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 |
|---|---|---|---|---|---|---|
47,901 | Standardize binary variable to create interaction term in regression? | It makes little sense to standardize dummy variables
It cannot be increased by a standard deviation so the regular interpretation for standardized coefficients does not apply
Moreover, the standard interpretation of the dummy variable, showing difference in average level of Y between two categories is lost
Your inter... | Standardize binary variable to create interaction term in regression? | It makes little sense to standardize dummy variables
It cannot be increased by a standard deviation so the regular interpretation for standardized coefficients does not apply
Moreover, the standard i | Standardize binary variable to create interaction term in regression?
It makes little sense to standardize dummy variables
It cannot be increased by a standard deviation so the regular interpretation for standardized coefficients does not apply
Moreover, the standard interpretation of the dummy variable, showing diffe... | Standardize binary variable to create interaction term in regression?
It makes little sense to standardize dummy variables
It cannot be increased by a standard deviation so the regular interpretation for standardized coefficients does not apply
Moreover, the standard i |
47,902 | Standardize binary variable to create interaction term in regression? | The "mean centring" procedure for categorical variables is different from the one you would use for continuous variable (in R: scale(x, center=T, scale=F).
You can mean center categorical variables by using an effect coding strategy instead of traditional dummy coding (0/1). The choice of the coding scheme does not rea... | Standardize binary variable to create interaction term in regression? | The "mean centring" procedure for categorical variables is different from the one you would use for continuous variable (in R: scale(x, center=T, scale=F).
You can mean center categorical variables by | Standardize binary variable to create interaction term in regression?
The "mean centring" procedure for categorical variables is different from the one you would use for continuous variable (in R: scale(x, center=T, scale=F).
You can mean center categorical variables by using an effect coding strategy instead of tradit... | Standardize binary variable to create interaction term in regression?
The "mean centring" procedure for categorical variables is different from the one you would use for continuous variable (in R: scale(x, center=T, scale=F).
You can mean center categorical variables by |
47,903 | Standardize binary variable to create interaction term in regression? | In terms of specifics on how to set up the interactions, this web page seems to be helpful: http://www.restore.ac.uk/srme/www/fac/soc/wie/research-new/srme/modules/mod3/11/
You don't need to standardize the variables (unless they are already standardized).
I'd set it up like this (limiting this list to a small number... | Standardize binary variable to create interaction term in regression? | In terms of specifics on how to set up the interactions, this web page seems to be helpful: http://www.restore.ac.uk/srme/www/fac/soc/wie/research-new/srme/modules/mod3/11/
You don't need to standard | Standardize binary variable to create interaction term in regression?
In terms of specifics on how to set up the interactions, this web page seems to be helpful: http://www.restore.ac.uk/srme/www/fac/soc/wie/research-new/srme/modules/mod3/11/
You don't need to standardize the variables (unless they are already standar... | Standardize binary variable to create interaction term in regression?
In terms of specifics on how to set up the interactions, this web page seems to be helpful: http://www.restore.ac.uk/srme/www/fac/soc/wie/research-new/srme/modules/mod3/11/
You don't need to standard |
47,904 | Standardize binary variable to create interaction term in regression? | By interaction term do you mean a higher order term?
So instead of:
fit <- lm(y ~ x1 + x2, data = mydata)
You can do
fit <- lm(y ~ x1 + x2 + x1*x2, data = mydata)
(looking back at this answer it's probably not what you looking for. I would suggest trying setting your sex variable to be -1 or 1, this way at least th... | Standardize binary variable to create interaction term in regression? | By interaction term do you mean a higher order term?
So instead of:
fit <- lm(y ~ x1 + x2, data = mydata)
You can do
fit <- lm(y ~ x1 + x2 + x1*x2, data = mydata)
(looking back at this answer it's | Standardize binary variable to create interaction term in regression?
By interaction term do you mean a higher order term?
So instead of:
fit <- lm(y ~ x1 + x2, data = mydata)
You can do
fit <- lm(y ~ x1 + x2 + x1*x2, data = mydata)
(looking back at this answer it's probably not what you looking for. I would sugges... | Standardize binary variable to create interaction term in regression?
By interaction term do you mean a higher order term?
So instead of:
fit <- lm(y ~ x1 + x2, data = mydata)
You can do
fit <- lm(y ~ x1 + x2 + x1*x2, data = mydata)
(looking back at this answer it's |
47,905 | Linear Properties of the Quantile Function | Hint: the probability that $X<Q_x(p)$ is $p$. Given that $Y$ is linearly related to $X$, can you now write a similar expression for $Y$? | Linear Properties of the Quantile Function | Hint: the probability that $X<Q_x(p)$ is $p$. Given that $Y$ is linearly related to $X$, can you now write a similar expression for $Y$? | Linear Properties of the Quantile Function
Hint: the probability that $X<Q_x(p)$ is $p$. Given that $Y$ is linearly related to $X$, can you now write a similar expression for $Y$? | Linear Properties of the Quantile Function
Hint: the probability that $X<Q_x(p)$ is $p$. Given that $Y$ is linearly related to $X$, can you now write a similar expression for $Y$? |
47,906 | How to perform a sensitivity analysis in Bayesian statistics? | A fairly standard approach to showing that your results were not heavily influenced by your choice of prior is simply to show that your results hold when choosing a different prior. For example, if you have an informed prior that suggests a certain result is more likely, you might want to also show your results also ho... | How to perform a sensitivity analysis in Bayesian statistics? | A fairly standard approach to showing that your results were not heavily influenced by your choice of prior is simply to show that your results hold when choosing a different prior. For example, if yo | How to perform a sensitivity analysis in Bayesian statistics?
A fairly standard approach to showing that your results were not heavily influenced by your choice of prior is simply to show that your results hold when choosing a different prior. For example, if you have an informed prior that suggests a certain result is... | How to perform a sensitivity analysis in Bayesian statistics?
A fairly standard approach to showing that your results were not heavily influenced by your choice of prior is simply to show that your results hold when choosing a different prior. For example, if yo |
47,907 | MLE for k-truncated poisson | Update: Now that the question is more specific, I've added the log likelihood and R code for left-trunction.
Right-truncation. For $Y_i$ distributed as a truncated Poisson with parameters $\lambda$ and truncation parameter $k$, the log likelihood of $n$ random samples is given by
$$\log(\lambda) \sum_{i=1}^n y_i -n \... | MLE for k-truncated poisson | Update: Now that the question is more specific, I've added the log likelihood and R code for left-trunction.
Right-truncation. For $Y_i$ distributed as a truncated Poisson with parameters $\lambda$ | MLE for k-truncated poisson
Update: Now that the question is more specific, I've added the log likelihood and R code for left-trunction.
Right-truncation. For $Y_i$ distributed as a truncated Poisson with parameters $\lambda$ and truncation parameter $k$, the log likelihood of $n$ random samples is given by
$$\log(\l... | MLE for k-truncated poisson
Update: Now that the question is more specific, I've added the log likelihood and R code for left-trunction.
Right-truncation. For $Y_i$ distributed as a truncated Poisson with parameters $\lambda$ |
47,908 | Logistic regression intercept representing baseline probability | The intercept might be interpreted as the estimated baseline log odds when all independent variables are set to 0, or the reference category in case of categorical variables. The probability when all independent variables are set to 0 is log(intercept)/(1+log(intercept)).
With a standardized continuous variable, the in... | Logistic regression intercept representing baseline probability | The intercept might be interpreted as the estimated baseline log odds when all independent variables are set to 0, or the reference category in case of categorical variables. The probability when all | Logistic regression intercept representing baseline probability
The intercept might be interpreted as the estimated baseline log odds when all independent variables are set to 0, or the reference category in case of categorical variables. The probability when all independent variables are set to 0 is log(intercept)/(1+... | Logistic regression intercept representing baseline probability
The intercept might be interpreted as the estimated baseline log odds when all independent variables are set to 0, or the reference category in case of categorical variables. The probability when all |
47,909 | Logistic regression intercept representing baseline probability | There is no simple interpretation in binary logistic models other than the intercept and slopes satisfy the property that the the average predicted probability equals the observed prevalence of $Y=1$ in the dataset used to fit the model. But I don't find it very useful to think about this in either linear models or lo... | Logistic regression intercept representing baseline probability | There is no simple interpretation in binary logistic models other than the intercept and slopes satisfy the property that the the average predicted probability equals the observed prevalence of $Y=1$ | Logistic regression intercept representing baseline probability
There is no simple interpretation in binary logistic models other than the intercept and slopes satisfy the property that the the average predicted probability equals the observed prevalence of $Y=1$ in the dataset used to fit the model. But I don't find ... | Logistic regression intercept representing baseline probability
There is no simple interpretation in binary logistic models other than the intercept and slopes satisfy the property that the the average predicted probability equals the observed prevalence of $Y=1$ |
47,910 | More efficient way to create Block Toeplitz matrix in R? [closed] | This question might be of interest to statisticians and data scientists because of the multidimensional array handling issues it presents.
Loops aren't the problem. Your goal likely is to perform the operation efficiently, regardless of how it might be done, and perhaps to do it in an obviously parallelizable way.
Exp... | More efficient way to create Block Toeplitz matrix in R? [closed] | This question might be of interest to statisticians and data scientists because of the multidimensional array handling issues it presents.
Loops aren't the problem. Your goal likely is to perform the | More efficient way to create Block Toeplitz matrix in R? [closed]
This question might be of interest to statisticians and data scientists because of the multidimensional array handling issues it presents.
Loops aren't the problem. Your goal likely is to perform the operation efficiently, regardless of how it might be ... | More efficient way to create Block Toeplitz matrix in R? [closed]
This question might be of interest to statisticians and data scientists because of the multidimensional array handling issues it presents.
Loops aren't the problem. Your goal likely is to perform the |
47,911 | More efficient way to create Block Toeplitz matrix in R? [closed] | The following function that takes as argument a list of blocks. However, this may be not the best solution: there still one loop present. And it needs some more work, since it doesn't do the transpose of the blocks lower diagonal (in my case I have symmetric matrices).
toeplitz.block <- function(blocks) {
l <- len... | More efficient way to create Block Toeplitz matrix in R? [closed] | The following function that takes as argument a list of blocks. However, this may be not the best solution: there still one loop present. And it needs some more work, since it doesn't do the transpose | More efficient way to create Block Toeplitz matrix in R? [closed]
The following function that takes as argument a list of blocks. However, this may be not the best solution: there still one loop present. And it needs some more work, since it doesn't do the transpose of the blocks lower diagonal (in my case I have symme... | More efficient way to create Block Toeplitz matrix in R? [closed]
The following function that takes as argument a list of blocks. However, this may be not the best solution: there still one loop present. And it needs some more work, since it doesn't do the transpose |
47,912 | How to compare 2 classifers using confusion matrix? | As @Enrique mentioned, there are many statistics you can calculate from a confusion matrix. To determine which ones are appropriate depends on the specific characteristics of your problem, such as the relative costs associated with true positives and false positives.
Chapter 11 of Applied Predictive Modeling gives a ve... | How to compare 2 classifers using confusion matrix? | As @Enrique mentioned, there are many statistics you can calculate from a confusion matrix. To determine which ones are appropriate depends on the specific characteristics of your problem, such as the | How to compare 2 classifers using confusion matrix?
As @Enrique mentioned, there are many statistics you can calculate from a confusion matrix. To determine which ones are appropriate depends on the specific characteristics of your problem, such as the relative costs associated with true positives and false positives.
... | How to compare 2 classifers using confusion matrix?
As @Enrique mentioned, there are many statistics you can calculate from a confusion matrix. To determine which ones are appropriate depends on the specific characteristics of your problem, such as the |
47,913 | How to compare 2 classifers using confusion matrix? | From the confusion matrices you can compute the sensitivity, specificity, accuracy, precision, among other performance metrics for each of the classifiers. Then you can evaluate them in terms of those metrics. Here you can find the definition of several metrics and how they can be computed. | How to compare 2 classifers using confusion matrix? | From the confusion matrices you can compute the sensitivity, specificity, accuracy, precision, among other performance metrics for each of the classifiers. Then you can evaluate them in terms of those | How to compare 2 classifers using confusion matrix?
From the confusion matrices you can compute the sensitivity, specificity, accuracy, precision, among other performance metrics for each of the classifiers. Then you can evaluate them in terms of those metrics. Here you can find the definition of several metrics and ho... | How to compare 2 classifers using confusion matrix?
From the confusion matrices you can compute the sensitivity, specificity, accuracy, precision, among other performance metrics for each of the classifiers. Then you can evaluate them in terms of those |
47,914 | How to compare 2 classifers using confusion matrix? | Confusion Matrix is a square matrix, which in the ideal case, its main diagonal must be valued and other sides must be none.
Confusion Matrix in binomial class:
------+-------
| TP | FP |
--------------
| FN | TN |
------+-------
Thus, there are two main methods to evaluate a predictor: Using accuracy or F1-scor... | How to compare 2 classifers using confusion matrix? | Confusion Matrix is a square matrix, which in the ideal case, its main diagonal must be valued and other sides must be none.
Confusion Matrix in binomial class:
------+-------
| TP | FP |
--------- | How to compare 2 classifers using confusion matrix?
Confusion Matrix is a square matrix, which in the ideal case, its main diagonal must be valued and other sides must be none.
Confusion Matrix in binomial class:
------+-------
| TP | FP |
--------------
| FN | TN |
------+-------
Thus, there are two main method... | How to compare 2 classifers using confusion matrix?
Confusion Matrix is a square matrix, which in the ideal case, its main diagonal must be valued and other sides must be none.
Confusion Matrix in binomial class:
------+-------
| TP | FP |
--------- |
47,915 | Testing for normality in non-normal distributions with zero skewness and zero excess kurtosis | Let my try to answer my own question:
If $X\sim\mathcal{N}\left(0,\sigma^2\right)$ (zero mean is w.l.o.g.). Define the centered moment $\mu_k=E\left(X-E X\right)^k$ and its empirical counterpart $m_k=\frac{1}{n}\sum_{i=1}^{n}\bigl(X_i-\overline{X}\bigr)^k$.
Analyze the two (method of moments) estimators
$$
b_1=\frac{m... | Testing for normality in non-normal distributions with zero skewness and zero excess kurtosis | Let my try to answer my own question:
If $X\sim\mathcal{N}\left(0,\sigma^2\right)$ (zero mean is w.l.o.g.). Define the centered moment $\mu_k=E\left(X-E X\right)^k$ and its empirical counterpart $m_k= | Testing for normality in non-normal distributions with zero skewness and zero excess kurtosis
Let my try to answer my own question:
If $X\sim\mathcal{N}\left(0,\sigma^2\right)$ (zero mean is w.l.o.g.). Define the centered moment $\mu_k=E\left(X-E X\right)^k$ and its empirical counterpart $m_k=\frac{1}{n}\sum_{i=1}^{n}\... | Testing for normality in non-normal distributions with zero skewness and zero excess kurtosis
Let my try to answer my own question:
If $X\sim\mathcal{N}\left(0,\sigma^2\right)$ (zero mean is w.l.o.g.). Define the centered moment $\mu_k=E\left(X-E X\right)^k$ and its empirical counterpart $m_k= |
47,916 | Testing for normality in non-normal distributions with zero skewness and zero excess kurtosis | It's not so much that the test would be without any power, but you might find it interesting to see what happens to the relative power as $n$ increases when you compare the "Jarque-Bera" with (say) Shapiro-Wilk. Actually, here you go:
The light grey line marks the type I error rate; below that (with a false null) we h... | Testing for normality in non-normal distributions with zero skewness and zero excess kurtosis | It's not so much that the test would be without any power, but you might find it interesting to see what happens to the relative power as $n$ increases when you compare the "Jarque-Bera" with (say) Sh | Testing for normality in non-normal distributions with zero skewness and zero excess kurtosis
It's not so much that the test would be without any power, but you might find it interesting to see what happens to the relative power as $n$ increases when you compare the "Jarque-Bera" with (say) Shapiro-Wilk. Actually, here... | Testing for normality in non-normal distributions with zero skewness and zero excess kurtosis
It's not so much that the test would be without any power, but you might find it interesting to see what happens to the relative power as $n$ increases when you compare the "Jarque-Bera" with (say) Sh |
47,917 | Why can't we use top-down methods in forecasting grouped time series? | The top-down methods implemented in the hts package were designed for hierarchical time series. If you want to define your own top-down method for some non-hierarchical time series, go right ahead. It's not wrong, it just hasn't been implemented in the hts package because there are much better solutions to the problem.... | Why can't we use top-down methods in forecasting grouped time series? | The top-down methods implemented in the hts package were designed for hierarchical time series. If you want to define your own top-down method for some non-hierarchical time series, go right ahead. It | Why can't we use top-down methods in forecasting grouped time series?
The top-down methods implemented in the hts package were designed for hierarchical time series. If you want to define your own top-down method for some non-hierarchical time series, go right ahead. It's not wrong, it just hasn't been implemented in t... | Why can't we use top-down methods in forecasting grouped time series?
The top-down methods implemented in the hts package were designed for hierarchical time series. If you want to define your own top-down method for some non-hierarchical time series, go right ahead. It |
47,918 | Why can't we use top-down methods in forecasting grouped time series? | Further to Rob's answer, let's look at your specific example:
Total
| |
A B
| | | |
AX AY BX BY
Total
| |
X Y
| | | |
AX BX AY BY
You have not one, but two hierarchies. For instance, the top one may group sales at product-location level first by l... | Why can't we use top-down methods in forecasting grouped time series? | Further to Rob's answer, let's look at your specific example:
Total
| |
A B
| | | |
AX AY BX BY
Total
| |
X Y
| | | |
AX BX AY B | Why can't we use top-down methods in forecasting grouped time series?
Further to Rob's answer, let's look at your specific example:
Total
| |
A B
| | | |
AX AY BX BY
Total
| |
X Y
| | | |
AX BX AY BY
You have not one, but two hierarchies. For inst... | Why can't we use top-down methods in forecasting grouped time series?
Further to Rob's answer, let's look at your specific example:
Total
| |
A B
| | | |
AX AY BX BY
Total
| |
X Y
| | | |
AX BX AY B |
47,919 | Bayesian importance sampling as an answer to a "paradox" by Wasserman | There exists a Bayesian approach to the numerical resolution of integrals and differential equations: it is called probabilistic numerics. Introducing a Gaussian process prior on $g$ leads to a posterior probability on $g$ itself and on integrals depending on $g$ once observations $g(x_1),\ldots,g(x_p)$ have been made.... | Bayesian importance sampling as an answer to a "paradox" by Wasserman | There exists a Bayesian approach to the numerical resolution of integrals and differential equations: it is called probabilistic numerics. Introducing a Gaussian process prior on $g$ leads to a poster | Bayesian importance sampling as an answer to a "paradox" by Wasserman
There exists a Bayesian approach to the numerical resolution of integrals and differential equations: it is called probabilistic numerics. Introducing a Gaussian process prior on $g$ leads to a posterior probability on $g$ itself and on integrals dep... | Bayesian importance sampling as an answer to a "paradox" by Wasserman
There exists a Bayesian approach to the numerical resolution of integrals and differential equations: it is called probabilistic numerics. Introducing a Gaussian process prior on $g$ leads to a poster |
47,920 | Linear regression - results depending on the scale of the inputs | Let's compare the two models.
The original one is clearly and well expressed in the question,
$$y = a + b_1x_1 + b_2 x_2 + b_{12}x_1x_2 + \epsilon.$$
Let's write the second model as
$$y = a^\prime + b^\prime_1 z_1 + b^\prime_2 z_2 + b^\prime_{12}z_1z_2 + \delta.$$
Because the values of the numbers 7.5, 47.5, etc. are o... | Linear regression - results depending on the scale of the inputs | Let's compare the two models.
The original one is clearly and well expressed in the question,
$$y = a + b_1x_1 + b_2 x_2 + b_{12}x_1x_2 + \epsilon.$$
Let's write the second model as
$$y = a^\prime + b | Linear regression - results depending on the scale of the inputs
Let's compare the two models.
The original one is clearly and well expressed in the question,
$$y = a + b_1x_1 + b_2 x_2 + b_{12}x_1x_2 + \epsilon.$$
Let's write the second model as
$$y = a^\prime + b^\prime_1 z_1 + b^\prime_2 z_2 + b^\prime_{12}z_1z_2 + ... | Linear regression - results depending on the scale of the inputs
Let's compare the two models.
The original one is clearly and well expressed in the question,
$$y = a + b_1x_1 + b_2 x_2 + b_{12}x_1x_2 + \epsilon.$$
Let's write the second model as
$$y = a^\prime + b |
47,921 | Linear regression - results depending on the scale of the inputs | The $p$-values did not change because of the rescaling, but because you added constants to the variable. The intercept is the expected value of y when all explanatory variables are 0. The $p$-value is the test of the hypothesis that that conditional mean is equal to 0. So if you add a constant to one or more of the exp... | Linear regression - results depending on the scale of the inputs | The $p$-values did not change because of the rescaling, but because you added constants to the variable. The intercept is the expected value of y when all explanatory variables are 0. The $p$-value is | Linear regression - results depending on the scale of the inputs
The $p$-values did not change because of the rescaling, but because you added constants to the variable. The intercept is the expected value of y when all explanatory variables are 0. The $p$-value is the test of the hypothesis that that conditional mean ... | Linear regression - results depending on the scale of the inputs
The $p$-values did not change because of the rescaling, but because you added constants to the variable. The intercept is the expected value of y when all explanatory variables are 0. The $p$-value is |
47,922 | Linear regression - results depending on the scale of the inputs | Just to show @Zhanxiong last comment:
# response variable
y <- rnorm(8, mean=17, sd=1.2)
# model 1
x1 <- c(1, -1, -1, -1, 1, 1, -1, 1)
x2 <- c(-1, 1, 1, -1, -1, 1, -1, 1)
fit1 <- lm(y ~ x1 + x2 + x1*x2)
# model 2 (factors transformation - no addtion)
z1 <- x1 * 7.5
z2 <- x2 * 11.5
fit2 <- lm(y ~ z1 + z2 + z1*z2)
... | Linear regression - results depending on the scale of the inputs | Just to show @Zhanxiong last comment:
# response variable
y <- rnorm(8, mean=17, sd=1.2)
# model 1
x1 <- c(1, -1, -1, -1, 1, 1, -1, 1)
x2 <- c(-1, 1, 1, -1, -1, 1, -1, 1)
fit1 <- lm(y ~ x1 + x2 + x1* | Linear regression - results depending on the scale of the inputs
Just to show @Zhanxiong last comment:
# response variable
y <- rnorm(8, mean=17, sd=1.2)
# model 1
x1 <- c(1, -1, -1, -1, 1, 1, -1, 1)
x2 <- c(-1, 1, 1, -1, -1, 1, -1, 1)
fit1 <- lm(y ~ x1 + x2 + x1*x2)
# model 2 (factors transformation - no addtion)
z1... | Linear regression - results depending on the scale of the inputs
Just to show @Zhanxiong last comment:
# response variable
y <- rnorm(8, mean=17, sd=1.2)
# model 1
x1 <- c(1, -1, -1, -1, 1, 1, -1, 1)
x2 <- c(-1, 1, 1, -1, -1, 1, -1, 1)
fit1 <- lm(y ~ x1 + x2 + x1* |
47,923 | How to specify logistic regression as transformed linear regression? | The short answer is "you don't". They don't correspond.
Logistic regression is not a transformed linear regression.
Even though $E(Y)$ ($=P(Y=1)$) may be written as $\text{logit}(X\beta)$, and so seemingly linearized, you can't transform the $y$ values to make a linear regression, nor can you fit a nonlinear least squa... | How to specify logistic regression as transformed linear regression? | The short answer is "you don't". They don't correspond.
Logistic regression is not a transformed linear regression.
Even though $E(Y)$ ($=P(Y=1)$) may be written as $\text{logit}(X\beta)$, and so seem | How to specify logistic regression as transformed linear regression?
The short answer is "you don't". They don't correspond.
Logistic regression is not a transformed linear regression.
Even though $E(Y)$ ($=P(Y=1)$) may be written as $\text{logit}(X\beta)$, and so seemingly linearized, you can't transform the $y$ value... | How to specify logistic regression as transformed linear regression?
The short answer is "you don't". They don't correspond.
Logistic regression is not a transformed linear regression.
Even though $E(Y)$ ($=P(Y=1)$) may be written as $\text{logit}(X\beta)$, and so seem |
47,924 | What is the purpose of the 'convenience' 1/2 fraction on the sum of squared errors? | As long as $E(w) \geq 0$ (which is true for this sum of squares), minimizing $(1/2)E(w)$ is equivalent to minimizing $E(w)$. As has been pointed out in the comments, the factor of $1/2$ disappears when you take the derivative of $E(w)$. | What is the purpose of the 'convenience' 1/2 fraction on the sum of squared errors? | As long as $E(w) \geq 0$ (which is true for this sum of squares), minimizing $(1/2)E(w)$ is equivalent to minimizing $E(w)$. As has been pointed out in the comments, the factor of $1/2$ disappears wh | What is the purpose of the 'convenience' 1/2 fraction on the sum of squared errors?
As long as $E(w) \geq 0$ (which is true for this sum of squares), minimizing $(1/2)E(w)$ is equivalent to minimizing $E(w)$. As has been pointed out in the comments, the factor of $1/2$ disappears when you take the derivative of $E(w)$... | What is the purpose of the 'convenience' 1/2 fraction on the sum of squared errors?
As long as $E(w) \geq 0$ (which is true for this sum of squares), minimizing $(1/2)E(w)$ is equivalent to minimizing $E(w)$. As has been pointed out in the comments, the factor of $1/2$ disappears wh |
47,925 | What is the purpose of the 'convenience' 1/2 fraction on the sum of squared errors? | It probably doesn't matter whether you use $\frac{1}{2}$ or $\frac{1}{n}$ for MSE because the denominator value of 2 and $n$ will never change for the dataset being evaluated. The scale of both methods will differ due to the magnitude of what's calculated, but nevertheless, you'll be dividing by a constant that never ... | What is the purpose of the 'convenience' 1/2 fraction on the sum of squared errors? | It probably doesn't matter whether you use $\frac{1}{2}$ or $\frac{1}{n}$ for MSE because the denominator value of 2 and $n$ will never change for the dataset being evaluated. The scale of both metho | What is the purpose of the 'convenience' 1/2 fraction on the sum of squared errors?
It probably doesn't matter whether you use $\frac{1}{2}$ or $\frac{1}{n}$ for MSE because the denominator value of 2 and $n$ will never change for the dataset being evaluated. The scale of both methods will differ due to the magnitude ... | What is the purpose of the 'convenience' 1/2 fraction on the sum of squared errors?
It probably doesn't matter whether you use $\frac{1}{2}$ or $\frac{1}{n}$ for MSE because the denominator value of 2 and $n$ will never change for the dataset being evaluated. The scale of both metho |
47,926 | How to compare performance of regression and classification? | Imagine this: You are trying to predict age of a population using some
features. It does not work that well. Then, you are reducing the
complexity of the problem. You only try to predict whether age is
above or below 20. This works well, using the same features. I simply
want to quantify the improvement gained ... | How to compare performance of regression and classification? | Imagine this: You are trying to predict age of a population using some
features. It does not work that well. Then, you are reducing the
complexity of the problem. You only try to predict whether a | How to compare performance of regression and classification?
Imagine this: You are trying to predict age of a population using some
features. It does not work that well. Then, you are reducing the
complexity of the problem. You only try to predict whether age is
above or below 20. This works well, using the same ... | How to compare performance of regression and classification?
Imagine this: You are trying to predict age of a population using some
features. It does not work that well. Then, you are reducing the
complexity of the problem. You only try to predict whether a |
47,927 | How to compare performance of regression and classification? | I would choose the metric depending on my problem. Ie, if my problem is to predict the age of a person, I'd choose RMSE, if my problem is to predict young or old, I'd choose accuracy.
After choosing the metric, you need to be able to use that metric with the models. Ie, if your problem is to predict young or old, then ... | How to compare performance of regression and classification? | I would choose the metric depending on my problem. Ie, if my problem is to predict the age of a person, I'd choose RMSE, if my problem is to predict young or old, I'd choose accuracy.
After choosing t | How to compare performance of regression and classification?
I would choose the metric depending on my problem. Ie, if my problem is to predict the age of a person, I'd choose RMSE, if my problem is to predict young or old, I'd choose accuracy.
After choosing the metric, you need to be able to use that metric with the ... | How to compare performance of regression and classification?
I would choose the metric depending on my problem. Ie, if my problem is to predict the age of a person, I'd choose RMSE, if my problem is to predict young or old, I'd choose accuracy.
After choosing t |
47,928 | Conditional mass function of minimum of two discrete uniform random variables given the maximum | Draw a $6$" $\times$ $6$" square and divide it into a $6\times 6$
array of $36$ one-inch
squares. Label the rows and columns with numbers $1$-$6$ and in each square, write down the values of $(X,Y)$, $U$, $V$, $S$ and $T$ in
each square . Then,
count! | Conditional mass function of minimum of two discrete uniform random variables given the maximum | Draw a $6$" $\times$ $6$" square and divide it into a $6\times 6$
array of $36$ one-inch
squares. Label the rows and columns with numbers $1$-$6$ and in each square, write down the values of $(X,Y)$, | Conditional mass function of minimum of two discrete uniform random variables given the maximum
Draw a $6$" $\times$ $6$" square and divide it into a $6\times 6$
array of $36$ one-inch
squares. Label the rows and columns with numbers $1$-$6$ and in each square, write down the values of $(X,Y)$, $U$, $V$, $S$ and $T$ in... | Conditional mass function of minimum of two discrete uniform random variables given the maximum
Draw a $6$" $\times$ $6$" square and divide it into a $6\times 6$
array of $36$ one-inch
squares. Label the rows and columns with numbers $1$-$6$ and in each square, write down the values of $(X,Y)$, |
47,929 | How to find factor that is making matrix singular | You can use an eigen-decomposition to find linear combinations of your columns that vanish, then remove enough columns participating in these linear combinations.
Here's a matrix with a vanishing column linear combination:
> M <- matrix(c(0, 0, 0, 1, 0, 1, 0, 1, 1, 1, -1, 0), nrow=4, byrow=TRUE)
> M
[,1] [,2] [,3... | How to find factor that is making matrix singular | You can use an eigen-decomposition to find linear combinations of your columns that vanish, then remove enough columns participating in these linear combinations.
Here's a matrix with a vanishing colu | How to find factor that is making matrix singular
You can use an eigen-decomposition to find linear combinations of your columns that vanish, then remove enough columns participating in these linear combinations.
Here's a matrix with a vanishing column linear combination:
> M <- matrix(c(0, 0, 0, 1, 0, 1, 0, 1, 1, 1, -... | How to find factor that is making matrix singular
You can use an eigen-decomposition to find linear combinations of your columns that vanish, then remove enough columns participating in these linear combinations.
Here's a matrix with a vanishing colu |
47,930 | Bonferroni Correction for Post Hoc Analysis in ANOVA & Regression | Preface: There are many different was to adjust for multiple comparisons. Olive Dunn proposed the Bonferroni adjustment in 1961, and the multiple comparisons literature (see, for example, Shaffer, 1995) has grown to a variety of family-wise error rate adjustment methods (of which Bonferroni is the simplest), and the mo... | Bonferroni Correction for Post Hoc Analysis in ANOVA & Regression | Preface: There are many different was to adjust for multiple comparisons. Olive Dunn proposed the Bonferroni adjustment in 1961, and the multiple comparisons literature (see, for example, Shaffer, 199 | Bonferroni Correction for Post Hoc Analysis in ANOVA & Regression
Preface: There are many different was to adjust for multiple comparisons. Olive Dunn proposed the Bonferroni adjustment in 1961, and the multiple comparisons literature (see, for example, Shaffer, 1995) has grown to a variety of family-wise error rate ad... | Bonferroni Correction for Post Hoc Analysis in ANOVA & Regression
Preface: There are many different was to adjust for multiple comparisons. Olive Dunn proposed the Bonferroni adjustment in 1961, and the multiple comparisons literature (see, for example, Shaffer, 199 |
47,931 | Simple way to cluster histograms | Use hierarchical clustering or DBSCAN.
They have one huge benefit over k-means: they work with arbitrary distance measures, and with histograms you might want to use like, for example, Jensen-Shannon divergence, etc. that are designed to capture the similarity of distributions. | Simple way to cluster histograms | Use hierarchical clustering or DBSCAN.
They have one huge benefit over k-means: they work with arbitrary distance measures, and with histograms you might want to use like, for example, Jensen-Shannon | Simple way to cluster histograms
Use hierarchical clustering or DBSCAN.
They have one huge benefit over k-means: they work with arbitrary distance measures, and with histograms you might want to use like, for example, Jensen-Shannon divergence, etc. that are designed to capture the similarity of distributions. | Simple way to cluster histograms
Use hierarchical clustering or DBSCAN.
They have one huge benefit over k-means: they work with arbitrary distance measures, and with histograms you might want to use like, for example, Jensen-Shannon |
47,932 | Simple way to cluster histograms | K-means could do this. K-means is an unsupervised clustering algorithm. Rewrite each histogram as a vector and use Euclidean distance.
This post goes into the assumptions of K-means: How to understand the drawbacks of K-means You might want to check these.
You have to determine the number of clusters yourself by estima... | Simple way to cluster histograms | K-means could do this. K-means is an unsupervised clustering algorithm. Rewrite each histogram as a vector and use Euclidean distance.
This post goes into the assumptions of K-means: How to understand | Simple way to cluster histograms
K-means could do this. K-means is an unsupervised clustering algorithm. Rewrite each histogram as a vector and use Euclidean distance.
This post goes into the assumptions of K-means: How to understand the drawbacks of K-means You might want to check these.
You have to determine the numb... | Simple way to cluster histograms
K-means could do this. K-means is an unsupervised clustering algorithm. Rewrite each histogram as a vector and use Euclidean distance.
This post goes into the assumptions of K-means: How to understand |
47,933 | Censored Binomial model - log likelihood | Let the constant probability of contamination be $p$, which is to be estimated, and let $X$ be a random variable equal to $1$ when contamination is observed, $0$ otherwise. Writing $q=1-p$ for the chance of no contamination in any individual unit, the probability of observing no contamination (that is, $X=0$) in a bat... | Censored Binomial model - log likelihood | Let the constant probability of contamination be $p$, which is to be estimated, and let $X$ be a random variable equal to $1$ when contamination is observed, $0$ otherwise. Writing $q=1-p$ for the ch | Censored Binomial model - log likelihood
Let the constant probability of contamination be $p$, which is to be estimated, and let $X$ be a random variable equal to $1$ when contamination is observed, $0$ otherwise. Writing $q=1-p$ for the chance of no contamination in any individual unit, the probability of observing n... | Censored Binomial model - log likelihood
Let the constant probability of contamination be $p$, which is to be estimated, and let $X$ be a random variable equal to $1$ when contamination is observed, $0$ otherwise. Writing $q=1-p$ for the ch |
47,934 | Censored Binomial model - log likelihood | Note that you can also pass the negative log likelihood to function mle2 in package bbmle, which has the advantage that you then also get 95% profile confidence intervals on your estimated $p$. For example :
n = c(7, 7, 7, 7, 7, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 13, 13, 13, 13, 15, 15,... | Censored Binomial model - log likelihood | Note that you can also pass the negative log likelihood to function mle2 in package bbmle, which has the advantage that you then also get 95% profile confidence intervals on your estimated $p$. For ex | Censored Binomial model - log likelihood
Note that you can also pass the negative log likelihood to function mle2 in package bbmle, which has the advantage that you then also get 95% profile confidence intervals on your estimated $p$. For example :
n = c(7, 7, 7, 7, 7, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12... | Censored Binomial model - log likelihood
Note that you can also pass the negative log likelihood to function mle2 in package bbmle, which has the advantage that you then also get 95% profile confidence intervals on your estimated $p$. For ex |
47,935 | Is it possible to calculate Q1, Median, Q3, StDev from already aggregated data? | You have mean, counts and StDev of the observations, so aggregated StDev is a matter of algebra. I'm sure you can figure it out easily.
The quantiles are trickier. Consider, Q1 of two samples. They form the bounds of the Q1 of the combined sample. If $Q1_1>Q1_2$, then it's easy to see that aggregated $Q1_2<Q1$ and $Q1<... | Is it possible to calculate Q1, Median, Q3, StDev from already aggregated data? | You have mean, counts and StDev of the observations, so aggregated StDev is a matter of algebra. I'm sure you can figure it out easily.
The quantiles are trickier. Consider, Q1 of two samples. They fo | Is it possible to calculate Q1, Median, Q3, StDev from already aggregated data?
You have mean, counts and StDev of the observations, so aggregated StDev is a matter of algebra. I'm sure you can figure it out easily.
The quantiles are trickier. Consider, Q1 of two samples. They form the bounds of the Q1 of the combined ... | Is it possible to calculate Q1, Median, Q3, StDev from already aggregated data?
You have mean, counts and StDev of the observations, so aggregated StDev is a matter of algebra. I'm sure you can figure it out easily.
The quantiles are trickier. Consider, Q1 of two samples. They fo |
47,936 | How can the "anti-correlation" between these two curves be shown? | One curve almost looks like the derivative of the other and sometimes such pairs of curves are plotted against each other with curved connections. For instance, for plotting velocity versus acceleration to see cycles better. Here is red versus blue for your toy data:
Arrows and annotations are sometimes added. I don't... | How can the "anti-correlation" between these two curves be shown? | One curve almost looks like the derivative of the other and sometimes such pairs of curves are plotted against each other with curved connections. For instance, for plotting velocity versus accelerati | How can the "anti-correlation" between these two curves be shown?
One curve almost looks like the derivative of the other and sometimes such pairs of curves are plotted against each other with curved connections. For instance, for plotting velocity versus acceleration to see cycles better. Here is red versus blue for y... | How can the "anti-correlation" between these two curves be shown?
One curve almost looks like the derivative of the other and sometimes such pairs of curves are plotted against each other with curved connections. For instance, for plotting velocity versus accelerati |
47,937 | How can the "anti-correlation" between these two curves be shown? | The plot of one measure against the other (@xan's answer) is a good idea, except that I don't think it makes sense to join the points in this way. It only makes sense if the order of observations is really important. My understanding is that the fact that they're anti-correlated doesn't have anything to do with their o... | How can the "anti-correlation" between these two curves be shown? | The plot of one measure against the other (@xan's answer) is a good idea, except that I don't think it makes sense to join the points in this way. It only makes sense if the order of observations is r | How can the "anti-correlation" between these two curves be shown?
The plot of one measure against the other (@xan's answer) is a good idea, except that I don't think it makes sense to join the points in this way. It only makes sense if the order of observations is really important. My understanding is that the fact tha... | How can the "anti-correlation" between these two curves be shown?
The plot of one measure against the other (@xan's answer) is a good idea, except that I don't think it makes sense to join the points in this way. It only makes sense if the order of observations is r |
47,938 | Difference between a random slope/intercept model and an ANCOVA with an interaction? | In this case, I think you want to use the model with the interaction. A random slopes/intercepts model would suggest that you have slopes/intercepts that can vary with regard to some random factor. For example, imagine that you have participants completing a task in which they make several responses to images under con... | Difference between a random slope/intercept model and an ANCOVA with an interaction? | In this case, I think you want to use the model with the interaction. A random slopes/intercepts model would suggest that you have slopes/intercepts that can vary with regard to some random factor. Fo | Difference between a random slope/intercept model and an ANCOVA with an interaction?
In this case, I think you want to use the model with the interaction. A random slopes/intercepts model would suggest that you have slopes/intercepts that can vary with regard to some random factor. For example, imagine that you have pa... | Difference between a random slope/intercept model and an ANCOVA with an interaction?
In this case, I think you want to use the model with the interaction. A random slopes/intercepts model would suggest that you have slopes/intercepts that can vary with regard to some random factor. Fo |
47,939 | Interaction between time-variant and time-invariant variable in FE model | Your intuition is fine. When you take the partial derivative with respect to $x_{1,it}$, then you get exactly what you were looking for.
$$\frac{\partial y_{it}}{\partial x_{1,it}} = \beta_1 + \eta z_i $$
This is particularly convenient if $z_i$ is a dummy variable. Wooldridge (2010) "Econometric Analysis of Cross Sect... | Interaction between time-variant and time-invariant variable in FE model | Your intuition is fine. When you take the partial derivative with respect to $x_{1,it}$, then you get exactly what you were looking for.
$$\frac{\partial y_{it}}{\partial x_{1,it}} = \beta_1 + \eta z_ | Interaction between time-variant and time-invariant variable in FE model
Your intuition is fine. When you take the partial derivative with respect to $x_{1,it}$, then you get exactly what you were looking for.
$$\frac{\partial y_{it}}{\partial x_{1,it}} = \beta_1 + \eta z_i $$
This is particularly convenient if $z_i$ i... | Interaction between time-variant and time-invariant variable in FE model
Your intuition is fine. When you take the partial derivative with respect to $x_{1,it}$, then you get exactly what you were looking for.
$$\frac{\partial y_{it}}{\partial x_{1,it}} = \beta_1 + \eta z_ |
47,940 | Classifer for unbalanced dataset? | Most classifiers in sklearn support unbalanced datasets, through the sample_weight parameter in the clf.fit methods. If you need to fit unbalanced data with a classifier that does not support this option, you can use sampling with replacement to enlarge the smaller class to match the larger one.
Here is an adapted ver... | Classifer for unbalanced dataset? | Most classifiers in sklearn support unbalanced datasets, through the sample_weight parameter in the clf.fit methods. If you need to fit unbalanced data with a classifier that does not support this op | Classifer for unbalanced dataset?
Most classifiers in sklearn support unbalanced datasets, through the sample_weight parameter in the clf.fit methods. If you need to fit unbalanced data with a classifier that does not support this option, you can use sampling with replacement to enlarge the smaller class to match the ... | Classifer for unbalanced dataset?
Most classifiers in sklearn support unbalanced datasets, through the sample_weight parameter in the clf.fit methods. If you need to fit unbalanced data with a classifier that does not support this op |
47,941 | Classifer for unbalanced dataset? | Linear SVM can handle unbalanced data sets just fine by using class-weights on the misclassification penalty. This functionality is available in any decent SVM implementation.
The objective function for class-weighted SVM is as follows:
$$\min_{\xi,\mathbf{w}} \frac{1}{2}\|\mathbf{w}\|^2 + C_{\mathcal{P}}\sum_{i\in\mat... | Classifer for unbalanced dataset? | Linear SVM can handle unbalanced data sets just fine by using class-weights on the misclassification penalty. This functionality is available in any decent SVM implementation.
The objective function f | Classifer for unbalanced dataset?
Linear SVM can handle unbalanced data sets just fine by using class-weights on the misclassification penalty. This functionality is available in any decent SVM implementation.
The objective function for class-weighted SVM is as follows:
$$\min_{\xi,\mathbf{w}} \frac{1}{2}\|\mathbf{w}\|... | Classifer for unbalanced dataset?
Linear SVM can handle unbalanced data sets just fine by using class-weights on the misclassification penalty. This functionality is available in any decent SVM implementation.
The objective function f |
47,942 | Matrix multiplication to find correlation matrix | I think the author is simply handy-wavy there. I believe it is assumed that the columns of $A$ have norm 1 and mean 0.
$A^TA$ is a Gram matrix. Given you are using random variables to construct $A$, $A^TA$ is approximately proportional to the covariance matrix (scale by $n$ the number of variables). The correlation mat... | Matrix multiplication to find correlation matrix | I think the author is simply handy-wavy there. I believe it is assumed that the columns of $A$ have norm 1 and mean 0.
$A^TA$ is a Gram matrix. Given you are using random variables to construct $A$, $ | Matrix multiplication to find correlation matrix
I think the author is simply handy-wavy there. I believe it is assumed that the columns of $A$ have norm 1 and mean 0.
$A^TA$ is a Gram matrix. Given you are using random variables to construct $A$, $A^TA$ is approximately proportional to the covariance matrix (scale by ... | Matrix multiplication to find correlation matrix
I think the author is simply handy-wavy there. I believe it is assumed that the columns of $A$ have norm 1 and mean 0.
$A^TA$ is a Gram matrix. Given you are using random variables to construct $A$, $ |
47,943 | Distances in PCA space [closed] | Bit late, but here we go:
The transformation spectra -> PC scores is typically set up to be a pure rotation. Thus Euclidean distance in PC score space equals Euclidean distance in original space as long as no PCs are discarded. Thus, neighbours stay neighbours.
For models that keep only some of the PCs, you can maybe ... | Distances in PCA space [closed] | Bit late, but here we go:
The transformation spectra -> PC scores is typically set up to be a pure rotation. Thus Euclidean distance in PC score space equals Euclidean distance in original space as lo | Distances in PCA space [closed]
Bit late, but here we go:
The transformation spectra -> PC scores is typically set up to be a pure rotation. Thus Euclidean distance in PC score space equals Euclidean distance in original space as long as no PCs are discarded. Thus, neighbours stay neighbours.
For models that keep only... | Distances in PCA space [closed]
Bit late, but here we go:
The transformation spectra -> PC scores is typically set up to be a pure rotation. Thus Euclidean distance in PC score space equals Euclidean distance in original space as lo |
47,944 | Distances in PCA space [closed] | Sounds like you want to know how to get from the PCA projection back to the original data for 1), and 2) what to do with nearest neighbors? Look at the PCA score coefficient matrix, which allows a back-projection. Regarding nearest neighbors after PCA, the focus commonly involves use of a doubly-centered Gram matrix ... | Distances in PCA space [closed] | Sounds like you want to know how to get from the PCA projection back to the original data for 1), and 2) what to do with nearest neighbors? Look at the PCA score coefficient matrix, which allows a ba | Distances in PCA space [closed]
Sounds like you want to know how to get from the PCA projection back to the original data for 1), and 2) what to do with nearest neighbors? Look at the PCA score coefficient matrix, which allows a back-projection. Regarding nearest neighbors after PCA, the focus commonly involves use o... | Distances in PCA space [closed]
Sounds like you want to know how to get from the PCA projection back to the original data for 1), and 2) what to do with nearest neighbors? Look at the PCA score coefficient matrix, which allows a ba |
47,945 | Distances in PCA space [closed] | I'm working with PCA coefficients now and probably you're done with your project by now but I think this might be helpful to others.
In PCA the higher dimensions present less deviation from the axis, so discarding them does not loos as much information. Nonetheless the distance of the points are not maintained but the ... | Distances in PCA space [closed] | I'm working with PCA coefficients now and probably you're done with your project by now but I think this might be helpful to others.
In PCA the higher dimensions present less deviation from the axis, | Distances in PCA space [closed]
I'm working with PCA coefficients now and probably you're done with your project by now but I think this might be helpful to others.
In PCA the higher dimensions present less deviation from the axis, so discarding them does not loos as much information. Nonetheless the distance of the po... | Distances in PCA space [closed]
I'm working with PCA coefficients now and probably you're done with your project by now but I think this might be helpful to others.
In PCA the higher dimensions present less deviation from the axis, |
47,946 | ARIMA vs. Random Forest | It has been awhile, but this is still getting upvotes. So ...
In the event, I did use a Random Forest, and it worked really well. A boosted tree did a bit better that that even. Also since then I know of several other similar projects that took the same route with success and very little fuss. While ARIMA is surely a p... | ARIMA vs. Random Forest | It has been awhile, but this is still getting upvotes. So ...
In the event, I did use a Random Forest, and it worked really well. A boosted tree did a bit better that that even. Also since then I know | ARIMA vs. Random Forest
It has been awhile, but this is still getting upvotes. So ...
In the event, I did use a Random Forest, and it worked really well. A boosted tree did a bit better that that even. Also since then I know of several other similar projects that took the same route with success and very little fuss. W... | ARIMA vs. Random Forest
It has been awhile, but this is still getting upvotes. So ...
In the event, I did use a Random Forest, and it worked really well. A boosted tree did a bit better that that even. Also since then I know |
47,947 | ARIMA vs. Random Forest | I don't see the argument that Random Forest is less work than ARIMA. I would present RF is more work.
Here is the process for ARIMA:
1. Detrend / transform your data
2. Test for stationarity
3. Run autocorrelation plots
4. Set your parameters for your model
5. Run a grid search if you want
6. Look at RMSE
Here is the p... | ARIMA vs. Random Forest | I don't see the argument that Random Forest is less work than ARIMA. I would present RF is more work.
Here is the process for ARIMA:
1. Detrend / transform your data
2. Test for stationarity
3. Run au | ARIMA vs. Random Forest
I don't see the argument that Random Forest is less work than ARIMA. I would present RF is more work.
Here is the process for ARIMA:
1. Detrend / transform your data
2. Test for stationarity
3. Run autocorrelation plots
4. Set your parameters for your model
5. Run a grid search if you want
6. Lo... | ARIMA vs. Random Forest
I don't see the argument that Random Forest is less work than ARIMA. I would present RF is more work.
Here is the process for ARIMA:
1. Detrend / transform your data
2. Test for stationarity
3. Run au |
47,948 | What does id (cluster) mean in gee? | I never used this kind of models but a quick Google search reveals that
Generalized Estimating Equations (GEE) (Liang and Zeger 1986) are a
general method for analyzing data collected in clusters where 1)
observations within a cluster may be correlated, 2) observations in
separate clusters are independent, 3) a ... | What does id (cluster) mean in gee? | I never used this kind of models but a quick Google search reveals that
Generalized Estimating Equations (GEE) (Liang and Zeger 1986) are a
general method for analyzing data collected in clusters w | What does id (cluster) mean in gee?
I never used this kind of models but a quick Google search reveals that
Generalized Estimating Equations (GEE) (Liang and Zeger 1986) are a
general method for analyzing data collected in clusters where 1)
observations within a cluster may be correlated, 2) observations in
sepa... | What does id (cluster) mean in gee?
I never used this kind of models but a quick Google search reveals that
Generalized Estimating Equations (GEE) (Liang and Zeger 1986) are a
general method for analyzing data collected in clusters w |
47,949 | Is there an estimator for the symmetry of a bimodal distribution? | By definition, a symmetric random variable $X$ is one for which there is a constant $\mu$ for which $X-\mu$ and $\mu-X$ are identically distributed. In terms of the distribution function $F$ this is equivalent to
$$\eqalign{
F(\mu+x) &= \Pr(X \le \mu+x) = \Pr(X-\mu \le x) \\
&= \Pr(\mu-X \le x) \\
&= \Pr(X-\mu \ge -... | Is there an estimator for the symmetry of a bimodal distribution? | By definition, a symmetric random variable $X$ is one for which there is a constant $\mu$ for which $X-\mu$ and $\mu-X$ are identically distributed. In terms of the distribution function $F$ this is | Is there an estimator for the symmetry of a bimodal distribution?
By definition, a symmetric random variable $X$ is one for which there is a constant $\mu$ for which $X-\mu$ and $\mu-X$ are identically distributed. In terms of the distribution function $F$ this is equivalent to
$$\eqalign{
F(\mu+x) &= \Pr(X \le \mu+x)... | Is there an estimator for the symmetry of a bimodal distribution?
By definition, a symmetric random variable $X$ is one for which there is a constant $\mu$ for which $X-\mu$ and $\mu-X$ are identically distributed. In terms of the distribution function $F$ this is |
47,950 | Is there an estimator for the symmetry of a bimodal distribution? | While I don't think that there is a single measure of symmetry for a bimodal distribution in general, for a special case of mixture of two normal distributions, perhaps, it is possible to use one or several bi-modality measures and statistical tests. Some mixture modeling software (especially, some R packages) might ha... | Is there an estimator for the symmetry of a bimodal distribution? | While I don't think that there is a single measure of symmetry for a bimodal distribution in general, for a special case of mixture of two normal distributions, perhaps, it is possible to use one or s | Is there an estimator for the symmetry of a bimodal distribution?
While I don't think that there is a single measure of symmetry for a bimodal distribution in general, for a special case of mixture of two normal distributions, perhaps, it is possible to use one or several bi-modality measures and statistical tests. Som... | Is there an estimator for the symmetry of a bimodal distribution?
While I don't think that there is a single measure of symmetry for a bimodal distribution in general, for a special case of mixture of two normal distributions, perhaps, it is possible to use one or s |
47,951 | Unequal sample size one way ANOVA [duplicate] | You have a total N = 148, distributed into 4 groups. If you had 37 in each group instead, you would have greater statistical power. Otherwise, a one-way ANOVA is just as valid here as anywhere else (given that the normal assumptions are met). (To understand this better, it may help to read my answer here: How should... | Unequal sample size one way ANOVA [duplicate] | You have a total N = 148, distributed into 4 groups. If you had 37 in each group instead, you would have greater statistical power. Otherwise, a one-way ANOVA is just as valid here as anywhere else | Unequal sample size one way ANOVA [duplicate]
You have a total N = 148, distributed into 4 groups. If you had 37 in each group instead, you would have greater statistical power. Otherwise, a one-way ANOVA is just as valid here as anywhere else (given that the normal assumptions are met). (To understand this better, ... | Unequal sample size one way ANOVA [duplicate]
You have a total N = 148, distributed into 4 groups. If you had 37 in each group instead, you would have greater statistical power. Otherwise, a one-way ANOVA is just as valid here as anywhere else |
47,952 | Degeneracy paradox | You would normally bet on the mode of the outcome distribution, not on the expected value. The mode corresponding to 98 flips is 0, so you would bet on 0.
The mode corresponding to a very large number $N$ of flips will be approximately $N \cdot 0.01$ (rounding will play a very small role for very large $N$), so you wo... | Degeneracy paradox | You would normally bet on the mode of the outcome distribution, not on the expected value. The mode corresponding to 98 flips is 0, so you would bet on 0.
The mode corresponding to a very large numbe | Degeneracy paradox
You would normally bet on the mode of the outcome distribution, not on the expected value. The mode corresponding to 98 flips is 0, so you would bet on 0.
The mode corresponding to a very large number $N$ of flips will be approximately $N \cdot 0.01$ (rounding will play a very small role for very la... | Degeneracy paradox
You would normally bet on the mode of the outcome distribution, not on the expected value. The mode corresponding to 98 flips is 0, so you would bet on 0.
The mode corresponding to a very large numbe |
47,953 | Density estimation and histograms | As @tristan comments, $m$ is a counter integer, while $n$ is the total number of data points in the sample, and $h$ is the histogram bin width. The formula is correct.
It may be easier to understand if you consider the case where you have $M$ bins and the same number of data points $\frac{n}{M}$ in each bin. Then your ... | Density estimation and histograms | As @tristan comments, $m$ is a counter integer, while $n$ is the total number of data points in the sample, and $h$ is the histogram bin width. The formula is correct.
It may be easier to understand i | Density estimation and histograms
As @tristan comments, $m$ is a counter integer, while $n$ is the total number of data points in the sample, and $h$ is the histogram bin width. The formula is correct.
It may be easier to understand if you consider the case where you have $M$ bins and the same number of data points $\f... | Density estimation and histograms
As @tristan comments, $m$ is a counter integer, while $n$ is the total number of data points in the sample, and $h$ is the histogram bin width. The formula is correct.
It may be easier to understand i |
47,954 | LSTM forgetting dependencies | In a typical LSTM block forget gate computes its value based on its inputs (from layer below) and weights associated with each input. Weights are trained, usually by gradient descent using gradients that are computed by backpropagation.
You can think of forget gate as simple logistic regression classifier that is trai... | LSTM forgetting dependencies | In a typical LSTM block forget gate computes its value based on its inputs (from layer below) and weights associated with each input. Weights are trained, usually by gradient descent using gradients t | LSTM forgetting dependencies
In a typical LSTM block forget gate computes its value based on its inputs (from layer below) and weights associated with each input. Weights are trained, usually by gradient descent using gradients that are computed by backpropagation.
You can think of forget gate as simple logistic regre... | LSTM forgetting dependencies
In a typical LSTM block forget gate computes its value based on its inputs (from layer below) and weights associated with each input. Weights are trained, usually by gradient descent using gradients t |
47,955 | LSTM forgetting dependencies | The forget gate simply applies to the stored cell value. If forget is closer to 1, he takes the stored value into account, otherwise, stored value will be ignored, he just gotta find a value for the forget value in which the error is minimized. | LSTM forgetting dependencies | The forget gate simply applies to the stored cell value. If forget is closer to 1, he takes the stored value into account, otherwise, stored value will be ignored, he just gotta find a value for the f | LSTM forgetting dependencies
The forget gate simply applies to the stored cell value. If forget is closer to 1, he takes the stored value into account, otherwise, stored value will be ignored, he just gotta find a value for the forget value in which the error is minimized. | LSTM forgetting dependencies
The forget gate simply applies to the stored cell value. If forget is closer to 1, he takes the stored value into account, otherwise, stored value will be ignored, he just gotta find a value for the f |
47,956 | What are the best empirical studies comparing causal inference with experimental, quasi-experimental, and non-experimental techniques? | The type of study you are referring to is called a within-study comparison. An early example that produced a lot of discussion is Dehejia and Wahba (1999; JASA) using Lalonde's (1986) NSW data in which they compared the results based on PSA to the randomized experimental benchmark. The Lalonde data set is now included ... | What are the best empirical studies comparing causal inference with experimental, quasi-experimental | The type of study you are referring to is called a within-study comparison. An early example that produced a lot of discussion is Dehejia and Wahba (1999; JASA) using Lalonde's (1986) NSW data in whic | What are the best empirical studies comparing causal inference with experimental, quasi-experimental, and non-experimental techniques?
The type of study you are referring to is called a within-study comparison. An early example that produced a lot of discussion is Dehejia and Wahba (1999; JASA) using Lalonde's (1986) N... | What are the best empirical studies comparing causal inference with experimental, quasi-experimental
The type of study you are referring to is called a within-study comparison. An early example that produced a lot of discussion is Dehejia and Wahba (1999; JASA) using Lalonde's (1986) NSW data in whic |
47,957 | What are the best empirical studies comparing causal inference with experimental, quasi-experimental, and non-experimental techniques? | In the medicine, the most recent and comprehensive work I'm aware of has been done by OMOP (the Observational Medical Outcomes Partnership). You'll find a lot of relevant research on their publications page, and I think the review paper, 'A systematic statistical approach to evaluating evidence from observational studi... | What are the best empirical studies comparing causal inference with experimental, quasi-experimental | In the medicine, the most recent and comprehensive work I'm aware of has been done by OMOP (the Observational Medical Outcomes Partnership). You'll find a lot of relevant research on their publication | What are the best empirical studies comparing causal inference with experimental, quasi-experimental, and non-experimental techniques?
In the medicine, the most recent and comprehensive work I'm aware of has been done by OMOP (the Observational Medical Outcomes Partnership). You'll find a lot of relevant research on th... | What are the best empirical studies comparing causal inference with experimental, quasi-experimental
In the medicine, the most recent and comprehensive work I'm aware of has been done by OMOP (the Observational Medical Outcomes Partnership). You'll find a lot of relevant research on their publication |
47,958 | What are the best empirical studies comparing causal inference with experimental, quasi-experimental, and non-experimental techniques? | Angus Deaton, the latest Economics Nobel Laureate, is interviewed in this link regarding his thoughts on RCTs as the gold standard. He's quite refreshingly skeptical pointing to, among other things, the typically small sample sizes in RCTs vs the nationally projectable estimates available from observational studies and... | What are the best empirical studies comparing causal inference with experimental, quasi-experimental | Angus Deaton, the latest Economics Nobel Laureate, is interviewed in this link regarding his thoughts on RCTs as the gold standard. He's quite refreshingly skeptical pointing to, among other things, t | What are the best empirical studies comparing causal inference with experimental, quasi-experimental, and non-experimental techniques?
Angus Deaton, the latest Economics Nobel Laureate, is interviewed in this link regarding his thoughts on RCTs as the gold standard. He's quite refreshingly skeptical pointing to, among ... | What are the best empirical studies comparing causal inference with experimental, quasi-experimental
Angus Deaton, the latest Economics Nobel Laureate, is interviewed in this link regarding his thoughts on RCTs as the gold standard. He's quite refreshingly skeptical pointing to, among other things, t |
47,959 | What are the best empirical studies comparing causal inference with experimental, quasi-experimental, and non-experimental techniques? | Have a look at ACIC causal inference competitions; have been going on for a couple of years by now. See for example https://statmodeling.stat.columbia.edu/2022/02/16/welcome-to-the-american-causal-inference-conference-2022-data-challenge/ and references therein. | What are the best empirical studies comparing causal inference with experimental, quasi-experimental | Have a look at ACIC causal inference competitions; have been going on for a couple of years by now. See for example https://statmodeling.stat.columbia.edu/2022/02/16/welcome-to-the-american-causal-inf | What are the best empirical studies comparing causal inference with experimental, quasi-experimental, and non-experimental techniques?
Have a look at ACIC causal inference competitions; have been going on for a couple of years by now. See for example https://statmodeling.stat.columbia.edu/2022/02/16/welcome-to-the-amer... | What are the best empirical studies comparing causal inference with experimental, quasi-experimental
Have a look at ACIC causal inference competitions; have been going on for a couple of years by now. See for example https://statmodeling.stat.columbia.edu/2022/02/16/welcome-to-the-american-causal-inf |
47,960 | Distribution of Ratio of 2 Chi-squared | With finite d.f.
the ratio of two independent chi-squared variates has a beta-prime distribution (also sometimes called a 'beta distribution of the second kind').
if you divide each of the chi-square variates by its df the ratio has an F-distribution.
Asymptotic arguments:
if $j\to\infty$, you can apply Slutsky's th... | Distribution of Ratio of 2 Chi-squared | With finite d.f.
the ratio of two independent chi-squared variates has a beta-prime distribution (also sometimes called a 'beta distribution of the second kind').
if you divide each of the chi-square | Distribution of Ratio of 2 Chi-squared
With finite d.f.
the ratio of two independent chi-squared variates has a beta-prime distribution (also sometimes called a 'beta distribution of the second kind').
if you divide each of the chi-square variates by its df the ratio has an F-distribution.
Asymptotic arguments:
if $... | Distribution of Ratio of 2 Chi-squared
With finite d.f.
the ratio of two independent chi-squared variates has a beta-prime distribution (also sometimes called a 'beta distribution of the second kind').
if you divide each of the chi-square |
47,961 | Fixed Regressor Conspiracy and Connection to Exchangeability | A regression model gives predictions of the response conditional on predictor values; so there's no problem in applying a model fitted to one set of predictor values fixed by design to another set of predictor values, even if the latter are randomly sampled from a population. With an experimental design matrix $X$, the... | Fixed Regressor Conspiracy and Connection to Exchangeability | A regression model gives predictions of the response conditional on predictor values; so there's no problem in applying a model fitted to one set of predictor values fixed by design to another set of | Fixed Regressor Conspiracy and Connection to Exchangeability
A regression model gives predictions of the response conditional on predictor values; so there's no problem in applying a model fitted to one set of predictor values fixed by design to another set of predictor values, even if the latter are randomly sampled f... | Fixed Regressor Conspiracy and Connection to Exchangeability
A regression model gives predictions of the response conditional on predictor values; so there's no problem in applying a model fitted to one set of predictor values fixed by design to another set of |
47,962 | Can I use likelihood-ratio test to compare two samples drawn from power-law distributions? | What you have certainly works. Another option that would only require you to only run the non-pooled model (where you estimate both $\hat\alpha_1$ and $\hat\alpha_2$) is the Wald test with the linear hpothesis
$$
H_o: \alpha_1 - \alpha_2 =0
$$
$$
H_1: \alpha_1 - \alpha_2 \neq 0
$$
If your sample size is large, then ... | Can I use likelihood-ratio test to compare two samples drawn from power-law distributions? | What you have certainly works. Another option that would only require you to only run the non-pooled model (where you estimate both $\hat\alpha_1$ and $\hat\alpha_2$) is the Wald test with the linear | Can I use likelihood-ratio test to compare two samples drawn from power-law distributions?
What you have certainly works. Another option that would only require you to only run the non-pooled model (where you estimate both $\hat\alpha_1$ and $\hat\alpha_2$) is the Wald test with the linear hpothesis
$$
H_o: \alpha_1 -... | Can I use likelihood-ratio test to compare two samples drawn from power-law distributions?
What you have certainly works. Another option that would only require you to only run the non-pooled model (where you estimate both $\hat\alpha_1$ and $\hat\alpha_2$) is the Wald test with the linear |
47,963 | Can I use likelihood-ratio test to compare two samples drawn from power-law distributions? | I can't do that by means of Kolmogorov-Smirnov test because my data are discrete,
Well, actually you could use a Komogorov-Smirnov test on discrete data as long as either:
(i) you don't use the distribution of the test statistic that assumes the data are continuous. You could, for example, run a permutation or random... | Can I use likelihood-ratio test to compare two samples drawn from power-law distributions? | I can't do that by means of Kolmogorov-Smirnov test because my data are discrete,
Well, actually you could use a Komogorov-Smirnov test on discrete data as long as either:
(i) you don't use the dist | Can I use likelihood-ratio test to compare two samples drawn from power-law distributions?
I can't do that by means of Kolmogorov-Smirnov test because my data are discrete,
Well, actually you could use a Komogorov-Smirnov test on discrete data as long as either:
(i) you don't use the distribution of the test statisti... | Can I use likelihood-ratio test to compare two samples drawn from power-law distributions?
I can't do that by means of Kolmogorov-Smirnov test because my data are discrete,
Well, actually you could use a Komogorov-Smirnov test on discrete data as long as either:
(i) you don't use the dist |
47,964 | Mann-Whitney U test and paired data | (I'm not sure I really follow your reasoning.) The Mann-Whitney U-test can be used with paired data. It will simply be less powerful. When you ignore the pairing, you are throwing a lot of information away.
I don't really understand this question.
The meaning of p-values here is the same as the meaning of p-value... | Mann-Whitney U test and paired data | (I'm not sure I really follow your reasoning.) The Mann-Whitney U-test can be used with paired data. It will simply be less powerful. When you ignore the pairing, you are throwing a lot of informat | Mann-Whitney U test and paired data
(I'm not sure I really follow your reasoning.) The Mann-Whitney U-test can be used with paired data. It will simply be less powerful. When you ignore the pairing, you are throwing a lot of information away.
I don't really understand this question.
The meaning of p-values here i... | Mann-Whitney U test and paired data
(I'm not sure I really follow your reasoning.) The Mann-Whitney U-test can be used with paired data. It will simply be less powerful. When you ignore the pairing, you are throwing a lot of informat |
47,965 | Mann-Whitney U test and paired data | I mean if it violate some assumptions of the test
Certainly it violates assumptions, because any machine that crops up in both samples will have scores for the two measures that are dependent (due to the impact of the unobserved $s_i$), when there is an explicit assumption is of independence.
This will impact the beh... | Mann-Whitney U test and paired data | I mean if it violate some assumptions of the test
Certainly it violates assumptions, because any machine that crops up in both samples will have scores for the two measures that are dependent (due to | Mann-Whitney U test and paired data
I mean if it violate some assumptions of the test
Certainly it violates assumptions, because any machine that crops up in both samples will have scores for the two measures that are dependent (due to the impact of the unobserved $s_i$), when there is an explicit assumption is of ind... | Mann-Whitney U test and paired data
I mean if it violate some assumptions of the test
Certainly it violates assumptions, because any machine that crops up in both samples will have scores for the two measures that are dependent (due to |
47,966 | Are the order statistics minimal sufficient for a location-scale family? | This question is Example 6.10, page 36, in Lehmann and Casella Theory of Point Estimation. If the density $f$ is unknown or simply outside exponential families, like the Cauchy distribution, the order statistics $$T(X) = (X_{(1)}, X_{(2)}, X_{(3)},..., X_{(n)})$$ is minimal sufficient.
The result is based on Theorem 6... | Are the order statistics minimal sufficient for a location-scale family? | This question is Example 6.10, page 36, in Lehmann and Casella Theory of Point Estimation. If the density $f$ is unknown or simply outside exponential families, like the Cauchy distribution, the order | Are the order statistics minimal sufficient for a location-scale family?
This question is Example 6.10, page 36, in Lehmann and Casella Theory of Point Estimation. If the density $f$ is unknown or simply outside exponential families, like the Cauchy distribution, the order statistics $$T(X) = (X_{(1)}, X_{(2)}, X_{(3)}... | Are the order statistics minimal sufficient for a location-scale family?
This question is Example 6.10, page 36, in Lehmann and Casella Theory of Point Estimation. If the density $f$ is unknown or simply outside exponential families, like the Cauchy distribution, the order |
47,967 | What is the proper name of a model that takes as input the output of another model? | The process is known as Cascaded classification/regression or Multi-stage classification/regression. It is a type of ensemble learning with some differences. You can find more in Wikipedia. | What is the proper name of a model that takes as input the output of another model? | The process is known as Cascaded classification/regression or Multi-stage classification/regression. It is a type of ensemble learning with some differences. You can find more in Wikipedia. | What is the proper name of a model that takes as input the output of another model?
The process is known as Cascaded classification/regression or Multi-stage classification/regression. It is a type of ensemble learning with some differences. You can find more in Wikipedia. | What is the proper name of a model that takes as input the output of another model?
The process is known as Cascaded classification/regression or Multi-stage classification/regression. It is a type of ensemble learning with some differences. You can find more in Wikipedia. |
47,968 | What is the proper name of a model that takes as input the output of another model? | I am not sure, if there exists at all a "proper name" for models that you describe, but I would call such models chain models, similarly to the chain procedures, introduced in the following paper: http://www.multxpert.com/doc/md2011.pdf. This term IMHO better reflects the nature of this type of models and, at the same ... | What is the proper name of a model that takes as input the output of another model? | I am not sure, if there exists at all a "proper name" for models that you describe, but I would call such models chain models, similarly to the chain procedures, introduced in the following paper: htt | What is the proper name of a model that takes as input the output of another model?
I am not sure, if there exists at all a "proper name" for models that you describe, but I would call such models chain models, similarly to the chain procedures, introduced in the following paper: http://www.multxpert.com/doc/md2011.pdf... | What is the proper name of a model that takes as input the output of another model?
I am not sure, if there exists at all a "proper name" for models that you describe, but I would call such models chain models, similarly to the chain procedures, introduced in the following paper: htt |
47,969 | Are prediction and distribution-fitting ever not the same thing? | one has also (perhaps implicitly) produced a good estimate of its marginal or conditional distribution.
Point predictions don't necessarily do this.
one must necessarily be able to generate good predictions.
Possibly -- depending on how we define "good"
I'm deliberately leaving the prediction and distribution-fitti... | Are prediction and distribution-fitting ever not the same thing? | one has also (perhaps implicitly) produced a good estimate of its marginal or conditional distribution.
Point predictions don't necessarily do this.
one must necessarily be able to generate good pre | Are prediction and distribution-fitting ever not the same thing?
one has also (perhaps implicitly) produced a good estimate of its marginal or conditional distribution.
Point predictions don't necessarily do this.
one must necessarily be able to generate good predictions.
Possibly -- depending on how we define "good... | Are prediction and distribution-fitting ever not the same thing?
one has also (perhaps implicitly) produced a good estimate of its marginal or conditional distribution.
Point predictions don't necessarily do this.
one must necessarily be able to generate good pre |
47,970 | Some questions about gam | Q1
Not really; in R's formula system, the intercept is implied and created when R parses the formula and builds the model matrix. If you want to suppress it, you need to add -1 or + 0 to the formula.
Q2
No, assuming $i$ is a grouping variable? mgcv::gam() will fit a spline that is equivalent to a random intercept in th... | Some questions about gam | Q1
Not really; in R's formula system, the intercept is implied and created when R parses the formula and builds the model matrix. If you want to suppress it, you need to add -1 or + 0 to the formula.
| Some questions about gam
Q1
Not really; in R's formula system, the intercept is implied and created when R parses the formula and builds the model matrix. If you want to suppress it, you need to add -1 or + 0 to the formula.
Q2
No, assuming $i$ is a grouping variable? mgcv::gam() will fit a spline that is equivalent to... | Some questions about gam
Q1
Not really; in R's formula system, the intercept is implied and created when R parses the formula and builds the model matrix. If you want to suppress it, you need to add -1 or + 0 to the formula.
|
47,971 | Initial value of the conditional variance in the GARCH process | I know of at least five ways of initializing the volatility process:
1) Set it equal to $\varepsilon_{t-1}^2$,
2) The sample variance,
3) Unconditional variance of the model ($\alpha_0/(1-\alpha_1 - \alpha_2)$),
4) Allow it it to be an parameter to be estimated,
5) Backcasting with an exponential filter.
The topic... | Initial value of the conditional variance in the GARCH process | I know of at least five ways of initializing the volatility process:
1) Set it equal to $\varepsilon_{t-1}^2$,
2) The sample variance,
3) Unconditional variance of the model ($\alpha_0/(1-\alpha_1 - | Initial value of the conditional variance in the GARCH process
I know of at least five ways of initializing the volatility process:
1) Set it equal to $\varepsilon_{t-1}^2$,
2) The sample variance,
3) Unconditional variance of the model ($\alpha_0/(1-\alpha_1 - \alpha_2)$),
4) Allow it it to be an parameter to be es... | Initial value of the conditional variance in the GARCH process
I know of at least five ways of initializing the volatility process:
1) Set it equal to $\varepsilon_{t-1}^2$,
2) The sample variance,
3) Unconditional variance of the model ($\alpha_0/(1-\alpha_1 - |
47,972 | Estimate an overall correlation from the correlations within subgroups | You will need 1. through 4. to compute the overall correlation for the two groups combined. The equation for this is: $$r_{xy} = \frac{n_1 sd_{x_1} sd_{y_1} r_{xy_1} + n_1 \delta_{x_1} \delta_{y_1} + n_2 sd_{x_2} sd_{y_2} r_{xy_2} + n_2 \delta_{x_2} \delta_{y_2}}{\sqrt{n_1 (sd_{x_1}^2 + \delta_{x_1}^2) + n_2 (sd_{x_2}^... | Estimate an overall correlation from the correlations within subgroups | You will need 1. through 4. to compute the overall correlation for the two groups combined. The equation for this is: $$r_{xy} = \frac{n_1 sd_{x_1} sd_{y_1} r_{xy_1} + n_1 \delta_{x_1} \delta_{y_1} + | Estimate an overall correlation from the correlations within subgroups
You will need 1. through 4. to compute the overall correlation for the two groups combined. The equation for this is: $$r_{xy} = \frac{n_1 sd_{x_1} sd_{y_1} r_{xy_1} + n_1 \delta_{x_1} \delta_{y_1} + n_2 sd_{x_2} sd_{y_2} r_{xy_2} + n_2 \delta_{x_2}... | Estimate an overall correlation from the correlations within subgroups
You will need 1. through 4. to compute the overall correlation for the two groups combined. The equation for this is: $$r_{xy} = \frac{n_1 sd_{x_1} sd_{y_1} r_{xy_1} + n_1 \delta_{x_1} \delta_{y_1} + |
47,973 | Converting Adjusted Odds Ratios to its RR counterpart | You can do this calculation for an adjusted OR (I presume from a logistic regression) to a RR, but the end result may not be useful for your goal of meta-analysis. The essential problem is that the adjusted OR $exp(\beta_1)$ from a logistic regression is not an "average" over the population. And so there's no way to ca... | Converting Adjusted Odds Ratios to its RR counterpart | You can do this calculation for an adjusted OR (I presume from a logistic regression) to a RR, but the end result may not be useful for your goal of meta-analysis. The essential problem is that the ad | Converting Adjusted Odds Ratios to its RR counterpart
You can do this calculation for an adjusted OR (I presume from a logistic regression) to a RR, but the end result may not be useful for your goal of meta-analysis. The essential problem is that the adjusted OR $exp(\beta_1)$ from a logistic regression is not an "ave... | Converting Adjusted Odds Ratios to its RR counterpart
You can do this calculation for an adjusted OR (I presume from a logistic regression) to a RR, but the end result may not be useful for your goal of meta-analysis. The essential problem is that the ad |
47,974 | Standard deviation of residuals from a linear regression | Yes, that's correct. You can also extract this result directly from the model object. For example:
2/sd(resid(mtcars_lm))
[1] 0.6674783 | Standard deviation of residuals from a linear regression | Yes, that's correct. You can also extract this result directly from the model object. For example:
2/sd(resid(mtcars_lm))
[1] 0.6674783 | Standard deviation of residuals from a linear regression
Yes, that's correct. You can also extract this result directly from the model object. For example:
2/sd(resid(mtcars_lm))
[1] 0.6674783 | Standard deviation of residuals from a linear regression
Yes, that's correct. You can also extract this result directly from the model object. For example:
2/sd(resid(mtcars_lm))
[1] 0.6674783 |
47,975 | Information theory without normalization | Just as unnormalized probabilities (likelihoods) can be compared but not turned into probabilities (without normalizing) — similarly, given log-likelihoods $\phi$ and $\psi$, you cannot calculate the KL divergence, but you can compare KL divergences.
For example, if you were trying to select a predictive unnormalized d... | Information theory without normalization | Just as unnormalized probabilities (likelihoods) can be compared but not turned into probabilities (without normalizing) — similarly, given log-likelihoods $\phi$ and $\psi$, you cannot calculate the | Information theory without normalization
Just as unnormalized probabilities (likelihoods) can be compared but not turned into probabilities (without normalizing) — similarly, given log-likelihoods $\phi$ and $\psi$, you cannot calculate the KL divergence, but you can compare KL divergences.
For example, if you were try... | Information theory without normalization
Just as unnormalized probabilities (likelihoods) can be compared but not turned into probabilities (without normalizing) — similarly, given log-likelihoods $\phi$ and $\psi$, you cannot calculate the |
47,976 | Dirac delta function in likelihood function | A typical example where mixed (continuous and finite support) distributions occur is in censoring: the simplest model is to have a continuous variable, say $x\sim\text{N}(\mu,\sigma^2)$ observed, unless it is larger than a fixed value, say $\omicron$, in which case the bound $\omicron$ is reported. In such a case, the ... | Dirac delta function in likelihood function | A typical example where mixed (continuous and finite support) distributions occur is in censoring: the simplest model is to have a continuous variable, say $x\sim\text{N}(\mu,\sigma^2)$ observed, unle | Dirac delta function in likelihood function
A typical example where mixed (continuous and finite support) distributions occur is in censoring: the simplest model is to have a continuous variable, say $x\sim\text{N}(\mu,\sigma^2)$ observed, unless it is larger than a fixed value, say $\omicron$, in which case the bound ... | Dirac delta function in likelihood function
A typical example where mixed (continuous and finite support) distributions occur is in censoring: the simplest model is to have a continuous variable, say $x\sim\text{N}(\mu,\sigma^2)$ observed, unle |
47,977 | Dirac delta function in likelihood function | The Dirac $\delta$ function is zero except for one point, where it is infinite. It is used with integrals, and is defined to have an integral of 1 if the special point is included. The effect of this is to "pull" specific point values through the integral.
This lets you assign probability weight to precise point values... | Dirac delta function in likelihood function | The Dirac $\delta$ function is zero except for one point, where it is infinite. It is used with integrals, and is defined to have an integral of 1 if the special point is included. The effect of this | Dirac delta function in likelihood function
The Dirac $\delta$ function is zero except for one point, where it is infinite. It is used with integrals, and is defined to have an integral of 1 if the special point is included. The effect of this is to "pull" specific point values through the integral.
This lets you assig... | Dirac delta function in likelihood function
The Dirac $\delta$ function is zero except for one point, where it is infinite. It is used with integrals, and is defined to have an integral of 1 if the special point is included. The effect of this |
47,978 | How does backpropagation learn convolution filters? | First of all: 1) The filters are the weights, 2) As justin pointed out, backpropagation is simply a method for computing gradients. You update the weights using another optimization algorithm such as gradient descent. Gradient descent updates the weights at each time step. It updates the weights using the negative grad... | How does backpropagation learn convolution filters? | First of all: 1) The filters are the weights, 2) As justin pointed out, backpropagation is simply a method for computing gradients. You update the weights using another optimization algorithm such as | How does backpropagation learn convolution filters?
First of all: 1) The filters are the weights, 2) As justin pointed out, backpropagation is simply a method for computing gradients. You update the weights using another optimization algorithm such as gradient descent. Gradient descent updates the weights at each time ... | How does backpropagation learn convolution filters?
First of all: 1) The filters are the weights, 2) As justin pointed out, backpropagation is simply a method for computing gradients. You update the weights using another optimization algorithm such as |
47,979 | Do these random variables satisfy Lindeberg's condition? | I will start with just some guidance, and perhaps return later to complete the answer.
Consider the first sequence of random variables, and note that $|X_n| = n$. In other words for given $n$, the absolute value of the random variable is a constant function (as is always the case for dichotomous random variables symmet... | Do these random variables satisfy Lindeberg's condition? | I will start with just some guidance, and perhaps return later to complete the answer.
Consider the first sequence of random variables, and note that $|X_n| = n$. In other words for given $n$, the abs | Do these random variables satisfy Lindeberg's condition?
I will start with just some guidance, and perhaps return later to complete the answer.
Consider the first sequence of random variables, and note that $|X_n| = n$. In other words for given $n$, the absolute value of the random variable is a constant function (as i... | Do these random variables satisfy Lindeberg's condition?
I will start with just some guidance, and perhaps return later to complete the answer.
Consider the first sequence of random variables, and note that $|X_n| = n$. In other words for given $n$, the abs |
47,980 | Using predicted probabilities as regressors | If you are interested in an approximation of the average partial effect you could just use a linear probability model in the first stage, i.e. do your instrumental variables estimation via 2SLS, for instance, in the usual way. However, due to the non-linearities involved this is not the efficient approach but it can gi... | Using predicted probabilities as regressors | If you are interested in an approximation of the average partial effect you could just use a linear probability model in the first stage, i.e. do your instrumental variables estimation via 2SLS, for i | Using predicted probabilities as regressors
If you are interested in an approximation of the average partial effect you could just use a linear probability model in the first stage, i.e. do your instrumental variables estimation via 2SLS, for instance, in the usual way. However, due to the non-linearities involved this... | Using predicted probabilities as regressors
If you are interested in an approximation of the average partial effect you could just use a linear probability model in the first stage, i.e. do your instrumental variables estimation via 2SLS, for i |
47,981 | Meaningful inference about data structure based on components with low variance in PCA | This sort of question did appear several times on CV (you have to browse through PCA clustering questions). The short answer to your question is yes, it makes sense inspecting junior dimensions in search for a structure (such as clusters) in your data. But why not? Often senior components explaining the lion's share of... | Meaningful inference about data structure based on components with low variance in PCA | This sort of question did appear several times on CV (you have to browse through PCA clustering questions). The short answer to your question is yes, it makes sense inspecting junior dimensions in sea | Meaningful inference about data structure based on components with low variance in PCA
This sort of question did appear several times on CV (you have to browse through PCA clustering questions). The short answer to your question is yes, it makes sense inspecting junior dimensions in search for a structure (such as clus... | Meaningful inference about data structure based on components with low variance in PCA
This sort of question did appear several times on CV (you have to browse through PCA clustering questions). The short answer to your question is yes, it makes sense inspecting junior dimensions in sea |
47,982 | Copulas with Regression | In my opinion the two methods (copula, regression) answer quite different questions. The copula approach is much more general than regression and one of the reasons why you have not seen regression models based on copulas, might be that using copulas is much harder than using regression. Two observations why this is so... | Copulas with Regression | In my opinion the two methods (copula, regression) answer quite different questions. The copula approach is much more general than regression and one of the reasons why you have not seen regression mo | Copulas with Regression
In my opinion the two methods (copula, regression) answer quite different questions. The copula approach is much more general than regression and one of the reasons why you have not seen regression models based on copulas, might be that using copulas is much harder than using regression. Two obs... | Copulas with Regression
In my opinion the two methods (copula, regression) answer quite different questions. The copula approach is much more general than regression and one of the reasons why you have not seen regression mo |
47,983 | Copulas with Regression | I have recently devised a curve fitting method for the relationship between two random variables based on copulas:
Regression by Integration demonstrated on Ångström-Prescott-type relations,
Renewable Energy,
Volume 127,
2018,
Pages 713-723,
ISSN 0960-1481,
https://doi.org/10.1016/j.renene.2018.05.004.
(http://www.sci... | Copulas with Regression | I have recently devised a curve fitting method for the relationship between two random variables based on copulas:
Regression by Integration demonstrated on Ångström-Prescott-type relations,
Renewabl | Copulas with Regression
I have recently devised a curve fitting method for the relationship between two random variables based on copulas:
Regression by Integration demonstrated on Ångström-Prescott-type relations,
Renewable Energy,
Volume 127,
2018,
Pages 713-723,
ISSN 0960-1481,
https://doi.org/10.1016/j.renene.2018... | Copulas with Regression
I have recently devised a curve fitting method for the relationship between two random variables based on copulas:
Regression by Integration demonstrated on Ångström-Prescott-type relations,
Renewabl |
47,984 | Can $p(Y|a,b)$ ever be equal to $p(Y|a) \cdot p(Y|b)$? | The equation in the question only happens in the trivial case where both $p(Y|a)$ and $p(Y|b)$ are point masses on a single $Y$ value. If it were true, then $\sum_Y p(Y|a) p(Y|b) = 1$. However, consider the following bound:
$$
\sum_Y p(Y|a) p(Y|b) \leq \sum_Y p(Y|a) \max_{Y'} p(Y'|b) = \max_{Y'} p(Y'|b)
$$
For the LH... | Can $p(Y|a,b)$ ever be equal to $p(Y|a) \cdot p(Y|b)$? | The equation in the question only happens in the trivial case where both $p(Y|a)$ and $p(Y|b)$ are point masses on a single $Y$ value. If it were true, then $\sum_Y p(Y|a) p(Y|b) = 1$. However, cons | Can $p(Y|a,b)$ ever be equal to $p(Y|a) \cdot p(Y|b)$?
The equation in the question only happens in the trivial case where both $p(Y|a)$ and $p(Y|b)$ are point masses on a single $Y$ value. If it were true, then $\sum_Y p(Y|a) p(Y|b) = 1$. However, consider the following bound:
$$
\sum_Y p(Y|a) p(Y|b) \leq \sum_Y p(Y... | Can $p(Y|a,b)$ ever be equal to $p(Y|a) \cdot p(Y|b)$?
The equation in the question only happens in the trivial case where both $p(Y|a)$ and $p(Y|b)$ are point masses on a single $Y$ value. If it were true, then $\sum_Y p(Y|a) p(Y|b) = 1$. However, cons |
47,985 | Can $p(Y|a,b)$ ever be equal to $p(Y|a) \cdot p(Y|b)$? | Sure, if $a$ and $b$ are assumed to be non-interacting events yielding independent information about $Y$ and $Y$ has a uniform prior, then this is the only reasonable way of combining information. Geoff Hinton calls this a product of experts. One caveat, if $Y$ doesn't have a uniform prior, then you'll double-count i... | Can $p(Y|a,b)$ ever be equal to $p(Y|a) \cdot p(Y|b)$? | Sure, if $a$ and $b$ are assumed to be non-interacting events yielding independent information about $Y$ and $Y$ has a uniform prior, then this is the only reasonable way of combining information. Ge | Can $p(Y|a,b)$ ever be equal to $p(Y|a) \cdot p(Y|b)$?
Sure, if $a$ and $b$ are assumed to be non-interacting events yielding independent information about $Y$ and $Y$ has a uniform prior, then this is the only reasonable way of combining information. Geoff Hinton calls this a product of experts. One caveat, if $Y$ d... | Can $p(Y|a,b)$ ever be equal to $p(Y|a) \cdot p(Y|b)$?
Sure, if $a$ and $b$ are assumed to be non-interacting events yielding independent information about $Y$ and $Y$ has a uniform prior, then this is the only reasonable way of combining information. Ge |
47,986 | Poisson confidence interval using the pivotal method | As far as I know, there isn't really a pivotal quantity for $\lambda$*, though it's possible to construct approximately pivotal quantities if $n\lambda$ isn't small. (I include $n$ there just in case you have multiple observations from the same Poisson. In many cases you'll just have the one count. From here on I'll ju... | Poisson confidence interval using the pivotal method | As far as I know, there isn't really a pivotal quantity for $\lambda$*, though it's possible to construct approximately pivotal quantities if $n\lambda$ isn't small. (I include $n$ there just in case | Poisson confidence interval using the pivotal method
As far as I know, there isn't really a pivotal quantity for $\lambda$*, though it's possible to construct approximately pivotal quantities if $n\lambda$ isn't small. (I include $n$ there just in case you have multiple observations from the same Poisson. In many cases... | Poisson confidence interval using the pivotal method
As far as I know, there isn't really a pivotal quantity for $\lambda$*, though it's possible to construct approximately pivotal quantities if $n\lambda$ isn't small. (I include $n$ there just in case |
47,987 | Wherefore Big Data? | it isn't only the data that is big, also the problem is big.
Indeed, the benefits of increasing the sampling size aren't great if you were computing the mean of terabytes of data. Only that nobody is interested in the 10th digit of the mean anyways...
More often than not, big data problems are more like a big amount ... | Wherefore Big Data? | it isn't only the data that is big, also the problem is big.
Indeed, the benefits of increasing the sampling size aren't great if you were computing the mean of terabytes of data. Only that nobody is | Wherefore Big Data?
it isn't only the data that is big, also the problem is big.
Indeed, the benefits of increasing the sampling size aren't great if you were computing the mean of terabytes of data. Only that nobody is interested in the 10th digit of the mean anyways...
More often than not, big data problems are mor... | Wherefore Big Data?
it isn't only the data that is big, also the problem is big.
Indeed, the benefits of increasing the sampling size aren't great if you were computing the mean of terabytes of data. Only that nobody is |
47,988 | Using K-fold cross validation to select a model's parameters | Cross-validation just gives you an estimate of your out-of-sample risk. It doesn't produce a better model. To get the most precise estimate of your coefficients, you should use all of your data. | Using K-fold cross validation to select a model's parameters | Cross-validation just gives you an estimate of your out-of-sample risk. It doesn't produce a better model. To get the most precise estimate of your coefficients, you should use all of your data. | Using K-fold cross validation to select a model's parameters
Cross-validation just gives you an estimate of your out-of-sample risk. It doesn't produce a better model. To get the most precise estimate of your coefficients, you should use all of your data. | Using K-fold cross validation to select a model's parameters
Cross-validation just gives you an estimate of your out-of-sample risk. It doesn't produce a better model. To get the most precise estimate of your coefficients, you should use all of your data. |
47,989 | Using K-fold cross validation to select a model's parameters | Actually, I've already understood how to do so. Just in case someone stumbles upon this question: cross validation can serve as a parameter tuning tool.To tune a model's parameters using K-fold cross validation you train and test each model K times against the K possible data combinations and average their out-of-sampl... | Using K-fold cross validation to select a model's parameters | Actually, I've already understood how to do so. Just in case someone stumbles upon this question: cross validation can serve as a parameter tuning tool.To tune a model's parameters using K-fold cross | Using K-fold cross validation to select a model's parameters
Actually, I've already understood how to do so. Just in case someone stumbles upon this question: cross validation can serve as a parameter tuning tool.To tune a model's parameters using K-fold cross validation you train and test each model K times against th... | Using K-fold cross validation to select a model's parameters
Actually, I've already understood how to do so. Just in case someone stumbles upon this question: cross validation can serve as a parameter tuning tool.To tune a model's parameters using K-fold cross |
47,990 | Affine equivariance and consistency | For a counterexample consider a one-dimensional sequence, $x_i\sim \mathcal N(0, \sigma^2)$, and let $S$ be the sample variance
$$S(x_1, \ldots, x_n) = \frac{1}{n}\sum_{i=1}x_i^2 - \left(\frac{1}{n}\sum_{i=1}^n x_i\right)^2.$$
Set $B=S$ except let $B=0$ whenever $|x_1+x_2+\cdots+x_n| \lt 1$. Although $S$ is equivarian... | Affine equivariance and consistency | For a counterexample consider a one-dimensional sequence, $x_i\sim \mathcal N(0, \sigma^2)$, and let $S$ be the sample variance
$$S(x_1, \ldots, x_n) = \frac{1}{n}\sum_{i=1}x_i^2 - \left(\frac{1}{n}\s | Affine equivariance and consistency
For a counterexample consider a one-dimensional sequence, $x_i\sim \mathcal N(0, \sigma^2)$, and let $S$ be the sample variance
$$S(x_1, \ldots, x_n) = \frac{1}{n}\sum_{i=1}x_i^2 - \left(\frac{1}{n}\sum_{i=1}^n x_i\right)^2.$$
Set $B=S$ except let $B=0$ whenever $|x_1+x_2+\cdots+x_n|... | Affine equivariance and consistency
For a counterexample consider a one-dimensional sequence, $x_i\sim \mathcal N(0, \sigma^2)$, and let $S$ be the sample variance
$$S(x_1, \ldots, x_n) = \frac{1}{n}\sum_{i=1}x_i^2 - \left(\frac{1}{n}\s |
47,991 | Affine equivariance and consistency | Let $B(X_n)$ be an affine equivariant estimator. Let $A(X_n) = B(X_n) k(B(X_n))$ where $k$ is a scalar function such as trace. Then $A$ is consistent but not affine equivariant. This is called a weak covariance functional. | Affine equivariance and consistency | Let $B(X_n)$ be an affine equivariant estimator. Let $A(X_n) = B(X_n) k(B(X_n))$ where $k$ is a scalar function such as trace. Then $A$ is consistent but not affine equivariant. This is called a we | Affine equivariance and consistency
Let $B(X_n)$ be an affine equivariant estimator. Let $A(X_n) = B(X_n) k(B(X_n))$ where $k$ is a scalar function such as trace. Then $A$ is consistent but not affine equivariant. This is called a weak covariance functional. | Affine equivariance and consistency
Let $B(X_n)$ be an affine equivariant estimator. Let $A(X_n) = B(X_n) k(B(X_n))$ where $k$ is a scalar function such as trace. Then $A$ is consistent but not affine equivariant. This is called a we |
47,992 | Resources to write a statistical analysis plan | While I would argue that a statistical analysis plan (SAP) is a really good idea for any study or experiment, it is in clinical research that you will find most guidance. That is because the field is heavily regulated and because it is in the interest of industry to describe best practices.
In most such research, ther... | Resources to write a statistical analysis plan | While I would argue that a statistical analysis plan (SAP) is a really good idea for any study or experiment, it is in clinical research that you will find most guidance. That is because the field is | Resources to write a statistical analysis plan
While I would argue that a statistical analysis plan (SAP) is a really good idea for any study or experiment, it is in clinical research that you will find most guidance. That is because the field is heavily regulated and because it is in the interest of industry to descr... | Resources to write a statistical analysis plan
While I would argue that a statistical analysis plan (SAP) is a really good idea for any study or experiment, it is in clinical research that you will find most guidance. That is because the field is |
47,993 | Expected value of dot product between a random unit vector in $\mathbb{R}^n$ and another given unit vector | Judging from the result, it appears the context implicitly supposes the distribution of $x$ is invariant under orthogonal transformations: I would call this a spherically-symmetric distribution.
(There are plenty of spherically-symmetric distributionns. Starting with any distribution $F$ in $\mathbb{R}^n$, define $\ti... | Expected value of dot product between a random unit vector in $\mathbb{R}^n$ and another given unit | Judging from the result, it appears the context implicitly supposes the distribution of $x$ is invariant under orthogonal transformations: I would call this a spherically-symmetric distribution.
(Ther | Expected value of dot product between a random unit vector in $\mathbb{R}^n$ and another given unit vector
Judging from the result, it appears the context implicitly supposes the distribution of $x$ is invariant under orthogonal transformations: I would call this a spherically-symmetric distribution.
(There are plenty ... | Expected value of dot product between a random unit vector in $\mathbb{R}^n$ and another given unit
Judging from the result, it appears the context implicitly supposes the distribution of $x$ is invariant under orthogonal transformations: I would call this a spherically-symmetric distribution.
(Ther |
47,994 | Many dependent variables, few samples: is this an example of "large $p$, small $n$" problem? | There are several combinations on the size of $n$ and $p$: small $n$ - large $p$, small $p$ - large $n$, large $n$ - large $p$ ... See Johnstone & Titterington, 2009, Statistical challenges of high-dimensional data for an overview.
In your case, it appears that you have a small $p=1$ and relatively small $n$, with high... | Many dependent variables, few samples: is this an example of "large $p$, small $n$" problem? | There are several combinations on the size of $n$ and $p$: small $n$ - large $p$, small $p$ - large $n$, large $n$ - large $p$ ... See Johnstone & Titterington, 2009, Statistical challenges of high-di | Many dependent variables, few samples: is this an example of "large $p$, small $n$" problem?
There are several combinations on the size of $n$ and $p$: small $n$ - large $p$, small $p$ - large $n$, large $n$ - large $p$ ... See Johnstone & Titterington, 2009, Statistical challenges of high-dimensional data for an overv... | Many dependent variables, few samples: is this an example of "large $p$, small $n$" problem?
There are several combinations on the size of $n$ and $p$: small $n$ - large $p$, small $p$ - large $n$, large $n$ - large $p$ ... See Johnstone & Titterington, 2009, Statistical challenges of high-di |
47,995 | Many dependent variables, few samples: is this an example of "large $p$, small $n$" problem? | [As I read it, the question is primarily about terminology, and @East's answer (good as it is) does not explicitly address it.]
Sometimes the distinction between dependent and independent variables is not so clear. As you are referring to MANOVA, you probably have $300$ variables measured for two groups. Technically, y... | Many dependent variables, few samples: is this an example of "large $p$, small $n$" problem? | [As I read it, the question is primarily about terminology, and @East's answer (good as it is) does not explicitly address it.]
Sometimes the distinction between dependent and independent variables is | Many dependent variables, few samples: is this an example of "large $p$, small $n$" problem?
[As I read it, the question is primarily about terminology, and @East's answer (good as it is) does not explicitly address it.]
Sometimes the distinction between dependent and independent variables is not so clear. As you are r... | Many dependent variables, few samples: is this an example of "large $p$, small $n$" problem?
[As I read it, the question is primarily about terminology, and @East's answer (good as it is) does not explicitly address it.]
Sometimes the distinction between dependent and independent variables is |
47,996 | Taylor's expansion on log likelihood | If one includes the notational dependency on $n$:
$$
\begin{align*}
\ell_n\left(\theta\right) & = \ell_n\left(\widehat{\theta}_n\right)+\frac{\partial\ell_n\left(\theta\right)}{\partial\theta}\Bigr|_{\theta=\widehat{\theta}_n}\left(\theta-\widehat{\theta}_n\right)+ o_n(|\widehat{\theta}_n - \theta|^2)
\\
\end{align*}
$... | Taylor's expansion on log likelihood | If one includes the notational dependency on $n$:
$$
\begin{align*}
\ell_n\left(\theta\right) & = \ell_n\left(\widehat{\theta}_n\right)+\frac{\partial\ell_n\left(\theta\right)}{\partial\theta}\Bigr|_{ | Taylor's expansion on log likelihood
If one includes the notational dependency on $n$:
$$
\begin{align*}
\ell_n\left(\theta\right) & = \ell_n\left(\widehat{\theta}_n\right)+\frac{\partial\ell_n\left(\theta\right)}{\partial\theta}\Bigr|_{\theta=\widehat{\theta}_n}\left(\theta-\widehat{\theta}_n\right)+ o_n(|\widehat{\th... | Taylor's expansion on log likelihood
If one includes the notational dependency on $n$:
$$
\begin{align*}
\ell_n\left(\theta\right) & = \ell_n\left(\widehat{\theta}_n\right)+\frac{\partial\ell_n\left(\theta\right)}{\partial\theta}\Bigr|_{ |
47,997 | Taylor's expansion on log likelihood | Strictly speaking, the likelihood function has two components: the observations and the parameters. It is typically seen as a function of the parameters when the sample is fixed but you can also study its behaviour as a random variable when you fix the parameters and see it as a function of the random variables (which ... | Taylor's expansion on log likelihood | Strictly speaking, the likelihood function has two components: the observations and the parameters. It is typically seen as a function of the parameters when the sample is fixed but you can also study | Taylor's expansion on log likelihood
Strictly speaking, the likelihood function has two components: the observations and the parameters. It is typically seen as a function of the parameters when the sample is fixed but you can also study its behaviour as a random variable when you fix the parameters and see it as a fun... | Taylor's expansion on log likelihood
Strictly speaking, the likelihood function has two components: the observations and the parameters. It is typically seen as a function of the parameters when the sample is fixed but you can also study |
47,998 | Taylor's expansion on log likelihood | Strictly speaking that expression doesn't make sense a priori. But It can be made precise. The log-likelihood is a random function (or a sequence of random functions if you're in the asymptotic setting) on the parameter space. So sure, for a given realization of that random function, one can write (for sample size $n$)... | Taylor's expansion on log likelihood | Strictly speaking that expression doesn't make sense a priori. But It can be made precise. The log-likelihood is a random function (or a sequence of random functions if you're in the asymptotic settin | Taylor's expansion on log likelihood
Strictly speaking that expression doesn't make sense a priori. But It can be made precise. The log-likelihood is a random function (or a sequence of random functions if you're in the asymptotic setting) on the parameter space. So sure, for a given realization of that random function... | Taylor's expansion on log likelihood
Strictly speaking that expression doesn't make sense a priori. But It can be made precise. The log-likelihood is a random function (or a sequence of random functions if you're in the asymptotic settin |
47,999 | When do kernel based method perform better than the regular | There is no easy answer to this question.
Typically you would start experimenting using the linear kernel function (to which I assume you refer as "regular"). If data is not linearly separable then there would be errors. So, if the performance is not satisfactory you would have to try some non-linear kernel functions.... | When do kernel based method perform better than the regular | There is no easy answer to this question.
Typically you would start experimenting using the linear kernel function (to which I assume you refer as "regular"). If data is not linearly separable then t | When do kernel based method perform better than the regular
There is no easy answer to this question.
Typically you would start experimenting using the linear kernel function (to which I assume you refer as "regular"). If data is not linearly separable then there would be errors. So, if the performance is not satisfac... | When do kernel based method perform better than the regular
There is no easy answer to this question.
Typically you would start experimenting using the linear kernel function (to which I assume you refer as "regular"). If data is not linearly separable then t |
48,000 | When do kernel based method perform better than the regular | Just some more thoughts to previous answer:
What is Kernel or Kernel Method ?
Kernel or Positive-definite kernel is a generalization of a positive-definite matrix.In linear algebra, a symmetric n × n real matrix M is said to be positive definite if zTMz is positive for every non-zero column vector z of n real numbers.... | When do kernel based method perform better than the regular | Just some more thoughts to previous answer:
What is Kernel or Kernel Method ?
Kernel or Positive-definite kernel is a generalization of a positive-definite matrix.In linear algebra, a symmetric n × n | When do kernel based method perform better than the regular
Just some more thoughts to previous answer:
What is Kernel or Kernel Method ?
Kernel or Positive-definite kernel is a generalization of a positive-definite matrix.In linear algebra, a symmetric n × n real matrix M is said to be positive definite if zTMz is po... | When do kernel based method perform better than the regular
Just some more thoughts to previous answer:
What is Kernel or Kernel Method ?
Kernel or Positive-definite kernel is a generalization of a positive-definite matrix.In linear algebra, a symmetric n × n |
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