idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
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38,801 | How to build an "equilibrated histogram"? | To follow up on @mbq's suggestion, here's the code to do that with R:
require(Hmisc)
x <- c(-2.153, -1.732, -1.699, -1.559, -1.355
, -1.306, -1.151, -1.129, -0.636, 0.4085
, 0.5408, 0.5731, 0.5842, 0.6206, 0.8175
, 0.8274, 0.8710, 1.3214, 1.5552, 2.2342
)
eqBins <- cut2(x, g = 4)
#what are the bins and how many in ea... | How to build an "equilibrated histogram"? | To follow up on @mbq's suggestion, here's the code to do that with R:
require(Hmisc)
x <- c(-2.153, -1.732, -1.699, -1.559, -1.355
, -1.306, -1.151, -1.129, -0.636, 0.4085
, 0.5408, 0.5731, 0.5842, 0 | How to build an "equilibrated histogram"?
To follow up on @mbq's suggestion, here's the code to do that with R:
require(Hmisc)
x <- c(-2.153, -1.732, -1.699, -1.559, -1.355
, -1.306, -1.151, -1.129, -0.636, 0.4085
, 0.5408, 0.5731, 0.5842, 0.6206, 0.8175
, 0.8274, 0.8710, 1.3214, 1.5552, 2.2342
)
eqBins <- cut2(x, g =... | How to build an "equilibrated histogram"?
To follow up on @mbq's suggestion, here's the code to do that with R:
require(Hmisc)
x <- c(-2.153, -1.732, -1.699, -1.559, -1.355
, -1.306, -1.151, -1.129, -0.636, 0.4085
, 0.5408, 0.5731, 0.5842, 0 |
38,802 | How to build an "equilibrated histogram"? | You are looking for quantiles; in R there is a function quantile that will calculate them for you; Hmisc R package provides cut2 function which explicitly calculates such "equilibrated bins". | How to build an "equilibrated histogram"? | You are looking for quantiles; in R there is a function quantile that will calculate them for you; Hmisc R package provides cut2 function which explicitly calculates such "equilibrated bins". | How to build an "equilibrated histogram"?
You are looking for quantiles; in R there is a function quantile that will calculate them for you; Hmisc R package provides cut2 function which explicitly calculates such "equilibrated bins". | How to build an "equilibrated histogram"?
You are looking for quantiles; in R there is a function quantile that will calculate them for you; Hmisc R package provides cut2 function which explicitly calculates such "equilibrated bins". |
38,803 | How to build an "equilibrated histogram"? | There's a histogram here, with R code that does approximately equal counts using the quantile function.
There's also the histogram function in the lattice package that comes with R. Compare:
library("lattice")
histogram(islands^(1/4)) # equal width
histogram(islands^(1/4),breaks=NULL,equal.widths=FALSE) # approx. equ... | How to build an "equilibrated histogram"? | There's a histogram here, with R code that does approximately equal counts using the quantile function.
There's also the histogram function in the lattice package that comes with R. Compare:
library(" | How to build an "equilibrated histogram"?
There's a histogram here, with R code that does approximately equal counts using the quantile function.
There's also the histogram function in the lattice package that comes with R. Compare:
library("lattice")
histogram(islands^(1/4)) # equal width
histogram(islands^(1/4),brea... | How to build an "equilibrated histogram"?
There's a histogram here, with R code that does approximately equal counts using the quantile function.
There's also the histogram function in the lattice package that comes with R. Compare:
library(" |
38,804 | Using weighted regression to obtain fit lines for which I only have summary data | Weight each mean by the number of points that went into computing it. You can later use the estimated standard deviations to test the hypothesis of homoscedasticity that this approach assumes. If the Ns are as small as in your example then this test probably wouldn't have much power unless the SDs vary greatly. | Using weighted regression to obtain fit lines for which I only have summary data | Weight each mean by the number of points that went into computing it. You can later use the estimated standard deviations to test the hypothesis of homoscedasticity that this approach assumes. If the | Using weighted regression to obtain fit lines for which I only have summary data
Weight each mean by the number of points that went into computing it. You can later use the estimated standard deviations to test the hypothesis of homoscedasticity that this approach assumes. If the Ns are as small as in your example then... | Using weighted regression to obtain fit lines for which I only have summary data
Weight each mean by the number of points that went into computing it. You can later use the estimated standard deviations to test the hypothesis of homoscedasticity that this approach assumes. If the |
38,805 | Using weighted regression to obtain fit lines for which I only have summary data | This is Analysis of Variance.
After all, consider one of the $y$'s, with standard deviation $s$, and let its predicted value (which depends on the corresponding $x$) be $f$. The original aim is to vary $f$ (within constraints depending on the model; often $f$ is required to be a linear function of $x$) to minimize the... | Using weighted regression to obtain fit lines for which I only have summary data | This is Analysis of Variance.
After all, consider one of the $y$'s, with standard deviation $s$, and let its predicted value (which depends on the corresponding $x$) be $f$. The original aim is to va | Using weighted regression to obtain fit lines for which I only have summary data
This is Analysis of Variance.
After all, consider one of the $y$'s, with standard deviation $s$, and let its predicted value (which depends on the corresponding $x$) be $f$. The original aim is to vary $f$ (within constraints depending on... | Using weighted regression to obtain fit lines for which I only have summary data
This is Analysis of Variance.
After all, consider one of the $y$'s, with standard deviation $s$, and let its predicted value (which depends on the corresponding $x$) be $f$. The original aim is to va |
38,806 | Using weighted regression to obtain fit lines for which I only have summary data | Let the disaggregrate model be:
$Y_{ia} = X_a \beta + \epsilon_i$
where
$\epsilon_i \sim N(0,\sigma^2)$
Your aggregate model is given by:
$Y_a = \frac{\sum_i(Y_{ia})}{n_a}$
where,
$n_a$ is the number of observations you have corresponding to the $a$ index.
Therefore, it follows that:
$Y_a = X_a \beta + \epsilon_a$
whe... | Using weighted regression to obtain fit lines for which I only have summary data | Let the disaggregrate model be:
$Y_{ia} = X_a \beta + \epsilon_i$
where
$\epsilon_i \sim N(0,\sigma^2)$
Your aggregate model is given by:
$Y_a = \frac{\sum_i(Y_{ia})}{n_a}$
where,
$n_a$ is the number | Using weighted regression to obtain fit lines for which I only have summary data
Let the disaggregrate model be:
$Y_{ia} = X_a \beta + \epsilon_i$
where
$\epsilon_i \sim N(0,\sigma^2)$
Your aggregate model is given by:
$Y_a = \frac{\sum_i(Y_{ia})}{n_a}$
where,
$n_a$ is the number of observations you have corresponding... | Using weighted regression to obtain fit lines for which I only have summary data
Let the disaggregrate model be:
$Y_{ia} = X_a \beta + \epsilon_i$
where
$\epsilon_i \sim N(0,\sigma^2)$
Your aggregate model is given by:
$Y_a = \frac{\sum_i(Y_{ia})}{n_a}$
where,
$n_a$ is the number |
38,807 | Using weighted regression to obtain fit lines for which I only have summary data | I think I would calculate normalized variables (z=(x-mean(x))/(sd(x)), and run the regression. Or you could work out a way to generate samples in a bootstrap. I'm not shure if this would be the textbook solution, but intuitively it should work. | Using weighted regression to obtain fit lines for which I only have summary data | I think I would calculate normalized variables (z=(x-mean(x))/(sd(x)), and run the regression. Or you could work out a way to generate samples in a bootstrap. I'm not shure if this would be the textbo | Using weighted regression to obtain fit lines for which I only have summary data
I think I would calculate normalized variables (z=(x-mean(x))/(sd(x)), and run the regression. Or you could work out a way to generate samples in a bootstrap. I'm not shure if this would be the textbook solution, but intuitively it should ... | Using weighted regression to obtain fit lines for which I only have summary data
I think I would calculate normalized variables (z=(x-mean(x))/(sd(x)), and run the regression. Or you could work out a way to generate samples in a bootstrap. I'm not shure if this would be the textbo |
38,808 | Does the concordance correlation coefficient make linearity or monotone assumptions? | The concordance correlation can be thought of as a measure of agreement. The question is: do two variables $x$ and $y$ (say) have identical values? If so, the concordance correlation will be 1. The question makes no sense unless the variables have the same units of measurement or more generally are recorded in the same... | Does the concordance correlation coefficient make linearity or monotone assumptions? | The concordance correlation can be thought of as a measure of agreement. The question is: do two variables $x$ and $y$ (say) have identical values? If so, the concordance correlation will be 1. The qu | Does the concordance correlation coefficient make linearity or monotone assumptions?
The concordance correlation can be thought of as a measure of agreement. The question is: do two variables $x$ and $y$ (say) have identical values? If so, the concordance correlation will be 1. The question makes no sense unless the va... | Does the concordance correlation coefficient make linearity or monotone assumptions?
The concordance correlation can be thought of as a measure of agreement. The question is: do two variables $x$ and $y$ (say) have identical values? If so, the concordance correlation will be 1. The qu |
38,809 | Does the concordance correlation coefficient make linearity or monotone assumptions? | This statistic measures a kind of correlation between two sets of data. Its calculation requires no assumptions about what their scatterplot looks like (if that's what you mean by "tendency"). | Does the concordance correlation coefficient make linearity or monotone assumptions? | This statistic measures a kind of correlation between two sets of data. Its calculation requires no assumptions about what their scatterplot looks like (if that's what you mean by "tendency"). | Does the concordance correlation coefficient make linearity or monotone assumptions?
This statistic measures a kind of correlation between two sets of data. Its calculation requires no assumptions about what their scatterplot looks like (if that's what you mean by "tendency"). | Does the concordance correlation coefficient make linearity or monotone assumptions?
This statistic measures a kind of correlation between two sets of data. Its calculation requires no assumptions about what their scatterplot looks like (if that's what you mean by "tendency"). |
38,810 | Does the concordance correlation coefficient make linearity or monotone assumptions? | If your data are a sample of pairs $(x_i,y_i)$ independently and identically distributed (according to some bivariate distribution); then the sample CCC is an estimate of the population CCC defined as
$$ \rho_c = \frac{2\sigma_{xy}}{\sigma_x^2+\sigma_y^2+(\mu_x-\mu_y)^2}$$ which:
equals $0$ when the covariance $\sig... | Does the concordance correlation coefficient make linearity or monotone assumptions? | If your data are a sample of pairs $(x_i,y_i)$ independently and identically distributed (according to some bivariate distribution); then the sample CCC is an estimate of the population CCC defined as | Does the concordance correlation coefficient make linearity or monotone assumptions?
If your data are a sample of pairs $(x_i,y_i)$ independently and identically distributed (according to some bivariate distribution); then the sample CCC is an estimate of the population CCC defined as
$$ \rho_c = \frac{2\sigma_{xy}}{\... | Does the concordance correlation coefficient make linearity or monotone assumptions?
If your data are a sample of pairs $(x_i,y_i)$ independently and identically distributed (according to some bivariate distribution); then the sample CCC is an estimate of the population CCC defined as |
38,811 | Why must a product of symmetric random variables be symmetric? | To say that a random variable $W$ "has a symmetric distribution around zero" is saying that $W$ and $-W$ have the same distribution.
Let $X$ be another random variable and set $Y=WX.$ By the rules of algebra, $-Y = -WX = (-W)X$ must have the same distribution as $WX$ provided $W$ and $X$ are independent. Therefore $Y... | Why must a product of symmetric random variables be symmetric? | To say that a random variable $W$ "has a symmetric distribution around zero" is saying that $W$ and $-W$ have the same distribution.
Let $X$ be another random variable and set $Y=WX.$ By the rules of | Why must a product of symmetric random variables be symmetric?
To say that a random variable $W$ "has a symmetric distribution around zero" is saying that $W$ and $-W$ have the same distribution.
Let $X$ be another random variable and set $Y=WX.$ By the rules of algebra, $-Y = -WX = (-W)X$ must have the same distribut... | Why must a product of symmetric random variables be symmetric?
To say that a random variable $W$ "has a symmetric distribution around zero" is saying that $W$ and $-W$ have the same distribution.
Let $X$ be another random variable and set $Y=WX.$ By the rules of |
38,812 | Why does R say 'cannot compute exact p-values with ties' when I can do it with pen and paper? | The wilcox.test from the standard stats library is limited to cases without ties because it uses an algorithm from the function pwilcox that assumes that there are no ties.
This algorithm is not your pen and paper solution which would become computation intensive for larger sample sizes.
The algorithm in pwilcox is not... | Why does R say 'cannot compute exact p-values with ties' when I can do it with pen and paper? | The wilcox.test from the standard stats library is limited to cases without ties because it uses an algorithm from the function pwilcox that assumes that there are no ties.
This algorithm is not your | Why does R say 'cannot compute exact p-values with ties' when I can do it with pen and paper?
The wilcox.test from the standard stats library is limited to cases without ties because it uses an algorithm from the function pwilcox that assumes that there are no ties.
This algorithm is not your pen and paper solution whi... | Why does R say 'cannot compute exact p-values with ties' when I can do it with pen and paper?
The wilcox.test from the standard stats library is limited to cases without ties because it uses an algorithm from the function pwilcox that assumes that there are no ties.
This algorithm is not your |
38,813 | How can I generate random observations from a concrete copula? | Copulas are usually defined via the joint cdf of the Uniform components,
$$ C(u_1,u_2,\dots,u_d)=\mathbb P[U_1\leq u_1,U_2\leq u_2,\dots,U_d\leq u_d]$$
Unfortunately, a value $C(u_1,u_2,\dots,u_d)$ computed from a random realisation from $C$ is not a Uniform variate as in the univariate case.
In the bivariate case, sim... | How can I generate random observations from a concrete copula? | Copulas are usually defined via the joint cdf of the Uniform components,
$$ C(u_1,u_2,\dots,u_d)=\mathbb P[U_1\leq u_1,U_2\leq u_2,\dots,U_d\leq u_d]$$
Unfortunately, a value $C(u_1,u_2,\dots,u_d)$ co | How can I generate random observations from a concrete copula?
Copulas are usually defined via the joint cdf of the Uniform components,
$$ C(u_1,u_2,\dots,u_d)=\mathbb P[U_1\leq u_1,U_2\leq u_2,\dots,U_d\leq u_d]$$
Unfortunately, a value $C(u_1,u_2,\dots,u_d)$ computed from a random realisation from $C$ is not a Unifor... | How can I generate random observations from a concrete copula?
Copulas are usually defined via the joint cdf of the Uniform components,
$$ C(u_1,u_2,\dots,u_d)=\mathbb P[U_1\leq u_1,U_2\leq u_2,\dots,U_d\leq u_d]$$
Unfortunately, a value $C(u_1,u_2,\dots,u_d)$ co |
38,814 | How can I generate random observations from a concrete copula? | A recipe for how to generate random observations depends on the level of detail. If you can generate from the copula, then you get points on $[0,1]\times [0,1]$. Then you transform the $X$ coordinate into $X$ and the $Y$ coordinate into $Y$ by using their respective inverse CDFs. | How can I generate random observations from a concrete copula? | A recipe for how to generate random observations depends on the level of detail. If you can generate from the copula, then you get points on $[0,1]\times [0,1]$. Then you transform the $X$ coordinate | How can I generate random observations from a concrete copula?
A recipe for how to generate random observations depends on the level of detail. If you can generate from the copula, then you get points on $[0,1]\times [0,1]$. Then you transform the $X$ coordinate into $X$ and the $Y$ coordinate into $Y$ by using their r... | How can I generate random observations from a concrete copula?
A recipe for how to generate random observations depends on the level of detail. If you can generate from the copula, then you get points on $[0,1]\times [0,1]$. Then you transform the $X$ coordinate |
38,815 | What is the reasoning behind expecting residuals in OLS regression to be normally distributed? | That author is writing nonsense. Just because errors are random doesn't mean that if you have a lot of them they will be normally distributed. It is absolutely not the case that OLS requires normally distributed residuals; its objective is "Least Squares", and minimizing the sum of squared deviations from the estimat... | What is the reasoning behind expecting residuals in OLS regression to be normally distributed? | That author is writing nonsense. Just because errors are random doesn't mean that if you have a lot of them they will be normally distributed. It is absolutely not the case that OLS requires normall | What is the reasoning behind expecting residuals in OLS regression to be normally distributed?
That author is writing nonsense. Just because errors are random doesn't mean that if you have a lot of them they will be normally distributed. It is absolutely not the case that OLS requires normally distributed residuals; ... | What is the reasoning behind expecting residuals in OLS regression to be normally distributed?
That author is writing nonsense. Just because errors are random doesn't mean that if you have a lot of them they will be normally distributed. It is absolutely not the case that OLS requires normall |
38,816 | Posterior distribution is impossible depending on which prior hyperparameters are used? | Let's work through the steps. To begin with we have
\begin{equation}
p(\delta|\beta)\,p(\beta|\alpha)\,p(\alpha) ,
\end{equation}
where
\begin{align}
p(\delta|\beta) &= \textsf{Bernoulli}(\delta|\beta) \\
p(\beta|\alpha) &= \textsf{Beta}(\beta|a_\alpha,b_\alpha) \\
p(\alpha) &= \textsf{Bernoulli}(\alpha|\rho) .
\end{a... | Posterior distribution is impossible depending on which prior hyperparameters are used? | Let's work through the steps. To begin with we have
\begin{equation}
p(\delta|\beta)\,p(\beta|\alpha)\,p(\alpha) ,
\end{equation}
where
\begin{align}
p(\delta|\beta) &= \textsf{Bernoulli}(\delta|\beta | Posterior distribution is impossible depending on which prior hyperparameters are used?
Let's work through the steps. To begin with we have
\begin{equation}
p(\delta|\beta)\,p(\beta|\alpha)\,p(\alpha) ,
\end{equation}
where
\begin{align}
p(\delta|\beta) &= \textsf{Bernoulli}(\delta|\beta) \\
p(\beta|\alpha) &= \textsf{... | Posterior distribution is impossible depending on which prior hyperparameters are used?
Let's work through the steps. To begin with we have
\begin{equation}
p(\delta|\beta)\,p(\beta|\alpha)\,p(\alpha) ,
\end{equation}
where
\begin{align}
p(\delta|\beta) &= \textsf{Bernoulli}(\delta|\beta |
38,817 | Posterior distribution is impossible depending on which prior hyperparameters are used? | The distribution you are describing is a mixture of Bernoulli distributions with the likelihood function
$$
p(\delta|\rho,\beta_0,\beta_1) = \rho \, \beta_0^{\delta} (1-\beta_0)^{1-\delta} + (1 - \rho) \, \beta_1^{\delta} (1-\beta_1)^{1-\delta}
$$
It does not have a closed-form solution, you need to use an optimization... | Posterior distribution is impossible depending on which prior hyperparameters are used? | The distribution you are describing is a mixture of Bernoulli distributions with the likelihood function
$$
p(\delta|\rho,\beta_0,\beta_1) = \rho \, \beta_0^{\delta} (1-\beta_0)^{1-\delta} + (1 - \rho | Posterior distribution is impossible depending on which prior hyperparameters are used?
The distribution you are describing is a mixture of Bernoulli distributions with the likelihood function
$$
p(\delta|\rho,\beta_0,\beta_1) = \rho \, \beta_0^{\delta} (1-\beta_0)^{1-\delta} + (1 - \rho) \, \beta_1^{\delta} (1-\beta_1... | Posterior distribution is impossible depending on which prior hyperparameters are used?
The distribution you are describing is a mixture of Bernoulli distributions with the likelihood function
$$
p(\delta|\rho,\beta_0,\beta_1) = \rho \, \beta_0^{\delta} (1-\beta_0)^{1-\delta} + (1 - \rho |
38,818 | Does $Cov(X,Y)<0$ and $Cov(X,Z)>0$ imply $Cov(Y,Z)<0$? | If the correlations between the three variables are $a$, $b$ and $c$, the eigenvalues of the correlation matrix satisfies
$$
\left|\begin{matrix}
1-\lambda & a & c \\
a & 1-\lambda & b \\
c & b & 1-\lambda
\end{matrix}\right|=0
$$
which after some algebra simplifies to
$$
(1-\lambda)^3 - (1-\lambda)(a^2+b^2+c^2)+2abc=0... | Does $Cov(X,Y)<0$ and $Cov(X,Z)>0$ imply $Cov(Y,Z)<0$? | If the correlations between the three variables are $a$, $b$ and $c$, the eigenvalues of the correlation matrix satisfies
$$
\left|\begin{matrix}
1-\lambda & a & c \\
a & 1-\lambda & b \\
c & b & 1-\l | Does $Cov(X,Y)<0$ and $Cov(X,Z)>0$ imply $Cov(Y,Z)<0$?
If the correlations between the three variables are $a$, $b$ and $c$, the eigenvalues of the correlation matrix satisfies
$$
\left|\begin{matrix}
1-\lambda & a & c \\
a & 1-\lambda & b \\
c & b & 1-\lambda
\end{matrix}\right|=0
$$
which after some algebra simplifie... | Does $Cov(X,Y)<0$ and $Cov(X,Z)>0$ imply $Cov(Y,Z)<0$?
If the correlations between the three variables are $a$, $b$ and $c$, the eigenvalues of the correlation matrix satisfies
$$
\left|\begin{matrix}
1-\lambda & a & c \\
a & 1-\lambda & b \\
c & b & 1-\l |
38,819 | Does $Cov(X,Y)<0$ and $Cov(X,Z)>0$ imply $Cov(Y,Z)<0$? | No (to the title question). Any positive semi-definite matrix is also a covariance matrix. So, if there exists a $\rho$ that the following matrix is PSD, then you've a contradiction:
$$\Sigma=\begin{bmatrix}1&-\rho&\rho\\-\rho &1&\rho\\\rho&\rho&1\end{bmatrix}$$
This occurs for $\rho=0.1$ for example. The conditions on... | Does $Cov(X,Y)<0$ and $Cov(X,Z)>0$ imply $Cov(Y,Z)<0$? | No (to the title question). Any positive semi-definite matrix is also a covariance matrix. So, if there exists a $\rho$ that the following matrix is PSD, then you've a contradiction:
$$\Sigma=\begin{b | Does $Cov(X,Y)<0$ and $Cov(X,Z)>0$ imply $Cov(Y,Z)<0$?
No (to the title question). Any positive semi-definite matrix is also a covariance matrix. So, if there exists a $\rho$ that the following matrix is PSD, then you've a contradiction:
$$\Sigma=\begin{bmatrix}1&-\rho&\rho\\-\rho &1&\rho\\\rho&\rho&1\end{bmatrix}$$
Th... | Does $Cov(X,Y)<0$ and $Cov(X,Z)>0$ imply $Cov(Y,Z)<0$?
No (to the title question). Any positive semi-definite matrix is also a covariance matrix. So, if there exists a $\rho$ that the following matrix is PSD, then you've a contradiction:
$$\Sigma=\begin{b |
38,820 | Is it possible to take the mean of a time series when it is not stationary? | I think you are operating under some faulty assumptions here. Regardless of stationarity, it is always possible to compute the sample mean of a set of data, and if you only want to know this sample mean then there is no inference problem to begin with. Stationarity of the time-series only becomes an issue when you wa... | Is it possible to take the mean of a time series when it is not stationary? | I think you are operating under some faulty assumptions here. Regardless of stationarity, it is always possible to compute the sample mean of a set of data, and if you only want to know this sample m | Is it possible to take the mean of a time series when it is not stationary?
I think you are operating under some faulty assumptions here. Regardless of stationarity, it is always possible to compute the sample mean of a set of data, and if you only want to know this sample mean then there is no inference problem to be... | Is it possible to take the mean of a time series when it is not stationary?
I think you are operating under some faulty assumptions here. Regardless of stationarity, it is always possible to compute the sample mean of a set of data, and if you only want to know this sample m |
38,821 | Do you do linear regression in logistic regression? | The simplest way of likening logistic regression to standard linear regression is using the latent variable interpretation. The logistic regression model can be described by considering the observable responses:
$$Y_i = \mathbb{I}(Y_i^* > 0),$$
based on the unobservable latent variables:
$$Y_i^* = \beta_0 + \beta_1 x_... | Do you do linear regression in logistic regression? | The simplest way of likening logistic regression to standard linear regression is using the latent variable interpretation. The logistic regression model can be described by considering the observabl | Do you do linear regression in logistic regression?
The simplest way of likening logistic regression to standard linear regression is using the latent variable interpretation. The logistic regression model can be described by considering the observable responses:
$$Y_i = \mathbb{I}(Y_i^* > 0),$$
based on the unobserva... | Do you do linear regression in logistic regression?
The simplest way of likening logistic regression to standard linear regression is using the latent variable interpretation. The logistic regression model can be described by considering the observabl |
38,822 | Do you do linear regression in logistic regression? | No, it is not done like this. Quoting my other answer
Logistic regression can be described as a linear combination
$$ \eta = \beta_0 + \beta_1 X_1 + ... + \beta_k X_k $$
that is passed through the link function $g$:
$$ g(E(Y)) = \eta $$
where the link function is a logit function
$$ E(Y|X,\beta) = p = \text{logit}^{-1... | Do you do linear regression in logistic regression? | No, it is not done like this. Quoting my other answer
Logistic regression can be described as a linear combination
$$ \eta = \beta_0 + \beta_1 X_1 + ... + \beta_k X_k $$
that is passed through the li | Do you do linear regression in logistic regression?
No, it is not done like this. Quoting my other answer
Logistic regression can be described as a linear combination
$$ \eta = \beta_0 + \beta_1 X_1 + ... + \beta_k X_k $$
that is passed through the link function $g$:
$$ g(E(Y)) = \eta $$
where the link function is a l... | Do you do linear regression in logistic regression?
No, it is not done like this. Quoting my other answer
Logistic regression can be described as a linear combination
$$ \eta = \beta_0 + \beta_1 X_1 + ... + \beta_k X_k $$
that is passed through the li |
38,823 | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$ | Definitely not the most efficient approach (doesn't condition on the sufficient statistic), but the mean is unbiased for $\theta$. And the variance of a single observation from the above distribution is 1/12. So, by the CLT:
$$ \sqrt{n} \left( \bar{X} - \theta \right) \rightarrow_d \mathcal{N}\left(0, 1/12 \right)$$
Wh... | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$ | Definitely not the most efficient approach (doesn't condition on the sufficient statistic), but the mean is unbiased for $\theta$. And the variance of a single observation from the above distribution | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$
Definitely not the most efficient approach (doesn't condition on the sufficient statistic), but the mean is unbiased for $\theta$. And the variance of a single observation from the above distribution is 1/12. So, by the CLT:
$$ ... | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$
Definitely not the most efficient approach (doesn't condition on the sufficient statistic), but the mean is unbiased for $\theta$. And the variance of a single observation from the above distribution |
38,824 | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$ | Frequentist interval via pivotal quantity: An improvement over the interval suggested by @AdamO but still suboptimal solution can be obtained using almost exactly the same method as the one I give here so I omit the details of the following derivation. The pdf of
$$
Z=\frac{X_{(1)}+X_{(n)}}2-\theta,
$$
is
$$
f(z)=n(1-... | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$ | Frequentist interval via pivotal quantity: An improvement over the interval suggested by @AdamO but still suboptimal solution can be obtained using almost exactly the same method as the one I give her | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$
Frequentist interval via pivotal quantity: An improvement over the interval suggested by @AdamO but still suboptimal solution can be obtained using almost exactly the same method as the one I give here so I omit the details of t... | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$
Frequentist interval via pivotal quantity: An improvement over the interval suggested by @AdamO but still suboptimal solution can be obtained using almost exactly the same method as the one I give her |
38,825 | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$ | I too will avoid analytic derivations of distributions
of $\max(X_i)$ and $\min(X_i)$ because I guess that is
the main point of this assignment. [Also see this page.]
However, results from a simulation for the case $n = 20, \theta = 10,$ based on
the sample midrange, are shown below.
set.seed(2021)
n=20
mr = replicate... | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$ | I too will avoid analytic derivations of distributions
of $\max(X_i)$ and $\min(X_i)$ because I guess that is
the main point of this assignment. [Also see this page.]
However, results from a simulati | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$
I too will avoid analytic derivations of distributions
of $\max(X_i)$ and $\min(X_i)$ because I guess that is
the main point of this assignment. [Also see this page.]
However, results from a simulation for the case $n = 20, \th... | Find a Confidence Interval for Data from a Uniform Distribution on $(\theta-.5,\theta+.5).$
I too will avoid analytic derivations of distributions
of $\max(X_i)$ and $\min(X_i)$ because I guess that is
the main point of this assignment. [Also see this page.]
However, results from a simulati |
38,826 | Why is AIC or BIC commonly used in model selections for time series forecasting? | What alternatives do we have in model selection for prediction?
The main ones are cross validation and information criteria.
Why are the latter attractive in the time series setting?
Information criteria are less computationally intensive. You only need to fit the model once to calculate an information criterion. Th... | Why is AIC or BIC commonly used in model selections for time series forecasting? | What alternatives do we have in model selection for prediction?
The main ones are cross validation and information criteria.
Why are the latter attractive in the time series setting?
Information cr | Why is AIC or BIC commonly used in model selections for time series forecasting?
What alternatives do we have in model selection for prediction?
The main ones are cross validation and information criteria.
Why are the latter attractive in the time series setting?
Information criteria are less computationally intensi... | Why is AIC or BIC commonly used in model selections for time series forecasting?
What alternatives do we have in model selection for prediction?
The main ones are cross validation and information criteria.
Why are the latter attractive in the time series setting?
Information cr |
38,827 | Why is AIC or BIC commonly used in model selections for time series forecasting? | First off, as Richard Hardy comments, information criteria do not assume we have the true model. Quite to the contrary. For instance, AIC estimates the Kullback-Leibler distance between the proposed model and the true data generating process (up to an offset), and picking the model with minimal AIC amounts to choosing ... | Why is AIC or BIC commonly used in model selections for time series forecasting? | First off, as Richard Hardy comments, information criteria do not assume we have the true model. Quite to the contrary. For instance, AIC estimates the Kullback-Leibler distance between the proposed m | Why is AIC or BIC commonly used in model selections for time series forecasting?
First off, as Richard Hardy comments, information criteria do not assume we have the true model. Quite to the contrary. For instance, AIC estimates the Kullback-Leibler distance between the proposed model and the true data generating proce... | Why is AIC or BIC commonly used in model selections for time series forecasting?
First off, as Richard Hardy comments, information criteria do not assume we have the true model. Quite to the contrary. For instance, AIC estimates the Kullback-Leibler distance between the proposed m |
38,828 | Why is AIC or BIC commonly used in model selections for time series forecasting? | First of all, sorry this was supposed to be a comment as opposed to an answer. The question has already been answered well. I just wanted to add that even though ICs aim at minimizing the distance to the true DGP, they might not always be able to do so. True DGP is unknown and there is no best way to identify the model... | Why is AIC or BIC commonly used in model selections for time series forecasting? | First of all, sorry this was supposed to be a comment as opposed to an answer. The question has already been answered well. I just wanted to add that even though ICs aim at minimizing the distance to | Why is AIC or BIC commonly used in model selections for time series forecasting?
First of all, sorry this was supposed to be a comment as opposed to an answer. The question has already been answered well. I just wanted to add that even though ICs aim at minimizing the distance to the true DGP, they might not always be ... | Why is AIC or BIC commonly used in model selections for time series forecasting?
First of all, sorry this was supposed to be a comment as opposed to an answer. The question has already been answered well. I just wanted to add that even though ICs aim at minimizing the distance to |
38,829 | Intuition for why likelihood function sometimes *is* a PDF | The purpose of this answer is to show that the situation is so rich and complicated that it's unlikely there exists any simple characterization of such distributional families.
I will first show, by construction, that there are many such families and they are flexible and varied. Then I will show that even this constr... | Intuition for why likelihood function sometimes *is* a PDF | The purpose of this answer is to show that the situation is so rich and complicated that it's unlikely there exists any simple characterization of such distributional families.
I will first show, by c | Intuition for why likelihood function sometimes *is* a PDF
The purpose of this answer is to show that the situation is so rich and complicated that it's unlikely there exists any simple characterization of such distributional families.
I will first show, by construction, that there are many such families and they are f... | Intuition for why likelihood function sometimes *is* a PDF
The purpose of this answer is to show that the situation is so rich and complicated that it's unlikely there exists any simple characterization of such distributional families.
I will first show, by c |
38,830 | Intuition for why likelihood function sometimes *is* a PDF | There is a similar question here:
What is the reason that a likelihood function is not a pdf?
Hopefully, that can be of help. In particular see whuber's comment regarding Fishers 1922 article:
Integrating to unity is beside the point. Fisher, in a 1922 paper On the Mathematical Foundations of Theoretical Statistics, o... | Intuition for why likelihood function sometimes *is* a PDF | There is a similar question here:
What is the reason that a likelihood function is not a pdf?
Hopefully, that can be of help. In particular see whuber's comment regarding Fishers 1922 article:
Integr | Intuition for why likelihood function sometimes *is* a PDF
There is a similar question here:
What is the reason that a likelihood function is not a pdf?
Hopefully, that can be of help. In particular see whuber's comment regarding Fishers 1922 article:
Integrating to unity is beside the point. Fisher, in a 1922 paper O... | Intuition for why likelihood function sometimes *is* a PDF
There is a similar question here:
What is the reason that a likelihood function is not a pdf?
Hopefully, that can be of help. In particular see whuber's comment regarding Fishers 1922 article:
Integr |
38,831 | Intuition for why likelihood function sometimes *is* a PDF | The mean of a normal distribution is a location parameter. In general, if $\theta$ is a location parameter, then:
$$\int_{\theta \in \mathbb{R}}g_{\theta}(x_0)\:d\theta = \int_{x \in \mathbb{R}}g_o(x_0 - x)\:dx = \int_{x \in \mathbb{R}}g_o(x)\:dx = 1$$
where $g_0$ is the pdf when $\theta = 0$. | Intuition for why likelihood function sometimes *is* a PDF | The mean of a normal distribution is a location parameter. In general, if $\theta$ is a location parameter, then:
$$\int_{\theta \in \mathbb{R}}g_{\theta}(x_0)\:d\theta = \int_{x \in \mathbb{R}}g_o(x_ | Intuition for why likelihood function sometimes *is* a PDF
The mean of a normal distribution is a location parameter. In general, if $\theta$ is a location parameter, then:
$$\int_{\theta \in \mathbb{R}}g_{\theta}(x_0)\:d\theta = \int_{x \in \mathbb{R}}g_o(x_0 - x)\:dx = \int_{x \in \mathbb{R}}g_o(x)\:dx = 1$$
where $g... | Intuition for why likelihood function sometimes *is* a PDF
The mean of a normal distribution is a location parameter. In general, if $\theta$ is a location parameter, then:
$$\int_{\theta \in \mathbb{R}}g_{\theta}(x_0)\:d\theta = \int_{x \in \mathbb{R}}g_o(x_ |
38,832 | Intuition for why likelihood function sometimes *is* a PDF | The likelihood is not a PDF
Is there an intuitive explanation for the fact that this particular likelihood function is a PDF?
This is a loaded question. It is based on a wrong premise. The likelihood function is not a PDF.
The condition $\int_{\theta \in \mathbb{R}} f(\theta) d\theta= 1$ is a necessary condition for ... | Intuition for why likelihood function sometimes *is* a PDF | The likelihood is not a PDF
Is there an intuitive explanation for the fact that this particular likelihood function is a PDF?
This is a loaded question. It is based on a wrong premise. The likelihoo | Intuition for why likelihood function sometimes *is* a PDF
The likelihood is not a PDF
Is there an intuitive explanation for the fact that this particular likelihood function is a PDF?
This is a loaded question. It is based on a wrong premise. The likelihood function is not a PDF.
The condition $\int_{\theta \in \mat... | Intuition for why likelihood function sometimes *is* a PDF
The likelihood is not a PDF
Is there an intuitive explanation for the fact that this particular likelihood function is a PDF?
This is a loaded question. It is based on a wrong premise. The likelihoo |
38,833 | Intuition for why likelihood function sometimes *is* a PDF | To quote from Ronald Fisher himself (and intersecting with the answer by NicoFish as well as the answer by WHuber to the earlier question):
when defining the likelihood function (as a function of the parameter $\theta$), Fisher (in his 1912 undergraduate memoir) warns against integrating it w.r.t. the parameter: “the ... | Intuition for why likelihood function sometimes *is* a PDF | To quote from Ronald Fisher himself (and intersecting with the answer by NicoFish as well as the answer by WHuber to the earlier question):
when defining the likelihood function (as a function of the | Intuition for why likelihood function sometimes *is* a PDF
To quote from Ronald Fisher himself (and intersecting with the answer by NicoFish as well as the answer by WHuber to the earlier question):
when defining the likelihood function (as a function of the parameter $\theta$), Fisher (in his 1912 undergraduate memoi... | Intuition for why likelihood function sometimes *is* a PDF
To quote from Ronald Fisher himself (and intersecting with the answer by NicoFish as well as the answer by WHuber to the earlier question):
when defining the likelihood function (as a function of the |
38,834 | Every second customer converts better. Can this be by chance? | With just the data you've posted, you can analyze it as a binomial distribution with each week being at trial. That would give a p-value of $2^{-11}$, or about $0.05\%$. You could also do a paired test, which would give you a binomial distribution with each pair of ids being a trial. Then there's a t-test on the total... | Every second customer converts better. Can this be by chance? | With just the data you've posted, you can analyze it as a binomial distribution with each week being at trial. That would give a p-value of $2^{-11}$, or about $0.05\%$. You could also do a paired te | Every second customer converts better. Can this be by chance?
With just the data you've posted, you can analyze it as a binomial distribution with each week being at trial. That would give a p-value of $2^{-11}$, or about $0.05\%$. You could also do a paired test, which would give you a binomial distribution with each... | Every second customer converts better. Can this be by chance?
With just the data you've posted, you can analyze it as a binomial distribution with each week being at trial. That would give a p-value of $2^{-11}$, or about $0.05\%$. You could also do a paired te |
38,835 | Every second customer converts better. Can this be by chance? | There are a number of statistical test you could use to answer a question like this, but the simplest is the 2×2 chi-square contingency test (example here). | Every second customer converts better. Can this be by chance? | There are a number of statistical test you could use to answer a question like this, but the simplest is the 2×2 chi-square contingency test (example here). | Every second customer converts better. Can this be by chance?
There are a number of statistical test you could use to answer a question like this, but the simplest is the 2×2 chi-square contingency test (example here). | Every second customer converts better. Can this be by chance?
There are a number of statistical test you could use to answer a question like this, but the simplest is the 2×2 chi-square contingency test (example here). |
38,836 | Every second customer converts better. Can this be by chance? | Just add up all the accepted applications with odd and even IDs across all the dates so that you have two numbers: $N_{odd}, N_{even}$. Call your random variable $X \in \{0,1\}$ where 0 means even and 1 means odd. The ex-ante probability of $X=0$ (given that odd/even is random) is $p=0.5$. You've got $N=N_{odd}+N_{even... | Every second customer converts better. Can this be by chance? | Just add up all the accepted applications with odd and even IDs across all the dates so that you have two numbers: $N_{odd}, N_{even}$. Call your random variable $X \in \{0,1\}$ where 0 means even and | Every second customer converts better. Can this be by chance?
Just add up all the accepted applications with odd and even IDs across all the dates so that you have two numbers: $N_{odd}, N_{even}$. Call your random variable $X \in \{0,1\}$ where 0 means even and 1 means odd. The ex-ante probability of $X=0$ (given that... | Every second customer converts better. Can this be by chance?
Just add up all the accepted applications with odd and even IDs across all the dates so that you have two numbers: $N_{odd}, N_{even}$. Call your random variable $X \in \{0,1\}$ where 0 means even and |
38,837 | Every second customer converts better. Can this be by chance? | At face value you have 11 samples in every case with B bigger than A. There are 2048 choices that could have happened (2^11) - viewing the choices as 11 binary bits makes it easy to visualise (assume bit is 0 for A>B and 1 for A<B). There is only one way B could always be bigger - the bit sequence 11111111111. But y... | Every second customer converts better. Can this be by chance? | At face value you have 11 samples in every case with B bigger than A. There are 2048 choices that could have happened (2^11) - viewing the choices as 11 binary bits makes it easy to visualise (assume | Every second customer converts better. Can this be by chance?
At face value you have 11 samples in every case with B bigger than A. There are 2048 choices that could have happened (2^11) - viewing the choices as 11 binary bits makes it easy to visualise (assume bit is 0 for A>B and 1 for A<B). There is only one way B... | Every second customer converts better. Can this be by chance?
At face value you have 11 samples in every case with B bigger than A. There are 2048 choices that could have happened (2^11) - viewing the choices as 11 binary bits makes it easy to visualise (assume |
38,838 | infinite coin toss probability | For a fair coin $n_H/n_T\to 1$, but it is not the case that $n_H-n_T\to 0$. We know by the central limit theorem that the approximate distribution of $n_H$ is $N(n/2, n/4)$, so that the approximate distribution of $n_H-n_T$ is $N(0, n)$.
That is, $n_H-n_T$ has mean zero, but has typical size $\sqrt{n}$.
The approximat... | infinite coin toss probability | For a fair coin $n_H/n_T\to 1$, but it is not the case that $n_H-n_T\to 0$. We know by the central limit theorem that the approximate distribution of $n_H$ is $N(n/2, n/4)$, so that the approximate d | infinite coin toss probability
For a fair coin $n_H/n_T\to 1$, but it is not the case that $n_H-n_T\to 0$. We know by the central limit theorem that the approximate distribution of $n_H$ is $N(n/2, n/4)$, so that the approximate distribution of $n_H-n_T$ is $N(0, n)$.
That is, $n_H-n_T$ has mean zero, but has typical ... | infinite coin toss probability
For a fair coin $n_H/n_T\to 1$, but it is not the case that $n_H-n_T\to 0$. We know by the central limit theorem that the approximate distribution of $n_H$ is $N(n/2, n/4)$, so that the approximate d |
38,839 | infinite coin toss probability | Your 1st case
You could make your expression $\mathbb{E}(X)=\frac{1}{2}2+\frac{1}{2}\frac{1}{2}=\frac{5}{4}$ more correctly like:
$$\mathbb{E}(M_{n+1})=\frac{1}{2}2\mathbb{E}(M_{n})+\frac{1}{2}\frac{1}{2}\mathbb{E}(M_{n})=\frac{5}{4} \mathbb{E}(M_{n})$$
and as a result:
$$\mathbb{E}(M_{n})= \left( \frac{5}{4} \right)^n... | infinite coin toss probability | Your 1st case
You could make your expression $\mathbb{E}(X)=\frac{1}{2}2+\frac{1}{2}\frac{1}{2}=\frac{5}{4}$ more correctly like:
$$\mathbb{E}(M_{n+1})=\frac{1}{2}2\mathbb{E}(M_{n})+\frac{1}{2}\frac{1 | infinite coin toss probability
Your 1st case
You could make your expression $\mathbb{E}(X)=\frac{1}{2}2+\frac{1}{2}\frac{1}{2}=\frac{5}{4}$ more correctly like:
$$\mathbb{E}(M_{n+1})=\frac{1}{2}2\mathbb{E}(M_{n})+\frac{1}{2}\frac{1}{2}\mathbb{E}(M_{n})=\frac{5}{4} \mathbb{E}(M_{n})$$
and as a result:
$$\mathbb{E}(M_{n}... | infinite coin toss probability
Your 1st case
You could make your expression $\mathbb{E}(X)=\frac{1}{2}2+\frac{1}{2}\frac{1}{2}=\frac{5}{4}$ more correctly like:
$$\mathbb{E}(M_{n+1})=\frac{1}{2}2\mathbb{E}(M_{n})+\frac{1}{2}\frac{1 |
38,840 | infinite coin toss probability | Some caution is warranted here: Starting with a 'fortune' of \$1,
the variance of the fortune under the model described becomes
infinite. The analyses for a finite number of
coin tosses are OK, but the original question asks for behavior
as $n \rightarrow \infty.$
It doesn't take a run of hugely many successive H's or ... | infinite coin toss probability | Some caution is warranted here: Starting with a 'fortune' of \$1,
the variance of the fortune under the model described becomes
infinite. The analyses for a finite number of
coin tosses are OK, but th | infinite coin toss probability
Some caution is warranted here: Starting with a 'fortune' of \$1,
the variance of the fortune under the model described becomes
infinite. The analyses for a finite number of
coin tosses are OK, but the original question asks for behavior
as $n \rightarrow \infty.$
It doesn't take a run of... | infinite coin toss probability
Some caution is warranted here: Starting with a 'fortune' of \$1,
the variance of the fortune under the model described becomes
infinite. The analyses for a finite number of
coin tosses are OK, but th |
38,841 | Calibration curve of XGBoost for binary classification | I'm not sure "the objective function of XGBoost is 'binary:logistic', the probabilities should be well calibrated" is correct: gradient boosting tends to push probability toward 0 and 1. Furthermore, you're applying weights, which should also skew your probabilities.
Because gradient boosting pushes probabilities outw... | Calibration curve of XGBoost for binary classification | I'm not sure "the objective function of XGBoost is 'binary:logistic', the probabilities should be well calibrated" is correct: gradient boosting tends to push probability toward 0 and 1. Furthermore, | Calibration curve of XGBoost for binary classification
I'm not sure "the objective function of XGBoost is 'binary:logistic', the probabilities should be well calibrated" is correct: gradient boosting tends to push probability toward 0 and 1. Furthermore, you're applying weights, which should also skew your probabiliti... | Calibration curve of XGBoost for binary classification
I'm not sure "the objective function of XGBoost is 'binary:logistic', the probabilities should be well calibrated" is correct: gradient boosting tends to push probability toward 0 and 1. Furthermore, |
38,842 | Calibration curve of XGBoost for binary classification | I'm not that familiar with gradient boosting, but I would assume that if you scale your minority class then your model will not be well calibrated. At the end of the day, it has learnt the distribution of the training data which does not reflect reality.
As for CalibratedClassifierCV, from reading the docs it seems tha... | Calibration curve of XGBoost for binary classification | I'm not that familiar with gradient boosting, but I would assume that if you scale your minority class then your model will not be well calibrated. At the end of the day, it has learnt the distributio | Calibration curve of XGBoost for binary classification
I'm not that familiar with gradient boosting, but I would assume that if you scale your minority class then your model will not be well calibrated. At the end of the day, it has learnt the distribution of the training data which does not reflect reality.
As for Cal... | Calibration curve of XGBoost for binary classification
I'm not that familiar with gradient boosting, but I would assume that if you scale your minority class then your model will not be well calibrated. At the end of the day, it has learnt the distributio |
38,843 | Calibration curve of XGBoost for binary classification | Answer:
No wonder your post-calibration plot looks like this. Look at your reliability curve pre-calibration. The more predicted score grows, the more actual positives it picks up. However, the best your model can do is to extract around 20% of actual positives (when the predicted score is over 0.9). The ideal calibrat... | Calibration curve of XGBoost for binary classification | Answer:
No wonder your post-calibration plot looks like this. Look at your reliability curve pre-calibration. The more predicted score grows, the more actual positives it picks up. However, the best y | Calibration curve of XGBoost for binary classification
Answer:
No wonder your post-calibration plot looks like this. Look at your reliability curve pre-calibration. The more predicted score grows, the more actual positives it picks up. However, the best your model can do is to extract around 20% of actual positives (wh... | Calibration curve of XGBoost for binary classification
Answer:
No wonder your post-calibration plot looks like this. Look at your reliability curve pre-calibration. The more predicted score grows, the more actual positives it picks up. However, the best y |
38,844 | Calculate variance without calculating the mean | The median absolute deviation is defined as
$$\text{MAD}(X) = \text{median} |X-\text{median}(X)|$$
and is considered an alternative to the standard deviation. But this is not the variance. In particular, it always exists, whether or not $X$ allows for moments. For instance, the MAD of a standard Cauchy is equal to one... | Calculate variance without calculating the mean | The median absolute deviation is defined as
$$\text{MAD}(X) = \text{median} |X-\text{median}(X)|$$
and is considered an alternative to the standard deviation. But this is not the variance. In particu | Calculate variance without calculating the mean
The median absolute deviation is defined as
$$\text{MAD}(X) = \text{median} |X-\text{median}(X)|$$
and is considered an alternative to the standard deviation. But this is not the variance. In particular, it always exists, whether or not $X$ allows for moments. For instan... | Calculate variance without calculating the mean
The median absolute deviation is defined as
$$\text{MAD}(X) = \text{median} |X-\text{median}(X)|$$
and is considered an alternative to the standard deviation. But this is not the variance. In particu |
38,845 | Calculate variance without calculating the mean | There is already a solution for this question on Math.stackexchange:
I summarize the answers:
You can use that the variance is $\overline{x^2} - \overline {x}^2$, which takes only one pass (computing the mean and the mean of the squares simultaneously), but can be more prone to roundoff error if the variance is small ... | Calculate variance without calculating the mean | There is already a solution for this question on Math.stackexchange:
I summarize the answers:
You can use that the variance is $\overline{x^2} - \overline {x}^2$, which takes only one pass (computing | Calculate variance without calculating the mean
There is already a solution for this question on Math.stackexchange:
I summarize the answers:
You can use that the variance is $\overline{x^2} - \overline {x}^2$, which takes only one pass (computing the mean and the mean of the squares simultaneously), but can be more p... | Calculate variance without calculating the mean
There is already a solution for this question on Math.stackexchange:
I summarize the answers:
You can use that the variance is $\overline{x^2} - \overline {x}^2$, which takes only one pass (computing |
38,846 | Why does gls model without random effects yield a similar fit to mixed effects model? | Random effects also model correlations. To explain this more formally, the model that both lme() and gls() are fitting is the following
$$y_i = X_i\beta + \varepsilon_i, \quad \varepsilon_i \sim \mathcal N(0, V_i),$$
where $y_i$ denotes the outcome variable for the $i$-th group (i.e., follicles in your example, and th... | Why does gls model without random effects yield a similar fit to mixed effects model? | Random effects also model correlations. To explain this more formally, the model that both lme() and gls() are fitting is the following
$$y_i = X_i\beta + \varepsilon_i, \quad \varepsilon_i \sim \mat | Why does gls model without random effects yield a similar fit to mixed effects model?
Random effects also model correlations. To explain this more formally, the model that both lme() and gls() are fitting is the following
$$y_i = X_i\beta + \varepsilon_i, \quad \varepsilon_i \sim \mathcal N(0, V_i),$$
where $y_i$ deno... | Why does gls model without random effects yield a similar fit to mixed effects model?
Random effects also model correlations. To explain this more formally, the model that both lme() and gls() are fitting is the following
$$y_i = X_i\beta + \varepsilon_i, \quad \varepsilon_i \sim \mat |
38,847 | Calculating total estimated degrees of freedom for a GAM | The correct terminology for the degrees of freedom that you need to compute is model degrees of freedom. You could also compute residual degrees of freedom.
The model degrees of freedom are indeed calculated by adding up the degrees of freedom used by the parametric and non-parametric (or smooth) terms in your model... | Calculating total estimated degrees of freedom for a GAM | The correct terminology for the degrees of freedom that you need to compute is model degrees of freedom. You could also compute residual degrees of freedom.
The model degrees of freedom are indeed | Calculating total estimated degrees of freedom for a GAM
The correct terminology for the degrees of freedom that you need to compute is model degrees of freedom. You could also compute residual degrees of freedom.
The model degrees of freedom are indeed calculated by adding up the degrees of freedom used by the para... | Calculating total estimated degrees of freedom for a GAM
The correct terminology for the degrees of freedom that you need to compute is model degrees of freedom. You could also compute residual degrees of freedom.
The model degrees of freedom are indeed |
38,848 | hard voting versus soft voting in ensemble based methods [duplicate] | Suppose you have probabilities:
0.45
0.45
0.90
Then hard voting would give you a score of 1/3 (1 vote in favour and 2 against), so it would classify as a "negative".
Soft voting would give you the average of the probabilities, which is 0.6, and would be a "positive".
Soft voting takes into account how certain each vot... | hard voting versus soft voting in ensemble based methods [duplicate] | Suppose you have probabilities:
0.45
0.45
0.90
Then hard voting would give you a score of 1/3 (1 vote in favour and 2 against), so it would classify as a "negative".
Soft voting would give you the ave | hard voting versus soft voting in ensemble based methods [duplicate]
Suppose you have probabilities:
0.45
0.45
0.90
Then hard voting would give you a score of 1/3 (1 vote in favour and 2 against), so it would classify as a "negative".
Soft voting would give you the average of the probabilities, which is 0.6, and would ... | hard voting versus soft voting in ensemble based methods [duplicate]
Suppose you have probabilities:
0.45
0.45
0.90
Then hard voting would give you a score of 1/3 (1 vote in favour and 2 against), so it would classify as a "negative".
Soft voting would give you the ave |
38,849 | What is a strict definition of U-shaped relationship? | The short answer to your question (as stated elsewhere) is that
there is no single mathematical definition of a U-shape. The comment by @whuber is the
best general definition that I have seen.
I do research on tests of U-shapes and for my presentation I have a slide with the title
"What does a U mean to you?", meaning ... | What is a strict definition of U-shaped relationship? | The short answer to your question (as stated elsewhere) is that
there is no single mathematical definition of a U-shape. The comment by @whuber is the
best general definition that I have seen.
I do re | What is a strict definition of U-shaped relationship?
The short answer to your question (as stated elsewhere) is that
there is no single mathematical definition of a U-shape. The comment by @whuber is the
best general definition that I have seen.
I do research on tests of U-shapes and for my presentation I have a slide... | What is a strict definition of U-shaped relationship?
The short answer to your question (as stated elsewhere) is that
there is no single mathematical definition of a U-shape. The comment by @whuber is the
best general definition that I have seen.
I do re |
38,850 | What is a strict definition of U-shaped relationship? | "U-shaped relationship" is not a mathematically precise term and there is no universally accepted definition. It usually means that the relationship is first decreasing and then increasing, or vice versa.
In other words, it means that the relationship is not monotonic (non-monotonic), but instead has exactly one extrem... | What is a strict definition of U-shaped relationship? | "U-shaped relationship" is not a mathematically precise term and there is no universally accepted definition. It usually means that the relationship is first decreasing and then increasing, or vice ve | What is a strict definition of U-shaped relationship?
"U-shaped relationship" is not a mathematically precise term and there is no universally accepted definition. It usually means that the relationship is first decreasing and then increasing, or vice versa.
In other words, it means that the relationship is not monoton... | What is a strict definition of U-shaped relationship?
"U-shaped relationship" is not a mathematically precise term and there is no universally accepted definition. It usually means that the relationship is first decreasing and then increasing, or vice ve |
38,851 | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty | Let's ignore the penalty term for a moment, while we explore the sensitivity of the solution to changes in a single observation. This has ramifications for all linear least-squares models, not just Ridge regression.
Notation
To simplify the notation, let $X$ be the model matrix, including a column of constant values (... | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty | Let's ignore the penalty term for a moment, while we explore the sensitivity of the solution to changes in a single observation. This has ramifications for all linear least-squares models, not just R | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty
Let's ignore the penalty term for a moment, while we explore the sensitivity of the solution to changes in a single observation. This has ramifications for all linear least-squares models, not just Ridge regression.
Notation
To simplify the nota... | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty
Let's ignore the penalty term for a moment, while we explore the sensitivity of the solution to changes in a single observation. This has ramifications for all linear least-squares models, not just R |
38,852 | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty | This can be most easily seen through Lagrange duality: there exists some $C$ so that $$\arg\min_{\beta \in \mathbb{R}^p} RSS + \lambda \sum_{i=0}^p \beta_i^2 = \arg\min_{\beta\in\mathbb{R}^p \, : \, \|\beta\|_2^2 \leq C} RSS.$$ Further, we know that larger $\lambda$ corresponds to smaller $C$. Therefore, increasing the... | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty | This can be most easily seen through Lagrange duality: there exists some $C$ so that $$\arg\min_{\beta \in \mathbb{R}^p} RSS + \lambda \sum_{i=0}^p \beta_i^2 = \arg\min_{\beta\in\mathbb{R}^p \, : \, \ | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty
This can be most easily seen through Lagrange duality: there exists some $C$ so that $$\arg\min_{\beta \in \mathbb{R}^p} RSS + \lambda \sum_{i=0}^p \beta_i^2 = \arg\min_{\beta\in\mathbb{R}^p \, : \, \|\beta\|_2^2 \leq C} RSS.$$ Further, we know t... | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty
This can be most easily seen through Lagrange duality: there exists some $C$ so that $$\arg\min_{\beta \in \mathbb{R}^p} RSS + \lambda \sum_{i=0}^p \beta_i^2 = \arg\min_{\beta\in\mathbb{R}^p \, : \, \ |
38,853 | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty | Here, we can see that a general increase in the β vector will decrease
RSS and increase the other term.
That is not strictly true. For example, check what happens to your $RSS$ if $p$ is $1$ and $y=0$ for all $n$ points as you increase $\beta$. | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty | Here, we can see that a general increase in the β vector will decrease
RSS and increase the other term.
That is not strictly true. For example, check what happens to your $RSS$ if $p$ is $1$ and $ | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty
Here, we can see that a general increase in the β vector will decrease
RSS and increase the other term.
That is not strictly true. For example, check what happens to your $RSS$ if $p$ is $1$ and $y=0$ for all $n$ points as you increase $\beta... | Ridge Regression -Increase in $\lambda$ leads to a decrease in flexibilty
Here, we can see that a general increase in the β vector will decrease
RSS and increase the other term.
That is not strictly true. For example, check what happens to your $RSS$ if $p$ is $1$ and $ |
38,854 | Percentage as dependent variable in multiple linear regression | You should not use linear regression here, nor should you transform your data with the logit transformation. You have a percentage variable in a sense, but that's just a way to display your data in a simplified manner. In another sense, you have a count of obese children out of a known total of kids. That is, you ha... | Percentage as dependent variable in multiple linear regression | You should not use linear regression here, nor should you transform your data with the logit transformation. You have a percentage variable in a sense, but that's just a way to display your data in a | Percentage as dependent variable in multiple linear regression
You should not use linear regression here, nor should you transform your data with the logit transformation. You have a percentage variable in a sense, but that's just a way to display your data in a simplified manner. In another sense, you have a count o... | Percentage as dependent variable in multiple linear regression
You should not use linear regression here, nor should you transform your data with the logit transformation. You have a percentage variable in a sense, but that's just a way to display your data in a |
38,855 | Percentage as dependent variable in multiple linear regression | In linear regression, there are several assumptions which include the normality and independence of the independent variables and the error terms. Based on theory, the expectation of dependent variable $E(Y)$ would be also normality distributed. However, in reality it's hard to see a perfectly bell-shaped distribution ... | Percentage as dependent variable in multiple linear regression | In linear regression, there are several assumptions which include the normality and independence of the independent variables and the error terms. Based on theory, the expectation of dependent variabl | Percentage as dependent variable in multiple linear regression
In linear regression, there are several assumptions which include the normality and independence of the independent variables and the error terms. Based on theory, the expectation of dependent variable $E(Y)$ would be also normality distributed. However, in... | Percentage as dependent variable in multiple linear regression
In linear regression, there are several assumptions which include the normality and independence of the independent variables and the error terms. Based on theory, the expectation of dependent variabl |
38,856 | Poisson Regression : expectation vs probability for each outcome | If you have a fitted expectation (as you show you can get), using that as a parameter in a Poisson pdf should work.
That is, given $\hat{\lambda}_i = E(Y=0|X=\boldsymbol{x}_i)$ you can use that to calculate whichever probability you want:
$$\hat{P}(Y=k|X=\boldsymbol{x}_i) = \frac{\exp(-\hat{\lambda}_i)\, \hat{\lambda}... | Poisson Regression : expectation vs probability for each outcome | If you have a fitted expectation (as you show you can get), using that as a parameter in a Poisson pdf should work.
That is, given $\hat{\lambda}_i = E(Y=0|X=\boldsymbol{x}_i)$ you can use that to ca | Poisson Regression : expectation vs probability for each outcome
If you have a fitted expectation (as you show you can get), using that as a parameter in a Poisson pdf should work.
That is, given $\hat{\lambda}_i = E(Y=0|X=\boldsymbol{x}_i)$ you can use that to calculate whichever probability you want:
$$\hat{P}(Y=k|X... | Poisson Regression : expectation vs probability for each outcome
If you have a fitted expectation (as you show you can get), using that as a parameter in a Poisson pdf should work.
That is, given $\hat{\lambda}_i = E(Y=0|X=\boldsymbol{x}_i)$ you can use that to ca |
38,857 | Poisson Regression : expectation vs probability for each outcome | Just to illustrate the canonical answer by Glen (and with the peace of mind of being late for the party to the point that the answer to the OP has rightfully been adjudicated), let me run a toy example in R, crediting this post also by Glen.
Instead of cars and money, I'll make it more picturesque, and focus on the nu... | Poisson Regression : expectation vs probability for each outcome | Just to illustrate the canonical answer by Glen (and with the peace of mind of being late for the party to the point that the answer to the OP has rightfully been adjudicated), let me run a toy exampl | Poisson Regression : expectation vs probability for each outcome
Just to illustrate the canonical answer by Glen (and with the peace of mind of being late for the party to the point that the answer to the OP has rightfully been adjudicated), let me run a toy example in R, crediting this post also by Glen.
Instead of c... | Poisson Regression : expectation vs probability for each outcome
Just to illustrate the canonical answer by Glen (and with the peace of mind of being late for the party to the point that the answer to the OP has rightfully been adjudicated), let me run a toy exampl |
38,858 | Batch Normalization shift/scale parameters defeat the point | I'm by no means an expert on this topic, but here are my thoughts.
I think the main point of batch normalization is that training of each network layer is unaffected by changes in scale of the preceding layers. The author wrote
$$BN(Wu) = BN((aW)u)$$
for any scalar $a$. Thus, also backpropagated error is unaffected as... | Batch Normalization shift/scale parameters defeat the point | I'm by no means an expert on this topic, but here are my thoughts.
I think the main point of batch normalization is that training of each network layer is unaffected by changes in scale of the precedi | Batch Normalization shift/scale parameters defeat the point
I'm by no means an expert on this topic, but here are my thoughts.
I think the main point of batch normalization is that training of each network layer is unaffected by changes in scale of the preceding layers. The author wrote
$$BN(Wu) = BN((aW)u)$$
for any ... | Batch Normalization shift/scale parameters defeat the point
I'm by no means an expert on this topic, but here are my thoughts.
I think the main point of batch normalization is that training of each network layer is unaffected by changes in scale of the precedi |
38,859 | Batch Normalization shift/scale parameters defeat the point | If someone passes by this question in the future, Ian Goodfellow and Yoshua Bengio and Aaron Courville's Deep Learning, Section 8.7.1 can give a good explanation:
Normalizing the mean and standard deviation of a unit can reduce the expressive power of the neural network containing that unit. To maintain the expressive... | Batch Normalization shift/scale parameters defeat the point | If someone passes by this question in the future, Ian Goodfellow and Yoshua Bengio and Aaron Courville's Deep Learning, Section 8.7.1 can give a good explanation:
Normalizing the mean and standard de | Batch Normalization shift/scale parameters defeat the point
If someone passes by this question in the future, Ian Goodfellow and Yoshua Bengio and Aaron Courville's Deep Learning, Section 8.7.1 can give a good explanation:
Normalizing the mean and standard deviation of a unit can reduce the expressive power of the neu... | Batch Normalization shift/scale parameters defeat the point
If someone passes by this question in the future, Ian Goodfellow and Yoshua Bengio and Aaron Courville's Deep Learning, Section 8.7.1 can give a good explanation:
Normalizing the mean and standard de |
38,860 | Batch Normalization shift/scale parameters defeat the point | Correct me if I am wrong...
The BN normalization values are calculated for each components to normalize the input to a neuron (i.e. elliptic distribution becomes circular). The resulting distribution may be too wide or too narrow and shifted to "activate" well the activation function. To my understanding, the gamma and... | Batch Normalization shift/scale parameters defeat the point | Correct me if I am wrong...
The BN normalization values are calculated for each components to normalize the input to a neuron (i.e. elliptic distribution becomes circular). The resulting distribution | Batch Normalization shift/scale parameters defeat the point
Correct me if I am wrong...
The BN normalization values are calculated for each components to normalize the input to a neuron (i.e. elliptic distribution becomes circular). The resulting distribution may be too wide or too narrow and shifted to "activate" well... | Batch Normalization shift/scale parameters defeat the point
Correct me if I am wrong...
The BN normalization values are calculated for each components to normalize the input to a neuron (i.e. elliptic distribution becomes circular). The resulting distribution |
38,861 | 2D convolution with depth | You already understand that the dimension of a single kernel is 3x3x3 and there are 5 kernels. So each kernel is a 2D window of 3x3 pixels and there are 3 components in each kernel, one corresponding to each color channel (R,G,B). When the kernel is placed at a particular location over the input image its 3 components ... | 2D convolution with depth | You already understand that the dimension of a single kernel is 3x3x3 and there are 5 kernels. So each kernel is a 2D window of 3x3 pixels and there are 3 components in each kernel, one corresponding | 2D convolution with depth
You already understand that the dimension of a single kernel is 3x3x3 and there are 5 kernels. So each kernel is a 2D window of 3x3 pixels and there are 3 components in each kernel, one corresponding to each color channel (R,G,B). When the kernel is placed at a particular location over the inp... | 2D convolution with depth
You already understand that the dimension of a single kernel is 3x3x3 and there are 5 kernels. So each kernel is a 2D window of 3x3 pixels and there are 3 components in each kernel, one corresponding |
38,862 | 2D convolution with depth | I had this exact same question and came here to find the answer. @farhanhubble, your answer was very helpful. Here is the math of the convolution drawn on top of images from the site you shared just for further clarification.
Images from the same site as farhanhubble's answer. | 2D convolution with depth | I had this exact same question and came here to find the answer. @farhanhubble, your answer was very helpful. Here is the math of the convolution drawn on top of images from the site you shared just f | 2D convolution with depth
I had this exact same question and came here to find the answer. @farhanhubble, your answer was very helpful. Here is the math of the convolution drawn on top of images from the site you shared just for further clarification.
Images from the same site as farhanhubble's answer. | 2D convolution with depth
I had this exact same question and came here to find the answer. @farhanhubble, your answer was very helpful. Here is the math of the convolution drawn on top of images from the site you shared just f |
38,863 | Which statistical test to use when there's no data of SD but just know the mean? | No, I do not believe there's any formal test that you can apply. (By the way, the usual term for such samples is pooled.)
You are right that repeating the blot will not tell you more about the biological variance in population. You could try to look up the variance obtained by others in similar, but non-pooled experime... | Which statistical test to use when there's no data of SD but just know the mean? | No, I do not believe there's any formal test that you can apply. (By the way, the usual term for such samples is pooled.)
You are right that repeating the blot will not tell you more about the biologi | Which statistical test to use when there's no data of SD but just know the mean?
No, I do not believe there's any formal test that you can apply. (By the way, the usual term for such samples is pooled.)
You are right that repeating the blot will not tell you more about the biological variance in population. You could t... | Which statistical test to use when there's no data of SD but just know the mean?
No, I do not believe there's any formal test that you can apply. (By the way, the usual term for such samples is pooled.)
You are right that repeating the blot will not tell you more about the biologi |
38,864 | Which statistical test to use when there's no data of SD but just know the mean? | There is probably not much to be done.
If there is some way to estimate the within-group variance of unpooled western blot results from previous data, you could use this to generate an estimate of the standard deviation for the pooled western blot and use that to run a test. But it does not sound like this is the case... | Which statistical test to use when there's no data of SD but just know the mean? | There is probably not much to be done.
If there is some way to estimate the within-group variance of unpooled western blot results from previous data, you could use this to generate an estimate of th | Which statistical test to use when there's no data of SD but just know the mean?
There is probably not much to be done.
If there is some way to estimate the within-group variance of unpooled western blot results from previous data, you could use this to generate an estimate of the standard deviation for the pooled wes... | Which statistical test to use when there's no data of SD but just know the mean?
There is probably not much to be done.
If there is some way to estimate the within-group variance of unpooled western blot results from previous data, you could use this to generate an estimate of th |
38,865 | Which statistical test to use when there's no data of SD but just know the mean? | I would look into nonparametric statistic to find a test that would work for your situation. These tests don't have as many parameters to match to make the test work and are designed to work for smaller sample sizes. | Which statistical test to use when there's no data of SD but just know the mean? | I would look into nonparametric statistic to find a test that would work for your situation. These tests don't have as many parameters to match to make the test work and are designed to work for sm | Which statistical test to use when there's no data of SD but just know the mean?
I would look into nonparametric statistic to find a test that would work for your situation. These tests don't have as many parameters to match to make the test work and are designed to work for smaller sample sizes. | Which statistical test to use when there's no data of SD but just know the mean?
I would look into nonparametric statistic to find a test that would work for your situation. These tests don't have as many parameters to match to make the test work and are designed to work for sm |
38,866 | How is a Chi-square distribution a gamma distribution if it only has one parameter? | Let's start with the p.d.f. of a gamma-distributed random variable $X$, where $\alpha$ is the shape parameter and $\beta$ is the rate parameter (the p.d.f. is a little bit different if $\beta$ is a scale parameter; both parameters are strictly positive):
$$
f_X(x) = \frac{x ^ {\alpha - 1} \beta ^ \alpha e ^ {-\beta x}}... | How is a Chi-square distribution a gamma distribution if it only has one parameter? | Let's start with the p.d.f. of a gamma-distributed random variable $X$, where $\alpha$ is the shape parameter and $\beta$ is the rate parameter (the p.d.f. is a little bit different if $\beta$ is a sc | How is a Chi-square distribution a gamma distribution if it only has one parameter?
Let's start with the p.d.f. of a gamma-distributed random variable $X$, where $\alpha$ is the shape parameter and $\beta$ is the rate parameter (the p.d.f. is a little bit different if $\beta$ is a scale parameter; both parameters are s... | How is a Chi-square distribution a gamma distribution if it only has one parameter?
Let's start with the p.d.f. of a gamma-distributed random variable $X$, where $\alpha$ is the shape parameter and $\beta$ is the rate parameter (the p.d.f. is a little bit different if $\beta$ is a sc |
38,867 | Pdf of $y = - \log(X)$ when $X$ is beta distributed The expected value of $Y$ | Let $Y=-\ln X,\quad X \sim Beta(\alpha,\beta)$ then
$$
F_Y(y)=P(Y<y)=P(-\ln X < y)=P(X > e^{-y})=\int_{e^{-y}}^1\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}dx
$$
$$
f_Y(y)=\frac{d}{dy}F_Y(y)=\frac{d}{dx}\int_{e^{-y}}^1\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alp... | Pdf of $y = - \log(X)$ when $X$ is beta distributed The expected value of $Y$ | Let $Y=-\ln X,\quad X \sim Beta(\alpha,\beta)$ then
$$
F_Y(y)=P(Y<y)=P(-\ln X < y)=P(X > e^{-y})=\int_{e^{-y}}^1\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}dx
$$ | Pdf of $y = - \log(X)$ when $X$ is beta distributed The expected value of $Y$
Let $Y=-\ln X,\quad X \sim Beta(\alpha,\beta)$ then
$$
F_Y(y)=P(Y<y)=P(-\ln X < y)=P(X > e^{-y})=\int_{e^{-y}}^1\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}dx
$$
$$
f_Y(y)=\frac{d}{dy}F_Y(y)=\frac{d}{dx}... | Pdf of $y = - \log(X)$ when $X$ is beta distributed The expected value of $Y$
Let $Y=-\ln X,\quad X \sim Beta(\alpha,\beta)$ then
$$
F_Y(y)=P(Y<y)=P(-\ln X < y)=P(X > e^{-y})=\int_{e^{-y}}^1\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}dx
$$ |
38,868 | Why is this not a valid moment generating function? | There's no playing about with series needed, no worrying about integrals (well not directly). This is "think about what MGFs are" kind of question.
Consider the definition of an MGF -- $\quad M_X(t)=E(e^{tX})$.
Given that definition, what should $M_X(0)$ be?
Do you notice the problem? | Why is this not a valid moment generating function? | There's no playing about with series needed, no worrying about integrals (well not directly). This is "think about what MGFs are" kind of question.
Consider the definition of an MGF -- $\quad M_X(t)=E | Why is this not a valid moment generating function?
There's no playing about with series needed, no worrying about integrals (well not directly). This is "think about what MGFs are" kind of question.
Consider the definition of an MGF -- $\quad M_X(t)=E(e^{tX})$.
Given that definition, what should $M_X(0)$ be?
Do you no... | Why is this not a valid moment generating function?
There's no playing about with series needed, no worrying about integrals (well not directly). This is "think about what MGFs are" kind of question.
Consider the definition of an MGF -- $\quad M_X(t)=E |
38,869 | What is the difference between MLP and RBF? | MLP: uses dot products (between inputs and weights) and sigmoidal activation functions (or other monotonic functions such as ReLU) and training is usually done through backpropagation for all layers (which can be as many as you want). This type of neural network is used in deep learning with the help of many techniques... | What is the difference between MLP and RBF? | MLP: uses dot products (between inputs and weights) and sigmoidal activation functions (or other monotonic functions such as ReLU) and training is usually done through backpropagation for all layers ( | What is the difference between MLP and RBF?
MLP: uses dot products (between inputs and weights) and sigmoidal activation functions (or other monotonic functions such as ReLU) and training is usually done through backpropagation for all layers (which can be as many as you want). This type of neural network is used in de... | What is the difference between MLP and RBF?
MLP: uses dot products (between inputs and weights) and sigmoidal activation functions (or other monotonic functions such as ReLU) and training is usually done through backpropagation for all layers ( |
38,870 | What is the difference between MLP and RBF? | RBF and MLP belong to a class of neural networks called feed-forward networks.
Hidden layer of RBF is different from MLP. It performs some computations. Each hidden unit act as a point in input space and activation/output for any instance depends on the distance between that point(Hidden Unit) and instance(Also a point... | What is the difference between MLP and RBF? | RBF and MLP belong to a class of neural networks called feed-forward networks.
Hidden layer of RBF is different from MLP. It performs some computations. Each hidden unit act as a point in input space | What is the difference between MLP and RBF?
RBF and MLP belong to a class of neural networks called feed-forward networks.
Hidden layer of RBF is different from MLP. It performs some computations. Each hidden unit act as a point in input space and activation/output for any instance depends on the distance between that ... | What is the difference between MLP and RBF?
RBF and MLP belong to a class of neural networks called feed-forward networks.
Hidden layer of RBF is different from MLP. It performs some computations. Each hidden unit act as a point in input space |
38,871 | Proof for Irreducible Error statement in ISLR page 19 [duplicate] | \begin{align*}
\mathbb{E}\left[(Y-\hat Y)^2\right]
&=\mathbb{E}\left[\left(f(X)+\epsilon-\hat{f}(X)\right)^2\right] \\
&=\mathbb{E}\left[\left(f(X)+\epsilon-\hat{f}(X)\right)
\left(f(X)+\epsilon-\hat{f}(X)\right)\right] \\
&=\mathbb{E}\left[\left(f(X)-\hat{f}(X)\right)
\lef... | Proof for Irreducible Error statement in ISLR page 19 [duplicate] | \begin{align*}
\mathbb{E}\left[(Y-\hat Y)^2\right]
&=\mathbb{E}\left[\left(f(X)+\epsilon-\hat{f}(X)\right)^2\right] \\
&=\mathbb{E}\left[\left(f(X)+\epsilon-\hat{f}(X)\right)
\ | Proof for Irreducible Error statement in ISLR page 19 [duplicate]
\begin{align*}
\mathbb{E}\left[(Y-\hat Y)^2\right]
&=\mathbb{E}\left[\left(f(X)+\epsilon-\hat{f}(X)\right)^2\right] \\
&=\mathbb{E}\left[\left(f(X)+\epsilon-\hat{f}(X)\right)
\left(f(X)+\epsilon-\hat{f}(X)\right)\right] \\
&=\ma... | Proof for Irreducible Error statement in ISLR page 19 [duplicate]
\begin{align*}
\mathbb{E}\left[(Y-\hat Y)^2\right]
&=\mathbb{E}\left[\left(f(X)+\epsilon-\hat{f}(X)\right)^2\right] \\
&=\mathbb{E}\left[\left(f(X)+\epsilon-\hat{f}(X)\right)
\ |
38,872 | Auto-regression versus linear regression of x(t)-with-t for modelling time series | Auto-regressive models (ARIMA) use previous values as predictors depending upon the form of the model and forecasts are adaptive in form generally responding to previous values. Models using time as a predictor can be understood as using previous values to estimate the model parameters (thus previous values do come in... | Auto-regression versus linear regression of x(t)-with-t for modelling time series | Auto-regressive models (ARIMA) use previous values as predictors depending upon the form of the model and forecasts are adaptive in form generally responding to previous values. Models using time as a | Auto-regression versus linear regression of x(t)-with-t for modelling time series
Auto-regressive models (ARIMA) use previous values as predictors depending upon the form of the model and forecasts are adaptive in form generally responding to previous values. Models using time as a predictor can be understood as using ... | Auto-regression versus linear regression of x(t)-with-t for modelling time series
Auto-regressive models (ARIMA) use previous values as predictors depending upon the form of the model and forecasts are adaptive in form generally responding to previous values. Models using time as a |
38,873 | Auto-regression versus linear regression of x(t)-with-t for modelling time series | My response is more from a practical perspective. I'm specifically going to address your second part of question: why can't we use machine learning techniques for time series?
Reason #1: there is NO empirical evidence that machine learning are known to be superior than simple statistical time series models. Why bother ... | Auto-regression versus linear regression of x(t)-with-t for modelling time series | My response is more from a practical perspective. I'm specifically going to address your second part of question: why can't we use machine learning techniques for time series?
Reason #1: there is NO e | Auto-regression versus linear regression of x(t)-with-t for modelling time series
My response is more from a practical perspective. I'm specifically going to address your second part of question: why can't we use machine learning techniques for time series?
Reason #1: there is NO empirical evidence that machine learnin... | Auto-regression versus linear regression of x(t)-with-t for modelling time series
My response is more from a practical perspective. I'm specifically going to address your second part of question: why can't we use machine learning techniques for time series?
Reason #1: there is NO e |
38,874 | Machine Learning Book (Python) | As mentioned in the page linked to by gung, the Elements of Statistical Learning is a great, in-depth reference. An Introduction to Statistical Learning is a more approachable book that covers many of the same topics and also provides examples in R.
I have also found Applied Predictive Modeling to be a really good, pra... | Machine Learning Book (Python) | As mentioned in the page linked to by gung, the Elements of Statistical Learning is a great, in-depth reference. An Introduction to Statistical Learning is a more approachable book that covers many of | Machine Learning Book (Python)
As mentioned in the page linked to by gung, the Elements of Statistical Learning is a great, in-depth reference. An Introduction to Statistical Learning is a more approachable book that covers many of the same topics and also provides examples in R.
I have also found Applied Predictive Mo... | Machine Learning Book (Python)
As mentioned in the page linked to by gung, the Elements of Statistical Learning is a great, in-depth reference. An Introduction to Statistical Learning is a more approachable book that covers many of |
38,875 | Machine Learning Book (Python) | I would recommend that you get two books:
General reference
I can definitely recommend Pattern Recognition and Machine Learning by Christopher Bishop. It is a bit older (2006) but you will obtain the foundation to dive in state of the art research papers.
The already mentioned Elements of Statistical Learning is also ... | Machine Learning Book (Python) | I would recommend that you get two books:
General reference
I can definitely recommend Pattern Recognition and Machine Learning by Christopher Bishop. It is a bit older (2006) but you will obtain the | Machine Learning Book (Python)
I would recommend that you get two books:
General reference
I can definitely recommend Pattern Recognition and Machine Learning by Christopher Bishop. It is a bit older (2006) but you will obtain the foundation to dive in state of the art research papers.
The already mentioned Elements o... | Machine Learning Book (Python)
I would recommend that you get two books:
General reference
I can definitely recommend Pattern Recognition and Machine Learning by Christopher Bishop. It is a bit older (2006) but you will obtain the |
38,876 | Machine Learning Book (Python) | Murphy's "Machine Learning" is a good book with many examples. They use MATLAB as a programming language, which means most of it can be easily run using Octave or translated to Python. | Machine Learning Book (Python) | Murphy's "Machine Learning" is a good book with many examples. They use MATLAB as a programming language, which means most of it can be easily run using Octave or translated to Python. | Machine Learning Book (Python)
Murphy's "Machine Learning" is a good book with many examples. They use MATLAB as a programming language, which means most of it can be easily run using Octave or translated to Python. | Machine Learning Book (Python)
Murphy's "Machine Learning" is a good book with many examples. They use MATLAB as a programming language, which means most of it can be easily run using Octave or translated to Python. |
38,877 | Machine Learning Book (Python) | A very recent book is Data Science from Scratch by Grus, which covers "k-nearest Neighbors, Naive Bayes, linear and logistic regression, decision trees, neural networks, and clustering." I haven't read it myself, but I have heard good things. | Machine Learning Book (Python) | A very recent book is Data Science from Scratch by Grus, which covers "k-nearest Neighbors, Naive Bayes, linear and logistic regression, decision trees, neural networks, and clustering." I haven't rea | Machine Learning Book (Python)
A very recent book is Data Science from Scratch by Grus, which covers "k-nearest Neighbors, Naive Bayes, linear and logistic regression, decision trees, neural networks, and clustering." I haven't read it myself, but I have heard good things. | Machine Learning Book (Python)
A very recent book is Data Science from Scratch by Grus, which covers "k-nearest Neighbors, Naive Bayes, linear and logistic regression, decision trees, neural networks, and clustering." I haven't rea |
38,878 | Machine Learning Book (Python) | A Course in Machine Learning by Hal Daumé III
(227 pages, 2013-2017)
aims to build intuition, with a nice blend of narrative, diagrams, equations and pseudocode (not python); the lively teacher comes through.
One can download the
whole book pdf,
or individual chapters:
Decision Trees
Limits of Learning
Geometry and Nea... | Machine Learning Book (Python) | A Course in Machine Learning by Hal Daumé III
(227 pages, 2013-2017)
aims to build intuition, with a nice blend of narrative, diagrams, equations and pseudocode (not python); the lively teacher comes | Machine Learning Book (Python)
A Course in Machine Learning by Hal Daumé III
(227 pages, 2013-2017)
aims to build intuition, with a nice blend of narrative, diagrams, equations and pseudocode (not python); the lively teacher comes through.
One can download the
whole book pdf,
or individual chapters:
Decision Trees
Limi... | Machine Learning Book (Python)
A Course in Machine Learning by Hal Daumé III
(227 pages, 2013-2017)
aims to build intuition, with a nice blend of narrative, diagrams, equations and pseudocode (not python); the lively teacher comes |
38,879 | Machine Learning Book (Python) | In my experience I recommend the book Practical Econometrics with Python. I love this book because links theory with real examples in Python. It goes from basic topics like OLS to advanced topics like VARMA. | Machine Learning Book (Python) | In my experience I recommend the book Practical Econometrics with Python. I love this book because links theory with real examples in Python. It goes from basic topics like OLS to advanced topics like | Machine Learning Book (Python)
In my experience I recommend the book Practical Econometrics with Python. I love this book because links theory with real examples in Python. It goes from basic topics like OLS to advanced topics like VARMA. | Machine Learning Book (Python)
In my experience I recommend the book Practical Econometrics with Python. I love this book because links theory with real examples in Python. It goes from basic topics like OLS to advanced topics like |
38,880 | Conditional vs. Marginal models | Either of the models you used are probably fine approaches -- and it's certainly reassuring that the results are similar.
Marginal models are population-average models whereas conditional models are subject-specific. As a result, there are subtle differences in interpretation. For example if you were studying the eff... | Conditional vs. Marginal models | Either of the models you used are probably fine approaches -- and it's certainly reassuring that the results are similar.
Marginal models are population-average models whereas conditional models are s | Conditional vs. Marginal models
Either of the models you used are probably fine approaches -- and it's certainly reassuring that the results are similar.
Marginal models are population-average models whereas conditional models are subject-specific. As a result, there are subtle differences in interpretation. For exam... | Conditional vs. Marginal models
Either of the models you used are probably fine approaches -- and it's certainly reassuring that the results are similar.
Marginal models are population-average models whereas conditional models are s |
38,881 | Conditional vs. Marginal models | One can view the marginal model as providing crude estimates of the regression coefficients [i.e. unadjusted for subjects] while the conditional model has regression coefficients that are assumed common to subjects and so the estimates are adjusted for subjects. In this sense, one is asking whether the subjects at hand... | Conditional vs. Marginal models | One can view the marginal model as providing crude estimates of the regression coefficients [i.e. unadjusted for subjects] while the conditional model has regression coefficients that are assumed comm | Conditional vs. Marginal models
One can view the marginal model as providing crude estimates of the regression coefficients [i.e. unadjusted for subjects] while the conditional model has regression coefficients that are assumed common to subjects and so the estimates are adjusted for subjects. In this sense, one is ask... | Conditional vs. Marginal models
One can view the marginal model as providing crude estimates of the regression coefficients [i.e. unadjusted for subjects] while the conditional model has regression coefficients that are assumed comm |
38,882 | Just "take the average" they say. It's not that straightforward, right? | This is not a direct answer to your question about loss functions, but I am a Statistician, and I use the jargon of my domain, not the jargon of machine learning. I will attempt to answer the question: "which statistic is the best estimator of the population mean?"
It's incorrect to say generally that the arithmetic me... | Just "take the average" they say. It's not that straightforward, right? | This is not a direct answer to your question about loss functions, but I am a Statistician, and I use the jargon of my domain, not the jargon of machine learning. I will attempt to answer the question | Just "take the average" they say. It's not that straightforward, right?
This is not a direct answer to your question about loss functions, but I am a Statistician, and I use the jargon of my domain, not the jargon of machine learning. I will attempt to answer the question: "which statistic is the best estimator of the ... | Just "take the average" they say. It's not that straightforward, right?
This is not a direct answer to your question about loss functions, but I am a Statistician, and I use the jargon of my domain, not the jargon of machine learning. I will attempt to answer the question |
38,883 | Just "take the average" they say. It's not that straightforward, right? | Piggy-backing / building off of Dalton's answer:
Your question, as posed, is incomplete. "Information" as a statistical concept is only defined with reference to an unknown parameter as well as some statistic (function of the full data)—including the degenerate case of the full data. Exactly no statistic of the data ha... | Just "take the average" they say. It's not that straightforward, right? | Piggy-backing / building off of Dalton's answer:
Your question, as posed, is incomplete. "Information" as a statistical concept is only defined with reference to an unknown parameter as well as some s | Just "take the average" they say. It's not that straightforward, right?
Piggy-backing / building off of Dalton's answer:
Your question, as posed, is incomplete. "Information" as a statistical concept is only defined with reference to an unknown parameter as well as some statistic (function of the full data)—including t... | Just "take the average" they say. It's not that straightforward, right?
Piggy-backing / building off of Dalton's answer:
Your question, as posed, is incomplete. "Information" as a statistical concept is only defined with reference to an unknown parameter as well as some s |
38,884 | Observed Vs Unobserved Variables [duplicate] | Observed variables are variables for which you have measurements in your dataset, whereas unobserved (or latent) variables are variables for which you don't.
When your analysis reveals correlations between observed variables, you might look for unobserved variables to explain the correlation, especially in cases where ... | Observed Vs Unobserved Variables [duplicate] | Observed variables are variables for which you have measurements in your dataset, whereas unobserved (or latent) variables are variables for which you don't.
When your analysis reveals correlations be | Observed Vs Unobserved Variables [duplicate]
Observed variables are variables for which you have measurements in your dataset, whereas unobserved (or latent) variables are variables for which you don't.
When your analysis reveals correlations between observed variables, you might look for unobserved variables to explai... | Observed Vs Unobserved Variables [duplicate]
Observed variables are variables for which you have measurements in your dataset, whereas unobserved (or latent) variables are variables for which you don't.
When your analysis reveals correlations be |
38,885 | Observed Vs Unobserved Variables [duplicate] | Are you doing research within economics?
Usually, the unobservable characteristics are those that are captured by the error term and bias your estimates, for instance through heteroskedasticity.
Let's me give you an example.
When you conduct a so called correspondence test in the labor market to study ethnic discrimina... | Observed Vs Unobserved Variables [duplicate] | Are you doing research within economics?
Usually, the unobservable characteristics are those that are captured by the error term and bias your estimates, for instance through heteroskedasticity.
Let's | Observed Vs Unobserved Variables [duplicate]
Are you doing research within economics?
Usually, the unobservable characteristics are those that are captured by the error term and bias your estimates, for instance through heteroskedasticity.
Let's me give you an example.
When you conduct a so called correspondence test i... | Observed Vs Unobserved Variables [duplicate]
Are you doing research within economics?
Usually, the unobservable characteristics are those that are captured by the error term and bias your estimates, for instance through heteroskedasticity.
Let's |
38,886 | Can percentages be averaged? | Specific advice is difficult here. We see no data and have only a broad idea of what is being done. What are the precise questions being answered?
Averaging percents is often going to be a bad idea, but the question needs to be made broader to allow good recommendations.
For example, if reducing time spent is of pri... | Can percentages be averaged? | Specific advice is difficult here. We see no data and have only a broad idea of what is being done. What are the precise questions being answered?
Averaging percents is often going to be a bad idea, | Can percentages be averaged?
Specific advice is difficult here. We see no data and have only a broad idea of what is being done. What are the precise questions being answered?
Averaging percents is often going to be a bad idea, but the question needs to be made broader to allow good recommendations.
For example, if ... | Can percentages be averaged?
Specific advice is difficult here. We see no data and have only a broad idea of what is being done. What are the precise questions being answered?
Averaging percents is often going to be a bad idea, |
38,887 | Can percentages be averaged? | I quite agree with Nick Cox's response, to give an example where one has to be wary, take compliance with a particular standard by an organisation with just two cost centres where the sample size in the first case is 10 and in the second it is 100. In the first case 8 out of 10 may be compliant, while in the second on... | Can percentages be averaged? | I quite agree with Nick Cox's response, to give an example where one has to be wary, take compliance with a particular standard by an organisation with just two cost centres where the sample size in t | Can percentages be averaged?
I quite agree with Nick Cox's response, to give an example where one has to be wary, take compliance with a particular standard by an organisation with just two cost centres where the sample size in the first case is 10 and in the second it is 100. In the first case 8 out of 10 may be comp... | Can percentages be averaged?
I quite agree with Nick Cox's response, to give an example where one has to be wary, take compliance with a particular standard by an organisation with just two cost centres where the sample size in t |
38,888 | Can percentages be averaged? | If Subject 1 went from 10 sec to 5 sec (50% improvement) while Subject 2 went from 10 sec to 7.5 sec (25% improvement) then it is valid to average the percentages in this case and say that the average improvement is about 37.5%. | Can percentages be averaged? | If Subject 1 went from 10 sec to 5 sec (50% improvement) while Subject 2 went from 10 sec to 7.5 sec (25% improvement) then it is valid to average the percentages in this case and say that the average | Can percentages be averaged?
If Subject 1 went from 10 sec to 5 sec (50% improvement) while Subject 2 went from 10 sec to 7.5 sec (25% improvement) then it is valid to average the percentages in this case and say that the average improvement is about 37.5%. | Can percentages be averaged?
If Subject 1 went from 10 sec to 5 sec (50% improvement) while Subject 2 went from 10 sec to 7.5 sec (25% improvement) then it is valid to average the percentages in this case and say that the average |
38,889 | In linear regression, what does $\beta_1 = 0$ really mean? | Part of the problem here is that your model is conceptually confused, as @CagdasOzgenc has correctly pointed out.(Edit: This has now been fixed.) I think I can address two of your specific questions.
If $\beta_1 = 0$ (note the absence of the 'hat'), and standard assumptions hold, then empirically, as we make ever m... | In linear regression, what does $\beta_1 = 0$ really mean? | Part of the problem here is that your model is conceptually confused, as @CagdasOzgenc has correctly pointed out.(Edit: This has now been fixed.) I think I can address two of your specific questions. | In linear regression, what does $\beta_1 = 0$ really mean?
Part of the problem here is that your model is conceptually confused, as @CagdasOzgenc has correctly pointed out.(Edit: This has now been fixed.) I think I can address two of your specific questions.
If $\beta_1 = 0$ (note the absence of the 'hat'), and sta... | In linear regression, what does $\beta_1 = 0$ really mean?
Part of the problem here is that your model is conceptually confused, as @CagdasOzgenc has correctly pointed out.(Edit: This has now been fixed.) I think I can address two of your specific questions. |
38,890 | In linear regression, what does $\beta_1 = 0$ really mean? | You are confusing the real data generation process with a model trying to make an approximation of this process (this is quite normal, I also banged my head to the wall several times regarding this matter).
First of all
$y = \beta_0 + \beta_1x_1 + \beta_2x_2$
is a deterministic model. Usually in regression, the model i... | In linear regression, what does $\beta_1 = 0$ really mean? | You are confusing the real data generation process with a model trying to make an approximation of this process (this is quite normal, I also banged my head to the wall several times regarding this ma | In linear regression, what does $\beta_1 = 0$ really mean?
You are confusing the real data generation process with a model trying to make an approximation of this process (this is quite normal, I also banged my head to the wall several times regarding this matter).
First of all
$y = \beta_0 + \beta_1x_1 + \beta_2x_2$
i... | In linear regression, what does $\beta_1 = 0$ really mean?
You are confusing the real data generation process with a model trying to make an approximation of this process (this is quite normal, I also banged my head to the wall several times regarding this ma |
38,891 | In linear regression, what does $\beta_1 = 0$ really mean? | Gung's answer is excellent, but I want to add an interpretation that I think goes underappreciated.
You wrote out the model as
$$
Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \varepsilon
$$
You didn't specify this, but presumably $\varepsilon$ is an error term with $\operatorname{\mathbb{E}}\left(\varepsilon\,|\,X\right) ... | In linear regression, what does $\beta_1 = 0$ really mean? | Gung's answer is excellent, but I want to add an interpretation that I think goes underappreciated.
You wrote out the model as
$$
Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \varepsilon
$$
You didn't sp | In linear regression, what does $\beta_1 = 0$ really mean?
Gung's answer is excellent, but I want to add an interpretation that I think goes underappreciated.
You wrote out the model as
$$
Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \varepsilon
$$
You didn't specify this, but presumably $\varepsilon$ is an error term wit... | In linear regression, what does $\beta_1 = 0$ really mean?
Gung's answer is excellent, but I want to add an interpretation that I think goes underappreciated.
You wrote out the model as
$$
Y = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \varepsilon
$$
You didn't sp |
38,892 | Probability that randomly chosen value from one distribution is greater than randomly chosen value from another distribution | Your bootstrap simulation suggests the A's and B's are independent; I will assume so.
Note that $P(A>B)=P(A-B>0)$.
In the case where A and B are also normal, $D=A-B \sim N(\mu_A-\mu_B,\sigma^2_A+\sigma^2_B)$
In that case
\begin{eqnarray*}
P(D>0) &=& P(\frac{D-\mu_D}{\sigma_D}>\frac{0-\mu_D}{\sigma_D}) \\
&=& P(Z>-\frac... | Probability that randomly chosen value from one distribution is greater than randomly chosen value f | Your bootstrap simulation suggests the A's and B's are independent; I will assume so.
Note that $P(A>B)=P(A-B>0)$.
In the case where A and B are also normal, $D=A-B \sim N(\mu_A-\mu_B,\sigma^2_A+\sigm | Probability that randomly chosen value from one distribution is greater than randomly chosen value from another distribution
Your bootstrap simulation suggests the A's and B's are independent; I will assume so.
Note that $P(A>B)=P(A-B>0)$.
In the case where A and B are also normal, $D=A-B \sim N(\mu_A-\mu_B,\sigma^2_A+... | Probability that randomly chosen value from one distribution is greater than randomly chosen value f
Your bootstrap simulation suggests the A's and B's are independent; I will assume so.
Note that $P(A>B)=P(A-B>0)$.
In the case where A and B are also normal, $D=A-B \sim N(\mu_A-\mu_B,\sigma^2_A+\sigm |
38,893 | Probability that randomly chosen value from one distribution is greater than randomly chosen value from another distribution | Answer: AUC
Probability that randomly chosen value from one distribution is greater than randomly chosen value from another distribution
This an eloquent description of the what the Area Under The Curve is estimating.
Check out the Wilcoxon-Mann-Whitney U (rank sum) test if you want a hypothesis-test (as others poi... | Probability that randomly chosen value from one distribution is greater than randomly chosen value f | Answer: AUC
Probability that randomly chosen value from one distribution is greater than randomly chosen value from another distribution
This an eloquent description of the what the Area Under The | Probability that randomly chosen value from one distribution is greater than randomly chosen value from another distribution
Answer: AUC
Probability that randomly chosen value from one distribution is greater than randomly chosen value from another distribution
This an eloquent description of the what the Area Under... | Probability that randomly chosen value from one distribution is greater than randomly chosen value f
Answer: AUC
Probability that randomly chosen value from one distribution is greater than randomly chosen value from another distribution
This an eloquent description of the what the Area Under The |
38,894 | AR(q) model with F-test | The regression model for the AR model that you specify departs from the assumptions of the classical regression model in that the explanatory variables are not deterministic or fixed. The explanatory variables are stochastic because they are lags of the dependent variable and hence depend on the stochastic term $\epsil... | AR(q) model with F-test | The regression model for the AR model that you specify departs from the assumptions of the classical regression model in that the explanatory variables are not deterministic or fixed. The explanatory | AR(q) model with F-test
The regression model for the AR model that you specify departs from the assumptions of the classical regression model in that the explanatory variables are not deterministic or fixed. The explanatory variables are stochastic because they are lags of the dependent variable and hence depend on the... | AR(q) model with F-test
The regression model for the AR model that you specify departs from the assumptions of the classical regression model in that the explanatory variables are not deterministic or fixed. The explanatory |
38,895 | AR(q) model with F-test | This answer can be found on Hamilton's "Time Series Analysis", page 215 (http://www.amazon.com/Series-Analysis-James-Douglas-Hamilton/dp/0691042896)
I am sure it is asymptotically valid if you include a constant term in your model:
Since $X_{t+j}, j=1,2,...$ are not independent from $\epsilon_t$, the F test does not ho... | AR(q) model with F-test | This answer can be found on Hamilton's "Time Series Analysis", page 215 (http://www.amazon.com/Series-Analysis-James-Douglas-Hamilton/dp/0691042896)
I am sure it is asymptotically valid if you include | AR(q) model with F-test
This answer can be found on Hamilton's "Time Series Analysis", page 215 (http://www.amazon.com/Series-Analysis-James-Douglas-Hamilton/dp/0691042896)
I am sure it is asymptotically valid if you include a constant term in your model:
Since $X_{t+j}, j=1,2,...$ are not independent from $\epsilon_t$... | AR(q) model with F-test
This answer can be found on Hamilton's "Time Series Analysis", page 215 (http://www.amazon.com/Series-Analysis-James-Douglas-Hamilton/dp/0691042896)
I am sure it is asymptotically valid if you include |
38,896 | How to describe a bin in a histogram? | This is partly a question of statistics terminology and partly one of English usage. (Clearly, some points may be irrelevant or need changing for anyone interested in this question but for some language other than English.)
Let's focus first on measured data.
To be completely clear in describing your bin to me you ha... | How to describe a bin in a histogram? | This is partly a question of statistics terminology and partly one of English usage. (Clearly, some points may be irrelevant or need changing for anyone interested in this question but for some langua | How to describe a bin in a histogram?
This is partly a question of statistics terminology and partly one of English usage. (Clearly, some points may be irrelevant or need changing for anyone interested in this question but for some language other than English.)
Let's focus first on measured data.
To be completely cle... | How to describe a bin in a histogram?
This is partly a question of statistics terminology and partly one of English usage. (Clearly, some points may be irrelevant or need changing for anyone interested in this question but for some langua |
38,897 | How to describe a bin in a histogram? | I agree with Nick's post - it's important advice for describing what a bin contains.
However, for this answer, I'll assume we're in a context where the content of the bins was already clear.
if I was simply referring to a particular bin**, I'd most likely refer to a bin by its ordinal position (first bin, second bin, .... | How to describe a bin in a histogram? | I agree with Nick's post - it's important advice for describing what a bin contains.
However, for this answer, I'll assume we're in a context where the content of the bins was already clear.
if I was | How to describe a bin in a histogram?
I agree with Nick's post - it's important advice for describing what a bin contains.
However, for this answer, I'll assume we're in a context where the content of the bins was already clear.
if I was simply referring to a particular bin**, I'd most likely refer to a bin by its ordi... | How to describe a bin in a histogram?
I agree with Nick's post - it's important advice for describing what a bin contains.
However, for this answer, I'll assume we're in a context where the content of the bins was already clear.
if I was |
38,898 | Correcting biased survey results | As Tim pointed out, you should use survey weighting.
In your case, more specifically, if all the auxiliary variables (your demographic variables) you want to use to make your sample match your population are qualitative variables you will use:
Post-stratification: If you have the full joint distribution of these vari... | Correcting biased survey results | As Tim pointed out, you should use survey weighting.
In your case, more specifically, if all the auxiliary variables (your demographic variables) you want to use to make your sample match your popula | Correcting biased survey results
As Tim pointed out, you should use survey weighting.
In your case, more specifically, if all the auxiliary variables (your demographic variables) you want to use to make your sample match your population are qualitative variables you will use:
Post-stratification: If you have the full... | Correcting biased survey results
As Tim pointed out, you should use survey weighting.
In your case, more specifically, if all the auxiliary variables (your demographic variables) you want to use to make your sample match your popula |
38,899 | Correcting biased survey results | The common thing to do in this kind of situation is to use survey weighting (or an intro here). A clear definition could be found on Wikipedia:
data should usually be weighted if the sample design does not give
each individual an equal chance of being selected. For instance, when
households have equal selection pr... | Correcting biased survey results | The common thing to do in this kind of situation is to use survey weighting (or an intro here). A clear definition could be found on Wikipedia:
data should usually be weighted if the sample design do | Correcting biased survey results
The common thing to do in this kind of situation is to use survey weighting (or an intro here). A clear definition could be found on Wikipedia:
data should usually be weighted if the sample design does not give
each individual an equal chance of being selected. For instance, when
h... | Correcting biased survey results
The common thing to do in this kind of situation is to use survey weighting (or an intro here). A clear definition could be found on Wikipedia:
data should usually be weighted if the sample design do |
38,900 | Correcting biased survey results | I follow both Raphael and Tim in their suggestions -- especially about the use of the R package survey. However, as Raphael suggested, these weighting techniques were developed for probability samples and it might not be your case.
If you are familiar to multilevel modeling and have quality auxiliary data to estimate t... | Correcting biased survey results | I follow both Raphael and Tim in their suggestions -- especially about the use of the R package survey. However, as Raphael suggested, these weighting techniques were developed for probability samples | Correcting biased survey results
I follow both Raphael and Tim in their suggestions -- especially about the use of the R package survey. However, as Raphael suggested, these weighting techniques were developed for probability samples and it might not be your case.
If you are familiar to multilevel modeling and have qua... | Correcting biased survey results
I follow both Raphael and Tim in their suggestions -- especially about the use of the R package survey. However, as Raphael suggested, these weighting techniques were developed for probability samples |
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