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 |
|---|---|---|---|---|---|---|
36,301 | Neuron saturation occurs only in last layer or all layers? | As the problem with quadratic loss is that it learns slowly when it reaches saturation and it does not happen in the case of cross-entropy because of the σ′ term in the formula of the quadratic cost. And as the σ′ term vanishes in the cross-entropy this prevents this slow learning at saturation in the cross-entropy.
S... | Neuron saturation occurs only in last layer or all layers? | As the problem with quadratic loss is that it learns slowly when it reaches saturation and it does not happen in the case of cross-entropy because of the σ′ term in the formula of the quadratic cost. | Neuron saturation occurs only in last layer or all layers?
As the problem with quadratic loss is that it learns slowly when it reaches saturation and it does not happen in the case of cross-entropy because of the σ′ term in the formula of the quadratic cost. And as the σ′ term vanishes in the cross-entropy this prevent... | Neuron saturation occurs only in last layer or all layers?
As the problem with quadratic loss is that it learns slowly when it reaches saturation and it does not happen in the case of cross-entropy because of the σ′ term in the formula of the quadratic cost. |
36,302 | Why divide RSS by n-2 to get RSE? | The reason is based on trying to get an unbiased estimator of the underlying error variance in the regression. In a simple linear regression with normal error terms it can be shown that:
$$\text{RSS}(\mathbf{x},\mathbf{Y}) \equiv \sum_{i=1}^n (Y_i - \hat{Y}_i) \sim \sigma^2 \cdot \text{Chi-Sq}(df = n-2).$$
That is, un... | Why divide RSS by n-2 to get RSE? | The reason is based on trying to get an unbiased estimator of the underlying error variance in the regression. In a simple linear regression with normal error terms it can be shown that:
$$\text{RSS} | Why divide RSS by n-2 to get RSE?
The reason is based on trying to get an unbiased estimator of the underlying error variance in the regression. In a simple linear regression with normal error terms it can be shown that:
$$\text{RSS}(\mathbf{x},\mathbf{Y}) \equiv \sum_{i=1}^n (Y_i - \hat{Y}_i) \sim \sigma^2 \cdot \tex... | Why divide RSS by n-2 to get RSE?
The reason is based on trying to get an unbiased estimator of the underlying error variance in the regression. In a simple linear regression with normal error terms it can be shown that:
$$\text{RSS} |
36,303 | Why divide RSS by n-2 to get RSE? | In linear regression, if you are observing the relationship between a single predictor and its response then the equation is of the form
$$Y = b_0 + b_1 X.$$
Here, $Y$ is the response variable and $X$ is the predictor variable; $b_1$ and $b_0$ are coefficients that need to be found. Now we have two values to be found a... | Why divide RSS by n-2 to get RSE? | In linear regression, if you are observing the relationship between a single predictor and its response then the equation is of the form
$$Y = b_0 + b_1 X.$$
Here, $Y$ is the response variable and $X$ | Why divide RSS by n-2 to get RSE?
In linear regression, if you are observing the relationship between a single predictor and its response then the equation is of the form
$$Y = b_0 + b_1 X.$$
Here, $Y$ is the response variable and $X$ is the predictor variable; $b_1$ and $b_0$ are coefficients that need to be found. No... | Why divide RSS by n-2 to get RSE?
In linear regression, if you are observing the relationship between a single predictor and its response then the equation is of the form
$$Y = b_0 + b_1 X.$$
Here, $Y$ is the response variable and $X$ |
36,304 | Non Linear Endogeneity | I agree that the distinctions here are not easy. I will try to shed some light with an example. It will drop the $t$ index you use as I believe it is immaterial to your problem.
Suppose our model is
$$
y_i=x_i\beta+u_i,
$$
where $x\sim N(0,1)$ and $u_i=x_i^2-1$. Then,
$$
cov(x,u)=E(x^3-x)-E(x)E(x^2-1)=0-0-0\cdot0=0,
$... | Non Linear Endogeneity | I agree that the distinctions here are not easy. I will try to shed some light with an example. It will drop the $t$ index you use as I believe it is immaterial to your problem.
Suppose our model is
$ | Non Linear Endogeneity
I agree that the distinctions here are not easy. I will try to shed some light with an example. It will drop the $t$ index you use as I believe it is immaterial to your problem.
Suppose our model is
$$
y_i=x_i\beta+u_i,
$$
where $x\sim N(0,1)$ and $u_i=x_i^2-1$. Then,
$$
cov(x,u)=E(x^3-x)-E(x)E(... | Non Linear Endogeneity
I agree that the distinctions here are not easy. I will try to shed some light with an example. It will drop the $t$ index you use as I believe it is immaterial to your problem.
Suppose our model is
$ |
36,305 | question about bayesian structural time series model | I can see several potential issues with your example:
You don't specify a seed so bsts will use the system clock and serial correlation between successive monte carlo runs will mess up your statistics
Your chosen metric, rsquare might not be what you think it is (see the help for summary.bsts)
Your model is not a gre... | question about bayesian structural time series model | I can see several potential issues with your example:
You don't specify a seed so bsts will use the system clock and serial correlation between successive monte carlo runs will mess up your statisti | question about bayesian structural time series model
I can see several potential issues with your example:
You don't specify a seed so bsts will use the system clock and serial correlation between successive monte carlo runs will mess up your statistics
Your chosen metric, rsquare might not be what you think it is (s... | question about bayesian structural time series model
I can see several potential issues with your example:
You don't specify a seed so bsts will use the system clock and serial correlation between successive monte carlo runs will mess up your statisti |
36,306 | Conjecture related to Kolmogorov 0-1 Law (for events) | If you want events $B_n$ that are independent in an interesting manner (not simply because $\mathbb{P}(B_n) = 0$ or $\mathbb{P}(B_n) = 1$) then the conjecture is false.
Here is a pedantic example. Suppose $(\Omega, \mathcal{F},\mathbb{P})$ is a suitably rich probability space.
Let $A \in \mathcal{F}$ be $\mathbb{P}$... | Conjecture related to Kolmogorov 0-1 Law (for events) | If you want events $B_n$ that are independent in an interesting manner (not simply because $\mathbb{P}(B_n) = 0$ or $\mathbb{P}(B_n) = 1$) then the conjecture is false.
Here is a pedantic example. Su | Conjecture related to Kolmogorov 0-1 Law (for events)
If you want events $B_n$ that are independent in an interesting manner (not simply because $\mathbb{P}(B_n) = 0$ or $\mathbb{P}(B_n) = 1$) then the conjecture is false.
Here is a pedantic example. Suppose $(\Omega, \mathcal{F},\mathbb{P})$ is a suitably rich probab... | Conjecture related to Kolmogorov 0-1 Law (for events)
If you want events $B_n$ that are independent in an interesting manner (not simply because $\mathbb{P}(B_n) = 0$ or $\mathbb{P}(B_n) = 1$) then the conjecture is false.
Here is a pedantic example. Su |
36,307 | What to do with categorical data when calculating standardized z-scores? | Are ordinal data treated the same as continuous data when calculating standardized z-scores?
No, they are not: When dealing with data on different measurement scales it is important that your analysis should not use mathematical operations that are not meaningful within that measurement scale. For ordinal data, only ... | What to do with categorical data when calculating standardized z-scores? | Are ordinal data treated the same as continuous data when calculating standardized z-scores?
No, they are not: When dealing with data on different measurement scales it is important that your analysi | What to do with categorical data when calculating standardized z-scores?
Are ordinal data treated the same as continuous data when calculating standardized z-scores?
No, they are not: When dealing with data on different measurement scales it is important that your analysis should not use mathematical operations that a... | What to do with categorical data when calculating standardized z-scores?
Are ordinal data treated the same as continuous data when calculating standardized z-scores?
No, they are not: When dealing with data on different measurement scales it is important that your analysi |
36,308 | What to do with categorical data when calculating standardized z-scores? | what is performed across a particular feature should be performed across all features to make sure they have a common scale in that sense what is done to a particular featuere has to be done to all the features is that right.
otherwise is it ok to find out a value that represents a feature typically
mean for a normally... | What to do with categorical data when calculating standardized z-scores? | what is performed across a particular feature should be performed across all features to make sure they have a common scale in that sense what is done to a particular featuere has to be done to all th | What to do with categorical data when calculating standardized z-scores?
what is performed across a particular feature should be performed across all features to make sure they have a common scale in that sense what is done to a particular featuere has to be done to all the features is that right.
otherwise is it ok to... | What to do with categorical data when calculating standardized z-scores?
what is performed across a particular feature should be performed across all features to make sure they have a common scale in that sense what is done to a particular featuere has to be done to all th |
36,309 | What to do with categorical data when calculating standardized z-scores? | You state that you need to standardise because otherwise some variables will dominate others. This seems surprising if what you want to do is to calculate correlations or do regression (which from your question seems likely). Similarly if you want to do variable reduction with principal component analysis. If you leave... | What to do with categorical data when calculating standardized z-scores? | You state that you need to standardise because otherwise some variables will dominate others. This seems surprising if what you want to do is to calculate correlations or do regression (which from you | What to do with categorical data when calculating standardized z-scores?
You state that you need to standardise because otherwise some variables will dominate others. This seems surprising if what you want to do is to calculate correlations or do regression (which from your question seems likely). Similarly if you want... | What to do with categorical data when calculating standardized z-scores?
You state that you need to standardise because otherwise some variables will dominate others. This seems surprising if what you want to do is to calculate correlations or do regression (which from you |
36,310 | Can ridge regression be used in the presence of categorical predictors? | You are correct to assume that a categorical variable is encoded as indicator function/vector in your design matrix $X$; this is standard. To that respect, usually one of the levels is omitted and subsequently treated as baseline (if not you would have surely a rank-deficient design matrix when incorporating an interce... | Can ridge regression be used in the presence of categorical predictors? | You are correct to assume that a categorical variable is encoded as indicator function/vector in your design matrix $X$; this is standard. To that respect, usually one of the levels is omitted and sub | Can ridge regression be used in the presence of categorical predictors?
You are correct to assume that a categorical variable is encoded as indicator function/vector in your design matrix $X$; this is standard. To that respect, usually one of the levels is omitted and subsequently treated as baseline (if not you would ... | Can ridge regression be used in the presence of categorical predictors?
You are correct to assume that a categorical variable is encoded as indicator function/vector in your design matrix $X$; this is standard. To that respect, usually one of the levels is omitted and sub |
36,311 | Can ridge regression be used in the presence of categorical predictors? | Yes you can, your beta_ridge will be a number that will pop up when gender is 1 and won't have an effect when gender is 0. If you have more than one category, in my experience, make them all binary. e.g. if you have apple, orange, pear, instead of saying 1,2,3 say is_apple = [0,1] is_orange=[0,1] is_pear=[0,1]. | Can ridge regression be used in the presence of categorical predictors? | Yes you can, your beta_ridge will be a number that will pop up when gender is 1 and won't have an effect when gender is 0. If you have more than one category, in my experience, make them all binary. e | Can ridge regression be used in the presence of categorical predictors?
Yes you can, your beta_ridge will be a number that will pop up when gender is 1 and won't have an effect when gender is 0. If you have more than one category, in my experience, make them all binary. e.g. if you have apple, orange, pear, instead of ... | Can ridge regression be used in the presence of categorical predictors?
Yes you can, your beta_ridge will be a number that will pop up when gender is 1 and won't have an effect when gender is 0. If you have more than one category, in my experience, make them all binary. e |
36,312 | Number of parameters in mixed model | fm1
fixed = distance ~ age, random = ~ 1 | Subject
K = 4: the coefficients for intercept, and age; variances for random intercept, and error term.
fm2
fixed = distance ~ age + Sex, random = ~ 1 | Subject
K = 5: the coefficients for intercept, age, and Sex; variances for random intercept, and error term.
fm3
fixed ... | Number of parameters in mixed model | fm1
fixed = distance ~ age, random = ~ 1 | Subject
K = 4: the coefficients for intercept, and age; variances for random intercept, and error term.
fm2
fixed = distance ~ age + Sex, random = ~ 1 | S | Number of parameters in mixed model
fm1
fixed = distance ~ age, random = ~ 1 | Subject
K = 4: the coefficients for intercept, and age; variances for random intercept, and error term.
fm2
fixed = distance ~ age + Sex, random = ~ 1 | Subject
K = 5: the coefficients for intercept, age, and Sex; variances for random int... | Number of parameters in mixed model
fm1
fixed = distance ~ age, random = ~ 1 | Subject
K = 4: the coefficients for intercept, and age; variances for random intercept, and error term.
fm2
fixed = distance ~ age + Sex, random = ~ 1 | S |
36,313 | np package kernel density estimation with Epanechnikov kernel | EDIT
This is explained in the FAQ:
I use plot() (npplot()) to plot, say, a density and the resulting plot looks
like an inverted density rather than a density
This can occur when the datadriven
bandwidth is dramatically undersmoothed. Data-driven (i.e., automatic) bandwidth
selection procedures are not guarantee... | np package kernel density estimation with Epanechnikov kernel | EDIT
This is explained in the FAQ:
I use plot() (npplot()) to plot, say, a density and the resulting plot looks
like an inverted density rather than a density
This can occur when the datadriven
b | np package kernel density estimation with Epanechnikov kernel
EDIT
This is explained in the FAQ:
I use plot() (npplot()) to plot, say, a density and the resulting plot looks
like an inverted density rather than a density
This can occur when the datadriven
bandwidth is dramatically undersmoothed. Data-driven (i.e.,... | np package kernel density estimation with Epanechnikov kernel
EDIT
This is explained in the FAQ:
I use plot() (npplot()) to plot, say, a density and the resulting plot looks
like an inverted density rather than a density
This can occur when the datadriven
b |
36,314 | What are Monte Carlo simulations? | Monte Carlo method (there is also Monte Carlo algorithm) is a general name for a broad class of algorithms that use random sampling to obtain numerical results. It is used to solve statistical problems by simulation. It was one of the first methods of computer simulation that was described (see here for broader introdu... | What are Monte Carlo simulations? | Monte Carlo method (there is also Monte Carlo algorithm) is a general name for a broad class of algorithms that use random sampling to obtain numerical results. It is used to solve statistical problem | What are Monte Carlo simulations?
Monte Carlo method (there is also Monte Carlo algorithm) is a general name for a broad class of algorithms that use random sampling to obtain numerical results. It is used to solve statistical problems by simulation. It was one of the first methods of computer simulation that was descr... | What are Monte Carlo simulations?
Monte Carlo method (there is also Monte Carlo algorithm) is a general name for a broad class of algorithms that use random sampling to obtain numerical results. It is used to solve statistical problem |
36,315 | What are Monte Carlo simulations? | A repeated experiment is a good start .
In finance, for example, we try to plan for a 30 year retirement, with investment returns that are not guaranteed. One model states that each year has a 10% return with 14% standard deviation.
Monte Carlo can help us understand the expected return by treating this return as a r... | What are Monte Carlo simulations? | A repeated experiment is a good start .
In finance, for example, we try to plan for a 30 year retirement, with investment returns that are not guaranteed. One model states that each year has a 10% re | What are Monte Carlo simulations?
A repeated experiment is a good start .
In finance, for example, we try to plan for a 30 year retirement, with investment returns that are not guaranteed. One model states that each year has a 10% return with 14% standard deviation.
Monte Carlo can help us understand the expected ret... | What are Monte Carlo simulations?
A repeated experiment is a good start .
In finance, for example, we try to plan for a 30 year retirement, with investment returns that are not guaranteed. One model states that each year has a 10% re |
36,316 | Grid fineness and overfitting when tuning $\lambda$ in LASSO, ridge, elastic net | Can I say something about the propensity to overfit in (A) versus (B)?
Provided that both grids cover a sufficient range, grid fineness doesn't really have anything to do with overfitting in this problem (though a coarse grid might underfit if it skips over a profitable interval). It's not as if testing too many value... | Grid fineness and overfitting when tuning $\lambda$ in LASSO, ridge, elastic net | Can I say something about the propensity to overfit in (A) versus (B)?
Provided that both grids cover a sufficient range, grid fineness doesn't really have anything to do with overfitting in this pro | Grid fineness and overfitting when tuning $\lambda$ in LASSO, ridge, elastic net
Can I say something about the propensity to overfit in (A) versus (B)?
Provided that both grids cover a sufficient range, grid fineness doesn't really have anything to do with overfitting in this problem (though a coarse grid might underf... | Grid fineness and overfitting when tuning $\lambda$ in LASSO, ridge, elastic net
Can I say something about the propensity to overfit in (A) versus (B)?
Provided that both grids cover a sufficient range, grid fineness doesn't really have anything to do with overfitting in this pro |
36,317 | How to use R anova() results to select best model? | Question 1:
Regarding the difference in p-values (or F-values) for Model 2 in the first ANOVA compared to the second, there is no explanation apparent to me. They should indeed be the same, since we are comparing at that point Model 1 to Model 2 in both instances. It would be very useful to get further input about the... | How to use R anova() results to select best model? | Question 1:
Regarding the difference in p-values (or F-values) for Model 2 in the first ANOVA compared to the second, there is no explanation apparent to me. They should indeed be the same, since we | How to use R anova() results to select best model?
Question 1:
Regarding the difference in p-values (or F-values) for Model 2 in the first ANOVA compared to the second, there is no explanation apparent to me. They should indeed be the same, since we are comparing at that point Model 1 to Model 2 in both instances. It ... | How to use R anova() results to select best model?
Question 1:
Regarding the difference in p-values (or F-values) for Model 2 in the first ANOVA compared to the second, there is no explanation apparent to me. They should indeed be the same, since we |
36,318 | How to use R anova() results to select best model? | Question 1: Note, that the anova commands you provided above are equivalent to giving anova() the full model.
If you do the command:
anova(m3) # where m3 is lm(mpg~disp+wt+am,mtcars)
anova(m4) # where m4 is lm(mpg~disp+wt+hp,mtcars)
you will see that the anova is really telling you the significance of each variable i... | How to use R anova() results to select best model? | Question 1: Note, that the anova commands you provided above are equivalent to giving anova() the full model.
If you do the command:
anova(m3) # where m3 is lm(mpg~disp+wt+am,mtcars)
anova(m4) # wher | How to use R anova() results to select best model?
Question 1: Note, that the anova commands you provided above are equivalent to giving anova() the full model.
If you do the command:
anova(m3) # where m3 is lm(mpg~disp+wt+am,mtcars)
anova(m4) # where m4 is lm(mpg~disp+wt+hp,mtcars)
you will see that the anova is rea... | How to use R anova() results to select best model?
Question 1: Note, that the anova commands you provided above are equivalent to giving anova() the full model.
If you do the command:
anova(m3) # where m3 is lm(mpg~disp+wt+am,mtcars)
anova(m4) # wher |
36,319 | How to use R anova() results to select best model? | For Question 3 I would like to add to the answer given by Antoni.
It is better to use cross-validation which is a direct method to choose among various models in forward stepwise, backward stepwise or best subset instead of being confused among which to use. This will not require you to use ANOVA() at all. ANOVA is bet... | How to use R anova() results to select best model? | For Question 3 I would like to add to the answer given by Antoni.
It is better to use cross-validation which is a direct method to choose among various models in forward stepwise, backward stepwise or | How to use R anova() results to select best model?
For Question 3 I would like to add to the answer given by Antoni.
It is better to use cross-validation which is a direct method to choose among various models in forward stepwise, backward stepwise or best subset instead of being confused among which to use. This will ... | How to use R anova() results to select best model?
For Question 3 I would like to add to the answer given by Antoni.
It is better to use cross-validation which is a direct method to choose among various models in forward stepwise, backward stepwise or |
36,320 | Cohen's d from regression coefficient? | Cohen's d is the mean difference divided by the (pooled) standard deviation of the data within the groups. Indeed, the coefficient for the dummy variable gives you the mean difference, but instead of dividing by the standard deviation of the dependent variable, you should divide by the (pooled) within-group standard de... | Cohen's d from regression coefficient? | Cohen's d is the mean difference divided by the (pooled) standard deviation of the data within the groups. Indeed, the coefficient for the dummy variable gives you the mean difference, but instead of | Cohen's d from regression coefficient?
Cohen's d is the mean difference divided by the (pooled) standard deviation of the data within the groups. Indeed, the coefficient for the dummy variable gives you the mean difference, but instead of dividing by the standard deviation of the dependent variable, you should divide b... | Cohen's d from regression coefficient?
Cohen's d is the mean difference divided by the (pooled) standard deviation of the data within the groups. Indeed, the coefficient for the dummy variable gives you the mean difference, but instead of |
36,321 | Confidence intervals derived from 'inverted hypothesis test' | The idea is that your hypothesis test is comparing the statistic from you data to some parameter $\theta_0$. The confidence interval consists of all possible values of $\theta_0$ that the test would not reject for the given data. Usually instead of finding every possible value we just invert the equation for the test... | Confidence intervals derived from 'inverted hypothesis test' | The idea is that your hypothesis test is comparing the statistic from you data to some parameter $\theta_0$. The confidence interval consists of all possible values of $\theta_0$ that the test would | Confidence intervals derived from 'inverted hypothesis test'
The idea is that your hypothesis test is comparing the statistic from you data to some parameter $\theta_0$. The confidence interval consists of all possible values of $\theta_0$ that the test would not reject for the given data. Usually instead of finding ... | Confidence intervals derived from 'inverted hypothesis test'
The idea is that your hypothesis test is comparing the statistic from you data to some parameter $\theta_0$. The confidence interval consists of all possible values of $\theta_0$ that the test would |
36,322 | Power of a Interaction term in R | I do not think there is an R function to calculate this power. My usual suggestion in such cases is to simulate (e.g., here or here or here). Specifically, pick a sample size $n$, simulate your covariate, your binary variable, the dependence of your outcome on both and the noise. Fit your model and assess whether $p<\a... | Power of a Interaction term in R | I do not think there is an R function to calculate this power. My usual suggestion in such cases is to simulate (e.g., here or here or here). Specifically, pick a sample size $n$, simulate your covari | Power of a Interaction term in R
I do not think there is an R function to calculate this power. My usual suggestion in such cases is to simulate (e.g., here or here or here). Specifically, pick a sample size $n$, simulate your covariate, your binary variable, the dependence of your outcome on both and the noise. Fit yo... | Power of a Interaction term in R
I do not think there is an R function to calculate this power. My usual suggestion in such cases is to simulate (e.g., here or here or here). Specifically, pick a sample size $n$, simulate your covari |
36,323 | Power of a Interaction term in R | The R package InteractionPoweR can do a power analysis for this sort of regression.
install.packages("devtools")
devtools::install_github("dbaranger/InteractionPoweR")
library(InteractionPoweR)
test_power<-power_interaction(
n.iter = 1000, # number of simulations per unique combination of input parameter... | Power of a Interaction term in R | The R package InteractionPoweR can do a power analysis for this sort of regression.
install.packages("devtools")
devtools::install_github("dbaranger/InteractionPoweR")
library(InteractionPoweR)
test | Power of a Interaction term in R
The R package InteractionPoweR can do a power analysis for this sort of regression.
install.packages("devtools")
devtools::install_github("dbaranger/InteractionPoweR")
library(InteractionPoweR)
test_power<-power_interaction(
n.iter = 1000, # number of simulations per uniq... | Power of a Interaction term in R
The R package InteractionPoweR can do a power analysis for this sort of regression.
install.packages("devtools")
devtools::install_github("dbaranger/InteractionPoweR")
library(InteractionPoweR)
test |
36,324 | How to use log probabilities for Gaussian Naive Bayes? [duplicate] | The third option is right one. In general, it is true that:
$$ \log(ab) = \log(a) + \log(b)$$
Plugging in the Naive Bayes equation, you get
$$ \log(P(\text{class }_i| \textbf{ data})) \propto \log(P(\text{class}_i)) + \sum_j \log(P(\textrm{data}_j|\text{class}_i))$$
This value may be negative. If your all of your terms... | How to use log probabilities for Gaussian Naive Bayes? [duplicate] | The third option is right one. In general, it is true that:
$$ \log(ab) = \log(a) + \log(b)$$
Plugging in the Naive Bayes equation, you get
$$ \log(P(\text{class }_i| \textbf{ data})) \propto \log(P(\ | How to use log probabilities for Gaussian Naive Bayes? [duplicate]
The third option is right one. In general, it is true that:
$$ \log(ab) = \log(a) + \log(b)$$
Plugging in the Naive Bayes equation, you get
$$ \log(P(\text{class }_i| \textbf{ data})) \propto \log(P(\text{class}_i)) + \sum_j \log(P(\textrm{data}_j|\text... | How to use log probabilities for Gaussian Naive Bayes? [duplicate]
The third option is right one. In general, it is true that:
$$ \log(ab) = \log(a) + \log(b)$$
Plugging in the Naive Bayes equation, you get
$$ \log(P(\text{class }_i| \textbf{ data})) \propto \log(P(\ |
36,325 | Lindeberg CLT for exponential independent variables | When $\mu_j=1/j$ (which satisfies the assumptions of the book), it seems that the sequence $\left(\sum^n_{i=1}(e_i-\mu_i)/\sqrt{\sum^n_{j=1}\mu_j^2}\right)_{n\geqslant 1}$ converges to a constant plus a Gumbel distribution (see Subsection 5.3 in A uniform asymptotic expansion for
weighted sums of exponentials by
J.S.H.... | Lindeberg CLT for exponential independent variables | When $\mu_j=1/j$ (which satisfies the assumptions of the book), it seems that the sequence $\left(\sum^n_{i=1}(e_i-\mu_i)/\sqrt{\sum^n_{j=1}\mu_j^2}\right)_{n\geqslant 1}$ converges to a constant plus | Lindeberg CLT for exponential independent variables
When $\mu_j=1/j$ (which satisfies the assumptions of the book), it seems that the sequence $\left(\sum^n_{i=1}(e_i-\mu_i)/\sqrt{\sum^n_{j=1}\mu_j^2}\right)_{n\geqslant 1}$ converges to a constant plus a Gumbel distribution (see Subsection 5.3 in A uniform asymptotic e... | Lindeberg CLT for exponential independent variables
When $\mu_j=1/j$ (which satisfies the assumptions of the book), it seems that the sequence $\left(\sum^n_{i=1}(e_i-\mu_i)/\sqrt{\sum^n_{j=1}\mu_j^2}\right)_{n\geqslant 1}$ converges to a constant plus |
36,326 | Are group effects in a mixed effects model assumed to have been picked from a normal distribution? | You are correct in saying that in standard linear mixed effects models, the random effects are assumed to be normally distributed. Thus, if this assumption holds (at least approximately), we can use what we know about normal distributions to help describe the distribution of the random effects, such as 95% of the rando... | Are group effects in a mixed effects model assumed to have been picked from a normal distribution? | You are correct in saying that in standard linear mixed effects models, the random effects are assumed to be normally distributed. Thus, if this assumption holds (at least approximately), we can use w | Are group effects in a mixed effects model assumed to have been picked from a normal distribution?
You are correct in saying that in standard linear mixed effects models, the random effects are assumed to be normally distributed. Thus, if this assumption holds (at least approximately), we can use what we know about nor... | Are group effects in a mixed effects model assumed to have been picked from a normal distribution?
You are correct in saying that in standard linear mixed effects models, the random effects are assumed to be normally distributed. Thus, if this assumption holds (at least approximately), we can use w |
36,327 | Limitations to generalized additive model (GAM) | As mentioned in the comments, a propensity to overfit is a limitation of GAMs. Another limitation is that the model will lose predictability when the smoothed variables have values outside of the range of training dataset. Essentially, you are sacrificing predictability outside of your data range for precision within... | Limitations to generalized additive model (GAM) | As mentioned in the comments, a propensity to overfit is a limitation of GAMs. Another limitation is that the model will lose predictability when the smoothed variables have values outside of the ran | Limitations to generalized additive model (GAM)
As mentioned in the comments, a propensity to overfit is a limitation of GAMs. Another limitation is that the model will lose predictability when the smoothed variables have values outside of the range of training dataset. Essentially, you are sacrificing predictability... | Limitations to generalized additive model (GAM)
As mentioned in the comments, a propensity to overfit is a limitation of GAMs. Another limitation is that the model will lose predictability when the smoothed variables have values outside of the ran |
36,328 | Limitations to generalized additive model (GAM) | Probably one of the biggest limitations to GAMs is that they cannot model complex regression paths that involve multiple responses or things like mediation paths. Structural equation modeling (SEM) can achieve this, but generally the tools for nonlinear methods don't seem as developed like GAMs.
GAMs also don't allow y... | Limitations to generalized additive model (GAM) | Probably one of the biggest limitations to GAMs is that they cannot model complex regression paths that involve multiple responses or things like mediation paths. Structural equation modeling (SEM) ca | Limitations to generalized additive model (GAM)
Probably one of the biggest limitations to GAMs is that they cannot model complex regression paths that involve multiple responses or things like mediation paths. Structural equation modeling (SEM) can achieve this, but generally the tools for nonlinear methods don't seem... | Limitations to generalized additive model (GAM)
Probably one of the biggest limitations to GAMs is that they cannot model complex regression paths that involve multiple responses or things like mediation paths. Structural equation modeling (SEM) ca |
36,329 | Sufficient Statistic for non-exponential family distribution | First this is an exponential family (as shown by the above excerpt from Brown, 1986) since the density writes down as
$$\exp\{\Phi_1(\theta) S_1({\mathbf x})+\Phi_2(\theta) S_2({\mathbf x})-\Psi(\theta)\}$$against a particular dominating measure.
That the two coefficients $\Phi_1(\theta)$ and $\Phi_2(\theta)$ are conne... | Sufficient Statistic for non-exponential family distribution | First this is an exponential family (as shown by the above excerpt from Brown, 1986) since the density writes down as
$$\exp\{\Phi_1(\theta) S_1({\mathbf x})+\Phi_2(\theta) S_2({\mathbf x})-\Psi(\thet | Sufficient Statistic for non-exponential family distribution
First this is an exponential family (as shown by the above excerpt from Brown, 1986) since the density writes down as
$$\exp\{\Phi_1(\theta) S_1({\mathbf x})+\Phi_2(\theta) S_2({\mathbf x})-\Psi(\theta)\}$$against a particular dominating measure.
That the two... | Sufficient Statistic for non-exponential family distribution
First this is an exponential family (as shown by the above excerpt from Brown, 1986) since the density writes down as
$$\exp\{\Phi_1(\theta) S_1({\mathbf x})+\Phi_2(\theta) S_2({\mathbf x})-\Psi(\thet |
36,330 | Sufficient Statistic for non-exponential family distribution | There are different definitions of "sufficient statistic(s)" and "exponential family" by different authors. I wish the theoretical statisticians can agree on a single set of definitions to avoid confusing the world.
Most of authors do not require the dimensionality of sufficient statistics to match that of parameters.... | Sufficient Statistic for non-exponential family distribution | There are different definitions of "sufficient statistic(s)" and "exponential family" by different authors. I wish the theoretical statisticians can agree on a single set of definitions to avoid conf | Sufficient Statistic for non-exponential family distribution
There are different definitions of "sufficient statistic(s)" and "exponential family" by different authors. I wish the theoretical statisticians can agree on a single set of definitions to avoid confusing the world.
Most of authors do not require the dimensi... | Sufficient Statistic for non-exponential family distribution
There are different definitions of "sufficient statistic(s)" and "exponential family" by different authors. I wish the theoretical statisticians can agree on a single set of definitions to avoid conf |
36,331 | What is a good way to test a simple Recurrent Neural Network | A very simple time series to validate the correctness of your code is the one caused by the function sin(x). It's periodic nature makes it a good test function imo. Just print out (or plot) the output activations of your network and compare it with the desired values to see the performance.
Alternatively you can just t... | What is a good way to test a simple Recurrent Neural Network | A very simple time series to validate the correctness of your code is the one caused by the function sin(x). It's periodic nature makes it a good test function imo. Just print out (or plot) the output | What is a good way to test a simple Recurrent Neural Network
A very simple time series to validate the correctness of your code is the one caused by the function sin(x). It's periodic nature makes it a good test function imo. Just print out (or plot) the output activations of your network and compare it with the desire... | What is a good way to test a simple Recurrent Neural Network
A very simple time series to validate the correctness of your code is the one caused by the function sin(x). It's periodic nature makes it a good test function imo. Just print out (or plot) the output |
36,332 | What is a good way to test a simple Recurrent Neural Network | There's a good list of tests in Hochreiter's paper here. Also check this and the next slide on Schmidhuber's presentation. | What is a good way to test a simple Recurrent Neural Network | There's a good list of tests in Hochreiter's paper here. Also check this and the next slide on Schmidhuber's presentation. | What is a good way to test a simple Recurrent Neural Network
There's a good list of tests in Hochreiter's paper here. Also check this and the next slide on Schmidhuber's presentation. | What is a good way to test a simple Recurrent Neural Network
There's a good list of tests in Hochreiter's paper here. Also check this and the next slide on Schmidhuber's presentation. |
36,333 | Hierarchical clustering with agnes - how to cut the tree? | Husson et al. (2010) propose an empirical criterion based on the between-cluster inertia gain (see section 3.2 of this paper). Basically, the optimal number of clusters q is the one for which the increase in between-cluster dissimilarity for q clusters to q+1 clusters is significantly less than the increase in between-... | Hierarchical clustering with agnes - how to cut the tree? | Husson et al. (2010) propose an empirical criterion based on the between-cluster inertia gain (see section 3.2 of this paper). Basically, the optimal number of clusters q is the one for which the incr | Hierarchical clustering with agnes - how to cut the tree?
Husson et al. (2010) propose an empirical criterion based on the between-cluster inertia gain (see section 3.2 of this paper). Basically, the optimal number of clusters q is the one for which the increase in between-cluster dissimilarity for q clusters to q+1 cl... | Hierarchical clustering with agnes - how to cut the tree?
Husson et al. (2010) propose an empirical criterion based on the between-cluster inertia gain (see section 3.2 of this paper). Basically, the optimal number of clusters q is the one for which the incr |
36,334 | Hierarchical clustering with agnes - how to cut the tree? | The academic research literature has a lot of information on this.
What I found more readable and easier to understand is section 8.1.3 from the book "Practical Data science with R" by N.Zumel and J.Mount.
From page 186 ff
PICKING THE NUMBER OF CLUSTERS
There are a number of heuristics and
rules-of-thumb for picking ... | Hierarchical clustering with agnes - how to cut the tree? | The academic research literature has a lot of information on this.
What I found more readable and easier to understand is section 8.1.3 from the book "Practical Data science with R" by N.Zumel and J. | Hierarchical clustering with agnes - how to cut the tree?
The academic research literature has a lot of information on this.
What I found more readable and easier to understand is section 8.1.3 from the book "Practical Data science with R" by N.Zumel and J.Mount.
From page 186 ff
PICKING THE NUMBER OF CLUSTERS
There ... | Hierarchical clustering with agnes - how to cut the tree?
The academic research literature has a lot of information on this.
What I found more readable and easier to understand is section 8.1.3 from the book "Practical Data science with R" by N.Zumel and J. |
36,335 | Efficient way to compute distances between centroids from distance matrix | Let the points be indexed $x_1, x_2, \ldots, x_n$, all of them in $\mathbb{R}^d$. Let $\mathcal{I}$ be the indexes for one cluster and $\mathcal{J}$ the indexes for another cluster. The centroids are
$$c_\mathcal{I} = \frac{1}{|\mathcal{I}|} \sum_{i\in\mathcal{I}} x_i,\ c_\mathcal{J} = \frac{1}{|\mathcal{J}|} \sum_{j... | Efficient way to compute distances between centroids from distance matrix | Let the points be indexed $x_1, x_2, \ldots, x_n$, all of them in $\mathbb{R}^d$. Let $\mathcal{I}$ be the indexes for one cluster and $\mathcal{J}$ the indexes for another cluster. The centroids ar | Efficient way to compute distances between centroids from distance matrix
Let the points be indexed $x_1, x_2, \ldots, x_n$, all of them in $\mathbb{R}^d$. Let $\mathcal{I}$ be the indexes for one cluster and $\mathcal{J}$ the indexes for another cluster. The centroids are
$$c_\mathcal{I} = \frac{1}{|\mathcal{I}|} \s... | Efficient way to compute distances between centroids from distance matrix
Let the points be indexed $x_1, x_2, \ldots, x_n$, all of them in $\mathbb{R}^d$. Let $\mathcal{I}$ be the indexes for one cluster and $\mathcal{J}$ the indexes for another cluster. The centroids ar |
36,336 | Why to report R squared? | Under conditions for instance explained here, $R^2$ measures the proportion of the variance in the dependent variable explained by the regression, which is a natural measure. Adjusted $R^2$ does not have this interpretation, as it modifies the $R^2$ value.
So while adjusted $R^2$ has the indisputable advantage of not i... | Why to report R squared? | Under conditions for instance explained here, $R^2$ measures the proportion of the variance in the dependent variable explained by the regression, which is a natural measure. Adjusted $R^2$ does not h | Why to report R squared?
Under conditions for instance explained here, $R^2$ measures the proportion of the variance in the dependent variable explained by the regression, which is a natural measure. Adjusted $R^2$ does not have this interpretation, as it modifies the $R^2$ value.
So while adjusted $R^2$ has the indisp... | Why to report R squared?
Under conditions for instance explained here, $R^2$ measures the proportion of the variance in the dependent variable explained by the regression, which is a natural measure. Adjusted $R^2$ does not h |
36,337 | Why to report R squared? | Adjusted R-squared is useful for comparing different regression models. This task cannot be accomplished by R-squared which, as Others have already said, has another informative goal, that is expressing the proportion of variance of the dependent variable that is explained by the regression model under investigation. | Why to report R squared? | Adjusted R-squared is useful for comparing different regression models. This task cannot be accomplished by R-squared which, as Others have already said, has another informative goal, that is expressi | Why to report R squared?
Adjusted R-squared is useful for comparing different regression models. This task cannot be accomplished by R-squared which, as Others have already said, has another informative goal, that is expressing the proportion of variance of the dependent variable that is explained by the regression mod... | Why to report R squared?
Adjusted R-squared is useful for comparing different regression models. This task cannot be accomplished by R-squared which, as Others have already said, has another informative goal, that is expressi |
36,338 | What is the honesty condition for regression trees? | The meaning of the "fully grown" condition, as the authors elaborate afterwards, is that each terminal node of the tree contains exactly one instance from the training data. The authors remark that this is a "theoretical convenience":
Meanwhile, (C) is a theoretical convenience that lets us simplify the
exposition. ... | What is the honesty condition for regression trees? | The meaning of the "fully grown" condition, as the authors elaborate afterwards, is that each terminal node of the tree contains exactly one instance from the training data. The authors remark that th | What is the honesty condition for regression trees?
The meaning of the "fully grown" condition, as the authors elaborate afterwards, is that each terminal node of the tree contains exactly one instance from the training data. The authors remark that this is a "theoretical convenience":
Meanwhile, (C) is a theoretical ... | What is the honesty condition for regression trees?
The meaning of the "fully grown" condition, as the authors elaborate afterwards, is that each terminal node of the tree contains exactly one instance from the training data. The authors remark that th |
36,339 | Simple Question about ROC Curve | The true positive rate is one minus the false negative rate; the false positive rate is one minus the true negative rate: so it doesn't really make any difference which one you pick from each pair to talk about. You're certainly not in any sense giving more importance to classification of positive cases over negative c... | Simple Question about ROC Curve | The true positive rate is one minus the false negative rate; the false positive rate is one minus the true negative rate: so it doesn't really make any difference which one you pick from each pair to | Simple Question about ROC Curve
The true positive rate is one minus the false negative rate; the false positive rate is one minus the true negative rate: so it doesn't really make any difference which one you pick from each pair to talk about. You're certainly not in any sense giving more importance to classification o... | Simple Question about ROC Curve
The true positive rate is one minus the false negative rate; the false positive rate is one minus the true negative rate: so it doesn't really make any difference which one you pick from each pair to |
36,340 | Can I safely use variable importance of a random forest in a paper? | It does not really seem that you can justify the "X, Y and Z are best predictors" sentence in this case. At least because all predictors are best for purpose, i.e. are they so specific that they can be used as the final truth in diagnosis, or are they so sensitive that given some predictor values, no cases will be miss... | Can I safely use variable importance of a random forest in a paper? | It does not really seem that you can justify the "X, Y and Z are best predictors" sentence in this case. At least because all predictors are best for purpose, i.e. are they so specific that they can b | Can I safely use variable importance of a random forest in a paper?
It does not really seem that you can justify the "X, Y and Z are best predictors" sentence in this case. At least because all predictors are best for purpose, i.e. are they so specific that they can be used as the final truth in diagnosis, or are they ... | Can I safely use variable importance of a random forest in a paper?
It does not really seem that you can justify the "X, Y and Z are best predictors" sentence in this case. At least because all predictors are best for purpose, i.e. are they so specific that they can b |
36,341 | Can I safely use variable importance of a random forest in a paper? | Would agree with everything coulminer answered above. Would add a few points that I'm not sure are useful:
- It might be hard to justify using RF over more traditional methods. You'd likely need to emphasize a combination of large number of variables + unknown number of interactions + non-linear effects to be convincin... | Can I safely use variable importance of a random forest in a paper? | Would agree with everything coulminer answered above. Would add a few points that I'm not sure are useful:
- It might be hard to justify using RF over more traditional methods. You'd likely need to em | Can I safely use variable importance of a random forest in a paper?
Would agree with everything coulminer answered above. Would add a few points that I'm not sure are useful:
- It might be hard to justify using RF over more traditional methods. You'd likely need to emphasize a combination of large number of variables +... | Can I safely use variable importance of a random forest in a paper?
Would agree with everything coulminer answered above. Would add a few points that I'm not sure are useful:
- It might be hard to justify using RF over more traditional methods. You'd likely need to em |
36,342 | Variance calculation RELU function (deep learning) | In terms of integrals you have:
$$
E[x^2] = \int_{-\infty}^{+\infty} \max(0,y)^2 p(y) dy
$$
where the part $y < 0$ does not contribute to the Integral
$$
= \int_{0}^{+\infty} y^2 p(y) dy
$$
which we can write as half the integral over the entire real domain ($y^2$ is symmetric around 0 and $p(y)$ is assumed to be symm... | Variance calculation RELU function (deep learning) | In terms of integrals you have:
$$
E[x^2] = \int_{-\infty}^{+\infty} \max(0,y)^2 p(y) dy
$$
where the part $y < 0$ does not contribute to the Integral
$$
= \int_{0}^{+\infty} y^2 p(y) dy
$$
which we | Variance calculation RELU function (deep learning)
In terms of integrals you have:
$$
E[x^2] = \int_{-\infty}^{+\infty} \max(0,y)^2 p(y) dy
$$
where the part $y < 0$ does not contribute to the Integral
$$
= \int_{0}^{+\infty} y^2 p(y) dy
$$
which we can write as half the integral over the entire real domain ($y^2$ is ... | Variance calculation RELU function (deep learning)
In terms of integrals you have:
$$
E[x^2] = \int_{-\infty}^{+\infty} \max(0,y)^2 p(y) dy
$$
where the part $y < 0$ does not contribute to the Integral
$$
= \int_{0}^{+\infty} y^2 p(y) dy
$$
which we |
36,343 | Conditional distribution for Exponential family | I have answered my own question. It turned out to be a rather obvious application of Bayes Rule only after making a somewhat arbitrary assumption. My question was not very clear, mostly due to my own tenuous understanding at that time.
However, this result is used quite a lot in machine learning literature involving in... | Conditional distribution for Exponential family | I have answered my own question. It turned out to be a rather obvious application of Bayes Rule only after making a somewhat arbitrary assumption. My question was not very clear, mostly due to my own | Conditional distribution for Exponential family
I have answered my own question. It turned out to be a rather obvious application of Bayes Rule only after making a somewhat arbitrary assumption. My question was not very clear, mostly due to my own tenuous understanding at that time.
However, this result is used quite a... | Conditional distribution for Exponential family
I have answered my own question. It turned out to be a rather obvious application of Bayes Rule only after making a somewhat arbitrary assumption. My question was not very clear, mostly due to my own |
36,344 | Explanation that the prior predictive (marginal) distribution follows from prior and sampling distributions | The equation follows from the definition of marginal distribution:
$$
p(y) = \int_\theta{p(y, \theta)}
$$
And, from factoring the joint probability of data and parameters into conditional probabilities, like so: $$p(y, \theta) = p(y|\theta)p(\theta)$$
(If this is confusing, divide both sides by $p(\theta)$ to get the ... | Explanation that the prior predictive (marginal) distribution follows from prior and sampling distri | The equation follows from the definition of marginal distribution:
$$
p(y) = \int_\theta{p(y, \theta)}
$$
And, from factoring the joint probability of data and parameters into conditional probabilitie | Explanation that the prior predictive (marginal) distribution follows from prior and sampling distributions
The equation follows from the definition of marginal distribution:
$$
p(y) = \int_\theta{p(y, \theta)}
$$
And, from factoring the joint probability of data and parameters into conditional probabilities, like so: ... | Explanation that the prior predictive (marginal) distribution follows from prior and sampling distri
The equation follows from the definition of marginal distribution:
$$
p(y) = \int_\theta{p(y, \theta)}
$$
And, from factoring the joint probability of data and parameters into conditional probabilitie |
36,345 | Explanation that the prior predictive (marginal) distribution follows from prior and sampling distributions | Yes it removes the conditionality. Prior Predictive is for predictive a new value BEFORE the sample has been gathered. The only information we have at this stage is our belief about the Prior, $p(\theta$) and sampling distribution i.e. $p(y_{new}|\theta)$.
After the sample has been gathered, we have new information i.e... | Explanation that the prior predictive (marginal) distribution follows from prior and sampling distri | Yes it removes the conditionality. Prior Predictive is for predictive a new value BEFORE the sample has been gathered. The only information we have at this stage is our belief about the Prior, $p(\the | Explanation that the prior predictive (marginal) distribution follows from prior and sampling distributions
Yes it removes the conditionality. Prior Predictive is for predictive a new value BEFORE the sample has been gathered. The only information we have at this stage is our belief about the Prior, $p(\theta$) and sam... | Explanation that the prior predictive (marginal) distribution follows from prior and sampling distri
Yes it removes the conditionality. Prior Predictive is for predictive a new value BEFORE the sample has been gathered. The only information we have at this stage is our belief about the Prior, $p(\the |
36,346 | Dealing with Postcode in Regression? | There are a couple approaches you can take here, from least to most laborious:
If there are few enough post codes in your study, it's probably fine to leave it as is. A general rule of thumb is that there should be ten events per variable (where each postal code is a different variable, plus all the other variables yo... | Dealing with Postcode in Regression? | There are a couple approaches you can take here, from least to most laborious:
If there are few enough post codes in your study, it's probably fine to leave it as is. A general rule of thumb is that | Dealing with Postcode in Regression?
There are a couple approaches you can take here, from least to most laborious:
If there are few enough post codes in your study, it's probably fine to leave it as is. A general rule of thumb is that there should be ten events per variable (where each postal code is a different vari... | Dealing with Postcode in Regression?
There are a couple approaches you can take here, from least to most laborious:
If there are few enough post codes in your study, it's probably fine to leave it as is. A general rule of thumb is that |
36,347 | cost function in logistic regression vs optimization algorithms | If I understand what you have done, you have done things out of order. Maximum likelihood or penalized maximum likelihood estimation is the gold standard method for estimating parameters in a regression model. The cost function does not apply to how one estimates risk. The cost function is applied to individual risk... | cost function in logistic regression vs optimization algorithms | If I understand what you have done, you have done things out of order. Maximum likelihood or penalized maximum likelihood estimation is the gold standard method for estimating parameters in a regress | cost function in logistic regression vs optimization algorithms
If I understand what you have done, you have done things out of order. Maximum likelihood or penalized maximum likelihood estimation is the gold standard method for estimating parameters in a regression model. The cost function does not apply to how one ... | cost function in logistic regression vs optimization algorithms
If I understand what you have done, you have done things out of order. Maximum likelihood or penalized maximum likelihood estimation is the gold standard method for estimating parameters in a regress |
36,348 | GLMM overfitting solutions | Computing the standard deviation (or variance of the coefficients) essentially means getting the fixed-effect estimates for each level (analogous to the BLUPs/conditional modes in a mixed model) and computing their variance. You can do this by appropriately setting contrasts to contr.sum (sum-to-zero contrasts) (in th... | GLMM overfitting solutions | Computing the standard deviation (or variance of the coefficients) essentially means getting the fixed-effect estimates for each level (analogous to the BLUPs/conditional modes in a mixed model) and c | GLMM overfitting solutions
Computing the standard deviation (or variance of the coefficients) essentially means getting the fixed-effect estimates for each level (analogous to the BLUPs/conditional modes in a mixed model) and computing their variance. You can do this by appropriately setting contrasts to contr.sum (su... | GLMM overfitting solutions
Computing the standard deviation (or variance of the coefficients) essentially means getting the fixed-effect estimates for each level (analogous to the BLUPs/conditional modes in a mixed model) and c |
36,349 | Why $Y$ should be transformed before the predictors? | Transforming X doesn't impact the shape of the conditional distribution, nor heteroskedasticity, so transforming X really only serves to deal with nonlinear relationships. (If you're fitting additive models it might serve to help with eliminating interaction, but even that's often best left to transforming Y)
An exampl... | Why $Y$ should be transformed before the predictors? | Transforming X doesn't impact the shape of the conditional distribution, nor heteroskedasticity, so transforming X really only serves to deal with nonlinear relationships. (If you're fitting additive | Why $Y$ should be transformed before the predictors?
Transforming X doesn't impact the shape of the conditional distribution, nor heteroskedasticity, so transforming X really only serves to deal with nonlinear relationships. (If you're fitting additive models it might serve to help with eliminating interaction, but eve... | Why $Y$ should be transformed before the predictors?
Transforming X doesn't impact the shape of the conditional distribution, nor heteroskedasticity, so transforming X really only serves to deal with nonlinear relationships. (If you're fitting additive |
36,350 | Why $Y$ should be transformed before the predictors? | Transforming Y initially is an anachronistic approach to data analysis. Our great-great-great grandfathers did that so why shouldn't we? Lots of reasons and your post reflecting that Gaussian assumptions are solely based on the errors from a model NOT the Y series is dead-on. | Why $Y$ should be transformed before the predictors? | Transforming Y initially is an anachronistic approach to data analysis. Our great-great-great grandfathers did that so why shouldn't we? Lots of reasons and your post reflecting that Gaussian assumpti | Why $Y$ should be transformed before the predictors?
Transforming Y initially is an anachronistic approach to data analysis. Our great-great-great grandfathers did that so why shouldn't we? Lots of reasons and your post reflecting that Gaussian assumptions are solely based on the errors from a model NOT the Y series is... | Why $Y$ should be transformed before the predictors?
Transforming Y initially is an anachronistic approach to data analysis. Our great-great-great grandfathers did that so why shouldn't we? Lots of reasons and your post reflecting that Gaussian assumpti |
36,351 | Interpretation of $\theta$ in negative binomial regression | $\theta$ is known as a dispersion parameter in GLM. But what does that really mean? Let me use an example to explain what the $\theta$ parameter is. Say you went to a party of mixed faculty members. You, as a statistician, looked for another statistician. Let $p$ be the probability of you succeeding in finding a statis... | Interpretation of $\theta$ in negative binomial regression | $\theta$ is known as a dispersion parameter in GLM. But what does that really mean? Let me use an example to explain what the $\theta$ parameter is. Say you went to a party of mixed faculty members. Y | Interpretation of $\theta$ in negative binomial regression
$\theta$ is known as a dispersion parameter in GLM. But what does that really mean? Let me use an example to explain what the $\theta$ parameter is. Say you went to a party of mixed faculty members. You, as a statistician, looked for another statistician. Let $... | Interpretation of $\theta$ in negative binomial regression
$\theta$ is known as a dispersion parameter in GLM. But what does that really mean? Let me use an example to explain what the $\theta$ parameter is. Say you went to a party of mixed faculty members. Y |
36,352 | Understanding scipy Kolmogorov-Smirnov test [duplicate] | For the KS test the p-value is itself distributed uniformly in [0,1] if the H0 is true (which it is if you test whether it your sample is from $U(0,1)$ and the random number generation works okay). It therefore must "vary wildly" between 0 and 1, in fact its standard deviation is $1/\sqrt{12}$ which is roughly 0.3.
Yo... | Understanding scipy Kolmogorov-Smirnov test [duplicate] | For the KS test the p-value is itself distributed uniformly in [0,1] if the H0 is true (which it is if you test whether it your sample is from $U(0,1)$ and the random number generation works okay). It | Understanding scipy Kolmogorov-Smirnov test [duplicate]
For the KS test the p-value is itself distributed uniformly in [0,1] if the H0 is true (which it is if you test whether it your sample is from $U(0,1)$ and the random number generation works okay). It therefore must "vary wildly" between 0 and 1, in fact its stand... | Understanding scipy Kolmogorov-Smirnov test [duplicate]
For the KS test the p-value is itself distributed uniformly in [0,1] if the H0 is true (which it is if you test whether it your sample is from $U(0,1)$ and the random number generation works okay). It |
36,353 | finding global minima of a random forest estimator | It's easy with scipy.optimize.fmin, evaluating your model on sampled points or rolling your own optimization routine. This is the bread and butter of how many models work, so it's worth learning about in great detail. See for instance, these lecture notes.
This probably isn't what you want though. The $r^2$ you cross-v... | finding global minima of a random forest estimator | It's easy with scipy.optimize.fmin, evaluating your model on sampled points or rolling your own optimization routine. This is the bread and butter of how many models work, so it's worth learning about | finding global minima of a random forest estimator
It's easy with scipy.optimize.fmin, evaluating your model on sampled points or rolling your own optimization routine. This is the bread and butter of how many models work, so it's worth learning about in great detail. See for instance, these lecture notes.
This probabl... | finding global minima of a random forest estimator
It's easy with scipy.optimize.fmin, evaluating your model on sampled points or rolling your own optimization routine. This is the bread and butter of how many models work, so it's worth learning about |
36,354 | why is adaboost predicting probabilities with so little standard deviation? | When using AdaBoost (and most other machine learning algorithms, such as Support Vector Machines), it is important to calibrate prediction scores. One popular method is Isotonic Regression, which I recommend for most machine learning tasks. If you pass the prediction scores from your AdaBoost model through an Isotonic ... | why is adaboost predicting probabilities with so little standard deviation? | When using AdaBoost (and most other machine learning algorithms, such as Support Vector Machines), it is important to calibrate prediction scores. One popular method is Isotonic Regression, which I re | why is adaboost predicting probabilities with so little standard deviation?
When using AdaBoost (and most other machine learning algorithms, such as Support Vector Machines), it is important to calibrate prediction scores. One popular method is Isotonic Regression, which I recommend for most machine learning tasks. If ... | why is adaboost predicting probabilities with so little standard deviation?
When using AdaBoost (and most other machine learning algorithms, such as Support Vector Machines), it is important to calibrate prediction scores. One popular method is Isotonic Regression, which I re |
36,355 | What exactly does a Type III test do? | It may help you to read my answer here: How to interpret type I (sequential) ANOVA and MANOVA to get a fuller sense of the types of sums of squares.
The distinction between the different types of SS only matters if you have more than one variable. It isn't clear that you do from your description, but imagine that yo... | What exactly does a Type III test do? | It may help you to read my answer here: How to interpret type I (sequential) ANOVA and MANOVA to get a fuller sense of the types of sums of squares.
The distinction between the different types of SS | What exactly does a Type III test do?
It may help you to read my answer here: How to interpret type I (sequential) ANOVA and MANOVA to get a fuller sense of the types of sums of squares.
The distinction between the different types of SS only matters if you have more than one variable. It isn't clear that you do from... | What exactly does a Type III test do?
It may help you to read my answer here: How to interpret type I (sequential) ANOVA and MANOVA to get a fuller sense of the types of sums of squares.
The distinction between the different types of SS |
36,356 | How do I investigate how long it takes one variable to affect another in a time series | Try looking at cross-correlations between the two series:
correlate series 1 with series 2;
correlate series 1 with series 2 lagged by 1 period;
correlate series 1 with series 2 lagged by 2 periods;
etc.
Explore as many lags as can reasonably be expected to be plausible in the context you are in. Also, depending on... | How do I investigate how long it takes one variable to affect another in a time series | Try looking at cross-correlations between the two series:
correlate series 1 with series 2;
correlate series 1 with series 2 lagged by 1 period;
correlate series 1 with series 2 lagged by 2 periods; | How do I investigate how long it takes one variable to affect another in a time series
Try looking at cross-correlations between the two series:
correlate series 1 with series 2;
correlate series 1 with series 2 lagged by 1 period;
correlate series 1 with series 2 lagged by 2 periods;
etc.
Explore as many lags as c... | How do I investigate how long it takes one variable to affect another in a time series
Try looking at cross-correlations between the two series:
correlate series 1 with series 2;
correlate series 1 with series 2 lagged by 1 period;
correlate series 1 with series 2 lagged by 2 periods; |
36,357 | How do I investigate how long it takes one variable to affect another in a time series | As often in regression, there are several ways to proceed, but the most straightforward one is probably given by the moving-average approach. That is, you assume a linear (linear-in-the-parameters) relationship between your target $y_t$ (change_in_rank) and your regressand $x_t$ (change_in_goodness) given by
$$
y_t = \... | How do I investigate how long it takes one variable to affect another in a time series | As often in regression, there are several ways to proceed, but the most straightforward one is probably given by the moving-average approach. That is, you assume a linear (linear-in-the-parameters) re | How do I investigate how long it takes one variable to affect another in a time series
As often in regression, there are several ways to proceed, but the most straightforward one is probably given by the moving-average approach. That is, you assume a linear (linear-in-the-parameters) relationship between your target $y... | How do I investigate how long it takes one variable to affect another in a time series
As often in regression, there are several ways to proceed, but the most straightforward one is probably given by the moving-average approach. That is, you assume a linear (linear-in-the-parameters) re |
36,358 | Notation: Deterministic Variable, Random Variable, Realization of Random Variable, Function | I think people usually use late-in-alphabet letters like $X, Y, Z$ for random variables -- and $x, y, z$ for realizations thereof. Then you have early-in-alphabet letters like $a, b, c$ that can be used for deterministic values, and later letters like $f, g, h$ for names of functions (also capitals like $F, G, H$ for c... | Notation: Deterministic Variable, Random Variable, Realization of Random Variable, Function | I think people usually use late-in-alphabet letters like $X, Y, Z$ for random variables -- and $x, y, z$ for realizations thereof. Then you have early-in-alphabet letters like $a, b, c$ that can be us | Notation: Deterministic Variable, Random Variable, Realization of Random Variable, Function
I think people usually use late-in-alphabet letters like $X, Y, Z$ for random variables -- and $x, y, z$ for realizations thereof. Then you have early-in-alphabet letters like $a, b, c$ that can be used for deterministic values,... | Notation: Deterministic Variable, Random Variable, Realization of Random Variable, Function
I think people usually use late-in-alphabet letters like $X, Y, Z$ for random variables -- and $x, y, z$ for realizations thereof. Then you have early-in-alphabet letters like $a, b, c$ that can be us |
36,359 | Number of distinct bootstrap samples | Let's ask the computer to generate some small examples. (The language is R.)
Take $n=5$. Begin by placing $n-1$ bars (represented as ones) randomly within $n + n-1 = 2n-1$ places; that is, by selecting an $n-1$ element subset of $\{1,2,\ldots,2n-1\}$:
n <- 5
set.seed(17)
y <- rep(0, 2*n-1)
y[sample.int(2*n-1, n-1, re... | Number of distinct bootstrap samples | Let's ask the computer to generate some small examples. (The language is R.)
Take $n=5$. Begin by placing $n-1$ bars (represented as ones) randomly within $n + n-1 = 2n-1$ places; that is, by select | Number of distinct bootstrap samples
Let's ask the computer to generate some small examples. (The language is R.)
Take $n=5$. Begin by placing $n-1$ bars (represented as ones) randomly within $n + n-1 = 2n-1$ places; that is, by selecting an $n-1$ element subset of $\{1,2,\ldots,2n-1\}$:
n <- 5
set.seed(17)
y <- rep(... | Number of distinct bootstrap samples
Let's ask the computer to generate some small examples. (The language is R.)
Take $n=5$. Begin by placing $n-1$ bars (represented as ones) randomly within $n + n-1 = 2n-1$ places; that is, by select |
36,360 | Number of distinct bootstrap samples | Notice that we want to take $n$ elements from the original set ($n$ distinct elements sample set) with replicates allowed.
Then suppose if I denote $x_1$: number of replicates of the first element, $x_2$: number of replicates of second element,...., $x_n$: number of replicates of $n$th element.
While we must take eleme... | Number of distinct bootstrap samples | Notice that we want to take $n$ elements from the original set ($n$ distinct elements sample set) with replicates allowed.
Then suppose if I denote $x_1$: number of replicates of the first element, $x | Number of distinct bootstrap samples
Notice that we want to take $n$ elements from the original set ($n$ distinct elements sample set) with replicates allowed.
Then suppose if I denote $x_1$: number of replicates of the first element, $x_2$: number of replicates of second element,...., $x_n$: number of replicates of $n... | Number of distinct bootstrap samples
Notice that we want to take $n$ elements from the original set ($n$ distinct elements sample set) with replicates allowed.
Then suppose if I denote $x_1$: number of replicates of the first element, $x |
36,361 | Advantage of multiple simulations in old-fashioned Monte Carlo? | As long as problems regarding pseudo-random number generation are avoided (see note at the end), the two approaches ($k$ simulations with $n$ draws vs. single simulation with sufficiently large $n$) are equivalent with respect to estimating the mean. Regarding memory, observe that, in the $k$ simulations case, you need... | Advantage of multiple simulations in old-fashioned Monte Carlo? | As long as problems regarding pseudo-random number generation are avoided (see note at the end), the two approaches ($k$ simulations with $n$ draws vs. single simulation with sufficiently large $n$) a | Advantage of multiple simulations in old-fashioned Monte Carlo?
As long as problems regarding pseudo-random number generation are avoided (see note at the end), the two approaches ($k$ simulations with $n$ draws vs. single simulation with sufficiently large $n$) are equivalent with respect to estimating the mean. Regar... | Advantage of multiple simulations in old-fashioned Monte Carlo?
As long as problems regarding pseudo-random number generation are avoided (see note at the end), the two approaches ($k$ simulations with $n$ draws vs. single simulation with sufficiently large $n$) a |
36,362 | Issues with fitting distribution to heavy-tailed data | I applied Tukey-Lambda PPCC to your data to get the following plot:
You see the local max at $\lambda=-0.47$, which implies heavier tails than normal distribution. For normal it's 0.14 and for Cauchy it's -1. So, if you go with this view, then your data's tail is not as "fat" as Cauchy's, but much fatter than normal.
... | Issues with fitting distribution to heavy-tailed data | I applied Tukey-Lambda PPCC to your data to get the following plot:
You see the local max at $\lambda=-0.47$, which implies heavier tails than normal distribution. For normal it's 0.14 and for Cauchy | Issues with fitting distribution to heavy-tailed data
I applied Tukey-Lambda PPCC to your data to get the following plot:
You see the local max at $\lambda=-0.47$, which implies heavier tails than normal distribution. For normal it's 0.14 and for Cauchy it's -1. So, if you go with this view, then your data's tail is n... | Issues with fitting distribution to heavy-tailed data
I applied Tukey-Lambda PPCC to your data to get the following plot:
You see the local max at $\lambda=-0.47$, which implies heavier tails than normal distribution. For normal it's 0.14 and for Cauchy |
36,363 | Issues with fitting distribution to heavy-tailed data | your data set is really very long-tailed. Consider making 1st some tests on less extreme data (to check this look e.g. to sampling kurtosis). L-moments are not suited for very-very-long tail data, e.g. for Cauchy distribution not all L-moments exist.
Bye Stephan | Issues with fitting distribution to heavy-tailed data | your data set is really very long-tailed. Consider making 1st some tests on less extreme data (to check this look e.g. to sampling kurtosis). L-moments are not suited for very-very-long tail data, e.g | Issues with fitting distribution to heavy-tailed data
your data set is really very long-tailed. Consider making 1st some tests on less extreme data (to check this look e.g. to sampling kurtosis). L-moments are not suited for very-very-long tail data, e.g. for Cauchy distribution not all L-moments exist.
Bye Stephan | Issues with fitting distribution to heavy-tailed data
your data set is really very long-tailed. Consider making 1st some tests on less extreme data (to check this look e.g. to sampling kurtosis). L-moments are not suited for very-very-long tail data, e.g |
36,364 | Issues with fitting distribution to heavy-tailed data | One way to do this is to construct an L-moment diagram with your data. There in an R package (with full manual) at http://cran.r-project.org/web/packages/lmom/index.html.
The L-moment diagram will plot the L-skewness and L-kurtosis of your data on a chart and show how different distribution-types would appear on that ... | Issues with fitting distribution to heavy-tailed data | One way to do this is to construct an L-moment diagram with your data. There in an R package (with full manual) at http://cran.r-project.org/web/packages/lmom/index.html.
The L-moment diagram will pl | Issues with fitting distribution to heavy-tailed data
One way to do this is to construct an L-moment diagram with your data. There in an R package (with full manual) at http://cran.r-project.org/web/packages/lmom/index.html.
The L-moment diagram will plot the L-skewness and L-kurtosis of your data on a chart and show ... | Issues with fitting distribution to heavy-tailed data
One way to do this is to construct an L-moment diagram with your data. There in an R package (with full manual) at http://cran.r-project.org/web/packages/lmom/index.html.
The L-moment diagram will pl |
36,365 | Issues with fitting distribution to heavy-tailed data | experimented a bit with a mixture of distributions but am still far from a satisfying result..
Data-set is the same as above (v1).
The maximum-likelihood function for the pareto is given by:
pareto.MLE <- function(X)
{
n <- length(X)
m <- min(X)
a <- n/sum(log(X)-log(m))
return( c(m,a) )
}
Now, I fit the upper... | Issues with fitting distribution to heavy-tailed data | experimented a bit with a mixture of distributions but am still far from a satisfying result..
Data-set is the same as above (v1).
The maximum-likelihood function for the pareto is given by:
pareto.M | Issues with fitting distribution to heavy-tailed data
experimented a bit with a mixture of distributions but am still far from a satisfying result..
Data-set is the same as above (v1).
The maximum-likelihood function for the pareto is given by:
pareto.MLE <- function(X)
{
n <- length(X)
m <- min(X)
a <- n/sum(log(... | Issues with fitting distribution to heavy-tailed data
experimented a bit with a mixture of distributions but am still far from a satisfying result..
Data-set is the same as above (v1).
The maximum-likelihood function for the pareto is given by:
pareto.M |
36,366 | Difference between rel error and xerror in rpart regression trees | The "rel error" is $1 - R^2$ Root mean squared error, similar to linear regression. This is the error on the observations used to estimate the model.
The "xerror" is related to the PRESS statistic. This is the error on the observations from cross validation data. | Difference between rel error and xerror in rpart regression trees | The "rel error" is $1 - R^2$ Root mean squared error, similar to linear regression. This is the error on the observations used to estimate the model.
The "xerror" is related to the PRESS statistic. Th | Difference between rel error and xerror in rpart regression trees
The "rel error" is $1 - R^2$ Root mean squared error, similar to linear regression. This is the error on the observations used to estimate the model.
The "xerror" is related to the PRESS statistic. This is the error on the observations from cross validat... | Difference between rel error and xerror in rpart regression trees
The "rel error" is $1 - R^2$ Root mean squared error, similar to linear regression. This is the error on the observations used to estimate the model.
The "xerror" is related to the PRESS statistic. Th |
36,367 | Difference between rel error and xerror in rpart regression trees | Can not comment but 1-r^2 is not called Root mean squared error, Root mean squared error is RMSE, The square root of MSE
r^2 is called coefficient of determination, which is SSR/SSTO
SSR = total error sum of (predicted value - mean value)^2
SSTO = total error sum of (real value - mean value)^2
can also be calculated by... | Difference between rel error and xerror in rpart regression trees | Can not comment but 1-r^2 is not called Root mean squared error, Root mean squared error is RMSE, The square root of MSE
r^2 is called coefficient of determination, which is SSR/SSTO
SSR = total error | Difference between rel error and xerror in rpart regression trees
Can not comment but 1-r^2 is not called Root mean squared error, Root mean squared error is RMSE, The square root of MSE
r^2 is called coefficient of determination, which is SSR/SSTO
SSR = total error sum of (predicted value - mean value)^2
SSTO = total ... | Difference between rel error and xerror in rpart regression trees
Can not comment but 1-r^2 is not called Root mean squared error, Root mean squared error is RMSE, The square root of MSE
r^2 is called coefficient of determination, which is SSR/SSTO
SSR = total error |
36,368 | Generating a correlated data matrix where both observations and variables are correlated | You can do the same thing that you did to the columns of the matrix to make them correlated, just do it to the rows instead. This will adjust the correlation on the observations within a column without affecting the correlations between the columns much:
L = chol(OCRMt)# Cholesky decomposition
p = dim(L)[1]
M2 <- t(L... | Generating a correlated data matrix where both observations and variables are correlated | You can do the same thing that you did to the columns of the matrix to make them correlated, just do it to the rows instead. This will adjust the correlation on the observations within a column witho | Generating a correlated data matrix where both observations and variables are correlated
You can do the same thing that you did to the columns of the matrix to make them correlated, just do it to the rows instead. This will adjust the correlation on the observations within a column without affecting the correlations b... | Generating a correlated data matrix where both observations and variables are correlated
You can do the same thing that you did to the columns of the matrix to make them correlated, just do it to the rows instead. This will adjust the correlation on the observations within a column witho |
36,369 | Creating and interpreting Bland-Altman plot | Have you looked at the Wikipedia entry I linked in your question?
You don't plot "the mean of the data", but for each data point measured in two ways, you plot the difference in the two measurements ($y$) against the average of the two measurements ($x$). Using R and some toy data:
> set.seed(1)
> measurements <- matri... | Creating and interpreting Bland-Altman plot | Have you looked at the Wikipedia entry I linked in your question?
You don't plot "the mean of the data", but for each data point measured in two ways, you plot the difference in the two measurements ( | Creating and interpreting Bland-Altman plot
Have you looked at the Wikipedia entry I linked in your question?
You don't plot "the mean of the data", but for each data point measured in two ways, you plot the difference in the two measurements ($y$) against the average of the two measurements ($x$). Using R and some toy... | Creating and interpreting Bland-Altman plot
Have you looked at the Wikipedia entry I linked in your question?
You don't plot "the mean of the data", but for each data point measured in two ways, you plot the difference in the two measurements ( |
36,370 | Creating and interpreting Bland-Altman plot | If you would like to do this in Python you can use this code
import matplotlib.pyplot as plt
import numpy as np
from numpy.random import random
%matplotlib inline
plt.style.use('ggplot')
I just added the last line because I like the ggplot style.
def plotblandaltman(x,y,title,sd_limit):
plt.figure(figsize=(20,8))
... | Creating and interpreting Bland-Altman plot | If you would like to do this in Python you can use this code
import matplotlib.pyplot as plt
import numpy as np
from numpy.random import random
%matplotlib inline
plt.style.use('ggplot')
I just added | Creating and interpreting Bland-Altman plot
If you would like to do this in Python you can use this code
import matplotlib.pyplot as plt
import numpy as np
from numpy.random import random
%matplotlib inline
plt.style.use('ggplot')
I just added the last line because I like the ggplot style.
def plotblandaltman(x,y,titl... | Creating and interpreting Bland-Altman plot
If you would like to do this in Python you can use this code
import matplotlib.pyplot as plt
import numpy as np
from numpy.random import random
%matplotlib inline
plt.style.use('ggplot')
I just added |
36,371 | Total expectation theorem for Poisson processes | heropup is right. The problem is that once you know that $X=A$, $X$ is not merely drawn from the exponential with rate $\lambda_A$ since you also know that the sampled value had to be small enough to win the comparison with the hypothetical sampled value from $B$.
So, the density given that $X=A$ is the renormalized p... | Total expectation theorem for Poisson processes | heropup is right. The problem is that once you know that $X=A$, $X$ is not merely drawn from the exponential with rate $\lambda_A$ since you also know that the sampled value had to be small enough to | Total expectation theorem for Poisson processes
heropup is right. The problem is that once you know that $X=A$, $X$ is not merely drawn from the exponential with rate $\lambda_A$ since you also know that the sampled value had to be small enough to win the comparison with the hypothetical sampled value from $B$.
So, th... | Total expectation theorem for Poisson processes
heropup is right. The problem is that once you know that $X=A$, $X$ is not merely drawn from the exponential with rate $\lambda_A$ since you also know that the sampled value had to be small enough to |
36,372 | Total expectation theorem for Poisson processes | \begin{align}
& \Pr( T_{A+B} > t \mid X=A) \\[10pt]
= {} & \frac{\Pr(T_{A+B} > t\ \&\ X=A)}{\Pr(X=A)} \\[10pt]
= {} & \frac{\Pr(t < T_A < T_B)}{\Pr(X=A)}, \tag 1
\end{align}
${}$
\begin{align}
\text{and } & \Pr(t < T_A < T_B) \\[10pt]
= {} & \int_t^\infty \left( \int_u^\infty
e^{-\lambda_A u} e^{-\lambda_B v} (\lambd... | Total expectation theorem for Poisson processes | \begin{align}
& \Pr( T_{A+B} > t \mid X=A) \\[10pt]
= {} & \frac{\Pr(T_{A+B} > t\ \&\ X=A)}{\Pr(X=A)} \\[10pt]
= {} & \frac{\Pr(t < T_A < T_B)}{\Pr(X=A)}, \tag 1
\end{align}
${}$
\begin{align}
\text{ | Total expectation theorem for Poisson processes
\begin{align}
& \Pr( T_{A+B} > t \mid X=A) \\[10pt]
= {} & \frac{\Pr(T_{A+B} > t\ \&\ X=A)}{\Pr(X=A)} \\[10pt]
= {} & \frac{\Pr(t < T_A < T_B)}{\Pr(X=A)}, \tag 1
\end{align}
${}$
\begin{align}
\text{and } & \Pr(t < T_A < T_B) \\[10pt]
= {} & \int_t^\infty \left( \int_u^\... | Total expectation theorem for Poisson processes
\begin{align}
& \Pr( T_{A+B} > t \mid X=A) \\[10pt]
= {} & \frac{\Pr(T_{A+B} > t\ \&\ X=A)}{\Pr(X=A)} \\[10pt]
= {} & \frac{\Pr(t < T_A < T_B)}{\Pr(X=A)}, \tag 1
\end{align}
${}$
\begin{align}
\text{ |
36,373 | Relation of Mahalanobis Distance to Log Likelihood | For completeness, I'll answer myself.
I eventually found the equation here, and as @whuber writes in his comment:
Define:
$k$: the multivariate dimension
$\mu$: the multivariate mean (a $k$-dimensional vector);
$\Sigma$: the $k\times k$ covariance matrix;
Then:
The squared Mahalanobis Distance is: $D^2=(x-\mu)^T\... | Relation of Mahalanobis Distance to Log Likelihood | For completeness, I'll answer myself.
I eventually found the equation here, and as @whuber writes in his comment:
Define:
$k$: the multivariate dimension
$\mu$: the multivariate mean (a $k$-dimens | Relation of Mahalanobis Distance to Log Likelihood
For completeness, I'll answer myself.
I eventually found the equation here, and as @whuber writes in his comment:
Define:
$k$: the multivariate dimension
$\mu$: the multivariate mean (a $k$-dimensional vector);
$\Sigma$: the $k\times k$ covariance matrix;
Then:
T... | Relation of Mahalanobis Distance to Log Likelihood
For completeness, I'll answer myself.
I eventually found the equation here, and as @whuber writes in his comment:
Define:
$k$: the multivariate dimension
$\mu$: the multivariate mean (a $k$-dimens |
36,374 | Is it valid to use Hartigans' dip test to reject uni-modal null hypothesis with large N? | It isn't so much that a hypothesis test has "too much power" with large n, it's that hypothesis tests don't seem to answer the question you're interested in.
Given the large sample size, the second plot looks to me like there is some suggestion of at least 2 modes.
"Qualitatively different" is basically "is this diff... | Is it valid to use Hartigans' dip test to reject uni-modal null hypothesis with large N? | It isn't so much that a hypothesis test has "too much power" with large n, it's that hypothesis tests don't seem to answer the question you're interested in.
Given the large sample size, the second p | Is it valid to use Hartigans' dip test to reject uni-modal null hypothesis with large N?
It isn't so much that a hypothesis test has "too much power" with large n, it's that hypothesis tests don't seem to answer the question you're interested in.
Given the large sample size, the second plot looks to me like there is s... | Is it valid to use Hartigans' dip test to reject uni-modal null hypothesis with large N?
It isn't so much that a hypothesis test has "too much power" with large n, it's that hypothesis tests don't seem to answer the question you're interested in.
Given the large sample size, the second p |
36,375 | Chi-square test: difference between goodness-of-fit test and test of independence | 1) A goodness of fit test is for testing whether a set of multinomial counts is distributed according to a prespecified (i.e. before you see the data!) set of population proportions.
2) A test of homogeneity tests whether two (or more) sets of multinomial counts come from different sets of population proportions.
3) A... | Chi-square test: difference between goodness-of-fit test and test of independence | 1) A goodness of fit test is for testing whether a set of multinomial counts is distributed according to a prespecified (i.e. before you see the data!) set of population proportions.
2) A test of hom | Chi-square test: difference between goodness-of-fit test and test of independence
1) A goodness of fit test is for testing whether a set of multinomial counts is distributed according to a prespecified (i.e. before you see the data!) set of population proportions.
2) A test of homogeneity tests whether two (or more) s... | Chi-square test: difference between goodness-of-fit test and test of independence
1) A goodness of fit test is for testing whether a set of multinomial counts is distributed according to a prespecified (i.e. before you see the data!) set of population proportions.
2) A test of hom |
36,376 | Chi-square test: difference between goodness-of-fit test and test of independence | There are 2 primary differences between a Pearson goodness of fit test and a Pearson test of independence:
The test of independence presumes that you have 2 random variables and you want to test their independence given the sample at hand. The goodness of fit test, on the other hand, works on 1 random variable at a ti... | Chi-square test: difference between goodness-of-fit test and test of independence | There are 2 primary differences between a Pearson goodness of fit test and a Pearson test of independence:
The test of independence presumes that you have 2 random variables and you want to test thei | Chi-square test: difference between goodness-of-fit test and test of independence
There are 2 primary differences between a Pearson goodness of fit test and a Pearson test of independence:
The test of independence presumes that you have 2 random variables and you want to test their independence given the sample at han... | Chi-square test: difference between goodness-of-fit test and test of independence
There are 2 primary differences between a Pearson goodness of fit test and a Pearson test of independence:
The test of independence presumes that you have 2 random variables and you want to test thei |
36,377 | Chi-square test: difference between goodness-of-fit test and test of independence | You may want to calculate the independence between two variables, A and B (test of independence) or if the distribution of A given B=B1 (first column) fits the distribution of A given B=B2 (second column). That is if P(A|B=B1)=P(A|B=B2). I'm taking as example the data table posted by @ColorStatistics in her answer.
The... | Chi-square test: difference between goodness-of-fit test and test of independence | You may want to calculate the independence between two variables, A and B (test of independence) or if the distribution of A given B=B1 (first column) fits the distribution of A given B=B2 (second col | Chi-square test: difference between goodness-of-fit test and test of independence
You may want to calculate the independence between two variables, A and B (test of independence) or if the distribution of A given B=B1 (first column) fits the distribution of A given B=B2 (second column). That is if P(A|B=B1)=P(A|B=B2). ... | Chi-square test: difference between goodness-of-fit test and test of independence
You may want to calculate the independence between two variables, A and B (test of independence) or if the distribution of A given B=B1 (first column) fits the distribution of A given B=B2 (second col |
36,378 | Machine learning techniques for spam detection, and in general for text classification | One basic technique is Naive Bayes, which is often used for spam filtering. Essentially, you look at the frequencies of words appearing in sentences that you have already judged to be spam, and also at the frequencies of those words appearing in sentences you have already judged to be not spam, and then use those frequ... | Machine learning techniques for spam detection, and in general for text classification | One basic technique is Naive Bayes, which is often used for spam filtering. Essentially, you look at the frequencies of words appearing in sentences that you have already judged to be spam, and also a | Machine learning techniques for spam detection, and in general for text classification
One basic technique is Naive Bayes, which is often used for spam filtering. Essentially, you look at the frequencies of words appearing in sentences that you have already judged to be spam, and also at the frequencies of those words ... | Machine learning techniques for spam detection, and in general for text classification
One basic technique is Naive Bayes, which is often used for spam filtering. Essentially, you look at the frequencies of words appearing in sentences that you have already judged to be spam, and also a |
36,379 | Is it ever okay to ignore heteroskedastic residuals and continue with analysis? | The spread of those residuals look okay to me. With that arrangement of x's what did you expect the spread should look like? (I'd worry more about transformations making your mean function nonlinear in x.)
Specifically, if I generate random data with a similar pattern of x's for which the y's have constant variance, t... | Is it ever okay to ignore heteroskedastic residuals and continue with analysis? | The spread of those residuals look okay to me. With that arrangement of x's what did you expect the spread should look like? (I'd worry more about transformations making your mean function nonlinear i | Is it ever okay to ignore heteroskedastic residuals and continue with analysis?
The spread of those residuals look okay to me. With that arrangement of x's what did you expect the spread should look like? (I'd worry more about transformations making your mean function nonlinear in x.)
Specifically, if I generate rando... | Is it ever okay to ignore heteroskedastic residuals and continue with analysis?
The spread of those residuals look okay to me. With that arrangement of x's what did you expect the spread should look like? (I'd worry more about transformations making your mean function nonlinear i |
36,380 | Permutation test for factor analysis | Several authors have explored rerandomization-based criteria for component retention. For example, Peres-Neto et al., (2009) implemented a rerandomization-based version of parallel analysis and of Monte Carlo parallel analysis (which uses a high centile of eigenvalues of random data, rather than the mean). Dray develop... | Permutation test for factor analysis | Several authors have explored rerandomization-based criteria for component retention. For example, Peres-Neto et al., (2009) implemented a rerandomization-based version of parallel analysis and of Mon | Permutation test for factor analysis
Several authors have explored rerandomization-based criteria for component retention. For example, Peres-Neto et al., (2009) implemented a rerandomization-based version of parallel analysis and of Monte Carlo parallel analysis (which uses a high centile of eigenvalues of random data... | Permutation test for factor analysis
Several authors have explored rerandomization-based criteria for component retention. For example, Peres-Neto et al., (2009) implemented a rerandomization-based version of parallel analysis and of Mon |
36,381 | How to split nodes in regression trees | You can compute variance with the same cost as the averages. Simply put you have to use an on-line version of algorithm for computing variance. It is crystal clear explained on an article of John D. Cook.
I used myself this online computing for a regression tree - split numeric method where I use an online computation ... | How to split nodes in regression trees | You can compute variance with the same cost as the averages. Simply put you have to use an on-line version of algorithm for computing variance. It is crystal clear explained on an article of John D. C | How to split nodes in regression trees
You can compute variance with the same cost as the averages. Simply put you have to use an on-line version of algorithm for computing variance. It is crystal clear explained on an article of John D. Cook.
I used myself this online computing for a regression tree - split numeric me... | How to split nodes in regression trees
You can compute variance with the same cost as the averages. Simply put you have to use an on-line version of algorithm for computing variance. It is crystal clear explained on an article of John D. C |
36,382 | Comparing dependent regression coefficients from models with different dependent variables | The tool you want is called seemingly unrelated regression (SUR). SUR is a way of estimating more than one regression equation on the same data at the same time. Obviously, one thing you can do is just run the two regressions separately. What would be wrong with that? Let's write your model as:
\begin{align}
DV_{1i... | Comparing dependent regression coefficients from models with different dependent variables | The tool you want is called seemingly unrelated regression (SUR). SUR is a way of estimating more than one regression equation on the same data at the same time. Obviously, one thing you can do is j | Comparing dependent regression coefficients from models with different dependent variables
The tool you want is called seemingly unrelated regression (SUR). SUR is a way of estimating more than one regression equation on the same data at the same time. Obviously, one thing you can do is just run the two regressions s... | Comparing dependent regression coefficients from models with different dependent variables
The tool you want is called seemingly unrelated regression (SUR). SUR is a way of estimating more than one regression equation on the same data at the same time. Obviously, one thing you can do is j |
36,383 | Difference between Random Forest and MART | First question
When you consider the prediction error through bias-variance decomposition you face the well-know compromise between how well you fit on average (bias) and how stable is you prediction model (variance).
One can improve the error by decreasing the variance or by increasing the bias. Either way you choose,... | Difference between Random Forest and MART | First question
When you consider the prediction error through bias-variance decomposition you face the well-know compromise between how well you fit on average (bias) and how stable is you prediction | Difference between Random Forest and MART
First question
When you consider the prediction error through bias-variance decomposition you face the well-know compromise between how well you fit on average (bias) and how stable is you prediction model (variance).
One can improve the error by decreasing the variance or by i... | Difference between Random Forest and MART
First question
When you consider the prediction error through bias-variance decomposition you face the well-know compromise between how well you fit on average (bias) and how stable is you prediction |
36,384 | Difference between Random Forest and MART | We can understand them from the training data.
Random forest is to learn different trees based on partial training data, where the training data sets are drawn from the raw data set and the attribute at each split is on random. It encourages the diversity of samples.
The MART or generally boosting usually train a sing... | Difference between Random Forest and MART | We can understand them from the training data.
Random forest is to learn different trees based on partial training data, where the training data sets are drawn from the raw data set and the attribute | Difference between Random Forest and MART
We can understand them from the training data.
Random forest is to learn different trees based on partial training data, where the training data sets are drawn from the raw data set and the attribute at each split is on random. It encourages the diversity of samples.
The MART ... | Difference between Random Forest and MART
We can understand them from the training data.
Random forest is to learn different trees based on partial training data, where the training data sets are drawn from the raw data set and the attribute |
36,385 | Deviate vs. Variable | My understanding is that a deviate/variate a particular outcome of random variable. If I generate a 100 draws from standard normal (a random variable) using my favorite PRNG, I get 100 standard normal deviates. | Deviate vs. Variable | My understanding is that a deviate/variate a particular outcome of random variable. If I generate a 100 draws from standard normal (a random variable) using my favorite PRNG, I get 100 standard normal | Deviate vs. Variable
My understanding is that a deviate/variate a particular outcome of random variable. If I generate a 100 draws from standard normal (a random variable) using my favorite PRNG, I get 100 standard normal deviates. | Deviate vs. Variable
My understanding is that a deviate/variate a particular outcome of random variable. If I generate a 100 draws from standard normal (a random variable) using my favorite PRNG, I get 100 standard normal |
36,386 | Deviate vs. Variable | The terms "variate" and "deviate" appear to be interchangeable:
a random variable is a measurable function from a probability space to a measurable space of values that the variable can take on. In that context, and in statistics, those values are known as a random variates, or occasionally random deviates[1]
[1] htt... | Deviate vs. Variable | The terms "variate" and "deviate" appear to be interchangeable:
a random variable is a measurable function from a probability space to a measurable space of values that the variable can take on. In t | Deviate vs. Variable
The terms "variate" and "deviate" appear to be interchangeable:
a random variable is a measurable function from a probability space to a measurable space of values that the variable can take on. In that context, and in statistics, those values are known as a random variates, or occasionally random... | Deviate vs. Variable
The terms "variate" and "deviate" appear to be interchangeable:
a random variable is a measurable function from a probability space to a measurable space of values that the variable can take on. In t |
36,387 | If the dependent variable is standardized by age and sex, does it still make sense to include these as controls in a multivariate regression? | This is permissible. If the standardization has been done on a sample similar to the one you are using, then you would expect that the parameter estimates for age and gender would be small, but I see no reason that cannot be included (except that it may make the model more complex than it needs to be).
The problem you... | If the dependent variable is standardized by age and sex, does it still make sense to include these | This is permissible. If the standardization has been done on a sample similar to the one you are using, then you would expect that the parameter estimates for age and gender would be small, but I see | If the dependent variable is standardized by age and sex, does it still make sense to include these as controls in a multivariate regression?
This is permissible. If the standardization has been done on a sample similar to the one you are using, then you would expect that the parameter estimates for age and gender woul... | If the dependent variable is standardized by age and sex, does it still make sense to include these
This is permissible. If the standardization has been done on a sample similar to the one you are using, then you would expect that the parameter estimates for age and gender would be small, but I see |
36,388 | If the dependent variable is standardized by age and sex, does it still make sense to include these as controls in a multivariate regression? | There are potential problems with this approach, besides the obvious question of why not just condition on age and sex in your model.
The standardization may have been done on a group of subjects that differ from your target group in a meaningful way, tilting its assessment of the age and sex effects
The standardizati... | If the dependent variable is standardized by age and sex, does it still make sense to include these | There are potential problems with this approach, besides the obvious question of why not just condition on age and sex in your model.
The standardization may have been done on a group of subjects tha | If the dependent variable is standardized by age and sex, does it still make sense to include these as controls in a multivariate regression?
There are potential problems with this approach, besides the obvious question of why not just condition on age and sex in your model.
The standardization may have been done on a... | If the dependent variable is standardized by age and sex, does it still make sense to include these
There are potential problems with this approach, besides the obvious question of why not just condition on age and sex in your model.
The standardization may have been done on a group of subjects tha |
36,389 | If the dependent variable is standardized by age and sex, does it still make sense to include these as controls in a multivariate regression? | If $\alpha_1$ and $\alpha_2$ are applied consistently to the entire sample, then you are correct in your thinking that it will bias the OLS estimates of age and gender but not of the adoption estimate, as it is just a linear transformation of scores based on Age and Gender. To boot if you know the value of the $\alpha$... | If the dependent variable is standardized by age and sex, does it still make sense to include these | If $\alpha_1$ and $\alpha_2$ are applied consistently to the entire sample, then you are correct in your thinking that it will bias the OLS estimates of age and gender but not of the adoption estimate | If the dependent variable is standardized by age and sex, does it still make sense to include these as controls in a multivariate regression?
If $\alpha_1$ and $\alpha_2$ are applied consistently to the entire sample, then you are correct in your thinking that it will bias the OLS estimates of age and gender but not of... | If the dependent variable is standardized by age and sex, does it still make sense to include these
If $\alpha_1$ and $\alpha_2$ are applied consistently to the entire sample, then you are correct in your thinking that it will bias the OLS estimates of age and gender but not of the adoption estimate |
36,390 | Does party package in R provide out-of-bag estimates of error for Random Forest models? | The caret package has a method for getting that. You can use train as the interface. For example:
> mod1 <- train(Species ~ .,
+ data = iris,
+ method = "cforest",
+ tuneGrid = data.frame(.mtry = 2),
+ trControl = trainControl(method = "oob"))
> mod1
150 sample... | Does party package in R provide out-of-bag estimates of error for Random Forest models? | The caret package has a method for getting that. You can use train as the interface. For example:
> mod1 <- train(Species ~ .,
+ data = iris,
+ method = "cforest",
+ | Does party package in R provide out-of-bag estimates of error for Random Forest models?
The caret package has a method for getting that. You can use train as the interface. For example:
> mod1 <- train(Species ~ .,
+ data = iris,
+ method = "cforest",
+ tuneGrid = data.frame... | Does party package in R provide out-of-bag estimates of error for Random Forest models?
The caret package has a method for getting that. You can use train as the interface. For example:
> mod1 <- train(Species ~ .,
+ data = iris,
+ method = "cforest",
+ |
36,391 | Does party package in R provide out-of-bag estimates of error for Random Forest models? | Perhaps to expand a bit on Momo's answer, party doesn't provide an OOB estimate by default, but the computation of it is not too difficult, according to the manual. In the predict function you can use the parameter OOB=T, and leave the parameter newdata with its default of NULL (i.e., using the training data).
Somethi... | Does party package in R provide out-of-bag estimates of error for Random Forest models? | Perhaps to expand a bit on Momo's answer, party doesn't provide an OOB estimate by default, but the computation of it is not too difficult, according to the manual. In the predict function you can use | Does party package in R provide out-of-bag estimates of error for Random Forest models?
Perhaps to expand a bit on Momo's answer, party doesn't provide an OOB estimate by default, but the computation of it is not too difficult, according to the manual. In the predict function you can use the parameter OOB=T, and leave ... | Does party package in R provide out-of-bag estimates of error for Random Forest models?
Perhaps to expand a bit on Momo's answer, party doesn't provide an OOB estimate by default, but the computation of it is not too difficult, according to the manual. In the predict function you can use |
36,392 | Bivariate Normal distribution and correlation | Yes.
By definition, the value of the CDF (call it $F_\rho$) at $(s,t)$ is the chance that the first component is less than or equal to $s$ and the second is less than or equal to $t$:
$$F_\rho(s,t) = \frac{1}{2 \pi \sqrt{1-\rho ^2}}\int_{-\infty}^t \int_{-\infty}^s e^{-\frac{\frac{x^2}{2}-\rho x y+\frac{y^2}{2}}{1-\r... | Bivariate Normal distribution and correlation | Yes.
By definition, the value of the CDF (call it $F_\rho$) at $(s,t)$ is the chance that the first component is less than or equal to $s$ and the second is less than or equal to $t$:
$$F_\rho(s,t) = | Bivariate Normal distribution and correlation
Yes.
By definition, the value of the CDF (call it $F_\rho$) at $(s,t)$ is the chance that the first component is less than or equal to $s$ and the second is less than or equal to $t$:
$$F_\rho(s,t) = \frac{1}{2 \pi \sqrt{1-\rho ^2}}\int_{-\infty}^t \int_{-\infty}^s e^{-\fr... | Bivariate Normal distribution and correlation
Yes.
By definition, the value of the CDF (call it $F_\rho$) at $(s,t)$ is the chance that the first component is less than or equal to $s$ and the second is less than or equal to $t$:
$$F_\rho(s,t) = |
36,393 | Bivariate Normal distribution and correlation | The answer would appear to be YES. I am not sure how this is usually proven (since the bivariate Normal cdf does not have a convenient closed form) ... but as a quick thought, I would appeal perhaps to graphical ideas. In particular, consider the contours of the zero mean bivariate Normal as $\rho$ increases, as per:
... | Bivariate Normal distribution and correlation | The answer would appear to be YES. I am not sure how this is usually proven (since the bivariate Normal cdf does not have a convenient closed form) ... but as a quick thought, I would appeal perhaps t | Bivariate Normal distribution and correlation
The answer would appear to be YES. I am not sure how this is usually proven (since the bivariate Normal cdf does not have a convenient closed form) ... but as a quick thought, I would appeal perhaps to graphical ideas. In particular, consider the contours of the zero mean b... | Bivariate Normal distribution and correlation
The answer would appear to be YES. I am not sure how this is usually proven (since the bivariate Normal cdf does not have a convenient closed form) ... but as a quick thought, I would appeal perhaps t |
36,394 | Good internal factor structure but poor Cronbach's $\alpha$? | You can calculate the reliability of your items from the CFA.
From your standardized solution, calculate:
(L1+...Lk)*2/[(L1+...Lk)*2+(Var(E1)+...+Var(Ek))]
This will give the composite reliability, which should be close to alpha.
It's harder to have good fit if you have high alpha, and it's harder to have high alpha i... | Good internal factor structure but poor Cronbach's $\alpha$? | You can calculate the reliability of your items from the CFA.
From your standardized solution, calculate:
(L1+...Lk)*2/[(L1+...Lk)*2+(Var(E1)+...+Var(Ek))]
This will give the composite reliability, w | Good internal factor structure but poor Cronbach's $\alpha$?
You can calculate the reliability of your items from the CFA.
From your standardized solution, calculate:
(L1+...Lk)*2/[(L1+...Lk)*2+(Var(E1)+...+Var(Ek))]
This will give the composite reliability, which should be close to alpha.
It's harder to have good fit... | Good internal factor structure but poor Cronbach's $\alpha$?
You can calculate the reliability of your items from the CFA.
From your standardized solution, calculate:
(L1+...Lk)*2/[(L1+...Lk)*2+(Var(E1)+...+Var(Ek))]
This will give the composite reliability, w |
36,395 | Good internal factor structure but poor Cronbach's $\alpha$? | If your instrument is evaluating two or more constructs, it is possible that your alpha could be low. I advise you to estimate one alpha for each sub-scale. | Good internal factor structure but poor Cronbach's $\alpha$? | If your instrument is evaluating two or more constructs, it is possible that your alpha could be low. I advise you to estimate one alpha for each sub-scale. | Good internal factor structure but poor Cronbach's $\alpha$?
If your instrument is evaluating two or more constructs, it is possible that your alpha could be low. I advise you to estimate one alpha for each sub-scale. | Good internal factor structure but poor Cronbach's $\alpha$?
If your instrument is evaluating two or more constructs, it is possible that your alpha could be low. I advise you to estimate one alpha for each sub-scale. |
36,396 | What is the intuition behind (M)ANCOVA and when/why should one use it? | To complete your scheme:
ANCOVA: ANOVA conducted to compare multiple (possibly only two) conditions on at least one independent variable while controlling for a set of continuous nuisance variables (possibly only one).
MANCOVA: MANOVA conducted to compare multiple (possibly only two) conditions on at least one ind... | What is the intuition behind (M)ANCOVA and when/why should one use it? | To complete your scheme:
ANCOVA: ANOVA conducted to compare multiple (possibly only two) conditions on at least one independent variable while controlling for a set of continuous nuisance variables | What is the intuition behind (M)ANCOVA and when/why should one use it?
To complete your scheme:
ANCOVA: ANOVA conducted to compare multiple (possibly only two) conditions on at least one independent variable while controlling for a set of continuous nuisance variables (possibly only one).
MANCOVA: MANOVA conducted... | What is the intuition behind (M)ANCOVA and when/why should one use it?
To complete your scheme:
ANCOVA: ANOVA conducted to compare multiple (possibly only two) conditions on at least one independent variable while controlling for a set of continuous nuisance variables |
36,397 | Approximating the relative quantities of coins in Canada | I would have to say that it would be extremely difficult for you to estimate the relative quantities of coins in circulation through any but an exhaustive (collecting a large portion of those coins simultaneously) survey.
The reason is because most businesses (I believe) hold a reasonably large portion of coins in st... | Approximating the relative quantities of coins in Canada | I would have to say that it would be extremely difficult for you to estimate the relative quantities of coins in circulation through any but an exhaustive (collecting a large portion of those coins si | Approximating the relative quantities of coins in Canada
I would have to say that it would be extremely difficult for you to estimate the relative quantities of coins in circulation through any but an exhaustive (collecting a large portion of those coins simultaneously) survey.
The reason is because most businesses (... | Approximating the relative quantities of coins in Canada
I would have to say that it would be extremely difficult for you to estimate the relative quantities of coins in circulation through any but an exhaustive (collecting a large portion of those coins si |
36,398 | Approximating the relative quantities of coins in Canada | The bigger problem is going to be part 1, not part 2.
It will be relatively easy to get a big sample of coins. But how do you know those coins are a random sample? Maybe people where you live use more of a particular coin than people in other parts of Canada. You certainly use money in a way that is not the same as ev... | Approximating the relative quantities of coins in Canada | The bigger problem is going to be part 1, not part 2.
It will be relatively easy to get a big sample of coins. But how do you know those coins are a random sample? Maybe people where you live use more | Approximating the relative quantities of coins in Canada
The bigger problem is going to be part 1, not part 2.
It will be relatively easy to get a big sample of coins. But how do you know those coins are a random sample? Maybe people where you live use more of a particular coin than people in other parts of Canada. Yo... | Approximating the relative quantities of coins in Canada
The bigger problem is going to be part 1, not part 2.
It will be relatively easy to get a big sample of coins. But how do you know those coins are a random sample? Maybe people where you live use more |
36,399 | Regression with rank order as dependent variable | Ordinal regression is ideal for this problem in my opinion. There is no problem other than computational burden caused by having as many unique $Y$ as there are observations. The R rms package's orm function solves the computational burden problem using a special sparse matrix representation. For an example see Whic... | Regression with rank order as dependent variable | Ordinal regression is ideal for this problem in my opinion. There is no problem other than computational burden caused by having as many unique $Y$ as there are observations. The R rms package's orm | Regression with rank order as dependent variable
Ordinal regression is ideal for this problem in my opinion. There is no problem other than computational burden caused by having as many unique $Y$ as there are observations. The R rms package's orm function solves the computational burden problem using a special spars... | Regression with rank order as dependent variable
Ordinal regression is ideal for this problem in my opinion. There is no problem other than computational burden caused by having as many unique $Y$ as there are observations. The R rms package's orm |
36,400 | Regression with rank order as dependent variable | In principle, you are right to worry that the response is bounded. In practice, with this kind of data, you are unlikely to get predictions beyond the observed range of the data. This won't be your fault, but just the effect of the high degree of unpredictability with firm-level data.
Put it this way: The worst you ca... | Regression with rank order as dependent variable | In principle, you are right to worry that the response is bounded. In practice, with this kind of data, you are unlikely to get predictions beyond the observed range of the data. This won't be your fa | Regression with rank order as dependent variable
In principle, you are right to worry that the response is bounded. In practice, with this kind of data, you are unlikely to get predictions beyond the observed range of the data. This won't be your fault, but just the effect of the high degree of unpredictability with fi... | Regression with rank order as dependent variable
In principle, you are right to worry that the response is bounded. In practice, with this kind of data, you are unlikely to get predictions beyond the observed range of the data. This won't be your fa |
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