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 |
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35,901 | Why does Off-Policy Monte Carlo Control only learn from the "Tails of Episodes"? | Actually, the answer to your first question (why does this method learn only from the tails of episodes?) is due to the second to last line.
The if statement basically says that the method will learn from a state-action pairs in an episode only if the action is near-greedy, otherwise $A_t$ likely would not equal $\pi(S... | Why does Off-Policy Monte Carlo Control only learn from the "Tails of Episodes"? | Actually, the answer to your first question (why does this method learn only from the tails of episodes?) is due to the second to last line.
The if statement basically says that the method will learn | Why does Off-Policy Monte Carlo Control only learn from the "Tails of Episodes"?
Actually, the answer to your first question (why does this method learn only from the tails of episodes?) is due to the second to last line.
The if statement basically says that the method will learn from a state-action pairs in an episode... | Why does Off-Policy Monte Carlo Control only learn from the "Tails of Episodes"?
Actually, the answer to your first question (why does this method learn only from the tails of episodes?) is due to the second to last line.
The if statement basically says that the method will learn |
35,902 | Can Bernoulli random variables be used to approximate more than just the normal distribution? | Bernoulli random variables can approximate (almost) any distribution to arbitrary accuracy: A sequence of Bernoulli values gives us a binary sequence, which can be interpreted as the binary representation of a real number. (This is not surprising, given that a real number is essentially just an infinite sequence of di... | Can Bernoulli random variables be used to approximate more than just the normal distribution? | Bernoulli random variables can approximate (almost) any distribution to arbitrary accuracy: A sequence of Bernoulli values gives us a binary sequence, which can be interpreted as the binary representa | Can Bernoulli random variables be used to approximate more than just the normal distribution?
Bernoulli random variables can approximate (almost) any distribution to arbitrary accuracy: A sequence of Bernoulli values gives us a binary sequence, which can be interpreted as the binary representation of a real number. (T... | Can Bernoulli random variables be used to approximate more than just the normal distribution?
Bernoulli random variables can approximate (almost) any distribution to arbitrary accuracy: A sequence of Bernoulli values gives us a binary sequence, which can be interpreted as the binary representa |
35,903 | Can Bernoulli random variables be used to approximate more than just the normal distribution? | Yes, depending on circumstances. Binomial and negative binomial (including geometric) distributions are defined in terms of Bernoulli random variables so the relationship of Bernoulli random variables to those distributions is not an approximation. Especially for large $n$ and small $p$,
$\mathsf{Binom}(n, p)$ is appro... | Can Bernoulli random variables be used to approximate more than just the normal distribution? | Yes, depending on circumstances. Binomial and negative binomial (including geometric) distributions are defined in terms of Bernoulli random variables so the relationship of Bernoulli random variables | Can Bernoulli random variables be used to approximate more than just the normal distribution?
Yes, depending on circumstances. Binomial and negative binomial (including geometric) distributions are defined in terms of Bernoulli random variables so the relationship of Bernoulli random variables to those distributions is... | Can Bernoulli random variables be used to approximate more than just the normal distribution?
Yes, depending on circumstances. Binomial and negative binomial (including geometric) distributions are defined in terms of Bernoulli random variables so the relationship of Bernoulli random variables |
35,904 | Can Bernoulli random variables be used to approximate more than just the normal distribution? | This will work in most cases, at least if the Bernoulli variables involved are independent and identically distributed, or at least exchangeable and identically distributed (Peres 1992), and an unlimited sequence of them is available. This answer will include discussion on algorithms to approximate a distribution using... | Can Bernoulli random variables be used to approximate more than just the normal distribution? | This will work in most cases, at least if the Bernoulli variables involved are independent and identically distributed, or at least exchangeable and identically distributed (Peres 1992), and an unlimi | Can Bernoulli random variables be used to approximate more than just the normal distribution?
This will work in most cases, at least if the Bernoulli variables involved are independent and identically distributed, or at least exchangeable and identically distributed (Peres 1992), and an unlimited sequence of them is av... | Can Bernoulli random variables be used to approximate more than just the normal distribution?
This will work in most cases, at least if the Bernoulli variables involved are independent and identically distributed, or at least exchangeable and identically distributed (Peres 1992), and an unlimi |
35,905 | In what sense did "average" ever come to mean a statistical quantity? | Consolidation of my comments to an answer.
This doesn't address the etymology, but the underlying concern of the OP as to whether "average" can mean median.
There are times when "average" is used to refer to medians. It's common to say "The average American household earns 50,000 per year," and intend that the median h... | In what sense did "average" ever come to mean a statistical quantity? | Consolidation of my comments to an answer.
This doesn't address the etymology, but the underlying concern of the OP as to whether "average" can mean median.
There are times when "average" is used to r | In what sense did "average" ever come to mean a statistical quantity?
Consolidation of my comments to an answer.
This doesn't address the etymology, but the underlying concern of the OP as to whether "average" can mean median.
There are times when "average" is used to refer to medians. It's common to say "The average A... | In what sense did "average" ever come to mean a statistical quantity?
Consolidation of my comments to an answer.
This doesn't address the etymology, but the underlying concern of the OP as to whether "average" can mean median.
There are times when "average" is used to r |
35,906 | In what sense did "average" ever come to mean a statistical quantity? | Average had for at least two centuries since its induction in European languages implied damage to shipping goods, Words with similar roots are noted among shipping peoples (English, Dutch) and other traders in the Southern Mediterranean (Italian, French). The root awar- is from Arabic meaning damaged.
King Louis the ... | In what sense did "average" ever come to mean a statistical quantity? | Average had for at least two centuries since its induction in European languages implied damage to shipping goods, Words with similar roots are noted among shipping peoples (English, Dutch) and other | In what sense did "average" ever come to mean a statistical quantity?
Average had for at least two centuries since its induction in European languages implied damage to shipping goods, Words with similar roots are noted among shipping peoples (English, Dutch) and other traders in the Southern Mediterranean (Italian, Fr... | In what sense did "average" ever come to mean a statistical quantity?
Average had for at least two centuries since its induction in European languages implied damage to shipping goods, Words with similar roots are noted among shipping peoples (English, Dutch) and other |
35,907 | In what sense did "average" ever come to mean a statistical quantity? | The word average came into English from Middle French avarie, a derivative of an Arabic word meaning “damaged merchandise.” Avarie originally meant damage sustained by a ship or its cargo, but came to mean the expenses of such damage. | In what sense did "average" ever come to mean a statistical quantity? | The word average came into English from Middle French avarie, a derivative of an Arabic word meaning “damaged merchandise.” Avarie originally meant damage sustained by a ship or its cargo, but came to | In what sense did "average" ever come to mean a statistical quantity?
The word average came into English from Middle French avarie, a derivative of an Arabic word meaning “damaged merchandise.” Avarie originally meant damage sustained by a ship or its cargo, but came to mean the expenses of such damage. | In what sense did "average" ever come to mean a statistical quantity?
The word average came into English from Middle French avarie, a derivative of an Arabic word meaning “damaged merchandise.” Avarie originally meant damage sustained by a ship or its cargo, but came to |
35,908 | McNemar or Fisher exact test for propensity score matched data? | This is definitely an ongoing debate in the literature, but at this point the evidence points to using paired analysis to compute standard errors and p-values. Although the goal of matching is to arrive at two samples that mimic a randomized control trial, not a paired-randomized control trial, matching does still indu... | McNemar or Fisher exact test for propensity score matched data? | This is definitely an ongoing debate in the literature, but at this point the evidence points to using paired analysis to compute standard errors and p-values. Although the goal of matching is to arri | McNemar or Fisher exact test for propensity score matched data?
This is definitely an ongoing debate in the literature, but at this point the evidence points to using paired analysis to compute standard errors and p-values. Although the goal of matching is to arrive at two samples that mimic a randomized control trial,... | McNemar or Fisher exact test for propensity score matched data?
This is definitely an ongoing debate in the literature, but at this point the evidence points to using paired analysis to compute standard errors and p-values. Although the goal of matching is to arri |
35,909 | Stacking without splitting data | You have to have real, out-of-sample, predictions as the input to your blender, otherwise your blender is not learning about, and thereby improving, prediction accuracy - but instead learning about, and thereby improving, in-sample estimation accuracy, which can lead to overfitting. This is why you cannot use the whol... | Stacking without splitting data | You have to have real, out-of-sample, predictions as the input to your blender, otherwise your blender is not learning about, and thereby improving, prediction accuracy - but instead learning about, a | Stacking without splitting data
You have to have real, out-of-sample, predictions as the input to your blender, otherwise your blender is not learning about, and thereby improving, prediction accuracy - but instead learning about, and thereby improving, in-sample estimation accuracy, which can lead to overfitting. Thi... | Stacking without splitting data
You have to have real, out-of-sample, predictions as the input to your blender, otherwise your blender is not learning about, and thereby improving, prediction accuracy - but instead learning about, a |
35,910 | Stacking without splitting data | Let me explain.
In stacking you split your data in two.
One holdout set (10 - 20%).
One training set (80 - 90%).
You train your base learners individually on the training set using the same cross-validation method for each. You must use the same cross-validation fold indices for all base learners. This is because you c... | Stacking without splitting data | Let me explain.
In stacking you split your data in two.
One holdout set (10 - 20%).
One training set (80 - 90%).
You train your base learners individually on the training set using the same cross-vali | Stacking without splitting data
Let me explain.
In stacking you split your data in two.
One holdout set (10 - 20%).
One training set (80 - 90%).
You train your base learners individually on the training set using the same cross-validation method for each. You must use the same cross-validation fold indices for all base... | Stacking without splitting data
Let me explain.
In stacking you split your data in two.
One holdout set (10 - 20%).
One training set (80 - 90%).
You train your base learners individually on the training set using the same cross-vali |
35,911 | GINI and AUC relationship | I had the same thoughts and I stumbled upon a nice presentation.
Let us use the following definitions:
Gini (mostly equal to the accuracy ratio "AR") is the ratio of the area between your curve and the diagonal and the area between the perfect model and the diagonal.
This definition on the CAP curve gives the usual Gin... | GINI and AUC relationship | I had the same thoughts and I stumbled upon a nice presentation.
Let us use the following definitions:
Gini (mostly equal to the accuracy ratio "AR") is the ratio of the area between your curve and th | GINI and AUC relationship
I had the same thoughts and I stumbled upon a nice presentation.
Let us use the following definitions:
Gini (mostly equal to the accuracy ratio "AR") is the ratio of the area between your curve and the diagonal and the area between the perfect model and the diagonal.
This definition on the CAP... | GINI and AUC relationship
I had the same thoughts and I stumbled upon a nice presentation.
Let us use the following definitions:
Gini (mostly equal to the accuracy ratio "AR") is the ratio of the area between your curve and th |
35,912 | In plain English what is the difference between a most powerful test and a uniformly most powerful test? | "Uniformly" means regardless of the values of the unobservable parameters. One test may be the most powerful one for a particular value of an unobservable parameter while a different test is the most powerful one for a different value of the parameter. A uniformly more powerful test remains the most powerful one regard... | In plain English what is the difference between a most powerful test and a uniformly most powerful t | "Uniformly" means regardless of the values of the unobservable parameters. One test may be the most powerful one for a particular value of an unobservable parameter while a different test is the most | In plain English what is the difference between a most powerful test and a uniformly most powerful test?
"Uniformly" means regardless of the values of the unobservable parameters. One test may be the most powerful one for a particular value of an unobservable parameter while a different test is the most powerful one fo... | In plain English what is the difference between a most powerful test and a uniformly most powerful t
"Uniformly" means regardless of the values of the unobservable parameters. One test may be the most powerful one for a particular value of an unobservable parameter while a different test is the most |
35,913 | In plain English what is the difference between a most powerful test and a uniformly most powerful test? | According to Mood [1]
Most powerful test (MP): a test $\delta$ of $H_0: \theta = \theta_0$ vs $H_1: \theta = \theta_1$ of size $\alpha$ is MP if it has the greatest power $\pi(\theta_1 \mid \delta)$ among all tests of size $\alpha$ or less, where $\pi$ is the power function. In words, $\delta$ has the greatest capacit... | In plain English what is the difference between a most powerful test and a uniformly most powerful t | According to Mood [1]
Most powerful test (MP): a test $\delta$ of $H_0: \theta = \theta_0$ vs $H_1: \theta = \theta_1$ of size $\alpha$ is MP if it has the greatest power $\pi(\theta_1 \mid \delta)$ | In plain English what is the difference between a most powerful test and a uniformly most powerful test?
According to Mood [1]
Most powerful test (MP): a test $\delta$ of $H_0: \theta = \theta_0$ vs $H_1: \theta = \theta_1$ of size $\alpha$ is MP if it has the greatest power $\pi(\theta_1 \mid \delta)$ among all tests... | In plain English what is the difference between a most powerful test and a uniformly most powerful t
According to Mood [1]
Most powerful test (MP): a test $\delta$ of $H_0: \theta = \theta_0$ vs $H_1: \theta = \theta_1$ of size $\alpha$ is MP if it has the greatest power $\pi(\theta_1 \mid \delta)$ |
35,914 | Expected minimum distance from a point with varying density | Consider the distance to the origin of $n$ independently distributed random variables $(X_i,Y_i)$ that have uniform distributions on the square $[-1,1]^2.$
Writing $R_i^2 = X_i^2+Y_i^2$ for the squared distance, Euclidean geometry shows us that
$$\Pr(R_i \le r \le 1) = \frac{1}{4} \pi\, r^2 $$
while (with a little mo... | Expected minimum distance from a point with varying density | Consider the distance to the origin of $n$ independently distributed random variables $(X_i,Y_i)$ that have uniform distributions on the square $[-1,1]^2.$
Writing $R_i^2 = X_i^2+Y_i^2$ for the squa | Expected minimum distance from a point with varying density
Consider the distance to the origin of $n$ independently distributed random variables $(X_i,Y_i)$ that have uniform distributions on the square $[-1,1]^2.$
Writing $R_i^2 = X_i^2+Y_i^2$ for the squared distance, Euclidean geometry shows us that
$$\Pr(R_i \le... | Expected minimum distance from a point with varying density
Consider the distance to the origin of $n$ independently distributed random variables $(X_i,Y_i)$ that have uniform distributions on the square $[-1,1]^2.$
Writing $R_i^2 = X_i^2+Y_i^2$ for the squa |
35,915 | Setting max_depth greater than the number of features in a Random Forest | There is no problem with setting the maximum depth of a Random Forest (or more specifically, of any tree) higher than the number of features.
For instance, you could have two features, Age and Sex. Then you could have a series of splits that first check whether Age>18, if so check whether Sex=M, if so check whether Age... | Setting max_depth greater than the number of features in a Random Forest | There is no problem with setting the maximum depth of a Random Forest (or more specifically, of any tree) higher than the number of features.
For instance, you could have two features, Age and Sex. Th | Setting max_depth greater than the number of features in a Random Forest
There is no problem with setting the maximum depth of a Random Forest (or more specifically, of any tree) higher than the number of features.
For instance, you could have two features, Age and Sex. Then you could have a series of splits that first... | Setting max_depth greater than the number of features in a Random Forest
There is no problem with setting the maximum depth of a Random Forest (or more specifically, of any tree) higher than the number of features.
For instance, you could have two features, Age and Sex. Th |
35,916 | Setting max_depth greater than the number of features in a Random Forest | A tree's(Decision Tree/ RF) nodes are only split based on the Information Gain or Gini impurity. The number of features is not a parameter(more precisely, hyperparameter). | Setting max_depth greater than the number of features in a Random Forest | A tree's(Decision Tree/ RF) nodes are only split based on the Information Gain or Gini impurity. The number of features is not a parameter(more precisely, hyperparameter). | Setting max_depth greater than the number of features in a Random Forest
A tree's(Decision Tree/ RF) nodes are only split based on the Information Gain or Gini impurity. The number of features is not a parameter(more precisely, hyperparameter). | Setting max_depth greater than the number of features in a Random Forest
A tree's(Decision Tree/ RF) nodes are only split based on the Information Gain or Gini impurity. The number of features is not a parameter(more precisely, hyperparameter). |
35,917 | feature extraction: freezing convolutional base vs. training on extracted features | I think you understand it pretty correctly. To address your questions:
(1.) Therefore it is essentially the same as the convolutional base in the first approach and the classifiers are identical as well so I think they should give us the same accuracy (and speed). What am I missing?
The methods are generally equal. Y... | feature extraction: freezing convolutional base vs. training on extracted features | I think you understand it pretty correctly. To address your questions:
(1.) Therefore it is essentially the same as the convolutional base in the first approach and the classifiers are identical as w | feature extraction: freezing convolutional base vs. training on extracted features
I think you understand it pretty correctly. To address your questions:
(1.) Therefore it is essentially the same as the convolutional base in the first approach and the classifiers are identical as well so I think they should give us th... | feature extraction: freezing convolutional base vs. training on extracted features
I think you understand it pretty correctly. To address your questions:
(1.) Therefore it is essentially the same as the convolutional base in the first approach and the classifiers are identical as w |
35,918 | feature extraction: freezing convolutional base vs. training on extracted features | For point (2), there seems to have been an issue in Keras where conv_base.trainable = False wasn't working prior to version 2.1.0 (see https://github.com/fchollet/deep-learning-with-python-notebooks/issues/21).
I.e. the code was inadvertently re-training at least part of the base network in the version of Keras that wa... | feature extraction: freezing convolutional base vs. training on extracted features | For point (2), there seems to have been an issue in Keras where conv_base.trainable = False wasn't working prior to version 2.1.0 (see https://github.com/fchollet/deep-learning-with-python-notebooks/i | feature extraction: freezing convolutional base vs. training on extracted features
For point (2), there seems to have been an issue in Keras where conv_base.trainable = False wasn't working prior to version 2.1.0 (see https://github.com/fchollet/deep-learning-with-python-notebooks/issues/21).
I.e. the code was inadvert... | feature extraction: freezing convolutional base vs. training on extracted features
For point (2), there seems to have been an issue in Keras where conv_base.trainable = False wasn't working prior to version 2.1.0 (see https://github.com/fchollet/deep-learning-with-python-notebooks/i |
35,919 | Maximum Likelihood estimation with a "dependent parameter"? | To deal with maximum likelihood estimation, you must formulate your problem in such a way that there is a clear range of allowable values for the unknown parameters. In your problem you need to formulate an allowable range of values for the parameter $\gamma$ for each possible value of the dependent parameter $\beta$.... | Maximum Likelihood estimation with a "dependent parameter"? | To deal with maximum likelihood estimation, you must formulate your problem in such a way that there is a clear range of allowable values for the unknown parameters. In your problem you need to formu | Maximum Likelihood estimation with a "dependent parameter"?
To deal with maximum likelihood estimation, you must formulate your problem in such a way that there is a clear range of allowable values for the unknown parameters. In your problem you need to formulate an allowable range of values for the parameter $\gamma$... | Maximum Likelihood estimation with a "dependent parameter"?
To deal with maximum likelihood estimation, you must formulate your problem in such a way that there is a clear range of allowable values for the unknown parameters. In your problem you need to formu |
35,920 | Maximum Likelihood estimation with a "dependent parameter"? | Part 1
Can you model $g(s)$ as a polynomial? If so, I believe you could turn $\gamma(\beta)$ into a closed form expression by analytically computing the integral. In general, if you can find some approximation of $g(s)$ that makes the integral analytically solvable, then you're back to ordinary maximum likelihood (with... | Maximum Likelihood estimation with a "dependent parameter"? | Part 1
Can you model $g(s)$ as a polynomial? If so, I believe you could turn $\gamma(\beta)$ into a closed form expression by analytically computing the integral. In general, if you can find some appr | Maximum Likelihood estimation with a "dependent parameter"?
Part 1
Can you model $g(s)$ as a polynomial? If so, I believe you could turn $\gamma(\beta)$ into a closed form expression by analytically computing the integral. In general, if you can find some approximation of $g(s)$ that makes the integral analytically sol... | Maximum Likelihood estimation with a "dependent parameter"?
Part 1
Can you model $g(s)$ as a polynomial? If so, I believe you could turn $\gamma(\beta)$ into a closed form expression by analytically computing the integral. In general, if you can find some appr |
35,921 | What is the justification of using taylor approximations inside expectation operators? | For your specific example, the first order Taylor approximation around $x_0=0, e^x = e^0 +e^0x+R_1 = 1+x+R_1$, so
$$E(e^x) = E(1+x) + E(R_1)$$
So the question is "what can we say about $E(R_1)$?
Well, we do not know as much as we would like to about the Taylor approximation -meaning, about the behavior of the remaind... | What is the justification of using taylor approximations inside expectation operators? | For your specific example, the first order Taylor approximation around $x_0=0, e^x = e^0 +e^0x+R_1 = 1+x+R_1$, so
$$E(e^x) = E(1+x) + E(R_1)$$
So the question is "what can we say about $E(R_1)$?
Wel | What is the justification of using taylor approximations inside expectation operators?
For your specific example, the first order Taylor approximation around $x_0=0, e^x = e^0 +e^0x+R_1 = 1+x+R_1$, so
$$E(e^x) = E(1+x) + E(R_1)$$
So the question is "what can we say about $E(R_1)$?
Well, we do not know as much as we w... | What is the justification of using taylor approximations inside expectation operators?
For your specific example, the first order Taylor approximation around $x_0=0, e^x = e^0 +e^0x+R_1 = 1+x+R_1$, so
$$E(e^x) = E(1+x) + E(R_1)$$
So the question is "what can we say about $E(R_1)$?
Wel |
35,922 | What is the justification of using taylor approximations inside expectation operators? | One situation in which this is used is asymptotics.
For example, suppose $\dfrac{X_n-\mu}{\sigma/\sqrt n}\sim N(0,1)$ and $g$ is a smooth function. Then
$$
\frac{g(X_n)-g(\mu)}{\left|g'(\mu)\right|\sigma/\sqrt n} \overset{\mathcal L} \longrightarrow N(0,1) \text{ as } n\to\infty,
$$
where $\text{“} \overset{\mathcal L}... | What is the justification of using taylor approximations inside expectation operators? | One situation in which this is used is asymptotics.
For example, suppose $\dfrac{X_n-\mu}{\sigma/\sqrt n}\sim N(0,1)$ and $g$ is a smooth function. Then
$$
\frac{g(X_n)-g(\mu)}{\left|g'(\mu)\right|\si | What is the justification of using taylor approximations inside expectation operators?
One situation in which this is used is asymptotics.
For example, suppose $\dfrac{X_n-\mu}{\sigma/\sqrt n}\sim N(0,1)$ and $g$ is a smooth function. Then
$$
\frac{g(X_n)-g(\mu)}{\left|g'(\mu)\right|\sigma/\sqrt n} \overset{\mathcal L}... | What is the justification of using taylor approximations inside expectation operators?
One situation in which this is used is asymptotics.
For example, suppose $\dfrac{X_n-\mu}{\sigma/\sqrt n}\sim N(0,1)$ and $g$ is a smooth function. Then
$$
\frac{g(X_n)-g(\mu)}{\left|g'(\mu)\right|\si |
35,923 | Why is my LDA performance a non-monotonic function of the amount of training data? | You discovered an interesting phenomenon.
LDA computations rely on inverting within-class scatter matrix $\mathbf S_W$. Usually LDA solution is presented as an eigenvalue decomposition of $\mathbf S_W^{-1}\mathbf S_B$, but scikit-learn never explicitly computes the scatter matrices and instead uses SVD of data matrices... | Why is my LDA performance a non-monotonic function of the amount of training data? | You discovered an interesting phenomenon.
LDA computations rely on inverting within-class scatter matrix $\mathbf S_W$. Usually LDA solution is presented as an eigenvalue decomposition of $\mathbf S_W | Why is my LDA performance a non-monotonic function of the amount of training data?
You discovered an interesting phenomenon.
LDA computations rely on inverting within-class scatter matrix $\mathbf S_W$. Usually LDA solution is presented as an eigenvalue decomposition of $\mathbf S_W^{-1}\mathbf S_B$, but scikit-learn n... | Why is my LDA performance a non-monotonic function of the amount of training data?
You discovered an interesting phenomenon.
LDA computations rely on inverting within-class scatter matrix $\mathbf S_W$. Usually LDA solution is presented as an eigenvalue decomposition of $\mathbf S_W |
35,924 | Gaussian process prior | A Gaussian Processes is considered a prior distribution on some unknown function $\mu(x)$ (in the context of regression). This is because you're assigning the GP a priori without exact knowledge as to the truth of $\mu(x)$. Learning a GP, and thus hyperparameters $\mathbf\theta$, is conditional on $\mathbf{X}$ in $k(\... | Gaussian process prior | A Gaussian Processes is considered a prior distribution on some unknown function $\mu(x)$ (in the context of regression). This is because you're assigning the GP a priori without exact knowledge as t | Gaussian process prior
A Gaussian Processes is considered a prior distribution on some unknown function $\mu(x)$ (in the context of regression). This is because you're assigning the GP a priori without exact knowledge as to the truth of $\mu(x)$. Learning a GP, and thus hyperparameters $\mathbf\theta$, is conditional ... | Gaussian process prior
A Gaussian Processes is considered a prior distribution on some unknown function $\mu(x)$ (in the context of regression). This is because you're assigning the GP a priori without exact knowledge as t |
35,925 | Are GAM models linear in the parameters? | Yes, GAMs are linear in the parameters. If we ignore the estimation of smoothness parameters, once we have created the bases for all the covariates we want to fit a smooth effects of, a GAM is just plain old GLM with coefficients for individual basis functions.
This is also true in smooth interactions. The ti(x1, x2) t... | Are GAM models linear in the parameters? | Yes, GAMs are linear in the parameters. If we ignore the estimation of smoothness parameters, once we have created the bases for all the covariates we want to fit a smooth effects of, a GAM is just pl | Are GAM models linear in the parameters?
Yes, GAMs are linear in the parameters. If we ignore the estimation of smoothness parameters, once we have created the bases for all the covariates we want to fit a smooth effects of, a GAM is just plain old GLM with coefficients for individual basis functions.
This is also true... | Are GAM models linear in the parameters?
Yes, GAMs are linear in the parameters. If we ignore the estimation of smoothness parameters, once we have created the bases for all the covariates we want to fit a smooth effects of, a GAM is just pl |
35,926 | What are the properties of MLE that make it more desirable than OLS? | As you move sufficiently far away from normality, all linear estimators may be arbitrarily bad.
Knowing that you can get the best of a bad lot (i.e. the best linear unbiased estimate) isn't much consolation.
If you can specify a suitable distributional model (ay, there's the rub),
maximizing the likelihood has both a d... | What are the properties of MLE that make it more desirable than OLS? | As you move sufficiently far away from normality, all linear estimators may be arbitrarily bad.
Knowing that you can get the best of a bad lot (i.e. the best linear unbiased estimate) isn't much conso | What are the properties of MLE that make it more desirable than OLS?
As you move sufficiently far away from normality, all linear estimators may be arbitrarily bad.
Knowing that you can get the best of a bad lot (i.e. the best linear unbiased estimate) isn't much consolation.
If you can specify a suitable distributiona... | What are the properties of MLE that make it more desirable than OLS?
As you move sufficiently far away from normality, all linear estimators may be arbitrarily bad.
Knowing that you can get the best of a bad lot (i.e. the best linear unbiased estimate) isn't much conso |
35,927 | What are the properties of MLE that make it more desirable than OLS? | In the case of normally distributed data, OLS converges with the MLE, a solution which is BLUE (in that point). Once out of normal, OLS isn't BLUE anymore (in the terms of Gauss-Markov theorem) - this is because OLS looks to minimize the SSR whereas GMT defines BLUE in terms of minimal SE. See more here.
Generally spea... | What are the properties of MLE that make it more desirable than OLS? | In the case of normally distributed data, OLS converges with the MLE, a solution which is BLUE (in that point). Once out of normal, OLS isn't BLUE anymore (in the terms of Gauss-Markov theorem) - this | What are the properties of MLE that make it more desirable than OLS?
In the case of normally distributed data, OLS converges with the MLE, a solution which is BLUE (in that point). Once out of normal, OLS isn't BLUE anymore (in the terms of Gauss-Markov theorem) - this is because OLS looks to minimize the SSR whereas G... | What are the properties of MLE that make it more desirable than OLS?
In the case of normally distributed data, OLS converges with the MLE, a solution which is BLUE (in that point). Once out of normal, OLS isn't BLUE anymore (in the terms of Gauss-Markov theorem) - this |
35,928 | Is building deep learning architectures a trial and error scheme? | At the current stage, the neural network architecture selection is driven much more by empirical results rater than solid mathematical theory. Moreover, the network architecture (depth, breadth, activation functions, connections) are not the only decisions you have to make; also the optimization algorithm and its param... | Is building deep learning architectures a trial and error scheme? | At the current stage, the neural network architecture selection is driven much more by empirical results rater than solid mathematical theory. Moreover, the network architecture (depth, breadth, activ | Is building deep learning architectures a trial and error scheme?
At the current stage, the neural network architecture selection is driven much more by empirical results rater than solid mathematical theory. Moreover, the network architecture (depth, breadth, activation functions, connections) are not the only decisio... | Is building deep learning architectures a trial and error scheme?
At the current stage, the neural network architecture selection is driven much more by empirical results rater than solid mathematical theory. Moreover, the network architecture (depth, breadth, activ |
35,929 | Do stepwise regression techniques increase a model's predictive power? | There are a variety of problems with stepwise selection. I discussed stepwise in my answer here: Algorithms for automatic model selection. In that answer, I did not primarily focus on the problems with inference, but on the fact that the coefficients are biased (the athletes trying out are analogous to variables). B... | Do stepwise regression techniques increase a model's predictive power? | There are a variety of problems with stepwise selection. I discussed stepwise in my answer here: Algorithms for automatic model selection. In that answer, I did not primarily focus on the problems w | Do stepwise regression techniques increase a model's predictive power?
There are a variety of problems with stepwise selection. I discussed stepwise in my answer here: Algorithms for automatic model selection. In that answer, I did not primarily focus on the problems with inference, but on the fact that the coefficie... | Do stepwise regression techniques increase a model's predictive power?
There are a variety of problems with stepwise selection. I discussed stepwise in my answer here: Algorithms for automatic model selection. In that answer, I did not primarily focus on the problems w |
35,930 | Do stepwise regression techniques increase a model's predictive power? | The exact effects will depend on the model and the "truth" which, of course, we can't know. You can look at the effects of stepwise in any particular case by crossvalidating or use a simple train and test approach. | Do stepwise regression techniques increase a model's predictive power? | The exact effects will depend on the model and the "truth" which, of course, we can't know. You can look at the effects of stepwise in any particular case by crossvalidating or use a simple train and | Do stepwise regression techniques increase a model's predictive power?
The exact effects will depend on the model and the "truth" which, of course, we can't know. You can look at the effects of stepwise in any particular case by crossvalidating or use a simple train and test approach. | Do stepwise regression techniques increase a model's predictive power?
The exact effects will depend on the model and the "truth" which, of course, we can't know. You can look at the effects of stepwise in any particular case by crossvalidating or use a simple train and |
35,931 | Quantile regression revealing different relationships at different quantiles: how? | The "true slope" in a normal linear model tells you how much the mean response changes thanks to a one point increase in $x$. By assuming normality and equal variance, all quantiles of the conditional distribution of the response move in line with that. Sometimes, these assumptions are very unrealistic: variance or ske... | Quantile regression revealing different relationships at different quantiles: how? | The "true slope" in a normal linear model tells you how much the mean response changes thanks to a one point increase in $x$. By assuming normality and equal variance, all quantiles of the conditional | Quantile regression revealing different relationships at different quantiles: how?
The "true slope" in a normal linear model tells you how much the mean response changes thanks to a one point increase in $x$. By assuming normality and equal variance, all quantiles of the conditional distribution of the response move in... | Quantile regression revealing different relationships at different quantiles: how?
The "true slope" in a normal linear model tells you how much the mean response changes thanks to a one point increase in $x$. By assuming normality and equal variance, all quantiles of the conditional |
35,932 | Inconsistent normality tests: Kolmogorov-Smirnov vs Shapiro-Wilk | There are innumerable ways a distribution can differ from a normal distribution. No test could capture all of them. As a result, each test differs in how it checks to see if your distribution matches the normal. For example, the KS test looks at the quantile where your empirical cumulative distribution function diff... | Inconsistent normality tests: Kolmogorov-Smirnov vs Shapiro-Wilk | There are innumerable ways a distribution can differ from a normal distribution. No test could capture all of them. As a result, each test differs in how it checks to see if your distribution matche | Inconsistent normality tests: Kolmogorov-Smirnov vs Shapiro-Wilk
There are innumerable ways a distribution can differ from a normal distribution. No test could capture all of them. As a result, each test differs in how it checks to see if your distribution matches the normal. For example, the KS test looks at the qu... | Inconsistent normality tests: Kolmogorov-Smirnov vs Shapiro-Wilk
There are innumerable ways a distribution can differ from a normal distribution. No test could capture all of them. As a result, each test differs in how it checks to see if your distribution matche |
35,933 | Inconsistent normality tests: Kolmogorov-Smirnov vs Shapiro-Wilk | You can't cherry pick normality tests based on the results. In this case, you either go with the rejection in any test conducted, or not use them at all. KS test is not very powerful, it's not a "specialized" normality test. If anything SW is probably more trustworthy in this case.
To me your QQ plot has signs of eith... | Inconsistent normality tests: Kolmogorov-Smirnov vs Shapiro-Wilk | You can't cherry pick normality tests based on the results. In this case, you either go with the rejection in any test conducted, or not use them at all. KS test is not very powerful, it's not a "spec | Inconsistent normality tests: Kolmogorov-Smirnov vs Shapiro-Wilk
You can't cherry pick normality tests based on the results. In this case, you either go with the rejection in any test conducted, or not use them at all. KS test is not very powerful, it's not a "specialized" normality test. If anything SW is probably mor... | Inconsistent normality tests: Kolmogorov-Smirnov vs Shapiro-Wilk
You can't cherry pick normality tests based on the results. In this case, you either go with the rejection in any test conducted, or not use them at all. KS test is not very powerful, it's not a "spec |
35,934 | Ask verification of a simple matrix result | Let $\mu = EX$ and $\Sigma = E(XX^T) - \mu\mu^T$. Then we need to show
$$
\mu^T (\Sigma + \mu\mu^T)^{-1}\mu \leq 1.
$$
Let $C = E(XX^T)$ so that $\Sigma = C - \mu\mu^T$. Using the matrix determinant lemma and the existence of $C^{-1}$ we can see that $\Sigma^{-1}$ exists exactly when $\mu^T C^{-1}\mu\neq 1$. If $\mu^T ... | Ask verification of a simple matrix result | Let $\mu = EX$ and $\Sigma = E(XX^T) - \mu\mu^T$. Then we need to show
$$
\mu^T (\Sigma + \mu\mu^T)^{-1}\mu \leq 1.
$$
Let $C = E(XX^T)$ so that $\Sigma = C - \mu\mu^T$. Using the matrix determinant l | Ask verification of a simple matrix result
Let $\mu = EX$ and $\Sigma = E(XX^T) - \mu\mu^T$. Then we need to show
$$
\mu^T (\Sigma + \mu\mu^T)^{-1}\mu \leq 1.
$$
Let $C = E(XX^T)$ so that $\Sigma = C - \mu\mu^T$. Using the matrix determinant lemma and the existence of $C^{-1}$ we can see that $\Sigma^{-1}$ exists exact... | Ask verification of a simple matrix result
Let $\mu = EX$ and $\Sigma = E(XX^T) - \mu\mu^T$. Then we need to show
$$
\mu^T (\Sigma + \mu\mu^T)^{-1}\mu \leq 1.
$$
Let $C = E(XX^T)$ so that $\Sigma = C - \mu\mu^T$. Using the matrix determinant l |
35,935 | Variance of Minimum and Maximum of 2 iid Normal | If you can convince yourself that
$$
\max(X,Y) \overset{d}{=} -\min(X,Y),
$$
then taking the variance on both sides will give you your answer.
Regarding the other part, you'll probably have to integrate by hand. | Variance of Minimum and Maximum of 2 iid Normal | If you can convince yourself that
$$
\max(X,Y) \overset{d}{=} -\min(X,Y),
$$
then taking the variance on both sides will give you your answer.
Regarding the other part, you'll probably have to integ | Variance of Minimum and Maximum of 2 iid Normal
If you can convince yourself that
$$
\max(X,Y) \overset{d}{=} -\min(X,Y),
$$
then taking the variance on both sides will give you your answer.
Regarding the other part, you'll probably have to integrate by hand. | Variance of Minimum and Maximum of 2 iid Normal
If you can convince yourself that
$$
\max(X,Y) \overset{d}{=} -\min(X,Y),
$$
then taking the variance on both sides will give you your answer.
Regarding the other part, you'll probably have to integ |
35,936 | Variance of Minimum and Maximum of 2 iid Normal | Doing it out the long way, which generalizes to more than 2 iid Normals, here are the integral calculations in MAPLE:
$EA^2 = $
2*int(z^2*1/sqrt(2*Pi)*int(exp(-x^2/2),x=-infinity..z)*1/sqrt(2*Pi)*exp(-z^2/2),z=-infinity..infinity);
which equals 1.
$EA = $
2*int(z*1/sqrt(2*Pi)*int(exp(-x^2/2),x=-infinity..z)*1/sqrt(2*P... | Variance of Minimum and Maximum of 2 iid Normal | Doing it out the long way, which generalizes to more than 2 iid Normals, here are the integral calculations in MAPLE:
$EA^2 = $
2*int(z^2*1/sqrt(2*Pi)*int(exp(-x^2/2),x=-infinity..z)*1/sqrt(2*Pi)*exp( | Variance of Minimum and Maximum of 2 iid Normal
Doing it out the long way, which generalizes to more than 2 iid Normals, here are the integral calculations in MAPLE:
$EA^2 = $
2*int(z^2*1/sqrt(2*Pi)*int(exp(-x^2/2),x=-infinity..z)*1/sqrt(2*Pi)*exp(-z^2/2),z=-infinity..infinity);
which equals 1.
$EA = $
2*int(z*1/sqrt(... | Variance of Minimum and Maximum of 2 iid Normal
Doing it out the long way, which generalizes to more than 2 iid Normals, here are the integral calculations in MAPLE:
$EA^2 = $
2*int(z^2*1/sqrt(2*Pi)*int(exp(-x^2/2),x=-infinity..z)*1/sqrt(2*Pi)*exp( |
35,937 | Variance of Minimum and Maximum of 2 iid Normal | Consider the standard normal case (since it's trivial to generalize). Let $Z = \max(X,Y)$.
$F_Z(z)=P(\max(X,Y)\leq z) = P(X\leq z,Y\leq z) = \Phi(z)^2$
hence obtain $f_Z(z)$ by differentiation.
As for expectation, note the following:
$\frac{d}{dx} \phi(x)\Phi(x) = -x\phi(x)\Phi(x) + \phi(x)^2$
Further note that $\phi(x... | Variance of Minimum and Maximum of 2 iid Normal | Consider the standard normal case (since it's trivial to generalize). Let $Z = \max(X,Y)$.
$F_Z(z)=P(\max(X,Y)\leq z) = P(X\leq z,Y\leq z) = \Phi(z)^2$
hence obtain $f_Z(z)$ by differentiation.
As for | Variance of Minimum and Maximum of 2 iid Normal
Consider the standard normal case (since it's trivial to generalize). Let $Z = \max(X,Y)$.
$F_Z(z)=P(\max(X,Y)\leq z) = P(X\leq z,Y\leq z) = \Phi(z)^2$
hence obtain $f_Z(z)$ by differentiation.
As for expectation, note the following:
$\frac{d}{dx} \phi(x)\Phi(x) = -x\phi(... | Variance of Minimum and Maximum of 2 iid Normal
Consider the standard normal case (since it's trivial to generalize). Let $Z = \max(X,Y)$.
$F_Z(z)=P(\max(X,Y)\leq z) = P(X\leq z,Y\leq z) = \Phi(z)^2$
hence obtain $f_Z(z)$ by differentiation.
As for |
35,938 | What is the objective of a variational autoencoder (VAE)? | Similar to Auto-encoders, the objective of a Variational Auto-encoder is to reconstruct the input.
The only difference is that AEs have direct links between encoder and decoder parts, but VAEs have a sampling layer which samples form a distribution (usually a Gaussian) and then feeds the generated samples to the decode... | What is the objective of a variational autoencoder (VAE)? | Similar to Auto-encoders, the objective of a Variational Auto-encoder is to reconstruct the input.
The only difference is that AEs have direct links between encoder and decoder parts, but VAEs have a | What is the objective of a variational autoencoder (VAE)?
Similar to Auto-encoders, the objective of a Variational Auto-encoder is to reconstruct the input.
The only difference is that AEs have direct links between encoder and decoder parts, but VAEs have a sampling layer which samples form a distribution (usually a Ga... | What is the objective of a variational autoencoder (VAE)?
Similar to Auto-encoders, the objective of a Variational Auto-encoder is to reconstruct the input.
The only difference is that AEs have direct links between encoder and decoder parts, but VAEs have a |
35,939 | What is the objective of a variational autoencoder (VAE)? | What do we want to compute and why?
I thought you are mostly interested in the inference issues here. Once we have the trained model we can inference any distribution using the model. For instance:
1. We can do sampling using the joint distribution $P(X,Z)$.
Why? It is a method for use to generate new data points.
H... | What is the objective of a variational autoencoder (VAE)? | What do we want to compute and why?
I thought you are mostly interested in the inference issues here. Once we have the trained model we can inference any distribution using the model. For instance: | What is the objective of a variational autoencoder (VAE)?
What do we want to compute and why?
I thought you are mostly interested in the inference issues here. Once we have the trained model we can inference any distribution using the model. For instance:
1. We can do sampling using the joint distribution $P(X,Z)$.
... | What is the objective of a variational autoencoder (VAE)?
What do we want to compute and why?
I thought you are mostly interested in the inference issues here. Once we have the trained model we can inference any distribution using the model. For instance: |
35,940 | What is the objective of a variational autoencoder (VAE)? | AVE's are networks that have two parts the encode and decoder. The encoder has a v shape so does the decoder. When they are placed together they have a shape like this ><. The interesting part is the latent vector, which is between the two v's. After you have trained the AVE's (say on faces) the decoder part can be dis... | What is the objective of a variational autoencoder (VAE)? | AVE's are networks that have two parts the encode and decoder. The encoder has a v shape so does the decoder. When they are placed together they have a shape like this ><. The interesting part is the | What is the objective of a variational autoencoder (VAE)?
AVE's are networks that have two parts the encode and decoder. The encoder has a v shape so does the decoder. When they are placed together they have a shape like this ><. The interesting part is the latent vector, which is between the two v's. After you have tr... | What is the objective of a variational autoencoder (VAE)?
AVE's are networks that have two parts the encode and decoder. The encoder has a v shape so does the decoder. When they are placed together they have a shape like this ><. The interesting part is the |
35,941 | What is the objective of a variational autoencoder (VAE)? | The objetive of an autoencoder is to learn an encoding of something (along with its decoding function). There are many uses for an encoding.
In a variational autoencoder what is learnt is the distribution of the encodings instead of the encoding function directly. A consequence of this is that you can sample many times... | What is the objective of a variational autoencoder (VAE)? | The objetive of an autoencoder is to learn an encoding of something (along with its decoding function). There are many uses for an encoding.
In a variational autoencoder what is learnt is the distribu | What is the objective of a variational autoencoder (VAE)?
The objetive of an autoencoder is to learn an encoding of something (along with its decoding function). There are many uses for an encoding.
In a variational autoencoder what is learnt is the distribution of the encodings instead of the encoding function directl... | What is the objective of a variational autoencoder (VAE)?
The objetive of an autoencoder is to learn an encoding of something (along with its decoding function). There are many uses for an encoding.
In a variational autoencoder what is learnt is the distribu |
35,942 | Why is the log derivative estimator considered of large variance? | Great question, but I'm not sure a good answer exists (at this point in time). Indeed, this estimator (alternatively called the REINFORCE estimator, the score function estimator [SFE], and the likelihood ratio estimator) is known to have very high variance, which is a major problem in RL, as well as in other problems (... | Why is the log derivative estimator considered of large variance? | Great question, but I'm not sure a good answer exists (at this point in time). Indeed, this estimator (alternatively called the REINFORCE estimator, the score function estimator [SFE], and the likelih | Why is the log derivative estimator considered of large variance?
Great question, but I'm not sure a good answer exists (at this point in time). Indeed, this estimator (alternatively called the REINFORCE estimator, the score function estimator [SFE], and the likelihood ratio estimator) is known to have very high varian... | Why is the log derivative estimator considered of large variance?
Great question, but I'm not sure a good answer exists (at this point in time). Indeed, this estimator (alternatively called the REINFORCE estimator, the score function estimator [SFE], and the likelih |
35,943 | Bayesian nonparametric answer to deep learning? | As the other answer notes, a common non-parametric Bayesian alternative to neural networks is the Gaussian Process.
(See also here).
However, the connection runs much deeper than that. Consider the class of models known as Bayesian Neural Networks (BNN). Such models are like regular deep neural networks except that eac... | Bayesian nonparametric answer to deep learning? | As the other answer notes, a common non-parametric Bayesian alternative to neural networks is the Gaussian Process.
(See also here).
However, the connection runs much deeper than that. Consider the cl | Bayesian nonparametric answer to deep learning?
As the other answer notes, a common non-parametric Bayesian alternative to neural networks is the Gaussian Process.
(See also here).
However, the connection runs much deeper than that. Consider the class of models known as Bayesian Neural Networks (BNN). Such models are l... | Bayesian nonparametric answer to deep learning?
As the other answer notes, a common non-parametric Bayesian alternative to neural networks is the Gaussian Process.
(See also here).
However, the connection runs much deeper than that. Consider the cl |
35,944 | Bayesian nonparametric answer to deep learning? | Hm I am not sure, but maybe deep gaussian processes might be one example of what you are looking for?
Deep Gaussian Processes
There is also more recent work on deep gaussian processes on scholar, but I am not knowledgeable enough to tell you what would be good to read:
https://scholar.google.de/scholar?as_ylo=2016&q=de... | Bayesian nonparametric answer to deep learning? | Hm I am not sure, but maybe deep gaussian processes might be one example of what you are looking for?
Deep Gaussian Processes
There is also more recent work on deep gaussian processes on scholar, but | Bayesian nonparametric answer to deep learning?
Hm I am not sure, but maybe deep gaussian processes might be one example of what you are looking for?
Deep Gaussian Processes
There is also more recent work on deep gaussian processes on scholar, but I am not knowledgeable enough to tell you what would be good to read:
ht... | Bayesian nonparametric answer to deep learning?
Hm I am not sure, but maybe deep gaussian processes might be one example of what you are looking for?
Deep Gaussian Processes
There is also more recent work on deep gaussian processes on scholar, but |
35,945 | Difference between: Offset and Weights? | Offset and weights are very different things. Offset is really a covariable included in a model with a fixed coefficient of $1$, which is not estimated.
They are mostly used with poisson models to represent exposure, see Should I use an offset for my Poisson GLM? , When to use an offset in a Poisson regression? and... | Difference between: Offset and Weights? | Offset and weights are very different things. Offset is really a covariable included in a model with a fixed coefficient of $1$, which is not estimated.
They are mostly used with poisson models to re | Difference between: Offset and Weights?
Offset and weights are very different things. Offset is really a covariable included in a model with a fixed coefficient of $1$, which is not estimated.
They are mostly used with poisson models to represent exposure, see Should I use an offset for my Poisson GLM? , When to use... | Difference between: Offset and Weights?
Offset and weights are very different things. Offset is really a covariable included in a model with a fixed coefficient of $1$, which is not estimated.
They are mostly used with poisson models to re |
35,946 | ANOVA for equivalence testing | Since you have only two groups, there is no need to perform an ANOVA. You can perform a TOST procedure by simply building a confidence interval (CI) for a two sample problem, say x and y, where x = MP2-MP1 in the control group and y = MP2-MP1 in the treated group. You need to pay attention to the usual assumptions for... | ANOVA for equivalence testing | Since you have only two groups, there is no need to perform an ANOVA. You can perform a TOST procedure by simply building a confidence interval (CI) for a two sample problem, say x and y, where x = M | ANOVA for equivalence testing
Since you have only two groups, there is no need to perform an ANOVA. You can perform a TOST procedure by simply building a confidence interval (CI) for a two sample problem, say x and y, where x = MP2-MP1 in the control group and y = MP2-MP1 in the treated group. You need to pay attentio... | ANOVA for equivalence testing
Since you have only two groups, there is no need to perform an ANOVA. You can perform a TOST procedure by simply building a confidence interval (CI) for a two sample problem, say x and y, where x = M |
35,947 | ANOVA for equivalence testing | Generally equivalence testing is very limited to simple models. I have, however, recently learned that the lsmeans package in r does equivalence testing for models more complex than a t-test. I have not used it, but I have used lsmeans and it is a very flexible tool. This might work, but you will have to check out the... | ANOVA for equivalence testing | Generally equivalence testing is very limited to simple models. I have, however, recently learned that the lsmeans package in r does equivalence testing for models more complex than a t-test. I have | ANOVA for equivalence testing
Generally equivalence testing is very limited to simple models. I have, however, recently learned that the lsmeans package in r does equivalence testing for models more complex than a t-test. I have not used it, but I have used lsmeans and it is a very flexible tool. This might work, but ... | ANOVA for equivalence testing
Generally equivalence testing is very limited to simple models. I have, however, recently learned that the lsmeans package in r does equivalence testing for models more complex than a t-test. I have |
35,948 | Why does the first eigenvector in PCA resemble the derivative of an underlying trend? | Let's ignore the mean-centering for a moment. One way to understand the data is to view each time series as being approximately a fixed multiple of an overall "trend," which itself is a time series $x=(x_1, x_2, \ldots, x_p)^\prime$ (with $p=7$ the number of time periods). I will refer to this below as "having a simi... | Why does the first eigenvector in PCA resemble the derivative of an underlying trend? | Let's ignore the mean-centering for a moment. One way to understand the data is to view each time series as being approximately a fixed multiple of an overall "trend," which itself is a time series $ | Why does the first eigenvector in PCA resemble the derivative of an underlying trend?
Let's ignore the mean-centering for a moment. One way to understand the data is to view each time series as being approximately a fixed multiple of an overall "trend," which itself is a time series $x=(x_1, x_2, \ldots, x_p)^\prime$ ... | Why does the first eigenvector in PCA resemble the derivative of an underlying trend?
Let's ignore the mean-centering for a moment. One way to understand the data is to view each time series as being approximately a fixed multiple of an overall "trend," which itself is a time series $ |
35,949 | Why does the first eigenvector in PCA resemble the derivative of an underlying trend? | Derivative of the data (~ first difference) removes the pointwise dependencies in the data which are due to nonstationarity (cf. ARIMA). What you then recover is approximately the stable stationary signal, which I guess the SVD is recovering. | Why does the first eigenvector in PCA resemble the derivative of an underlying trend? | Derivative of the data (~ first difference) removes the pointwise dependencies in the data which are due to nonstationarity (cf. ARIMA). What you then recover is approximately the stable stationary si | Why does the first eigenvector in PCA resemble the derivative of an underlying trend?
Derivative of the data (~ first difference) removes the pointwise dependencies in the data which are due to nonstationarity (cf. ARIMA). What you then recover is approximately the stable stationary signal, which I guess the SVD is rec... | Why does the first eigenvector in PCA resemble the derivative of an underlying trend?
Derivative of the data (~ first difference) removes the pointwise dependencies in the data which are due to nonstationarity (cf. ARIMA). What you then recover is approximately the stable stationary si |
35,950 | Logistic regression with lasso versus PCA? | I answered PCA allows you to do feature selection outside of the fit and transform and therefore give more flexibility in the hyper parameter search.
PCA can be used as a dimensionality reduction technique if you drop Principal Components based on a heuristic, but it offers no feature selection, as the Principal Comp... | Logistic regression with lasso versus PCA? | I answered PCA allows you to do feature selection outside of the fit and transform and therefore give more flexibility in the hyper parameter search.
PCA can be used as a dimensionality reduction te | Logistic regression with lasso versus PCA?
I answered PCA allows you to do feature selection outside of the fit and transform and therefore give more flexibility in the hyper parameter search.
PCA can be used as a dimensionality reduction technique if you drop Principal Components based on a heuristic, but it offers ... | Logistic regression with lasso versus PCA?
I answered PCA allows you to do feature selection outside of the fit and transform and therefore give more flexibility in the hyper parameter search.
PCA can be used as a dimensionality reduction te |
35,951 | Logistic regression with lasso versus PCA? | PCA while reducing the number of features does not care about the class labels. The only thing that it cares about is preserving the maximum variance which may not always be optimal for classification task.
L1-Reg on the other hand pushes those features towards zero that do not have much correlation with the class lab... | Logistic regression with lasso versus PCA? | PCA while reducing the number of features does not care about the class labels. The only thing that it cares about is preserving the maximum variance which may not always be optimal for classification | Logistic regression with lasso versus PCA?
PCA while reducing the number of features does not care about the class labels. The only thing that it cares about is preserving the maximum variance which may not always be optimal for classification task.
L1-Reg on the other hand pushes those features towards zero that do n... | Logistic regression with lasso versus PCA?
PCA while reducing the number of features does not care about the class labels. The only thing that it cares about is preserving the maximum variance which may not always be optimal for classification |
35,952 | Show that $Y_1+Y_2$ have distribution skew-normal | Reparameterizing the skew in terms of $\delta=\lambda/\sqrt{1+\lambda^2}$ and using the mgf of the skew normal (see below), since $Y_1$ and $Y_2$ are independent, $Z=Y_1 + Y_2$ has mgf
\begin{align}
M_Z(t) &= M_{Y_1}(t)M_{Y_2}(t) \\
&= 2e^{\mu_1 t +\sigma_1^2 t^2/2}\Phi(\sigma_1\delta t)e^{\mu_2t +\sigma_2^2 t^2... | Show that $Y_1+Y_2$ have distribution skew-normal | Reparameterizing the skew in terms of $\delta=\lambda/\sqrt{1+\lambda^2}$ and using the mgf of the skew normal (see below), since $Y_1$ and $Y_2$ are independent, $Z=Y_1 + Y_2$ has mgf
\begin{align}
M | Show that $Y_1+Y_2$ have distribution skew-normal
Reparameterizing the skew in terms of $\delta=\lambda/\sqrt{1+\lambda^2}$ and using the mgf of the skew normal (see below), since $Y_1$ and $Y_2$ are independent, $Z=Y_1 + Y_2$ has mgf
\begin{align}
M_Z(t) &= M_{Y_1}(t)M_{Y_2}(t) \\
&= 2e^{\mu_1 t +\sigma_1^2 t^2... | Show that $Y_1+Y_2$ have distribution skew-normal
Reparameterizing the skew in terms of $\delta=\lambda/\sqrt{1+\lambda^2}$ and using the mgf of the skew normal (see below), since $Y_1$ and $Y_2$ are independent, $Z=Y_1 + Y_2$ has mgf
\begin{align}
M |
35,953 | How is a negative binomial regression model different from OLS with a logged outcome variable? | Assuming you've already identified the predictor(s) of interest/importance, the considerations for a model (in approximate order of importance) would be:
a. do you want to model the conditional mean or something else (e.g. a quantile? some more robust location-estimate than the mean? ...)?
b. What is the anticipated ... | How is a negative binomial regression model different from OLS with a logged outcome variable? | Assuming you've already identified the predictor(s) of interest/importance, the considerations for a model (in approximate order of importance) would be:
a. do you want to model the conditional mean | How is a negative binomial regression model different from OLS with a logged outcome variable?
Assuming you've already identified the predictor(s) of interest/importance, the considerations for a model (in approximate order of importance) would be:
a. do you want to model the conditional mean or something else (e.g. a... | How is a negative binomial regression model different from OLS with a logged outcome variable?
Assuming you've already identified the predictor(s) of interest/importance, the considerations for a model (in approximate order of importance) would be:
a. do you want to model the conditional mean |
35,954 | Required sample size and degrees of freedom for a VAR | One should not expect there to be a hard-and-fast rule for such thresholds. The reason is that the precision of the estimates does not only depend on the ratio between parameters and observations, but also, for example, the "signal-to-noise ratio". That is, if the variance of the errors driving the process is large rel... | Required sample size and degrees of freedom for a VAR | One should not expect there to be a hard-and-fast rule for such thresholds. The reason is that the precision of the estimates does not only depend on the ratio between parameters and observations, but | Required sample size and degrees of freedom for a VAR
One should not expect there to be a hard-and-fast rule for such thresholds. The reason is that the precision of the estimates does not only depend on the ratio between parameters and observations, but also, for example, the "signal-to-noise ratio". That is, if the v... | Required sample size and degrees of freedom for a VAR
One should not expect there to be a hard-and-fast rule for such thresholds. The reason is that the precision of the estimates does not only depend on the ratio between parameters and observations, but |
35,955 | Generate a random set of numbers with fixed sum and desired means and variances? | Multivariate logit-normal distribution can be considered as a generalization of the Dirichlet distribution that you have in mind. It is parametrized by a vector of $D-1$ means $\boldsymbol{\mu}$ for $D$-dimensional vector $\mathbf{x}$ and covariance matrix $\boldsymbol{\Sigma}$,
$$
f_X( \mathbf{x}; \boldsymbol{\mu} , ... | Generate a random set of numbers with fixed sum and desired means and variances? | Multivariate logit-normal distribution can be considered as a generalization of the Dirichlet distribution that you have in mind. It is parametrized by a vector of $D-1$ means $\boldsymbol{\mu}$ for | Generate a random set of numbers with fixed sum and desired means and variances?
Multivariate logit-normal distribution can be considered as a generalization of the Dirichlet distribution that you have in mind. It is parametrized by a vector of $D-1$ means $\boldsymbol{\mu}$ for $D$-dimensional vector $\mathbf{x}$ and... | Generate a random set of numbers with fixed sum and desired means and variances?
Multivariate logit-normal distribution can be considered as a generalization of the Dirichlet distribution that you have in mind. It is parametrized by a vector of $D-1$ means $\boldsymbol{\mu}$ for |
35,956 | Generate a random set of numbers with fixed sum and desired means and variances? | Wikipedia (https://en.wikipedia.org/wiki/Dirichlet_distribution) describes the means and variances of the n individual x's in terms of the n exponenents (written as a's) and their sum. The means will not be changed if all the a's are multiplied by some constant b. However the variances will change, at least somewhat.... | Generate a random set of numbers with fixed sum and desired means and variances? | Wikipedia (https://en.wikipedia.org/wiki/Dirichlet_distribution) describes the means and variances of the n individual x's in terms of the n exponenents (written as a's) and their sum. The means will | Generate a random set of numbers with fixed sum and desired means and variances?
Wikipedia (https://en.wikipedia.org/wiki/Dirichlet_distribution) describes the means and variances of the n individual x's in terms of the n exponenents (written as a's) and their sum. The means will not be changed if all the a's are mult... | Generate a random set of numbers with fixed sum and desired means and variances?
Wikipedia (https://en.wikipedia.org/wiki/Dirichlet_distribution) describes the means and variances of the n individual x's in terms of the n exponenents (written as a's) and their sum. The means will |
35,957 | Hypothesis testing with an unusual alternative hypothesis ($H_1:\mu = \mu_1$) | It's standard to look at point null vs point alternative when first introducing the Neyman Pearson lemma, so if you've seen that you'll probably have seen this simple-alternative case done already.
Very little is different in the simple-null-simple-alternative case from the simple-null case, you are just in a situatio... | Hypothesis testing with an unusual alternative hypothesis ($H_1:\mu = \mu_1$) | It's standard to look at point null vs point alternative when first introducing the Neyman Pearson lemma, so if you've seen that you'll probably have seen this simple-alternative case done already.
V | Hypothesis testing with an unusual alternative hypothesis ($H_1:\mu = \mu_1$)
It's standard to look at point null vs point alternative when first introducing the Neyman Pearson lemma, so if you've seen that you'll probably have seen this simple-alternative case done already.
Very little is different in the simple-null... | Hypothesis testing with an unusual alternative hypothesis ($H_1:\mu = \mu_1$)
It's standard to look at point null vs point alternative when first introducing the Neyman Pearson lemma, so if you've seen that you'll probably have seen this simple-alternative case done already.
V |
35,958 | Hypothesis testing with an unusual alternative hypothesis ($H_1:\mu = \mu_1$) | I'm not as expert at statistics as some people here so please bear with me; but I'd like to throw in my two cents worth. As I understand the example problem you quote, it is basically asking if a sample mean of 57 is more likely, or has a lower (more statistically significant) p-value, than a sample mean of 60, if the ... | Hypothesis testing with an unusual alternative hypothesis ($H_1:\mu = \mu_1$) | I'm not as expert at statistics as some people here so please bear with me; but I'd like to throw in my two cents worth. As I understand the example problem you quote, it is basically asking if a samp | Hypothesis testing with an unusual alternative hypothesis ($H_1:\mu = \mu_1$)
I'm not as expert at statistics as some people here so please bear with me; but I'd like to throw in my two cents worth. As I understand the example problem you quote, it is basically asking if a sample mean of 57 is more likely, or has a low... | Hypothesis testing with an unusual alternative hypothesis ($H_1:\mu = \mu_1$)
I'm not as expert at statistics as some people here so please bear with me; but I'd like to throw in my two cents worth. As I understand the example problem you quote, it is basically asking if a samp |
35,959 | Manually calculating logistic regression coefficient | You are right that although you should be able to calculate the OLS coefficient estimate in logit space, you can't do it directly because the logit, $g(y) = \log \frac{p}{1-p}$, goes either to $-\infty$ for $y=0$ or $\infty$ for $y=1$. An added difficulty is that the variance in this model depends on $x$.
The likelihoo... | Manually calculating logistic regression coefficient | You are right that although you should be able to calculate the OLS coefficient estimate in logit space, you can't do it directly because the logit, $g(y) = \log \frac{p}{1-p}$, goes either to $-\inft | Manually calculating logistic regression coefficient
You are right that although you should be able to calculate the OLS coefficient estimate in logit space, you can't do it directly because the logit, $g(y) = \log \frac{p}{1-p}$, goes either to $-\infty$ for $y=0$ or $\infty$ for $y=1$. An added difficulty is that the... | Manually calculating logistic regression coefficient
You are right that although you should be able to calculate the OLS coefficient estimate in logit space, you can't do it directly because the logit, $g(y) = \log \frac{p}{1-p}$, goes either to $-\inft |
35,960 | Can I use output of classifier A as feature for classifier B? | Yes, using outputs of one model as inputs of another is possible, and this concept is used to some extent in some approaches.
If you stack models on top of each other, in the application case the model at position $n$ in the chain essentially does preprocessing/feature derivation for the model at position $n+1$. One a... | Can I use output of classifier A as feature for classifier B? | Yes, using outputs of one model as inputs of another is possible, and this concept is used to some extent in some approaches.
If you stack models on top of each other, in the application case the mod | Can I use output of classifier A as feature for classifier B?
Yes, using outputs of one model as inputs of another is possible, and this concept is used to some extent in some approaches.
If you stack models on top of each other, in the application case the model at position $n$ in the chain essentially does preproces... | Can I use output of classifier A as feature for classifier B?
Yes, using outputs of one model as inputs of another is possible, and this concept is used to some extent in some approaches.
If you stack models on top of each other, in the application case the mod |
35,961 | Variance explained - equivalent statistics for categorical data? | For generalised linear models (GLMs) we use deviance values as a generalisation of the scaled sums-of-squares used in regression (see related answer here). Instead of the coefficient of determination used in linear regression, we would use McFadden's $R^2$ value, which is given by:
$$R^2_\text{GLM} = \frac{\hat{\ell}_... | Variance explained - equivalent statistics for categorical data? | For generalised linear models (GLMs) we use deviance values as a generalisation of the scaled sums-of-squares used in regression (see related answer here). Instead of the coefficient of determination | Variance explained - equivalent statistics for categorical data?
For generalised linear models (GLMs) we use deviance values as a generalisation of the scaled sums-of-squares used in regression (see related answer here). Instead of the coefficient of determination used in linear regression, we would use McFadden's $R^... | Variance explained - equivalent statistics for categorical data?
For generalised linear models (GLMs) we use deviance values as a generalisation of the scaled sums-of-squares used in regression (see related answer here). Instead of the coefficient of determination |
35,962 | Variance explained - equivalent statistics for categorical data? | If you’re comfortable using $R^2$ as a crude approximation for proportion of variance explained, knowing that it is not true in the nonlinear case like in a logistic regression (but you decide how good it is for your work as an easy-to-compute estimate that your audience thinks it understands), then it doesn’t matter i... | Variance explained - equivalent statistics for categorical data? | If you’re comfortable using $R^2$ as a crude approximation for proportion of variance explained, knowing that it is not true in the nonlinear case like in a logistic regression (but you decide how goo | Variance explained - equivalent statistics for categorical data?
If you’re comfortable using $R^2$ as a crude approximation for proportion of variance explained, knowing that it is not true in the nonlinear case like in a logistic regression (but you decide how good it is for your work as an easy-to-compute estimate th... | Variance explained - equivalent statistics for categorical data?
If you’re comfortable using $R^2$ as a crude approximation for proportion of variance explained, knowing that it is not true in the nonlinear case like in a logistic regression (but you decide how goo |
35,963 | Variance explained - equivalent statistics for categorical data? | Given that you only have one dependent and one independent variable, you can perform correlation tests tailored for categorical variables and report the correlation statistics (and their p-values). This article is a good overview of your options.
That said, alternatively you can simply perform cross validation and repo... | Variance explained - equivalent statistics for categorical data? | Given that you only have one dependent and one independent variable, you can perform correlation tests tailored for categorical variables and report the correlation statistics (and their p-values). Th | Variance explained - equivalent statistics for categorical data?
Given that you only have one dependent and one independent variable, you can perform correlation tests tailored for categorical variables and report the correlation statistics (and their p-values). This article is a good overview of your options.
That sai... | Variance explained - equivalent statistics for categorical data?
Given that you only have one dependent and one independent variable, you can perform correlation tests tailored for categorical variables and report the correlation statistics (and their p-values). Th |
35,964 | Variance explained - equivalent statistics for categorical data? | Based on Dave's and Vasilis' response (I'm sorry Ben my non-statistical brain could not quite absorb your answer - which is on me) I wrote and roughly validated a python function to produce a grid of "R2 equivalents".
It worries me a bit that I'm needing to write code to do what I would have thought was a common task. ... | Variance explained - equivalent statistics for categorical data? | Based on Dave's and Vasilis' response (I'm sorry Ben my non-statistical brain could not quite absorb your answer - which is on me) I wrote and roughly validated a python function to produce a grid of | Variance explained - equivalent statistics for categorical data?
Based on Dave's and Vasilis' response (I'm sorry Ben my non-statistical brain could not quite absorb your answer - which is on me) I wrote and roughly validated a python function to produce a grid of "R2 equivalents".
It worries me a bit that I'm needing ... | Variance explained - equivalent statistics for categorical data?
Based on Dave's and Vasilis' response (I'm sorry Ben my non-statistical brain could not quite absorb your answer - which is on me) I wrote and roughly validated a python function to produce a grid of |
35,965 | Intuition of Bayesian normalizing constant | I needed to do this for a course I'm preparing, so I created this demonstration website: A demonstration of Bayes' theorem as "selecting a subset" in the binomial case (make sure to hide the toolbars, bottom right). Basically, if you show the joint distribution -- which is just $p(y\mid\theta)p(\theta)$ -- you can see ... | Intuition of Bayesian normalizing constant | I needed to do this for a course I'm preparing, so I created this demonstration website: A demonstration of Bayes' theorem as "selecting a subset" in the binomial case (make sure to hide the toolbars, | Intuition of Bayesian normalizing constant
I needed to do this for a course I'm preparing, so I created this demonstration website: A demonstration of Bayes' theorem as "selecting a subset" in the binomial case (make sure to hide the toolbars, bottom right). Basically, if you show the joint distribution -- which is jus... | Intuition of Bayesian normalizing constant
I needed to do this for a course I'm preparing, so I created this demonstration website: A demonstration of Bayes' theorem as "selecting a subset" in the binomial case (make sure to hide the toolbars, |
35,966 | Intuition of Bayesian normalizing constant | Besides the interpretations you mention, you can think of the normalizing constant as the value of the prior predictive distribution at the observed x. If the prior predictive is discrete then this is a probability mass, and if the prior predictive is continuous it is a probability density.
The prior predictive is in ... | Intuition of Bayesian normalizing constant | Besides the interpretations you mention, you can think of the normalizing constant as the value of the prior predictive distribution at the observed x. If the prior predictive is discrete then this is | Intuition of Bayesian normalizing constant
Besides the interpretations you mention, you can think of the normalizing constant as the value of the prior predictive distribution at the observed x. If the prior predictive is discrete then this is a probability mass, and if the prior predictive is continuous it is a probab... | Intuition of Bayesian normalizing constant
Besides the interpretations you mention, you can think of the normalizing constant as the value of the prior predictive distribution at the observed x. If the prior predictive is discrete then this is |
35,967 | Intuition of Bayesian normalizing constant | Richard's 3-d graphic was very helpful. What I need, however, is something I can paste as graph in a manuscript. After some searching, I located this image from Westfall and Henning, Understanding Advanced Statistical Methods, Chapman & Hall/CRC, 2013.
Relabeling the axes as the binomial probability p on the left and ... | Intuition of Bayesian normalizing constant | Richard's 3-d graphic was very helpful. What I need, however, is something I can paste as graph in a manuscript. After some searching, I located this image from Westfall and Henning, Understanding Adv | Intuition of Bayesian normalizing constant
Richard's 3-d graphic was very helpful. What I need, however, is something I can paste as graph in a manuscript. After some searching, I located this image from Westfall and Henning, Understanding Advanced Statistical Methods, Chapman & Hall/CRC, 2013.
Relabeling the axes as ... | Intuition of Bayesian normalizing constant
Richard's 3-d graphic was very helpful. What I need, however, is something I can paste as graph in a manuscript. After some searching, I located this image from Westfall and Henning, Understanding Adv |
35,968 | Why does the reconstruction error of truncated SVD equal the sum of squared singular values? | Let $$X = U\Sigma V^\prime$$ be the SVD of the $n\times r$ matrix $X$. Let $||\quad ||$ be any matrix norm that is left- and right-invariant under orthogonal transformations (reflections and rotations); that is, whenever $P$ is an $n\times n$ orthogonal matrix or $Q$ is an $r\times r$ orthogonal matrix, then
$$||P^\pr... | Why does the reconstruction error of truncated SVD equal the sum of squared singular values? | Let $$X = U\Sigma V^\prime$$ be the SVD of the $n\times r$ matrix $X$. Let $||\quad ||$ be any matrix norm that is left- and right-invariant under orthogonal transformations (reflections and rotation | Why does the reconstruction error of truncated SVD equal the sum of squared singular values?
Let $$X = U\Sigma V^\prime$$ be the SVD of the $n\times r$ matrix $X$. Let $||\quad ||$ be any matrix norm that is left- and right-invariant under orthogonal transformations (reflections and rotations); that is, whenever $P$ i... | Why does the reconstruction error of truncated SVD equal the sum of squared singular values?
Let $$X = U\Sigma V^\prime$$ be the SVD of the $n\times r$ matrix $X$. Let $||\quad ||$ be any matrix norm that is left- and right-invariant under orthogonal transformations (reflections and rotation |
35,969 | Differentiating the RSS w.r.t. $\beta$ in Linear Model | Note that $-y^TX - y^TX + \beta^T(X^TX + X^TX) \ne -2\beta^T(y^TX + X^TX)$. You pulled an extra $\beta^T$ out from the first term. So you were wrong with the algebra there.
You have,
$$RSS = y^Ty - \beta^TX^Ty - y^TX \beta + \beta^TX^TX\beta.$$
Notice that $-y^TX \beta$ is a scalar quantity, thus $y^TX\beta = (y^TX \... | Differentiating the RSS w.r.t. $\beta$ in Linear Model | Note that $-y^TX - y^TX + \beta^T(X^TX + X^TX) \ne -2\beta^T(y^TX + X^TX)$. You pulled an extra $\beta^T$ out from the first term. So you were wrong with the algebra there.
You have,
$$RSS = y^Ty - \ | Differentiating the RSS w.r.t. $\beta$ in Linear Model
Note that $-y^TX - y^TX + \beta^T(X^TX + X^TX) \ne -2\beta^T(y^TX + X^TX)$. You pulled an extra $\beta^T$ out from the first term. So you were wrong with the algebra there.
You have,
$$RSS = y^Ty - \beta^TX^Ty - y^TX \beta + \beta^TX^TX\beta.$$
Notice that $-y^TX ... | Differentiating the RSS w.r.t. $\beta$ in Linear Model
Note that $-y^TX - y^TX + \beta^T(X^TX + X^TX) \ne -2\beta^T(y^TX + X^TX)$. You pulled an extra $\beta^T$ out from the first term. So you were wrong with the algebra there.
You have,
$$RSS = y^Ty - \ |
35,970 | Is removing duplicate data necessary for Gaussian Process Regression (GPR)? | The duplicate data add no additional information, and rank-deficiency in the kernel matrix is fatal to the process. Removing them has literally no inferential consequence.
That said, numerically, the kernel matrix $K$ will occasionally become numerically singular if some points are too close together (but not necessar... | Is removing duplicate data necessary for Gaussian Process Regression (GPR)? | The duplicate data add no additional information, and rank-deficiency in the kernel matrix is fatal to the process. Removing them has literally no inferential consequence.
That said, numerically, the | Is removing duplicate data necessary for Gaussian Process Regression (GPR)?
The duplicate data add no additional information, and rank-deficiency in the kernel matrix is fatal to the process. Removing them has literally no inferential consequence.
That said, numerically, the kernel matrix $K$ will occasionally become ... | Is removing duplicate data necessary for Gaussian Process Regression (GPR)?
The duplicate data add no additional information, and rank-deficiency in the kernel matrix is fatal to the process. Removing them has literally no inferential consequence.
That said, numerically, the |
35,971 | Sampling 100000 times from normal distribution in R : strange distribution of samples' standard deviation | You've got a bug; you're taking sd of moy rather than sam. I bet your code is also pretty slow; a more R-like method would be as follows.
N <- 100000
n <- 10
d <- matrix(rnorm(N*n), nrow=10)
m <- colMeans(d)
s <- apply(d, 2, sd)
hist(s, 10000) | Sampling 100000 times from normal distribution in R : strange distribution of samples' standard devi | You've got a bug; you're taking sd of moy rather than sam. I bet your code is also pretty slow; a more R-like method would be as follows.
N <- 100000
n <- 10
d <- matrix(rnorm(N*n), nrow=10)
m <- colM | Sampling 100000 times from normal distribution in R : strange distribution of samples' standard deviation
You've got a bug; you're taking sd of moy rather than sam. I bet your code is also pretty slow; a more R-like method would be as follows.
N <- 100000
n <- 10
d <- matrix(rnorm(N*n), nrow=10)
m <- colMeans(d)
s <- a... | Sampling 100000 times from normal distribution in R : strange distribution of samples' standard devi
You've got a bug; you're taking sd of moy rather than sam. I bet your code is also pretty slow; a more R-like method would be as follows.
N <- 100000
n <- 10
d <- matrix(rnorm(N*n), nrow=10)
m <- colM |
35,972 | When (and why) is a conditional logistic regression equivalent to a Cox proportional hazards model? | Both conditional logistic regression and survival analysis are forms of semi-parametric inference where complex relationships between unmeasured risk factors (such as a baseline hazard function or unmeasured risk factors) are controlled by organizing data into risk sets.
Formally, a risk set in survival analysis is a ... | When (and why) is a conditional logistic regression equivalent to a Cox proportional hazards model? | Both conditional logistic regression and survival analysis are forms of semi-parametric inference where complex relationships between unmeasured risk factors (such as a baseline hazard function or unm | When (and why) is a conditional logistic regression equivalent to a Cox proportional hazards model?
Both conditional logistic regression and survival analysis are forms of semi-parametric inference where complex relationships between unmeasured risk factors (such as a baseline hazard function or unmeasured risk factors... | When (and why) is a conditional logistic regression equivalent to a Cox proportional hazards model?
Both conditional logistic regression and survival analysis are forms of semi-parametric inference where complex relationships between unmeasured risk factors (such as a baseline hazard function or unm |
35,973 | When (and why) is a conditional logistic regression equivalent to a Cox proportional hazards model? | The data structure is with all observed event times tied at the same time and only happens with one particular handling of these ties - namely the one that ensures that the test matches up with the standard log-rank test. This has an underlying assumption that tied event times truly indicate events happening at the exa... | When (and why) is a conditional logistic regression equivalent to a Cox proportional hazards model? | The data structure is with all observed event times tied at the same time and only happens with one particular handling of these ties - namely the one that ensures that the test matches up with the st | When (and why) is a conditional logistic regression equivalent to a Cox proportional hazards model?
The data structure is with all observed event times tied at the same time and only happens with one particular handling of these ties - namely the one that ensures that the test matches up with the standard log-rank test... | When (and why) is a conditional logistic regression equivalent to a Cox proportional hazards model?
The data structure is with all observed event times tied at the same time and only happens with one particular handling of these ties - namely the one that ensures that the test matches up with the st |
35,974 | Sample mean and variance independence in the case of correlated observations | The sample mean of a multivariate normal vector $\mathbf{X}=(X_1, X_2, \ldots, X_n)$ is a function of
$$M = X_1+X_2+\cdots X_n$$
and the sample variance is a function of the residual vector with components
$$Z_i = -X_1 - X_2 - \cdots - X_{i-1} + (n-1)X_i - X_{i+1} - \cdots - X_n,$$
$i=1, 2, \ldots, n$.
Let $\Sigma$ be... | Sample mean and variance independence in the case of correlated observations | The sample mean of a multivariate normal vector $\mathbf{X}=(X_1, X_2, \ldots, X_n)$ is a function of
$$M = X_1+X_2+\cdots X_n$$
and the sample variance is a function of the residual vector with comp | Sample mean and variance independence in the case of correlated observations
The sample mean of a multivariate normal vector $\mathbf{X}=(X_1, X_2, \ldots, X_n)$ is a function of
$$M = X_1+X_2+\cdots X_n$$
and the sample variance is a function of the residual vector with components
$$Z_i = -X_1 - X_2 - \cdots - X_{i-1... | Sample mean and variance independence in the case of correlated observations
The sample mean of a multivariate normal vector $\mathbf{X}=(X_1, X_2, \ldots, X_n)$ is a function of
$$M = X_1+X_2+\cdots X_n$$
and the sample variance is a function of the residual vector with comp |
35,975 | Sample mean and variance independence in the case of correlated observations | Only if $\Sigma$ is diagonal it appears. (see edit). I'm going to look at the covariance between $[\bar{X}, \ldots, \bar{X}]'$ and $X - [\bar{X}, \ldots, \bar{X}]'$. It's easier to work with the unsquared deviates rather than the sum of squared deviates.
Let $X \sim \text{Normal}(\mu, \Sigma)$, where $X, \mu \in R^n$.... | Sample mean and variance independence in the case of correlated observations | Only if $\Sigma$ is diagonal it appears. (see edit). I'm going to look at the covariance between $[\bar{X}, \ldots, \bar{X}]'$ and $X - [\bar{X}, \ldots, \bar{X}]'$. It's easier to work with the unsq | Sample mean and variance independence in the case of correlated observations
Only if $\Sigma$ is diagonal it appears. (see edit). I'm going to look at the covariance between $[\bar{X}, \ldots, \bar{X}]'$ and $X - [\bar{X}, \ldots, \bar{X}]'$. It's easier to work with the unsquared deviates rather than the sum of squar... | Sample mean and variance independence in the case of correlated observations
Only if $\Sigma$ is diagonal it appears. (see edit). I'm going to look at the covariance between $[\bar{X}, \ldots, \bar{X}]'$ and $X - [\bar{X}, \ldots, \bar{X}]'$. It's easier to work with the unsq |
35,976 | Cross-correlation between two seasonal series | The essential question is, what problem are you trying to solve?
If you intend to build a good model for the data (and later use the model for hypothesis testing, forecasting or whatever), you need to account for all the patterns there are. If there is seasonality, you should include seasonal patterns in your model. If... | Cross-correlation between two seasonal series | The essential question is, what problem are you trying to solve?
If you intend to build a good model for the data (and later use the model for hypothesis testing, forecasting or whatever), you need to | Cross-correlation between two seasonal series
The essential question is, what problem are you trying to solve?
If you intend to build a good model for the data (and later use the model for hypothesis testing, forecasting or whatever), you need to account for all the patterns there are. If there is seasonality, you shou... | Cross-correlation between two seasonal series
The essential question is, what problem are you trying to solve?
If you intend to build a good model for the data (and later use the model for hypothesis testing, forecasting or whatever), you need to |
35,977 | Cross-correlation between two seasonal series | If you are regressing two (unrelated) time series with seasonality you might get what is called spurious correlation. An example of that is available here
"it is important to consider whether significant trends exist in the series;
if we ignore a common trend, we may be estimating a spurious regression, in which both $... | Cross-correlation between two seasonal series | If you are regressing two (unrelated) time series with seasonality you might get what is called spurious correlation. An example of that is available here
"it is important to consider whether signific | Cross-correlation between two seasonal series
If you are regressing two (unrelated) time series with seasonality you might get what is called spurious correlation. An example of that is available here
"it is important to consider whether significant trends exist in the series;
if we ignore a common trend, we may be est... | Cross-correlation between two seasonal series
If you are regressing two (unrelated) time series with seasonality you might get what is called spurious correlation. An example of that is available here
"it is important to consider whether signific |
35,978 | Cross-correlation between two seasonal series | The link to the "more formal solution" seems to go to a generic uni page. Could you you update it to point to the article, if still available? | Cross-correlation between two seasonal series | The link to the "more formal solution" seems to go to a generic uni page. Could you you update it to point to the article, if still available? | Cross-correlation between two seasonal series
The link to the "more formal solution" seems to go to a generic uni page. Could you you update it to point to the article, if still available? | Cross-correlation between two seasonal series
The link to the "more formal solution" seems to go to a generic uni page. Could you you update it to point to the article, if still available? |
35,979 | Neural networks with complex weights | Gradient Descent will work in your case. You can use Theano. It supports differentiation with respect to complex variables. You can check this link if you need more information. One important note is that your error function needs to return a scalar.
I'm not quit sure that other algorithms like Conjugate gradient or qu... | Neural networks with complex weights | Gradient Descent will work in your case. You can use Theano. It supports differentiation with respect to complex variables. You can check this link if you need more information. One important note is | Neural networks with complex weights
Gradient Descent will work in your case. You can use Theano. It supports differentiation with respect to complex variables. You can check this link if you need more information. One important note is that your error function needs to return a scalar.
I'm not quit sure that other alg... | Neural networks with complex weights
Gradient Descent will work in your case. You can use Theano. It supports differentiation with respect to complex variables. You can check this link if you need more information. One important note is |
35,980 | Neural networks with complex weights | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
The differences between complex-valued and real-valued... | Neural networks with complex weights | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
| Neural networks with complex weights
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
The differences b... | Neural networks with complex weights
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35,981 | randomForest vs randomForestSRC discrepancies | One of the causes of the packages producing different results is the way nodesize is implemented internally. In randomForest, the value appears to be a strict lower bound. In randomForestSRC, while we (unfortunately) don't document the subtlety, we will not attempt to split a node without at least 2 * nodesize replica... | randomForest vs randomForestSRC discrepancies | One of the causes of the packages producing different results is the way nodesize is implemented internally. In randomForest, the value appears to be a strict lower bound. In randomForestSRC, while w | randomForest vs randomForestSRC discrepancies
One of the causes of the packages producing different results is the way nodesize is implemented internally. In randomForest, the value appears to be a strict lower bound. In randomForestSRC, while we (unfortunately) don't document the subtlety, we will not attempt to spli... | randomForest vs randomForestSRC discrepancies
One of the causes of the packages producing different results is the way nodesize is implemented internally. In randomForest, the value appears to be a strict lower bound. In randomForestSRC, while w |
35,982 | randomForest vs randomForestSRC discrepancies | I like randomForestSRC a lot. It has some really nice plots and diagnostics.
There are a lot of choices of how to implement the algorithm. For example, looking at the help pages, rfsrc has a splitrule, where "The default rule is weighted mean-squared error splitting mse". How does randomForest do it? They each can cont... | randomForest vs randomForestSRC discrepancies | I like randomForestSRC a lot. It has some really nice plots and diagnostics.
There are a lot of choices of how to implement the algorithm. For example, looking at the help pages, rfsrc has a splitrule | randomForest vs randomForestSRC discrepancies
I like randomForestSRC a lot. It has some really nice plots and diagnostics.
There are a lot of choices of how to implement the algorithm. For example, looking at the help pages, rfsrc has a splitrule, where "The default rule is weighted mean-squared error splitting mse". H... | randomForest vs randomForestSRC discrepancies
I like randomForestSRC a lot. It has some really nice plots and diagnostics.
There are a lot of choices of how to implement the algorithm. For example, looking at the help pages, rfsrc has a splitrule |
35,983 | Bibliography for linear models | I have read Christensen's books and would include it too. Certainly you'd want to include the following (in no particular order).
P. McCullagh, John A. Nelder. Generalized Linear Models, Second Edition. Volume 37 of Chapman & Hall/CRC Monographs on Statistics & Applied Probability. CRC Press, 1989.
Norman R. Drape... | Bibliography for linear models | I have read Christensen's books and would include it too. Certainly you'd want to include the following (in no particular order).
P. McCullagh, John A. Nelder. Generalized Linear Models, Second Ed | Bibliography for linear models
I have read Christensen's books and would include it too. Certainly you'd want to include the following (in no particular order).
P. McCullagh, John A. Nelder. Generalized Linear Models, Second Edition. Volume 37 of Chapman & Hall/CRC Monographs on Statistics & Applied Probability. C... | Bibliography for linear models
I have read Christensen's books and would include it too. Certainly you'd want to include the following (in no particular order).
P. McCullagh, John A. Nelder. Generalized Linear Models, Second Ed |
35,984 | Derivation of bias-variance decomposition expression for K-nearest neighbor regression | The previous answer is wrong in the part of bias derivation. I think the correct and full answer should be the following.
Firstly, I should note that for kNN we use an important assumption that all $X_i$ are fixed in the training set, i.e. $\mathcal{T}=(x_i,Y_i)_{i=1}^N$ (all randomness arises from the $Y_i$). Hence, h... | Derivation of bias-variance decomposition expression for K-nearest neighbor regression | The previous answer is wrong in the part of bias derivation. I think the correct and full answer should be the following.
Firstly, I should note that for kNN we use an important assumption that all $X | Derivation of bias-variance decomposition expression for K-nearest neighbor regression
The previous answer is wrong in the part of bias derivation. I think the correct and full answer should be the following.
Firstly, I should note that for kNN we use an important assumption that all $X_i$ are fixed in the training set... | Derivation of bias-variance decomposition expression for K-nearest neighbor regression
The previous answer is wrong in the part of bias derivation. I think the correct and full answer should be the following.
Firstly, I should note that for kNN we use an important assumption that all $X |
35,985 | Derivation of bias-variance decomposition expression for K-nearest neighbor regression | Edit
This is incorrect as others have noted, see correct answer below this one.
OLD
Let the label of $x$ be given by $Y(x) = f(x) + \epsilon$. Let the nearest neighbors of $x_0$ be $x_i$. Then the variance of this estimate is:
\begin{align}
variance &= var \left( \frac{1}{k} \sum_i^k Y(x_i) \right) \\
&= \frac{1}{... | Derivation of bias-variance decomposition expression for K-nearest neighbor regression | Edit
This is incorrect as others have noted, see correct answer below this one.
OLD
Let the label of $x$ be given by $Y(x) = f(x) + \epsilon$. Let the nearest neighbors of $x_0$ be $x_i$. Then the | Derivation of bias-variance decomposition expression for K-nearest neighbor regression
Edit
This is incorrect as others have noted, see correct answer below this one.
OLD
Let the label of $x$ be given by $Y(x) = f(x) + \epsilon$. Let the nearest neighbors of $x_0$ be $x_i$. Then the variance of this estimate is:
\b... | Derivation of bias-variance decomposition expression for K-nearest neighbor regression
Edit
This is incorrect as others have noted, see correct answer below this one.
OLD
Let the label of $x$ be given by $Y(x) = f(x) + \epsilon$. Let the nearest neighbors of $x_0$ be $x_i$. Then the |
35,986 | Why do statisticians prove asymptotic normality? | It is for example useful to do so in order to be able to quantify the sampling uncertainty of an estimator, or the null distribution of a test.
Recall that normal random variables take 95% of their realizations in the interval $\mu\pm1.96\sigma$. So if you can demonstrate that (typically, a scaled version of) an estima... | Why do statisticians prove asymptotic normality? | It is for example useful to do so in order to be able to quantify the sampling uncertainty of an estimator, or the null distribution of a test.
Recall that normal random variables take 95% of their re | Why do statisticians prove asymptotic normality?
It is for example useful to do so in order to be able to quantify the sampling uncertainty of an estimator, or the null distribution of a test.
Recall that normal random variables take 95% of their realizations in the interval $\mu\pm1.96\sigma$. So if you can demonstrat... | Why do statisticians prove asymptotic normality?
It is for example useful to do so in order to be able to quantify the sampling uncertainty of an estimator, or the null distribution of a test.
Recall that normal random variables take 95% of their re |
35,987 | Why do statisticians prove asymptotic normality? | Knowing the distribution of the random variable is knowing everything that is knowable about it. The estimators are random variables because they're functions of random samples. Therefore, it's not just a tradition but an ultimate goal of statistical analysis to establish the probability distribution of the metric. Oft... | Why do statisticians prove asymptotic normality? | Knowing the distribution of the random variable is knowing everything that is knowable about it. The estimators are random variables because they're functions of random samples. Therefore, it's not ju | Why do statisticians prove asymptotic normality?
Knowing the distribution of the random variable is knowing everything that is knowable about it. The estimators are random variables because they're functions of random samples. Therefore, it's not just a tradition but an ultimate goal of statistical analysis to establis... | Why do statisticians prove asymptotic normality?
Knowing the distribution of the random variable is knowing everything that is knowable about it. The estimators are random variables because they're functions of random samples. Therefore, it's not ju |
35,988 | A possible mistake in a conditional probability derivation | I may also be mistaken but see no difficulty with the decomposition.
When $t\ge 0$,
\begin{align*}
P(R(\cos(\pi U)&+\varrho)\ge t,\cos(\pi U)+\varrho\ge 0)\\
&+P(R(\cos(\pi U)+\varrho)\ge t,\cos(\pi U)+\varrho\le 0)\\
&=P(R(\cos(\pi U)+\varrho)\ge t,\cos(\pi U)+\varrho\ge 0)
\end{align*}
since the second term is zero,... | A possible mistake in a conditional probability derivation | I may also be mistaken but see no difficulty with the decomposition.
When $t\ge 0$,
\begin{align*}
P(R(\cos(\pi U)&+\varrho)\ge t,\cos(\pi U)+\varrho\ge 0)\\
&+P(R(\cos(\pi U)+\varrho)\ge t,\cos(\pi | A possible mistake in a conditional probability derivation
I may also be mistaken but see no difficulty with the decomposition.
When $t\ge 0$,
\begin{align*}
P(R(\cos(\pi U)&+\varrho)\ge t,\cos(\pi U)+\varrho\ge 0)\\
&+P(R(\cos(\pi U)+\varrho)\ge t,\cos(\pi U)+\varrho\le 0)\\
&=P(R(\cos(\pi U)+\varrho)\ge t,\cos(\pi U... | A possible mistake in a conditional probability derivation
I may also be mistaken but see no difficulty with the decomposition.
When $t\ge 0$,
\begin{align*}
P(R(\cos(\pi U)&+\varrho)\ge t,\cos(\pi U)+\varrho\ge 0)\\
&+P(R(\cos(\pi U)+\varrho)\ge t,\cos(\pi |
35,989 | What does the "1" in the R formula "y ~ 1" mean? | In most R regression packages, y ~ 1 means "fit an intercept only".
So in the case of linear regression, this is exactly the same as mean(y). For glm's, it depends on the link function.
Also, y ~ x - 1 means "regress x on y, but leave out the intercept". | What does the "1" in the R formula "y ~ 1" mean? | In most R regression packages, y ~ 1 means "fit an intercept only".
So in the case of linear regression, this is exactly the same as mean(y). For glm's, it depends on the link function.
Also, y ~ x | What does the "1" in the R formula "y ~ 1" mean?
In most R regression packages, y ~ 1 means "fit an intercept only".
So in the case of linear regression, this is exactly the same as mean(y). For glm's, it depends on the link function.
Also, y ~ x - 1 means "regress x on y, but leave out the intercept". | What does the "1" in the R formula "y ~ 1" mean?
In most R regression packages, y ~ 1 means "fit an intercept only".
So in the case of linear regression, this is exactly the same as mean(y). For glm's, it depends on the link function.
Also, y ~ x |
35,990 | What does the "1" in the R formula "y ~ 1" mean? | That means intercept only model. You can use model.matrix to find out. Try the following codes:
library(boot)
library(reshape)
dataset <- data.frame(Person = c(rep("A", 20), rep("B", 10)), Success = c(rbinom(20, 1, 0.25), rbinom(10, 1, 0.75)))
Aggregated <- cast(Person ~ ., data = dataset, value = "Success", fun = l... | What does the "1" in the R formula "y ~ 1" mean? | That means intercept only model. You can use model.matrix to find out. Try the following codes:
library(boot)
library(reshape)
dataset <- data.frame(Person = c(rep("A", 20), rep("B", 10)), Success = | What does the "1" in the R formula "y ~ 1" mean?
That means intercept only model. You can use model.matrix to find out. Try the following codes:
library(boot)
library(reshape)
dataset <- data.frame(Person = c(rep("A", 20), rep("B", 10)), Success = c(rbinom(20, 1, 0.25), rbinom(10, 1, 0.75)))
Aggregated <- cast(Perso... | What does the "1" in the R formula "y ~ 1" mean?
That means intercept only model. You can use model.matrix to find out. Try the following codes:
library(boot)
library(reshape)
dataset <- data.frame(Person = c(rep("A", 20), rep("B", 10)), Success = |
35,991 | Skewness, kurtosis and how many standard deviations values are from the mean | You can always calculate how many SDs values are from the mean by just plugging in sample values, (value $-$ mean)/SD, and then binning and counting.
Precise numerical facts such as you cite for the normal (Gaussian) in general depend on knowing one or more of the density, distribution or quantile functions, numerical... | Skewness, kurtosis and how many standard deviations values are from the mean | You can always calculate how many SDs values are from the mean by just plugging in sample values, (value $-$ mean)/SD, and then binning and counting.
Precise numerical facts such as you cite for the | Skewness, kurtosis and how many standard deviations values are from the mean
You can always calculate how many SDs values are from the mean by just plugging in sample values, (value $-$ mean)/SD, and then binning and counting.
Precise numerical facts such as you cite for the normal (Gaussian) in general depend on know... | Skewness, kurtosis and how many standard deviations values are from the mean
You can always calculate how many SDs values are from the mean by just plugging in sample values, (value $-$ mean)/SD, and then binning and counting.
Precise numerical facts such as you cite for the |
35,992 | Skewness, kurtosis and how many standard deviations values are from the mean | Here is a precise answer that shows that the median absolute deviation from the mean is not necessarily related to kurtosis.
Consider the family of distributions of $X = \mu + \sigma Z$, where $Z$ has the discrete distribution
$Z = -0.5$, with probability (wp) $.25$
$ = +0.5$, wp $.25$
$ = -1.2$, wp $.25 - \the... | Skewness, kurtosis and how many standard deviations values are from the mean | Here is a precise answer that shows that the median absolute deviation from the mean is not necessarily related to kurtosis.
Consider the family of distributions of $X = \mu + \sigma Z$, where $Z$ has | Skewness, kurtosis and how many standard deviations values are from the mean
Here is a precise answer that shows that the median absolute deviation from the mean is not necessarily related to kurtosis.
Consider the family of distributions of $X = \mu + \sigma Z$, where $Z$ has the discrete distribution
$Z = -0.5$, wi... | Skewness, kurtosis and how many standard deviations values are from the mean
Here is a precise answer that shows that the median absolute deviation from the mean is not necessarily related to kurtosis.
Consider the family of distributions of $X = \mu + \sigma Z$, where $Z$ has |
35,993 | OLS in terms of means and sample size | The approach is correct, but there's a slight numerical error: there are only $100$ females, not $200$. The mean heights for males and females can be converted to sums via
$$\text{Sum of male heights} = 100 \times 175$$
and
$$\text{Sum of female heights} = 100 \times 165.$$
Therefore the sum of all heights is
$$\text{... | OLS in terms of means and sample size | The approach is correct, but there's a slight numerical error: there are only $100$ females, not $200$. The mean heights for males and females can be converted to sums via
$$\text{Sum of male heights | OLS in terms of means and sample size
The approach is correct, but there's a slight numerical error: there are only $100$ females, not $200$. The mean heights for males and females can be converted to sums via
$$\text{Sum of male heights} = 100 \times 175$$
and
$$\text{Sum of female heights} = 100 \times 165.$$
Theref... | OLS in terms of means and sample size
The approach is correct, but there's a slight numerical error: there are only $100$ females, not $200$. The mean heights for males and females can be converted to sums via
$$\text{Sum of male heights |
35,994 | OLS in terms of means and sample size | Im quite confused. What does $u$ mean? Are these residuals? If so, then
$\mathbf{X'X}$ = $\begin{bmatrix}
200 & 100 \\ 100 & 100
\end{bmatrix}$
since
$\mathbf{X} = \frac{\partial{y}}{\partial\beta} =
\left[\begin{array}{cccc|cccc}
\frac{\partial{y_1}}{\partial\beta_1} & \frac{\partial{y_2}}{\partial\beta_1} & ... & \... | OLS in terms of means and sample size | Im quite confused. What does $u$ mean? Are these residuals? If so, then
$\mathbf{X'X}$ = $\begin{bmatrix}
200 & 100 \\ 100 & 100
\end{bmatrix}$
since
$\mathbf{X} = \frac{\partial{y}}{\partial\beta} = | OLS in terms of means and sample size
Im quite confused. What does $u$ mean? Are these residuals? If so, then
$\mathbf{X'X}$ = $\begin{bmatrix}
200 & 100 \\ 100 & 100
\end{bmatrix}$
since
$\mathbf{X} = \frac{\partial{y}}{\partial\beta} =
\left[\begin{array}{cccc|cccc}
\frac{\partial{y_1}}{\partial\beta_1} & \frac{\pa... | OLS in terms of means and sample size
Im quite confused. What does $u$ mean? Are these residuals? If so, then
$\mathbf{X'X}$ = $\begin{bmatrix}
200 & 100 \\ 100 & 100
\end{bmatrix}$
since
$\mathbf{X} = \frac{\partial{y}}{\partial\beta} = |
35,995 | What is the necessary condition for a unbiased estimator to be UMVUE? | Let us show that there can be a UMVUE which is not a sufficient statistic.
First of all, if the estimator $T$ takes (say) value $0$ on all samples, then clearly $T$ is a UMVUE of $0$, which latter can be considered a (constant) function of $\theta$. On the other hand, this estimator $T$ is clearly not sufficient in ge... | What is the necessary condition for a unbiased estimator to be UMVUE? | Let us show that there can be a UMVUE which is not a sufficient statistic.
First of all, if the estimator $T$ takes (say) value $0$ on all samples, then clearly $T$ is a UMVUE of $0$, which latter ca | What is the necessary condition for a unbiased estimator to be UMVUE?
Let us show that there can be a UMVUE which is not a sufficient statistic.
First of all, if the estimator $T$ takes (say) value $0$ on all samples, then clearly $T$ is a UMVUE of $0$, which latter can be considered a (constant) function of $\theta$.... | What is the necessary condition for a unbiased estimator to be UMVUE?
Let us show that there can be a UMVUE which is not a sufficient statistic.
First of all, if the estimator $T$ takes (say) value $0$ on all samples, then clearly $T$ is a UMVUE of $0$, which latter ca |
35,996 | What is the necessary condition for a unbiased estimator to be UMVUE? | On Uniformly Minimum Variance Unbiased Estimation when no Complete Sufficient Statistics Exist by L. Bondesson gives some examples of UMVUEs which are not complete sufficient statistics, including the following one:
Let $X_1, \ldots, X_n$ be independent observations of a random variable $X = \mu + \sigma Y$, where $\mu... | What is the necessary condition for a unbiased estimator to be UMVUE? | On Uniformly Minimum Variance Unbiased Estimation when no Complete Sufficient Statistics Exist by L. Bondesson gives some examples of UMVUEs which are not complete sufficient statistics, including the | What is the necessary condition for a unbiased estimator to be UMVUE?
On Uniformly Minimum Variance Unbiased Estimation when no Complete Sufficient Statistics Exist by L. Bondesson gives some examples of UMVUEs which are not complete sufficient statistics, including the following one:
Let $X_1, \ldots, X_n$ be independ... | What is the necessary condition for a unbiased estimator to be UMVUE?
On Uniformly Minimum Variance Unbiased Estimation when no Complete Sufficient Statistics Exist by L. Bondesson gives some examples of UMVUEs which are not complete sufficient statistics, including the |
35,997 | Non-normally distributed data - Box-Cox transformation? | The data are highly skewed & take just a few discrete values: the within-pair differences must consist of predominantly noughts & ones; no transformation will make them look much like normal variates. This is typical of count data where counts are fairly low.
If you assume that counts for each individual $j$ follow a d... | Non-normally distributed data - Box-Cox transformation? | The data are highly skewed & take just a few discrete values: the within-pair differences must consist of predominantly noughts & ones; no transformation will make them look much like normal variates. | Non-normally distributed data - Box-Cox transformation?
The data are highly skewed & take just a few discrete values: the within-pair differences must consist of predominantly noughts & ones; no transformation will make them look much like normal variates. This is typical of count data where counts are fairly low.
If y... | Non-normally distributed data - Box-Cox transformation?
The data are highly skewed & take just a few discrete values: the within-pair differences must consist of predominantly noughts & ones; no transformation will make them look much like normal variates. |
35,998 | Email and IP String preprocessing for classification task | This is a really interesting question! String vectorization is an area of active research right now, and a there's a ton of interesting approaches out there.
First of all, ip addresses are hierarchical, and can be split by decimals into 4 categorical variables, each with 256 levels (watch out for IPv4 vs IPv6 though)! ... | Email and IP String preprocessing for classification task | This is a really interesting question! String vectorization is an area of active research right now, and a there's a ton of interesting approaches out there.
First of all, ip addresses are hierarchica | Email and IP String preprocessing for classification task
This is a really interesting question! String vectorization is an area of active research right now, and a there's a ton of interesting approaches out there.
First of all, ip addresses are hierarchical, and can be split by decimals into 4 categorical variables, ... | Email and IP String preprocessing for classification task
This is a really interesting question! String vectorization is an area of active research right now, and a there's a ton of interesting approaches out there.
First of all, ip addresses are hierarchica |
35,999 | Spearman's correlation as a parameter | There's an interpretation given in some work on copulas.
e.g. see p 15 of Embrechts et al (2001) [1], which has for the Spearman correlation of $(X,Y)^T$:
$\rho_S(X,Y)=3(\mathbb{P}\{(X-\tilde{X})(Y-Y')>0\}-\mathbb{P}\{(X-\tilde{X})(Y-Y')<0\})$
where $(X, Y)^T$, $(\tilde{X},\tilde{Y})^T$
and $(X',Y')^T$ are independent ... | Spearman's correlation as a parameter | There's an interpretation given in some work on copulas.
e.g. see p 15 of Embrechts et al (2001) [1], which has for the Spearman correlation of $(X,Y)^T$:
$\rho_S(X,Y)=3(\mathbb{P}\{(X-\tilde{X})(Y-Y' | Spearman's correlation as a parameter
There's an interpretation given in some work on copulas.
e.g. see p 15 of Embrechts et al (2001) [1], which has for the Spearman correlation of $(X,Y)^T$:
$\rho_S(X,Y)=3(\mathbb{P}\{(X-\tilde{X})(Y-Y')>0\}-\mathbb{P}\{(X-\tilde{X})(Y-Y')<0\})$
where $(X, Y)^T$, $(\tilde{X},\tilde{Y... | Spearman's correlation as a parameter
There's an interpretation given in some work on copulas.
e.g. see p 15 of Embrechts et al (2001) [1], which has for the Spearman correlation of $(X,Y)^T$:
$\rho_S(X,Y)=3(\mathbb{P}\{(X-\tilde{X})(Y-Y' |
36,000 | Difference between "computational statistics" and "statistical computing"? | Roughly speaking, Computational Statistics refers to the statistical topics that require heavy computation, while Statistical Computing refers to the computational/numerical methods that can be applied to statistics.
We apply the computational methods from Statistical Computing to implement the statistical methods fr... | Difference between "computational statistics" and "statistical computing"? | Roughly speaking, Computational Statistics refers to the statistical topics that require heavy computation, while Statistical Computing refers to the computational/numerical methods that can be applie | Difference between "computational statistics" and "statistical computing"?
Roughly speaking, Computational Statistics refers to the statistical topics that require heavy computation, while Statistical Computing refers to the computational/numerical methods that can be applied to statistics.
We apply the computational... | Difference between "computational statistics" and "statistical computing"?
Roughly speaking, Computational Statistics refers to the statistical topics that require heavy computation, while Statistical Computing refers to the computational/numerical methods that can be applie |
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