idx int64 1 56k | question stringlengths 15 155 | answer stringlengths 2 29.2k ⌀ | question_cut stringlengths 15 100 | answer_cut stringlengths 2 200 ⌀ | conversation stringlengths 47 29.3k | conversation_cut stringlengths 47 301 |
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41,601 | Beginner level: Help in learning Kalman Smoother (Part 1) [closed] | Let me take a few steps back. EM algorithm is not required in a Kalman Filter if the design matrices(A, B, Q, R, etc) are known. They are known only if you know which physical system you are modelling from the beginning. If not then you will have to estimate these matrices. Filtering and smoothing operations are perfor... | Beginner level: Help in learning Kalman Smoother (Part 1) [closed] | Let me take a few steps back. EM algorithm is not required in a Kalman Filter if the design matrices(A, B, Q, R, etc) are known. They are known only if you know which physical system you are modelling | Beginner level: Help in learning Kalman Smoother (Part 1) [closed]
Let me take a few steps back. EM algorithm is not required in a Kalman Filter if the design matrices(A, B, Q, R, etc) are known. They are known only if you know which physical system you are modelling from the beginning. If not then you will have to est... | Beginner level: Help in learning Kalman Smoother (Part 1) [closed]
Let me take a few steps back. EM algorithm is not required in a Kalman Filter if the design matrices(A, B, Q, R, etc) are known. They are known only if you know which physical system you are modelling |
41,602 | finding out 2D transformation given a list of sample points | This is almost least squares regression except for the constraint on the parameter. In fact, by viewing $(x,y)\sim x + iy = z$ and $(x^\prime, y^\prime) = x^\prime + i y^\prime = z^\prime$ as complex numbers, we may write the model as
$$z_j^\prime = \alpha + \beta z_j + \varepsilon_j$$
where $\alpha = \Delta_x + i\Del... | finding out 2D transformation given a list of sample points | This is almost least squares regression except for the constraint on the parameter. In fact, by viewing $(x,y)\sim x + iy = z$ and $(x^\prime, y^\prime) = x^\prime + i y^\prime = z^\prime$ as complex | finding out 2D transformation given a list of sample points
This is almost least squares regression except for the constraint on the parameter. In fact, by viewing $(x,y)\sim x + iy = z$ and $(x^\prime, y^\prime) = x^\prime + i y^\prime = z^\prime$ as complex numbers, we may write the model as
$$z_j^\prime = \alpha + ... | finding out 2D transformation given a list of sample points
This is almost least squares regression except for the constraint on the parameter. In fact, by viewing $(x,y)\sim x + iy = z$ and $(x^\prime, y^\prime) = x^\prime + i y^\prime = z^\prime$ as complex |
41,603 | Parallel regression assumption | The parallel regression assumption (aka proportional regression assumption) in ordinal logistic regression says that the coefficients that describe the odds of being in the lowest category vs. all higher categories of the response variable are the same as those that describe the odds between the second lowest category ... | Parallel regression assumption | The parallel regression assumption (aka proportional regression assumption) in ordinal logistic regression says that the coefficients that describe the odds of being in the lowest category vs. all hig | Parallel regression assumption
The parallel regression assumption (aka proportional regression assumption) in ordinal logistic regression says that the coefficients that describe the odds of being in the lowest category vs. all higher categories of the response variable are the same as those that describe the odds betw... | Parallel regression assumption
The parallel regression assumption (aka proportional regression assumption) in ordinal logistic regression says that the coefficients that describe the odds of being in the lowest category vs. all hig |
41,604 | Interpreting case influence statistics (leverage, studentized residuals, and Cook's distance) | No, the fact that you have a large studentized residual does not necessarily mean that the observation is an outlier. (Although some define outlier as simply a large residual, in which case it would be by definition.)
"Influential" is somewhat ambiguous. One could think of leverage as a measure of influence, or of ... | Interpreting case influence statistics (leverage, studentized residuals, and Cook's distance) | No, the fact that you have a large studentized residual does not necessarily mean that the observation is an outlier. (Although some define outlier as simply a large residual, in which case it would | Interpreting case influence statistics (leverage, studentized residuals, and Cook's distance)
No, the fact that you have a large studentized residual does not necessarily mean that the observation is an outlier. (Although some define outlier as simply a large residual, in which case it would be by definition.)
"Infl... | Interpreting case influence statistics (leverage, studentized residuals, and Cook's distance)
No, the fact that you have a large studentized residual does not necessarily mean that the observation is an outlier. (Although some define outlier as simply a large residual, in which case it would |
41,605 | How to handle data normalization in kNN when new test data is received | Your validation process and the reasoning behind is entirely correct.
Using the same reasoning / model building process: After you have selected a k by validation, you build the final model using all the training data $X$ and calculate the mean and variance based only on $X$ since these values are also part of the mode... | How to handle data normalization in kNN when new test data is received | Your validation process and the reasoning behind is entirely correct.
Using the same reasoning / model building process: After you have selected a k by validation, you build the final model using all | How to handle data normalization in kNN when new test data is received
Your validation process and the reasoning behind is entirely correct.
Using the same reasoning / model building process: After you have selected a k by validation, you build the final model using all the training data $X$ and calculate the mean and ... | How to handle data normalization in kNN when new test data is received
Your validation process and the reasoning behind is entirely correct.
Using the same reasoning / model building process: After you have selected a k by validation, you build the final model using all |
41,606 | What to do when parallel regression assumption violated | It looks to me that you are looking for the "partial" version of proportional odds. Reference:
R. S. Society, “Partial Proportional Odds Models for Ordinal Response Variables,” vol. 39, no. 2, pp. 205–217, 1999.
If in the "standard" ordered logit (or proportional odds), the cumulative probability is modeled as
$$
P(Y ... | What to do when parallel regression assumption violated | It looks to me that you are looking for the "partial" version of proportional odds. Reference:
R. S. Society, “Partial Proportional Odds Models for Ordinal Response Variables,” vol. 39, no. 2, pp. 20 | What to do when parallel regression assumption violated
It looks to me that you are looking for the "partial" version of proportional odds. Reference:
R. S. Society, “Partial Proportional Odds Models for Ordinal Response Variables,” vol. 39, no. 2, pp. 205–217, 1999.
If in the "standard" ordered logit (or proportional... | What to do when parallel regression assumption violated
It looks to me that you are looking for the "partial" version of proportional odds. Reference:
R. S. Society, “Partial Proportional Odds Models for Ordinal Response Variables,” vol. 39, no. 2, pp. 20 |
41,607 | Solving formula for confidence intervals on two proportions | If you did want to create a confidence interval for the difference of two proportions, there are better procedures than the one you are using: you might consider making use of the Wilson procedure. The Wikipedia article on binomial confidence intervals mentions this and several other possibilities, but does not show th... | Solving formula for confidence intervals on two proportions | If you did want to create a confidence interval for the difference of two proportions, there are better procedures than the one you are using: you might consider making use of the Wilson procedure. Th | Solving formula for confidence intervals on two proportions
If you did want to create a confidence interval for the difference of two proportions, there are better procedures than the one you are using: you might consider making use of the Wilson procedure. The Wikipedia article on binomial confidence intervals mention... | Solving formula for confidence intervals on two proportions
If you did want to create a confidence interval for the difference of two proportions, there are better procedures than the one you are using: you might consider making use of the Wilson procedure. Th |
41,608 | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | The intuition behind the argument saying that the optimal policy is independent of initial state is the following:
The optimal policy is defined by a function that selects an action for every possible state and actions in different states are independent.
Formally speaking, for an unknown initial distribution, the val... | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | The intuition behind the argument saying that the optimal policy is independent of initial state is the following:
The optimal policy is defined by a function that selects an action for every possible | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
The intuition behind the argument saying that the optimal policy is independent of initial state is the following:
The optimal policy is defined by a function that selects an action for every possible state and actions in diff... | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
The intuition behind the argument saying that the optimal policy is independent of initial state is the following:
The optimal policy is defined by a function that selects an action for every possible |
41,609 | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | why is it that $π^∗$ has the property that its the optimal policy for all states?
This is actually because of Markov property of environment. Markov property states that the history of previous states and actions leading to state $s$ does not affect $R(s)$ and $P_{sa}(s')$. So in any state s, the optimal policy for th... | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | why is it that $π^∗$ has the property that its the optimal policy for all states?
This is actually because of Markov property of environment. Markov property states that the history of previous state | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
why is it that $π^∗$ has the property that its the optimal policy for all states?
This is actually because of Markov property of environment. Markov property states that the history of previous states and actions leading to s... | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
why is it that $π^∗$ has the property that its the optimal policy for all states?
This is actually because of Markov property of environment. Markov property states that the history of previous state |
41,610 | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | I didn't dive in the details for quite some time, but it appears very intuitive to me that $π^∗$ is valid for all states if you take practical examples.
The case of solving a maze is one of these: the agent is (potentially randomly positioned) in a maze and is trying to get out of it. It gets rewarded only when it find... | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | I didn't dive in the details for quite some time, but it appears very intuitive to me that $π^∗$ is valid for all states if you take practical examples.
The case of solving a maze is one of these: the | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
I didn't dive in the details for quite some time, but it appears very intuitive to me that $π^∗$ is valid for all states if you take practical examples.
The case of solving a maze is one of these: the agent is (potentially ran... | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
I didn't dive in the details for quite some time, but it appears very intuitive to me that $π^∗$ is valid for all states if you take practical examples.
The case of solving a maze is one of these: the |
41,611 | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | In my opinion, any policy that achieves the optimal value is an optimal policy. Since the optimal value function for a given MDP is unique, this optimal value function actually defines a equivalent class over the policy space, i.e., those whose value is optimal are actually equivalent. In other words, although optimal ... | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | In my opinion, any policy that achieves the optimal value is an optimal policy. Since the optimal value function for a given MDP is unique, this optimal value function actually defines a equivalent cl | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
In my opinion, any policy that achieves the optimal value is an optimal policy. Since the optimal value function for a given MDP is unique, this optimal value function actually defines a equivalent class over the policy space,... | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
In my opinion, any policy that achieves the optimal value is an optimal policy. Since the optimal value function for a given MDP is unique, this optimal value function actually defines a equivalent cl |
41,612 | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | Perhaps, one should view the optimization process taken over the space of functions span by the admissible policies, however explicitly described by the MDP definition.
probabilities could always be forced to follow the strict more explicit notation of functions of many variables from general mathematics. All moments ... | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | Perhaps, one should view the optimization process taken over the space of functions span by the admissible policies, however explicitly described by the MDP definition.
probabilities could always be f | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
Perhaps, one should view the optimization process taken over the space of functions span by the admissible policies, however explicitly described by the MDP definition.
probabilities could always be forced to follow the strict... | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
Perhaps, one should view the optimization process taken over the space of functions span by the admissible policies, however explicitly described by the MDP definition.
probabilities could always be f |
41,613 | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | By definition, $\pi^*$ is optimal if $V_{\pi^*}(s) = \max_{\pi} V_{\pi}(s)$ for all $s$. So isn't just definition? | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state? | By definition, $\pi^*$ is optimal if $V_{\pi^*}(s) = \max_{\pi} V_{\pi}(s)$ for all $s$. So isn't just definition? | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
By definition, $\pi^*$ is optimal if $V_{\pi^*}(s) = \max_{\pi} V_{\pi}(s)$ for all $s$. So isn't just definition? | Why is the optimal policy in Markov Decision Process (MDP), independent of the initial state?
By definition, $\pi^*$ is optimal if $V_{\pi^*}(s) = \max_{\pi} V_{\pi}(s)$ for all $s$. So isn't just definition? |
41,614 | Do I need more than one random slope? | It does make sense, but you have to be a little bit careful in setting up the model. The way you've written the model,
S ~ X1 + X2 + X3 + (X1|biome) + (X2|biome) + (X3|biome)
it implicitly incorporates an intercept term with each random effect slope. You could write it as
S ~ X1 + X2 + X3 + (1|biome) + (X1+0|biome... | Do I need more than one random slope? | It does make sense, but you have to be a little bit careful in setting up the model. The way you've written the model,
S ~ X1 + X2 + X3 + (X1|biome) + (X2|biome) + (X3|biome)
it implicitly incorpor | Do I need more than one random slope?
It does make sense, but you have to be a little bit careful in setting up the model. The way you've written the model,
S ~ X1 + X2 + X3 + (X1|biome) + (X2|biome) + (X3|biome)
it implicitly incorporates an intercept term with each random effect slope. You could write it as
S ~ ... | Do I need more than one random slope?
It does make sense, but you have to be a little bit careful in setting up the model. The way you've written the model,
S ~ X1 + X2 + X3 + (X1|biome) + (X2|biome) + (X3|biome)
it implicitly incorpor |
41,615 | What makes constant function an estimator? | I think it's not so much a question of ''what makes constant function'' an estimator but ''what makes estimator an estimator''.
First, from mathematical point of view an estimator is a function of special kind, it's a random variable, that fulfills some requirements - it's a statistic, which means it has to be independ... | What makes constant function an estimator? | I think it's not so much a question of ''what makes constant function'' an estimator but ''what makes estimator an estimator''.
First, from mathematical point of view an estimator is a function of spe | What makes constant function an estimator?
I think it's not so much a question of ''what makes constant function'' an estimator but ''what makes estimator an estimator''.
First, from mathematical point of view an estimator is a function of special kind, it's a random variable, that fulfills some requirements - it's a s... | What makes constant function an estimator?
I think it's not so much a question of ''what makes constant function'' an estimator but ''what makes estimator an estimator''.
First, from mathematical point of view an estimator is a function of spe |
41,616 | What makes constant function an estimator? | An estimator is simply some function of a potential sample of data that seeks to estimate an unknown population parameter. It's a recipe or a formula. Your constant is an estimator that does not depend on the data at all: the estimate that is produces will always be the same.
There's an infinite number of estimators, a... | What makes constant function an estimator? | An estimator is simply some function of a potential sample of data that seeks to estimate an unknown population parameter. It's a recipe or a formula. Your constant is an estimator that does not depen | What makes constant function an estimator?
An estimator is simply some function of a potential sample of data that seeks to estimate an unknown population parameter. It's a recipe or a formula. Your constant is an estimator that does not depend on the data at all: the estimate that is produces will always be the same.
... | What makes constant function an estimator?
An estimator is simply some function of a potential sample of data that seeks to estimate an unknown population parameter. It's a recipe or a formula. Your constant is an estimator that does not depen |
41,617 | What makes constant function an estimator? | Constant estimators/predictors have a use as benchmarks against which one judges the performance of "proper" estimators/predictors.
A standard example is in the context of binary logistic regression, where we attempt to estimate conditional probabilities, exploiting the information that possibly resides in the regresso... | What makes constant function an estimator? | Constant estimators/predictors have a use as benchmarks against which one judges the performance of "proper" estimators/predictors.
A standard example is in the context of binary logistic regression, | What makes constant function an estimator?
Constant estimators/predictors have a use as benchmarks against which one judges the performance of "proper" estimators/predictors.
A standard example is in the context of binary logistic regression, where we attempt to estimate conditional probabilities, exploiting the inform... | What makes constant function an estimator?
Constant estimators/predictors have a use as benchmarks against which one judges the performance of "proper" estimators/predictors.
A standard example is in the context of binary logistic regression, |
41,618 | Residual variance for glmer | One possible interpretation of the logistic regression model is to state that there is an underlying score $$ y^*_i = x_i'\beta + \epsilon_i, $$ with the observed variable being $$y_i = \left\{ \begin{array}{ll} 1, & y^*_i > 0 \\ 0, & y^*_i \le 0\end{array}\right.$$ This would be the way logistic regression would be in... | Residual variance for glmer | One possible interpretation of the logistic regression model is to state that there is an underlying score $$ y^*_i = x_i'\beta + \epsilon_i, $$ with the observed variable being $$y_i = \left\{ \begin | Residual variance for glmer
One possible interpretation of the logistic regression model is to state that there is an underlying score $$ y^*_i = x_i'\beta + \epsilon_i, $$ with the observed variable being $$y_i = \left\{ \begin{array}{ll} 1, & y^*_i > 0 \\ 0, & y^*_i \le 0\end{array}\right.$$ This would be the way log... | Residual variance for glmer
One possible interpretation of the logistic regression model is to state that there is an underlying score $$ y^*_i = x_i'\beta + \epsilon_i, $$ with the observed variable being $$y_i = \left\{ \begin |
41,619 | Show the shortest confidence interval of a normal distribution | The length of the interval is
$$\left(\bar{X}-Z_{(1-k)\alpha}\frac{\sigma}{\sqrt{n}}\right) - \left(\bar{X}-Z_{1 - k\alpha}\frac{\sigma}{\sqrt{n}}\right) = \left(Z_{1 - k\alpha} - Z_{(1-k)\alpha}\right)\frac{\sigma}{\sqrt{n}}. $$
Because when $k$ is varied $\sigma/\sqrt{n}$ remains constant, this is minimized provided ... | Show the shortest confidence interval of a normal distribution | The length of the interval is
$$\left(\bar{X}-Z_{(1-k)\alpha}\frac{\sigma}{\sqrt{n}}\right) - \left(\bar{X}-Z_{1 - k\alpha}\frac{\sigma}{\sqrt{n}}\right) = \left(Z_{1 - k\alpha} - Z_{(1-k)\alpha}\righ | Show the shortest confidence interval of a normal distribution
The length of the interval is
$$\left(\bar{X}-Z_{(1-k)\alpha}\frac{\sigma}{\sqrt{n}}\right) - \left(\bar{X}-Z_{1 - k\alpha}\frac{\sigma}{\sqrt{n}}\right) = \left(Z_{1 - k\alpha} - Z_{(1-k)\alpha}\right)\frac{\sigma}{\sqrt{n}}. $$
Because when $k$ is varied ... | Show the shortest confidence interval of a normal distribution
The length of the interval is
$$\left(\bar{X}-Z_{(1-k)\alpha}\frac{\sigma}{\sqrt{n}}\right) - \left(\bar{X}-Z_{1 - k\alpha}\frac{\sigma}{\sqrt{n}}\right) = \left(Z_{1 - k\alpha} - Z_{(1-k)\alpha}\righ |
41,620 | Are eigenvectors obtained in Kernel PCA orthogonal? | Yes, they are orthogonal. To see that your last expression equals zero, write it in the vector notation: $$\sum_{i,j=1}^n \alpha_i \beta_j K_{ij} = \boldsymbol \alpha^\top \boldsymbol K \boldsymbol \beta=0,$$ because $\boldsymbol \alpha$ and $\boldsymbol \beta$ are two different eigenvectors of $\boldsymbol K$. It is a... | Are eigenvectors obtained in Kernel PCA orthogonal? | Yes, they are orthogonal. To see that your last expression equals zero, write it in the vector notation: $$\sum_{i,j=1}^n \alpha_i \beta_j K_{ij} = \boldsymbol \alpha^\top \boldsymbol K \boldsymbol \b | Are eigenvectors obtained in Kernel PCA orthogonal?
Yes, they are orthogonal. To see that your last expression equals zero, write it in the vector notation: $$\sum_{i,j=1}^n \alpha_i \beta_j K_{ij} = \boldsymbol \alpha^\top \boldsymbol K \boldsymbol \beta=0,$$ because $\boldsymbol \alpha$ and $\boldsymbol \beta$ are tw... | Are eigenvectors obtained in Kernel PCA orthogonal?
Yes, they are orthogonal. To see that your last expression equals zero, write it in the vector notation: $$\sum_{i,j=1}^n \alpha_i \beta_j K_{ij} = \boldsymbol \alpha^\top \boldsymbol K \boldsymbol \b |
41,621 | Are there problems with arbitrary application of bootstrap? | The term "bootstrap" covers many (somewhat related) things -- enough to fill a number of books (which indeed it does). Some things are more prone to problems with naive application than others.
If you're just applying resampling directly to observations, the most obvious problem you might encounter with a naive appli... | Are there problems with arbitrary application of bootstrap? | The term "bootstrap" covers many (somewhat related) things -- enough to fill a number of books (which indeed it does). Some things are more prone to problems with naive application than others.
If y | Are there problems with arbitrary application of bootstrap?
The term "bootstrap" covers many (somewhat related) things -- enough to fill a number of books (which indeed it does). Some things are more prone to problems with naive application than others.
If you're just applying resampling directly to observations, the... | Are there problems with arbitrary application of bootstrap?
The term "bootstrap" covers many (somewhat related) things -- enough to fill a number of books (which indeed it does). Some things are more prone to problems with naive application than others.
If y |
41,622 | What’s wrong with this way of fitting time-dependent coefficients in a Cox regression? | I am not very familiar with the mechanics of survival analysis data management in R, but I think I can explain why this happens and show an example using Stata.
The hazard of the risk changes the instant the variable changes as time flows (no delays or anticipation allowed), though it remains constant in the intervals ... | What’s wrong with this way of fitting time-dependent coefficients in a Cox regression? | I am not very familiar with the mechanics of survival analysis data management in R, but I think I can explain why this happens and show an example using Stata.
The hazard of the risk changes the inst | What’s wrong with this way of fitting time-dependent coefficients in a Cox regression?
I am not very familiar with the mechanics of survival analysis data management in R, but I think I can explain why this happens and show an example using Stata.
The hazard of the risk changes the instant the variable changes as time ... | What’s wrong with this way of fitting time-dependent coefficients in a Cox regression?
I am not very familiar with the mechanics of survival analysis data management in R, but I think I can explain why this happens and show an example using Stata.
The hazard of the risk changes the inst |
41,623 | What’s wrong with this way of fitting time-dependent coefficients in a Cox regression? | As a side note, you can archive your initial model in a "non-invalid" way in R by using the tt function. See the last example in the Using Time Dependent Covariates and Time Dependent Coefficients in the Cox Model vignette of the Survival package (at least the last example in version 2.41-3). | What’s wrong with this way of fitting time-dependent coefficients in a Cox regression? | As a side note, you can archive your initial model in a "non-invalid" way in R by using the tt function. See the last example in the Using Time Dependent Covariates and Time Dependent Coefficients in | What’s wrong with this way of fitting time-dependent coefficients in a Cox regression?
As a side note, you can archive your initial model in a "non-invalid" way in R by using the tt function. See the last example in the Using Time Dependent Covariates and Time Dependent Coefficients in the Cox Model vignette of the Sur... | What’s wrong with this way of fitting time-dependent coefficients in a Cox regression?
As a side note, you can archive your initial model in a "non-invalid" way in R by using the tt function. See the last example in the Using Time Dependent Covariates and Time Dependent Coefficients in |
41,624 | ROC-AUC and Precision-Recall for random classifiers in class imbalanced problems | Area under the ROC curve is insensitive to class balance. Area under the PR curve, on the other hand, is highly influenced by this. | ROC-AUC and Precision-Recall for random classifiers in class imbalanced problems | Area under the ROC curve is insensitive to class balance. Area under the PR curve, on the other hand, is highly influenced by this. | ROC-AUC and Precision-Recall for random classifiers in class imbalanced problems
Area under the ROC curve is insensitive to class balance. Area under the PR curve, on the other hand, is highly influenced by this. | ROC-AUC and Precision-Recall for random classifiers in class imbalanced problems
Area under the ROC curve is insensitive to class balance. Area under the PR curve, on the other hand, is highly influenced by this. |
41,625 | Obtaining an estimator for z given an estimator for log z | I don't necessarily have a problem with you exponentiating your predicted values. You just need to realize that if the former was an expectation, the result is no longer an expectiation. Specifically, a regression model is intended to give the expected value of $Y$ at each point in $X$ ($E(Y|X=x_i)$). An expected va... | Obtaining an estimator for z given an estimator for log z | I don't necessarily have a problem with you exponentiating your predicted values. You just need to realize that if the former was an expectation, the result is no longer an expectiation. Specificall | Obtaining an estimator for z given an estimator for log z
I don't necessarily have a problem with you exponentiating your predicted values. You just need to realize that if the former was an expectation, the result is no longer an expectiation. Specifically, a regression model is intended to give the expected value o... | Obtaining an estimator for z given an estimator for log z
I don't necessarily have a problem with you exponentiating your predicted values. You just need to realize that if the former was an expectation, the result is no longer an expectiation. Specificall |
41,626 | Stochastic Programming with MCMC | PyMC2 can be combined with the LP solver of your choice to solve stochastic LP problems like this one. Here is code to do it for this very simple case. I've left a note on how I would change this for a more complex LP.
c1 = pm.Normal('c1', mu=2, tau=.5**-2)
c2 = -3
b1 = pm.Normal('b1', mu=0, tau=3.**-2)
@pm.determinis... | Stochastic Programming with MCMC | PyMC2 can be combined with the LP solver of your choice to solve stochastic LP problems like this one. Here is code to do it for this very simple case. I've left a note on how I would change this for | Stochastic Programming with MCMC
PyMC2 can be combined with the LP solver of your choice to solve stochastic LP problems like this one. Here is code to do it for this very simple case. I've left a note on how I would change this for a more complex LP.
c1 = pm.Normal('c1', mu=2, tau=.5**-2)
c2 = -3
b1 = pm.Normal('b1', ... | Stochastic Programming with MCMC
PyMC2 can be combined with the LP solver of your choice to solve stochastic LP problems like this one. Here is code to do it for this very simple case. I've left a note on how I would change this for |
41,627 | Feature Normalization/Standardization before or after Feature Selection? | Before.
In fact, it's the "feature selection process" you mention that is pretty much the reason why you want to have your features standardized in the first place. | Feature Normalization/Standardization before or after Feature Selection? | Before.
In fact, it's the "feature selection process" you mention that is pretty much the reason why you want to have your features standardized in the first place. | Feature Normalization/Standardization before or after Feature Selection?
Before.
In fact, it's the "feature selection process" you mention that is pretty much the reason why you want to have your features standardized in the first place. | Feature Normalization/Standardization before or after Feature Selection?
Before.
In fact, it's the "feature selection process" you mention that is pretty much the reason why you want to have your features standardized in the first place. |
41,628 | GLM analogue of weighted least squares | Fit an MLE by maximizing
$$
l(\mathbf{\theta};\mathbf{y})=\sum_{i=1}^Nl{\left(\theta;y_i\right)}
$$
where $l$ is the log-likelihood. Fitting an MLE with inverse-probability (i.e. frequency) weights entails modifying the log-likelihood to:
$$
l(\mathbf{\theta};\mathbf{y})=\sum_{i=1}^Nw_i~l{\left(\theta;y_i\right)}.
$$
I... | GLM analogue of weighted least squares | Fit an MLE by maximizing
$$
l(\mathbf{\theta};\mathbf{y})=\sum_{i=1}^Nl{\left(\theta;y_i\right)}
$$
where $l$ is the log-likelihood. Fitting an MLE with inverse-probability (i.e. frequency) weights en | GLM analogue of weighted least squares
Fit an MLE by maximizing
$$
l(\mathbf{\theta};\mathbf{y})=\sum_{i=1}^Nl{\left(\theta;y_i\right)}
$$
where $l$ is the log-likelihood. Fitting an MLE with inverse-probability (i.e. frequency) weights entails modifying the log-likelihood to:
$$
l(\mathbf{\theta};\mathbf{y})=\sum_{i=1... | GLM analogue of weighted least squares
Fit an MLE by maximizing
$$
l(\mathbf{\theta};\mathbf{y})=\sum_{i=1}^Nl{\left(\theta;y_i\right)}
$$
where $l$ is the log-likelihood. Fitting an MLE with inverse-probability (i.e. frequency) weights en |
41,629 | Matrix Factorization algorithms for Recommender Systems | Matrix factorisation is part of Numerical Linear Algebra (NLA). The following are some useful books in NLA and Data Mining / Statistical Learning.
The classic in NLA is Golub & Van Loan's Matrix Computations. Van Loan's webpage lists his books in and links to others.
A modern approach that's great for self-study, ... | Matrix Factorization algorithms for Recommender Systems | Matrix factorisation is part of Numerical Linear Algebra (NLA). The following are some useful books in NLA and Data Mining / Statistical Learning.
The classic in NLA is Golub & Van Loan's Matrix C | Matrix Factorization algorithms for Recommender Systems
Matrix factorisation is part of Numerical Linear Algebra (NLA). The following are some useful books in NLA and Data Mining / Statistical Learning.
The classic in NLA is Golub & Van Loan's Matrix Computations. Van Loan's webpage lists his books in and links to... | Matrix Factorization algorithms for Recommender Systems
Matrix factorisation is part of Numerical Linear Algebra (NLA). The following are some useful books in NLA and Data Mining / Statistical Learning.
The classic in NLA is Golub & Van Loan's Matrix C |
41,630 | Matrix Factorization algorithms for Recommender Systems | You can have a look to http://dl.acm.org/citation.cfm?id=2043956. For more details you can look into Recommender system handbook.
I am working on similar problems. For more details or discussion mail me @ pranav.waila[at]gmail[dot]com. | Matrix Factorization algorithms for Recommender Systems | You can have a look to http://dl.acm.org/citation.cfm?id=2043956. For more details you can look into Recommender system handbook.
I am working on similar problems. For more details or discussion mail | Matrix Factorization algorithms for Recommender Systems
You can have a look to http://dl.acm.org/citation.cfm?id=2043956. For more details you can look into Recommender system handbook.
I am working on similar problems. For more details or discussion mail me @ pranav.waila[at]gmail[dot]com. | Matrix Factorization algorithms for Recommender Systems
You can have a look to http://dl.acm.org/citation.cfm?id=2043956. For more details you can look into Recommender system handbook.
I am working on similar problems. For more details or discussion mail |
41,631 | What does "Conditioning on the margins of ____" mean? | Margins
Margins here refers to the values on the edges (margins!) of the table, that is, the total number of reds, total number of blacks, total number of drawn, and total number of not drawn. The related term marginal distribution refers to the distribution of a single variable obtained from a joint distribution of se... | What does "Conditioning on the margins of ____" mean? | Margins
Margins here refers to the values on the edges (margins!) of the table, that is, the total number of reds, total number of blacks, total number of drawn, and total number of not drawn. The rel | What does "Conditioning on the margins of ____" mean?
Margins
Margins here refers to the values on the edges (margins!) of the table, that is, the total number of reds, total number of blacks, total number of drawn, and total number of not drawn. The related term marginal distribution refers to the distribution of a si... | What does "Conditioning on the margins of ____" mean?
Margins
Margins here refers to the values on the edges (margins!) of the table, that is, the total number of reds, total number of blacks, total number of drawn, and total number of not drawn. The rel |
41,632 | Clustering without a distance matrix | Inverted lists work very well for sparse data. This is what e.g. Lucene uses.
I don't know how extensible scikit-learn is. A lot of the code in it seems to be written in Cython, so it is Python-like code compiled via C. This would make it harder to extend.
ELKI, the data mining tool I am contributing a lot to, has an -... | Clustering without a distance matrix | Inverted lists work very well for sparse data. This is what e.g. Lucene uses.
I don't know how extensible scikit-learn is. A lot of the code in it seems to be written in Cython, so it is Python-like c | Clustering without a distance matrix
Inverted lists work very well for sparse data. This is what e.g. Lucene uses.
I don't know how extensible scikit-learn is. A lot of the code in it seems to be written in Cython, so it is Python-like code compiled via C. This would make it harder to extend.
ELKI, the data mining tool... | Clustering without a distance matrix
Inverted lists work very well for sparse data. This is what e.g. Lucene uses.
I don't know how extensible scikit-learn is. A lot of the code in it seems to be written in Cython, so it is Python-like c |
41,633 | GLMM and two slopes | The problem is the number of abscissas, or nodes, you've selected for the Adaptive Gaussian Quadrature (AGQ) approximation of the log-likelihood, specified by nAGQ. The default value is 1 (equivalent to the Laplacian approximation).
The glmer function's Details section (page 29 in the lme4 help page) states:
The mo... | GLMM and two slopes | The problem is the number of abscissas, or nodes, you've selected for the Adaptive Gaussian Quadrature (AGQ) approximation of the log-likelihood, specified by nAGQ. The default value is 1 (equivalent | GLMM and two slopes
The problem is the number of abscissas, or nodes, you've selected for the Adaptive Gaussian Quadrature (AGQ) approximation of the log-likelihood, specified by nAGQ. The default value is 1 (equivalent to the Laplacian approximation).
The glmer function's Details section (page 29 in the lme4 help p... | GLMM and two slopes
The problem is the number of abscissas, or nodes, you've selected for the Adaptive Gaussian Quadrature (AGQ) approximation of the log-likelihood, specified by nAGQ. The default value is 1 (equivalent |
41,634 | Intuition behind the Calinski-Harabasz Index | Some simple intuition: $[B(k)/(k-1)]/[W(k)/(n-k)]$ is analogous to an F-ratio in ANOVA; $B(k)$ and $W(k)$ are between- and within-cluster sums of squares for the $k$ clusters.
$B(k)$ has $k-1$ degrees of freedom, while $W(k)$ has $n-k$ degrees of freedom.
As $k$ grows, if the clusters were all actually just from the ... | Intuition behind the Calinski-Harabasz Index | Some simple intuition: $[B(k)/(k-1)]/[W(k)/(n-k)]$ is analogous to an F-ratio in ANOVA; $B(k)$ and $W(k)$ are between- and within-cluster sums of squares for the $k$ clusters.
$B(k)$ has $k-1$ degree | Intuition behind the Calinski-Harabasz Index
Some simple intuition: $[B(k)/(k-1)]/[W(k)/(n-k)]$ is analogous to an F-ratio in ANOVA; $B(k)$ and $W(k)$ are between- and within-cluster sums of squares for the $k$ clusters.
$B(k)$ has $k-1$ degrees of freedom, while $W(k)$ has $n-k$ degrees of freedom.
As $k$ grows, if ... | Intuition behind the Calinski-Harabasz Index
Some simple intuition: $[B(k)/(k-1)]/[W(k)/(n-k)]$ is analogous to an F-ratio in ANOVA; $B(k)$ and $W(k)$ are between- and within-cluster sums of squares for the $k$ clusters.
$B(k)$ has $k-1$ degree |
41,635 | If $X$ is lognormally distributed, what is the distribution of $1 / (1 + X)$? | To close this one:
We want the pdf of the variable
$$Z = g(X) =\frac 1{X+1} \Rightarrow X = g^{-1}(Z)=\frac 1Z -1 \Rightarrow \frac {\partial g^{-1}(z)}{\partial z}=-\frac 1{z^2}$$
Note that by construction $0 \leq Z \leq 1$.
The density of the log-normal is known. Applying the change-of-variable formula
we obtain
$$f_... | If $X$ is lognormally distributed, what is the distribution of $1 / (1 + X)$? | To close this one:
We want the pdf of the variable
$$Z = g(X) =\frac 1{X+1} \Rightarrow X = g^{-1}(Z)=\frac 1Z -1 \Rightarrow \frac {\partial g^{-1}(z)}{\partial z}=-\frac 1{z^2}$$
Note that by constr | If $X$ is lognormally distributed, what is the distribution of $1 / (1 + X)$?
To close this one:
We want the pdf of the variable
$$Z = g(X) =\frac 1{X+1} \Rightarrow X = g^{-1}(Z)=\frac 1Z -1 \Rightarrow \frac {\partial g^{-1}(z)}{\partial z}=-\frac 1{z^2}$$
Note that by construction $0 \leq Z \leq 1$.
The density of t... | If $X$ is lognormally distributed, what is the distribution of $1 / (1 + X)$?
To close this one:
We want the pdf of the variable
$$Z = g(X) =\frac 1{X+1} \Rightarrow X = g^{-1}(Z)=\frac 1Z -1 \Rightarrow \frac {\partial g^{-1}(z)}{\partial z}=-\frac 1{z^2}$$
Note that by constr |
41,636 | Given two sets, how can I say statistically if they are similar/different | null hypothesis in this case is that [...] they are different
-- That's not how null hypotheses work. You need something you can calculate the distribution of a test statistic under; generally that's no effect/no difference (whence, "null").
similarity ... whether or not the two groups were sampled from the same pop... | Given two sets, how can I say statistically if they are similar/different | null hypothesis in this case is that [...] they are different
-- That's not how null hypotheses work. You need something you can calculate the distribution of a test statistic under; generally that's | Given two sets, how can I say statistically if they are similar/different
null hypothesis in this case is that [...] they are different
-- That's not how null hypotheses work. You need something you can calculate the distribution of a test statistic under; generally that's no effect/no difference (whence, "null").
s... | Given two sets, how can I say statistically if they are similar/different
null hypothesis in this case is that [...] they are different
-- That's not how null hypotheses work. You need something you can calculate the distribution of a test statistic under; generally that's |
41,637 | Given two sets, how can I say statistically if they are similar/different | I would like to compare these two sets of data samples and determine
whether the null hypothesis holds...Therefore how might one determine
how similar/different two groups are...
These are two different things. It is not clear to me that you should be using hypothesis testing at all if you already know the distrib... | Given two sets, how can I say statistically if they are similar/different | I would like to compare these two sets of data samples and determine
whether the null hypothesis holds...Therefore how might one determine
how similar/different two groups are...
These are two di | Given two sets, how can I say statistically if they are similar/different
I would like to compare these two sets of data samples and determine
whether the null hypothesis holds...Therefore how might one determine
how similar/different two groups are...
These are two different things. It is not clear to me that you... | Given two sets, how can I say statistically if they are similar/different
I would like to compare these two sets of data samples and determine
whether the null hypothesis holds...Therefore how might one determine
how similar/different two groups are...
These are two di |
41,638 | Given two sets, how can I say statistically if they are similar/different | On average how many observations per item?
I agree with your naive approach and wanting something more robust. There are many robust nonparametric tests to compare two samples, such as permutation tests or Wilcoxon Rank Sum Tests. You can compare these to the results of the two-sample t-tests and look more closely at... | Given two sets, how can I say statistically if they are similar/different | On average how many observations per item?
I agree with your naive approach and wanting something more robust. There are many robust nonparametric tests to compare two samples, such as permutation te | Given two sets, how can I say statistically if they are similar/different
On average how many observations per item?
I agree with your naive approach and wanting something more robust. There are many robust nonparametric tests to compare two samples, such as permutation tests or Wilcoxon Rank Sum Tests. You can compa... | Given two sets, how can I say statistically if they are similar/different
On average how many observations per item?
I agree with your naive approach and wanting something more robust. There are many robust nonparametric tests to compare two samples, such as permutation te |
41,639 | What is the meaning of these terms in image processing? | In image processing, computer vision and related fields, an image moment is a certain particular weighted average (moment) of the image pixels' intensities, or a function of such moments, usually chosen to have some attractive property or interpretation.
Local and global characteristics. Example, suppose we want to smo... | What is the meaning of these terms in image processing? | In image processing, computer vision and related fields, an image moment is a certain particular weighted average (moment) of the image pixels' intensities, or a function of such moments, usually chos | What is the meaning of these terms in image processing?
In image processing, computer vision and related fields, an image moment is a certain particular weighted average (moment) of the image pixels' intensities, or a function of such moments, usually chosen to have some attractive property or interpretation.
Local and... | What is the meaning of these terms in image processing?
In image processing, computer vision and related fields, an image moment is a certain particular weighted average (moment) of the image pixels' intensities, or a function of such moments, usually chos |
41,640 | Normal Approximation of the sum of correlated Bernoulli Random Variables | If the number of variables is sufficiently large and the correlation is bounded away from 1, then there are Central Limit Theorems that apply (e.g., also see versions of the CLT for stationary processes).
So if your $i$-th variable has parameter $p_i$, then the variance of $X_i$ is
$p_i(1-p_i)$ and :
$$\text{Cov}(X_i... | Normal Approximation of the sum of correlated Bernoulli Random Variables | If the number of variables is sufficiently large and the correlation is bounded away from 1, then there are Central Limit Theorems that apply (e.g., also see versions of the CLT for stationary process | Normal Approximation of the sum of correlated Bernoulli Random Variables
If the number of variables is sufficiently large and the correlation is bounded away from 1, then there are Central Limit Theorems that apply (e.g., also see versions of the CLT for stationary processes).
So if your $i$-th variable has parameter $... | Normal Approximation of the sum of correlated Bernoulli Random Variables
If the number of variables is sufficiently large and the correlation is bounded away from 1, then there are Central Limit Theorems that apply (e.g., also see versions of the CLT for stationary process |
41,641 | Kernel smoothing for Edgeworth expansion | I don't know what estimator you are considering but what you propose has certainly been done before.
Horowitz (1998) investigates whether the bootstrap can be used for asymptotic refinements of median regression. He faces the same problem as you given that the objective function has an embedded indicator
$$\widehat{\be... | Kernel smoothing for Edgeworth expansion | I don't know what estimator you are considering but what you propose has certainly been done before.
Horowitz (1998) investigates whether the bootstrap can be used for asymptotic refinements of median | Kernel smoothing for Edgeworth expansion
I don't know what estimator you are considering but what you propose has certainly been done before.
Horowitz (1998) investigates whether the bootstrap can be used for asymptotic refinements of median regression. He faces the same problem as you given that the objective function... | Kernel smoothing for Edgeworth expansion
I don't know what estimator you are considering but what you propose has certainly been done before.
Horowitz (1998) investigates whether the bootstrap can be used for asymptotic refinements of median |
41,642 | How good is Monte Carlo Simulation when the variable distribution is unknown? | In practice simulation is almost always done without full certainty about the distribution.
For example, one might consider a lognormal distribution for some random variable... but what if it was a little lighter tailed (gamma, say)? Or heavier tailed (inverse gamma, say)? Or much heavier tailed (inverse Gaussian, say)... | How good is Monte Carlo Simulation when the variable distribution is unknown? | In practice simulation is almost always done without full certainty about the distribution.
For example, one might consider a lognormal distribution for some random variable... but what if it was a li | How good is Monte Carlo Simulation when the variable distribution is unknown?
In practice simulation is almost always done without full certainty about the distribution.
For example, one might consider a lognormal distribution for some random variable... but what if it was a little lighter tailed (gamma, say)? Or heavi... | How good is Monte Carlo Simulation when the variable distribution is unknown?
In practice simulation is almost always done without full certainty about the distribution.
For example, one might consider a lognormal distribution for some random variable... but what if it was a li |
41,643 | Does the slope of a regression between observed and predicted values always equal the $R^2$ of the original model? | I always get exactly the same value for the overall model fit $R^2$ and slope of the observed versus predicted regression.
This will be true provided a constant term is included in the overall model. Why?
$R^2$ measures the variance of the fit $\hat Y$ relative to the variance of $Y$ (provided the model includes a c... | Does the slope of a regression between observed and predicted values always equal the $R^2$ of the o | I always get exactly the same value for the overall model fit $R^2$ and slope of the observed versus predicted regression.
This will be true provided a constant term is included in the overall model. | Does the slope of a regression between observed and predicted values always equal the $R^2$ of the original model?
I always get exactly the same value for the overall model fit $R^2$ and slope of the observed versus predicted regression.
This will be true provided a constant term is included in the overall model. Why... | Does the slope of a regression between observed and predicted values always equal the $R^2$ of the o
I always get exactly the same value for the overall model fit $R^2$ and slope of the observed versus predicted regression.
This will be true provided a constant term is included in the overall model. |
41,644 | Does the slope of a regression between observed and predicted values always equal the $R^2$ of the original model? | $R^2$ doesn't have to be equal to $\beta$. you either have a rare coincidence, or reading the same field from the fit object somehow. | Does the slope of a regression between observed and predicted values always equal the $R^2$ of the o | $R^2$ doesn't have to be equal to $\beta$. you either have a rare coincidence, or reading the same field from the fit object somehow. | Does the slope of a regression between observed and predicted values always equal the $R^2$ of the original model?
$R^2$ doesn't have to be equal to $\beta$. you either have a rare coincidence, or reading the same field from the fit object somehow. | Does the slope of a regression between observed and predicted values always equal the $R^2$ of the o
$R^2$ doesn't have to be equal to $\beta$. you either have a rare coincidence, or reading the same field from the fit object somehow. |
41,645 | Modelling slopes over time | Instead of running the models for your two periods separately
$$y_1 = \alpha_1 + \beta_1 X_1 + u_1$$
$$y_2 = \alpha_2 + \beta_2 X_2 + u_2$$
you can combine them as
$$y_t = (X_1\cdot d_1)\beta_1 + (X_2\cdot d_2)\beta_2 + d_1\cdot u_1 + d_2\cdot u_2$$
where $d_1$ is a dummy which equals one for the period pre day 150 and... | Modelling slopes over time | Instead of running the models for your two periods separately
$$y_1 = \alpha_1 + \beta_1 X_1 + u_1$$
$$y_2 = \alpha_2 + \beta_2 X_2 + u_2$$
you can combine them as
$$y_t = (X_1\cdot d_1)\beta_1 + (X_2 | Modelling slopes over time
Instead of running the models for your two periods separately
$$y_1 = \alpha_1 + \beta_1 X_1 + u_1$$
$$y_2 = \alpha_2 + \beta_2 X_2 + u_2$$
you can combine them as
$$y_t = (X_1\cdot d_1)\beta_1 + (X_2\cdot d_2)\beta_2 + d_1\cdot u_1 + d_2\cdot u_2$$
where $d_1$ is a dummy which equals one for... | Modelling slopes over time
Instead of running the models for your two periods separately
$$y_1 = \alpha_1 + \beta_1 X_1 + u_1$$
$$y_2 = \alpha_2 + \beta_2 X_2 + u_2$$
you can combine them as
$$y_t = (X_1\cdot d_1)\beta_1 + (X_2 |
41,646 | The between estimator in panel data | As you said correctly, the between estimator takes the individual effects model
$$y_{it} = \alpha_i + x'_{it}\beta + \epsilon_{it}$$
and averages out the time component resulting in the regression
$$\overline{y}_{i.} = \alpha + \overline{x}'_{i.} + (\alpha_i - \alpha + \overline{\epsilon}_{i.})$$
where bars indicate av... | The between estimator in panel data | As you said correctly, the between estimator takes the individual effects model
$$y_{it} = \alpha_i + x'_{it}\beta + \epsilon_{it}$$
and averages out the time component resulting in the regression
$$\ | The between estimator in panel data
As you said correctly, the between estimator takes the individual effects model
$$y_{it} = \alpha_i + x'_{it}\beta + \epsilon_{it}$$
and averages out the time component resulting in the regression
$$\overline{y}_{i.} = \alpha + \overline{x}'_{i.} + (\alpha_i - \alpha + \overline{\eps... | The between estimator in panel data
As you said correctly, the between estimator takes the individual effects model
$$y_{it} = \alpha_i + x'_{it}\beta + \epsilon_{it}$$
and averages out the time component resulting in the regression
$$\ |
41,647 | Clustering in Cox proportional hazards model MLM vs. sandwich estimator | Neglecting the clustering is, I think, usual practice when analysing multicentre clinical trials based on a time-to-event outcome. That is, the standard Cox model is used. In that case, the treatment effect has a population-averaged (marginal) interpretation, the effect being averaged over all centres. It is known that... | Clustering in Cox proportional hazards model MLM vs. sandwich estimator | Neglecting the clustering is, I think, usual practice when analysing multicentre clinical trials based on a time-to-event outcome. That is, the standard Cox model is used. In that case, the treatment | Clustering in Cox proportional hazards model MLM vs. sandwich estimator
Neglecting the clustering is, I think, usual practice when analysing multicentre clinical trials based on a time-to-event outcome. That is, the standard Cox model is used. In that case, the treatment effect has a population-averaged (marginal) inte... | Clustering in Cox proportional hazards model MLM vs. sandwich estimator
Neglecting the clustering is, I think, usual practice when analysing multicentre clinical trials based on a time-to-event outcome. That is, the standard Cox model is used. In that case, the treatment |
41,648 | The Conjugate Beta Prior proof | To go from the third to fourth row just ignore factors that are constant with respect to $\pi$. That is, let
$$\binom{n}{y}\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}=k$$
so
$$p(\pi|y)=k \cdot \pi^y (1-\pi)^{(n-y)}\pi^{(\alpha-1)}(1-\pi)^{(\beta-1)}$$
or
$$p(\pi|y) \propto \pi^y(1-\pi)^{(n-y)}\pi^{(\alpha... | The Conjugate Beta Prior proof | To go from the third to fourth row just ignore factors that are constant with respect to $\pi$. That is, let
$$\binom{n}{y}\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}=k$$
so
$$p(\pi|y)=k | The Conjugate Beta Prior proof
To go from the third to fourth row just ignore factors that are constant with respect to $\pi$. That is, let
$$\binom{n}{y}\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}=k$$
so
$$p(\pi|y)=k \cdot \pi^y (1-\pi)^{(n-y)}\pi^{(\alpha-1)}(1-\pi)^{(\beta-1)}$$
or
$$p(\pi|y) \propto \... | The Conjugate Beta Prior proof
To go from the third to fourth row just ignore factors that are constant with respect to $\pi$. That is, let
$$\binom{n}{y}\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}=k$$
so
$$p(\pi|y)=k |
41,649 | Compare maxima of two Gaussian samples | I found the answer, thanks mainly to whuber's comment. Let $M_y$ and $M_x$ denote the sample maxima of $Y$ and $X$, respectively. Scaled versions of $M_y$ and $M_x$ are both Gumbel distributed [source 1], and the difference between two Gumbels is logistic [source 2]. Specifically, we have
$$\sqrt{2 \ln n}~(M_y-M_x-c) \... | Compare maxima of two Gaussian samples | I found the answer, thanks mainly to whuber's comment. Let $M_y$ and $M_x$ denote the sample maxima of $Y$ and $X$, respectively. Scaled versions of $M_y$ and $M_x$ are both Gumbel distributed [source | Compare maxima of two Gaussian samples
I found the answer, thanks mainly to whuber's comment. Let $M_y$ and $M_x$ denote the sample maxima of $Y$ and $X$, respectively. Scaled versions of $M_y$ and $M_x$ are both Gumbel distributed [source 1], and the difference between two Gumbels is logistic [source 2]. Specifically,... | Compare maxima of two Gaussian samples
I found the answer, thanks mainly to whuber's comment. Let $M_y$ and $M_x$ denote the sample maxima of $Y$ and $X$, respectively. Scaled versions of $M_y$ and $M_x$ are both Gumbel distributed [source |
41,650 | Large difference between Mann-Whitney test and Wilcoxon signed rank test significance | In scipy.stats, the Mann-Whitney U test compares two populations:
Computes the Mann-Whitney rank test on samples x and y.
but the Wilcoxon test compares two PAIRED populations:
The Wilcoxon signed-rank test tests the null hypothesis that two
related paired samples come from the same distribution. In particular,
... | Large difference between Mann-Whitney test and Wilcoxon signed rank test significance | In scipy.stats, the Mann-Whitney U test compares two populations:
Computes the Mann-Whitney rank test on samples x and y.
but the Wilcoxon test compares two PAIRED populations:
The Wilcoxon signed- | Large difference between Mann-Whitney test and Wilcoxon signed rank test significance
In scipy.stats, the Mann-Whitney U test compares two populations:
Computes the Mann-Whitney rank test on samples x and y.
but the Wilcoxon test compares two PAIRED populations:
The Wilcoxon signed-rank test tests the null hypothesi... | Large difference between Mann-Whitney test and Wilcoxon signed rank test significance
In scipy.stats, the Mann-Whitney U test compares two populations:
Computes the Mann-Whitney rank test on samples x and y.
but the Wilcoxon test compares two PAIRED populations:
The Wilcoxon signed- |
41,651 | Visual display of multiple comparisons test | iv <- c("A","B","C","D")
dv <- c(1.2,2.3,4.5,6.7)
gp <- c(1,1,1,2)
par(mai=c(1,1,0,0))
plot(dv, gp, axes=F, xlab="Average time", ylab="Grouping based on
\n mean comparison",
ylim=c(0,3), xlim=c(0,7), pch=16)
text(dv, gp-.2, iv)
axis(side=2, label=c("i", "ii"), at=c(1,2))
axis(side=1)
abline(h=... | Visual display of multiple comparisons test | iv <- c("A","B","C","D")
dv <- c(1.2,2.3,4.5,6.7)
gp <- c(1,1,1,2)
par(mai=c(1,1,0,0))
plot(dv, gp, axes=F, xlab="Average time", ylab="Grouping based on
\n mean comparison",
| Visual display of multiple comparisons test
iv <- c("A","B","C","D")
dv <- c(1.2,2.3,4.5,6.7)
gp <- c(1,1,1,2)
par(mai=c(1,1,0,0))
plot(dv, gp, axes=F, xlab="Average time", ylab="Grouping based on
\n mean comparison",
ylim=c(0,3), xlim=c(0,7), pch=16)
text(dv, gp-.2, iv)
axis(side=2, label=c("... | Visual display of multiple comparisons test
iv <- c("A","B","C","D")
dv <- c(1.2,2.3,4.5,6.7)
gp <- c(1,1,1,2)
par(mai=c(1,1,0,0))
plot(dv, gp, axes=F, xlab="Average time", ylab="Grouping based on
\n mean comparison",
|
41,652 | Visual display of multiple comparisons test | Based on your question and follow-up comments, I'd start with a dot-plot. They're quick and easy (even in Excel). Here's s sample with your data:
This chart type scales well, handles large numbers of data points well and is very easy to understand-even to a non-tech audience. | Visual display of multiple comparisons test | Based on your question and follow-up comments, I'd start with a dot-plot. They're quick and easy (even in Excel). Here's s sample with your data:
This chart type scales well, handles large numbers o | Visual display of multiple comparisons test
Based on your question and follow-up comments, I'd start with a dot-plot. They're quick and easy (even in Excel). Here's s sample with your data:
This chart type scales well, handles large numbers of data points well and is very easy to understand-even to a non-tech audienc... | Visual display of multiple comparisons test
Based on your question and follow-up comments, I'd start with a dot-plot. They're quick and easy (even in Excel). Here's s sample with your data:
This chart type scales well, handles large numbers o |
41,653 | Visual display of multiple comparisons test | Point is, that your dataset is too small (4 groups 5 values each). The means obtained from such data are not very accurate representative values for each group - and therefore you should not run ANOVA to make inference about differences among group.
One thing is to be understandable to the audience but more important ... | Visual display of multiple comparisons test | Point is, that your dataset is too small (4 groups 5 values each). The means obtained from such data are not very accurate representative values for each group - and therefore you should not run ANOVA | Visual display of multiple comparisons test
Point is, that your dataset is too small (4 groups 5 values each). The means obtained from such data are not very accurate representative values for each group - and therefore you should not run ANOVA to make inference about differences among group.
One thing is to be unders... | Visual display of multiple comparisons test
Point is, that your dataset is too small (4 groups 5 values each). The means obtained from such data are not very accurate representative values for each group - and therefore you should not run ANOVA |
41,654 | Using the eigenvalues from PCA in k-nearest-neighbours | There are many options you could pursue, I can suggest a few.
First, if you already have a training set, and assuming the training set is large enough, you could learn a distance metric instead of using PCA weights-based interpretation.
See Mahalanobis Distance as an example of distance metric learning.
The main idea i... | Using the eigenvalues from PCA in k-nearest-neighbours | There are many options you could pursue, I can suggest a few.
First, if you already have a training set, and assuming the training set is large enough, you could learn a distance metric instead of usi | Using the eigenvalues from PCA in k-nearest-neighbours
There are many options you could pursue, I can suggest a few.
First, if you already have a training set, and assuming the training set is large enough, you could learn a distance metric instead of using PCA weights-based interpretation.
See Mahalanobis Distance as ... | Using the eigenvalues from PCA in k-nearest-neighbours
There are many options you could pursue, I can suggest a few.
First, if you already have a training set, and assuming the training set is large enough, you could learn a distance metric instead of usi |
41,655 | "One-tailed" Levene Test | Browne-Forsythe simply performs ANOVA on $z_{ij}=|y_{ij} - \tilde{y}_j|$, where $y_{ij}$ is the $i$th observation in group $j$. Groups with larger spread in $y$ will have larger mean $z$. (Levene is similar, but the $z$'s are defined in terms of the deviations from the group mean instead of the group median.)
If you on... | "One-tailed" Levene Test | Browne-Forsythe simply performs ANOVA on $z_{ij}=|y_{ij} - \tilde{y}_j|$, where $y_{ij}$ is the $i$th observation in group $j$. Groups with larger spread in $y$ will have larger mean $z$. (Levene is s | "One-tailed" Levene Test
Browne-Forsythe simply performs ANOVA on $z_{ij}=|y_{ij} - \tilde{y}_j|$, where $y_{ij}$ is the $i$th observation in group $j$. Groups with larger spread in $y$ will have larger mean $z$. (Levene is similar, but the $z$'s are defined in terms of the deviations from the group mean instead of the... | "One-tailed" Levene Test
Browne-Forsythe simply performs ANOVA on $z_{ij}=|y_{ij} - \tilde{y}_j|$, where $y_{ij}$ is the $i$th observation in group $j$. Groups with larger spread in $y$ will have larger mean $z$. (Levene is s |
41,656 | "One-tailed" Levene Test | I know this is 2 years later, but you may want to investigate the software you are using to see if it's already a one-way test. In R there are 3 packages that contain Brown-Forsythe (which is simply Levene's, but uses median instead of mean), and all of them are the one-sided test.
You can test this yourself just by l... | "One-tailed" Levene Test | I know this is 2 years later, but you may want to investigate the software you are using to see if it's already a one-way test. In R there are 3 packages that contain Brown-Forsythe (which is simply | "One-tailed" Levene Test
I know this is 2 years later, but you may want to investigate the software you are using to see if it's already a one-way test. In R there are 3 packages that contain Brown-Forsythe (which is simply Levene's, but uses median instead of mean), and all of them are the one-sided test.
You can tes... | "One-tailed" Levene Test
I know this is 2 years later, but you may want to investigate the software you are using to see if it's already a one-way test. In R there are 3 packages that contain Brown-Forsythe (which is simply |
41,657 | Assigning values to missing data for use in binary logistic regression in SAS | In general, dealing with missing input values is always problematic. To my best knowledge, none of the existing methods can deal with it without introducing some bias to the model, so you have to consider this during your research. There are at least few possible options:
ignore data with missing values (which I do be... | Assigning values to missing data for use in binary logistic regression in SAS | In general, dealing with missing input values is always problematic. To my best knowledge, none of the existing methods can deal with it without introducing some bias to the model, so you have to cons | Assigning values to missing data for use in binary logistic regression in SAS
In general, dealing with missing input values is always problematic. To my best knowledge, none of the existing methods can deal with it without introducing some bias to the model, so you have to consider this during your research. There are ... | Assigning values to missing data for use in binary logistic regression in SAS
In general, dealing with missing input values is always problematic. To my best knowledge, none of the existing methods can deal with it without introducing some bias to the model, so you have to cons |
41,658 | Confidence interval for Uniform($\theta$, $\theta + a$) | It's possible to construct silly confidence intervals that are technically valid, i.e. have the claimed coverage, but it's not obligatory. So I don't know why doing so should demonstrate a flaw in "confidence interval methodology" (whatever that is)—you could have constructed a more appropriate one for your purposes.
A... | Confidence interval for Uniform($\theta$, $\theta + a$) | It's possible to construct silly confidence intervals that are technically valid, i.e. have the claimed coverage, but it's not obligatory. So I don't know why doing so should demonstrate a flaw in "co | Confidence interval for Uniform($\theta$, $\theta + a$)
It's possible to construct silly confidence intervals that are technically valid, i.e. have the claimed coverage, but it's not obligatory. So I don't know why doing so should demonstrate a flaw in "confidence interval methodology" (whatever that is)—you could have... | Confidence interval for Uniform($\theta$, $\theta + a$)
It's possible to construct silly confidence intervals that are technically valid, i.e. have the claimed coverage, but it's not obligatory. So I don't know why doing so should demonstrate a flaw in "co |
41,659 | Entropy of (Sum of Gaussians) versus Sum of (Entropy of Gaussians) | The issue is that you are working with a differential entropy for continuous random variables, which doesn't share all the nice properties of Shannon's entropy for discrete random variables and can behave counter to intuition. In particular, differential entropy can be negative!
The following might help to get a feel f... | Entropy of (Sum of Gaussians) versus Sum of (Entropy of Gaussians) | The issue is that you are working with a differential entropy for continuous random variables, which doesn't share all the nice properties of Shannon's entropy for discrete random variables and can be | Entropy of (Sum of Gaussians) versus Sum of (Entropy of Gaussians)
The issue is that you are working with a differential entropy for continuous random variables, which doesn't share all the nice properties of Shannon's entropy for discrete random variables and can behave counter to intuition. In particular, differentia... | Entropy of (Sum of Gaussians) versus Sum of (Entropy of Gaussians)
The issue is that you are working with a differential entropy for continuous random variables, which doesn't share all the nice properties of Shannon's entropy for discrete random variables and can be |
41,660 | McNemar-Bowker test for comparison of two measures | McNemar-Bowker test of symmetry of k X k contingency table is inherently 2-sided: the alternative hypothesis is undirected. So, in general case it cannot be used to test a one sided alternative that subdiagonal frequencies are larger/smaller than superdiagonal frequencies. But since in your case the differences are con... | McNemar-Bowker test for comparison of two measures | McNemar-Bowker test of symmetry of k X k contingency table is inherently 2-sided: the alternative hypothesis is undirected. So, in general case it cannot be used to test a one sided alternative that s | McNemar-Bowker test for comparison of two measures
McNemar-Bowker test of symmetry of k X k contingency table is inherently 2-sided: the alternative hypothesis is undirected. So, in general case it cannot be used to test a one sided alternative that subdiagonal frequencies are larger/smaller than superdiagonal frequenc... | McNemar-Bowker test for comparison of two measures
McNemar-Bowker test of symmetry of k X k contingency table is inherently 2-sided: the alternative hypothesis is undirected. So, in general case it cannot be used to test a one sided alternative that s |
41,661 | What to do when an independent variable is not significant, but it definitely should be! | One thing you can do is look at the effect sizes and build a confidence interval around them. “Very” significant coefficients do not necessarily represent strong effects and test results depend a lot on the sample size. Therefore, a failure to reject the null hypothesis is not as such evidence that your results are inc... | What to do when an independent variable is not significant, but it definitely should be! | One thing you can do is look at the effect sizes and build a confidence interval around them. “Very” significant coefficients do not necessarily represent strong effects and test results depend a lot | What to do when an independent variable is not significant, but it definitely should be!
One thing you can do is look at the effect sizes and build a confidence interval around them. “Very” significant coefficients do not necessarily represent strong effects and test results depend a lot on the sample size. Therefore, ... | What to do when an independent variable is not significant, but it definitely should be!
One thing you can do is look at the effect sizes and build a confidence interval around them. “Very” significant coefficients do not necessarily represent strong effects and test results depend a lot |
41,662 | What to do when an independent variable is not significant, but it definitely should be! | After reading more I found out that the values are not significant because the fixed effects model does not work well with the data. I have a high heterskedasticity, which is not fixed by transforming the data into its log form. So far, the OLS with Robust Error adjustment for heteroskedasticity is what is giving us t... | What to do when an independent variable is not significant, but it definitely should be! | After reading more I found out that the values are not significant because the fixed effects model does not work well with the data. I have a high heterskedasticity, which is not fixed by transforming | What to do when an independent variable is not significant, but it definitely should be!
After reading more I found out that the values are not significant because the fixed effects model does not work well with the data. I have a high heterskedasticity, which is not fixed by transforming the data into its log form. S... | What to do when an independent variable is not significant, but it definitely should be!
After reading more I found out that the values are not significant because the fixed effects model does not work well with the data. I have a high heterskedasticity, which is not fixed by transforming |
41,663 | Are these variables mutually independent? | Even assuming the $X_i$ and $Y_j$ are independent of $N$ and of one another, $S_X$ and $S_Y$ will be positively correlated and (therefore) not independent. We can calculate that correlation given the distribution of $N$.
To illustrate, let $N$ take on either the value $1$ or $12$ with equal probability. Then half the... | Are these variables mutually independent? | Even assuming the $X_i$ and $Y_j$ are independent of $N$ and of one another, $S_X$ and $S_Y$ will be positively correlated and (therefore) not independent. We can calculate that correlation given the | Are these variables mutually independent?
Even assuming the $X_i$ and $Y_j$ are independent of $N$ and of one another, $S_X$ and $S_Y$ will be positively correlated and (therefore) not independent. We can calculate that correlation given the distribution of $N$.
To illustrate, let $N$ take on either the value $1$ or $... | Are these variables mutually independent?
Even assuming the $X_i$ and $Y_j$ are independent of $N$ and of one another, $S_X$ and $S_Y$ will be positively correlated and (therefore) not independent. We can calculate that correlation given the |
41,664 | ANOVA and Regression give opposite results in R | Essentially, the question is, how come that one coefficient in the linear model is significantly different from 0, but ANOVA shows no significant effect and vice versa.
For this, let's consider a simpler example.
set.seed( 123 )
data <- data.frame( x= rnorm( 100 ), g= rep( letters[1:10], each= 10 ) )
data$x[ data$g == ... | ANOVA and Regression give opposite results in R | Essentially, the question is, how come that one coefficient in the linear model is significantly different from 0, but ANOVA shows no significant effect and vice versa.
For this, let's consider a simp | ANOVA and Regression give opposite results in R
Essentially, the question is, how come that one coefficient in the linear model is significantly different from 0, but ANOVA shows no significant effect and vice versa.
For this, let's consider a simpler example.
set.seed( 123 )
data <- data.frame( x= rnorm( 100 ), g= rep... | ANOVA and Regression give opposite results in R
Essentially, the question is, how come that one coefficient in the linear model is significantly different from 0, but ANOVA shows no significant effect and vice versa.
For this, let's consider a simp |
41,665 | ANOVA and Regression give opposite results in R | What is going on here is a multiple comparisons issue. You have 10 df for interaction so you can look at 10 independent interaction effects (although I suspect the 10 you are actually looking at, i.e. the 10 regression effects, are not independent).
The 10 df interaction test will be significant at some level if and... | ANOVA and Regression give opposite results in R | What is going on here is a multiple comparisons issue. You have 10 df for interaction so you can look at 10 independent interaction effects (although I suspect the 10 you are actually looking at, i.e | ANOVA and Regression give opposite results in R
What is going on here is a multiple comparisons issue. You have 10 df for interaction so you can look at 10 independent interaction effects (although I suspect the 10 you are actually looking at, i.e. the 10 regression effects, are not independent).
The 10 df interacti... | ANOVA and Regression give opposite results in R
What is going on here is a multiple comparisons issue. You have 10 df for interaction so you can look at 10 independent interaction effects (although I suspect the 10 you are actually looking at, i.e |
41,666 | How many observations per subject are necessary to fit a random slope in a mixed model? | In a basic mixed effects model,
$$E[Y_{it}|X_{it}] = \alpha + \sigma_i^2 + \beta X_{it}$$
clusters having just one observation contribute influence to both the estimated variance of the random effect and the slope of the fixed effect. This is because the random intercept is never actually estimated. While some numeric... | How many observations per subject are necessary to fit a random slope in a mixed model? | In a basic mixed effects model,
$$E[Y_{it}|X_{it}] = \alpha + \sigma_i^2 + \beta X_{it}$$
clusters having just one observation contribute influence to both the estimated variance of the random effect | How many observations per subject are necessary to fit a random slope in a mixed model?
In a basic mixed effects model,
$$E[Y_{it}|X_{it}] = \alpha + \sigma_i^2 + \beta X_{it}$$
clusters having just one observation contribute influence to both the estimated variance of the random effect and the slope of the fixed effe... | How many observations per subject are necessary to fit a random slope in a mixed model?
In a basic mixed effects model,
$$E[Y_{it}|X_{it}] = \alpha + \sigma_i^2 + \beta X_{it}$$
clusters having just one observation contribute influence to both the estimated variance of the random effect |
41,667 | The dangers of stepwise variable selection in regression | The figure shows the distribution of estimated slope parameters over all models, not just those which were significantly different from zero. The spike at zero represents all the models where the slope was deemed insignificant, and so a zero-slope model was used. The point is to demonstrate that the variable-selection ... | The dangers of stepwise variable selection in regression | The figure shows the distribution of estimated slope parameters over all models, not just those which were significantly different from zero. The spike at zero represents all the models where the slop | The dangers of stepwise variable selection in regression
The figure shows the distribution of estimated slope parameters over all models, not just those which were significantly different from zero. The spike at zero represents all the models where the slope was deemed insignificant, and so a zero-slope model was used.... | The dangers of stepwise variable selection in regression
The figure shows the distribution of estimated slope parameters over all models, not just those which were significantly different from zero. The spike at zero represents all the models where the slop |
41,668 | Standard error of the quotient of two estimates (Wald estimators) using the delta method | Here's an example in Stata of how to create the ratio and test a hypothesis using nlcom:
. webuse regress
. regress y x1 x2 x3
Source | SS df MS Number of obs = 148
-------------+------------------------------ F( 3, 144) = 96.12
Model | 3259.3561 3 ... | Standard error of the quotient of two estimates (Wald estimators) using the delta method | Here's an example in Stata of how to create the ratio and test a hypothesis using nlcom:
. webuse regress
. regress y x1 x2 x3
Source | SS df MS Number of obs = | Standard error of the quotient of two estimates (Wald estimators) using the delta method
Here's an example in Stata of how to create the ratio and test a hypothesis using nlcom:
. webuse regress
. regress y x1 x2 x3
Source | SS df MS Number of obs = 148
-------------+---------... | Standard error of the quotient of two estimates (Wald estimators) using the delta method
Here's an example in Stata of how to create the ratio and test a hypothesis using nlcom:
. webuse regress
. regress y x1 x2 x3
Source | SS df MS Number of obs = |
41,669 | Standard error of the quotient of two estimates (Wald estimators) using the delta method | Another method of looking at the ratio is due to Fieller. An excellent post is at: How to compute the confidence interval of the ratio of two normal means | Standard error of the quotient of two estimates (Wald estimators) using the delta method | Another method of looking at the ratio is due to Fieller. An excellent post is at: How to compute the confidence interval of the ratio of two normal means | Standard error of the quotient of two estimates (Wald estimators) using the delta method
Another method of looking at the ratio is due to Fieller. An excellent post is at: How to compute the confidence interval of the ratio of two normal means | Standard error of the quotient of two estimates (Wald estimators) using the delta method
Another method of looking at the ratio is due to Fieller. An excellent post is at: How to compute the confidence interval of the ratio of two normal means |
41,670 | What is the reasoning behind the formulae for the different standard errors of measurement? | Okay, I'm not sure if anyone's still checking this (since it's from a year ago), but here's my response to your question: "I can follow the definitions but the article doesn't really explain why the calculations should be different in the different contexts. Is anyone able to provide an explanation?"
The calculations a... | What is the reasoning behind the formulae for the different standard errors of measurement? | Okay, I'm not sure if anyone's still checking this (since it's from a year ago), but here's my response to your question: "I can follow the definitions but the article doesn't really explain why the c | What is the reasoning behind the formulae for the different standard errors of measurement?
Okay, I'm not sure if anyone's still checking this (since it's from a year ago), but here's my response to your question: "I can follow the definitions but the article doesn't really explain why the calculations should be differ... | What is the reasoning behind the formulae for the different standard errors of measurement?
Okay, I'm not sure if anyone's still checking this (since it's from a year ago), but here's my response to your question: "I can follow the definitions but the article doesn't really explain why the c |
41,671 | What is the reasoning behind the formulae for the different standard errors of measurement? | Only the first of those is a standard error of measurement. Calling all three of them by the same name only confuses things. See sec 3.8 Errors of Measurement, Estimation, and Prediction, pp 66-69 in Lord & Novick's Statistical Theories of Mental Test Scores, Addison-Wesley, 1968. | What is the reasoning behind the formulae for the different standard errors of measurement? | Only the first of those is a standard error of measurement. Calling all three of them by the same name only confuses things. See sec 3.8 Errors of Measurement, Estimation, and Prediction, pp 66-69 in | What is the reasoning behind the formulae for the different standard errors of measurement?
Only the first of those is a standard error of measurement. Calling all three of them by the same name only confuses things. See sec 3.8 Errors of Measurement, Estimation, and Prediction, pp 66-69 in Lord & Novick's Statistical ... | What is the reasoning behind the formulae for the different standard errors of measurement?
Only the first of those is a standard error of measurement. Calling all three of them by the same name only confuses things. See sec 3.8 Errors of Measurement, Estimation, and Prediction, pp 66-69 in |
41,672 | What is the reasoning behind the formulae for the different standard errors of measurement? | You are measuring 3 differents things, why do you expect to have only one standard error ? Intuitively, you will probably have differents means. This is the same thing for standard error.
With your exemple you can see that in the thrid process you take 2 measures, this is likely to be more reliable then taking only on... | What is the reasoning behind the formulae for the different standard errors of measurement? | You are measuring 3 differents things, why do you expect to have only one standard error ? Intuitively, you will probably have differents means. This is the same thing for standard error.
With your e | What is the reasoning behind the formulae for the different standard errors of measurement?
You are measuring 3 differents things, why do you expect to have only one standard error ? Intuitively, you will probably have differents means. This is the same thing for standard error.
With your exemple you can see that in t... | What is the reasoning behind the formulae for the different standard errors of measurement?
You are measuring 3 differents things, why do you expect to have only one standard error ? Intuitively, you will probably have differents means. This is the same thing for standard error.
With your e |
41,673 | Iterative proportional fitting in R | This is old, but here we go:
As @Henrico wrote, it seems that what are you trying to achieve is indeed raking. Instead of using survey::rake you might fit the distribution to the marginals "by hand" using a Poisson GLM, as suggested by @DWin. To get the right frequencies you need to use an offset.
Let
$n_{ij}$ be yo... | Iterative proportional fitting in R | This is old, but here we go:
As @Henrico wrote, it seems that what are you trying to achieve is indeed raking. Instead of using survey::rake you might fit the distribution to the marginals "by hand" u | Iterative proportional fitting in R
This is old, but here we go:
As @Henrico wrote, it seems that what are you trying to achieve is indeed raking. Instead of using survey::rake you might fit the distribution to the marginals "by hand" using a Poisson GLM, as suggested by @DWin. To get the right frequencies you need to ... | Iterative proportional fitting in R
This is old, but here we go:
As @Henrico wrote, it seems that what are you trying to achieve is indeed raking. Instead of using survey::rake you might fit the distribution to the marginals "by hand" u |
41,674 | Iterative proportional fitting in R | Make a glm fit to the marginals with Poisson errors (yielding a log-linear model) and then use predict on expand.grid data.frame from the the row and column values based of the second sample. (There's no particular advantage that I can see in using IPF to estimate a log-linear model of this sort.)
require(reshape2)
L... | Iterative proportional fitting in R | Make a glm fit to the marginals with Poisson errors (yielding a log-linear model) and then use predict on expand.grid data.frame from the the row and column values based of the second sample. (There | Iterative proportional fitting in R
Make a glm fit to the marginals with Poisson errors (yielding a log-linear model) and then use predict on expand.grid data.frame from the the row and column values based of the second sample. (There's no particular advantage that I can see in using IPF to estimate a log-linear mode... | Iterative proportional fitting in R
Make a glm fit to the marginals with Poisson errors (yielding a log-linear model) and then use predict on expand.grid data.frame from the the row and column values based of the second sample. (There |
41,675 | Equivalence between single sample cross-validation index and the Akaike information criterion for prediction | You only ever really need to fit the full model once for cross validation. You can use the results from the full run to work out the residuals from predicting a subset. Now suppose you consider a specific group of observations, say $m$, where $n-m\geq p$ where $n$ is the number of samples and $p$ is the number of bet... | Equivalence between single sample cross-validation index and the Akaike information criterion for pr | You only ever really need to fit the full model once for cross validation. You can use the results from the full run to work out the residuals from predicting a subset. Now suppose you consider a sp | Equivalence between single sample cross-validation index and the Akaike information criterion for prediction
You only ever really need to fit the full model once for cross validation. You can use the results from the full run to work out the residuals from predicting a subset. Now suppose you consider a specific grou... | Equivalence between single sample cross-validation index and the Akaike information criterion for pr
You only ever really need to fit the full model once for cross validation. You can use the results from the full run to work out the residuals from predicting a subset. Now suppose you consider a sp |
41,676 | Sampling from the normal-gamma distribution in R | Sampling from normal-gamma distribution is easy, and in fact the algorithm is described on Wikipedia:
Generation of random variates is straightforward:
Sample $\tau$ from a gamma distribution with parameters $\alpha$ and $\beta$
Sample $x$ from a normal distribution with mean $\mu$ and variance $1/(\lambda \tau)$
W... | Sampling from the normal-gamma distribution in R | Sampling from normal-gamma distribution is easy, and in fact the algorithm is described on Wikipedia:
Generation of random variates is straightforward:
Sample $\tau$ from a gamma distribution with p | Sampling from the normal-gamma distribution in R
Sampling from normal-gamma distribution is easy, and in fact the algorithm is described on Wikipedia:
Generation of random variates is straightforward:
Sample $\tau$ from a gamma distribution with parameters $\alpha$ and $\beta$
Sample $x$ from a normal distribution wi... | Sampling from the normal-gamma distribution in R
Sampling from normal-gamma distribution is easy, and in fact the algorithm is described on Wikipedia:
Generation of random variates is straightforward:
Sample $\tau$ from a gamma distribution with p |
41,677 | Sampling from the normal-gamma distribution in R | Use the rigamma() function from the pscl package on CRAN. You can also take a look at the ghyp package. | Sampling from the normal-gamma distribution in R | Use the rigamma() function from the pscl package on CRAN. You can also take a look at the ghyp package. | Sampling from the normal-gamma distribution in R
Use the rigamma() function from the pscl package on CRAN. You can also take a look at the ghyp package. | Sampling from the normal-gamma distribution in R
Use the rigamma() function from the pscl package on CRAN. You can also take a look at the ghyp package. |
41,678 | Sampling from the normal-gamma distribution in R | Ok this is very late, but did you can approximate samples from the normal inverse gamma bivariate distribution using a Gibbs Sampler where the two alternating sampling distributions would be the normal distribution and the inverse gamma distribution. Did you look into that? | Sampling from the normal-gamma distribution in R | Ok this is very late, but did you can approximate samples from the normal inverse gamma bivariate distribution using a Gibbs Sampler where the two alternating sampling distributions would be the norma | Sampling from the normal-gamma distribution in R
Ok this is very late, but did you can approximate samples from the normal inverse gamma bivariate distribution using a Gibbs Sampler where the two alternating sampling distributions would be the normal distribution and the inverse gamma distribution. Did you look into th... | Sampling from the normal-gamma distribution in R
Ok this is very late, but did you can approximate samples from the normal inverse gamma bivariate distribution using a Gibbs Sampler where the two alternating sampling distributions would be the norma |
41,679 | Random forests vs boosting | I would use whichever one performed better out of sample.
So far, I've found in impossible to tell which model will be better for a novel problem a priori. | Random forests vs boosting | I would use whichever one performed better out of sample.
So far, I've found in impossible to tell which model will be better for a novel problem a priori. | Random forests vs boosting
I would use whichever one performed better out of sample.
So far, I've found in impossible to tell which model will be better for a novel problem a priori. | Random forests vs boosting
I would use whichever one performed better out of sample.
So far, I've found in impossible to tell which model will be better for a novel problem a priori. |
41,680 | Random forests vs boosting | Just to start, a quick thought.
Random Forest can run in parallel and they are much faster to train, Boosting is an iterative algorithm instead. However, Boosting might converge early iteration-wise.
Boosting might overfit when there are many noisy features but Random Forest also does.
On the other hand, their targe... | Random forests vs boosting | Just to start, a quick thought.
Random Forest can run in parallel and they are much faster to train, Boosting is an iterative algorithm instead. However, Boosting might converge early iteration-wise. | Random forests vs boosting
Just to start, a quick thought.
Random Forest can run in parallel and they are much faster to train, Boosting is an iterative algorithm instead. However, Boosting might converge early iteration-wise.
Boosting might overfit when there are many noisy features but Random Forest also does.
On ... | Random forests vs boosting
Just to start, a quick thought.
Random Forest can run in parallel and they are much faster to train, Boosting is an iterative algorithm instead. However, Boosting might converge early iteration-wise. |
41,681 | Sample size effects on R squared | My guess is that you only ran these simulations once. If you run them a few times the results will vary. You might get a smaller coefficient for the first one. But in general the reason for this pattern is because the true correlation for the underlying population is 0 and your simulations are following the law of l... | Sample size effects on R squared | My guess is that you only ran these simulations once. If you run them a few times the results will vary. You might get a smaller coefficient for the first one. But in general the reason for this pa | Sample size effects on R squared
My guess is that you only ran these simulations once. If you run them a few times the results will vary. You might get a smaller coefficient for the first one. But in general the reason for this pattern is because the true correlation for the underlying population is 0 and your simul... | Sample size effects on R squared
My guess is that you only ran these simulations once. If you run them a few times the results will vary. You might get a smaller coefficient for the first one. But in general the reason for this pa |
41,682 | covariance of RVs under a nonlinear transformation | Let's tackle this in relatively simple steps.
First, covariances are found in terms of expectations; e.g.,
$$\eqalign{
\text{Cov}(x_1x_2, x_1x_3) & = E[(x_1x_2 - E[x_1x_2])(x_1x_3 - E[x_1x_3])] \\
& = E[x_1^2x_2x_3] - E[x_1x_2E[x_1x_3]] - E[E[x_1x_2]x_1x_3] + E[x_1x_2]E[x_1x_3] \\
& = E[x_1^2x_2x_3] - E[x_1x_2]E[x_1... | covariance of RVs under a nonlinear transformation | Let's tackle this in relatively simple steps.
First, covariances are found in terms of expectations; e.g.,
$$\eqalign{
\text{Cov}(x_1x_2, x_1x_3) & = E[(x_1x_2 - E[x_1x_2])(x_1x_3 - E[x_1x_3])] \\
& | covariance of RVs under a nonlinear transformation
Let's tackle this in relatively simple steps.
First, covariances are found in terms of expectations; e.g.,
$$\eqalign{
\text{Cov}(x_1x_2, x_1x_3) & = E[(x_1x_2 - E[x_1x_2])(x_1x_3 - E[x_1x_3])] \\
& = E[x_1^2x_2x_3] - E[x_1x_2E[x_1x_3]] - E[E[x_1x_2]x_1x_3] + E[x_1x_2... | covariance of RVs under a nonlinear transformation
Let's tackle this in relatively simple steps.
First, covariances are found in terms of expectations; e.g.,
$$\eqalign{
\text{Cov}(x_1x_2, x_1x_3) & = E[(x_1x_2 - E[x_1x_2])(x_1x_3 - E[x_1x_3])] \\
& |
41,683 | Standard deviation of a ratio (percentage change) | If you don't know the distribution, the usual approach would be via Taylor expansion.
e.g. see here or top of p 6 here
or
http://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables
(You have to recognize that the two sample means are themselves random variables to apply it.)
---
Ed... | Standard deviation of a ratio (percentage change) | If you don't know the distribution, the usual approach would be via Taylor expansion.
e.g. see here or top of p 6 here
or
http://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_o | Standard deviation of a ratio (percentage change)
If you don't know the distribution, the usual approach would be via Taylor expansion.
e.g. see here or top of p 6 here
or
http://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_of_random_variables
(You have to recognize that the two sample means ar... | Standard deviation of a ratio (percentage change)
If you don't know the distribution, the usual approach would be via Taylor expansion.
e.g. see here or top of p 6 here
or
http://en.wikipedia.org/wiki/Taylor_expansions_for_the_moments_of_functions_o |
41,684 | Standard deviation of a ratio (percentage change) | The Taylor series method yields an estimator for the variance which can then be used to estimate a symmetric confidence interval based upon a crude normality assumption of your ratio.
A more general approach which directly yields a (possibly unsymmetric) confidence interval for the ratio of two random variables is MOVE... | Standard deviation of a ratio (percentage change) | The Taylor series method yields an estimator for the variance which can then be used to estimate a symmetric confidence interval based upon a crude normality assumption of your ratio.
A more general a | Standard deviation of a ratio (percentage change)
The Taylor series method yields an estimator for the variance which can then be used to estimate a symmetric confidence interval based upon a crude normality assumption of your ratio.
A more general approach which directly yields a (possibly unsymmetric) confidence inte... | Standard deviation of a ratio (percentage change)
The Taylor series method yields an estimator for the variance which can then be used to estimate a symmetric confidence interval based upon a crude normality assumption of your ratio.
A more general a |
41,685 | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order? | David's answer is incorrect because they are not i.i.d. The probability the nth draw is greater than all previous draws and the probability the n-1th draw is greater than all previous draws are not independent.
Here is some python code to calculate the exact probability recursively:
# Recursively finds the number of c... | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order? | David's answer is incorrect because they are not i.i.d. The probability the nth draw is greater than all previous draws and the probability the n-1th draw is greater than all previous draws are not in | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order?
David's answer is incorrect because they are not i.i.d. The probability the nth draw is greater than all previous draws and the probability the n-1th draw is greater than all previous draws are not independent.
Here is som... | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order?
David's answer is incorrect because they are not i.i.d. The probability the nth draw is greater than all previous draws and the probability the n-1th draw is greater than all previous draws are not in |
41,686 | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order? | Recursive Solution
Let the first number chosen be $k$ (with probability $1/m$). The chance that all are in order now equals the chance that the remaining $n-1$ are (a) in order and (b) equal or exceed $k$. Subtracting $k-1$ from all of the remaining numbers puts them in one-to-one correspondence with the possible way... | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order? | Recursive Solution
Let the first number chosen be $k$ (with probability $1/m$). The chance that all are in order now equals the chance that the remaining $n-1$ are (a) in order and (b) equal or excee | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order?
Recursive Solution
Let the first number chosen be $k$ (with probability $1/m$). The chance that all are in order now equals the chance that the remaining $n-1$ are (a) in order and (b) equal or exceed $k$. Subtracting $k-... | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order?
Recursive Solution
Let the first number chosen be $k$ (with probability $1/m$). The chance that all are in order now equals the chance that the remaining $n-1$ are (a) in order and (b) equal or excee |
41,687 | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order? | Unless I got something wrong, I think it might me possible to bring down you $n,m$ problem to a $2,n$ problem, which you have resolved for $n=10$ and of which the solution is more generally $p = \frac{n+1}{2n}$
$P[X_m \ge X_{m-1} \ge ... \ge X_1]=$
$P[X_m \ge \max (X_{m-1},X_{m-2} ... X_1);X_{m-1} \ge \max (X_{m-2},X... | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order? | Unless I got something wrong, I think it might me possible to bring down you $n,m$ problem to a $2,n$ problem, which you have resolved for $n=10$ and of which the solution is more generally $p = \frac | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order?
Unless I got something wrong, I think it might me possible to bring down you $n,m$ problem to a $2,n$ problem, which you have resolved for $n=10$ and of which the solution is more generally $p = \frac{n+1}{2n}$
$P[X_m \ge X... | What is the probability of randomly selecting n random numbers in the range 1-m in sorted order?
Unless I got something wrong, I think it might me possible to bring down you $n,m$ problem to a $2,n$ problem, which you have resolved for $n=10$ and of which the solution is more generally $p = \frac |
41,688 | How to get conditional variance from Schur complement? | Assume WLOG that everything is mean $0$. If you know the formula for the inverse of a matrix in block form, then it should be as simple as checking that
$$(X, Y)^T V^{-1} (X, Y) = X^T A^{-1} X + (Y - \mu_{Y | X})^T S^{-1} (Y - \mu_{Y|X}) + c$$
where $S$ is the Schur complement and $c$ is a constant and $\mu_{Y|X} = B^T... | How to get conditional variance from Schur complement? | Assume WLOG that everything is mean $0$. If you know the formula for the inverse of a matrix in block form, then it should be as simple as checking that
$$(X, Y)^T V^{-1} (X, Y) = X^T A^{-1} X + (Y - | How to get conditional variance from Schur complement?
Assume WLOG that everything is mean $0$. If you know the formula for the inverse of a matrix in block form, then it should be as simple as checking that
$$(X, Y)^T V^{-1} (X, Y) = X^T A^{-1} X + (Y - \mu_{Y | X})^T S^{-1} (Y - \mu_{Y|X}) + c$$
where $S$ is the Schu... | How to get conditional variance from Schur complement?
Assume WLOG that everything is mean $0$. If you know the formula for the inverse of a matrix in block form, then it should be as simple as checking that
$$(X, Y)^T V^{-1} (X, Y) = X^T A^{-1} X + (Y - |
41,689 | How to get conditional variance from Schur complement? | The detailed derivation may be found in:
von Mises, Richard (1964). Mathematical theory of probability and statistics. Chapter VIII.9.3. Academic Press.
I have added this reference to the Wikipedia article that you linked to, for the benefit of anyone else who stumbles over this question. | How to get conditional variance from Schur complement? | The detailed derivation may be found in:
von Mises, Richard (1964). Mathematical theory of probability and statistics. Chapter VIII.9.3. Academic Press.
I have added this reference to the Wikipedi | How to get conditional variance from Schur complement?
The detailed derivation may be found in:
von Mises, Richard (1964). Mathematical theory of probability and statistics. Chapter VIII.9.3. Academic Press.
I have added this reference to the Wikipedia article that you linked to, for the benefit of anyone else who ... | How to get conditional variance from Schur complement?
The detailed derivation may be found in:
von Mises, Richard (1964). Mathematical theory of probability and statistics. Chapter VIII.9.3. Academic Press.
I have added this reference to the Wikipedi |
41,690 | What is the difference between the 'Pivot table' and 'Contingency table'? | You may create a contingency table using a software tool called pivot table :)
A contingency table is a crosstable with rows, columns and data related to each of the row/column combination. You may draw such a table on a piece of paper, you may use an OLAP cube as the source of data etc. As this site says, a contingenc... | What is the difference between the 'Pivot table' and 'Contingency table'? | You may create a contingency table using a software tool called pivot table :)
A contingency table is a crosstable with rows, columns and data related to each of the row/column combination. You may dr | What is the difference between the 'Pivot table' and 'Contingency table'?
You may create a contingency table using a software tool called pivot table :)
A contingency table is a crosstable with rows, columns and data related to each of the row/column combination. You may draw such a table on a piece of paper, you may u... | What is the difference between the 'Pivot table' and 'Contingency table'?
You may create a contingency table using a software tool called pivot table :)
A contingency table is a crosstable with rows, columns and data related to each of the row/column combination. You may dr |
41,691 | Where can I use Chebychev's inequality? | I once convinced a very tall woman to have dinner with me by using Chebychev's inequality to argue that her high partner height threshold would lead to many lonely nights spent with her cat. Alcohol was involved, so arithmetic mistakes were made that exaggerated the main thrust of the conclusion.
It didn't last.
In thi... | Where can I use Chebychev's inequality? | I once convinced a very tall woman to have dinner with me by using Chebychev's inequality to argue that her high partner height threshold would lead to many lonely nights spent with her cat. Alcohol w | Where can I use Chebychev's inequality?
I once convinced a very tall woman to have dinner with me by using Chebychev's inequality to argue that her high partner height threshold would lead to many lonely nights spent with her cat. Alcohol was involved, so arithmetic mistakes were made that exaggerated the main thrust o... | Where can I use Chebychev's inequality?
I once convinced a very tall woman to have dinner with me by using Chebychev's inequality to argue that her high partner height threshold would lead to many lonely nights spent with her cat. Alcohol w |
41,692 | Where can I use Chebychev's inequality? | While it's most often used for establishing bounds in various things, here's an example of it being used for constructing intervals for a real problem:
http://www.sciencedirect.com/science/article/pii/S0001457504000910
(It's not actually necessary for this problem; tighter bounds can be obtained.) | Where can I use Chebychev's inequality? | While it's most often used for establishing bounds in various things, here's an example of it being used for constructing intervals for a real problem:
http://www.sciencedirect.com/science/article/pii | Where can I use Chebychev's inequality?
While it's most often used for establishing bounds in various things, here's an example of it being used for constructing intervals for a real problem:
http://www.sciencedirect.com/science/article/pii/S0001457504000910
(It's not actually necessary for this problem; tighter bounds... | Where can I use Chebychev's inequality?
While it's most often used for establishing bounds in various things, here's an example of it being used for constructing intervals for a real problem:
http://www.sciencedirect.com/science/article/pii |
41,693 | Finding mean when class sizes are unequal | Number of students:
11+10+7+4+4+3+1 = 21+11+8 = 40
Total number of absent days across all students:
(0+6)/2*11+(6+10)/2*10+(10+14)/2*7+(14+20)/2*4+
(20+28)/2*4+(28+30)/2*3+(38+40)/2*1 = 487
Mean number of absent days per student:
(Total number of absent days) / ( Number of students) = 487 / 40 = 12.175
Hence the m... | Finding mean when class sizes are unequal | Number of students:
11+10+7+4+4+3+1 = 21+11+8 = 40
Total number of absent days across all students:
(0+6)/2*11+(6+10)/2*10+(10+14)/2*7+(14+20)/2*4+
(20+28)/2*4+(28+30)/2*3+(38+40)/2*1 = 487
Mean n | Finding mean when class sizes are unequal
Number of students:
11+10+7+4+4+3+1 = 21+11+8 = 40
Total number of absent days across all students:
(0+6)/2*11+(6+10)/2*10+(10+14)/2*7+(14+20)/2*4+
(20+28)/2*4+(28+30)/2*3+(38+40)/2*1 = 487
Mean number of absent days per student:
(Total number of absent days) / ( Number of ... | Finding mean when class sizes are unequal
Number of students:
11+10+7+4+4+3+1 = 21+11+8 = 40
Total number of absent days across all students:
(0+6)/2*11+(6+10)/2*10+(10+14)/2*7+(14+20)/2*4+
(20+28)/2*4+(28+30)/2*3+(38+40)/2*1 = 487
Mean n |
41,694 | Finding mean when class sizes are unequal | The point of this question is to show that not all means need to be evenly weighted to be summed. It may lose some granularity in the final answer, but the estimate still holds true. In some ways, you can think of it as a mean of means.
So the theory is:
class mean = sum(mean(range_of_days_absent) * number_of_students... | Finding mean when class sizes are unequal | The point of this question is to show that not all means need to be evenly weighted to be summed. It may lose some granularity in the final answer, but the estimate still holds true. In some ways, yo | Finding mean when class sizes are unequal
The point of this question is to show that not all means need to be evenly weighted to be summed. It may lose some granularity in the final answer, but the estimate still holds true. In some ways, you can think of it as a mean of means.
So the theory is:
class mean = sum(mean(... | Finding mean when class sizes are unequal
The point of this question is to show that not all means need to be evenly weighted to be summed. It may lose some granularity in the final answer, but the estimate still holds true. In some ways, yo |
41,695 | Finding mean when class sizes are unequal | The method used will be the same as used for continuous grouped data with equal intervals.
First of all we will find the class marks of each interval with equal to lower limit + upper limit divided by 2. Then we will multiply the frequencies with their respective class marks.
Will will add all these products and divid... | Finding mean when class sizes are unequal | The method used will be the same as used for continuous grouped data with equal intervals.
First of all we will find the class marks of each interval with equal to lower limit + upper limit divided by | Finding mean when class sizes are unequal
The method used will be the same as used for continuous grouped data with equal intervals.
First of all we will find the class marks of each interval with equal to lower limit + upper limit divided by 2. Then we will multiply the frequencies with their respective class marks.
... | Finding mean when class sizes are unequal
The method used will be the same as used for continuous grouped data with equal intervals.
First of all we will find the class marks of each interval with equal to lower limit + upper limit divided by |
41,696 | Finding mean when class sizes are unequal | The overlap of classes in your example is a bit confusing, so I'll use some slightly modified values
No. of days 0-5 6-9 10-13 14-19 20-27 28-30 38- 40
No. of students 11 10 7 4 4 3 1
An approximation to the problem, not assuming symmetry within the classes ... | Finding mean when class sizes are unequal | The overlap of classes in your example is a bit confusing, so I'll use some slightly modified values
No. of days 0-5 6-9 10-13 14-19 20-27 28-30 38- 40
No. of students 11 | Finding mean when class sizes are unequal
The overlap of classes in your example is a bit confusing, so I'll use some slightly modified values
No. of days 0-5 6-9 10-13 14-19 20-27 28-30 38- 40
No. of students 11 10 7 4 4 3 1
An approximation to the problem,... | Finding mean when class sizes are unequal
The overlap of classes in your example is a bit confusing, so I'll use some slightly modified values
No. of days 0-5 6-9 10-13 14-19 20-27 28-30 38- 40
No. of students 11 |
41,697 | Is output of Deamer deconvolution not a density? | It seems you are really putting my library at a test!!
The output of deamer is an estimation of a density
Several things:
Although, theoretically, any np density estimation (with deconvolution or not) ensures that the resulting estimate integrates to 1 (by use of a kernel which is itself a density), you have (necessar... | Is output of Deamer deconvolution not a density? | It seems you are really putting my library at a test!!
The output of deamer is an estimation of a density
Several things:
Although, theoretically, any np density estimation (with deconvolution or not | Is output of Deamer deconvolution not a density?
It seems you are really putting my library at a test!!
The output of deamer is an estimation of a density
Several things:
Although, theoretically, any np density estimation (with deconvolution or not) ensures that the resulting estimate integrates to 1 (by use of a kern... | Is output of Deamer deconvolution not a density?
It seems you are really putting my library at a test!!
The output of deamer is an estimation of a density
Several things:
Although, theoretically, any np density estimation (with deconvolution or not |
41,698 | Beta distribution from mean and quantile | If you really have to do it with pesky Excel:
Create cells with quantile probability $p$, quantile value $q$, mean $m$.
Create a cell with some initial $\alpha$ value. Create a cell with formula $\beta=\left(\frac{1-m}{m}\right)\alpha$.
Create a cell with formula $\mathrm{abs}(q - \mathrm{beta.inv}(p, \alpha,\beta))$.... | Beta distribution from mean and quantile | If you really have to do it with pesky Excel:
Create cells with quantile probability $p$, quantile value $q$, mean $m$.
Create a cell with some initial $\alpha$ value. Create a cell with formula $\be | Beta distribution from mean and quantile
If you really have to do it with pesky Excel:
Create cells with quantile probability $p$, quantile value $q$, mean $m$.
Create a cell with some initial $\alpha$ value. Create a cell with formula $\beta=\left(\frac{1-m}{m}\right)\alpha$.
Create a cell with formula $\mathrm{abs}(... | Beta distribution from mean and quantile
If you really have to do it with pesky Excel:
Create cells with quantile probability $p$, quantile value $q$, mean $m$.
Create a cell with some initial $\alpha$ value. Create a cell with formula $\be |
41,699 | Mean centering for PCA in a 2D array...across rows or cols? | Usually, each row is an "observation" (in your case image), and each column is a variable (in your case pixel value). Therefore, you should center and scale the columns before doing PCA.
Also, lots of good PCA libraries already exist, such as sklearn.decomposition.PCA, which can save you a lot of effort re-inventing t... | Mean centering for PCA in a 2D array...across rows or cols? | Usually, each row is an "observation" (in your case image), and each column is a variable (in your case pixel value). Therefore, you should center and scale the columns before doing PCA.
Also, lots o | Mean centering for PCA in a 2D array...across rows or cols?
Usually, each row is an "observation" (in your case image), and each column is a variable (in your case pixel value). Therefore, you should center and scale the columns before doing PCA.
Also, lots of good PCA libraries already exist, such as sklearn.decompos... | Mean centering for PCA in a 2D array...across rows or cols?
Usually, each row is an "observation" (in your case image), and each column is a variable (in your case pixel value). Therefore, you should center and scale the columns before doing PCA.
Also, lots o |
41,700 | Mean centering for PCA in a 2D array...across rows or cols? | It depends on the way the data is set up. For instance, if you would calculate the covariance matrix as $\Sigma=\frac{1}{n}X'X$ where $X$ is the de-meaned data, then you would want to remove the mean from each column, and vice-versa. | Mean centering for PCA in a 2D array...across rows or cols? | It depends on the way the data is set up. For instance, if you would calculate the covariance matrix as $\Sigma=\frac{1}{n}X'X$ where $X$ is the de-meaned data, then you would want to remove the mean | Mean centering for PCA in a 2D array...across rows or cols?
It depends on the way the data is set up. For instance, if you would calculate the covariance matrix as $\Sigma=\frac{1}{n}X'X$ where $X$ is the de-meaned data, then you would want to remove the mean from each column, and vice-versa. | Mean centering for PCA in a 2D array...across rows or cols?
It depends on the way the data is set up. For instance, if you would calculate the covariance matrix as $\Sigma=\frac{1}{n}X'X$ where $X$ is the de-meaned data, then you would want to remove the mean |
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