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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39,801 | Choosing between best two attributes with the same information gain when building decision tree | The decision tree method is a step-wise approach to model building, so variables entered earlier have a direct impact on which variables are entered later. Can you try both trees, and see which creates a better separation at your terminal nodes and/or builds a tree that fits more closely with the theoretical understand... | Choosing between best two attributes with the same information gain when building decision tree | The decision tree method is a step-wise approach to model building, so variables entered earlier have a direct impact on which variables are entered later. Can you try both trees, and see which create | Choosing between best two attributes with the same information gain when building decision tree
The decision tree method is a step-wise approach to model building, so variables entered earlier have a direct impact on which variables are entered later. Can you try both trees, and see which creates a better separation at... | Choosing between best two attributes with the same information gain when building decision tree
The decision tree method is a step-wise approach to model building, so variables entered earlier have a direct impact on which variables are entered later. Can you try both trees, and see which create |
39,802 | Choosing between best two attributes with the same information gain when building decision tree | I work with the text classification problem, and for the classification I am using decision tree classifiers(ID3, Random forests etc). So I can give you an example that is related to text classification. In classification we are going to deal with the different words as the attributes, and you can reduce the features u... | Choosing between best two attributes with the same information gain when building decision tree | I work with the text classification problem, and for the classification I am using decision tree classifiers(ID3, Random forests etc). So I can give you an example that is related to text classificati | Choosing between best two attributes with the same information gain when building decision tree
I work with the text classification problem, and for the classification I am using decision tree classifiers(ID3, Random forests etc). So I can give you an example that is related to text classification. In classification we... | Choosing between best two attributes with the same information gain when building decision tree
I work with the text classification problem, and for the classification I am using decision tree classifiers(ID3, Random forests etc). So I can give you an example that is related to text classificati |
39,803 | McKay's bivariate Gamma distribution | You can create whole families of joint distributions on $(X,Y)$ such that $X\sim \Gamma(a_1,\alpha)$ and $Y\sim \Gamma(a_2,\beta)$ by using copulas like
$$
F_{(X,Y)}(x,y) = \mathbb{P}(X\le x,Y\le y) =
\dfrac{F_X(x)F_Y(y)}{1+\varrho (1-F_X(x))(1-F_Y(y))}
$$
for $-1\le \varrho \le 1$. The joint distribution is continuous... | McKay's bivariate Gamma distribution | You can create whole families of joint distributions on $(X,Y)$ such that $X\sim \Gamma(a_1,\alpha)$ and $Y\sim \Gamma(a_2,\beta)$ by using copulas like
$$
F_{(X,Y)}(x,y) = \mathbb{P}(X\le x,Y\le y) = | McKay's bivariate Gamma distribution
You can create whole families of joint distributions on $(X,Y)$ such that $X\sim \Gamma(a_1,\alpha)$ and $Y\sim \Gamma(a_2,\beta)$ by using copulas like
$$
F_{(X,Y)}(x,y) = \mathbb{P}(X\le x,Y\le y) =
\dfrac{F_X(x)F_Y(y)}{1+\varrho (1-F_X(x))(1-F_Y(y))}
$$
for $-1\le \varrho \le 1$.... | McKay's bivariate Gamma distribution
You can create whole families of joint distributions on $(X,Y)$ such that $X\sim \Gamma(a_1,\alpha)$ and $Y\sim \Gamma(a_2,\beta)$ by using copulas like
$$
F_{(X,Y)}(x,y) = \mathbb{P}(X\le x,Y\le y) = |
39,804 | McKay's bivariate Gamma distribution | For McKay's distribution $X$ is a Gamma variate that is the sum of a subset of squares taken from the other, $Y$, which is the sum of a larger set of squares. Implying that $Y>X$ with probability $1$. See McKay's original paper:
McKay, A. T. (1934) Sampling from batches. Journal of the Royal Statistical Society—Suppl... | McKay's bivariate Gamma distribution | For McKay's distribution $X$ is a Gamma variate that is the sum of a subset of squares taken from the other, $Y$, which is the sum of a larger set of squares. Implying that $Y>X$ with probability $1$. | McKay's bivariate Gamma distribution
For McKay's distribution $X$ is a Gamma variate that is the sum of a subset of squares taken from the other, $Y$, which is the sum of a larger set of squares. Implying that $Y>X$ with probability $1$. See McKay's original paper:
McKay, A. T. (1934) Sampling from batches. Journal o... | McKay's bivariate Gamma distribution
For McKay's distribution $X$ is a Gamma variate that is the sum of a subset of squares taken from the other, $Y$, which is the sum of a larger set of squares. Implying that $Y>X$ with probability $1$. |
39,805 | Detecting changes in distribution of multiple variables | This sounds like a job for the Mahalanobis Distance. You would apply this by estimating the population covariance and mean vector using the previous $(k-1)$ samples (assuming $k > N$, where $N$ is the dimension of your observed vectors), and then computing the Mahalanobis Distance. A google search reveals a lot of hits... | Detecting changes in distribution of multiple variables | This sounds like a job for the Mahalanobis Distance. You would apply this by estimating the population covariance and mean vector using the previous $(k-1)$ samples (assuming $k > N$, where $N$ is the | Detecting changes in distribution of multiple variables
This sounds like a job for the Mahalanobis Distance. You would apply this by estimating the population covariance and mean vector using the previous $(k-1)$ samples (assuming $k > N$, where $N$ is the dimension of your observed vectors), and then computing the Mah... | Detecting changes in distribution of multiple variables
This sounds like a job for the Mahalanobis Distance. You would apply this by estimating the population covariance and mean vector using the previous $(k-1)$ samples (assuming $k > N$, where $N$ is the |
39,806 | Detecting changes in distribution of multiple variables | This sounds to me more like a change point problem. You should investigate the bcp package, http://cran.r-project.org/web/packages/bcp/index.html, as well as strucchange, http://cran.r-project.org/web/packages/strucchange/index.html
John Emerson maintains the bcp package, and when I met him, he seemed very willing to e... | Detecting changes in distribution of multiple variables | This sounds to me more like a change point problem. You should investigate the bcp package, http://cran.r-project.org/web/packages/bcp/index.html, as well as strucchange, http://cran.r-project.org/web | Detecting changes in distribution of multiple variables
This sounds to me more like a change point problem. You should investigate the bcp package, http://cran.r-project.org/web/packages/bcp/index.html, as well as strucchange, http://cran.r-project.org/web/packages/strucchange/index.html
John Emerson maintains the bcp ... | Detecting changes in distribution of multiple variables
This sounds to me more like a change point problem. You should investigate the bcp package, http://cran.r-project.org/web/packages/bcp/index.html, as well as strucchange, http://cran.r-project.org/web |
39,807 | Chi square statistic and contingency coefficients | There are two standard accounts of $\chi^2$ as applied to a multinomial distribution, both of which show why the denominator should not be squared:
After expanding the logarithm through second order and doing some algebra, the likelihood ratio statistic for the multinomial distribution, $-2\log(l) = 2\sum_{i=1}^k n_i ... | Chi square statistic and contingency coefficients | There are two standard accounts of $\chi^2$ as applied to a multinomial distribution, both of which show why the denominator should not be squared:
After expanding the logarithm through second order | Chi square statistic and contingency coefficients
There are two standard accounts of $\chi^2$ as applied to a multinomial distribution, both of which show why the denominator should not be squared:
After expanding the logarithm through second order and doing some algebra, the likelihood ratio statistic for the multino... | Chi square statistic and contingency coefficients
There are two standard accounts of $\chi^2$ as applied to a multinomial distribution, both of which show why the denominator should not be squared:
After expanding the logarithm through second order |
39,808 | Chi square statistic and contingency coefficients | I don't know what is "more true" or "less true" form of normalization. Any normalization that makes a particular sense will do. The quantity $(n_{ij}-E_{ij})^2/E_{ij}$ is called squared standardized residual in a cell, and overall Chi-square of a table is the sum of those across all cells. Chi-square can be viewed ther... | Chi square statistic and contingency coefficients | I don't know what is "more true" or "less true" form of normalization. Any normalization that makes a particular sense will do. The quantity $(n_{ij}-E_{ij})^2/E_{ij}$ is called squared standardized r | Chi square statistic and contingency coefficients
I don't know what is "more true" or "less true" form of normalization. Any normalization that makes a particular sense will do. The quantity $(n_{ij}-E_{ij})^2/E_{ij}$ is called squared standardized residual in a cell, and overall Chi-square of a table is the sum of tho... | Chi square statistic and contingency coefficients
I don't know what is "more true" or "less true" form of normalization. Any normalization that makes a particular sense will do. The quantity $(n_{ij}-E_{ij})^2/E_{ij}$ is called squared standardized r |
39,809 | Chi square statistic and contingency coefficients | An intuitive explanation for why
$$
d_{ij}=(n_{ij}−E_{ij})^2/E_{ij}^2
$$
would not be a good test statistic is to notice that it doesn't scale with the size of the sample: Suppose we increase our sample by 10 fold. Then if the null-hypothesis is false, we should be much more likely to reject -- that is, the test ... | Chi square statistic and contingency coefficients | An intuitive explanation for why
$$
d_{ij}=(n_{ij}−E_{ij})^2/E_{ij}^2
$$
would not be a good test statistic is to notice that it doesn't scale with the size of the sample: Suppose we increase our | Chi square statistic and contingency coefficients
An intuitive explanation for why
$$
d_{ij}=(n_{ij}−E_{ij})^2/E_{ij}^2
$$
would not be a good test statistic is to notice that it doesn't scale with the size of the sample: Suppose we increase our sample by 10 fold. Then if the null-hypothesis is false, we should b... | Chi square statistic and contingency coefficients
An intuitive explanation for why
$$
d_{ij}=(n_{ij}−E_{ij})^2/E_{ij}^2
$$
would not be a good test statistic is to notice that it doesn't scale with the size of the sample: Suppose we increase our |
39,810 | Chi square statistic and contingency coefficients | About the contingency coefficient, I want to take the opportunity to leave a remark.
Consider the "theoretical chi-square statistic"
$$
v= \sum_{i,j}\frac{(p_{i,j}-p_ip_j)^2}{p_ip_j}
$$
and the theoretical contigency coefficient
$$
C_0 = \sqrt{\frac{v}{v+n}}.
$$
It is called the effect size in power calculations. As ... | Chi square statistic and contingency coefficients | About the contingency coefficient, I want to take the opportunity to leave a remark.
Consider the "theoretical chi-square statistic"
$$
v= \sum_{i,j}\frac{(p_{i,j}-p_ip_j)^2}{p_ip_j}
$$
and the theor | Chi square statistic and contingency coefficients
About the contingency coefficient, I want to take the opportunity to leave a remark.
Consider the "theoretical chi-square statistic"
$$
v= \sum_{i,j}\frac{(p_{i,j}-p_ip_j)^2}{p_ip_j}
$$
and the theoretical contigency coefficient
$$
C_0 = \sqrt{\frac{v}{v+n}}.
$$
It is... | Chi square statistic and contingency coefficients
About the contingency coefficient, I want to take the opportunity to leave a remark.
Consider the "theoretical chi-square statistic"
$$
v= \sum_{i,j}\frac{(p_{i,j}-p_ip_j)^2}{p_ip_j}
$$
and the theor |
39,811 | Average number of failures before success | I disagree with the answer given by @probabilityislogic. There is no geometric distribution here at all, truncated or otherwise.
One of 13 agents can answer the question. The worker calls the agents one by one until the worker reaches the agent who can answer the question. Let us number the agents as $1$, $2, \ldots... | Average number of failures before success | I disagree with the answer given by @probabilityislogic. There is no geometric distribution here at all, truncated or otherwise.
One of 13 agents can answer the question. The worker calls the agents | Average number of failures before success
I disagree with the answer given by @probabilityislogic. There is no geometric distribution here at all, truncated or otherwise.
One of 13 agents can answer the question. The worker calls the agents one by one until the worker reaches the agent who can answer the question. L... | Average number of failures before success
I disagree with the answer given by @probabilityislogic. There is no geometric distribution here at all, truncated or otherwise.
One of 13 agents can answer the question. The worker calls the agents |
39,812 | Average number of failures before success | What you require is a truncated geometric distribution. The usual geometric distribution has a support of $0,1,\dots$ (i.e. to infinity) whereas your one has support only up to $12$.
So you need to renormalise the pdf.
$$Pr(X=x|p)=Ap(1-p)^x\implies A^{-1}=\sum_{x=0}^{12}p(1-p)^x=1-(1-p)^{13}$$
$$\implies Pr(X=x|p)=\fr... | Average number of failures before success | What you require is a truncated geometric distribution. The usual geometric distribution has a support of $0,1,\dots$ (i.e. to infinity) whereas your one has support only up to $12$.
So you need to r | Average number of failures before success
What you require is a truncated geometric distribution. The usual geometric distribution has a support of $0,1,\dots$ (i.e. to infinity) whereas your one has support only up to $12$.
So you need to renormalise the pdf.
$$Pr(X=x|p)=Ap(1-p)^x\implies A^{-1}=\sum_{x=0}^{12}p(1-p)... | Average number of failures before success
What you require is a truncated geometric distribution. The usual geometric distribution has a support of $0,1,\dots$ (i.e. to infinity) whereas your one has support only up to $12$.
So you need to r |
39,813 | How to solve Chuck-a-Luck puzzle | All three methods should produce the same result for the house edge. To check yours, you should find
Number 1s Prob
0 125/216
1 75/216
2 15/216
3 1/216
which would make your expected winnings (for a stake of 1)
$$\frac{-1\times 125 + 1\times 75 + 2 \times 15 + 3 \tim... | How to solve Chuck-a-Luck puzzle | All three methods should produce the same result for the house edge. To check yours, you should find
Number 1s Prob
0 125/216
1 75/216
2 15/216
3 1/2 | How to solve Chuck-a-Luck puzzle
All three methods should produce the same result for the house edge. To check yours, you should find
Number 1s Prob
0 125/216
1 75/216
2 15/216
3 1/216
which would make your expected winnings (for a stake of 1)
$$\frac{-1\times 125 + ... | How to solve Chuck-a-Luck puzzle
All three methods should produce the same result for the house edge. To check yours, you should find
Number 1s Prob
0 125/216
1 75/216
2 15/216
3 1/2 |
39,814 | How to compare group means for two samples with t-test using R? | 1) Yes, it looks to me like you used the t.test function correctly.
2) Since your p-value is 0.9998 all you can say is that you don't have enough evidence to reject your null hypothesis (that the means are equal). In other words, your evidence (the scores from both group) doesn't show that the mean of the first group i... | How to compare group means for two samples with t-test using R? | 1) Yes, it looks to me like you used the t.test function correctly.
2) Since your p-value is 0.9998 all you can say is that you don't have enough evidence to reject your null hypothesis (that the mean | How to compare group means for two samples with t-test using R?
1) Yes, it looks to me like you used the t.test function correctly.
2) Since your p-value is 0.9998 all you can say is that you don't have enough evidence to reject your null hypothesis (that the means are equal). In other words, your evidence (the scores ... | How to compare group means for two samples with t-test using R?
1) Yes, it looks to me like you used the t.test function correctly.
2) Since your p-value is 0.9998 all you can say is that you don't have enough evidence to reject your null hypothesis (that the mean |
39,815 | How to compare group means for two samples with t-test using R? | The hypothesis stated is clearly one-sided and the problem is set up in a Neyman-Pearsonian manner. Those features might be deliberate, or it might be an accidental consequence of the fact that N-P is the dominant paradigm in many introductory statistics texts. If it is deliberate then the results of the analysis shoul... | How to compare group means for two samples with t-test using R? | The hypothesis stated is clearly one-sided and the problem is set up in a Neyman-Pearsonian manner. Those features might be deliberate, or it might be an accidental consequence of the fact that N-P is | How to compare group means for two samples with t-test using R?
The hypothesis stated is clearly one-sided and the problem is set up in a Neyman-Pearsonian manner. Those features might be deliberate, or it might be an accidental consequence of the fact that N-P is the dominant paradigm in many introductory statistics t... | How to compare group means for two samples with t-test using R?
The hypothesis stated is clearly one-sided and the problem is set up in a Neyman-Pearsonian manner. Those features might be deliberate, or it might be an accidental consequence of the fact that N-P is |
39,816 | Presenting logistic model fit graphically | First I want to reiterate what others have said in the comments: this is in general a difficult problem with no easy solution.
With that in mind, one way of assessing model fit with logistic regression models when your primary interest is simply predictive power is to plot ROCs. Here's an example (with caveats) using t... | Presenting logistic model fit graphically | First I want to reiterate what others have said in the comments: this is in general a difficult problem with no easy solution.
With that in mind, one way of assessing model fit with logistic regressio | Presenting logistic model fit graphically
First I want to reiterate what others have said in the comments: this is in general a difficult problem with no easy solution.
With that in mind, one way of assessing model fit with logistic regression models when your primary interest is simply predictive power is to plot ROC... | Presenting logistic model fit graphically
First I want to reiterate what others have said in the comments: this is in general a difficult problem with no easy solution.
With that in mind, one way of assessing model fit with logistic regressio |
39,817 | Presenting logistic model fit graphically | A good reference for the kind of graphs you seem to want is John Fox, "Effect Displays for Multinomial and Proportional-Odds Logit Models," in Sociological Methodology (2006). See the citations to his earlier work. He implements these techniques in R and S-Plus in his book that accompanies his text on linear regression... | Presenting logistic model fit graphically | A good reference for the kind of graphs you seem to want is John Fox, "Effect Displays for Multinomial and Proportional-Odds Logit Models," in Sociological Methodology (2006). See the citations to his | Presenting logistic model fit graphically
A good reference for the kind of graphs you seem to want is John Fox, "Effect Displays for Multinomial and Proportional-Odds Logit Models," in Sociological Methodology (2006). See the citations to his earlier work. He implements these techniques in R and S-Plus in his book tha... | Presenting logistic model fit graphically
A good reference for the kind of graphs you seem to want is John Fox, "Effect Displays for Multinomial and Proportional-Odds Logit Models," in Sociological Methodology (2006). See the citations to his |
39,818 | How to interpret KPSS results? | You specified a null hypothesis of trend stationarity. That is the data follow a straight line time trend with stationary errors. The p-value is 0.1, so the null hypothesis is not rejected at the usual 5% level. | How to interpret KPSS results? | You specified a null hypothesis of trend stationarity. That is the data follow a straight line time trend with stationary errors. The p-value is 0.1, so the null hypothesis is not rejected at the usua | How to interpret KPSS results?
You specified a null hypothesis of trend stationarity. That is the data follow a straight line time trend with stationary errors. The p-value is 0.1, so the null hypothesis is not rejected at the usual 5% level. | How to interpret KPSS results?
You specified a null hypothesis of trend stationarity. That is the data follow a straight line time trend with stationary errors. The p-value is 0.1, so the null hypothesis is not rejected at the usua |
39,819 | Testing normality | I would not rely on p-values for any test of normality (or for much else, frankly). Look at the graphs.
You can, a priori, say that EVERY distribution is non-normal. If you have a large dataset the nonnormality will be statistically significant. The questions are HOW non-normal? Non-normal in what ways? and What are th... | Testing normality | I would not rely on p-values for any test of normality (or for much else, frankly). Look at the graphs.
You can, a priori, say that EVERY distribution is non-normal. If you have a large dataset the no | Testing normality
I would not rely on p-values for any test of normality (or for much else, frankly). Look at the graphs.
You can, a priori, say that EVERY distribution is non-normal. If you have a large dataset the nonnormality will be statistically significant. The questions are HOW non-normal? Non-normal in what way... | Testing normality
I would not rely on p-values for any test of normality (or for much else, frankly). Look at the graphs.
You can, a priori, say that EVERY distribution is non-normal. If you have a large dataset the no |
39,820 | Testing normality | All I can say is that your eyes are one of your better EDA tools. If your data (with 500,000 observations) doesn't look normal, then there's no reason to even perform a statistical test for normality. Especially with that many data points, any slight deviation from normality should make you fail the test.
It looks like... | Testing normality | All I can say is that your eyes are one of your better EDA tools. If your data (with 500,000 observations) doesn't look normal, then there's no reason to even perform a statistical test for normality. | Testing normality
All I can say is that your eyes are one of your better EDA tools. If your data (with 500,000 observations) doesn't look normal, then there's no reason to even perform a statistical test for normality. Especially with that many data points, any slight deviation from normality should make you fail the t... | Testing normality
All I can say is that your eyes are one of your better EDA tools. If your data (with 500,000 observations) doesn't look normal, then there's no reason to even perform a statistical test for normality. |
39,821 | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-mean imputation) | Some years ago, I thought it might be a good idea to apply person-mean imputation (person-mean substitution or case-mean imputation) in case of item non-response. Nowadays, however, it seems obvious to me that this approach assumes that all scale items share similar characteristics (similar variance, standard deviation... | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-me | Some years ago, I thought it might be a good idea to apply person-mean imputation (person-mean substitution or case-mean imputation) in case of item non-response. Nowadays, however, it seems obvious t | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-mean imputation)
Some years ago, I thought it might be a good idea to apply person-mean imputation (person-mean substitution or case-mean imputation) in case of item non-response. Nowadays, however, it seems obvious to me ... | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-me
Some years ago, I thought it might be a good idea to apply person-mean imputation (person-mean substitution or case-mean imputation) in case of item non-response. Nowadays, however, it seems obvious t |
39,822 | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-mean imputation) | Person-mean imputation with an minimum-item threshold is a simple strategy for retaining scale scores where participants miss the occasional response.
Some general principles
If missing data is minimal (e.g., less than 5% of participants are missing 1 item on a 10 item scale), the method of dealing with missing data i... | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-me | Person-mean imputation with an minimum-item threshold is a simple strategy for retaining scale scores where participants miss the occasional response.
Some general principles
If missing data is minim | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-mean imputation)
Person-mean imputation with an minimum-item threshold is a simple strategy for retaining scale scores where participants miss the occasional response.
Some general principles
If missing data is minimal (e... | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-me
Person-mean imputation with an minimum-item threshold is a simple strategy for retaining scale scores where participants miss the occasional response.
Some general principles
If missing data is minim |
39,823 | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-mean imputation) | one more piece of advice: make sure the full 6-item composite scale is reliable & that none of the included items reduces scale reliability. If those conditions aren't satisfied, you shouldn't be averaging them even in cases where data are complete. If these conditions are satisfied, then using a subset of items for ca... | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-me | one more piece of advice: make sure the full 6-item composite scale is reliable & that none of the included items reduces scale reliability. If those conditions aren't satisfied, you shouldn't be aver | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-mean imputation)
one more piece of advice: make sure the full 6-item composite scale is reliable & that none of the included items reduces scale reliability. If those conditions aren't satisfied, you shouldn't be averaging... | Appropriateness of calculating scale means based on available non-missing responses (i.e., person-me
one more piece of advice: make sure the full 6-item composite scale is reliable & that none of the included items reduces scale reliability. If those conditions aren't satisfied, you shouldn't be aver |
39,824 | Rank transformed 2-way ANOVA | The proportional odds (PO) ordinal logistic model is a generalization of the Wilcoxon and Kruskal-Wallis tests, allowing for covariates, interactions, and anything else you can do in a regression model for a univariate response. A two-way ANOVA on ranks is not based on strong statistical principles.
One of many comput... | Rank transformed 2-way ANOVA | The proportional odds (PO) ordinal logistic model is a generalization of the Wilcoxon and Kruskal-Wallis tests, allowing for covariates, interactions, and anything else you can do in a regression mode | Rank transformed 2-way ANOVA
The proportional odds (PO) ordinal logistic model is a generalization of the Wilcoxon and Kruskal-Wallis tests, allowing for covariates, interactions, and anything else you can do in a regression model for a univariate response. A two-way ANOVA on ranks is not based on strong statistical p... | Rank transformed 2-way ANOVA
The proportional odds (PO) ordinal logistic model is a generalization of the Wilcoxon and Kruskal-Wallis tests, allowing for covariates, interactions, and anything else you can do in a regression mode |
39,825 | What's the minimum number of individuals in a group for repeated-measures ANOVA? | Performing the ANOVA assumes that the nature and amount of variation in the hypothetical population represented by the one patient are the same as the variation in the hypothetical population represented by the ten other patients, and that all were obtained randomly and independently from their populations.
Under these... | What's the minimum number of individuals in a group for repeated-measures ANOVA? | Performing the ANOVA assumes that the nature and amount of variation in the hypothetical population represented by the one patient are the same as the variation in the hypothetical population represen | What's the minimum number of individuals in a group for repeated-measures ANOVA?
Performing the ANOVA assumes that the nature and amount of variation in the hypothetical population represented by the one patient are the same as the variation in the hypothetical population represented by the ten other patients, and that... | What's the minimum number of individuals in a group for repeated-measures ANOVA?
Performing the ANOVA assumes that the nature and amount of variation in the hypothetical population represented by the one patient are the same as the variation in the hypothetical population represen |
39,826 | How do I generate a QQ-Plot for data fitted using fitdistr? | Try the following code:
simdata <- qexp(ppoints(length(data)), rate = params$estimate)
qqplot(data, simdata)
(Inspired by the base R implementation of qqnorm)
PS When using non-base R functions, you should state what library they come from. I had to Google to discover that fitdistr is from MASS. | How do I generate a QQ-Plot for data fitted using fitdistr? | Try the following code:
simdata <- qexp(ppoints(length(data)), rate = params$estimate)
qqplot(data, simdata)
(Inspired by the base R implementation of qqnorm)
PS When using non-base R functions, you | How do I generate a QQ-Plot for data fitted using fitdistr?
Try the following code:
simdata <- qexp(ppoints(length(data)), rate = params$estimate)
qqplot(data, simdata)
(Inspired by the base R implementation of qqnorm)
PS When using non-base R functions, you should state what library they come from. I had to Google t... | How do I generate a QQ-Plot for data fitted using fitdistr?
Try the following code:
simdata <- qexp(ppoints(length(data)), rate = params$estimate)
qqplot(data, simdata)
(Inspired by the base R implementation of qqnorm)
PS When using non-base R functions, you |
39,827 | How do I generate a QQ-Plot for data fitted using fitdistr? | You could also look Wessa P., (2008), Maximum-likelihood Exponential Distribution Fitting which has a nice, documented online execution of the problem with the R source code attached to the actual plot. | How do I generate a QQ-Plot for data fitted using fitdistr? | You could also look Wessa P., (2008), Maximum-likelihood Exponential Distribution Fitting which has a nice, documented online execution of the problem with the R source code attached to the actual plo | How do I generate a QQ-Plot for data fitted using fitdistr?
You could also look Wessa P., (2008), Maximum-likelihood Exponential Distribution Fitting which has a nice, documented online execution of the problem with the R source code attached to the actual plot. | How do I generate a QQ-Plot for data fitted using fitdistr?
You could also look Wessa P., (2008), Maximum-likelihood Exponential Distribution Fitting which has a nice, documented online execution of the problem with the R source code attached to the actual plo |
39,828 | Annotating graphs in R [closed] | A quick and dirty way to paste some text and numerical results along the labels of your legend is to simply rename the factor levels. For instance,
df <- data.frame(x=rnorm(100), y=rnorm(100), f=gl(2,50))
df$f2 <- df$f
levels(df$f2) <- paste(levels(df$f), tapply(df$y, df$f, mean), sep=": ")
p <- ggplot(data=df) + geom_... | Annotating graphs in R [closed] | A quick and dirty way to paste some text and numerical results along the labels of your legend is to simply rename the factor levels. For instance,
df <- data.frame(x=rnorm(100), y=rnorm(100), f=gl(2, | Annotating graphs in R [closed]
A quick and dirty way to paste some text and numerical results along the labels of your legend is to simply rename the factor levels. For instance,
df <- data.frame(x=rnorm(100), y=rnorm(100), f=gl(2,50))
df$f2 <- df$f
levels(df$f2) <- paste(levels(df$f), tapply(df$y, df$f, mean), sep=":... | Annotating graphs in R [closed]
A quick and dirty way to paste some text and numerical results along the labels of your legend is to simply rename the factor levels. For instance,
df <- data.frame(x=rnorm(100), y=rnorm(100), f=gl(2, |
39,829 | Annotating graphs in R [closed] | I'm sure that there's a more elegant way to do this but you can try this :
Data <- data.frame(serie1 = cumsum(rnorm(100)), serie2 = rnorm(100), temps = as.Date(1:100, origin = "2000-01-01"))
label <- c("In traffic [max 2.0G bps] [Avg 1.1G bps] [95% 1.8G bps] [Min 569.4M bps]", "Out traffic [max 2.0G bps] [Avg 672M bp... | Annotating graphs in R [closed] | I'm sure that there's a more elegant way to do this but you can try this :
Data <- data.frame(serie1 = cumsum(rnorm(100)), serie2 = rnorm(100), temps = as.Date(1:100, origin = "2000-01-01"))
label < | Annotating graphs in R [closed]
I'm sure that there's a more elegant way to do this but you can try this :
Data <- data.frame(serie1 = cumsum(rnorm(100)), serie2 = rnorm(100), temps = as.Date(1:100, origin = "2000-01-01"))
label <- c("In traffic [max 2.0G bps] [Avg 1.1G bps] [95% 1.8G bps] [Min 569.4M bps]", "Out tra... | Annotating graphs in R [closed]
I'm sure that there's a more elegant way to do this but you can try this :
Data <- data.frame(serie1 = cumsum(rnorm(100)), serie2 = rnorm(100), temps = as.Date(1:100, origin = "2000-01-01"))
label < |
39,830 | Annotating graphs in R [closed] | Try the directlabels package for some more control. As far as the caption goes, if you're automatically generating the graphs, you might simply use it as caption text (in LaTeX or HTML or what have you). Hope this helps. | Annotating graphs in R [closed] | Try the directlabels package for some more control. As far as the caption goes, if you're automatically generating the graphs, you might simply use it as caption text (in LaTeX or HTML or what have y | Annotating graphs in R [closed]
Try the directlabels package for some more control. As far as the caption goes, if you're automatically generating the graphs, you might simply use it as caption text (in LaTeX or HTML or what have you). Hope this helps. | Annotating graphs in R [closed]
Try the directlabels package for some more control. As far as the caption goes, if you're automatically generating the graphs, you might simply use it as caption text (in LaTeX or HTML or what have y |
39,831 | Checking factor / covariate independence in ANCOVA | One approach is to see if the covariate is correlated with the predictor variables. That is, if the ANCOVA is given by:
predicted ~ covariate + predictor1*predictor2*predictor3
Then first assess whether the covariate and the various predictor effects/interactions are correlated:
covariate ~ predictor1*predictor2*predi... | Checking factor / covariate independence in ANCOVA | One approach is to see if the covariate is correlated with the predictor variables. That is, if the ANCOVA is given by:
predicted ~ covariate + predictor1*predictor2*predictor3
Then first assess whet | Checking factor / covariate independence in ANCOVA
One approach is to see if the covariate is correlated with the predictor variables. That is, if the ANCOVA is given by:
predicted ~ covariate + predictor1*predictor2*predictor3
Then first assess whether the covariate and the various predictor effects/interactions are ... | Checking factor / covariate independence in ANCOVA
One approach is to see if the covariate is correlated with the predictor variables. That is, if the ANCOVA is given by:
predicted ~ covariate + predictor1*predictor2*predictor3
Then first assess whet |
39,832 | Checking factor / covariate independence in ANCOVA | EDIT As pointed out in the comments below, this is an answer to a different question, one about ensuring that the effect of the covariate is the same in all groups.
You have to check for interactions of the covariate and the factors. So if the ANCOVA model was
lm1 <- lm(outcome ~ covariate + factor1*factor2*factor3)
... | Checking factor / covariate independence in ANCOVA | EDIT As pointed out in the comments below, this is an answer to a different question, one about ensuring that the effect of the covariate is the same in all groups.
You have to check for interactions | Checking factor / covariate independence in ANCOVA
EDIT As pointed out in the comments below, this is an answer to a different question, one about ensuring that the effect of the covariate is the same in all groups.
You have to check for interactions of the covariate and the factors. So if the ANCOVA model was
lm1 <- ... | Checking factor / covariate independence in ANCOVA
EDIT As pointed out in the comments below, this is an answer to a different question, one about ensuring that the effect of the covariate is the same in all groups.
You have to check for interactions |
39,833 | Mixed effects log-linear models | Log-linear or Poisson model are part of generalized linear models. Look at the lme4 package which allows for mixed-effects modeling, with family=poisson().
Here is an example of use:
> data(homerun, package="Zelig")
> with(homerun, table(homeruns, month))
month
homeruns April August July June March May Septembe... | Mixed effects log-linear models | Log-linear or Poisson model are part of generalized linear models. Look at the lme4 package which allows for mixed-effects modeling, with family=poisson().
Here is an example of use:
> data(homerun, p | Mixed effects log-linear models
Log-linear or Poisson model are part of generalized linear models. Look at the lme4 package which allows for mixed-effects modeling, with family=poisson().
Here is an example of use:
> data(homerun, package="Zelig")
> with(homerun, table(homeruns, month))
month
homeruns April Aug... | Mixed effects log-linear models
Log-linear or Poisson model are part of generalized linear models. Look at the lme4 package which allows for mixed-effects modeling, with family=poisson().
Here is an example of use:
> data(homerun, p |
39,834 | Log graph question | Yes. Quantiles can be transformed under any monotonically increasing transformation.
To see this, suppose $Y$ is the random variable and $q_{0.16}$ is the 16% quantile. Then
$$
\text{Pr}(Y\le q_{0.16}) = \text{Pr}(\log(Y)\le\log(q_{0.16})) = 0.16.
$$
Generally, if $f$ is monotonic and increasing then
$$
\text{Pr}(Y\le... | Log graph question | Yes. Quantiles can be transformed under any monotonically increasing transformation.
To see this, suppose $Y$ is the random variable and $q_{0.16}$ is the 16% quantile. Then
$$
\text{Pr}(Y\le q_{0.16 | Log graph question
Yes. Quantiles can be transformed under any monotonically increasing transformation.
To see this, suppose $Y$ is the random variable and $q_{0.16}$ is the 16% quantile. Then
$$
\text{Pr}(Y\le q_{0.16}) = \text{Pr}(\log(Y)\le\log(q_{0.16})) = 0.16.
$$
Generally, if $f$ is monotonic and increasing the... | Log graph question
Yes. Quantiles can be transformed under any monotonically increasing transformation.
To see this, suppose $Y$ is the random variable and $q_{0.16}$ is the 16% quantile. Then
$$
\text{Pr}(Y\le q_{0.16 |
39,835 | Log graph question | Yes.
When you say that "X is the 16% quantile", what it means is that 16% of the sample have a lower value than X. The log of any number smaller than X is smaller than log(X) and the log of any number greater than X is greater than log(X), so the ordering is not changed. | Log graph question | Yes.
When you say that "X is the 16% quantile", what it means is that 16% of the sample have a lower value than X. The log of any number smaller than X is smaller than log(X) and the log of any number | Log graph question
Yes.
When you say that "X is the 16% quantile", what it means is that 16% of the sample have a lower value than X. The log of any number smaller than X is smaller than log(X) and the log of any number greater than X is greater than log(X), so the ordering is not changed. | Log graph question
Yes.
When you say that "X is the 16% quantile", what it means is that 16% of the sample have a lower value than X. The log of any number smaller than X is smaller than log(X) and the log of any number |
39,836 | Difference between Norm of Residuals and what is a "good" Norm of Residual | So, I would recommend using standard method for comparing nested models. In your case, you consider two alternative models, the cubic fit being the more "complex" one. An F- or $\chi^2$-test tells you whether the residual sum of squares or deviance significantly decrease when you add further terms. It is very like comp... | Difference between Norm of Residuals and what is a "good" Norm of Residual | So, I would recommend using standard method for comparing nested models. In your case, you consider two alternative models, the cubic fit being the more "complex" one. An F- or $\chi^2$-test tells you | Difference between Norm of Residuals and what is a "good" Norm of Residual
So, I would recommend using standard method for comparing nested models. In your case, you consider two alternative models, the cubic fit being the more "complex" one. An F- or $\chi^2$-test tells you whether the residual sum of squares or devia... | Difference between Norm of Residuals and what is a "good" Norm of Residual
So, I would recommend using standard method for comparing nested models. In your case, you consider two alternative models, the cubic fit being the more "complex" one. An F- or $\chi^2$-test tells you |
39,837 | How to explain such a big difference between parametric and non parametric test (and other questions)? | Because the null hypothesis your McNemar test tests, is not the same as the one tested by the $\chi^2$ test. The McNemar test actually tests whether the probability of 1-2 equals that of 2-1 (say first number is the row and second the column). If you switch columns, then you get a completely different outcome. The $\ch... | How to explain such a big difference between parametric and non parametric test (and other questions | Because the null hypothesis your McNemar test tests, is not the same as the one tested by the $\chi^2$ test. The McNemar test actually tests whether the probability of 1-2 equals that of 2-1 (say firs | How to explain such a big difference between parametric and non parametric test (and other questions)?
Because the null hypothesis your McNemar test tests, is not the same as the one tested by the $\chi^2$ test. The McNemar test actually tests whether the probability of 1-2 equals that of 2-1 (say first number is the r... | How to explain such a big difference between parametric and non parametric test (and other questions
Because the null hypothesis your McNemar test tests, is not the same as the one tested by the $\chi^2$ test. The McNemar test actually tests whether the probability of 1-2 equals that of 2-1 (say firs |
39,838 | Asymptotic power | The definition above (a fixed alternative, sample size going to infinity) is more precisely related to the consistency (or not) of a hypothesis test. That is, a test is consistent against a fixed alternative if the power function approaches 1 at that alternative.
Asymptotic power is something different. As Joris rema... | Asymptotic power | The definition above (a fixed alternative, sample size going to infinity) is more precisely related to the consistency (or not) of a hypothesis test. That is, a test is consistent against a fixed alt | Asymptotic power
The definition above (a fixed alternative, sample size going to infinity) is more precisely related to the consistency (or not) of a hypothesis test. That is, a test is consistent against a fixed alternative if the power function approaches 1 at that alternative.
Asymptotic power is something differen... | Asymptotic power
The definition above (a fixed alternative, sample size going to infinity) is more precisely related to the consistency (or not) of a hypothesis test. That is, a test is consistent against a fixed alt |
39,839 | Asymptotic power | As I understood it, the asymptotic power is the hypothetical power when the effect size goes to zero and the sample size to infinity. Basically it should be 0 or 1, indicating whether the test cannot or can distinguish an arbitrary small deviation from the null hypothesis when the sample size is sufficiently large. | Asymptotic power | As I understood it, the asymptotic power is the hypothetical power when the effect size goes to zero and the sample size to infinity. Basically it should be 0 or 1, indicating whether the test cannot | Asymptotic power
As I understood it, the asymptotic power is the hypothetical power when the effect size goes to zero and the sample size to infinity. Basically it should be 0 or 1, indicating whether the test cannot or can distinguish an arbitrary small deviation from the null hypothesis when the sample size is suffic... | Asymptotic power
As I understood it, the asymptotic power is the hypothetical power when the effect size goes to zero and the sample size to infinity. Basically it should be 0 or 1, indicating whether the test cannot |
39,840 | Asymptotic power | Yes, you are right. I would only replace "there exists a sample size n that the selected test would reject the null hypothesis" with "for every e>0 there exists a sample size n_0 such that the probability to reject the null hypothesis is greater than 1-e for all n>n_0". | Asymptotic power | Yes, you are right. I would only replace "there exists a sample size n that the selected test would reject the null hypothesis" with "for every e>0 there exists a sample size n_0 such that the probabi | Asymptotic power
Yes, you are right. I would only replace "there exists a sample size n that the selected test would reject the null hypothesis" with "for every e>0 there exists a sample size n_0 such that the probability to reject the null hypothesis is greater than 1-e for all n>n_0". | Asymptotic power
Yes, you are right. I would only replace "there exists a sample size n that the selected test would reject the null hypothesis" with "for every e>0 there exists a sample size n_0 such that the probabi |
39,841 | Why does AIC formula in R appear to use one extra parameter than expected? | > -2*logLik(fit)+2*(length(fit$coef)+1)
[1] 3332.282
(you forgot; you have 6 parameter because $\sigma_{\epsilon}$ also has to be estimated! | Why does AIC formula in R appear to use one extra parameter than expected? | > -2*logLik(fit)+2*(length(fit$coef)+1)
[1] 3332.282
(you forgot; you have 6 parameter because $\sigma_{\epsilon}$ also has to be estimated! | Why does AIC formula in R appear to use one extra parameter than expected?
> -2*logLik(fit)+2*(length(fit$coef)+1)
[1] 3332.282
(you forgot; you have 6 parameter because $\sigma_{\epsilon}$ also has to be estimated! | Why does AIC formula in R appear to use one extra parameter than expected?
> -2*logLik(fit)+2*(length(fit$coef)+1)
[1] 3332.282
(you forgot; you have 6 parameter because $\sigma_{\epsilon}$ also has to be estimated! |
39,842 | Working through a clustering problem | You can try Latent Semantic Analysis, which basically provides a way to represent in a reduced space your news feeds and any term (in your case, keyword appearing in the title). As it relies on Singular Value Decomposition, I suppose you may then be able to check if there exists a particular association between those t... | Working through a clustering problem | You can try Latent Semantic Analysis, which basically provides a way to represent in a reduced space your news feeds and any term (in your case, keyword appearing in the title). As it relies on Singul | Working through a clustering problem
You can try Latent Semantic Analysis, which basically provides a way to represent in a reduced space your news feeds and any term (in your case, keyword appearing in the title). As it relies on Singular Value Decomposition, I suppose you may then be able to check if there exists a p... | Working through a clustering problem
You can try Latent Semantic Analysis, which basically provides a way to represent in a reduced space your news feeds and any term (in your case, keyword appearing in the title). As it relies on Singul |
39,843 | Working through a clustering problem | The traditional solution to this problem is to use the vector representation for the news stories and then cluster the vectors. The vectors are arrays where each entry represents a word or word class. The value associated to each word will be the tf-idf weight. This value goes up the more frequent the word in the do... | Working through a clustering problem | The traditional solution to this problem is to use the vector representation for the news stories and then cluster the vectors. The vectors are arrays where each entry represents a word or word class | Working through a clustering problem
The traditional solution to this problem is to use the vector representation for the news stories and then cluster the vectors. The vectors are arrays where each entry represents a word or word class. The value associated to each word will be the tf-idf weight. This value goes up... | Working through a clustering problem
The traditional solution to this problem is to use the vector representation for the news stories and then cluster the vectors. The vectors are arrays where each entry represents a word or word class |
39,844 | Working through a clustering problem | This problem you are asking about is known as text mining!
There are a few things you need to consider. For example in your question you mentioned using keywords in titles. One may ask "why not the text in the article rather than just the title?" which brings me to the first consideration: What data do you limit yourse... | Working through a clustering problem | This problem you are asking about is known as text mining!
There are a few things you need to consider. For example in your question you mentioned using keywords in titles. One may ask "why not the te | Working through a clustering problem
This problem you are asking about is known as text mining!
There are a few things you need to consider. For example in your question you mentioned using keywords in titles. One may ask "why not the text in the article rather than just the title?" which brings me to the first conside... | Working through a clustering problem
This problem you are asking about is known as text mining!
There are a few things you need to consider. For example in your question you mentioned using keywords in titles. One may ask "why not the te |
39,845 | Working through a clustering problem | I would start with a frequency distribution. Collect for a big corpus the word-frequencies and select smartly the words that are keywords (not misspellings, with a very low frequency, and not stop words like "and", "or")
Then when you have a number of new feeds, compare their distribution with the distribution that you... | Working through a clustering problem | I would start with a frequency distribution. Collect for a big corpus the word-frequencies and select smartly the words that are keywords (not misspellings, with a very low frequency, and not stop wor | Working through a clustering problem
I would start with a frequency distribution. Collect for a big corpus the word-frequencies and select smartly the words that are keywords (not misspellings, with a very low frequency, and not stop words like "and", "or")
Then when you have a number of new feeds, compare their distri... | Working through a clustering problem
I would start with a frequency distribution. Collect for a big corpus the word-frequencies and select smartly the words that are keywords (not misspellings, with a very low frequency, and not stop wor |
39,846 | Let $X_1,\dots, X_n$ be random sample from $Bernoulli(p)$. Which estimator is better? | You have been given the MLE as well as the posterior mean under a Beta prior (see e.g. here).
As you write (in slightly different notation with $\theta=p$, $k=n\bar X$, $\alpha_0=\alpha$, $\hat \theta_1=\hat p_1$, $R=MSE$ etc.),
$$
\begin{align*}
R(\theta, \hat{\theta}_2) =& \, V_\theta (\hat{\theta}_2) + ({bias}_\t... | Let $X_1,\dots, X_n$ be random sample from $Bernoulli(p)$. Which estimator is better? | You have been given the MLE as well as the posterior mean under a Beta prior (see e.g. here).
As you write (in slightly different notation with $\theta=p$, $k=n\bar X$, $\alpha_0=\alpha$, $\hat \theta | Let $X_1,\dots, X_n$ be random sample from $Bernoulli(p)$. Which estimator is better?
You have been given the MLE as well as the posterior mean under a Beta prior (see e.g. here).
As you write (in slightly different notation with $\theta=p$, $k=n\bar X$, $\alpha_0=\alpha$, $\hat \theta_1=\hat p_1$, $R=MSE$ etc.),
$$
\... | Let $X_1,\dots, X_n$ be random sample from $Bernoulli(p)$. Which estimator is better?
You have been given the MLE as well as the posterior mean under a Beta prior (see e.g. here).
As you write (in slightly different notation with $\theta=p$, $k=n\bar X$, $\alpha_0=\alpha$, $\hat \theta |
39,847 | Interpreting zero odds ratio from Fisher's exact test | If in a 2-by-2 table, we let $n_{11}$ be the counts in (1,1) cell, $n_{12}$ the counts in the (1,2) cell, $n_{21}$ those of cell (2,1) cell and $n_{22}$ those of cell (2,2), then the sample odds ratio is defined by
$$
\hat\theta = \frac{n_{11}n_{22}}{n_{12}n_{21}}.
$$
If any of $n_{ij}$ is zero, then $\hat\theta$ equal... | Interpreting zero odds ratio from Fisher's exact test | If in a 2-by-2 table, we let $n_{11}$ be the counts in (1,1) cell, $n_{12}$ the counts in the (1,2) cell, $n_{21}$ those of cell (2,1) cell and $n_{22}$ those of cell (2,2), then the sample odds ratio | Interpreting zero odds ratio from Fisher's exact test
If in a 2-by-2 table, we let $n_{11}$ be the counts in (1,1) cell, $n_{12}$ the counts in the (1,2) cell, $n_{21}$ those of cell (2,1) cell and $n_{22}$ those of cell (2,2), then the sample odds ratio is defined by
$$
\hat\theta = \frac{n_{11}n_{22}}{n_{12}n_{21}}.
... | Interpreting zero odds ratio from Fisher's exact test
If in a 2-by-2 table, we let $n_{11}$ be the counts in (1,1) cell, $n_{12}$ the counts in the (1,2) cell, $n_{21}$ those of cell (2,1) cell and $n_{22}$ those of cell (2,2), then the sample odds ratio |
39,848 | Why are generalized additive models (GAMs) so popular in ecology? | TL;DR: GAMs are useful models when specific functional relationships are not hypothesized.
Ecology as a science (and like many other sciences, particularly population sciences) often has hypotheses for its statistical models which do not take specific functional forms. For example, causal relationships may be specified... | Why are generalized additive models (GAMs) so popular in ecology? | TL;DR: GAMs are useful models when specific functional relationships are not hypothesized.
Ecology as a science (and like many other sciences, particularly population sciences) often has hypotheses fo | Why are generalized additive models (GAMs) so popular in ecology?
TL;DR: GAMs are useful models when specific functional relationships are not hypothesized.
Ecology as a science (and like many other sciences, particularly population sciences) often has hypotheses for its statistical models which do not take specific fu... | Why are generalized additive models (GAMs) so popular in ecology?
TL;DR: GAMs are useful models when specific functional relationships are not hypothesized.
Ecology as a science (and like many other sciences, particularly population sciences) often has hypotheses fo |
39,849 | Pearl, Causal Inference in Statistics Q3.5.1 (Backdoor criterion) | No you were right to begin with, you can control for any variable along the back door path so long as it doesn’t open up new such paths.
You can try it for yourself for the specific diagram here (set Z to adjusted and some other one to see only the causal path remain colored): http://dagitty.net/dags.html?id=331 | Pearl, Causal Inference in Statistics Q3.5.1 (Backdoor criterion) | No you were right to begin with, you can control for any variable along the back door path so long as it doesn’t open up new such paths.
You can try it for yourself for the specific diagram here (set | Pearl, Causal Inference in Statistics Q3.5.1 (Backdoor criterion)
No you were right to begin with, you can control for any variable along the back door path so long as it doesn’t open up new such paths.
You can try it for yourself for the specific diagram here (set Z to adjusted and some other one to see only the causa... | Pearl, Causal Inference in Statistics Q3.5.1 (Backdoor criterion)
No you were right to begin with, you can control for any variable along the back door path so long as it doesn’t open up new such paths.
You can try it for yourself for the specific diagram here (set |
39,850 | Sample size recommended for a chi-square test for the variance | If you are using a chi-squared test of $H_0: \sigma^2 = 64$ against
$H_a: \sigma^2 > 64,$ then the sample size required depends on how much
greater than $64$ is important to you. Let's say you want have
probability $.90$ of detecting if the actual variance is $\sigma^2 = 100$ or more.
That is, you want the 'power' of t... | Sample size recommended for a chi-square test for the variance | If you are using a chi-squared test of $H_0: \sigma^2 = 64$ against
$H_a: \sigma^2 > 64,$ then the sample size required depends on how much
greater than $64$ is important to you. Let's say you want ha | Sample size recommended for a chi-square test for the variance
If you are using a chi-squared test of $H_0: \sigma^2 = 64$ against
$H_a: \sigma^2 > 64,$ then the sample size required depends on how much
greater than $64$ is important to you. Let's say you want have
probability $.90$ of detecting if the actual variance ... | Sample size recommended for a chi-square test for the variance
If you are using a chi-squared test of $H_0: \sigma^2 = 64$ against
$H_a: \sigma^2 > 64,$ then the sample size required depends on how much
greater than $64$ is important to you. Let's say you want ha |
39,851 | Sample size recommended for a chi-square test for the variance | The test "works as advertized" even for very small sample sizes but is pretty sensitive to the assumption of normality (e.g. it is generally sensitive to changes in kurtosis).
It is notably more level-sensitive than a t-test is to the assumption of normality, and in this case large samples will not tend to mitigate the... | Sample size recommended for a chi-square test for the variance | The test "works as advertized" even for very small sample sizes but is pretty sensitive to the assumption of normality (e.g. it is generally sensitive to changes in kurtosis).
It is notably more level | Sample size recommended for a chi-square test for the variance
The test "works as advertized" even for very small sample sizes but is pretty sensitive to the assumption of normality (e.g. it is generally sensitive to changes in kurtosis).
It is notably more level-sensitive than a t-test is to the assumption of normalit... | Sample size recommended for a chi-square test for the variance
The test "works as advertized" even for very small sample sizes but is pretty sensitive to the assumption of normality (e.g. it is generally sensitive to changes in kurtosis).
It is notably more level |
39,852 | What is the meaning of isotropic gaussian blobs , which are generated by sklearn.datasets.makeblobs? | The make_blobs() function draws samples from a special Gaussian mixture model. A general Gaussian mixture model with $k$ clusters has a density of the form
$$
p(x) = \sum_{i=1}^k \pi_i \mathcal{N}(\mu_i, \Sigma_i)
$$
where $\pi_i \ge 0$ are the weights of each cluster with $\sum_{i=1}^k \pi_i = 1$, $\mu_i$ are the clu... | What is the meaning of isotropic gaussian blobs , which are generated by sklearn.datasets.makeblobs? | The make_blobs() function draws samples from a special Gaussian mixture model. A general Gaussian mixture model with $k$ clusters has a density of the form
$$
p(x) = \sum_{i=1}^k \pi_i \mathcal{N}(\m | What is the meaning of isotropic gaussian blobs , which are generated by sklearn.datasets.makeblobs?
The make_blobs() function draws samples from a special Gaussian mixture model. A general Gaussian mixture model with $k$ clusters has a density of the form
$$
p(x) = \sum_{i=1}^k \pi_i \mathcal{N}(\mu_i, \Sigma_i)
$$
w... | What is the meaning of isotropic gaussian blobs , which are generated by sklearn.datasets.makeblobs?
The make_blobs() function draws samples from a special Gaussian mixture model. A general Gaussian mixture model with $k$ clusters has a density of the form
$$
p(x) = \sum_{i=1}^k \pi_i \mathcal{N}(\m |
39,853 | What is the meaning of isotropic gaussian blobs , which are generated by sklearn.datasets.makeblobs? | Thanks Talsup, for giving the nice explanation. I am sharing the things in the nutshell.
The code snippet for understanding the make_blobs() is here. make_blobs_notebook | What is the meaning of isotropic gaussian blobs , which are generated by sklearn.datasets.makeblobs? | Thanks Talsup, for giving the nice explanation. I am sharing the things in the nutshell.
The code snippet for understanding the make_blobs() is here. make_blobs_notebook | What is the meaning of isotropic gaussian blobs , which are generated by sklearn.datasets.makeblobs?
Thanks Talsup, for giving the nice explanation. I am sharing the things in the nutshell.
The code snippet for understanding the make_blobs() is here. make_blobs_notebook | What is the meaning of isotropic gaussian blobs , which are generated by sklearn.datasets.makeblobs?
Thanks Talsup, for giving the nice explanation. I am sharing the things in the nutshell.
The code snippet for understanding the make_blobs() is here. make_blobs_notebook |
39,854 | Negative Binomial as Gamma-Poisson Mixture or Compound Logarithmic Poisson: can this correspondence be generalized to other distributions? | Let
$$
M_\theta(z)=E e^{z\theta}
$$
denote the moment generating function of $\theta$. Since $X|\theta\sim \operatorname{Poisson}(\theta)$, the probability generating function (pgf) of $X|\theta$ is
$$
G_{X|\theta}(z)=Ez^X|\theta = e^{\theta(z-1)}.
$$
Hence, using the law of total expectation, the pgf of $X$ is
$$
G_X... | Negative Binomial as Gamma-Poisson Mixture or Compound Logarithmic Poisson: can this correspondence | Let
$$
M_\theta(z)=E e^{z\theta}
$$
denote the moment generating function of $\theta$. Since $X|\theta\sim \operatorname{Poisson}(\theta)$, the probability generating function (pgf) of $X|\theta$ is
| Negative Binomial as Gamma-Poisson Mixture or Compound Logarithmic Poisson: can this correspondence be generalized to other distributions?
Let
$$
M_\theta(z)=E e^{z\theta}
$$
denote the moment generating function of $\theta$. Since $X|\theta\sim \operatorname{Poisson}(\theta)$, the probability generating function (pgf... | Negative Binomial as Gamma-Poisson Mixture or Compound Logarithmic Poisson: can this correspondence
Let
$$
M_\theta(z)=E e^{z\theta}
$$
denote the moment generating function of $\theta$. Since $X|\theta\sim \operatorname{Poisson}(\theta)$, the probability generating function (pgf) of $X|\theta$ is
|
39,855 | Negative Binomial as Gamma-Poisson Mixture or Compound Logarithmic Poisson: can this correspondence be generalized to other distributions? | At the moment this is just a partial answer. I'm going to start by deriving the requirement for these two mixture forms to be equal. This will get you some of the way by at least showing the equation that your functions must solve. At the moment I can't see a full solution.
We can see from the first mixture that the... | Negative Binomial as Gamma-Poisson Mixture or Compound Logarithmic Poisson: can this correspondence | At the moment this is just a partial answer. I'm going to start by deriving the requirement for these two mixture forms to be equal. This will get you some of the way by at least showing the equatio | Negative Binomial as Gamma-Poisson Mixture or Compound Logarithmic Poisson: can this correspondence be generalized to other distributions?
At the moment this is just a partial answer. I'm going to start by deriving the requirement for these two mixture forms to be equal. This will get you some of the way by at least ... | Negative Binomial as Gamma-Poisson Mixture or Compound Logarithmic Poisson: can this correspondence
At the moment this is just a partial answer. I'm going to start by deriving the requirement for these two mixture forms to be equal. This will get you some of the way by at least showing the equatio |
39,856 | Why are the residuals in $\mathbb{R}^{n-p}$? | $\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\tr}{\mathrm{tr}}$
$\newcommand{\real}{\mathbb{R}}$
$\newcommand{\eps}{\epsilon}$
Write the linear model in the matrix form
\begin{align*}
y = X\beta + \epsilon
\end{align*}
where $y \in \mathbb{R}^n, X = \begin{pmatrix} x_1 & \cdots & x_{p}\end{pmatrix} \in \mathbb{R}^{... | Why are the residuals in $\mathbb{R}^{n-p}$? | $\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\tr}{\mathrm{tr}}$
$\newcommand{\real}{\mathbb{R}}$
$\newcommand{\eps}{\epsilon}$
Write the linear model in the matrix form
\begin{align*}
y = X\beta + | Why are the residuals in $\mathbb{R}^{n-p}$?
$\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\tr}{\mathrm{tr}}$
$\newcommand{\real}{\mathbb{R}}$
$\newcommand{\eps}{\epsilon}$
Write the linear model in the matrix form
\begin{align*}
y = X\beta + \epsilon
\end{align*}
where $y \in \mathbb{R}^n, X = \begin{pmatrix} x_1 &... | Why are the residuals in $\mathbb{R}^{n-p}$?
$\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\tr}{\mathrm{tr}}$
$\newcommand{\real}{\mathbb{R}}$
$\newcommand{\eps}{\epsilon}$
Write the linear model in the matrix form
\begin{align*}
y = X\beta + |
39,857 | Why are the residuals in $\mathbb{R}^{n-p}$? | Orthogonal component
The space for the residuals is the orthogonal complement of the space spanned by $X$.
This is because the residual $\mathbf{y}-\mathbf{\hat{y}}$ is a vector that is perpendicular to the fit $\mathbf{\hat{y}}$. In the geometric representation it is a perpendicular projection of $y$ into the space sp... | Why are the residuals in $\mathbb{R}^{n-p}$? | Orthogonal component
The space for the residuals is the orthogonal complement of the space spanned by $X$.
This is because the residual $\mathbf{y}-\mathbf{\hat{y}}$ is a vector that is perpendicular | Why are the residuals in $\mathbb{R}^{n-p}$?
Orthogonal component
The space for the residuals is the orthogonal complement of the space spanned by $X$.
This is because the residual $\mathbf{y}-\mathbf{\hat{y}}$ is a vector that is perpendicular to the fit $\mathbf{\hat{y}}$. In the geometric representation it is a perp... | Why are the residuals in $\mathbb{R}^{n-p}$?
Orthogonal component
The space for the residuals is the orthogonal complement of the space spanned by $X$.
This is because the residual $\mathbf{y}-\mathbf{\hat{y}}$ is a vector that is perpendicular |
39,858 | Why are the residuals in $\mathbb{R}^{n-p}$? | I think the constraints on the residuals mean that they can be represented by $n-p$ numbers. Take a trivial case,
$$
y = \bar{y} + \varepsilon
$$
Here, there will be n error terms, but they will sum to zero. So, if you tell me $n-1$ error terms, I can work out the other one.
I think there is something analogous for the... | Why are the residuals in $\mathbb{R}^{n-p}$? | I think the constraints on the residuals mean that they can be represented by $n-p$ numbers. Take a trivial case,
$$
y = \bar{y} + \varepsilon
$$
Here, there will be n error terms, but they will sum t | Why are the residuals in $\mathbb{R}^{n-p}$?
I think the constraints on the residuals mean that they can be represented by $n-p$ numbers. Take a trivial case,
$$
y = \bar{y} + \varepsilon
$$
Here, there will be n error terms, but they will sum to zero. So, if you tell me $n-1$ error terms, I can work out the other one.... | Why are the residuals in $\mathbb{R}^{n-p}$?
I think the constraints on the residuals mean that they can be represented by $n-p$ numbers. Take a trivial case,
$$
y = \bar{y} + \varepsilon
$$
Here, there will be n error terms, but they will sum t |
39,859 | Is $\mathrm{binomial}(n, p)$ family both full and curved for $n$ fixed? | From Casella & Berger, p.137-38:
When $n$ is given, (1) is an exponential family. It is full as the parameter space $(0,1)$ is the largest possible. In the sense that
$$\log p - \log (1-p)$$
ranges $(-\infty,+\infty)$ when $p\in(0,1)$. Rewriting it as
$$ f(x|p)=\tbinom{n}{x}\exp \left[x\log p+(n-x)\log(1-p)\right]\tag... | Is $\mathrm{binomial}(n, p)$ family both full and curved for $n$ fixed? | From Casella & Berger, p.137-38:
When $n$ is given, (1) is an exponential family. It is full as the parameter space $(0,1)$ is the largest possible. In the sense that
$$\log p - \log (1-p)$$
ranges $ | Is $\mathrm{binomial}(n, p)$ family both full and curved for $n$ fixed?
From Casella & Berger, p.137-38:
When $n$ is given, (1) is an exponential family. It is full as the parameter space $(0,1)$ is the largest possible. In the sense that
$$\log p - \log (1-p)$$
ranges $(-\infty,+\infty)$ when $p\in(0,1)$. Rewriting i... | Is $\mathrm{binomial}(n, p)$ family both full and curved for $n$ fixed?
From Casella & Berger, p.137-38:
When $n$ is given, (1) is an exponential family. It is full as the parameter space $(0,1)$ is the largest possible. In the sense that
$$\log p - \log (1-p)$$
ranges $ |
39,860 | Rejection threshold of the Benjamini-Hochberg procedure | As you sense, there is no fixed p-value cutoff for the Benjamini-Hochberg control of false discovery rate. The cutoff depends on the specific distribution of p-values among the $m$ hypotheses that you are evaluating together. You put them in increasing order and count up in $k$ from the lowest p-value $(k=1)$. You agre... | Rejection threshold of the Benjamini-Hochberg procedure | As you sense, there is no fixed p-value cutoff for the Benjamini-Hochberg control of false discovery rate. The cutoff depends on the specific distribution of p-values among the $m$ hypotheses that you | Rejection threshold of the Benjamini-Hochberg procedure
As you sense, there is no fixed p-value cutoff for the Benjamini-Hochberg control of false discovery rate. The cutoff depends on the specific distribution of p-values among the $m$ hypotheses that you are evaluating together. You put them in increasing order and c... | Rejection threshold of the Benjamini-Hochberg procedure
As you sense, there is no fixed p-value cutoff for the Benjamini-Hochberg control of false discovery rate. The cutoff depends on the specific distribution of p-values among the $m$ hypotheses that you |
39,861 | Rejection threshold of the Benjamini-Hochberg procedure | I'm not certain of the formal validity of this approach, but you could calculate the corresponding FWER as given by the Hochberg method.
The Benjamini-Hochberg procedure for controlling the False Discovery Rate is (I'm going to quote Wikipedia)
... we have $H_1 \ldots H_m$ null hypotheses tested and $P_1 \ldots P_m$ t... | Rejection threshold of the Benjamini-Hochberg procedure | I'm not certain of the formal validity of this approach, but you could calculate the corresponding FWER as given by the Hochberg method.
The Benjamini-Hochberg procedure for controlling the False Disc | Rejection threshold of the Benjamini-Hochberg procedure
I'm not certain of the formal validity of this approach, but you could calculate the corresponding FWER as given by the Hochberg method.
The Benjamini-Hochberg procedure for controlling the False Discovery Rate is (I'm going to quote Wikipedia)
... we have $H_1 \... | Rejection threshold of the Benjamini-Hochberg procedure
I'm not certain of the formal validity of this approach, but you could calculate the corresponding FWER as given by the Hochberg method.
The Benjamini-Hochberg procedure for controlling the False Disc |
39,862 | Rejection threshold of the Benjamini-Hochberg procedure | After some thought, I believe that the unadjusted p-value of the last (rank-wise) significant test after the BH-procedure comes closest to a significance threshold.
An example:
Do the BH-procedure:
Some p-values: $0.0001,0.0234,0.3354,0.0021,0.5211,0.9123,0.0008,0.0293,0.0500, 1.000$
Order them: $0.0001, 0.0008, 0.00... | Rejection threshold of the Benjamini-Hochberg procedure | After some thought, I believe that the unadjusted p-value of the last (rank-wise) significant test after the BH-procedure comes closest to a significance threshold.
An example:
Do the BH-procedure:
S | Rejection threshold of the Benjamini-Hochberg procedure
After some thought, I believe that the unadjusted p-value of the last (rank-wise) significant test after the BH-procedure comes closest to a significance threshold.
An example:
Do the BH-procedure:
Some p-values: $0.0001,0.0234,0.3354,0.0021,0.5211,0.9123,0.0008,... | Rejection threshold of the Benjamini-Hochberg procedure
After some thought, I believe that the unadjusted p-value of the last (rank-wise) significant test after the BH-procedure comes closest to a significance threshold.
An example:
Do the BH-procedure:
S |
39,863 | Obtaining the correct Log-likelihood function | The loglikelihood is correct. Your problem is the derivatives.
The loglikelihood is not differentiable with respect to $\mu$ where $\mu=x_i$ (and in particular, not at the MLE). The second derivative with respect to $\mu$ is zero everywhere that it is defined. So you can't just do calculus.
For any fixed $\lambda$, the... | Obtaining the correct Log-likelihood function | The loglikelihood is correct. Your problem is the derivatives.
The loglikelihood is not differentiable with respect to $\mu$ where $\mu=x_i$ (and in particular, not at the MLE). The second derivative | Obtaining the correct Log-likelihood function
The loglikelihood is correct. Your problem is the derivatives.
The loglikelihood is not differentiable with respect to $\mu$ where $\mu=x_i$ (and in particular, not at the MLE). The second derivative with respect to $\mu$ is zero everywhere that it is defined. So you can't ... | Obtaining the correct Log-likelihood function
The loglikelihood is correct. Your problem is the derivatives.
The loglikelihood is not differentiable with respect to $\mu$ where $\mu=x_i$ (and in particular, not at the MLE). The second derivative |
39,864 | Which categorical variable encoding should I use if I want the model intercept to have the interpretation of the global mean? | If the data are balanced, then deviation coding should work.
Let's look at a simple example:
set.seed(13)
dt <- expand.grid(X1 = LETTERS[1:3], reps = 1:5)
X <- model.matrix(~ X1, dt)
betas <- c(3, 1, 2)
dt$Y <- X %*% betas + rnorm(nrow(dt))
mean(dt$Y)
[1] 4.11413
So we would like the intercept to be 4.11413
If we fi... | Which categorical variable encoding should I use if I want the model intercept to have the interpret | If the data are balanced, then deviation coding should work.
Let's look at a simple example:
set.seed(13)
dt <- expand.grid(X1 = LETTERS[1:3], reps = 1:5)
X <- model.matrix(~ X1, dt)
betas <- c(3, 1, | Which categorical variable encoding should I use if I want the model intercept to have the interpretation of the global mean?
If the data are balanced, then deviation coding should work.
Let's look at a simple example:
set.seed(13)
dt <- expand.grid(X1 = LETTERS[1:3], reps = 1:5)
X <- model.matrix(~ X1, dt)
betas <- c(... | Which categorical variable encoding should I use if I want the model intercept to have the interpret
If the data are balanced, then deviation coding should work.
Let's look at a simple example:
set.seed(13)
dt <- expand.grid(X1 = LETTERS[1:3], reps = 1:5)
X <- model.matrix(~ X1, dt)
betas <- c(3, 1, |
39,865 | What are the steps to simulate data for a linear model with random slopes and random intercepts | The general approach to simulating data for a mixed model is as follows:
Create the variable(s) for the fixed effects
Create the variable(s) for the group(s)
The fixed effects coefficients will be supplied / given and these will be a column vector $\beta$
Create a model matrix, $X$, for the fixed effects
Simulate the ... | What are the steps to simulate data for a linear model with random slopes and random intercepts | The general approach to simulating data for a mixed model is as follows:
Create the variable(s) for the fixed effects
Create the variable(s) for the group(s)
The fixed effects coefficients will be su | What are the steps to simulate data for a linear model with random slopes and random intercepts
The general approach to simulating data for a mixed model is as follows:
Create the variable(s) for the fixed effects
Create the variable(s) for the group(s)
The fixed effects coefficients will be supplied / given and these... | What are the steps to simulate data for a linear model with random slopes and random intercepts
The general approach to simulating data for a mixed model is as follows:
Create the variable(s) for the fixed effects
Create the variable(s) for the group(s)
The fixed effects coefficients will be su |
39,866 | Probability that linear combination of normal random variables exceeds a value | If you're looking for an analytic solution, this is what you should do. Start by defining a new variable
$$ X = 2Y_1 + 4Y_2 - 3Y_3$$
A sum of jointly normal random variables is also normal, even if the variables are not independent (see Wikipedia article here). All that remains is to compute the mean and variance.
The ... | Probability that linear combination of normal random variables exceeds a value | If you're looking for an analytic solution, this is what you should do. Start by defining a new variable
$$ X = 2Y_1 + 4Y_2 - 3Y_3$$
A sum of jointly normal random variables is also normal, even if th | Probability that linear combination of normal random variables exceeds a value
If you're looking for an analytic solution, this is what you should do. Start by defining a new variable
$$ X = 2Y_1 + 4Y_2 - 3Y_3$$
A sum of jointly normal random variables is also normal, even if the variables are not independent (see Wiki... | Probability that linear combination of normal random variables exceeds a value
If you're looking for an analytic solution, this is what you should do. Start by defining a new variable
$$ X = 2Y_1 + 4Y_2 - 3Y_3$$
A sum of jointly normal random variables is also normal, even if th |
39,867 | Probability that linear combination of normal random variables exceeds a value | You can obtain the mean and variance of the resulting linear combination or you can also do a simulation and obtain the results.
Via simulation.
> covmat=matrix(c(2,0.5,-0.75,0.5,2,0,-0.75,0,8),nrow=3)
> is.positive.definite(x, tol=1e-8)
[1] TRUE
> means=c(10,12,14)
> weights=c(2,4,-3)
> mat=mvrnorm(10^7,means,covmat)
... | Probability that linear combination of normal random variables exceeds a value | You can obtain the mean and variance of the resulting linear combination or you can also do a simulation and obtain the results.
Via simulation.
> covmat=matrix(c(2,0.5,-0.75,0.5,2,0,-0.75,0,8),nrow=3 | Probability that linear combination of normal random variables exceeds a value
You can obtain the mean and variance of the resulting linear combination or you can also do a simulation and obtain the results.
Via simulation.
> covmat=matrix(c(2,0.5,-0.75,0.5,2,0,-0.75,0,8),nrow=3)
> is.positive.definite(x, tol=1e-8)
[1]... | Probability that linear combination of normal random variables exceeds a value
You can obtain the mean and variance of the resulting linear combination or you can also do a simulation and obtain the results.
Via simulation.
> covmat=matrix(c(2,0.5,-0.75,0.5,2,0,-0.75,0,8),nrow=3 |
39,868 | reparameterization trick in VAEs, How should we do this? | We do this simply because the variance of a random variable is constrained to be positive (i.e. $\sigma^2 \in \mathbb{R}^+$) and so if we were to try to learn the variance we would have to constrain somehow the output of a neural network to be positive. A simple way around this is to learn the logarithm instead since $... | reparameterization trick in VAEs, How should we do this? | We do this simply because the variance of a random variable is constrained to be positive (i.e. $\sigma^2 \in \mathbb{R}^+$) and so if we were to try to learn the variance we would have to constrain s | reparameterization trick in VAEs, How should we do this?
We do this simply because the variance of a random variable is constrained to be positive (i.e. $\sigma^2 \in \mathbb{R}^+$) and so if we were to try to learn the variance we would have to constrain somehow the output of a neural network to be positive. A simple ... | reparameterization trick in VAEs, How should we do this?
We do this simply because the variance of a random variable is constrained to be positive (i.e. $\sigma^2 \in \mathbb{R}^+$) and so if we were to try to learn the variance we would have to constrain s |
39,869 | power analysis for binomial test | In order to find 'power', you need to have a specific alternative in mind.
Suppose your null hypothesis is $H_0: p = 0.5$ vs. $H_a: p > 0.5,$ where
$p = P(\mathrm{Female}).$ Also suppose you have $n = 64$ and you want the power
of a test at level $\alpha = 0.05$ against the specific alternative $p = 0.6.$
For an exact ... | power analysis for binomial test | In order to find 'power', you need to have a specific alternative in mind.
Suppose your null hypothesis is $H_0: p = 0.5$ vs. $H_a: p > 0.5,$ where
$p = P(\mathrm{Female}).$ Also suppose you have $n = | power analysis for binomial test
In order to find 'power', you need to have a specific alternative in mind.
Suppose your null hypothesis is $H_0: p = 0.5$ vs. $H_a: p > 0.5,$ where
$p = P(\mathrm{Female}).$ Also suppose you have $n = 64$ and you want the power
of a test at level $\alpha = 0.05$ against the specific alt... | power analysis for binomial test
In order to find 'power', you need to have a specific alternative in mind.
Suppose your null hypothesis is $H_0: p = 0.5$ vs. $H_a: p > 0.5,$ where
$p = P(\mathrm{Female}).$ Also suppose you have $n = |
39,870 | power analysis for binomial test | It's also important to ask yourself why you are calculating the power. Since you already have the data, you are calculating a 'post-hoc' power statistic (as opposed to a pre-hoc power statistic). It's worth nothing that a number of authors have criticised the use of post-hoc power statistics from a Frequentist point ... | power analysis for binomial test | It's also important to ask yourself why you are calculating the power. Since you already have the data, you are calculating a 'post-hoc' power statistic (as opposed to a pre-hoc power statistic). It | power analysis for binomial test
It's also important to ask yourself why you are calculating the power. Since you already have the data, you are calculating a 'post-hoc' power statistic (as opposed to a pre-hoc power statistic). It's worth nothing that a number of authors have criticised the use of post-hoc power sta... | power analysis for binomial test
It's also important to ask yourself why you are calculating the power. Since you already have the data, you are calculating a 'post-hoc' power statistic (as opposed to a pre-hoc power statistic). It |
39,871 | Can adding a random intercept change the fixed effect estimates in a regression model? | Yes.
This is an example of Simpson's paradox. There are already plenty of resources online explaining Simpson's paradox, so I won't go into it here.
To see this in action, let's look at simulated behavioural data where
Participants produce a response, $y$, in response to varying stimuli, $x$.
Participants' intercepts ... | Can adding a random intercept change the fixed effect estimates in a regression model? | Yes.
This is an example of Simpson's paradox. There are already plenty of resources online explaining Simpson's paradox, so I won't go into it here.
To see this in action, let's look at simulated beha | Can adding a random intercept change the fixed effect estimates in a regression model?
Yes.
This is an example of Simpson's paradox. There are already plenty of resources online explaining Simpson's paradox, so I won't go into it here.
To see this in action, let's look at simulated behavioural data where
Participants ... | Can adding a random intercept change the fixed effect estimates in a regression model?
Yes.
This is an example of Simpson's paradox. There are already plenty of resources online explaining Simpson's paradox, so I won't go into it here.
To see this in action, let's look at simulated beha |
39,872 | Can adding a random intercept change the fixed effect estimates in a regression model? | The accepted answer is absolutely correct but the degree of change in the fixed effect with the inclusion of the random effect can be very sensitive to the number of repeated measurements per subject. For instance, in my own analysis (data plotted below), the inclusion of a random intercept does not change the slope of... | Can adding a random intercept change the fixed effect estimates in a regression model? | The accepted answer is absolutely correct but the degree of change in the fixed effect with the inclusion of the random effect can be very sensitive to the number of repeated measurements per subject. | Can adding a random intercept change the fixed effect estimates in a regression model?
The accepted answer is absolutely correct but the degree of change in the fixed effect with the inclusion of the random effect can be very sensitive to the number of repeated measurements per subject. For instance, in my own analysis... | Can adding a random intercept change the fixed effect estimates in a regression model?
The accepted answer is absolutely correct but the degree of change in the fixed effect with the inclusion of the random effect can be very sensitive to the number of repeated measurements per subject. |
39,873 | Cross entropy vs KL divergence: What's minimized directly in practice? | Let $q$ be the density of your true data-generating process and $f_\theta$ be your model-density.
Then $$KL(q||f_\theta) = \int q(x) log\left(\frac{q(x)}{f_\theta(x)}\right)dx = -\int q(x) \log(f_\theta(x))dx + \int q(x) \log(q(x)) dx$$
The first term is the Cross Entropy $H(q, f_\theta)$ and the second term is the (di... | Cross entropy vs KL divergence: What's minimized directly in practice? | Let $q$ be the density of your true data-generating process and $f_\theta$ be your model-density.
Then $$KL(q||f_\theta) = \int q(x) log\left(\frac{q(x)}{f_\theta(x)}\right)dx = -\int q(x) \log(f_\the | Cross entropy vs KL divergence: What's minimized directly in practice?
Let $q$ be the density of your true data-generating process and $f_\theta$ be your model-density.
Then $$KL(q||f_\theta) = \int q(x) log\left(\frac{q(x)}{f_\theta(x)}\right)dx = -\int q(x) \log(f_\theta(x))dx + \int q(x) \log(q(x)) dx$$
The first te... | Cross entropy vs KL divergence: What's minimized directly in practice?
Let $q$ be the density of your true data-generating process and $f_\theta$ be your model-density.
Then $$KL(q||f_\theta) = \int q(x) log\left(\frac{q(x)}{f_\theta(x)}\right)dx = -\int q(x) \log(f_\the |
39,874 | Deviance for Gamma GLM | The general derivation of the deviance for a GLM family is given in Section 5.4 of Dunn and Smyth (2018) (the book that you mentioned in a previous post).
You can insert the form of the gamma density to get the result, but the density has to parametrized in the right way.
A common way to write the gamma density is as
$... | Deviance for Gamma GLM | The general derivation of the deviance for a GLM family is given in Section 5.4 of Dunn and Smyth (2018) (the book that you mentioned in a previous post).
You can insert the form of the gamma density | Deviance for Gamma GLM
The general derivation of the deviance for a GLM family is given in Section 5.4 of Dunn and Smyth (2018) (the book that you mentioned in a previous post).
You can insert the form of the gamma density to get the result, but the density has to parametrized in the right way.
A common way to write th... | Deviance for Gamma GLM
The general derivation of the deviance for a GLM family is given in Section 5.4 of Dunn and Smyth (2018) (the book that you mentioned in a previous post).
You can insert the form of the gamma density |
39,875 | What is the expected value of x log(x) of the gamma distribution? | $$\begin{align}E[X\log X]&=\int_0^\infty \log x \frac{\lambda^\alpha}{\Gamma(\alpha)}x^\alpha e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}\int_0^\infty\log x\frac{\lambda^{\alpha+1}}{\Gamma(\alpha+1)}x^{(\alpha+1)-1}e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}E[\log Y]\end{al... | What is the expected value of x log(x) of the gamma distribution? | $$\begin{align}E[X\log X]&=\int_0^\infty \log x \frac{\lambda^\alpha}{\Gamma(\alpha)}x^\alpha e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}\int_0^\infty\log x\frac{\lambda^{\alpha | What is the expected value of x log(x) of the gamma distribution?
$$\begin{align}E[X\log X]&=\int_0^\infty \log x \frac{\lambda^\alpha}{\Gamma(\alpha)}x^\alpha e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}\int_0^\infty\log x\frac{\lambda^{\alpha+1}}{\Gamma(\alpha+1)}x^{(\alpha+1)-1}e^{-\lambda x}dx... | What is the expected value of x log(x) of the gamma distribution?
$$\begin{align}E[X\log X]&=\int_0^\infty \log x \frac{\lambda^\alpha}{\Gamma(\alpha)}x^\alpha e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}\int_0^\infty\log x\frac{\lambda^{\alpha |
39,876 | How could one (tentatively) estimate the actual number of COVID-19 infections in an area, using hard data like age-adjusted death rate? | Death rate
Using death rate may work when people that die have most likely acquired the virus locally.
However, for the moment, many people die because they acquired the virus during a visit to another area, or because they got it from another person that acquired the virus abroad.
Thus this death-rate based number i... | How could one (tentatively) estimate the actual number of COVID-19 infections in an area, using hard | Death rate
Using death rate may work when people that die have most likely acquired the virus locally.
However, for the moment, many people die because they acquired the virus during a visit to anoth | How could one (tentatively) estimate the actual number of COVID-19 infections in an area, using hard data like age-adjusted death rate?
Death rate
Using death rate may work when people that die have most likely acquired the virus locally.
However, for the moment, many people die because they acquired the virus during ... | How could one (tentatively) estimate the actual number of COVID-19 infections in an area, using hard
Death rate
Using death rate may work when people that die have most likely acquired the virus locally.
However, for the moment, many people die because they acquired the virus during a visit to anoth |
39,877 | How could one (tentatively) estimate the actual number of COVID-19 infections in an area, using hard data like age-adjusted death rate? | It seems that there is a blog post that also deals with this question.
If you have deaths in your region, you can use that to guess the number of true current cases. We know approximately how long it takes for that person to go from catching the virus to dying on average (17.3 days). That means the person who died on... | How could one (tentatively) estimate the actual number of COVID-19 infections in an area, using hard | It seems that there is a blog post that also deals with this question.
If you have deaths in your region, you can use that to guess the number of true current cases. We know approximately how long i | How could one (tentatively) estimate the actual number of COVID-19 infections in an area, using hard data like age-adjusted death rate?
It seems that there is a blog post that also deals with this question.
If you have deaths in your region, you can use that to guess the number of true current cases. We know approxim... | How could one (tentatively) estimate the actual number of COVID-19 infections in an area, using hard
It seems that there is a blog post that also deals with this question.
If you have deaths in your region, you can use that to guess the number of true current cases. We know approximately how long i |
39,878 | Is there any difference between hypothesis testing $A\beta = a$ and $\beta = A^{-1}a$? | If you wish to perform a constrained linear regression with linear equality constraints of the form $A\beta = a$, then tests such as the likelihood ratio test are more naturally formulated in this way. This is because the constrained OLS estimator $\beta^*$ can be written in terms of the unconstrained estimator $\hat{... | Is there any difference between hypothesis testing $A\beta = a$ and $\beta = A^{-1}a$? | If you wish to perform a constrained linear regression with linear equality constraints of the form $A\beta = a$, then tests such as the likelihood ratio test are more naturally formulated in this way | Is there any difference between hypothesis testing $A\beta = a$ and $\beta = A^{-1}a$?
If you wish to perform a constrained linear regression with linear equality constraints of the form $A\beta = a$, then tests such as the likelihood ratio test are more naturally formulated in this way. This is because the constraine... | Is there any difference between hypothesis testing $A\beta = a$ and $\beta = A^{-1}a$?
If you wish to perform a constrained linear regression with linear equality constraints of the form $A\beta = a$, then tests such as the likelihood ratio test are more naturally formulated in this way |
39,879 | Is there any difference between hypothesis testing $A\beta = a$ and $\beta = A^{-1}a$? | It is often the case that $A$ is not a square matrix, and so not invertible. However, assuming that $A$ is actually invertible, the statement $A \beta = a$ is equivalent to $\beta = A^{-1} a$, and so your formulation of the test is legitimate (though unusual). Regardless of which form you use to state the hypotheses,... | Is there any difference between hypothesis testing $A\beta = a$ and $\beta = A^{-1}a$? | It is often the case that $A$ is not a square matrix, and so not invertible. However, assuming that $A$ is actually invertible, the statement $A \beta = a$ is equivalent to $\beta = A^{-1} a$, and so | Is there any difference between hypothesis testing $A\beta = a$ and $\beta = A^{-1}a$?
It is often the case that $A$ is not a square matrix, and so not invertible. However, assuming that $A$ is actually invertible, the statement $A \beta = a$ is equivalent to $\beta = A^{-1} a$, and so your formulation of the test is ... | Is there any difference between hypothesis testing $A\beta = a$ and $\beta = A^{-1}a$?
It is often the case that $A$ is not a square matrix, and so not invertible. However, assuming that $A$ is actually invertible, the statement $A \beta = a$ is equivalent to $\beta = A^{-1} a$, and so |
39,880 | Is independence subjective? | I'm a bit late and good things have already been written, however I didn't see the following mentioned:
For person A coin tosses are independent and identically distributed, hence they are exchangeable.
For person B coin tosses are neither independent nor identically distributed but they are exchangeable."
This doesn... | Is independence subjective? | I'm a bit late and good things have already been written, however I didn't see the following mentioned:
For person A coin tosses are independent and identically distributed, hence they are exchangeab | Is independence subjective?
I'm a bit late and good things have already been written, however I didn't see the following mentioned:
For person A coin tosses are independent and identically distributed, hence they are exchangeable.
For person B coin tosses are neither independent nor identically distributed but they ar... | Is independence subjective?
I'm a bit late and good things have already been written, however I didn't see the following mentioned:
For person A coin tosses are independent and identically distributed, hence they are exchangeab |
39,881 | Is independence subjective? | It seems me that ultimately the answer to your question boil down in the concept of probability that you accept.
Let me recover the Tim answer’s
In probability theory, independence of two events is defined as
$P(A∩B)=P(A)P(B)$
There is nothing subjective about the definition, it is a possible property of random variab... | Is independence subjective? | It seems me that ultimately the answer to your question boil down in the concept of probability that you accept.
Let me recover the Tim answer’s
In probability theory, independence of two events is d | Is independence subjective?
It seems me that ultimately the answer to your question boil down in the concept of probability that you accept.
Let me recover the Tim answer’s
In probability theory, independence of two events is defined as
$P(A∩B)=P(A)P(B)$
There is nothing subjective about the definition, it is a possib... | Is independence subjective?
It seems me that ultimately the answer to your question boil down in the concept of probability that you accept.
Let me recover the Tim answer’s
In probability theory, independence of two events is d |
39,882 | Is independence subjective? | In probability theory, independence of two events is defined as
$$
P(A \cap B) = P(A)\,P(B)
$$
There is nothing subjective about the definition, it is a possible property of random variables. For real life data, because of things like sampling bias, measurement error, insufficient sample size, numerical precision etc.,... | Is independence subjective? | In probability theory, independence of two events is defined as
$$
P(A \cap B) = P(A)\,P(B)
$$
There is nothing subjective about the definition, it is a possible property of random variables. For real | Is independence subjective?
In probability theory, independence of two events is defined as
$$
P(A \cap B) = P(A)\,P(B)
$$
There is nothing subjective about the definition, it is a possible property of random variables. For real life data, because of things like sampling bias, measurement error, insufficient sample siz... | Is independence subjective?
In probability theory, independence of two events is defined as
$$
P(A \cap B) = P(A)\,P(B)
$$
There is nothing subjective about the definition, it is a possible property of random variables. For real |
39,883 | Is independence subjective? | Up to the last paragraph, the setting appears to be entirely Bayesian. I will first respond to the questions raised up to (but excluding) the last paragraph from a Bayesian perspective.
...for the same physical process of tossing fair coins in a row, one person thinks that the tosses are independent and the other does... | Is independence subjective? | Up to the last paragraph, the setting appears to be entirely Bayesian. I will first respond to the questions raised up to (but excluding) the last paragraph from a Bayesian perspective.
...for the sa | Is independence subjective?
Up to the last paragraph, the setting appears to be entirely Bayesian. I will first respond to the questions raised up to (but excluding) the last paragraph from a Bayesian perspective.
...for the same physical process of tossing fair coins in a row, one person thinks that the tosses are in... | Is independence subjective?
Up to the last paragraph, the setting appears to be entirely Bayesian. I will first respond to the questions raised up to (but excluding) the last paragraph from a Bayesian perspective.
...for the sa |
39,884 | Is independence subjective? | The coin tosses that each of the two subjects will conduct will always be independent and identically distributed random variables (and thus exchangeable) no matter what they think. The underlying properties and distributions are not subject to our beliefs about them.
For instance, if you fill up a room with 1 million ... | Is independence subjective? | The coin tosses that each of the two subjects will conduct will always be independent and identically distributed random variables (and thus exchangeable) no matter what they think. The underlying pro | Is independence subjective?
The coin tosses that each of the two subjects will conduct will always be independent and identically distributed random variables (and thus exchangeable) no matter what they think. The underlying properties and distributions are not subject to our beliefs about them.
For instance, if you fi... | Is independence subjective?
The coin tosses that each of the two subjects will conduct will always be independent and identically distributed random variables (and thus exchangeable) no matter what they think. The underlying pro |
39,885 | Does $X\stackrel{d}\to X_1$ and $Y\stackrel{d}\to Y_1$ imply $X+Y\stackrel{d}\to X_1+Y_1$? | What $X+Y$ and $X_1+Y_1$ converge to depends upon the joint distributions of $(X,Y)$ and $(X_1,Y_1)$; what $X, X_1, Y, Y_1$ converge to individually depends upon the marginal distributions. You can have the same marginal distributions with different joint distributions. Because the joint distributions may be differen... | Does $X\stackrel{d}\to X_1$ and $Y\stackrel{d}\to Y_1$ imply $X+Y\stackrel{d}\to X_1+Y_1$? | What $X+Y$ and $X_1+Y_1$ converge to depends upon the joint distributions of $(X,Y)$ and $(X_1,Y_1)$; what $X, X_1, Y, Y_1$ converge to individually depends upon the marginal distributions. You can h | Does $X\stackrel{d}\to X_1$ and $Y\stackrel{d}\to Y_1$ imply $X+Y\stackrel{d}\to X_1+Y_1$?
What $X+Y$ and $X_1+Y_1$ converge to depends upon the joint distributions of $(X,Y)$ and $(X_1,Y_1)$; what $X, X_1, Y, Y_1$ converge to individually depends upon the marginal distributions. You can have the same marginal distrib... | Does $X\stackrel{d}\to X_1$ and $Y\stackrel{d}\to Y_1$ imply $X+Y\stackrel{d}\to X_1+Y_1$?
What $X+Y$ and $X_1+Y_1$ converge to depends upon the joint distributions of $(X,Y)$ and $(X_1,Y_1)$; what $X, X_1, Y, Y_1$ converge to individually depends upon the marginal distributions. You can h |
39,886 | Approximating the median of a $\Gamma(\alpha,1)$ distribution with $0<\alpha<1$ | Let $\mu_\alpha$ be the median of a $\Gamma(\alpha)$ distribution. This means the area under the density $$f_{\Gamma(\alpha)}(x) = \frac{x^{\alpha-1}}{\Gamma(\alpha)} \,e^{-x}$$ between $x=0$ and $x=\mu_\alpha$ equals $1/2.$ A graph of $f_{\Gamma(\alpha)}$ is sketched here in black (for $\alpha=0.3$), understanding t... | Approximating the median of a $\Gamma(\alpha,1)$ distribution with $0<\alpha<1$ | Let $\mu_\alpha$ be the median of a $\Gamma(\alpha)$ distribution. This means the area under the density $$f_{\Gamma(\alpha)}(x) = \frac{x^{\alpha-1}}{\Gamma(\alpha)} \,e^{-x}$$ between $x=0$ and $x= | Approximating the median of a $\Gamma(\alpha,1)$ distribution with $0<\alpha<1$
Let $\mu_\alpha$ be the median of a $\Gamma(\alpha)$ distribution. This means the area under the density $$f_{\Gamma(\alpha)}(x) = \frac{x^{\alpha-1}}{\Gamma(\alpha)} \,e^{-x}$$ between $x=0$ and $x=\mu_\alpha$ equals $1/2.$ A graph of $f... | Approximating the median of a $\Gamma(\alpha,1)$ distribution with $0<\alpha<1$
Let $\mu_\alpha$ be the median of a $\Gamma(\alpha)$ distribution. This means the area under the density $$f_{\Gamma(\alpha)}(x) = \frac{x^{\alpha-1}}{\Gamma(\alpha)} \,e^{-x}$$ between $x=0$ and $x= |
39,887 | Approximating the median of a $\Gamma(\alpha,1)$ distribution with $0<\alpha<1$ | The other answer by whuber gives some nice simple bounds for the median. In this answer I will give an alternative closed form approximation based on finite Newton iteration to the true quantile. My answer uses the lower incomplete Gamma function $\gamma$ (so perhaps you consider that cheating), but it does not requi... | Approximating the median of a $\Gamma(\alpha,1)$ distribution with $0<\alpha<1$ | The other answer by whuber gives some nice simple bounds for the median. In this answer I will give an alternative closed form approximation based on finite Newton iteration to the true quantile. My | Approximating the median of a $\Gamma(\alpha,1)$ distribution with $0<\alpha<1$
The other answer by whuber gives some nice simple bounds for the median. In this answer I will give an alternative closed form approximation based on finite Newton iteration to the true quantile. My answer uses the lower incomplete Gamma ... | Approximating the median of a $\Gamma(\alpha,1)$ distribution with $0<\alpha<1$
The other answer by whuber gives some nice simple bounds for the median. In this answer I will give an alternative closed form approximation based on finite Newton iteration to the true quantile. My |
39,888 | Using all Metropolis-Hastings proposals to estimate an integral | Recycling proposed values in a Metropolis-Hastings algorithm goes under the name of Rao-Blackwellisation. For instance, we made such a proposal in
Casella and Robert (1996) Rao-Blackwellisation of sampling schemes.
Douc and Robert (2013) A vanilla Rao-Blackwellisation of Metropolis-Hastings algorithms
Note that the... | Using all Metropolis-Hastings proposals to estimate an integral | Recycling proposed values in a Metropolis-Hastings algorithm goes under the name of Rao-Blackwellisation. For instance, we made such a proposal in
Casella and Robert (1996) Rao-Blackwellisation of s | Using all Metropolis-Hastings proposals to estimate an integral
Recycling proposed values in a Metropolis-Hastings algorithm goes under the name of Rao-Blackwellisation. For instance, we made such a proposal in
Casella and Robert (1996) Rao-Blackwellisation of sampling schemes.
Douc and Robert (2013) A vanilla Rao-B... | Using all Metropolis-Hastings proposals to estimate an integral
Recycling proposed values in a Metropolis-Hastings algorithm goes under the name of Rao-Blackwellisation. For instance, we made such a proposal in
Casella and Robert (1996) Rao-Blackwellisation of s |
39,889 | Using all Metropolis-Hastings proposals to estimate an integral | This looks similar to "local" importance sampling. In the literature, constructing estimators of this sort seem to termed as "waste-recyling", and quick search yields a few papers:
This thesis by Murray
PNAS paper by Freknel here
Does waste-recyling help by Delmas and Jourdain here
Their use in parallel MCMC here
Fur... | Using all Metropolis-Hastings proposals to estimate an integral | This looks similar to "local" importance sampling. In the literature, constructing estimators of this sort seem to termed as "waste-recyling", and quick search yields a few papers:
This thesis by Mur | Using all Metropolis-Hastings proposals to estimate an integral
This looks similar to "local" importance sampling. In the literature, constructing estimators of this sort seem to termed as "waste-recyling", and quick search yields a few papers:
This thesis by Murray
PNAS paper by Freknel here
Does waste-recyling help ... | Using all Metropolis-Hastings proposals to estimate an integral
This looks similar to "local" importance sampling. In the literature, constructing estimators of this sort seem to termed as "waste-recyling", and quick search yields a few papers:
This thesis by Mur |
39,890 | Which likelihood function is used in linear regression? | As you say, in this case the choice doesn't matter; the maximization over $w$ is the same. But you touch on a good point.
Typically for regression problems we prefer to write the version with the conditioning on $x$ because we think of regression as modeling the conditional distribution $p(y | x)$ while ignoring the de... | Which likelihood function is used in linear regression? | As you say, in this case the choice doesn't matter; the maximization over $w$ is the same. But you touch on a good point.
Typically for regression problems we prefer to write the version with the cond | Which likelihood function is used in linear regression?
As you say, in this case the choice doesn't matter; the maximization over $w$ is the same. But you touch on a good point.
Typically for regression problems we prefer to write the version with the conditioning on $x$ because we think of regression as modeling the c... | Which likelihood function is used in linear regression?
As you say, in this case the choice doesn't matter; the maximization over $w$ is the same. But you touch on a good point.
Typically for regression problems we prefer to write the version with the cond |
39,891 | Which likelihood function is used in linear regression? | Linear regression is about how $x$ influences/varies with $y$, the outcome or response variable. The model equation is
$$ Y_i =\beta_0 + \beta_1 x_{i1} + \dotsm + \beta_p x_{ip}+\epsilon_i
$$ say, and how $x$ is distributed doesn't by itself give information about the $\beta$'s. That's why your second form of likeliho... | Which likelihood function is used in linear regression? | Linear regression is about how $x$ influences/varies with $y$, the outcome or response variable. The model equation is
$$ Y_i =\beta_0 + \beta_1 x_{i1} + \dotsm + \beta_p x_{ip}+\epsilon_i
$$ say, an | Which likelihood function is used in linear regression?
Linear regression is about how $x$ influences/varies with $y$, the outcome or response variable. The model equation is
$$ Y_i =\beta_0 + \beta_1 x_{i1} + \dotsm + \beta_p x_{ip}+\epsilon_i
$$ say, and how $x$ is distributed doesn't by itself give information abou... | Which likelihood function is used in linear regression?
Linear regression is about how $x$ influences/varies with $y$, the outcome or response variable. The model equation is
$$ Y_i =\beta_0 + \beta_1 x_{i1} + \dotsm + \beta_p x_{ip}+\epsilon_i
$$ say, an |
39,892 | Which likelihood function is used in linear regression? | That's a good question since the difference is a bit subtle - hopefully this helps.
The difference between in calculating $p(y|x,w)$ and $p(y,x|w)$ lies in whether you model X as being fixed or random. In the first, you're looking for the joint probability of X and Y conditioned on w, whereas on the 2nd one, you want ... | Which likelihood function is used in linear regression? | That's a good question since the difference is a bit subtle - hopefully this helps.
The difference between in calculating $p(y|x,w)$ and $p(y,x|w)$ lies in whether you model X as being fixed or rando | Which likelihood function is used in linear regression?
That's a good question since the difference is a bit subtle - hopefully this helps.
The difference between in calculating $p(y|x,w)$ and $p(y,x|w)$ lies in whether you model X as being fixed or random. In the first, you're looking for the joint probability of X a... | Which likelihood function is used in linear regression?
That's a good question since the difference is a bit subtle - hopefully this helps.
The difference between in calculating $p(y|x,w)$ and $p(y,x|w)$ lies in whether you model X as being fixed or rando |
39,893 | Can we use two convolutional layers without a activation layer in middle of them? | Since convolution is a linear operator, two consecutive convolutional layers can be realized by a single convolutional layer if there is no non-linear activation layer between them. Then, it is trivial to train one extra layer which has no positive effect on the representational power of your model. | Can we use two convolutional layers without a activation layer in middle of them? | Since convolution is a linear operator, two consecutive convolutional layers can be realized by a single convolutional layer if there is no non-linear activation layer between them. Then, it is trivia | Can we use two convolutional layers without a activation layer in middle of them?
Since convolution is a linear operator, two consecutive convolutional layers can be realized by a single convolutional layer if there is no non-linear activation layer between them. Then, it is trivial to train one extra layer which has n... | Can we use two convolutional layers without a activation layer in middle of them?
Since convolution is a linear operator, two consecutive convolutional layers can be realized by a single convolutional layer if there is no non-linear activation layer between them. Then, it is trivia |
39,894 | Can we use two convolutional layers without a activation layer in middle of them? | Stacking convolutions is equivalent to convolution, but with different filter size. It might be that the effectively bigger filter gives you different results, but there exists appropriately sized conv layer that is equivalent wiith two conv layers (if the boundary conditions are right) | Can we use two convolutional layers without a activation layer in middle of them? | Stacking convolutions is equivalent to convolution, but with different filter size. It might be that the effectively bigger filter gives you different results, but there exists appropriately sized con | Can we use two convolutional layers without a activation layer in middle of them?
Stacking convolutions is equivalent to convolution, but with different filter size. It might be that the effectively bigger filter gives you different results, but there exists appropriately sized conv layer that is equivalent wiith two c... | Can we use two convolutional layers without a activation layer in middle of them?
Stacking convolutions is equivalent to convolution, but with different filter size. It might be that the effectively bigger filter gives you different results, but there exists appropriately sized con |
39,895 | What's an example of an estimator that is $o_p(n^{-1/2})$? | Estimators of percentiles can act like this.
The idea is that when a great deal of probability is concentrated in small neighborhoods around a percentile, then the sample percentile will tend to come extremely close to the true percentile. (Indeed, an extreme case occurs when a distribution has an atom at a percentile... | What's an example of an estimator that is $o_p(n^{-1/2})$? | Estimators of percentiles can act like this.
The idea is that when a great deal of probability is concentrated in small neighborhoods around a percentile, then the sample percentile will tend to come | What's an example of an estimator that is $o_p(n^{-1/2})$?
Estimators of percentiles can act like this.
The idea is that when a great deal of probability is concentrated in small neighborhoods around a percentile, then the sample percentile will tend to come extremely close to the true percentile. (Indeed, an extreme ... | What's an example of an estimator that is $o_p(n^{-1/2})$?
Estimators of percentiles can act like this.
The idea is that when a great deal of probability is concentrated in small neighborhoods around a percentile, then the sample percentile will tend to come |
39,896 | What's an example of an estimator that is $o_p(n^{-1/2})$? | The range for $n$ i.i.d. uniform random variables on $[0, 1]$ has a $\mbox{Beta}(n-1, 2)$ distribution, thus its variance is $\frac{2(n-1)}{(n+1)^2 (n+2)}$. Thus this estimator is $O(n^{-1})$ and thus $o(n^{-1/2})$.
At the other extreme, the range for $n$ i.i.d. exponential random variables has a variance equal to $\fr... | What's an example of an estimator that is $o_p(n^{-1/2})$? | The range for $n$ i.i.d. uniform random variables on $[0, 1]$ has a $\mbox{Beta}(n-1, 2)$ distribution, thus its variance is $\frac{2(n-1)}{(n+1)^2 (n+2)}$. Thus this estimator is $O(n^{-1})$ and thus | What's an example of an estimator that is $o_p(n^{-1/2})$?
The range for $n$ i.i.d. uniform random variables on $[0, 1]$ has a $\mbox{Beta}(n-1, 2)$ distribution, thus its variance is $\frac{2(n-1)}{(n+1)^2 (n+2)}$. Thus this estimator is $O(n^{-1})$ and thus $o(n^{-1/2})$.
At the other extreme, the range for $n$ i.i.d... | What's an example of an estimator that is $o_p(n^{-1/2})$?
The range for $n$ i.i.d. uniform random variables on $[0, 1]$ has a $\mbox{Beta}(n-1, 2)$ distribution, thus its variance is $\frac{2(n-1)}{(n+1)^2 (n+2)}$. Thus this estimator is $O(n^{-1})$ and thus |
39,897 | What's an example of an estimator that is $o_p(n^{-1/2})$? | Here's an example that doesn't involve quantiles/extremes:
Suppose $(X,Y)$ are bivariate Normal with correlation $\rho$ and consider the usual estimator $$\hat\rho=\frac{\sum_i (X_i-\bar X) (Y_i-\bar Y)}{\sqrt{\sum_i (X_i-\bar X)^2\sum_j(Y_j-\bar Y)^2}}$$
Now suppose we want $\rho^2$ and we estimate it with $\hat\rho^... | What's an example of an estimator that is $o_p(n^{-1/2})$? | Here's an example that doesn't involve quantiles/extremes:
Suppose $(X,Y)$ are bivariate Normal with correlation $\rho$ and consider the usual estimator $$\hat\rho=\frac{\sum_i (X_i-\bar X) (Y_i-\bar | What's an example of an estimator that is $o_p(n^{-1/2})$?
Here's an example that doesn't involve quantiles/extremes:
Suppose $(X,Y)$ are bivariate Normal with correlation $\rho$ and consider the usual estimator $$\hat\rho=\frac{\sum_i (X_i-\bar X) (Y_i-\bar Y)}{\sqrt{\sum_i (X_i-\bar X)^2\sum_j(Y_j-\bar Y)^2}}$$
Now ... | What's an example of an estimator that is $o_p(n^{-1/2})$?
Here's an example that doesn't involve quantiles/extremes:
Suppose $(X,Y)$ are bivariate Normal with correlation $\rho$ and consider the usual estimator $$\hat\rho=\frac{\sum_i (X_i-\bar X) (Y_i-\bar |
39,898 | Linking generative, discriminative models to supervised and unsupervised learning | Well first I would nitpick your definition. Consider a conditional GAN which given a category $x$ generates a realistic image $y$ in category $x$ -- that is, it models $P(y|x)$. This is arguably not a discriminative model, but it is according to your definition.
One counterexample to claim 1 would be a model like Neura... | Linking generative, discriminative models to supervised and unsupervised learning | Well first I would nitpick your definition. Consider a conditional GAN which given a category $x$ generates a realistic image $y$ in category $x$ -- that is, it models $P(y|x)$. This is arguably not a | Linking generative, discriminative models to supervised and unsupervised learning
Well first I would nitpick your definition. Consider a conditional GAN which given a category $x$ generates a realistic image $y$ in category $x$ -- that is, it models $P(y|x)$. This is arguably not a discriminative model, but it is accor... | Linking generative, discriminative models to supervised and unsupervised learning
Well first I would nitpick your definition. Consider a conditional GAN which given a category $x$ generates a realistic image $y$ in category $x$ -- that is, it models $P(y|x)$. This is arguably not a |
39,899 | Linking generative, discriminative models to supervised and unsupervised learning | A note about the generative-discriminative dichotomy
Although your formal generative-vs-discriminative definition is widely used and very useful [1], I agree with @shimao that it's somewhat of a false dichotomy [2] which starts falling apart when you consider that it's in fact a spectrum which you can even smoothly int... | Linking generative, discriminative models to supervised and unsupervised learning | A note about the generative-discriminative dichotomy
Although your formal generative-vs-discriminative definition is widely used and very useful [1], I agree with @shimao that it's somewhat of a false | Linking generative, discriminative models to supervised and unsupervised learning
A note about the generative-discriminative dichotomy
Although your formal generative-vs-discriminative definition is widely used and very useful [1], I agree with @shimao that it's somewhat of a false dichotomy [2] which starts falling ap... | Linking generative, discriminative models to supervised and unsupervised learning
A note about the generative-discriminative dichotomy
Although your formal generative-vs-discriminative definition is widely used and very useful [1], I agree with @shimao that it's somewhat of a false |
39,900 | Regression as mutual information minimization | I don't completely understand your notation; my apologies if I misunderstand. But I think MI is too weak for regression, because it only enforces dependence between variables, rather than any specific relationship (as regression requires). (In fact, it is this generality that makes MI better for detecting hidden statis... | Regression as mutual information minimization | I don't completely understand your notation; my apologies if I misunderstand. But I think MI is too weak for regression, because it only enforces dependence between variables, rather than any specific | Regression as mutual information minimization
I don't completely understand your notation; my apologies if I misunderstand. But I think MI is too weak for regression, because it only enforces dependence between variables, rather than any specific relationship (as regression requires). (In fact, it is this generality th... | Regression as mutual information minimization
I don't completely understand your notation; my apologies if I misunderstand. But I think MI is too weak for regression, because it only enforces dependence between variables, rather than any specific |
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