen, and if so, retrieve that one.
Why is that challenging?
Let us first take a step back and assume that we were only looking at points in Euclidean spac... | Can a neural network tell if it has seen an image before? | I suppose you are not interested in only recognizing exactly the same image [1]. Instead, you want to know if an image is extremely similar to one you have already seen, and if so, retrieve that one.
| Can a neural network tell if it has seen an image before?
I suppose you are not interested in only recognizing exactly the same image [1]. Instead, you want to know if an image is extremely similar to one you have already seen, and if so, retrieve that one.
Why is that challenging?
Let us first take a step back and as... | Can a neural network tell if it has seen an image before?
I suppose you are not interested in only recognizing exactly the same image [1]. Instead, you want to know if an image is extremely similar to one you have already seen, and if so, retrieve that one.
|
39,970 | Can a neural network tell if it has seen an image before? | You can overfit neural network with training images. If you propagate training image (known images) through the network you should get predicted value very close to the expected. For the main training task you can use convolutional autoencoders. Convolutional autoencoder suppose to learn features specific for the train... | Can a neural network tell if it has seen an image before? | You can overfit neural network with training images. If you propagate training image (known images) through the network you should get predicted value very close to the expected. For the main training | Can a neural network tell if it has seen an image before?
You can overfit neural network with training images. If you propagate training image (known images) through the network you should get predicted value very close to the expected. For the main training task you can use convolutional autoencoders. Convolutional au... | Can a neural network tell if it has seen an image before?
You can overfit neural network with training images. If you propagate training image (known images) through the network you should get predicted value very close to the expected. For the main training |
39,971 | Can a neural network tell if it has seen an image before? | My guess would be that it should be possible, but that it would require two distinct substructures - one to respond to the image and one to classify as old/new. I would do some reading on models of memory such as the ones you find on this page.
Also, I'm not sure how robust such a model would be. | Can a neural network tell if it has seen an image before? | My guess would be that it should be possible, but that it would require two distinct substructures - one to respond to the image and one to classify as old/new. I would do some reading on models of me | Can a neural network tell if it has seen an image before?
My guess would be that it should be possible, but that it would require two distinct substructures - one to respond to the image and one to classify as old/new. I would do some reading on models of memory such as the ones you find on this page.
Also, I'm not sur... | Can a neural network tell if it has seen an image before?
My guess would be that it should be possible, but that it would require two distinct substructures - one to respond to the image and one to classify as old/new. I would do some reading on models of me |
39,972 | Can a neural network tell if it has seen an image before? | I'm not aware of a paper but I would suggest the following scheme -
1) Take an image input and distort till you are ok with noise
2) Feed both image and its distortions to a siamese network and return True
3) Feed image and all other images and return False
4) Repeat
This will give you a prediction on when the images a... | Can a neural network tell if it has seen an image before? | I'm not aware of a paper but I would suggest the following scheme -
1) Take an image input and distort till you are ok with noise
2) Feed both image and its distortions to a siamese network and return | Can a neural network tell if it has seen an image before?
I'm not aware of a paper but I would suggest the following scheme -
1) Take an image input and distort till you are ok with noise
2) Feed both image and its distortions to a siamese network and return True
3) Feed image and all other images and return False
4) R... | Can a neural network tell if it has seen an image before?
I'm not aware of a paper but I would suggest the following scheme -
1) Take an image input and distort till you are ok with noise
2) Feed both image and its distortions to a siamese network and return |
39,973 | How to prove this regularized matrix is invertible? | $X^TX$ is either PD or PSD.* If it's PD, it's already invertible. If it's PSD, its smallest eigenvalue is $0$. For any $\lambda>0$, you're making the smallest eigenvalue inside the parentheses positive.
*Consider $(Xa)^2$; we can write $||Xa||^2_2\ge0\implies a^TX^TXa\ge 0$. The LHS is nonnegative, so the RHS must also... | How to prove this regularized matrix is invertible? | $X^TX$ is either PD or PSD.* If it's PD, it's already invertible. If it's PSD, its smallest eigenvalue is $0$. For any $\lambda>0$, you're making the smallest eigenvalue inside the parentheses positiv | How to prove this regularized matrix is invertible?
$X^TX$ is either PD or PSD.* If it's PD, it's already invertible. If it's PSD, its smallest eigenvalue is $0$. For any $\lambda>0$, you're making the smallest eigenvalue inside the parentheses positive.
*Consider $(Xa)^2$; we can write $||Xa||^2_2\ge0\implies a^TX^TXa... | How to prove this regularized matrix is invertible?
$X^TX$ is either PD or PSD.* If it's PD, it's already invertible. If it's PSD, its smallest eigenvalue is $0$. For any $\lambda>0$, you're making the smallest eigenvalue inside the parentheses positiv |
39,974 | How to prove this regularized matrix is invertible? | We are seeing regularization is adding a diagonal elements in $X^TX$, then, $X^TX$ is the full rank matrix. Full rank is invertible matrix. | How to prove this regularized matrix is invertible? | We are seeing regularization is adding a diagonal elements in $X^TX$, then, $X^TX$ is the full rank matrix. Full rank is invertible matrix. | How to prove this regularized matrix is invertible?
We are seeing regularization is adding a diagonal elements in $X^TX$, then, $X^TX$ is the full rank matrix. Full rank is invertible matrix. | How to prove this regularized matrix is invertible?
We are seeing regularization is adding a diagonal elements in $X^TX$, then, $X^TX$ is the full rank matrix. Full rank is invertible matrix. |
39,975 | How to prove this regularized matrix is invertible? | Iff $\vec{u}=\langle1, 0, 0, \ldots, 0\rangle$ is a null vector for $X$ (meaning the first column of $X$ is 0), the matrix $X^TX + \lambda M$ will be singular.
In the $3x3$ case,
$$\lambda M\vec{u} = \left[
\begin{matrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{matrix}
\right]
\begin{bmatrix} 1\\0\\0 \end{bmatrix} = ... | How to prove this regularized matrix is invertible? | Iff $\vec{u}=\langle1, 0, 0, \ldots, 0\rangle$ is a null vector for $X$ (meaning the first column of $X$ is 0), the matrix $X^TX + \lambda M$ will be singular.
In the $3x3$ case,
$$\lambda M\vec{u} = | How to prove this regularized matrix is invertible?
Iff $\vec{u}=\langle1, 0, 0, \ldots, 0\rangle$ is a null vector for $X$ (meaning the first column of $X$ is 0), the matrix $X^TX + \lambda M$ will be singular.
In the $3x3$ case,
$$\lambda M\vec{u} = \left[
\begin{matrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{matr... | How to prove this regularized matrix is invertible?
Iff $\vec{u}=\langle1, 0, 0, \ldots, 0\rangle$ is a null vector for $X$ (meaning the first column of $X$ is 0), the matrix $X^TX + \lambda M$ will be singular.
In the $3x3$ case,
$$\lambda M\vec{u} = |
39,976 | What are .RDX and .RDB files for R? [closed] | Actually, the link you give leads to just two files. The first is a pdf file, as I'm sure you know. The second is a zip file containing an R package. You are supposed to unzip the file, then copy the folder "crhaz" to the "library" folder of your R installation. Then you run R and type library(crhaz) at the R prompt.
... | What are .RDX and .RDB files for R? [closed] | Actually, the link you give leads to just two files. The first is a pdf file, as I'm sure you know. The second is a zip file containing an R package. You are supposed to unzip the file, then copy the | What are .RDX and .RDB files for R? [closed]
Actually, the link you give leads to just two files. The first is a pdf file, as I'm sure you know. The second is a zip file containing an R package. You are supposed to unzip the file, then copy the folder "crhaz" to the "library" folder of your R installation. Then you run... | What are .RDX and .RDB files for R? [closed]
Actually, the link you give leads to just two files. The first is a pdf file, as I'm sure you know. The second is a zip file containing an R package. You are supposed to unzip the file, then copy the |
39,977 | Is there a mathematical expression that shows how LASSO shrinks coefficients (including some to zero)? | Firstly, I think it's worth noting that the description of what ridge does assumes that the data matrix is orthonormal.
Secondly, the answer to your question is yes under those circumstances. The details may be found in "Elements of Statistical Learning" on p. 69 bis (section 3.4.3) . The short story is that
$ \bet... | Is there a mathematical expression that shows how LASSO shrinks coefficients (including some to zero | Firstly, I think it's worth noting that the description of what ridge does assumes that the data matrix is orthonormal.
Secondly, the answer to your question is yes under those circumstances. The de | Is there a mathematical expression that shows how LASSO shrinks coefficients (including some to zero)?
Firstly, I think it's worth noting that the description of what ridge does assumes that the data matrix is orthonormal.
Secondly, the answer to your question is yes under those circumstances. The details may be foun... | Is there a mathematical expression that shows how LASSO shrinks coefficients (including some to zero
Firstly, I think it's worth noting that the description of what ridge does assumes that the data matrix is orthonormal.
Secondly, the answer to your question is yes under those circumstances. The de |
39,978 | Is there a mathematical expression that shows how LASSO shrinks coefficients (including some to zero)? | The question can be answered when one assumes an orthogonal matrix of predictors. Then the right singular vector matrix $\mathbf V$ equals the identity matrix $\mathbf I$, and the derivation below holds only for this case.
Consider the lasso problem
$$
\min_{\mathbf \beta} \frac12 ||\mathbf X \mathbf \beta - \mathbf y|... | Is there a mathematical expression that shows how LASSO shrinks coefficients (including some to zero | The question can be answered when one assumes an orthogonal matrix of predictors. Then the right singular vector matrix $\mathbf V$ equals the identity matrix $\mathbf I$, and the derivation below hol | Is there a mathematical expression that shows how LASSO shrinks coefficients (including some to zero)?
The question can be answered when one assumes an orthogonal matrix of predictors. Then the right singular vector matrix $\mathbf V$ equals the identity matrix $\mathbf I$, and the derivation below holds only for this ... | Is there a mathematical expression that shows how LASSO shrinks coefficients (including some to zero
The question can be answered when one assumes an orthogonal matrix of predictors. Then the right singular vector matrix $\mathbf V$ equals the identity matrix $\mathbf I$, and the derivation below hol |
39,979 | What is the value of linear dimensionality reduction in the presence of nonlinear alternatives? | I'll expand on some of the points mentioned in the comments, and add a few more.
Computational complexity. PCA is more efficient in terms of both time and memory than more complicated nonlinear dimensionality reduction (NLDR) techniques. This is an important issue when working with large datasets, for which NLDR techni... | What is the value of linear dimensionality reduction in the presence of nonlinear alternatives? | I'll expand on some of the points mentioned in the comments, and add a few more.
Computational complexity. PCA is more efficient in terms of both time and memory than more complicated nonlinear dimens | What is the value of linear dimensionality reduction in the presence of nonlinear alternatives?
I'll expand on some of the points mentioned in the comments, and add a few more.
Computational complexity. PCA is more efficient in terms of both time and memory than more complicated nonlinear dimensionality reduction (NLDR... | What is the value of linear dimensionality reduction in the presence of nonlinear alternatives?
I'll expand on some of the points mentioned in the comments, and add a few more.
Computational complexity. PCA is more efficient in terms of both time and memory than more complicated nonlinear dimens |
39,980 | Covariance matrix for a linear combination of correlated Gaussian random variables | If $X$ and $Y$ are correlated (univariate) normal random variables and $Z = AX+BY+C$, then the linearity of expectation and the bilinearity of the covariance function gives us that
\begin{align}
E[Z] &= AE[X] + BE[Y] + C,\tag{1}\\
\operatorname{cov}(Z,X) &= \operatorname{cov}(AX+BY+C,X) = A\operatorname{var}(X) + B\op... | Covariance matrix for a linear combination of correlated Gaussian random variables | If $X$ and $Y$ are correlated (univariate) normal random variables and $Z = AX+BY+C$, then the linearity of expectation and the bilinearity of the covariance function gives us that
\begin{align}
E[Z] | Covariance matrix for a linear combination of correlated Gaussian random variables
If $X$ and $Y$ are correlated (univariate) normal random variables and $Z = AX+BY+C$, then the linearity of expectation and the bilinearity of the covariance function gives us that
\begin{align}
E[Z] &= AE[X] + BE[Y] + C,\tag{1}\\
\oper... | Covariance matrix for a linear combination of correlated Gaussian random variables
If $X$ and $Y$ are correlated (univariate) normal random variables and $Z = AX+BY+C$, then the linearity of expectation and the bilinearity of the covariance function gives us that
\begin{align}
E[Z] |
39,981 | Covariance matrix for a linear combination of correlated Gaussian random variables | Covariance matrix $\Sigma_{xy}$ can be written:
$$ \Sigma_{xy} = \left[ \begin{array}{cc} c_{xx} & c_{xy} \\ c_{yx} & c_{yy} \end{array} \right] $$
Where $c_{ab}$ denotes the covariance between $a$ and $b$. (Note symmetric so $c_{ab} = c_{ba}$.)
The covariance matrix for $X$,$Y$, and $Z$ would be:
$$ \Sigma_{xyz} = \l... | Covariance matrix for a linear combination of correlated Gaussian random variables | Covariance matrix $\Sigma_{xy}$ can be written:
$$ \Sigma_{xy} = \left[ \begin{array}{cc} c_{xx} & c_{xy} \\ c_{yx} & c_{yy} \end{array} \right] $$
Where $c_{ab}$ denotes the covariance between $a$ a | Covariance matrix for a linear combination of correlated Gaussian random variables
Covariance matrix $\Sigma_{xy}$ can be written:
$$ \Sigma_{xy} = \left[ \begin{array}{cc} c_{xx} & c_{xy} \\ c_{yx} & c_{yy} \end{array} \right] $$
Where $c_{ab}$ denotes the covariance between $a$ and $b$. (Note symmetric so $c_{ab} = ... | Covariance matrix for a linear combination of correlated Gaussian random variables
Covariance matrix $\Sigma_{xy}$ can be written:
$$ \Sigma_{xy} = \left[ \begin{array}{cc} c_{xx} & c_{xy} \\ c_{yx} & c_{yy} \end{array} \right] $$
Where $c_{ab}$ denotes the covariance between $a$ a |
39,982 | Is support vector clustering a method for implementing k-means, or is it a different clustering algorithm? | The algorithms are completely different. The only common thing between them is that they both are clustering algorithms.
K-means searches for K centers, and attachment of points to them, such that:
each point is attached to the closest center
each center is the average (center of gravity) of all points attached to it... | Is support vector clustering a method for implementing k-means, or is it a different clustering algo | The algorithms are completely different. The only common thing between them is that they both are clustering algorithms.
K-means searches for K centers, and attachment of points to them, such that:
| Is support vector clustering a method for implementing k-means, or is it a different clustering algorithm?
The algorithms are completely different. The only common thing between them is that they both are clustering algorithms.
K-means searches for K centers, and attachment of points to them, such that:
each point is... | Is support vector clustering a method for implementing k-means, or is it a different clustering algo
The algorithms are completely different. The only common thing between them is that they both are clustering algorithms.
K-means searches for K centers, and attachment of points to them, such that:
|
39,983 | Is support vector clustering a method for implementing k-means, or is it a different clustering algorithm? | They are different.
The high level difference is that k-means doesn't use a kernel, and so has black and white boundaries between clusters, where SVC uses a Gaussian kernel. The significance of the kernel is that it allows for a smooth transition at the boundary rather than a hard change.
The paper introducing Support ... | Is support vector clustering a method for implementing k-means, or is it a different clustering algo | They are different.
The high level difference is that k-means doesn't use a kernel, and so has black and white boundaries between clusters, where SVC uses a Gaussian kernel. The significance of the ke | Is support vector clustering a method for implementing k-means, or is it a different clustering algorithm?
They are different.
The high level difference is that k-means doesn't use a kernel, and so has black and white boundaries between clusters, where SVC uses a Gaussian kernel. The significance of the kernel is that ... | Is support vector clustering a method for implementing k-means, or is it a different clustering algo
They are different.
The high level difference is that k-means doesn't use a kernel, and so has black and white boundaries between clusters, where SVC uses a Gaussian kernel. The significance of the ke |
39,984 | Robustness of GLM to link function | If you're fitting only nominal categorical predictors (and models of full order), the link function will be of essentially no consequence --- in the sense that it doesn't alter the fit.
Here's an example using log and identity links with a Poisson glm. First the data (y is the response, a count, and x1f and x2f have th... | Robustness of GLM to link function | If you're fitting only nominal categorical predictors (and models of full order), the link function will be of essentially no consequence --- in the sense that it doesn't alter the fit.
Here's an exam | Robustness of GLM to link function
If you're fitting only nominal categorical predictors (and models of full order), the link function will be of essentially no consequence --- in the sense that it doesn't alter the fit.
Here's an example using log and identity links with a Poisson glm. First the data (y is the respons... | Robustness of GLM to link function
If you're fitting only nominal categorical predictors (and models of full order), the link function will be of essentially no consequence --- in the sense that it doesn't alter the fit.
Here's an exam |
39,985 | Most suitable optimizer for the Gaussain process likelihood function | I think this is an open-ended question because a lot will depend on the actual dataset you are optimising against, how close your first candidate solution $s_0$ is to a local optimum and if you are interested / are able to use derivative information or not.
I have used R's standard optim function and generally I have f... | Most suitable optimizer for the Gaussain process likelihood function | I think this is an open-ended question because a lot will depend on the actual dataset you are optimising against, how close your first candidate solution $s_0$ is to a local optimum and if you are in | Most suitable optimizer for the Gaussain process likelihood function
I think this is an open-ended question because a lot will depend on the actual dataset you are optimising against, how close your first candidate solution $s_0$ is to a local optimum and if you are interested / are able to use derivative information o... | Most suitable optimizer for the Gaussain process likelihood function
I think this is an open-ended question because a lot will depend on the actual dataset you are optimising against, how close your first candidate solution $s_0$ is to a local optimum and if you are in |
39,986 | If two traits have known correlation, can you predict probability they'll "align" for a random pair? | No - knowing the correlation (and even linear regression formula) between two traits is not enough to predict the probability that a higher BMI will have a higher blood pressure.
See Anscombe's quartet for a visual example of four dissimilar distributions with identical correlations and fitted linear regression lines t... | If two traits have known correlation, can you predict probability they'll "align" for a random pair? | No - knowing the correlation (and even linear regression formula) between two traits is not enough to predict the probability that a higher BMI will have a higher blood pressure.
See Anscombe's quarte | If two traits have known correlation, can you predict probability they'll "align" for a random pair?
No - knowing the correlation (and even linear regression formula) between two traits is not enough to predict the probability that a higher BMI will have a higher blood pressure.
See Anscombe's quartet for a visual exam... | If two traits have known correlation, can you predict probability they'll "align" for a random pair?
No - knowing the correlation (and even linear regression formula) between two traits is not enough to predict the probability that a higher BMI will have a higher blood pressure.
See Anscombe's quarte |
39,987 | If two traits have known correlation, can you predict probability they'll "align" for a random pair? | If two traits have known correlation, can you predict probability they'll “align” for a random pair?
It depends on which population correlation you look at.
For the Pearson correlation you mention ($\rho$), the answer is "no", at least not without additional assumptions. (RobertF's answer is correct)
If instead you kn... | If two traits have known correlation, can you predict probability they'll "align" for a random pair? | If two traits have known correlation, can you predict probability they'll “align” for a random pair?
It depends on which population correlation you look at.
For the Pearson correlation you mention ($ | If two traits have known correlation, can you predict probability they'll "align" for a random pair?
If two traits have known correlation, can you predict probability they'll “align” for a random pair?
It depends on which population correlation you look at.
For the Pearson correlation you mention ($\rho$), the answer ... | If two traits have known correlation, can you predict probability they'll "align" for a random pair?
If two traits have known correlation, can you predict probability they'll “align” for a random pair?
It depends on which population correlation you look at.
For the Pearson correlation you mention ($ |
39,988 | If two traits have known correlation, can you predict probability they'll "align" for a random pair? | I recommend increasing the variables you are measuring. Age, gender, location etc. weight them in your formula to lower the probability of false negatives. Maximize your ROC curve. It would be interesting to see a model that keeps the same correlation given datasets over different decades. | If two traits have known correlation, can you predict probability they'll "align" for a random pair? | I recommend increasing the variables you are measuring. Age, gender, location etc. weight them in your formula to lower the probability of false negatives. Maximize your ROC curve. It would be interes | If two traits have known correlation, can you predict probability they'll "align" for a random pair?
I recommend increasing the variables you are measuring. Age, gender, location etc. weight them in your formula to lower the probability of false negatives. Maximize your ROC curve. It would be interesting to see a model... | If two traits have known correlation, can you predict probability they'll "align" for a random pair?
I recommend increasing the variables you are measuring. Age, gender, location etc. weight them in your formula to lower the probability of false negatives. Maximize your ROC curve. It would be interes |
39,989 | Is the logarithmic transformation sufficient to tame every distribution? | The answer is NO. You can construct such distributions that are untamed by log following the example of Log-Cauchy distribution. | Is the logarithmic transformation sufficient to tame every distribution? | The answer is NO. You can construct such distributions that are untamed by log following the example of Log-Cauchy distribution. | Is the logarithmic transformation sufficient to tame every distribution?
The answer is NO. You can construct such distributions that are untamed by log following the example of Log-Cauchy distribution. | Is the logarithmic transformation sufficient to tame every distribution?
The answer is NO. You can construct such distributions that are untamed by log following the example of Log-Cauchy distribution. |
39,990 | Is the logarithmic transformation sufficient to tame every distribution? | No. Consider the following situations:
If your distribution is discrete and only takes a few different values, taking logs generally doesn't really "tame" anything. e.g. 99.9% chance of 2.8 and 0.1% chance of ten billion, take natural logs and you have 99.9% chance of 1.03 and 0.1% chance of 23(ish). Take logs again a... | Is the logarithmic transformation sufficient to tame every distribution? | No. Consider the following situations:
If your distribution is discrete and only takes a few different values, taking logs generally doesn't really "tame" anything. e.g. 99.9% chance of 2.8 and 0.1% | Is the logarithmic transformation sufficient to tame every distribution?
No. Consider the following situations:
If your distribution is discrete and only takes a few different values, taking logs generally doesn't really "tame" anything. e.g. 99.9% chance of 2.8 and 0.1% chance of ten billion, take natural logs and yo... | Is the logarithmic transformation sufficient to tame every distribution?
No. Consider the following situations:
If your distribution is discrete and only takes a few different values, taking logs generally doesn't really "tame" anything. e.g. 99.9% chance of 2.8 and 0.1% |
39,991 | Why does the sum of residuals equal 0 from a graphical perspective? | To fit $\hat y = b_1 + b_2 x$ by OLS we minimise the residual sum of squares, $RSS = \sum_{i=1}^n e_i^2$. As the question states, we can do this analytically by setting
$$\frac{\partial RSS}{\partial b_1}=0 , \qquad \frac{\partial RSS}{\partial b_2}=0 $$
and solving the resulting normal equations. Note that from the no... | Why does the sum of residuals equal 0 from a graphical perspective? | To fit $\hat y = b_1 + b_2 x$ by OLS we minimise the residual sum of squares, $RSS = \sum_{i=1}^n e_i^2$. As the question states, we can do this analytically by setting
$$\frac{\partial RSS}{\partial | Why does the sum of residuals equal 0 from a graphical perspective?
To fit $\hat y = b_1 + b_2 x$ by OLS we minimise the residual sum of squares, $RSS = \sum_{i=1}^n e_i^2$. As the question states, we can do this analytically by setting
$$\frac{\partial RSS}{\partial b_1}=0 , \qquad \frac{\partial RSS}{\partial b_2}=0 ... | Why does the sum of residuals equal 0 from a graphical perspective?
To fit $\hat y = b_1 + b_2 x$ by OLS we minimise the residual sum of squares, $RSS = \sum_{i=1}^n e_i^2$. As the question states, we can do this analytically by setting
$$\frac{\partial RSS}{\partial |
39,992 | Why does the sum of residuals equal 0 from a graphical perspective? | One geometrical answer that might or might not resonate with you is that
two legs of any right triangle are perpendicular.
It applies here because the "constant" or "intercept" in the regression results in projecting the response $y$ onto the vector $\hat{y}=\bar{y}(1,1,\ldots,1)$, which is one leg, and the residual... | Why does the sum of residuals equal 0 from a graphical perspective? | One geometrical answer that might or might not resonate with you is that
two legs of any right triangle are perpendicular.
It applies here because the "constant" or "intercept" in the regression re | Why does the sum of residuals equal 0 from a graphical perspective?
One geometrical answer that might or might not resonate with you is that
two legs of any right triangle are perpendicular.
It applies here because the "constant" or "intercept" in the regression results in projecting the response $y$ onto the vector... | Why does the sum of residuals equal 0 from a graphical perspective?
One geometrical answer that might or might not resonate with you is that
two legs of any right triangle are perpendicular.
It applies here because the "constant" or "intercept" in the regression re |
39,993 | Why does the sum of residuals equal 0 from a graphical perspective? | You must keep in mind the difference between the true values of the parameters and your estimates of them. Graphically, you can think of the situation as the true line and the estimated line.
The usual way of returning the estimated line from the data just happens to correspond to the sum of residuals being zero. The... | Why does the sum of residuals equal 0 from a graphical perspective? | You must keep in mind the difference between the true values of the parameters and your estimates of them. Graphically, you can think of the situation as the true line and the estimated line.
The us | Why does the sum of residuals equal 0 from a graphical perspective?
You must keep in mind the difference between the true values of the parameters and your estimates of them. Graphically, you can think of the situation as the true line and the estimated line.
The usual way of returning the estimated line from the dat... | Why does the sum of residuals equal 0 from a graphical perspective?
You must keep in mind the difference between the true values of the parameters and your estimates of them. Graphically, you can think of the situation as the true line and the estimated line.
The us |
39,994 | How to check for the distribution stability? | First, about measuring the fit.
The Kolmogorov–Smirnov test is for a one-dimensional distribution. Though it was extended to multivariate data, it wasn't designed for time series.
I'm not sure how do you use your time series data.
If you are interested in just the probability of an event to happen, you can use the test... | How to check for the distribution stability? | First, about measuring the fit.
The Kolmogorov–Smirnov test is for a one-dimensional distribution. Though it was extended to multivariate data, it wasn't designed for time series.
I'm not sure how do | How to check for the distribution stability?
First, about measuring the fit.
The Kolmogorov–Smirnov test is for a one-dimensional distribution. Though it was extended to multivariate data, it wasn't designed for time series.
I'm not sure how do you use your time series data.
If you are interested in just the probabilit... | How to check for the distribution stability?
First, about measuring the fit.
The Kolmogorov–Smirnov test is for a one-dimensional distribution. Though it was extended to multivariate data, it wasn't designed for time series.
I'm not sure how do |
39,995 | How to check for the distribution stability? | In the limit as the quality of your random sampling method approaches perfect randomness, and in the limit as number of samples approach infinity, then the distributions of the training and testing samples will become identical.
But after considering your comment to dsaxton, it appears that you are in a special case wh... | How to check for the distribution stability? | In the limit as the quality of your random sampling method approaches perfect randomness, and in the limit as number of samples approach infinity, then the distributions of the training and testing sa | How to check for the distribution stability?
In the limit as the quality of your random sampling method approaches perfect randomness, and in the limit as number of samples approach infinity, then the distributions of the training and testing samples will become identical.
But after considering your comment to dsaxton,... | How to check for the distribution stability?
In the limit as the quality of your random sampling method approaches perfect randomness, and in the limit as number of samples approach infinity, then the distributions of the training and testing sa |
39,996 | How to check for the distribution stability? | If you selected training and test sets randomly from your data, you shouldn't have to worry about equal distributions.
More importantly, you test your model, which you generated from the training set, on the test set and by doing this you verify that you can use your model to predict other values than the ones from the... | How to check for the distribution stability? | If you selected training and test sets randomly from your data, you shouldn't have to worry about equal distributions.
More importantly, you test your model, which you generated from the training set, | How to check for the distribution stability?
If you selected training and test sets randomly from your data, you shouldn't have to worry about equal distributions.
More importantly, you test your model, which you generated from the training set, on the test set and by doing this you verify that you can use your model t... | How to check for the distribution stability?
If you selected training and test sets randomly from your data, you shouldn't have to worry about equal distributions.
More importantly, you test your model, which you generated from the training set, |
39,997 | How to check for the distribution stability? | It looks like you're generally looking at the question of whether your time series is stationarity. The other answers discuss great ways to handle time series while doing machine learning.
Separately from the ML lens, if you just want to understand the stationarity of your data, you can use the Augmented Dickey-Fuller ... | How to check for the distribution stability? | It looks like you're generally looking at the question of whether your time series is stationarity. The other answers discuss great ways to handle time series while doing machine learning.
Separately | How to check for the distribution stability?
It looks like you're generally looking at the question of whether your time series is stationarity. The other answers discuss great ways to handle time series while doing machine learning.
Separately from the ML lens, if you just want to understand the stationarity of your d... | How to check for the distribution stability?
It looks like you're generally looking at the question of whether your time series is stationarity. The other answers discuss great ways to handle time series while doing machine learning.
Separately |
39,998 | Oscillating validation accuracy for a convolutional neural network? | This is likely due to the ordering of your dataset. If there's many observations of the same class in a sequence the weights of the network will move too far in the direction of classifying this class.
A common cause is if you balance the classes in your dataset by resampling observations and appending them to the data... | Oscillating validation accuracy for a convolutional neural network? | This is likely due to the ordering of your dataset. If there's many observations of the same class in a sequence the weights of the network will move too far in the direction of classifying this class | Oscillating validation accuracy for a convolutional neural network?
This is likely due to the ordering of your dataset. If there's many observations of the same class in a sequence the weights of the network will move too far in the direction of classifying this class.
A common cause is if you balance the classes in yo... | Oscillating validation accuracy for a convolutional neural network?
This is likely due to the ordering of your dataset. If there's many observations of the same class in a sequence the weights of the network will move too far in the direction of classifying this class |
39,999 | Oscillating validation accuracy for a convolutional neural network? | I had the same issue in the past and found out that the learning rate usually is the cause of oscillation.
Try lowering your learning rate or using learning rate decay and keep training until the curve converges. | Oscillating validation accuracy for a convolutional neural network? | I had the same issue in the past and found out that the learning rate usually is the cause of oscillation.
Try lowering your learning rate or using learning rate decay and keep training until the cur | Oscillating validation accuracy for a convolutional neural network?
I had the same issue in the past and found out that the learning rate usually is the cause of oscillation.
Try lowering your learning rate or using learning rate decay and keep training until the curve converges. | Oscillating validation accuracy for a convolutional neural network?
I had the same issue in the past and found out that the learning rate usually is the cause of oscillation.
Try lowering your learning rate or using learning rate decay and keep training until the cur |
40,000 | Oscillating validation accuracy for a convolutional neural network? | Probably the learning rate is too high. The system is overfitting almost immediately, as overall the accuracy is falling after the first epoch.
If you want to find the sweet spot using early stopping you surely need a lower learning rate to extend your choice.
In addition, as suggested in other answers, I would use a l... | Oscillating validation accuracy for a convolutional neural network? | Probably the learning rate is too high. The system is overfitting almost immediately, as overall the accuracy is falling after the first epoch.
If you want to find the sweet spot using early stopping | Oscillating validation accuracy for a convolutional neural network?
Probably the learning rate is too high. The system is overfitting almost immediately, as overall the accuracy is falling after the first epoch.
If you want to find the sweet spot using early stopping you surely need a lower learning rate to extend your... | Oscillating validation accuracy for a convolutional neural network?
Probably the learning rate is too high. The system is overfitting almost immediately, as overall the accuracy is falling after the first epoch.
If you want to find the sweet spot using early stopping |
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