idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
|---|---|---|---|---|---|---|
39,401 | What is the difference between t-SNE and plain SNE? | Actually, I find that because the t-Distribution is a long tail distribution, it prevents the crowding problem (which is one of the disadvantages of SNE). | What is the difference between t-SNE and plain SNE? | Actually, I find that because the t-Distribution is a long tail distribution, it prevents the crowding problem (which is one of the disadvantages of SNE). | What is the difference between t-SNE and plain SNE?
Actually, I find that because the t-Distribution is a long tail distribution, it prevents the crowding problem (which is one of the disadvantages of SNE). | What is the difference between t-SNE and plain SNE?
Actually, I find that because the t-Distribution is a long tail distribution, it prevents the crowding problem (which is one of the disadvantages of SNE). |
39,402 | What is the difference between t-SNE and plain SNE? | You can also watch this lecture from 17:48 to 20:12 to hear the reason with a great example from the author of t-SNE. | What is the difference between t-SNE and plain SNE? | You can also watch this lecture from 17:48 to 20:12 to hear the reason with a great example from the author of t-SNE. | What is the difference between t-SNE and plain SNE?
You can also watch this lecture from 17:48 to 20:12 to hear the reason with a great example from the author of t-SNE. | What is the difference between t-SNE and plain SNE?
You can also watch this lecture from 17:48 to 20:12 to hear the reason with a great example from the author of t-SNE. |
39,403 | What is the difference between t-SNE and plain SNE? | The cluster structure produced by tSNE tend to be more separated, to have more stable shape; and be more repeatable. | What is the difference between t-SNE and plain SNE? | The cluster structure produced by tSNE tend to be more separated, to have more stable shape; and be more repeatable. | What is the difference between t-SNE and plain SNE?
The cluster structure produced by tSNE tend to be more separated, to have more stable shape; and be more repeatable. | What is the difference between t-SNE and plain SNE?
The cluster structure produced by tSNE tend to be more separated, to have more stable shape; and be more repeatable. |
39,404 | What is the difference between t-SNE and plain SNE? | We are learning a topological structure here. So mapping the neighbors in the lower dimension is the necessary and fundamental objective of SNE. Note that, in lower dimension we don't have much space to accommodate all the neighbors.
for motivation note that, we can accommodate maximum $n+1$ equidistant points in a $n$... | What is the difference between t-SNE and plain SNE? | We are learning a topological structure here. So mapping the neighbors in the lower dimension is the necessary and fundamental objective of SNE. Note that, in lower dimension we don't have much space | What is the difference between t-SNE and plain SNE?
We are learning a topological structure here. So mapping the neighbors in the lower dimension is the necessary and fundamental objective of SNE. Note that, in lower dimension we don't have much space to accommodate all the neighbors.
for motivation note that, we can a... | What is the difference between t-SNE and plain SNE?
We are learning a topological structure here. So mapping the neighbors in the lower dimension is the necessary and fundamental objective of SNE. Note that, in lower dimension we don't have much space |
39,405 | Neural network softmax activation | The internet has told me that when using Softmax combined with cross entropy, Step 1 simply becomes $\frac{\partial E} {\partial z_j} = o_j - t_j$ where $t$ is a one-hot encoded target output vector. Is this correct?
Yes. Before going through the proof, let me change the notation to avoid careless mistakes in translat... | Neural network softmax activation | The internet has told me that when using Softmax combined with cross entropy, Step 1 simply becomes $\frac{\partial E} {\partial z_j} = o_j - t_j$ where $t$ is a one-hot encoded target output vector. | Neural network softmax activation
The internet has told me that when using Softmax combined with cross entropy, Step 1 simply becomes $\frac{\partial E} {\partial z_j} = o_j - t_j$ where $t$ is a one-hot encoded target output vector. Is this correct?
Yes. Before going through the proof, let me change the notation to a... | Neural network softmax activation
The internet has told me that when using Softmax combined with cross entropy, Step 1 simply becomes $\frac{\partial E} {\partial z_j} = o_j - t_j$ where $t$ is a one-hot encoded target output vector. |
39,406 | Is a minimal sufficient statistic also a complete statistic | Examples of minimal sufficient statistic which are not complete are aplenty.
A simple instance is $X\sim U (\theta,\theta+1)$ where $\theta\in \mathbb R$.
It is not difficult to show $X$ is a minimal sufficient statistic for $\theta$. However, $$E_{\theta}(\sin 2\pi X)=\int_{\theta}^{\theta+1} \sin (2\pi x)\,\mathrm{d... | Is a minimal sufficient statistic also a complete statistic | Examples of minimal sufficient statistic which are not complete are aplenty.
A simple instance is $X\sim U (\theta,\theta+1)$ where $\theta\in \mathbb R$.
It is not difficult to show $X$ is a minimal | Is a minimal sufficient statistic also a complete statistic
Examples of minimal sufficient statistic which are not complete are aplenty.
A simple instance is $X\sim U (\theta,\theta+1)$ where $\theta\in \mathbb R$.
It is not difficult to show $X$ is a minimal sufficient statistic for $\theta$. However, $$E_{\theta}(\s... | Is a minimal sufficient statistic also a complete statistic
Examples of minimal sufficient statistic which are not complete are aplenty.
A simple instance is $X\sim U (\theta,\theta+1)$ where $\theta\in \mathbb R$.
It is not difficult to show $X$ is a minimal |
39,407 | Is a minimal sufficient statistic also a complete statistic | Consider $N(\theta,\theta)$ where $\theta>0$.Of course $\dfrac{1}{n}\sum_{i=1}^n X_i$ is minimal sufficient but not complete. To see why it is not complete, find $a$ and $b$ such that:
$$E\Big(a\sum_{i=1}^n (X_i-\overline{X})^2 \Big)=E\Big(b\sum_{i=1}^nX_i^2\Big)=\theta^2$$
and therefore $E\Big(a\sum_{i=1}^n (X_i-\over... | Is a minimal sufficient statistic also a complete statistic | Consider $N(\theta,\theta)$ where $\theta>0$.Of course $\dfrac{1}{n}\sum_{i=1}^n X_i$ is minimal sufficient but not complete. To see why it is not complete, find $a$ and $b$ such that:
$$E\Big(a\sum_{ | Is a minimal sufficient statistic also a complete statistic
Consider $N(\theta,\theta)$ where $\theta>0$.Of course $\dfrac{1}{n}\sum_{i=1}^n X_i$ is minimal sufficient but not complete. To see why it is not complete, find $a$ and $b$ such that:
$$E\Big(a\sum_{i=1}^n (X_i-\overline{X})^2 \Big)=E\Big(b\sum_{i=1}^nX_i^2\B... | Is a minimal sufficient statistic also a complete statistic
Consider $N(\theta,\theta)$ where $\theta>0$.Of course $\dfrac{1}{n}\sum_{i=1}^n X_i$ is minimal sufficient but not complete. To see why it is not complete, find $a$ and $b$ such that:
$$E\Big(a\sum_{ |
39,408 | Is a minimal sufficient statistic also a complete statistic | In the Cauchy distribution with unknown location,
$$f(x;\mu) = \frac{1}{\pi} \, \frac{1}{1+(x-\mu)^2}$$
for a sample $(X_1,\ldots,X_n)$
the order statistic $(X_{(1)},\ldots,X_{(n)})$ is minimal sufficient, but it is incomplete since $$\mathbb{E}_\mu[\phi(X_{(i)} - X_{(j)})]\qquad i\ne j$$is constant in $\mu$ for bounde... | Is a minimal sufficient statistic also a complete statistic | In the Cauchy distribution with unknown location,
$$f(x;\mu) = \frac{1}{\pi} \, \frac{1}{1+(x-\mu)^2}$$
for a sample $(X_1,\ldots,X_n)$
the order statistic $(X_{(1)},\ldots,X_{(n)})$ is minimal suffic | Is a minimal sufficient statistic also a complete statistic
In the Cauchy distribution with unknown location,
$$f(x;\mu) = \frac{1}{\pi} \, \frac{1}{1+(x-\mu)^2}$$
for a sample $(X_1,\ldots,X_n)$
the order statistic $(X_{(1)},\ldots,X_{(n)})$ is minimal sufficient, but it is incomplete since $$\mathbb{E}_\mu[\phi(X_{(i... | Is a minimal sufficient statistic also a complete statistic
In the Cauchy distribution with unknown location,
$$f(x;\mu) = \frac{1}{\pi} \, \frac{1}{1+(x-\mu)^2}$$
for a sample $(X_1,\ldots,X_n)$
the order statistic $(X_{(1)},\ldots,X_{(n)})$ is minimal suffic |
39,409 | what is the “learning” that takes place in Naive Bayes? | Different from the nearest neighbor algorithm, the Naive Bayes algorithm is not a lazy method; A real learning takes place for Naive Bayes. The parameters that are learned in Naive Bayes are the prior probabilities of different classes, as well as the likelihood of different features for each class. In the test phase, ... | what is the “learning” that takes place in Naive Bayes? | Different from the nearest neighbor algorithm, the Naive Bayes algorithm is not a lazy method; A real learning takes place for Naive Bayes. The parameters that are learned in Naive Bayes are the prior | what is the “learning” that takes place in Naive Bayes?
Different from the nearest neighbor algorithm, the Naive Bayes algorithm is not a lazy method; A real learning takes place for Naive Bayes. The parameters that are learned in Naive Bayes are the prior probabilities of different classes, as well as the likelihood o... | what is the “learning” that takes place in Naive Bayes?
Different from the nearest neighbor algorithm, the Naive Bayes algorithm is not a lazy method; A real learning takes place for Naive Bayes. The parameters that are learned in Naive Bayes are the prior |
39,410 | what is the “learning” that takes place in Naive Bayes? | I did not see any parameters in the Naive Bayes Classifier.
I think we do not need to learn parameters like what we do in Neural networks, but we learn(or calculate) the prior probability from the training data. Then we apply the prior we calculated from training data to make predictions in test data. | what is the “learning” that takes place in Naive Bayes? | I did not see any parameters in the Naive Bayes Classifier.
I think we do not need to learn parameters like what we do in Neural networks, but we learn(or calculate) the prior probability from the tra | what is the “learning” that takes place in Naive Bayes?
I did not see any parameters in the Naive Bayes Classifier.
I think we do not need to learn parameters like what we do in Neural networks, but we learn(or calculate) the prior probability from the training data. Then we apply the prior we calculated from training ... | what is the “learning” that takes place in Naive Bayes?
I did not see any parameters in the Naive Bayes Classifier.
I think we do not need to learn parameters like what we do in Neural networks, but we learn(or calculate) the prior probability from the tra |
39,411 | Boruta 'all-relevant' feature selection vs Random Forest 'variables of importance' | Boruta and random forrest differences
Boruta algorithm uses randomization on top of results obtained from variable importance obtained from random forest to determine the truly important and statistically valid results. For details of the difference please refer to Section 2 of the article:
Kursa, Miron B., and Witold ... | Boruta 'all-relevant' feature selection vs Random Forest 'variables of importance' | Boruta and random forrest differences
Boruta algorithm uses randomization on top of results obtained from variable importance obtained from random forest to determine the truly important and statistic | Boruta 'all-relevant' feature selection vs Random Forest 'variables of importance'
Boruta and random forrest differences
Boruta algorithm uses randomization on top of results obtained from variable importance obtained from random forest to determine the truly important and statistically valid results. For details of th... | Boruta 'all-relevant' feature selection vs Random Forest 'variables of importance'
Boruta and random forrest differences
Boruta algorithm uses randomization on top of results obtained from variable importance obtained from random forest to determine the truly important and statistic |
39,412 | Boruta 'all-relevant' feature selection vs Random Forest 'variables of importance' | Basic Idea of Boruta Algorithm
Perform shuffling of predictors' values and join them with the original predictors and then build random forest on the merged dataset. Then make comparison of original variables with the randomised variables to measure variable importance. Only variables having higher importance than that... | Boruta 'all-relevant' feature selection vs Random Forest 'variables of importance' | Basic Idea of Boruta Algorithm
Perform shuffling of predictors' values and join them with the original predictors and then build random forest on the merged dataset. Then make comparison of original v | Boruta 'all-relevant' feature selection vs Random Forest 'variables of importance'
Basic Idea of Boruta Algorithm
Perform shuffling of predictors' values and join them with the original predictors and then build random forest on the merged dataset. Then make comparison of original variables with the randomised variable... | Boruta 'all-relevant' feature selection vs Random Forest 'variables of importance'
Basic Idea of Boruta Algorithm
Perform shuffling of predictors' values and join them with the original predictors and then build random forest on the merged dataset. Then make comparison of original v |
39,413 | Regression Bounded Between -1 and 1 | You can always use beta regression (Ferrari and Cribari-Neto, 2004). It's a model for response variable bounded in $(0, 1)$, but you can easily transform your variable by taking $\frac{Y+1}{2}$ (I know you said you do not want to transform, but it's a really basic transformation).
Moreover, such model still makes perfe... | Regression Bounded Between -1 and 1 | You can always use beta regression (Ferrari and Cribari-Neto, 2004). It's a model for response variable bounded in $(0, 1)$, but you can easily transform your variable by taking $\frac{Y+1}{2}$ (I kno | Regression Bounded Between -1 and 1
You can always use beta regression (Ferrari and Cribari-Neto, 2004). It's a model for response variable bounded in $(0, 1)$, but you can easily transform your variable by taking $\frac{Y+1}{2}$ (I know you said you do not want to transform, but it's a really basic transformation).
Mo... | Regression Bounded Between -1 and 1
You can always use beta regression (Ferrari and Cribari-Neto, 2004). It's a model for response variable bounded in $(0, 1)$, but you can easily transform your variable by taking $\frac{Y+1}{2}$ (I kno |
39,414 | Regression Bounded Between -1 and 1 | There now exist a proposed advancement of classical beta regression, referred to a 'Boosted Beta Regression', that may be of value in this meta-analysis with locational data.
Per this 2013 research article available online to quote, in part:
In the statistical literature, beta regression has been established as a powe... | Regression Bounded Between -1 and 1 | There now exist a proposed advancement of classical beta regression, referred to a 'Boosted Beta Regression', that may be of value in this meta-analysis with locational data.
Per this 2013 research ar | Regression Bounded Between -1 and 1
There now exist a proposed advancement of classical beta regression, referred to a 'Boosted Beta Regression', that may be of value in this meta-analysis with locational data.
Per this 2013 research article available online to quote, in part:
In the statistical literature, beta regre... | Regression Bounded Between -1 and 1
There now exist a proposed advancement of classical beta regression, referred to a 'Boosted Beta Regression', that may be of value in this meta-analysis with locational data.
Per this 2013 research ar |
39,415 | Why is Gaussian Copula's Tail Dependence Zero? | Consider a bivariate Gaussian copula $C(\cdot)$.
Because of the radial symmetry of a Gaussian copula we can consider just the lower tail dependence. We know that the lower tail dependence for this copula is:
$$\begin{align}
\lambda&=\lim_{\,\,q\to 0^{+}} \frac{\partial C(q,q)}{\partial q}\\
&=\lim_{\,\,q\to 0^{+}} \tex... | Why is Gaussian Copula's Tail Dependence Zero? | Consider a bivariate Gaussian copula $C(\cdot)$.
Because of the radial symmetry of a Gaussian copula we can consider just the lower tail dependence. We know that the lower tail dependence for this cop | Why is Gaussian Copula's Tail Dependence Zero?
Consider a bivariate Gaussian copula $C(\cdot)$.
Because of the radial symmetry of a Gaussian copula we can consider just the lower tail dependence. We know that the lower tail dependence for this copula is:
$$\begin{align}
\lambda&=\lim_{\,\,q\to 0^{+}} \frac{\partial C(q... | Why is Gaussian Copula's Tail Dependence Zero?
Consider a bivariate Gaussian copula $C(\cdot)$.
Because of the radial symmetry of a Gaussian copula we can consider just the lower tail dependence. We know that the lower tail dependence for this cop |
39,416 | Why is Gaussian Copula's Tail Dependence Zero? | For a non-technical, intuitive view of what the tail index is telling you, we can look at simulation and compute sample estimates of the quantity $P[F(Y) > q | F(X) > q]$ as $q$ increases.
Here the original correlation is $0.96$, but as we get further into the upper tail of $X$ the correlation decreases, and the propo... | Why is Gaussian Copula's Tail Dependence Zero? | For a non-technical, intuitive view of what the tail index is telling you, we can look at simulation and compute sample estimates of the quantity $P[F(Y) > q | F(X) > q]$ as $q$ increases.
Here the o | Why is Gaussian Copula's Tail Dependence Zero?
For a non-technical, intuitive view of what the tail index is telling you, we can look at simulation and compute sample estimates of the quantity $P[F(Y) > q | F(X) > q]$ as $q$ increases.
Here the original correlation is $0.96$, but as we get further into the upper tail ... | Why is Gaussian Copula's Tail Dependence Zero?
For a non-technical, intuitive view of what the tail index is telling you, we can look at simulation and compute sample estimates of the quantity $P[F(Y) > q | F(X) > q]$ as $q$ increases.
Here the o |
39,417 | What is the equivalent in R of scikit-learn's `LogisticRegression` with `penalty="l2"` | Your question is how to run L2 regularized logistic regression in R.
My another detailed answer can be found here. Regularization methods for logistic regression
For implementation, there are more than one way of doing this.
Method 1, use glmnet(data,label,family="binomial", alpha=0, lambda=1), Details can be found ... | What is the equivalent in R of scikit-learn's `LogisticRegression` with `penalty="l2"` | Your question is how to run L2 regularized logistic regression in R.
My another detailed answer can be found here. Regularization methods for logistic regression
For implementation, there are more th | What is the equivalent in R of scikit-learn's `LogisticRegression` with `penalty="l2"`
Your question is how to run L2 regularized logistic regression in R.
My another detailed answer can be found here. Regularization methods for logistic regression
For implementation, there are more than one way of doing this.
Metho... | What is the equivalent in R of scikit-learn's `LogisticRegression` with `penalty="l2"`
Your question is how to run L2 regularized logistic regression in R.
My another detailed answer can be found here. Regularization methods for logistic regression
For implementation, there are more th |
39,418 | What is an example use of Auto differentiation such as implemented in Tensorflow and why is it important? | In auto differentiation systems mostly an operator (e.g. addition, subtraction) is defined together with its differentiation.
So after you write a function by stacking a series of operators, the system can figure out by itself how the corresponding derivatives should be computed, usually by using computation graphs an... | What is an example use of Auto differentiation such as implemented in Tensorflow and why is it impor | In auto differentiation systems mostly an operator (e.g. addition, subtraction) is defined together with its differentiation.
So after you write a function by stacking a series of operators, the syst | What is an example use of Auto differentiation such as implemented in Tensorflow and why is it important?
In auto differentiation systems mostly an operator (e.g. addition, subtraction) is defined together with its differentiation.
So after you write a function by stacking a series of operators, the system can figure ... | What is an example use of Auto differentiation such as implemented in Tensorflow and why is it impor
In auto differentiation systems mostly an operator (e.g. addition, subtraction) is defined together with its differentiation.
So after you write a function by stacking a series of operators, the syst |
39,419 | sampling/importance resampling - why resample? | SIR uses two ideas. The first idea is importance sampling. The main idea is that you draw from one probability distribution (in your case, it's the uniform), in order to get information about another. You do this by drawing from one distribution, then weighting the samples. Generally you are trying to get information a... | sampling/importance resampling - why resample? | SIR uses two ideas. The first idea is importance sampling. The main idea is that you draw from one probability distribution (in your case, it's the uniform), in order to get information about another. | sampling/importance resampling - why resample?
SIR uses two ideas. The first idea is importance sampling. The main idea is that you draw from one probability distribution (in your case, it's the uniform), in order to get information about another. You do this by drawing from one distribution, then weighting the samples... | sampling/importance resampling - why resample?
SIR uses two ideas. The first idea is importance sampling. The main idea is that you draw from one probability distribution (in your case, it's the uniform), in order to get information about another. |
39,420 | If $\text{Var}(X) < \infty$, is $\text{Var}(XY) < \infty$ for $0 \le Y \le 1$? | I've unaccepted kjetil's answer since, as was pointed out in the comments, it assumes $X$ and $Y$ are independent.
The following answer should work when $X$ and $Y$ are dependent, by using whuber's suggestion:
\begin{align}
\text{Var}(XY) &= E((XY)^2) - E(XY)^2 \\
&\le E(X^2Y^2) \\
&\le E(X^2)\sup(Y^2) \\
&= E(X^2) \\
... | If $\text{Var}(X) < \infty$, is $\text{Var}(XY) < \infty$ for $0 \le Y \le 1$? | I've unaccepted kjetil's answer since, as was pointed out in the comments, it assumes $X$ and $Y$ are independent.
The following answer should work when $X$ and $Y$ are dependent, by using whuber's su | If $\text{Var}(X) < \infty$, is $\text{Var}(XY) < \infty$ for $0 \le Y \le 1$?
I've unaccepted kjetil's answer since, as was pointed out in the comments, it assumes $X$ and $Y$ are independent.
The following answer should work when $X$ and $Y$ are dependent, by using whuber's suggestion:
\begin{align}
\text{Var}(XY) &=... | If $\text{Var}(X) < \infty$, is $\text{Var}(XY) < \infty$ for $0 \le Y \le 1$?
I've unaccepted kjetil's answer since, as was pointed out in the comments, it assumes $X$ and $Y$ are independent.
The following answer should work when $X$ and $Y$ are dependent, by using whuber's su |
39,421 | If $\text{Var}(X) < \infty$, is $\text{Var}(XY) < \infty$ for $0 \le Y \le 1$? | You need to use the formula
$$ \DeclareMathOperator{\Var}{\mathbb{V}} \DeclareMathOperator{\E}{\mathbb{E}}
\Var (XY) = \E (\Var (XY | Y)) + \Var (\E (XY | Y))
$$
where $\Var $ is the variance operator. Take it term for term, write $\mu=\E X, \sigma^2=\Var X$, $\E (XY | Y= y) = \E (yX) = y \E (X) =\mu y$ with varia... | If $\text{Var}(X) < \infty$, is $\text{Var}(XY) < \infty$ for $0 \le Y \le 1$? | You need to use the formula
$$ \DeclareMathOperator{\Var}{\mathbb{V}} \DeclareMathOperator{\E}{\mathbb{E}}
\Var (XY) = \E (\Var (XY | Y)) + \Var (\E (XY | Y))
$$
where $\Var $ is the variance ope | If $\text{Var}(X) < \infty$, is $\text{Var}(XY) < \infty$ for $0 \le Y \le 1$?
You need to use the formula
$$ \DeclareMathOperator{\Var}{\mathbb{V}} \DeclareMathOperator{\E}{\mathbb{E}}
\Var (XY) = \E (\Var (XY | Y)) + \Var (\E (XY | Y))
$$
where $\Var $ is the variance operator. Take it term for term, write $\mu=... | If $\text{Var}(X) < \infty$, is $\text{Var}(XY) < \infty$ for $0 \le Y \le 1$?
You need to use the formula
$$ \DeclareMathOperator{\Var}{\mathbb{V}} \DeclareMathOperator{\E}{\mathbb{E}}
\Var (XY) = \E (\Var (XY | Y)) + \Var (\E (XY | Y))
$$
where $\Var $ is the variance ope |
39,422 | Forecasting Time Series: Stationary vs Non-Stationary | [W]hat is the difference between forecasting using the original non-stationary series and the forecasting using the now stationary differenced series?
(Here I deliberately left out the qualification that the series can be transformed to a stationary series using first differencing and that the OP is interested in fore... | Forecasting Time Series: Stationary vs Non-Stationary | [W]hat is the difference between forecasting using the original non-stationary series and the forecasting using the now stationary differenced series?
(Here I deliberately left out the qualification | Forecasting Time Series: Stationary vs Non-Stationary
[W]hat is the difference between forecasting using the original non-stationary series and the forecasting using the now stationary differenced series?
(Here I deliberately left out the qualification that the series can be transformed to a stationary series using fi... | Forecasting Time Series: Stationary vs Non-Stationary
[W]hat is the difference between forecasting using the original non-stationary series and the forecasting using the now stationary differenced series?
(Here I deliberately left out the qualification |
39,423 | Forecasting Time Series: Stationary vs Non-Stationary | In your case there's no difference. ARIMA(p,1,q) is the same as ARMA(p,q) on the differenced series. ARIMA can model non-stationary series, ARMA cannot. So, for ARMA you do differencing before feeding the series into it.
See how ARIMA(p,D,q) models are defined at MATLAB web site, for instance:
$$\phi(L)(1-L)^Dy_t=c+\th... | Forecasting Time Series: Stationary vs Non-Stationary | In your case there's no difference. ARIMA(p,1,q) is the same as ARMA(p,q) on the differenced series. ARIMA can model non-stationary series, ARMA cannot. So, for ARMA you do differencing before feeding | Forecasting Time Series: Stationary vs Non-Stationary
In your case there's no difference. ARIMA(p,1,q) is the same as ARMA(p,q) on the differenced series. ARIMA can model non-stationary series, ARMA cannot. So, for ARMA you do differencing before feeding the series into it.
See how ARIMA(p,D,q) models are defined at MA... | Forecasting Time Series: Stationary vs Non-Stationary
In your case there's no difference. ARIMA(p,1,q) is the same as ARMA(p,q) on the differenced series. ARIMA can model non-stationary series, ARMA cannot. So, for ARMA you do differencing before feeding |
39,424 | Meta analysis across studies with multiple response measures | I'll try to address a few different components of your question
You seem to be under the impression that you want to meta-analyze means of some bird-related variable, but after reading...
I want to look at if, across birds, these response measures increase along with another variable that I have for each bird
it inst... | Meta analysis across studies with multiple response measures | I'll try to address a few different components of your question
You seem to be under the impression that you want to meta-analyze means of some bird-related variable, but after reading...
I want to l | Meta analysis across studies with multiple response measures
I'll try to address a few different components of your question
You seem to be under the impression that you want to meta-analyze means of some bird-related variable, but after reading...
I want to look at if, across birds, these response measures increase a... | Meta analysis across studies with multiple response measures
I'll try to address a few different components of your question
You seem to be under the impression that you want to meta-analyze means of some bird-related variable, but after reading...
I want to l |
39,425 | Meta analysis across studies with multiple response measures | I think you need to use the metafor package and look at the function rma.mv and its documentation. It would also be worth looking at his website which has many examples. Perhaps you would like to try and then ask again? The author does post on Cross Validated but it may also be worth your while trying to post on the R-... | Meta analysis across studies with multiple response measures | I think you need to use the metafor package and look at the function rma.mv and its documentation. It would also be worth looking at his website which has many examples. Perhaps you would like to try | Meta analysis across studies with multiple response measures
I think you need to use the metafor package and look at the function rma.mv and its documentation. It would also be worth looking at his website which has many examples. Perhaps you would like to try and then ask again? The author does post on Cross Validated... | Meta analysis across studies with multiple response measures
I think you need to use the metafor package and look at the function rma.mv and its documentation. It would also be worth looking at his website which has many examples. Perhaps you would like to try |
39,426 | Meta analysis across studies with multiple response measures | You can use the standard mean difference for pooling different scales. for further reading check : 'metaanalysis with r' a book writen by the author of the package 'meta' pg 25
However, in many settings different studies use different outcome
scales, e.g. different depression scales or quality of life scales. In
s... | Meta analysis across studies with multiple response measures | You can use the standard mean difference for pooling different scales. for further reading check : 'metaanalysis with r' a book writen by the author of the package 'meta' pg 25
However, in many setti | Meta analysis across studies with multiple response measures
You can use the standard mean difference for pooling different scales. for further reading check : 'metaanalysis with r' a book writen by the author of the package 'meta' pg 25
However, in many settings different studies use different outcome
scales, e.g. ... | Meta analysis across studies with multiple response measures
You can use the standard mean difference for pooling different scales. for further reading check : 'metaanalysis with r' a book writen by the author of the package 'meta' pg 25
However, in many setti |
39,427 | Which parameter should be considered as "scale" parameter for Gamma distribution? | A scale parameter is defined as a parameter $\sigma$ such that,
if $X\sim F(x;\mu,\sigma)$, then $\sigma^{-1}X \sim F(x;\mu,1)$, that
is, when $\sigma^{-1}X$ is independent from $\sigma$.
This means $\theta$ is a scale parameter for the Gamma $\text{Ga}(\kappa,\theta)$ distribution.
When looking at the exponential... | Which parameter should be considered as "scale" parameter for Gamma distribution? | A scale parameter is defined as a parameter $\sigma$ such that,
if $X\sim F(x;\mu,\sigma)$, then $\sigma^{-1}X \sim F(x;\mu,1)$, that
is, when $\sigma^{-1}X$ is independent from $\sigma$.
This me | Which parameter should be considered as "scale" parameter for Gamma distribution?
A scale parameter is defined as a parameter $\sigma$ such that,
if $X\sim F(x;\mu,\sigma)$, then $\sigma^{-1}X \sim F(x;\mu,1)$, that
is, when $\sigma^{-1}X$ is independent from $\sigma$.
This means $\theta$ is a scale parameter for ... | Which parameter should be considered as "scale" parameter for Gamma distribution?
A scale parameter is defined as a parameter $\sigma$ such that,
if $X\sim F(x;\mu,\sigma)$, then $\sigma^{-1}X \sim F(x;\mu,1)$, that
is, when $\sigma^{-1}X$ is independent from $\sigma$.
This me |
39,428 | Which parameter should be considered as "scale" parameter for Gamma distribution? | Comparing (2) and (3), it is that φ=1/k should be called as scale
parameter!
According the equation you wrote it seems that both of your phis (you really need to use different letters for parameters) are part of the scale parameter. Why don't you write it in standard form so that the density is
$$f(x) \propto \exp(... | Which parameter should be considered as "scale" parameter for Gamma distribution? | Comparing (2) and (3), it is that φ=1/k should be called as scale
parameter!
According the equation you wrote it seems that both of your phis (you really need to use different letters for parameter | Which parameter should be considered as "scale" parameter for Gamma distribution?
Comparing (2) and (3), it is that φ=1/k should be called as scale
parameter!
According the equation you wrote it seems that both of your phis (you really need to use different letters for parameters) are part of the scale parameter. W... | Which parameter should be considered as "scale" parameter for Gamma distribution?
Comparing (2) and (3), it is that φ=1/k should be called as scale
parameter!
According the equation you wrote it seems that both of your phis (you really need to use different letters for parameter |
39,429 | Which parameter should be considered as "scale" parameter for Gamma distribution? | Both alternatives are (as mentioned prior) given here, one with $\frac{x}{\theta }$, where $\theta$ is indeed a scale parameter, and $\beta x$, where $\beta$ is a rate scale parameter, the reciprocal of $\theta$. $\theta$ is the scale factor. Similarly for exponential distributions, $\frac{x}{\theta }$, where $\theta$... | Which parameter should be considered as "scale" parameter for Gamma distribution? | Both alternatives are (as mentioned prior) given here, one with $\frac{x}{\theta }$, where $\theta$ is indeed a scale parameter, and $\beta x$, where $\beta$ is a rate scale parameter, the reciprocal | Which parameter should be considered as "scale" parameter for Gamma distribution?
Both alternatives are (as mentioned prior) given here, one with $\frac{x}{\theta }$, where $\theta$ is indeed a scale parameter, and $\beta x$, where $\beta$ is a rate scale parameter, the reciprocal of $\theta$. $\theta$ is the scale fac... | Which parameter should be considered as "scale" parameter for Gamma distribution?
Both alternatives are (as mentioned prior) given here, one with $\frac{x}{\theta }$, where $\theta$ is indeed a scale parameter, and $\beta x$, where $\beta$ is a rate scale parameter, the reciprocal |
39,430 | Multiple regression - how to calculate the predicted value after feature normalization? | You should perform feature normalization only on features - so only on your input vector $x$. Not on output $y$ or $\theta$. When you trained a model using feature normalization, then you should apply that normalization every time you make a prediction. Also it is expected that you have different $\theta$ and cost func... | Multiple regression - how to calculate the predicted value after feature normalization? | You should perform feature normalization only on features - so only on your input vector $x$. Not on output $y$ or $\theta$. When you trained a model using feature normalization, then you should apply | Multiple regression - how to calculate the predicted value after feature normalization?
You should perform feature normalization only on features - so only on your input vector $x$. Not on output $y$ or $\theta$. When you trained a model using feature normalization, then you should apply that normalization every time y... | Multiple regression - how to calculate the predicted value after feature normalization?
You should perform feature normalization only on features - so only on your input vector $x$. Not on output $y$ or $\theta$. When you trained a model using feature normalization, then you should apply |
39,431 | Multiple regression - how to calculate the predicted value after feature normalization? | EDIT: I've made a few changes to the code based on @Yurii's answer.
Okay it seems that after a little fiddling about and checking out this answer, I got it to work:
As mentioned, I used
[scaledX,avgX,stdX]=feature_scale(X)
to scale the features. I then used gradient descent to get the vector theta, which is used to mak... | Multiple regression - how to calculate the predicted value after feature normalization? | EDIT: I've made a few changes to the code based on @Yurii's answer.
Okay it seems that after a little fiddling about and checking out this answer, I got it to work:
As mentioned, I used
[scaledX,avgX, | Multiple regression - how to calculate the predicted value after feature normalization?
EDIT: I've made a few changes to the code based on @Yurii's answer.
Okay it seems that after a little fiddling about and checking out this answer, I got it to work:
As mentioned, I used
[scaledX,avgX,stdX]=feature_scale(X)
to scale ... | Multiple regression - how to calculate the predicted value after feature normalization?
EDIT: I've made a few changes to the code based on @Yurii's answer.
Okay it seems that after a little fiddling about and checking out this answer, I got it to work:
As mentioned, I used
[scaledX,avgX, |
39,432 | Multiple regression - how to calculate the predicted value after feature normalization? | I understand the concepts explained in the previous 2 answers i.e. after we do feature scaling and calculate the intercept (θ0) and the slope (θ1), we get a hypothesis function (h(x)) which uses the scaled down features (Assuming univariate/single variable linear regression)
h(x) = θ0 + θ1x' -- (1)
where
x' = (x-μ)/... | Multiple regression - how to calculate the predicted value after feature normalization? | I understand the concepts explained in the previous 2 answers i.e. after we do feature scaling and calculate the intercept (θ0) and the slope (θ1), we get a hypothesis function (h(x)) which uses the s | Multiple regression - how to calculate the predicted value after feature normalization?
I understand the concepts explained in the previous 2 answers i.e. after we do feature scaling and calculate the intercept (θ0) and the slope (θ1), we get a hypothesis function (h(x)) which uses the scaled down features (Assuming un... | Multiple regression - how to calculate the predicted value after feature normalization?
I understand the concepts explained in the previous 2 answers i.e. after we do feature scaling and calculate the intercept (θ0) and the slope (θ1), we get a hypothesis function (h(x)) which uses the s |
39,433 | When is a Naive Bayes Model not Bayesian? | Informally, to be 'Bayesian' about a model (Naive Bayes just names a class of discrete mixture models) is to use Bayes theorem to infer the values of its parameters or other quantities of interest. To be 'Frequentist' about the same model is, roughly, and among other things, to use the sampling distribution of estima... | When is a Naive Bayes Model not Bayesian? | Informally, to be 'Bayesian' about a model (Naive Bayes just names a class of discrete mixture models) is to use Bayes theorem to infer the values of its parameters or other quantities of interest. | When is a Naive Bayes Model not Bayesian?
Informally, to be 'Bayesian' about a model (Naive Bayes just names a class of discrete mixture models) is to use Bayes theorem to infer the values of its parameters or other quantities of interest. To be 'Frequentist' about the same model is, roughly, and among other things, ... | When is a Naive Bayes Model not Bayesian?
Informally, to be 'Bayesian' about a model (Naive Bayes just names a class of discrete mixture models) is to use Bayes theorem to infer the values of its parameters or other quantities of interest. |
39,434 | Help with Taylor expansion of log likelihood function | You should convince yourself that $f(x)-f(y)\approx f'(x)(x-y)$ is just another way to express Taylor expansion (under appropriate regularity assumptions).
Then, using linearity of the derivation, you can generalize it to the sum. | Help with Taylor expansion of log likelihood function | You should convince yourself that $f(x)-f(y)\approx f'(x)(x-y)$ is just another way to express Taylor expansion (under appropriate regularity assumptions).
Then, using linearity of the derivation, you | Help with Taylor expansion of log likelihood function
You should convince yourself that $f(x)-f(y)\approx f'(x)(x-y)$ is just another way to express Taylor expansion (under appropriate regularity assumptions).
Then, using linearity of the derivation, you can generalize it to the sum. | Help with Taylor expansion of log likelihood function
You should convince yourself that $f(x)-f(y)\approx f'(x)(x-y)$ is just another way to express Taylor expansion (under appropriate regularity assumptions).
Then, using linearity of the derivation, you |
39,435 | Help with Taylor expansion of log likelihood function | I think your problem is just the notation problem
Let us start from the beginning;
Suppose$X_1, X_2,...,X_n$ are i.i.d random variables with probability density function $f(x;\gamma)$
Usually, people will use $\theta$ here, I will keep it for later use.
The likelihood function is $L(\gamma;x)=f(x_1;\gamma)f(x_2;\gamma)... | Help with Taylor expansion of log likelihood function | I think your problem is just the notation problem
Let us start from the beginning;
Suppose$X_1, X_2,...,X_n$ are i.i.d random variables with probability density function $f(x;\gamma)$
Usually, people | Help with Taylor expansion of log likelihood function
I think your problem is just the notation problem
Let us start from the beginning;
Suppose$X_1, X_2,...,X_n$ are i.i.d random variables with probability density function $f(x;\gamma)$
Usually, people will use $\theta$ here, I will keep it for later use.
The likeliho... | Help with Taylor expansion of log likelihood function
I think your problem is just the notation problem
Let us start from the beginning;
Suppose$X_1, X_2,...,X_n$ are i.i.d random variables with probability density function $f(x;\gamma)$
Usually, people |
39,436 | What exactly is the standard error of the intercept in multiple regression analysis? | The standard error of the the intercept allows you to test whether or not the estimated intercept is statistically significant from a specified(hypothesized) value ...normally 0.0 . If you test against 0.0 and fail to reject then you can then re-estimate your model without the intercept term being present. | What exactly is the standard error of the intercept in multiple regression analysis? | The standard error of the the intercept allows you to test whether or not the estimated intercept is statistically significant from a specified(hypothesized) value ...normally 0.0 . If you test agains | What exactly is the standard error of the intercept in multiple regression analysis?
The standard error of the the intercept allows you to test whether or not the estimated intercept is statistically significant from a specified(hypothesized) value ...normally 0.0 . If you test against 0.0 and fail to reject then you c... | What exactly is the standard error of the intercept in multiple regression analysis?
The standard error of the the intercept allows you to test whether or not the estimated intercept is statistically significant from a specified(hypothesized) value ...normally 0.0 . If you test agains |
39,437 | What exactly is the standard error of the intercept in multiple regression analysis? | Your characterization of how multiple regression works is inaccurate. Your version implies fitting a simple linear regression for each variable in turn (and presumably using each of those slopes as the coefficient for that variable in the multiple regression model). This notion leaves you with the problem of how to d... | What exactly is the standard error of the intercept in multiple regression analysis? | Your characterization of how multiple regression works is inaccurate. Your version implies fitting a simple linear regression for each variable in turn (and presumably using each of those slopes as t | What exactly is the standard error of the intercept in multiple regression analysis?
Your characterization of how multiple regression works is inaccurate. Your version implies fitting a simple linear regression for each variable in turn (and presumably using each of those slopes as the coefficient for that variable in... | What exactly is the standard error of the intercept in multiple regression analysis?
Your characterization of how multiple regression works is inaccurate. Your version implies fitting a simple linear regression for each variable in turn (and presumably using each of those slopes as t |
39,438 | Joint distribution of AR(1) model | Let us write the joint density as
\begin{equation}
p(y_1,\ldots,y_T) = p(y_1)\,p(y_2\mid y_1) \, p(y_3 \mid y_2,y_1) \ldots \, p(y_T \mid y_{T-1},\ldots y_1).
\end{equation}
Furthermore, since the process is AR(1), the past values influence future values only via the latest value, i.e., we have the Markov property $p(y... | Joint distribution of AR(1) model | Let us write the joint density as
\begin{equation}
p(y_1,\ldots,y_T) = p(y_1)\,p(y_2\mid y_1) \, p(y_3 \mid y_2,y_1) \ldots \, p(y_T \mid y_{T-1},\ldots y_1).
\end{equation}
Furthermore, since the pro | Joint distribution of AR(1) model
Let us write the joint density as
\begin{equation}
p(y_1,\ldots,y_T) = p(y_1)\,p(y_2\mid y_1) \, p(y_3 \mid y_2,y_1) \ldots \, p(y_T \mid y_{T-1},\ldots y_1).
\end{equation}
Furthermore, since the process is AR(1), the past values influence future values only via the latest value, i.e.... | Joint distribution of AR(1) model
Let us write the joint density as
\begin{equation}
p(y_1,\ldots,y_T) = p(y_1)\,p(y_2\mid y_1) \, p(y_3 \mid y_2,y_1) \ldots \, p(y_T \mid y_{T-1},\ldots y_1).
\end{equation}
Furthermore, since the pro |
39,439 | Joint distribution of AR(1) model | The title of the question points towards a functional specification of the form
$$y_t= \phi_0 - \phi_1\,\phi_0 + \phi_1\,y_{t-1} + e_t,~e_t \sim N(0,\sigma^2), t>1, |\phi_1|<1$$
with the error term $e_t$ being i.i.d.
Given the assumption on the initial available observation (which does not necessarily represent the beg... | Joint distribution of AR(1) model | The title of the question points towards a functional specification of the form
$$y_t= \phi_0 - \phi_1\,\phi_0 + \phi_1\,y_{t-1} + e_t,~e_t \sim N(0,\sigma^2), t>1, |\phi_1|<1$$
with the error term $e | Joint distribution of AR(1) model
The title of the question points towards a functional specification of the form
$$y_t= \phi_0 - \phi_1\,\phi_0 + \phi_1\,y_{t-1} + e_t,~e_t \sim N(0,\sigma^2), t>1, |\phi_1|<1$$
with the error term $e_t$ being i.i.d.
Given the assumption on the initial available observation (which does... | Joint distribution of AR(1) model
The title of the question points towards a functional specification of the form
$$y_t= \phi_0 - \phi_1\,\phi_0 + \phi_1\,y_{t-1} + e_t,~e_t \sim N(0,\sigma^2), t>1, |\phi_1|<1$$
with the error term $e |
39,440 | What does improper learning mean in the context of statistical learning theory and machine learning? | In statistical learning theory, the standard batch learning problem is defined in terms of a distribution $P$ over some space $\mathcal{Z}$ belonging to some set of distributions $\mathcal{P}$, a hypothesis class $\mathcal{H}$ and a loss function $\ell$, which assigns a (say) nonnegative real ("a loss") to pairs $(P,h)... | What does improper learning mean in the context of statistical learning theory and machine learning? | In statistical learning theory, the standard batch learning problem is defined in terms of a distribution $P$ over some space $\mathcal{Z}$ belonging to some set of distributions $\mathcal{P}$, a hypo | What does improper learning mean in the context of statistical learning theory and machine learning?
In statistical learning theory, the standard batch learning problem is defined in terms of a distribution $P$ over some space $\mathcal{Z}$ belonging to some set of distributions $\mathcal{P}$, a hypothesis class $\math... | What does improper learning mean in the context of statistical learning theory and machine learning?
In statistical learning theory, the standard batch learning problem is defined in terms of a distribution $P$ over some space $\mathcal{Z}$ belonging to some set of distributions $\mathcal{P}$, a hypo |
39,441 | Minimization of the Sum of Absolute Deviations | So that this has some form of answer, let's look at what the OP has discovered, and then put some additional clarity/detail into that (I'm going to go into more detail than I normally would for a self-study question because of the issues with the question, which I think require some explanation):
The issue here is that... | Minimization of the Sum of Absolute Deviations | So that this has some form of answer, let's look at what the OP has discovered, and then put some additional clarity/detail into that (I'm going to go into more detail than I normally would for a self | Minimization of the Sum of Absolute Deviations
So that this has some form of answer, let's look at what the OP has discovered, and then put some additional clarity/detail into that (I'm going to go into more detail than I normally would for a self-study question because of the issues with the question, which I think re... | Minimization of the Sum of Absolute Deviations
So that this has some form of answer, let's look at what the OP has discovered, and then put some additional clarity/detail into that (I'm going to go into more detail than I normally would for a self |
39,442 | What is a multivariate random variable? | Yes, throwing two dice is a multivariate random variable. Specifically, you will get two independent and identically distributed discrete variables (assuming fair dice).
Throwing two dice and adding the results gets you a univariate random variable, with possible values between 2 and 12.
Picking a person at random and ... | What is a multivariate random variable? | Yes, throwing two dice is a multivariate random variable. Specifically, you will get two independent and identically distributed discrete variables (assuming fair dice).
Throwing two dice and adding t | What is a multivariate random variable?
Yes, throwing two dice is a multivariate random variable. Specifically, you will get two independent and identically distributed discrete variables (assuming fair dice).
Throwing two dice and adding the results gets you a univariate random variable, with possible values between 2... | What is a multivariate random variable?
Yes, throwing two dice is a multivariate random variable. Specifically, you will get two independent and identically distributed discrete variables (assuming fair dice).
Throwing two dice and adding t |
39,443 | What is a multivariate random variable? | With multivariate distributions, correlations between variables are important. If there is no correlation between to variable, then you basically have two univariate distributions.
Throwing two die would be a multivariate distribution, but would probably have a correlation of zero (the exceptions to this are interesti... | What is a multivariate random variable? | With multivariate distributions, correlations between variables are important. If there is no correlation between to variable, then you basically have two univariate distributions.
Throwing two die w | What is a multivariate random variable?
With multivariate distributions, correlations between variables are important. If there is no correlation between to variable, then you basically have two univariate distributions.
Throwing two die would be a multivariate distribution, but would probably have a correlation of ze... | What is a multivariate random variable?
With multivariate distributions, correlations between variables are important. If there is no correlation between to variable, then you basically have two univariate distributions.
Throwing two die w |
39,444 | Estimation of unit-root AR(1) model with OLS | It is generally assumed that the explanatory variables have finite moments at least up to second order. In this case, as the explanatory variable is a random walk, its variance is not finite. This makes the matrix $Q=\hbox{plim } X′X/n$ not finite, with the consequences discussed below.
The explanatory variable $x_{t-1... | Estimation of unit-root AR(1) model with OLS | It is generally assumed that the explanatory variables have finite moments at least up to second order. In this case, as the explanatory variable is a random walk, its variance is not finite. This mak | Estimation of unit-root AR(1) model with OLS
It is generally assumed that the explanatory variables have finite moments at least up to second order. In this case, as the explanatory variable is a random walk, its variance is not finite. This makes the matrix $Q=\hbox{plim } X′X/n$ not finite, with the consequences disc... | Estimation of unit-root AR(1) model with OLS
It is generally assumed that the explanatory variables have finite moments at least up to second order. In this case, as the explanatory variable is a random walk, its variance is not finite. This mak |
39,445 | Estimation of unit-root AR(1) model with OLS | One of the key assumptions I would list among standard OLS assumptions is that there is no weak LLN for the "average of the $X'X$-matrix" $1/T\sum_tx_{t-1}^2$. Instead, we have weak convergence to a functional of Brownian motion provided we scale by $T^2$, viz.
$$
T^{-2}\sum_{t=1}^Tx^2_{t-1}\Rightarrow\sigma^2\int_0^1W... | Estimation of unit-root AR(1) model with OLS | One of the key assumptions I would list among standard OLS assumptions is that there is no weak LLN for the "average of the $X'X$-matrix" $1/T\sum_tx_{t-1}^2$. Instead, we have weak convergence to a f | Estimation of unit-root AR(1) model with OLS
One of the key assumptions I would list among standard OLS assumptions is that there is no weak LLN for the "average of the $X'X$-matrix" $1/T\sum_tx_{t-1}^2$. Instead, we have weak convergence to a functional of Brownian motion provided we scale by $T^2$, viz.
$$
T^{-2}\sum... | Estimation of unit-root AR(1) model with OLS
One of the key assumptions I would list among standard OLS assumptions is that there is no weak LLN for the "average of the $X'X$-matrix" $1/T\sum_tx_{t-1}^2$. Instead, we have weak convergence to a f |
39,446 | Comparing (hidden) regression coefficients in simple linear regression? | EDIT to respond to the altered question:
Again you phrase your hypothesis based on parameters outside this model, which makes it a little uncertain what exactly you are going at. But interpreting your hypotheses to be referring to the marginal effect of each of the three topics being covered in one text, I think what y... | Comparing (hidden) regression coefficients in simple linear regression? | EDIT to respond to the altered question:
Again you phrase your hypothesis based on parameters outside this model, which makes it a little uncertain what exactly you are going at. But interpreting your | Comparing (hidden) regression coefficients in simple linear regression?
EDIT to respond to the altered question:
Again you phrase your hypothesis based on parameters outside this model, which makes it a little uncertain what exactly you are going at. But interpreting your hypotheses to be referring to the marginal effe... | Comparing (hidden) regression coefficients in simple linear regression?
EDIT to respond to the altered question:
Again you phrase your hypothesis based on parameters outside this model, which makes it a little uncertain what exactly you are going at. But interpreting your |
39,447 | Comparing (hidden) regression coefficients in simple linear regression? | To test if $\beta_1=\beta_3$, we need to "reframe" the regression a bit.
$$Y=\beta_0+\beta_1x_1+\beta_2x_2$$
$$Y=\beta_0+\beta_1x_1+\beta_2(1-x_1-x_3)$$
$$Y=\beta_0+\beta_2+(\beta_1-\beta_2)x_1-\beta_2x_3$$
Now $\beta_1=\beta_3$ is equivalent to $\beta_1-\beta_2=-\beta_2$, becoming $\beta_1=0$. So just test if $\beta_1... | Comparing (hidden) regression coefficients in simple linear regression? | To test if $\beta_1=\beta_3$, we need to "reframe" the regression a bit.
$$Y=\beta_0+\beta_1x_1+\beta_2x_2$$
$$Y=\beta_0+\beta_1x_1+\beta_2(1-x_1-x_3)$$
$$Y=\beta_0+\beta_2+(\beta_1-\beta_2)x_1-\beta_ | Comparing (hidden) regression coefficients in simple linear regression?
To test if $\beta_1=\beta_3$, we need to "reframe" the regression a bit.
$$Y=\beta_0+\beta_1x_1+\beta_2x_2$$
$$Y=\beta_0+\beta_1x_1+\beta_2(1-x_1-x_3)$$
$$Y=\beta_0+\beta_2+(\beta_1-\beta_2)x_1-\beta_2x_3$$
Now $\beta_1=\beta_3$ is equivalent to $\... | Comparing (hidden) regression coefficients in simple linear regression?
To test if $\beta_1=\beta_3$, we need to "reframe" the regression a bit.
$$Y=\beta_0+\beta_1x_1+\beta_2x_2$$
$$Y=\beta_0+\beta_1x_1+\beta_2(1-x_1-x_3)$$
$$Y=\beta_0+\beta_2+(\beta_1-\beta_2)x_1-\beta_ |
39,448 | Comparing (hidden) regression coefficients in simple linear regression? | It's not possible to do all 3 pairwise comparisons of regression coefficients that you wish, because you only really have two regression coefficients.
The value of $\beta_3$ in a linear regression would be the effect of a change in $x_3$ on $Y$ with $x_1$ and $x_2$ held constant. That is not possible in your situation ... | Comparing (hidden) regression coefficients in simple linear regression? | It's not possible to do all 3 pairwise comparisons of regression coefficients that you wish, because you only really have two regression coefficients.
The value of $\beta_3$ in a linear regression wou | Comparing (hidden) regression coefficients in simple linear regression?
It's not possible to do all 3 pairwise comparisons of regression coefficients that you wish, because you only really have two regression coefficients.
The value of $\beta_3$ in a linear regression would be the effect of a change in $x_3$ on $Y$ wit... | Comparing (hidden) regression coefficients in simple linear regression?
It's not possible to do all 3 pairwise comparisons of regression coefficients that you wish, because you only really have two regression coefficients.
The value of $\beta_3$ in a linear regression wou |
39,449 | Comparing (hidden) regression coefficients in simple linear regression? | My suggestion is to interpret the t-statistics of $x_1$ and $x_2$ in relation to $x_3$, because the latter is the reference group.
That will answer 2. and 3. Test the equality of $x_{1}$ and $x_2$ separately with an appropriate test which will answer 1. | Comparing (hidden) regression coefficients in simple linear regression? | My suggestion is to interpret the t-statistics of $x_1$ and $x_2$ in relation to $x_3$, because the latter is the reference group.
That will answer 2. and 3. Test the equality of $x_{1}$ and $x_2$ se | Comparing (hidden) regression coefficients in simple linear regression?
My suggestion is to interpret the t-statistics of $x_1$ and $x_2$ in relation to $x_3$, because the latter is the reference group.
That will answer 2. and 3. Test the equality of $x_{1}$ and $x_2$ separately with an appropriate test which will ans... | Comparing (hidden) regression coefficients in simple linear regression?
My suggestion is to interpret the t-statistics of $x_1$ and $x_2$ in relation to $x_3$, because the latter is the reference group.
That will answer 2. and 3. Test the equality of $x_{1}$ and $x_2$ se |
39,450 | What does an error in ANOVA indicate? | Many models are based on a model for the dependent variable of the form "population mean + variation about the mean". Indeed, t-tests, one and two way ANOVA, multiple regression are all examples of this.
In the case of a two-way ANOVA with interaction, the model (in simplest terms) looks like this:
$$y_{ijk}=μ_{ij}+ε_{... | What does an error in ANOVA indicate? | Many models are based on a model for the dependent variable of the form "population mean + variation about the mean". Indeed, t-tests, one and two way ANOVA, multiple regression are all examples of th | What does an error in ANOVA indicate?
Many models are based on a model for the dependent variable of the form "population mean + variation about the mean". Indeed, t-tests, one and two way ANOVA, multiple regression are all examples of this.
In the case of a two-way ANOVA with interaction, the model (in simplest terms)... | What does an error in ANOVA indicate?
Many models are based on a model for the dependent variable of the form "population mean + variation about the mean". Indeed, t-tests, one and two way ANOVA, multiple regression are all examples of th |
39,451 | What does an error in ANOVA indicate? | I just wanted to add some information to @Glen_b's nice answer (+1). Perhaps, the OP knows that, but I would still clarify the terminology a little to the best of my knowledge/understanding.
$SS(error)$ represents error (residual) sum-of-squares and usually is referred to as $SSE$. Consequently, $df(error)$ represents ... | What does an error in ANOVA indicate? | I just wanted to add some information to @Glen_b's nice answer (+1). Perhaps, the OP knows that, but I would still clarify the terminology a little to the best of my knowledge/understanding.
$SS(error | What does an error in ANOVA indicate?
I just wanted to add some information to @Glen_b's nice answer (+1). Perhaps, the OP knows that, but I would still clarify the terminology a little to the best of my knowledge/understanding.
$SS(error)$ represents error (residual) sum-of-squares and usually is referred to as $SSE$.... | What does an error in ANOVA indicate?
I just wanted to add some information to @Glen_b's nice answer (+1). Perhaps, the OP knows that, but I would still clarify the terminology a little to the best of my knowledge/understanding.
$SS(error |
39,452 | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R? | In short, the discrepancy is because you're not doing it correctly.
The least squares estimates of regression coefficients (which are ML at the normal) are $(X'X)^{-1}X'y$, where $X$ consists of a column of 1's beside a column of the independent variable. (In practice, you don't actually compute the inverse.)
However,... | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R? | In short, the discrepancy is because you're not doing it correctly.
The least squares estimates of regression coefficients (which are ML at the normal) are $(X'X)^{-1}X'y$, where $X$ consists of a co | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R?
In short, the discrepancy is because you're not doing it correctly.
The least squares estimates of regression coefficients (which are ML at the normal) are $(X'X)^{-1}X'y$, where $X$ consists of a column of 1's beside a column of the... | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R?
In short, the discrepancy is because you're not doing it correctly.
The least squares estimates of regression coefficients (which are ML at the normal) are $(X'X)^{-1}X'y$, where $X$ consists of a co |
39,453 | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R? | Thanks a lot!
I understood the problem and your solution.
But I don't get how this could be computed using the $\textbf{X}$ 2xn matrix.
I don't know how to solve this for $\beta_0$ and $\beta_1$:
$$ \beta = \begin{bmatrix} \beta_0 \\ \beta_1 \end{bmatrix} = \mathbf{X}^T y [\mathbf{X}^T \mathbf{X}]^{-1}= \frac{\begin{bm... | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R? | Thanks a lot!
I understood the problem and your solution.
But I don't get how this could be computed using the $\textbf{X}$ 2xn matrix.
I don't know how to solve this for $\beta_0$ and $\beta_1$:
$$ \ | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R?
Thanks a lot!
I understood the problem and your solution.
But I don't get how this could be computed using the $\textbf{X}$ 2xn matrix.
I don't know how to solve this for $\beta_0$ and $\beta_1$:
$$ \beta = \begin{bmatrix} \beta_0 \\ ... | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R?
Thanks a lot!
I understood the problem and your solution.
But I don't get how this could be computed using the $\textbf{X}$ 2xn matrix.
I don't know how to solve this for $\beta_0$ and $\beta_1$:
$$ \ |
39,454 | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R? | The expression $\hat{\beta} = [X^TX]^{-1}X^Ty$ is a product of matrices. Matrix products mostly does not commute and the term $[X^TX]^{-1}$ is a matrix inverse. You can find information about this by doing a search on Linear Algebra.
Further the ones are always needed since the expression derives from $y = [1, x][\beta... | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R? | The expression $\hat{\beta} = [X^TX]^{-1}X^Ty$ is a product of matrices. Matrix products mostly does not commute and the term $[X^TX]^{-1}$ is a matrix inverse. You can find information about this by | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R?
The expression $\hat{\beta} = [X^TX]^{-1}X^Ty$ is a product of matrices. Matrix products mostly does not commute and the term $[X^TX]^{-1}$ is a matrix inverse. You can find information about this by doing a search on Linear Algebra.
... | Derivation of MLE of linear regression: and now? Why is there discrepancy to lm in R?
The expression $\hat{\beta} = [X^TX]^{-1}X^Ty$ is a product of matrices. Matrix products mostly does not commute and the term $[X^TX]^{-1}$ is a matrix inverse. You can find information about this by |
39,455 | generate a time series comprising seasonal, trend and remainder components in R | One possibility is to generate the data upon the state-space representation of the basic structural time series model described in Harvey (1989).
Harvey, A. C. (1989)
Forecasting, Structural Time Series Models and the Kalman Filter.
Cambridge University Press.
The basic structural model is defined as follows:
\... | generate a time series comprising seasonal, trend and remainder components in R | One possibility is to generate the data upon the state-space representation of the basic structural time series model described in Harvey (1989).
Harvey, A. C. (1989)
Forecasting, Structural Time | generate a time series comprising seasonal, trend and remainder components in R
One possibility is to generate the data upon the state-space representation of the basic structural time series model described in Harvey (1989).
Harvey, A. C. (1989)
Forecasting, Structural Time Series Models and the Kalman Filter.
... | generate a time series comprising seasonal, trend and remainder components in R
One possibility is to generate the data upon the state-space representation of the basic structural time series model described in Harvey (1989).
Harvey, A. C. (1989)
Forecasting, Structural Time |
39,456 | Why does Tableau's Box/Whisker plot show outliers automatically and how can I get rid of it? | The usual (and original) definition of a box and whisker plot does include outliers (indeed, Tukey had two kinds of outlying points, which these days are often not distinguished).
Specifically, the ends of the whiskers in the Tukey boxplot go at the nearest observations inside the inner fences, which are generally at t... | Why does Tableau's Box/Whisker plot show outliers automatically and how can I get rid of it? | The usual (and original) definition of a box and whisker plot does include outliers (indeed, Tukey had two kinds of outlying points, which these days are often not distinguished).
Specifically, the en | Why does Tableau's Box/Whisker plot show outliers automatically and how can I get rid of it?
The usual (and original) definition of a box and whisker plot does include outliers (indeed, Tukey had two kinds of outlying points, which these days are often not distinguished).
Specifically, the ends of the whiskers in the T... | Why does Tableau's Box/Whisker plot show outliers automatically and how can I get rid of it?
The usual (and original) definition of a box and whisker plot does include outliers (indeed, Tukey had two kinds of outlying points, which these days are often not distinguished).
Specifically, the en |
39,457 | Why does Tableau's Box/Whisker plot show outliers automatically and how can I get rid of it? | Tableau offers two options - Schematic box plot which is often referred to as Tukey box plot and skeletal box plot. Latter has whiskers extending from minimum to maximum. Former whiskers extending to the nearest data points within 1.5 IQR from the hinges. There is an option to toggle whether to show all points in the v... | Why does Tableau's Box/Whisker plot show outliers automatically and how can I get rid of it? | Tableau offers two options - Schematic box plot which is often referred to as Tukey box plot and skeletal box plot. Latter has whiskers extending from minimum to maximum. Former whiskers extending to | Why does Tableau's Box/Whisker plot show outliers automatically and how can I get rid of it?
Tableau offers two options - Schematic box plot which is often referred to as Tukey box plot and skeletal box plot. Latter has whiskers extending from minimum to maximum. Former whiskers extending to the nearest data points wit... | Why does Tableau's Box/Whisker plot show outliers automatically and how can I get rid of it?
Tableau offers two options - Schematic box plot which is often referred to as Tukey box plot and skeletal box plot. Latter has whiskers extending from minimum to maximum. Former whiskers extending to |
39,458 | Inconsistency between R and SAS for MLE on Weibull | First remind that the fitdistr function (from the MASS package) is a
very general function that can work with nearly any distribution.
The warnings come from non-allowed parameter values (e.g. negative
scale or shape) met during the optimisation unconstrained by default.
It seems a good idea here to try a specific MLE... | Inconsistency between R and SAS for MLE on Weibull | First remind that the fitdistr function (from the MASS package) is a
very general function that can work with nearly any distribution.
The warnings come from non-allowed parameter values (e.g. negati | Inconsistency between R and SAS for MLE on Weibull
First remind that the fitdistr function (from the MASS package) is a
very general function that can work with nearly any distribution.
The warnings come from non-allowed parameter values (e.g. negative
scale or shape) met during the optimisation unconstrained by defau... | Inconsistency between R and SAS for MLE on Weibull
First remind that the fitdistr function (from the MASS package) is a
very general function that can work with nearly any distribution.
The warnings come from non-allowed parameter values (e.g. negati |
39,459 | Inconsistency between R and SAS for MLE on Weibull | An optimization function shouldn't be expected to give identical answers to a similar function in a different package - or even to the same function with different options.
I tried a variety of different optimizers and starting places in fitdistr. They generally gave really similar results, of which the SAS and the res... | Inconsistency between R and SAS for MLE on Weibull | An optimization function shouldn't be expected to give identical answers to a similar function in a different package - or even to the same function with different options.
I tried a variety of differ | Inconsistency between R and SAS for MLE on Weibull
An optimization function shouldn't be expected to give identical answers to a similar function in a different package - or even to the same function with different options.
I tried a variety of different optimizers and starting places in fitdistr. They generally gave r... | Inconsistency between R and SAS for MLE on Weibull
An optimization function shouldn't be expected to give identical answers to a similar function in a different package - or even to the same function with different options.
I tried a variety of differ |
39,460 | Inconsistency between R and SAS for MLE on Weibull | While the SAS output is better than the R output, the unpleasant fact is that both perform rather poorly. To see this, note that the gradient at the reported solution should disappear to 0 ... whereas for both the R and SAS results, this is not the case.
In particular, let $X \sim Weibull(b,c)$ with pdf $f(x)$:
(sourc... | Inconsistency between R and SAS for MLE on Weibull | While the SAS output is better than the R output, the unpleasant fact is that both perform rather poorly. To see this, note that the gradient at the reported solution should disappear to 0 ... whereas | Inconsistency between R and SAS for MLE on Weibull
While the SAS output is better than the R output, the unpleasant fact is that both perform rather poorly. To see this, note that the gradient at the reported solution should disappear to 0 ... whereas for both the R and SAS results, this is not the case.
In particular,... | Inconsistency between R and SAS for MLE on Weibull
While the SAS output is better than the R output, the unpleasant fact is that both perform rather poorly. To see this, note that the gradient at the reported solution should disappear to 0 ... whereas |
39,461 | How to know if "best fit line" really represents known set of data? | I want to know if my line represents those datapoints
The problem with such an idea is there are many ways that a fitted model might be unrepresentative of the data. As a result, a single measure won't really capture the ways in which a model can fail to be representative.
This is why regression diagnostics consist ... | How to know if "best fit line" really represents known set of data? | I want to know if my line represents those datapoints
The problem with such an idea is there are many ways that a fitted model might be unrepresentative of the data. As a result, a single measure wo | How to know if "best fit line" really represents known set of data?
I want to know if my line represents those datapoints
The problem with such an idea is there are many ways that a fitted model might be unrepresentative of the data. As a result, a single measure won't really capture the ways in which a model can fai... | How to know if "best fit line" really represents known set of data?
I want to know if my line represents those datapoints
The problem with such an idea is there are many ways that a fitted model might be unrepresentative of the data. As a result, a single measure wo |
39,462 | How to know if "best fit line" really represents known set of data? | There are a several ways you could do this. First recall that the linear best fit line is the line which minimizes the sum of squared residuals (see least squares):
$$\sum_{i=1}^{n}{r_i^2}$$
where $r_i$ is the residual for data point $i$, and $n$ is the number of data points. A residual is the distance between a point ... | How to know if "best fit line" really represents known set of data? | There are a several ways you could do this. First recall that the linear best fit line is the line which minimizes the sum of squared residuals (see least squares):
$$\sum_{i=1}^{n}{r_i^2}$$
where $r_ | How to know if "best fit line" really represents known set of data?
There are a several ways you could do this. First recall that the linear best fit line is the line which minimizes the sum of squared residuals (see least squares):
$$\sum_{i=1}^{n}{r_i^2}$$
where $r_i$ is the residual for data point $i$, and $n$ is th... | How to know if "best fit line" really represents known set of data?
There are a several ways you could do this. First recall that the linear best fit line is the line which minimizes the sum of squared residuals (see least squares):
$$\sum_{i=1}^{n}{r_i^2}$$
where $r_ |
39,463 | From SAS to R - what are "must" packages for reporting | The problem with R is that there are so many ways to construct great reports, and so many R packages that are helpful for this task. One approach, though getting out of date, is shown in http://biostat.app.vumc.org/wiki/pub/Main/StatReport/summary.pdf . Note that some of the functions there have been updated as shown ... | From SAS to R - what are "must" packages for reporting | The problem with R is that there are so many ways to construct great reports, and so many R packages that are helpful for this task. One approach, though getting out of date, is shown in http://biost | From SAS to R - what are "must" packages for reporting
The problem with R is that there are so many ways to construct great reports, and so many R packages that are helpful for this task. One approach, though getting out of date, is shown in http://biostat.app.vumc.org/wiki/pub/Main/StatReport/summary.pdf . Note that ... | From SAS to R - what are "must" packages for reporting
The problem with R is that there are so many ways to construct great reports, and so many R packages that are helpful for this task. One approach, though getting out of date, is shown in http://biost |
39,464 | From SAS to R - what are "must" packages for reporting | It seems like reporting and filtering/splitting data by variables are two orthogonal tasks. And people usually use different packages for those.
For managing the data there are few really popular packages: dplyr and data.table.
For reporting tables one package that stands out for me is stargazer
Here are some demonstra... | From SAS to R - what are "must" packages for reporting | It seems like reporting and filtering/splitting data by variables are two orthogonal tasks. And people usually use different packages for those.
For managing the data there are few really popular pack | From SAS to R - what are "must" packages for reporting
It seems like reporting and filtering/splitting data by variables are two orthogonal tasks. And people usually use different packages for those.
For managing the data there are few really popular packages: dplyr and data.table.
For reporting tables one package that... | From SAS to R - what are "must" packages for reporting
It seems like reporting and filtering/splitting data by variables are two orthogonal tasks. And people usually use different packages for those.
For managing the data there are few really popular pack |
39,465 | From SAS to R - what are "must" packages for reporting | In 2020 the reporter package was released, which operates much like proc report. You get the data and statistics you want using other R packages, and then send the resulting data frame into reporter. Like this:
library(reporter)
# Create temporary path
tmp <- file.path(tempdir(), "example3.pdf")
# Read in prepared ... | From SAS to R - what are "must" packages for reporting | In 2020 the reporter package was released, which operates much like proc report. You get the data and statistics you want using other R packages, and then send the resulting data frame into reporter. | From SAS to R - what are "must" packages for reporting
In 2020 the reporter package was released, which operates much like proc report. You get the data and statistics you want using other R packages, and then send the resulting data frame into reporter. Like this:
library(reporter)
# Create temporary path
tmp <- fi... | From SAS to R - what are "must" packages for reporting
In 2020 the reporter package was released, which operates much like proc report. You get the data and statistics you want using other R packages, and then send the resulting data frame into reporter. |
39,466 | How to infer correlations from correlations | Given corr(A,B) and corr(A,C) you can obtain bounds on corr(B,C) (and similar such calculations involving more variables), but the bounds are in general quite wide. Indeed, typically such calculations aren't very informative at all.
Specifically, by looking at the relationship between the ordinary pairwise correlation ... | How to infer correlations from correlations | Given corr(A,B) and corr(A,C) you can obtain bounds on corr(B,C) (and similar such calculations involving more variables), but the bounds are in general quite wide. Indeed, typically such calculations | How to infer correlations from correlations
Given corr(A,B) and corr(A,C) you can obtain bounds on corr(B,C) (and similar such calculations involving more variables), but the bounds are in general quite wide. Indeed, typically such calculations aren't very informative at all.
Specifically, by looking at the relationshi... | How to infer correlations from correlations
Given corr(A,B) and corr(A,C) you can obtain bounds on corr(B,C) (and similar such calculations involving more variables), but the bounds are in general quite wide. Indeed, typically such calculations |
39,467 | Rstan Stan model for a simple mixture of normals | Because all of the parameters of this distribution are known, and we merely want to draw samples from this distribution, coding the model in rstan is straightforward. Note that this is, by far, one of the least efficient paths to sampling from this particular model, in terms of time that I spent coding it (15 minutes).... | Rstan Stan model for a simple mixture of normals | Because all of the parameters of this distribution are known, and we merely want to draw samples from this distribution, coding the model in rstan is straightforward. Note that this is, by far, one of | Rstan Stan model for a simple mixture of normals
Because all of the parameters of this distribution are known, and we merely want to draw samples from this distribution, coding the model in rstan is straightforward. Note that this is, by far, one of the least efficient paths to sampling from this particular model, in t... | Rstan Stan model for a simple mixture of normals
Because all of the parameters of this distribution are known, and we merely want to draw samples from this distribution, coding the model in rstan is straightforward. Note that this is, by far, one of |
39,468 | Forecasting a seasonal time series in R | As regards the comparison of models, the idea proposed by @forecaster can be helpful since you have a relatively long series.
Regarding your last question about how to obtain forecasts for the trend component rather than for the whole series: as far as I know there is no package on CRAN that decomposes a fitted ARIMA m... | Forecasting a seasonal time series in R | As regards the comparison of models, the idea proposed by @forecaster can be helpful since you have a relatively long series.
Regarding your last question about how to obtain forecasts for the trend c | Forecasting a seasonal time series in R
As regards the comparison of models, the idea proposed by @forecaster can be helpful since you have a relatively long series.
Regarding your last question about how to obtain forecasts for the trend component rather than for the whole series: as far as I know there is no package ... | Forecasting a seasonal time series in R
As regards the comparison of models, the idea proposed by @forecaster can be helpful since you have a relatively long series.
Regarding your last question about how to obtain forecasts for the trend c |
39,469 | Exact central confidence interval for a correlation | You can get the same values with:
cor.test(cd4$baseline, cd4$oneyear, method = "pearson", conf.level = 0.9)
The method used to obtain such interval is explained in:
http://stat.ethz.ch/R-manual/R-patched/library/stats/html/cor.test.html
If method is "pearson", the test statistic is based on Pearson's product moment c... | Exact central confidence interval for a correlation | You can get the same values with:
cor.test(cd4$baseline, cd4$oneyear, method = "pearson", conf.level = 0.9)
The method used to obtain such interval is explained in:
http://stat.ethz.ch/R-manual/R-pat | Exact central confidence interval for a correlation
You can get the same values with:
cor.test(cd4$baseline, cd4$oneyear, method = "pearson", conf.level = 0.9)
The method used to obtain such interval is explained in:
http://stat.ethz.ch/R-manual/R-patched/library/stats/html/cor.test.html
If method is "pearson", the t... | Exact central confidence interval for a correlation
You can get the same values with:
cor.test(cd4$baseline, cd4$oneyear, method = "pearson", conf.level = 0.9)
The method used to obtain such interval is explained in:
http://stat.ethz.ch/R-manual/R-pat |
39,470 | Exact central confidence interval for a correlation | You already mentioned that the Fisher transformation is not correct in your code. You first have to transform r to a z value (atanh part), then you add and subtract the standard error with the appropriate multiplier to get the correct confidence (as you did correctly). Finally, you have to transform the whole thing bac... | Exact central confidence interval for a correlation | You already mentioned that the Fisher transformation is not correct in your code. You first have to transform r to a z value (atanh part), then you add and subtract the standard error with the appropr | Exact central confidence interval for a correlation
You already mentioned that the Fisher transformation is not correct in your code. You first have to transform r to a z value (atanh part), then you add and subtract the standard error with the appropriate multiplier to get the correct confidence (as you did correctly)... | Exact central confidence interval for a correlation
You already mentioned that the Fisher transformation is not correct in your code. You first have to transform r to a z value (atanh part), then you add and subtract the standard error with the appropr |
39,471 | Nonlinear regression | You have several problems there. The biggest problems are easiest to see if you reparameterize your fitted function from
$y = a/(1 + b x^{-c})$
to
$y = 1/(\frac{1}{a} + \frac{b}{a} x^{-c})$
$\quad = 1/(a_1 + b_1 x^{-c})$
(this gives the same model fit, just some of the parameters are different from your expression of t... | Nonlinear regression | You have several problems there. The biggest problems are easiest to see if you reparameterize your fitted function from
$y = a/(1 + b x^{-c})$
to
$y = 1/(\frac{1}{a} + \frac{b}{a} x^{-c})$
$\quad = 1 | Nonlinear regression
You have several problems there. The biggest problems are easiest to see if you reparameterize your fitted function from
$y = a/(1 + b x^{-c})$
to
$y = 1/(\frac{1}{a} + \frac{b}{a} x^{-c})$
$\quad = 1/(a_1 + b_1 x^{-c})$
(this gives the same model fit, just some of the parameters are different from... | Nonlinear regression
You have several problems there. The biggest problems are easiest to see if you reparameterize your fitted function from
$y = a/(1 + b x^{-c})$
to
$y = 1/(\frac{1}{a} + \frac{b}{a} x^{-c})$
$\quad = 1 |
39,472 | Logistic regression: maximum likelihood vs misclassification | One could possible estimate the logistic regression model by minimizing the classification error, but there is usually no reason why to do so! Why do you want to do it?
But, such questions have been asked here before, so I will not rewrite an answer, very good answers can be found to this question: Logistic regressio... | Logistic regression: maximum likelihood vs misclassification | One could possible estimate the logistic regression model by minimizing the classification error, but there is usually no reason why to do so! Why do you want to do it?
But, such questions have been | Logistic regression: maximum likelihood vs misclassification
One could possible estimate the logistic regression model by minimizing the classification error, but there is usually no reason why to do so! Why do you want to do it?
But, such questions have been asked here before, so I will not rewrite an answer, very go... | Logistic regression: maximum likelihood vs misclassification
One could possible estimate the logistic regression model by minimizing the classification error, but there is usually no reason why to do so! Why do you want to do it?
But, such questions have been |
39,473 | Logistic regression: maximum likelihood vs misclassification | In a word, yes, but it wouldn't be logistic regression anymore.
The logistic regression loss function (i.e., the negative log likelihood), is essentially a regression in the log odds. Changing that to a least squares loss function will make it linear regression, which will lose three things: (i) the interpretation of t... | Logistic regression: maximum likelihood vs misclassification | In a word, yes, but it wouldn't be logistic regression anymore.
The logistic regression loss function (i.e., the negative log likelihood), is essentially a regression in the log odds. Changing that to | Logistic regression: maximum likelihood vs misclassification
In a word, yes, but it wouldn't be logistic regression anymore.
The logistic regression loss function (i.e., the negative log likelihood), is essentially a regression in the log odds. Changing that to a least squares loss function will make it linear regressi... | Logistic regression: maximum likelihood vs misclassification
In a word, yes, but it wouldn't be logistic regression anymore.
The logistic regression loss function (i.e., the negative log likelihood), is essentially a regression in the log odds. Changing that to |
39,474 | Logistic regression: maximum likelihood vs misclassification | I believe you can reduce the probability of misclassification by introducing a class of random effects mixing distributions. Using this approach, you can develop a full-likelihood model, which includes effects of misclassification [Ref 1, 2, 3].
However, there are two major challenges in applying this modelling approa... | Logistic regression: maximum likelihood vs misclassification | I believe you can reduce the probability of misclassification by introducing a class of random effects mixing distributions. Using this approach, you can develop a full-likelihood model, which include | Logistic regression: maximum likelihood vs misclassification
I believe you can reduce the probability of misclassification by introducing a class of random effects mixing distributions. Using this approach, you can develop a full-likelihood model, which includes effects of misclassification [Ref 1, 2, 3].
However, the... | Logistic regression: maximum likelihood vs misclassification
I believe you can reduce the probability of misclassification by introducing a class of random effects mixing distributions. Using this approach, you can develop a full-likelihood model, which include |
39,475 | Gibbs sampling how to sample from the conditional probability? Bayesian model | Since we are calculating the joint distribution, we'll assume that our initial sample is $ x = P(D=0,I=0,G=0,L=0,S=0) $ .
To calculate the next sample, we'll need to sample each variable from the conditional distribution.
$ P(D\mid G,I,S,L) $,from the conditional independencies in the Bayes net, simplifies to just sa... | Gibbs sampling how to sample from the conditional probability? Bayesian model | Since we are calculating the joint distribution, we'll assume that our initial sample is $ x = P(D=0,I=0,G=0,L=0,S=0) $ .
To calculate the next sample, we'll need to sample each variable from the cond | Gibbs sampling how to sample from the conditional probability? Bayesian model
Since we are calculating the joint distribution, we'll assume that our initial sample is $ x = P(D=0,I=0,G=0,L=0,S=0) $ .
To calculate the next sample, we'll need to sample each variable from the conditional distribution.
$ P(D\mid G,I,S,L)... | Gibbs sampling how to sample from the conditional probability? Bayesian model
Since we are calculating the joint distribution, we'll assume that our initial sample is $ x = P(D=0,I=0,G=0,L=0,S=0) $ .
To calculate the next sample, we'll need to sample each variable from the cond |
39,476 | Gibbs sampling how to sample from the conditional probability? Bayesian model | If the conditional distribution cannot be generated directly from standard random generators, you can apply a Metroplolis-Hastings schema within the Gibbs sampler.
draw one sample, $D_t$, from $P(D|G_{t-1},I_{t-1},S_{t-1},L_{t-1})$ using Metropolis sampler with initial value $D_{t-1}$
draw one sample, $G_t$, from $P(G... | Gibbs sampling how to sample from the conditional probability? Bayesian model | If the conditional distribution cannot be generated directly from standard random generators, you can apply a Metroplolis-Hastings schema within the Gibbs sampler.
draw one sample, $D_t$, from $P(D|G | Gibbs sampling how to sample from the conditional probability? Bayesian model
If the conditional distribution cannot be generated directly from standard random generators, you can apply a Metroplolis-Hastings schema within the Gibbs sampler.
draw one sample, $D_t$, from $P(D|G_{t-1},I_{t-1},S_{t-1},L_{t-1})$ using Met... | Gibbs sampling how to sample from the conditional probability? Bayesian model
If the conditional distribution cannot be generated directly from standard random generators, you can apply a Metroplolis-Hastings schema within the Gibbs sampler.
draw one sample, $D_t$, from $P(D|G |
39,477 | Comparing estimators of location of the Cauchy distribution | Cauchy distributions have infinite mean and infinite variance. Because of this fact, the laws of large numbers and central limit theorems does not apply.
This demonstration is designed to give some intuition about what happens as you add additional observations to your sample when those samples are Cauchy. Eventually, ... | Comparing estimators of location of the Cauchy distribution | Cauchy distributions have infinite mean and infinite variance. Because of this fact, the laws of large numbers and central limit theorems does not apply.
This demonstration is designed to give some in | Comparing estimators of location of the Cauchy distribution
Cauchy distributions have infinite mean and infinite variance. Because of this fact, the laws of large numbers and central limit theorems does not apply.
This demonstration is designed to give some intuition about what happens as you add additional observation... | Comparing estimators of location of the Cauchy distribution
Cauchy distributions have infinite mean and infinite variance. Because of this fact, the laws of large numbers and central limit theorems does not apply.
This demonstration is designed to give some in |
39,478 | Comparing estimators of location of the Cauchy distribution | You can also estimate the location of the Cauchy using heavy-tail(s) Lambert W x F distributions (Disclaimer: I am the author.) since both are symmetric around $c$ (location of Cauchy) and $\mu_x$ (mean of the input X ~ F), respectively. In fact, I give an example of estimating the location of a Cauchy in the paper and... | Comparing estimators of location of the Cauchy distribution | You can also estimate the location of the Cauchy using heavy-tail(s) Lambert W x F distributions (Disclaimer: I am the author.) since both are symmetric around $c$ (location of Cauchy) and $\mu_x$ (me | Comparing estimators of location of the Cauchy distribution
You can also estimate the location of the Cauchy using heavy-tail(s) Lambert W x F distributions (Disclaimer: I am the author.) since both are symmetric around $c$ (location of Cauchy) and $\mu_x$ (mean of the input X ~ F), respectively. In fact, I give an exa... | Comparing estimators of location of the Cauchy distribution
You can also estimate the location of the Cauchy using heavy-tail(s) Lambert W x F distributions (Disclaimer: I am the author.) since both are symmetric around $c$ (location of Cauchy) and $\mu_x$ (me |
39,479 | Reporting from a likelihood ratio test | For a likelihood ratio test, the degrees of freedom are equal to the difference in number of parameters for the two models. In this case, df = 1, and so $\chi^2(1)=11.96$, $p=0.0005$. | Reporting from a likelihood ratio test | For a likelihood ratio test, the degrees of freedom are equal to the difference in number of parameters for the two models. In this case, df = 1, and so $\chi^2(1)=11.96$, $p=0.0005$. | Reporting from a likelihood ratio test
For a likelihood ratio test, the degrees of freedom are equal to the difference in number of parameters for the two models. In this case, df = 1, and so $\chi^2(1)=11.96$, $p=0.0005$. | Reporting from a likelihood ratio test
For a likelihood ratio test, the degrees of freedom are equal to the difference in number of parameters for the two models. In this case, df = 1, and so $\chi^2(1)=11.96$, $p=0.0005$. |
39,480 | Invariance property of maximum likelihood estimator? | You seem to be confusing what it means for a parameter transformation to occur. In general, the values of the likelihood functions do not change. To illustrate, let $L(\theta; x)$ be a likelihood function and let $\lambda = g(\theta)$ where $g$ is one-to-one. Then the likelihood function parameterized in terms of $\lam... | Invariance property of maximum likelihood estimator? | You seem to be confusing what it means for a parameter transformation to occur. In general, the values of the likelihood functions do not change. To illustrate, let $L(\theta; x)$ be a likelihood func | Invariance property of maximum likelihood estimator?
You seem to be confusing what it means for a parameter transformation to occur. In general, the values of the likelihood functions do not change. To illustrate, let $L(\theta; x)$ be a likelihood function and let $\lambda = g(\theta)$ where $g$ is one-to-one. Then th... | Invariance property of maximum likelihood estimator?
You seem to be confusing what it means for a parameter transformation to occur. In general, the values of the likelihood functions do not change. To illustrate, let $L(\theta; x)$ be a likelihood func |
39,481 | Bayesian regularized NNs over classical NNs | The key problem with neural nets tends to be preventing over-fitting. Bayesian regularisation (which restricts the magnitude of the weights) is one approach to this, structural stabilisation (i.e. restricting the number of hidden nodes and/or weights is another). Neither approach is a panacea, and generally a combina... | Bayesian regularized NNs over classical NNs | The key problem with neural nets tends to be preventing over-fitting. Bayesian regularisation (which restricts the magnitude of the weights) is one approach to this, structural stabilisation (i.e. re | Bayesian regularized NNs over classical NNs
The key problem with neural nets tends to be preventing over-fitting. Bayesian regularisation (which restricts the magnitude of the weights) is one approach to this, structural stabilisation (i.e. restricting the number of hidden nodes and/or weights is another). Neither ap... | Bayesian regularized NNs over classical NNs
The key problem with neural nets tends to be preventing over-fitting. Bayesian regularisation (which restricts the magnitude of the weights) is one approach to this, structural stabilisation (i.e. re |
39,482 | Bayesian regularized NNs over classical NNs | You use BRANNs for the same purposes as regular ANNs, typically classification and regression. As Dikran Marsupial says, the are better because they are more robust against overfitting, and allows you to work with higher number of neurons without running into overfitting. Besides, it provides you with error bars on the... | Bayesian regularized NNs over classical NNs | You use BRANNs for the same purposes as regular ANNs, typically classification and regression. As Dikran Marsupial says, the are better because they are more robust against overfitting, and allows you | Bayesian regularized NNs over classical NNs
You use BRANNs for the same purposes as regular ANNs, typically classification and regression. As Dikran Marsupial says, the are better because they are more robust against overfitting, and allows you to work with higher number of neurons without running into overfitting. Bes... | Bayesian regularized NNs over classical NNs
You use BRANNs for the same purposes as regular ANNs, typically classification and regression. As Dikran Marsupial says, the are better because they are more robust against overfitting, and allows you |
39,483 | Are two Pearson correlation coefficients different? | Just in case that someone (else) has to perform a comparison of correlation coefficients on multiple pairs of variables, here's a function based on rg255's helpful reply to copy:
cor.diff.test = function(x1, x2, y1, y2, method="pearson") {
cor1 = cor.test(x1, x2, method=method)
cor2 = cor.test(y1, y2, method=method... | Are two Pearson correlation coefficients different? | Just in case that someone (else) has to perform a comparison of correlation coefficients on multiple pairs of variables, here's a function based on rg255's helpful reply to copy:
cor.diff.test = funct | Are two Pearson correlation coefficients different?
Just in case that someone (else) has to perform a comparison of correlation coefficients on multiple pairs of variables, here's a function based on rg255's helpful reply to copy:
cor.diff.test = function(x1, x2, y1, y2, method="pearson") {
cor1 = cor.test(x1, x2, me... | Are two Pearson correlation coefficients different?
Just in case that someone (else) has to perform a comparison of correlation coefficients on multiple pairs of variables, here's a function based on rg255's helpful reply to copy:
cor.diff.test = funct |
39,484 | Are two Pearson correlation coefficients different? | Once the Fisher's z transformations are done it is just a case of obtaining p-values
# Correlations
cor.test (df1$a, df1$b, method = "p")
cor.test (df2$a, df2$b, method = "p")
# function to do fisher transformations
fisher.z<- function (r1,r2,n1,n2) ((0.5*log((1+r1)/(1-r1)))-(0.5*log((1+r2)/(1-r2))))/((1/(n1-3))+... | Are two Pearson correlation coefficients different? | Once the Fisher's z transformations are done it is just a case of obtaining p-values
# Correlations
cor.test (df1$a, df1$b, method = "p")
cor.test (df2$a, df2$b, method = "p")
# function to do fi | Are two Pearson correlation coefficients different?
Once the Fisher's z transformations are done it is just a case of obtaining p-values
# Correlations
cor.test (df1$a, df1$b, method = "p")
cor.test (df2$a, df2$b, method = "p")
# function to do fisher transformations
fisher.z<- function (r1,r2,n1,n2) ((0.5*log((1... | Are two Pearson correlation coefficients different?
Once the Fisher's z transformations are done it is just a case of obtaining p-values
# Correlations
cor.test (df1$a, df1$b, method = "p")
cor.test (df2$a, df2$b, method = "p")
# function to do fi |
39,485 | Are two Pearson correlation coefficients different? | In case people are still looking for an easy way to compare two $r$ Pearson correlations.
There is a function called paired.r in the package psych for R exactly for that.
Usage:
paired.r(xy, xz, yz=NULL, n, n2=NULL,twotailed=TRUE)
Or as a simple example: For r1 and r2 and a sample size of n just do:
paired.r(r1,r2,n=... | Are two Pearson correlation coefficients different? | In case people are still looking for an easy way to compare two $r$ Pearson correlations.
There is a function called paired.r in the package psych for R exactly for that.
Usage:
paired.r(xy, xz, yz=N | Are two Pearson correlation coefficients different?
In case people are still looking for an easy way to compare two $r$ Pearson correlations.
There is a function called paired.r in the package psych for R exactly for that.
Usage:
paired.r(xy, xz, yz=NULL, n, n2=NULL,twotailed=TRUE)
Or as a simple example: For r1 and ... | Are two Pearson correlation coefficients different?
In case people are still looking for an easy way to compare two $r$ Pearson correlations.
There is a function called paired.r in the package psych for R exactly for that.
Usage:
paired.r(xy, xz, yz=N |
39,486 | How to check the features which are selected by LASSO | Use the coef function on the glmnet model.
You will need to choose a lambda value, as different lambdas will give you different feature sets. Typically this is done through cross-validation.
/edit: For example, using cv.glment:
library(glmnet)
x <- model.matrix(Sepal.Length~., iris)[,-1]
y <- iris$Sepal.Length
mod <- ... | How to check the features which are selected by LASSO | Use the coef function on the glmnet model.
You will need to choose a lambda value, as different lambdas will give you different feature sets. Typically this is done through cross-validation.
/edit: F | How to check the features which are selected by LASSO
Use the coef function on the glmnet model.
You will need to choose a lambda value, as different lambdas will give you different feature sets. Typically this is done through cross-validation.
/edit: For example, using cv.glment:
library(glmnet)
x <- model.matrix(Sep... | How to check the features which are selected by LASSO
Use the coef function on the glmnet model.
You will need to choose a lambda value, as different lambdas will give you different feature sets. Typically this is done through cross-validation.
/edit: F |
39,487 | Covariance Matrices | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
Look at the corrplot package for R. It has several op... | Covariance Matrices | Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
| Covariance Matrices
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
Look at the corrplot package for R... | Covariance Matrices
Want to improve this post? Provide detailed answers to this question, including citations and an explanation of why your answer is correct. Answers without enough detail may be edited or deleted.
|
39,488 | Covariance Matrices | I'm not sure about any R packages that do this, but my favourite approach (and I feel a much more informative approach) to visualization of covariance matrices are using correlation networks. Basically, I trim the covariance matrix using the Glasso algorithm (I'll explain why later), then use a force-directed algorithm... | Covariance Matrices | I'm not sure about any R packages that do this, but my favourite approach (and I feel a much more informative approach) to visualization of covariance matrices are using correlation networks. Basicall | Covariance Matrices
I'm not sure about any R packages that do this, but my favourite approach (and I feel a much more informative approach) to visualization of covariance matrices are using correlation networks. Basically, I trim the covariance matrix using the Glasso algorithm (I'll explain why later), then use a forc... | Covariance Matrices
I'm not sure about any R packages that do this, but my favourite approach (and I feel a much more informative approach) to visualization of covariance matrices are using correlation networks. Basicall |
39,489 | Covariance Matrices | If you're using shiny and have the ability to make interactive tools, then just develop the covariance plot as a ggplot object so that you can add hover, dblclick, and brush tools for zooming and showing pop-up information.
Here's an intro to creating different correlation like plots in R with ggplot2: GGPlot2 CorrMat... | Covariance Matrices | If you're using shiny and have the ability to make interactive tools, then just develop the covariance plot as a ggplot object so that you can add hover, dblclick, and brush tools for zooming and show | Covariance Matrices
If you're using shiny and have the ability to make interactive tools, then just develop the covariance plot as a ggplot object so that you can add hover, dblclick, and brush tools for zooming and showing pop-up information.
Here's an intro to creating different correlation like plots in R with ggpl... | Covariance Matrices
If you're using shiny and have the ability to make interactive tools, then just develop the covariance plot as a ggplot object so that you can add hover, dblclick, and brush tools for zooming and show |
39,490 | Can I probe cross-level interactions without random slope in a hierarchical linear model? | Having random slopes at level 1 is not a necessary condition for examining cross-level interactions. All that is necessary is that you have 2 predictors that vary at different levels, and their interaction.
EDIT: I looked over the Hofmann paper posted in the comments and I think I see the source of confusion here.
Hofm... | Can I probe cross-level interactions without random slope in a hierarchical linear model? | Having random slopes at level 1 is not a necessary condition for examining cross-level interactions. All that is necessary is that you have 2 predictors that vary at different levels, and their intera | Can I probe cross-level interactions without random slope in a hierarchical linear model?
Having random slopes at level 1 is not a necessary condition for examining cross-level interactions. All that is necessary is that you have 2 predictors that vary at different levels, and their interaction.
EDIT: I looked over the... | Can I probe cross-level interactions without random slope in a hierarchical linear model?
Having random slopes at level 1 is not a necessary condition for examining cross-level interactions. All that is necessary is that you have 2 predictors that vary at different levels, and their intera |
39,491 | Why are these proper kernels and how to deduce that they are? | In a machine learning context (i.e. "kernel methods"), the key requirement for a kernel is that it must be symmetric and positive-definite, that is, if $K$ is a kernel matrix, then for any (column) vector $x$ of the appropriate length, $x^{T}Kx$ must be a positive real number. This restriction is in place mostly due to... | Why are these proper kernels and how to deduce that they are? | In a machine learning context (i.e. "kernel methods"), the key requirement for a kernel is that it must be symmetric and positive-definite, that is, if $K$ is a kernel matrix, then for any (column) ve | Why are these proper kernels and how to deduce that they are?
In a machine learning context (i.e. "kernel methods"), the key requirement for a kernel is that it must be symmetric and positive-definite, that is, if $K$ is a kernel matrix, then for any (column) vector $x$ of the appropriate length, $x^{T}Kx$ must be a po... | Why are these proper kernels and how to deduce that they are?
In a machine learning context (i.e. "kernel methods"), the key requirement for a kernel is that it must be symmetric and positive-definite, that is, if $K$ is a kernel matrix, then for any (column) ve |
39,492 | Why are these proper kernels and how to deduce that they are? | One easy way is to think of a kernel function as a positive-semidefinite (PSD) matrix. Then you can use PSD tricks.
For example, if $K$ is PSD, then $-K$ is NSD and thus not PSD. On the other hand, if $K$ is PSD, $c K$ is PSD if $c$ is a scalar and $c > 0$. Thus, $K_4$ and $K_3$ are not generally kernels and $K_5$ is. | Why are these proper kernels and how to deduce that they are? | One easy way is to think of a kernel function as a positive-semidefinite (PSD) matrix. Then you can use PSD tricks.
For example, if $K$ is PSD, then $-K$ is NSD and thus not PSD. On the other hand, if | Why are these proper kernels and how to deduce that they are?
One easy way is to think of a kernel function as a positive-semidefinite (PSD) matrix. Then you can use PSD tricks.
For example, if $K$ is PSD, then $-K$ is NSD and thus not PSD. On the other hand, if $K$ is PSD, $c K$ is PSD if $c$ is a scalar and $c > 0$.... | Why are these proper kernels and how to deduce that they are?
One easy way is to think of a kernel function as a positive-semidefinite (PSD) matrix. Then you can use PSD tricks.
For example, if $K$ is PSD, then $-K$ is NSD and thus not PSD. On the other hand, if |
39,493 | Difference between a 2 factor ANOVA and mixed effects model | I'm absolutely not a specialist, but this is my contribution:
In your ANOVA model, you treated both 'recipe' and 'temperature' as fixed factors, which can be thought of in terms of differences.
In your linear mixed model, you treated 'temperature' as a random factor, which is defined by a distribution and whose valu... | Difference between a 2 factor ANOVA and mixed effects model | I'm absolutely not a specialist, but this is my contribution:
In your ANOVA model, you treated both 'recipe' and 'temperature' as fixed factors, which can be thought of in terms of differences.
In | Difference between a 2 factor ANOVA and mixed effects model
I'm absolutely not a specialist, but this is my contribution:
In your ANOVA model, you treated both 'recipe' and 'temperature' as fixed factors, which can be thought of in terms of differences.
In your linear mixed model, you treated 'temperature' as a rand... | Difference between a 2 factor ANOVA and mixed effects model
I'm absolutely not a specialist, but this is my contribution:
In your ANOVA model, you treated both 'recipe' and 'temperature' as fixed factors, which can be thought of in terms of differences.
In |
39,494 | Difference between a 2 factor ANOVA and mixed effects model | Very briefly: In a two factor ANOVA (or, more generally, in a model that can be analyzed with lm in R) variables are controlled for. That is, it asks "Holding other independent variables constant, what is the linear relationship of each independent variable with the dependent variable?" Such models have a number of ass... | Difference between a 2 factor ANOVA and mixed effects model | Very briefly: In a two factor ANOVA (or, more generally, in a model that can be analyzed with lm in R) variables are controlled for. That is, it asks "Holding other independent variables constant, wha | Difference between a 2 factor ANOVA and mixed effects model
Very briefly: In a two factor ANOVA (or, more generally, in a model that can be analyzed with lm in R) variables are controlled for. That is, it asks "Holding other independent variables constant, what is the linear relationship of each independent variable wi... | Difference between a 2 factor ANOVA and mixed effects model
Very briefly: In a two factor ANOVA (or, more generally, in a model that can be analyzed with lm in R) variables are controlled for. That is, it asks "Holding other independent variables constant, wha |
39,495 | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in particular)? | I don't think the currently accepted answer is correct. What it describes is identification, not standardization. The unstandardized coefficients is what comes directly out of the estimation procedure. The standardized coefficients recast regression coefficients and covariance in metrics of correlations, and unique var... | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in | I don't think the currently accepted answer is correct. What it describes is identification, not standardization. The unstandardized coefficients is what comes directly out of the estimation procedure | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in particular)?
I don't think the currently accepted answer is correct. What it describes is identification, not standardization. The unstandardized coefficients is what comes directly out of the estimation procedure. The ... | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in
I don't think the currently accepted answer is correct. What it describes is identification, not standardization. The unstandardized coefficients is what comes directly out of the estimation procedure |
39,496 | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in particular)? | Two weeks later, I see that no one has answered the question. However extensive Google searching revealed the following.
Assume multivariate data where the $p$ variables supposedly indicate the presence of one or more underlying factors. A simple measurement model involves relationships of the following type:
$$x_i=\la... | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in | Two weeks later, I see that no one has answered the question. However extensive Google searching revealed the following.
Assume multivariate data where the $p$ variables supposedly indicate the presen | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in particular)?
Two weeks later, I see that no one has answered the question. However extensive Google searching revealed the following.
Assume multivariate data where the $p$ variables supposedly indicate the presence of ... | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in
Two weeks later, I see that no one has answered the question. However extensive Google searching revealed the following.
Assume multivariate data where the $p$ variables supposedly indicate the presen |
39,497 | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in particular)? | I think from my simple experience in Amos that unstandardized we request one of the parameters should be 1 to give as the result of factor loading to help determine which parameter needs to dropped first and to keep many parameter that need to be estimated.
According to standardized: Here we got the real value of param... | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in | I think from my simple experience in Amos that unstandardized we request one of the parameters should be 1 to give as the result of factor loading to help determine which parameter needs to dropped fi | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in particular)?
I think from my simple experience in Amos that unstandardized we request one of the parameters should be 1 to give as the result of factor loading to help determine which parameter needs to dropped first an... | What is the difference between standardized and unstandardized estimates in SEM (thinking of AMOS in
I think from my simple experience in Amos that unstandardized we request one of the parameters should be 1 to give as the result of factor loading to help determine which parameter needs to dropped fi |
39,498 | How to detect outliers in skewed data set? | Bottom line is that the decision to remove data from your dataset is a subject-matter decision, not a statistical decision. The statistics help you to identify outliers given what you believe about the dataset.
A very readable applied treatment of outliers is given in
B. Iglewicz and D. C. Hoaglin, How to Detect and H... | How to detect outliers in skewed data set? | Bottom line is that the decision to remove data from your dataset is a subject-matter decision, not a statistical decision. The statistics help you to identify outliers given what you believe about th | How to detect outliers in skewed data set?
Bottom line is that the decision to remove data from your dataset is a subject-matter decision, not a statistical decision. The statistics help you to identify outliers given what you believe about the dataset.
A very readable applied treatment of outliers is given in
B. Igle... | How to detect outliers in skewed data set?
Bottom line is that the decision to remove data from your dataset is a subject-matter decision, not a statistical decision. The statistics help you to identify outliers given what you believe about th |
39,499 | How to detect outliers in skewed data set? | Flagging outlier is not a subject-matter decision but a statistical one. Outliers have a precise, objective definition: they are observations that do not follow the pattern of the majority of the data. Such observations need to be set apart at the onset of any analysis simply because their distance from the bulk of the... | How to detect outliers in skewed data set? | Flagging outlier is not a subject-matter decision but a statistical one. Outliers have a precise, objective definition: they are observations that do not follow the pattern of the majority of the data | How to detect outliers in skewed data set?
Flagging outlier is not a subject-matter decision but a statistical one. Outliers have a precise, objective definition: they are observations that do not follow the pattern of the majority of the data. Such observations need to be set apart at the onset of any analysis simply ... | How to detect outliers in skewed data set?
Flagging outlier is not a subject-matter decision but a statistical one. Outliers have a precise, objective definition: they are observations that do not follow the pattern of the majority of the data |
39,500 | How to detect outliers in skewed data set? | Bacon answered this question a few centuries ago in Novum Organum. To paraphrase: To do science is to search for repeated patterns. To detect anomalies is to identify values that do not follow repeated patterns. "For whoever knows the ways of Nature will more" easily notice her deviations and, on the other hand, whoe... | How to detect outliers in skewed data set? | Bacon answered this question a few centuries ago in Novum Organum. To paraphrase: To do science is to search for repeated patterns. To detect anomalies is to identify values that do not follow repeat | How to detect outliers in skewed data set?
Bacon answered this question a few centuries ago in Novum Organum. To paraphrase: To do science is to search for repeated patterns. To detect anomalies is to identify values that do not follow repeated patterns. "For whoever knows the ways of Nature will more" easily notice ... | How to detect outliers in skewed data set?
Bacon answered this question a few centuries ago in Novum Organum. To paraphrase: To do science is to search for repeated patterns. To detect anomalies is to identify values that do not follow repeat |
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