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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50,601 | creating multiple categorical variable with specified degree of association (correlation) matrix [duplicate] | If the variables are just dichotomous you can treat them as binomial. Then the job becomes easier. The package bindata can simulate multivariate distribution with specified correlation. Just small example from the manual
amat <- cbind(c(1/2,1/5,1/6),c(1/5,1/2,1/6),c(1/6,1/6,1/2))
require(bindata)
out <- rmvbin(n... | creating multiple categorical variable with specified degree of association (correlation) matrix [du | If the variables are just dichotomous you can treat them as binomial. Then the job becomes easier. The package bindata can simulate multivariate distribution with specified correlation. Just small exa | creating multiple categorical variable with specified degree of association (correlation) matrix [duplicate]
If the variables are just dichotomous you can treat them as binomial. Then the job becomes easier. The package bindata can simulate multivariate distribution with specified correlation. Just small example from t... | creating multiple categorical variable with specified degree of association (correlation) matrix [du
If the variables are just dichotomous you can treat them as binomial. Then the job becomes easier. The package bindata can simulate multivariate distribution with specified correlation. Just small exa |
50,602 | How would you model this random effects structure? | This is very similar to a classic split plot design.
Forest type is a fixed effect with four levels. It is a bit of a stretch to think of forest type being randomly assigned to sites. Instead, sites are a random effect nested in forest type.
Treatment is a fixed effect with three levels.
Plots are a random effect n... | How would you model this random effects structure? | This is very similar to a classic split plot design.
Forest type is a fixed effect with four levels. It is a bit of a stretch to think of forest type being randomly assigned to sites. Instead, sit | How would you model this random effects structure?
This is very similar to a classic split plot design.
Forest type is a fixed effect with four levels. It is a bit of a stretch to think of forest type being randomly assigned to sites. Instead, sites are a random effect nested in forest type.
Treatment is a fixed ef... | How would you model this random effects structure?
This is very similar to a classic split plot design.
Forest type is a fixed effect with four levels. It is a bit of a stretch to think of forest type being randomly assigned to sites. Instead, sit |
50,603 | Estimating total variation distance from a given distribution | This reference demonstrates how to test that the distributions $p$ and $q$ have $||p-q||_{TV}\leq \max\left(\frac{\epsilon^2}{32\sqrt[3]{n}},\frac{\epsilon}{4\sqrt{n}}\right)$ with probability at least $1-\delta$ for your choices of $\epsilon$ and $\delta$, with number of samples $O(n^{2/3}\epsilon^{-4}\log n)$.
In you... | Estimating total variation distance from a given distribution | This reference demonstrates how to test that the distributions $p$ and $q$ have $||p-q||_{TV}\leq \max\left(\frac{\epsilon^2}{32\sqrt[3]{n}},\frac{\epsilon}{4\sqrt{n}}\right)$ with probability at leas | Estimating total variation distance from a given distribution
This reference demonstrates how to test that the distributions $p$ and $q$ have $||p-q||_{TV}\leq \max\left(\frac{\epsilon^2}{32\sqrt[3]{n}},\frac{\epsilon}{4\sqrt{n}}\right)$ with probability at least $1-\delta$ for your choices of $\epsilon$ and $\delta$, ... | Estimating total variation distance from a given distribution
This reference demonstrates how to test that the distributions $p$ and $q$ have $||p-q||_{TV}\leq \max\left(\frac{\epsilon^2}{32\sqrt[3]{n}},\frac{\epsilon}{4\sqrt{n}}\right)$ with probability at leas |
50,604 | Independence of a linear and a quadratic form | Use Craig's Theorem. Consider the quadratic form on b. If two random variables are independent, then any univariate functions of those random variables are likewise independent. The quadratic forms are independent, ergo the linear form on b and the quadratic form on A are likewise independent. | Independence of a linear and a quadratic form | Use Craig's Theorem. Consider the quadratic form on b. If two random variables are independent, then any univariate functions of those random variables are likewise independent. The quadratic forms | Independence of a linear and a quadratic form
Use Craig's Theorem. Consider the quadratic form on b. If two random variables are independent, then any univariate functions of those random variables are likewise independent. The quadratic forms are independent, ergo the linear form on b and the quadratic form on A ar... | Independence of a linear and a quadratic form
Use Craig's Theorem. Consider the quadratic form on b. If two random variables are independent, then any univariate functions of those random variables are likewise independent. The quadratic forms |
50,605 | Independence of a linear and a quadratic form | Starting with the univariate case $X=X_1$, we find the correlation:
$\rho(bX,AX^2)=bA\rho(X,X^2)=bA\dfrac{\mathrm{Cov}(X,X^2)}{ \sigma_X \sigma_{X^2}} =bA\dfrac{E[(X-\mu_X)(X^2-\mu_{X^2})]}{ \sigma_X\sigma_{X^2}}$
With $\mu_x=0$, $\sigma_x=\sigma$, and for the expectation we know the distributions of $X$ and $X^2$ (Nor... | Independence of a linear and a quadratic form | Starting with the univariate case $X=X_1$, we find the correlation:
$\rho(bX,AX^2)=bA\rho(X,X^2)=bA\dfrac{\mathrm{Cov}(X,X^2)}{ \sigma_X \sigma_{X^2}} =bA\dfrac{E[(X-\mu_X)(X^2-\mu_{X^2})]}{ \sigma_X\ | Independence of a linear and a quadratic form
Starting with the univariate case $X=X_1$, we find the correlation:
$\rho(bX,AX^2)=bA\rho(X,X^2)=bA\dfrac{\mathrm{Cov}(X,X^2)}{ \sigma_X \sigma_{X^2}} =bA\dfrac{E[(X-\mu_X)(X^2-\mu_{X^2})]}{ \sigma_X\sigma_{X^2}}$
With $\mu_x=0$, $\sigma_x=\sigma$, and for the expectation w... | Independence of a linear and a quadratic form
Starting with the univariate case $X=X_1$, we find the correlation:
$\rho(bX,AX^2)=bA\rho(X,X^2)=bA\dfrac{\mathrm{Cov}(X,X^2)}{ \sigma_X \sigma_{X^2}} =bA\dfrac{E[(X-\mu_X)(X^2-\mu_{X^2})]}{ \sigma_X\ |
50,606 | Why is it necessary to use ML estimation instead of REML to compare multilevel linear models? | (RE)ML estimation is an iterative process. ML estimate the variances as if the fixed parameters are known, so doesn't account for the degrees of freedom lost in their estimation. REML adjusts for the uncertainty about the fixed parameters.
So you generally cannot use REML to compare models, because whatever difference ... | Why is it necessary to use ML estimation instead of REML to compare multilevel linear models? | (RE)ML estimation is an iterative process. ML estimate the variances as if the fixed parameters are known, so doesn't account for the degrees of freedom lost in their estimation. REML adjusts for the | Why is it necessary to use ML estimation instead of REML to compare multilevel linear models?
(RE)ML estimation is an iterative process. ML estimate the variances as if the fixed parameters are known, so doesn't account for the degrees of freedom lost in their estimation. REML adjusts for the uncertainty about the fixe... | Why is it necessary to use ML estimation instead of REML to compare multilevel linear models?
(RE)ML estimation is an iterative process. ML estimate the variances as if the fixed parameters are known, so doesn't account for the degrees of freedom lost in their estimation. REML adjusts for the |
50,607 | Meaning of Qqnorm plot in R | The x-coordinate of the points is the value this point would have if it were drawn from the standard normal distribution (preserving it's current quantile). That is to say, if it currently is at the median of the sample, it's corresponding value in the standard normal distribution would be 0 (2.5th percentile would be ... | Meaning of Qqnorm plot in R | The x-coordinate of the points is the value this point would have if it were drawn from the standard normal distribution (preserving it's current quantile). That is to say, if it currently is at the m | Meaning of Qqnorm plot in R
The x-coordinate of the points is the value this point would have if it were drawn from the standard normal distribution (preserving it's current quantile). That is to say, if it currently is at the median of the sample, it's corresponding value in the standard normal distribution would be 0... | Meaning of Qqnorm plot in R
The x-coordinate of the points is the value this point would have if it were drawn from the standard normal distribution (preserving it's current quantile). That is to say, if it currently is at the m |
50,608 | Is there an equivalent to Lower bound of Wilson score confidence interval for variables with more outcome | It's easy to think of the following 'workaround' which adapts a multi-ranking system to the 'upvote/downvote' solution discussed in the linked article:
Let's say you have the popular 5 star rating system. So we have a number of votes, each having a value of: 1, 2, 3, 4 or 5.
To 'convert' these ratings to up/down votes,... | Is there an equivalent to Lower bound of Wilson score confidence interval for variables with more ou | It's easy to think of the following 'workaround' which adapts a multi-ranking system to the 'upvote/downvote' solution discussed in the linked article:
Let's say you have the popular 5 star rating sys | Is there an equivalent to Lower bound of Wilson score confidence interval for variables with more outcome
It's easy to think of the following 'workaround' which adapts a multi-ranking system to the 'upvote/downvote' solution discussed in the linked article:
Let's say you have the popular 5 star rating system. So we hav... | Is there an equivalent to Lower bound of Wilson score confidence interval for variables with more ou
It's easy to think of the following 'workaround' which adapts a multi-ranking system to the 'upvote/downvote' solution discussed in the linked article:
Let's say you have the popular 5 star rating sys |
50,609 | Setting up a naive tensor product B-spline example | I know this is an old question, but it's a good one, and I thought about it recently myself.
You select the rows of the Kronecker product matrix in which the row indices of each $B_x$ and $B_z$ element are the same. For example, each element of $B_x \otimes B_z$ is (I am going to use $x$ to mean an element of the $B_x$... | Setting up a naive tensor product B-spline example | I know this is an old question, but it's a good one, and I thought about it recently myself.
You select the rows of the Kronecker product matrix in which the row indices of each $B_x$ and $B_z$ elemen | Setting up a naive tensor product B-spline example
I know this is an old question, but it's a good one, and I thought about it recently myself.
You select the rows of the Kronecker product matrix in which the row indices of each $B_x$ and $B_z$ element are the same. For example, each element of $B_x \otimes B_z$ is (I ... | Setting up a naive tensor product B-spline example
I know this is an old question, but it's a good one, and I thought about it recently myself.
You select the rows of the Kronecker product matrix in which the row indices of each $B_x$ and $B_z$ elemen |
50,610 | Similarity measures for point processes | I think the most direct way to determine if your measurements are similar is to compute the wait time distribution for each measurement.
The wait time is just the time elapsed between events, where an event is, presumably, signal equals $1$. This will give you a series of wait times. You can then plot these as a distri... | Similarity measures for point processes | I think the most direct way to determine if your measurements are similar is to compute the wait time distribution for each measurement.
The wait time is just the time elapsed between events, where an | Similarity measures for point processes
I think the most direct way to determine if your measurements are similar is to compute the wait time distribution for each measurement.
The wait time is just the time elapsed between events, where an event is, presumably, signal equals $1$. This will give you a series of wait ti... | Similarity measures for point processes
I think the most direct way to determine if your measurements are similar is to compute the wait time distribution for each measurement.
The wait time is just the time elapsed between events, where an |
50,611 | Similarity measures for point processes | i would try coherence
another idea is to apply haar wavelet analysis and compare the measurements in frequency space | Similarity measures for point processes | i would try coherence
another idea is to apply haar wavelet analysis and compare the measurements in frequency space | Similarity measures for point processes
i would try coherence
another idea is to apply haar wavelet analysis and compare the measurements in frequency space | Similarity measures for point processes
i would try coherence
another idea is to apply haar wavelet analysis and compare the measurements in frequency space |
50,612 | Why is Sampling Importance Resampling (SIR) better than Importance Sampling (IS)? | The main point is that most of the time, SIS/SIR is used in a sequential setting and this is why one needs to reallocate particles to best deal with the next time integration. When performing IS, it is quite common that all of the weightings are attributed to only a very small subpart of the particles whereas they char... | Why is Sampling Importance Resampling (SIR) better than Importance Sampling (IS)? | The main point is that most of the time, SIS/SIR is used in a sequential setting and this is why one needs to reallocate particles to best deal with the next time integration. When performing IS, it i | Why is Sampling Importance Resampling (SIR) better than Importance Sampling (IS)?
The main point is that most of the time, SIS/SIR is used in a sequential setting and this is why one needs to reallocate particles to best deal with the next time integration. When performing IS, it is quite common that all of the weighti... | Why is Sampling Importance Resampling (SIR) better than Importance Sampling (IS)?
The main point is that most of the time, SIS/SIR is used in a sequential setting and this is why one needs to reallocate particles to best deal with the next time integration. When performing IS, it i |
50,613 | lme4: glmer problems with offset() | I don't think you do want to add sampling_effort as a offset at all.
You are estimating the logit of marked over unmarked catch probabilities - call that quantity $\eta$ - and you think, quite reasonably, that the quantity $N=$ marked+unmarked will be a function of how long the traps remain undamaged, i.e. a functio... | lme4: glmer problems with offset() | I don't think you do want to add sampling_effort as a offset at all.
You are estimating the logit of marked over unmarked catch probabilities - call that quantity $\eta$ - and you think, quite reas | lme4: glmer problems with offset()
I don't think you do want to add sampling_effort as a offset at all.
You are estimating the logit of marked over unmarked catch probabilities - call that quantity $\eta$ - and you think, quite reasonably, that the quantity $N=$ marked+unmarked will be a function of how long the tra... | lme4: glmer problems with offset()
I don't think you do want to add sampling_effort as a offset at all.
You are estimating the logit of marked over unmarked catch probabilities - call that quantity $\eta$ - and you think, quite reas |
50,614 | Could you explain how gradient boosting algorithm works? | For implementation checkout:
https://github.com/2pc/libgbdt.git
For algorithm detail there is another graphical depiction which explained better:
http://www.lifesciencessociety.org/CSB2006/toc/PDF/43.2006.pdf
and this:
which in the protein folding case, translate to this: | Could you explain how gradient boosting algorithm works? | For implementation checkout:
https://github.com/2pc/libgbdt.git
For algorithm detail there is another graphical depiction which explained better:
http://www.lifesciencessociety.org/CSB2006/toc/PDF/43. | Could you explain how gradient boosting algorithm works?
For implementation checkout:
https://github.com/2pc/libgbdt.git
For algorithm detail there is another graphical depiction which explained better:
http://www.lifesciencessociety.org/CSB2006/toc/PDF/43.2006.pdf
and this:
which in the protein folding case, transla... | Could you explain how gradient boosting algorithm works?
For implementation checkout:
https://github.com/2pc/libgbdt.git
For algorithm detail there is another graphical depiction which explained better:
http://www.lifesciencessociety.org/CSB2006/toc/PDF/43. |
50,615 | Could you explain how gradient boosting algorithm works? | From the FAQ in the appendix of an article I wrote with Jeremy Howard, called How to explain gradient boosting:
"Instead of creating a single powerful model, boosting combines multiple simple models into a single composite model. The idea is that, as we introduce more and more simple models, the overall model becomes s... | Could you explain how gradient boosting algorithm works? | From the FAQ in the appendix of an article I wrote with Jeremy Howard, called How to explain gradient boosting:
"Instead of creating a single powerful model, boosting combines multiple simple models i | Could you explain how gradient boosting algorithm works?
From the FAQ in the appendix of an article I wrote with Jeremy Howard, called How to explain gradient boosting:
"Instead of creating a single powerful model, boosting combines multiple simple models into a single composite model. The idea is that, as we introduce... | Could you explain how gradient boosting algorithm works?
From the FAQ in the appendix of an article I wrote with Jeremy Howard, called How to explain gradient boosting:
"Instead of creating a single powerful model, boosting combines multiple simple models i |
50,616 | Spline – basis functions | This looks like a truncated power basis. The answer is b) although $h_5(X)$ will only be non-zero if $X$ is greater than $\xi_1$ and similarly for $h_6(X)$ and $\xi_2$ | Spline – basis functions | This looks like a truncated power basis. The answer is b) although $h_5(X)$ will only be non-zero if $X$ is greater than $\xi_1$ and similarly for $h_6(X)$ and $\xi_2$ | Spline – basis functions
This looks like a truncated power basis. The answer is b) although $h_5(X)$ will only be non-zero if $X$ is greater than $\xi_1$ and similarly for $h_6(X)$ and $\xi_2$ | Spline – basis functions
This looks like a truncated power basis. The answer is b) although $h_5(X)$ will only be non-zero if $X$ is greater than $\xi_1$ and similarly for $h_6(X)$ and $\xi_2$ |
50,617 | Is the p-value equivalent to the false alarm value in the Bayesian rule? | They're related in that selecting the critical value for p amounts to selecting a false alarm rate, under the assumption that the null hypothesis is true. The difference between the two is what is observed and known prior to the test.
It's conventional to denote the probability of making a type I error, i.e. a false po... | Is the p-value equivalent to the false alarm value in the Bayesian rule? | They're related in that selecting the critical value for p amounts to selecting a false alarm rate, under the assumption that the null hypothesis is true. The difference between the two is what is obs | Is the p-value equivalent to the false alarm value in the Bayesian rule?
They're related in that selecting the critical value for p amounts to selecting a false alarm rate, under the assumption that the null hypothesis is true. The difference between the two is what is observed and known prior to the test.
It's convent... | Is the p-value equivalent to the false alarm value in the Bayesian rule?
They're related in that selecting the critical value for p amounts to selecting a false alarm rate, under the assumption that the null hypothesis is true. The difference between the two is what is obs |
50,618 | Demonstrate difference in growth over time | 1) log, I think log would work.
(motivation, error scales with size of measurement, also your error bars might have hit negative values, ie normal distribution is not right and less powerfull)
m<-lm(y~factor(Tx),data=df1) #p ~ 0.05
m<-lm(log(y)~factor(Tx),data=df1) #p ~ 0.015
and equaly for
m<-lm(y~factor(Day)+facto... | Demonstrate difference in growth over time | 1) log, I think log would work.
(motivation, error scales with size of measurement, also your error bars might have hit negative values, ie normal distribution is not right and less powerfull)
m<-lm( | Demonstrate difference in growth over time
1) log, I think log would work.
(motivation, error scales with size of measurement, also your error bars might have hit negative values, ie normal distribution is not right and less powerfull)
m<-lm(y~factor(Tx),data=df1) #p ~ 0.05
m<-lm(log(y)~factor(Tx),data=df1) #p ~ 0.015... | Demonstrate difference in growth over time
1) log, I think log would work.
(motivation, error scales with size of measurement, also your error bars might have hit negative values, ie normal distribution is not right and less powerfull)
m<-lm( |
50,619 | Demonstrate difference in growth over time | OK so we gave up and did it properly physically, and it worked fine.
Nonetheless this remains a borderline case so will keep it up here for a while. According to my 'GraphPadPrisn' manual 'AIC' is the way to approach this sort of problem so will keep this open until I verify for myself. Maybe one of those inter-generat... | Demonstrate difference in growth over time | OK so we gave up and did it properly physically, and it worked fine.
Nonetheless this remains a borderline case so will keep it up here for a while. According to my 'GraphPadPrisn' manual 'AIC' is the | Demonstrate difference in growth over time
OK so we gave up and did it properly physically, and it worked fine.
Nonetheless this remains a borderline case so will keep it up here for a while. According to my 'GraphPadPrisn' manual 'AIC' is the way to approach this sort of problem so will keep this open until I verify f... | Demonstrate difference in growth over time
OK so we gave up and did it properly physically, and it worked fine.
Nonetheless this remains a borderline case so will keep it up here for a while. According to my 'GraphPadPrisn' manual 'AIC' is the |
50,620 | How to use a mathematical model for data analysis in R | You can fit this equation to your data using non-linear regression.
I'd give it a try with nls. The crucial aspect of using nls is to provide sensible starting values. An example code could look something like
nls.mod <- nls(Wa~Wma*(1+a*(Na + alpha*N))^(-b), data = dataset, start = list(a = 1, b = 1, alpha = 1)). – CO... | How to use a mathematical model for data analysis in R | You can fit this equation to your data using non-linear regression.
I'd give it a try with nls. The crucial aspect of using nls is to provide sensible starting values. An example code could look somet | How to use a mathematical model for data analysis in R
You can fit this equation to your data using non-linear regression.
I'd give it a try with nls. The crucial aspect of using nls is to provide sensible starting values. An example code could look something like
nls.mod <- nls(Wa~Wma*(1+a*(Na + alpha*N))^(-b), data ... | How to use a mathematical model for data analysis in R
You can fit this equation to your data using non-linear regression.
I'd give it a try with nls. The crucial aspect of using nls is to provide sensible starting values. An example code could look somet |
50,621 | How to use a mathematical model for data analysis in R | The mean-yield per plant is strictly positive, which means we can deal with its logarithm (as a real number). The deterministic version of the model can be usefully rewritten as:
$$\ln W_{A} = \ln W_{mA} - b_A \ln (1 + a_{A} (N_{A}+\alpha N_{B})).$$
The obvious stochastic analogy would be the non-linear regression:
$... | How to use a mathematical model for data analysis in R | The mean-yield per plant is strictly positive, which means we can deal with its logarithm (as a real number). The deterministic version of the model can be usefully rewritten as:
$$\ln W_{A} = \ln W_ | How to use a mathematical model for data analysis in R
The mean-yield per plant is strictly positive, which means we can deal with its logarithm (as a real number). The deterministic version of the model can be usefully rewritten as:
$$\ln W_{A} = \ln W_{mA} - b_A \ln (1 + a_{A} (N_{A}+\alpha N_{B})).$$
The obvious st... | How to use a mathematical model for data analysis in R
The mean-yield per plant is strictly positive, which means we can deal with its logarithm (as a real number). The deterministic version of the model can be usefully rewritten as:
$$\ln W_{A} = \ln W_ |
50,622 | Bias and variance estimation with boostrap | I think the formulas for Bootstrap are as follows, although I can't seem to find a proper reference:
Let $h = \frac{1}{K} \sum_{k=1}^{K}{\hat{y}_k}$
be the mean predicted outcome from the $K$ bootstrap resamples.
Then:
Bias is $y - h$
Variance is $\frac{1}{K-1}\sum_{k=1}^{K} (y_k - h)^2$ | Bias and variance estimation with boostrap | I think the formulas for Bootstrap are as follows, although I can't seem to find a proper reference:
Let $h = \frac{1}{K} \sum_{k=1}^{K}{\hat{y}_k}$
be the mean predicted outcome from the $K$ bootstr | Bias and variance estimation with boostrap
I think the formulas for Bootstrap are as follows, although I can't seem to find a proper reference:
Let $h = \frac{1}{K} \sum_{k=1}^{K}{\hat{y}_k}$
be the mean predicted outcome from the $K$ bootstrap resamples.
Then:
Bias is $y - h$
Variance is $\frac{1}{K-1}\sum_{k=1}^{K... | Bias and variance estimation with boostrap
I think the formulas for Bootstrap are as follows, although I can't seem to find a proper reference:
Let $h = \frac{1}{K} \sum_{k=1}^{K}{\hat{y}_k}$
be the mean predicted outcome from the $K$ bootstr |
50,623 | How to report Kruskal-Wallis test? | As @Germaniawerks remarked above, if you only have two groups (managers vs juniors) you should use ranksum (aka Mann-Whitney-Wilcoxon) test and there is no need for Kruskal-Wallis. If you have more than two groups, then Kruskal-Wallis will tell you if they are significantly different, but if you want to know which pair... | How to report Kruskal-Wallis test? | As @Germaniawerks remarked above, if you only have two groups (managers vs juniors) you should use ranksum (aka Mann-Whitney-Wilcoxon) test and there is no need for Kruskal-Wallis. If you have more th | How to report Kruskal-Wallis test?
As @Germaniawerks remarked above, if you only have two groups (managers vs juniors) you should use ranksum (aka Mann-Whitney-Wilcoxon) test and there is no need for Kruskal-Wallis. If you have more than two groups, then Kruskal-Wallis will tell you if they are significantly different,... | How to report Kruskal-Wallis test?
As @Germaniawerks remarked above, if you only have two groups (managers vs juniors) you should use ranksum (aka Mann-Whitney-Wilcoxon) test and there is no need for Kruskal-Wallis. If you have more th |
50,624 | How to report Kruskal-Wallis test? | Following above very useful comments I like to add that the median should be reported instead of the mean.
The statement "Managers are more likely to arrive late than juniors (H=14.338, p<.01)" is incomplete. The only thing it says is that there is a difference between the groups. It does not specify where the differe... | How to report Kruskal-Wallis test? | Following above very useful comments I like to add that the median should be reported instead of the mean.
The statement "Managers are more likely to arrive late than juniors (H=14.338, p<.01)" is in | How to report Kruskal-Wallis test?
Following above very useful comments I like to add that the median should be reported instead of the mean.
The statement "Managers are more likely to arrive late than juniors (H=14.338, p<.01)" is incomplete. The only thing it says is that there is a difference between the groups. It... | How to report Kruskal-Wallis test?
Following above very useful comments I like to add that the median should be reported instead of the mean.
The statement "Managers are more likely to arrive late than juniors (H=14.338, p<.01)" is in |
50,625 | Maximizing returns - A Bayesian approach | As a general principle, risk-responsiveness is incorporated into an economic analysis through the shape of the utility function. In your analysis you have a utility function $U$ that operates on the total return $\boldsymbol{a} \cdot \boldsymbol{r}$ to produce the utility of that return. Since your utility function i... | Maximizing returns - A Bayesian approach | As a general principle, risk-responsiveness is incorporated into an economic analysis through the shape of the utility function. In your analysis you have a utility function $U$ that operates on the | Maximizing returns - A Bayesian approach
As a general principle, risk-responsiveness is incorporated into an economic analysis through the shape of the utility function. In your analysis you have a utility function $U$ that operates on the total return $\boldsymbol{a} \cdot \boldsymbol{r}$ to produce the utility of th... | Maximizing returns - A Bayesian approach
As a general principle, risk-responsiveness is incorporated into an economic analysis through the shape of the utility function. In your analysis you have a utility function $U$ that operates on the |
50,626 | Image classification using histogram | In the paper you sited the histogram considered the three color values of each pixel. They used HSV instead of RGB. Here's a package which can do that transform, but probably get this working with RGB first.
Each bin of the histogram considered each possible combination of colors, not just a single one. Think of it as... | Image classification using histogram | In the paper you sited the histogram considered the three color values of each pixel. They used HSV instead of RGB. Here's a package which can do that transform, but probably get this working with RGB | Image classification using histogram
In the paper you sited the histogram considered the three color values of each pixel. They used HSV instead of RGB. Here's a package which can do that transform, but probably get this working with RGB first.
Each bin of the histogram considered each possible combination of colors, ... | Image classification using histogram
In the paper you sited the histogram considered the three color values of each pixel. They used HSV instead of RGB. Here's a package which can do that transform, but probably get this working with RGB |
50,627 | Prediction interval for number of biased coin tosses to get 2 consecutive heads | This answer implements an approach along the lines of whuber's comment, where $p$ is estimated naively from the 10 tosses made at the start, and then 'plugged-in' to $P(F_{HH}|p)$ to get the prediction interval.
The approach does not explicitly account for uncertainty in $p$, which leads to poor performance in some ca... | Prediction interval for number of biased coin tosses to get 2 consecutive heads | This answer implements an approach along the lines of whuber's comment, where $p$ is estimated naively from the 10 tosses made at the start, and then 'plugged-in' to $P(F_{HH}|p)$ to get the predictio | Prediction interval for number of biased coin tosses to get 2 consecutive heads
This answer implements an approach along the lines of whuber's comment, where $p$ is estimated naively from the 10 tosses made at the start, and then 'plugged-in' to $P(F_{HH}|p)$ to get the prediction interval.
The approach does not expli... | Prediction interval for number of biased coin tosses to get 2 consecutive heads
This answer implements an approach along the lines of whuber's comment, where $p$ is estimated naively from the 10 tosses made at the start, and then 'plugged-in' to $P(F_{HH}|p)$ to get the predictio |
50,628 | Oversampling correction for multinomial logistic regression | Off the cuff, I presume one could proceed as in logistic regression: a generalisation to $K>2$ categories and base category $K$ would be to set the $i$-th correction term to be
$$\log \frac{(r_i p_K)}{(r_K p_i)}$$
corresponding to the $i$ vs $K$ contrast. For $K=2$, $p_1$ is as before and $p_K = p_2 = 1-p_1$, so it ... | Oversampling correction for multinomial logistic regression | Off the cuff, I presume one could proceed as in logistic regression: a generalisation to $K>2$ categories and base category $K$ would be to set the $i$-th correction term to be
$$\log \frac{(r_i p_K) | Oversampling correction for multinomial logistic regression
Off the cuff, I presume one could proceed as in logistic regression: a generalisation to $K>2$ categories and base category $K$ would be to set the $i$-th correction term to be
$$\log \frac{(r_i p_K)}{(r_K p_i)}$$
corresponding to the $i$ vs $K$ contrast. F... | Oversampling correction for multinomial logistic regression
Off the cuff, I presume one could proceed as in logistic regression: a generalisation to $K>2$ categories and base category $K$ would be to set the $i$-th correction term to be
$$\log \frac{(r_i p_K) |
50,629 | What to do with my data? | If I may make some suggestions:
I would not implement a confidence interval because most students don't really understand what it is anyway. An inter-quartile range would be more appropriate instead
Most(?) professors cook their grades to have a normal distribution, so the presence of the normal distribution should no... | What to do with my data? | If I may make some suggestions:
I would not implement a confidence interval because most students don't really understand what it is anyway. An inter-quartile range would be more appropriate instead
| What to do with my data?
If I may make some suggestions:
I would not implement a confidence interval because most students don't really understand what it is anyway. An inter-quartile range would be more appropriate instead
Most(?) professors cook their grades to have a normal distribution, so the presence of the norm... | What to do with my data?
If I may make some suggestions:
I would not implement a confidence interval because most students don't really understand what it is anyway. An inter-quartile range would be more appropriate instead
|
50,630 | What to do with my data? | I've seen students' scores of different kinds. The distribution often exhibits one or more tresholds reflecting what they might want or have to achieve. And even with untresheld and continuous scores the distributions are not normal but rather skewed towards higher scores.
You should test the normality assumption.
As ... | What to do with my data? | I've seen students' scores of different kinds. The distribution often exhibits one or more tresholds reflecting what they might want or have to achieve. And even with untresheld and continuous scores | What to do with my data?
I've seen students' scores of different kinds. The distribution often exhibits one or more tresholds reflecting what they might want or have to achieve. And even with untresheld and continuous scores the distributions are not normal but rather skewed towards higher scores.
You should test the n... | What to do with my data?
I've seen students' scores of different kinds. The distribution often exhibits one or more tresholds reflecting what they might want or have to achieve. And even with untresheld and continuous scores |
50,631 | Is Bayesian structural equation modelling better than maximum likelihood with smaller sample sizes? | This question is very broad. It first of all really depends on the model you want to test, in which a higher complexity would decrease the validity of an ML-SEM model (but probably also of a BSEM model). I would say, as a starter, try both and experience/see which difference you get. To give you a gross insight in the ... | Is Bayesian structural equation modelling better than maximum likelihood with smaller sample sizes? | This question is very broad. It first of all really depends on the model you want to test, in which a higher complexity would decrease the validity of an ML-SEM model (but probably also of a BSEM mode | Is Bayesian structural equation modelling better than maximum likelihood with smaller sample sizes?
This question is very broad. It first of all really depends on the model you want to test, in which a higher complexity would decrease the validity of an ML-SEM model (but probably also of a BSEM model). I would say, as ... | Is Bayesian structural equation modelling better than maximum likelihood with smaller sample sizes?
This question is very broad. It first of all really depends on the model you want to test, in which a higher complexity would decrease the validity of an ML-SEM model (but probably also of a BSEM mode |
50,632 | What would be a parametric model with properties similar to the Theil-Sen estimator? | I beleive, the S estimator[1] (and it's algorithm, FastS[2]) is the closest parametric equivalent to the Theil-Sen estimator.
This is because the S estimator explicitly adds a parametric assumption on the distribution of the residuals (through the tuning constant $c$) to get better efficiency at uncontaminated samples.... | What would be a parametric model with properties similar to the Theil-Sen estimator? | I beleive, the S estimator[1] (and it's algorithm, FastS[2]) is the closest parametric equivalent to the Theil-Sen estimator.
This is because the S estimator explicitly adds a parametric assumption on | What would be a parametric model with properties similar to the Theil-Sen estimator?
I beleive, the S estimator[1] (and it's algorithm, FastS[2]) is the closest parametric equivalent to the Theil-Sen estimator.
This is because the S estimator explicitly adds a parametric assumption on the distribution of the residuals ... | What would be a parametric model with properties similar to the Theil-Sen estimator?
I beleive, the S estimator[1] (and it's algorithm, FastS[2]) is the closest parametric equivalent to the Theil-Sen estimator.
This is because the S estimator explicitly adds a parametric assumption on |
50,633 | What would be a parametric model with properties similar to the Theil-Sen estimator? | One possibility consists of using flexible error distributions. This is, you have a model
$$y_j = x_j^{\top}\beta + \epsilon_j,$$
where $\epsilon_j\sim F$, and $F$ is a flexible distribution. So, for instance, in order to produce a model that is (relatively) robust to the presence of outliers and skewness, a possible c... | What would be a parametric model with properties similar to the Theil-Sen estimator? | One possibility consists of using flexible error distributions. This is, you have a model
$$y_j = x_j^{\top}\beta + \epsilon_j,$$
where $\epsilon_j\sim F$, and $F$ is a flexible distribution. So, for | What would be a parametric model with properties similar to the Theil-Sen estimator?
One possibility consists of using flexible error distributions. This is, you have a model
$$y_j = x_j^{\top}\beta + \epsilon_j,$$
where $\epsilon_j\sim F$, and $F$ is a flexible distribution. So, for instance, in order to produce a mod... | What would be a parametric model with properties similar to the Theil-Sen estimator?
One possibility consists of using flexible error distributions. This is, you have a model
$$y_j = x_j^{\top}\beta + \epsilon_j,$$
where $\epsilon_j\sim F$, and $F$ is a flexible distribution. So, for |
50,634 | Conditional expectations by conditioning on functions of random variables | Injectivity is sufficient (so long as your function is measurable)
Let's assume that $f$ is a measureable function, so that all relevant random variables and events are well-defined. Now, to give this some more structure, suppose we are working in a probability space $(\Omega, \mathscr{S}, \mathbb{P})$ so that $X: \Om... | Conditional expectations by conditioning on functions of random variables | Injectivity is sufficient (so long as your function is measurable)
Let's assume that $f$ is a measureable function, so that all relevant random variables and events are well-defined. Now, to give thi | Conditional expectations by conditioning on functions of random variables
Injectivity is sufficient (so long as your function is measurable)
Let's assume that $f$ is a measureable function, so that all relevant random variables and events are well-defined. Now, to give this some more structure, suppose we are working ... | Conditional expectations by conditioning on functions of random variables
Injectivity is sufficient (so long as your function is measurable)
Let's assume that $f$ is a measureable function, so that all relevant random variables and events are well-defined. Now, to give thi |
50,635 | Conditional expectations by conditioning on functions of random variables | if $Z=f(X)$ and $f$ is injective function so $$\sigma(Z)=\sigma(X)$$
since $Z=f(X)$ so $$\sigma(Z) \subset \sigma(X)$$
and because $f$ is injective so $X=f^{-1}(Z)=g(Z)$ so $$\sigma(X) \subset \sigma(Z)$$
so $\sigma(Z)=\sigma(X)$ or $\sigma(f(X))=\sigma(X)$
now $$E(Y|X)=E(Y|\sigma(X))=E(Y|\sigma(f(X)))=E(Y|f(X))$$ | Conditional expectations by conditioning on functions of random variables | if $Z=f(X)$ and $f$ is injective function so $$\sigma(Z)=\sigma(X)$$
since $Z=f(X)$ so $$\sigma(Z) \subset \sigma(X)$$
and because $f$ is injective so $X=f^{-1}(Z)=g(Z)$ so $$\sigma(X) \subset \sigma | Conditional expectations by conditioning on functions of random variables
if $Z=f(X)$ and $f$ is injective function so $$\sigma(Z)=\sigma(X)$$
since $Z=f(X)$ so $$\sigma(Z) \subset \sigma(X)$$
and because $f$ is injective so $X=f^{-1}(Z)=g(Z)$ so $$\sigma(X) \subset \sigma(Z)$$
so $\sigma(Z)=\sigma(X)$ or $\sigma(f(X... | Conditional expectations by conditioning on functions of random variables
if $Z=f(X)$ and $f$ is injective function so $$\sigma(Z)=\sigma(X)$$
since $Z=f(X)$ so $$\sigma(Z) \subset \sigma(X)$$
and because $f$ is injective so $X=f^{-1}(Z)=g(Z)$ so $$\sigma(X) \subset \sigma |
50,636 | Conditional expectations by conditioning on functions of random variables | Assuming $f$ is measurable I think the weakest condition is:
Whenever $\mathbb{E}(Y|X=x_1) \ne \mathbb{E}(Y|X=x_2), f(x_1) \ne f(x_2)$. | Conditional expectations by conditioning on functions of random variables | Assuming $f$ is measurable I think the weakest condition is:
Whenever $\mathbb{E}(Y|X=x_1) \ne \mathbb{E}(Y|X=x_2), f(x_1) \ne f(x_2)$. | Conditional expectations by conditioning on functions of random variables
Assuming $f$ is measurable I think the weakest condition is:
Whenever $\mathbb{E}(Y|X=x_1) \ne \mathbb{E}(Y|X=x_2), f(x_1) \ne f(x_2)$. | Conditional expectations by conditioning on functions of random variables
Assuming $f$ is measurable I think the weakest condition is:
Whenever $\mathbb{E}(Y|X=x_1) \ne \mathbb{E}(Y|X=x_2), f(x_1) \ne f(x_2)$. |
50,637 | Comparing multiclass classification algorithms for a particular application | I am simply copy-pasting the answers I got from Alexandre Passos on Metaoptimize. It would really help if someone here can add more to it.
Any binary classifier can be used for multiclass with the 1-vs-all reduction, or the all-vs-all reduction. This list seems to cover most
of the common multiclass algorithms.
Log... | Comparing multiclass classification algorithms for a particular application | I am simply copy-pasting the answers I got from Alexandre Passos on Metaoptimize. It would really help if someone here can add more to it.
Any binary classifier can be used for multiclass with the 1 | Comparing multiclass classification algorithms for a particular application
I am simply copy-pasting the answers I got from Alexandre Passos on Metaoptimize. It would really help if someone here can add more to it.
Any binary classifier can be used for multiclass with the 1-vs-all reduction, or the all-vs-all reducti... | Comparing multiclass classification algorithms for a particular application
I am simply copy-pasting the answers I got from Alexandre Passos on Metaoptimize. It would really help if someone here can add more to it.
Any binary classifier can be used for multiclass with the 1 |
50,638 | Taking the log of variables | You might find this display interesting:
These are residuals from a linear regression with ten x-variables (IVs), a skewed error distribution (but one with all moments finite, to which the CLT definitely applies!), and 1000 observations (i.e. the data was simulated).
It's a normal qqplot, which if the residuals are clo... | Taking the log of variables | You might find this display interesting:
These are residuals from a linear regression with ten x-variables (IVs), a skewed error distribution (but one with all moments finite, to which the CLT definit | Taking the log of variables
You might find this display interesting:
These are residuals from a linear regression with ten x-variables (IVs), a skewed error distribution (but one with all moments finite, to which the CLT definitely applies!), and 1000 observations (i.e. the data was simulated).
It's a normal qqplot, wh... | Taking the log of variables
You might find this display interesting:
These are residuals from a linear regression with ten x-variables (IVs), a skewed error distribution (but one with all moments finite, to which the CLT definit |
50,639 | Taking the log of variables | Per your comment, Lindberg-Feller CLT requires independence (but not identically distributed), along with a finite means and variance. Are you sure that the "Y can't be [independent] by definition but this is the case for all regressions" part doesn't kill your argument? Just because it's true by definition doesn't mea... | Taking the log of variables | Per your comment, Lindberg-Feller CLT requires independence (but not identically distributed), along with a finite means and variance. Are you sure that the "Y can't be [independent] by definition but | Taking the log of variables
Per your comment, Lindberg-Feller CLT requires independence (but not identically distributed), along with a finite means and variance. Are you sure that the "Y can't be [independent] by definition but this is the case for all regressions" part doesn't kill your argument? Just because it's tr... | Taking the log of variables
Per your comment, Lindberg-Feller CLT requires independence (but not identically distributed), along with a finite means and variance. Are you sure that the "Y can't be [independent] by definition but |
50,640 | How to analyze this incomplete block design in R? | I think you're exactly right.
Set up data like your example:
d <- expand.grid(Site=factor(1:10),rep=1:5)
d <- transform(d,Clone=factor(LETTERS[(as.numeric(Site)+1) %/% 2]))
library(lme4)
## could use development version of lme4 to simulate, but will do
## it by hand
beta <- c(2,1,3,-2,2) ## clone effects (intercept + ... | How to analyze this incomplete block design in R? | I think you're exactly right.
Set up data like your example:
d <- expand.grid(Site=factor(1:10),rep=1:5)
d <- transform(d,Clone=factor(LETTERS[(as.numeric(Site)+1) %/% 2]))
library(lme4)
## could use | How to analyze this incomplete block design in R?
I think you're exactly right.
Set up data like your example:
d <- expand.grid(Site=factor(1:10),rep=1:5)
d <- transform(d,Clone=factor(LETTERS[(as.numeric(Site)+1) %/% 2]))
library(lme4)
## could use development version of lme4 to simulate, but will do
## it by hand
bet... | How to analyze this incomplete block design in R?
I think you're exactly right.
Set up data like your example:
d <- expand.grid(Site=factor(1:10),rep=1:5)
d <- transform(d,Clone=factor(LETTERS[(as.numeric(Site)+1) %/% 2]))
library(lme4)
## could use |
50,641 | Sampling technique to estimate how many toxic waste sites are in a country? | Your approach seems reasonable, especially your choice to stratify your sampling. This will make it more efficient provided you can easily delineate the different industrial zones.
I don't have a book to recommend you, but you could model your uncertainty using the Poisson distribution, with the $\lambda =$ No. of Tox... | Sampling technique to estimate how many toxic waste sites are in a country? | Your approach seems reasonable, especially your choice to stratify your sampling. This will make it more efficient provided you can easily delineate the different industrial zones.
I don't have a boo | Sampling technique to estimate how many toxic waste sites are in a country?
Your approach seems reasonable, especially your choice to stratify your sampling. This will make it more efficient provided you can easily delineate the different industrial zones.
I don't have a book to recommend you, but you could model your... | Sampling technique to estimate how many toxic waste sites are in a country?
Your approach seems reasonable, especially your choice to stratify your sampling. This will make it more efficient provided you can easily delineate the different industrial zones.
I don't have a boo |
50,642 | If $X_{n+1}$ is a martingale subject to $Y_0,\ldots,Y_n$, then is it a martingale with respect to $Y_0^2,\ldots,Y_n^2$? | Prove or disprove
$
E \left( X_{n+1} | Y_0^2,\ldots,Y_n^2 \right) = X_n
$
I am thinking that if $F=\sigma \left(Y_0,\ldots,Y_n \right)$ and
$G=\sigma \left(Y_0^2,\ldots,Y_n^2 \right)$, I need to prove that F = G? Is this correct?
Actually, $G\subseteq F$ and the inclusion can be strict.
Then I can do something li... | If $X_{n+1}$ is a martingale subject to $Y_0,\ldots,Y_n$, then is it a martingale with respect to $Y | Prove or disprove
$
E \left( X_{n+1} | Y_0^2,\ldots,Y_n^2 \right) = X_n
$
I am thinking that if $F=\sigma \left(Y_0,\ldots,Y_n \right)$ and
$G=\sigma \left(Y_0^2,\ldots,Y_n^2 \right)$, I need to p | If $X_{n+1}$ is a martingale subject to $Y_0,\ldots,Y_n$, then is it a martingale with respect to $Y_0^2,\ldots,Y_n^2$?
Prove or disprove
$
E \left( X_{n+1} | Y_0^2,\ldots,Y_n^2 \right) = X_n
$
I am thinking that if $F=\sigma \left(Y_0,\ldots,Y_n \right)$ and
$G=\sigma \left(Y_0^2,\ldots,Y_n^2 \right)$, I need to p... | If $X_{n+1}$ is a martingale subject to $Y_0,\ldots,Y_n$, then is it a martingale with respect to $Y
Prove or disprove
$
E \left( X_{n+1} | Y_0^2,\ldots,Y_n^2 \right) = X_n
$
I am thinking that if $F=\sigma \left(Y_0,\ldots,Y_n \right)$ and
$G=\sigma \left(Y_0^2,\ldots,Y_n^2 \right)$, I need to p |
50,643 | If $X_{n+1}$ is a martingale subject to $Y_0,\ldots,Y_n$, then is it a martingale with respect to $Y_0^2,\ldots,Y_n^2$? | $G \subseteq F$
If we know $Y_0, Y_1, ...$, then we know $Y_0^2, Y_1^2, ...$ The converse is not true.
$E[X|G] = E[E[X|F]|G]$, but $E[X|F] \ne E[E[X|G]|F]$.
$E[X|G] = E[E[X|F]|G]$ allows you to prove the converse of your conjecture is true. | If $X_{n+1}$ is a martingale subject to $Y_0,\ldots,Y_n$, then is it a martingale with respect to $Y | $G \subseteq F$
If we know $Y_0, Y_1, ...$, then we know $Y_0^2, Y_1^2, ...$ The converse is not true.
$E[X|G] = E[E[X|F]|G]$, but $E[X|F] \ne E[E[X|G]|F]$.
$E[X|G] = E[E[X|F]|G]$ allows you to prove | If $X_{n+1}$ is a martingale subject to $Y_0,\ldots,Y_n$, then is it a martingale with respect to $Y_0^2,\ldots,Y_n^2$?
$G \subseteq F$
If we know $Y_0, Y_1, ...$, then we know $Y_0^2, Y_1^2, ...$ The converse is not true.
$E[X|G] = E[E[X|F]|G]$, but $E[X|F] \ne E[E[X|G]|F]$.
$E[X|G] = E[E[X|F]|G]$ allows you to prove ... | If $X_{n+1}$ is a martingale subject to $Y_0,\ldots,Y_n$, then is it a martingale with respect to $Y
$G \subseteq F$
If we know $Y_0, Y_1, ...$, then we know $Y_0^2, Y_1^2, ...$ The converse is not true.
$E[X|G] = E[E[X|F]|G]$, but $E[X|F] \ne E[E[X|G]|F]$.
$E[X|G] = E[E[X|F]|G]$ allows you to prove |
50,644 | Quantitative results of cluster analysis | How are the data sets related? IF both data sets are drawn from the same distribution (they describe the same problem) than you can use the labeled set as a "test set" for the clustering. Basically you treat the clustering algorithm as a classifier. The only problem is that you must find a match between the output of t... | Quantitative results of cluster analysis | How are the data sets related? IF both data sets are drawn from the same distribution (they describe the same problem) than you can use the labeled set as a "test set" for the clustering. Basically yo | Quantitative results of cluster analysis
How are the data sets related? IF both data sets are drawn from the same distribution (they describe the same problem) than you can use the labeled set as a "test set" for the clustering. Basically you treat the clustering algorithm as a classifier. The only problem is that you ... | Quantitative results of cluster analysis
How are the data sets related? IF both data sets are drawn from the same distribution (they describe the same problem) than you can use the labeled set as a "test set" for the clustering. Basically yo |
50,645 | Is adjusted R-squared appropriate to compare models with different response variables? | I believe using R2 or adjusted R2 is okay in your case. Fact is that we should not use RSE(Residue Standard Error) when the scale is different. This is because both R2 and adjusted R2 are normalized quantity having maximum value of 1, but the RSE is not normalized.
Ref : https://datastoriesweb.wordpress.com/2017/01/15/... | Is adjusted R-squared appropriate to compare models with different response variables? | I believe using R2 or adjusted R2 is okay in your case. Fact is that we should not use RSE(Residue Standard Error) when the scale is different. This is because both R2 and adjusted R2 are normalized q | Is adjusted R-squared appropriate to compare models with different response variables?
I believe using R2 or adjusted R2 is okay in your case. Fact is that we should not use RSE(Residue Standard Error) when the scale is different. This is because both R2 and adjusted R2 are normalized quantity having maximum value of 1... | Is adjusted R-squared appropriate to compare models with different response variables?
I believe using R2 or adjusted R2 is okay in your case. Fact is that we should not use RSE(Residue Standard Error) when the scale is different. This is because both R2 and adjusted R2 are normalized q |
50,646 | Probability of pairwise difference of samples from distribution with finite support | OP wrote:
Let $X_1, X_2, \dots, X_N$ be i.i.d. continuous random variables with support $[0, 1]$.
What is a reasonable bound on the probability that some pair of random variables is less than ϵ apart? ... interested in the cases of a uniform distribution ... I only need a reasonable probability bound, not the exact... | Probability of pairwise difference of samples from distribution with finite support | OP wrote:
Let $X_1, X_2, \dots, X_N$ be i.i.d. continuous random variables with support $[0, 1]$.
What is a reasonable bound on the probability that some pair of random variables is less than ϵ ap | Probability of pairwise difference of samples from distribution with finite support
OP wrote:
Let $X_1, X_2, \dots, X_N$ be i.i.d. continuous random variables with support $[0, 1]$.
What is a reasonable bound on the probability that some pair of random variables is less than ϵ apart? ... interested in the cases of ... | Probability of pairwise difference of samples from distribution with finite support
OP wrote:
Let $X_1, X_2, \dots, X_N$ be i.i.d. continuous random variables with support $[0, 1]$.
What is a reasonable bound on the probability that some pair of random variables is less than ϵ ap |
50,647 | How does this remove autocorrelation? | As @Analyst pointed out the inclusion of lagged dependent variables excludes one source of regression error autocorrelation. The autocorrelation can still be present if the lags of dependent variables are included. Here is the mathematical illustration. Suppose the true model is the following
$$Y_t=\alpha+\beta_0X_t+\b... | How does this remove autocorrelation? | As @Analyst pointed out the inclusion of lagged dependent variables excludes one source of regression error autocorrelation. The autocorrelation can still be present if the lags of dependent variables | How does this remove autocorrelation?
As @Analyst pointed out the inclusion of lagged dependent variables excludes one source of regression error autocorrelation. The autocorrelation can still be present if the lags of dependent variables are included. Here is the mathematical illustration. Suppose the true model is th... | How does this remove autocorrelation?
As @Analyst pointed out the inclusion of lagged dependent variables excludes one source of regression error autocorrelation. The autocorrelation can still be present if the lags of dependent variables |
50,648 | How does this remove autocorrelation? | Often not including lagged values of dependent variable or independent variables will induce autocorrelation structure in residuals when these values should have been included. | How does this remove autocorrelation? | Often not including lagged values of dependent variable or independent variables will induce autocorrelation structure in residuals when these values should have been included. | How does this remove autocorrelation?
Often not including lagged values of dependent variable or independent variables will induce autocorrelation structure in residuals when these values should have been included. | How does this remove autocorrelation?
Often not including lagged values of dependent variable or independent variables will induce autocorrelation structure in residuals when these values should have been included. |
50,649 | Can I perform Cox regression on left truncated records? | First a disclaimer: I've never had to use the time start/end variable in this way and although I'm familiar with mixed effects models I have never really had to use them IRL. Feel free to correct me if I've made a mistake
The problem consists out of two things as I see it:
One person can occur multiple times. This put... | Can I perform Cox regression on left truncated records? | First a disclaimer: I've never had to use the time start/end variable in this way and although I'm familiar with mixed effects models I have never really had to use them IRL. Feel free to correct me i | Can I perform Cox regression on left truncated records?
First a disclaimer: I've never had to use the time start/end variable in this way and although I'm familiar with mixed effects models I have never really had to use them IRL. Feel free to correct me if I've made a mistake
The problem consists out of two things as ... | Can I perform Cox regression on left truncated records?
First a disclaimer: I've never had to use the time start/end variable in this way and although I'm familiar with mixed effects models I have never really had to use them IRL. Feel free to correct me i |
50,650 | Computing the steady state probability vector of a random walk on $\{0, 1, \dots, n\}$ | You can solve your second equation to give $\pi(1)=\frac{a}{1-p} \pi(0)$, then your third to give $\pi(2)=\frac{ap}{(1-p)^2} \pi(0)$, then your first to give $\pi(x)=\frac{a}{p}\left(\frac{p}{1-p}\right)^x \pi(0)$ for $0 \lt x \lt n$, and finally your fourth to give $\pi(n)=\frac{p}{b} \pi(n-1) = \frac{a}{b}\left(\frac... | Computing the steady state probability vector of a random walk on $\{0, 1, \dots, n\}$ | You can solve your second equation to give $\pi(1)=\frac{a}{1-p} \pi(0)$, then your third to give $\pi(2)=\frac{ap}{(1-p)^2} \pi(0)$, then your first to give $\pi(x)=\frac{a}{p}\left(\frac{p}{1-p}\rig | Computing the steady state probability vector of a random walk on $\{0, 1, \dots, n\}$
You can solve your second equation to give $\pi(1)=\frac{a}{1-p} \pi(0)$, then your third to give $\pi(2)=\frac{ap}{(1-p)^2} \pi(0)$, then your first to give $\pi(x)=\frac{a}{p}\left(\frac{p}{1-p}\right)^x \pi(0)$ for $0 \lt x \lt n$... | Computing the steady state probability vector of a random walk on $\{0, 1, \dots, n\}$
You can solve your second equation to give $\pi(1)=\frac{a}{1-p} \pi(0)$, then your third to give $\pi(2)=\frac{ap}{(1-p)^2} \pi(0)$, then your first to give $\pi(x)=\frac{a}{p}\left(\frac{p}{1-p}\rig |
50,651 | Computing the steady state probability vector of a random walk on $\{0, 1, \dots, n\}$ | Here are the generic steps to solve these problems. The first computations may be difficult or time consuming, so do not hesitate to use online symbolic equation solvers like wolframalpha to gain time.
Start by solving the case $n=3$ to compute explicitly $\pi_3$ and try to guess what will be the general form of $\pi_... | Computing the steady state probability vector of a random walk on $\{0, 1, \dots, n\}$ | Here are the generic steps to solve these problems. The first computations may be difficult or time consuming, so do not hesitate to use online symbolic equation solvers like wolframalpha to gain time | Computing the steady state probability vector of a random walk on $\{0, 1, \dots, n\}$
Here are the generic steps to solve these problems. The first computations may be difficult or time consuming, so do not hesitate to use online symbolic equation solvers like wolframalpha to gain time.
Start by solving the case $n=3... | Computing the steady state probability vector of a random walk on $\{0, 1, \dots, n\}$
Here are the generic steps to solve these problems. The first computations may be difficult or time consuming, so do not hesitate to use online symbolic equation solvers like wolframalpha to gain time |
50,652 | How to compare rates of occurence in consecutive time series count data? | To keep things really simple, you could consider using a simple mean/standard deviation inspired ratio, a bit like a z-score?
If you assume that the counts for two days, $X_1$ and $X_2$ are Poisson random samples with $\lambda_1$ and $\lambda_2$ respectively, then the change in word count follows a Skellam distribution... | How to compare rates of occurence in consecutive time series count data? | To keep things really simple, you could consider using a simple mean/standard deviation inspired ratio, a bit like a z-score?
If you assume that the counts for two days, $X_1$ and $X_2$ are Poisson ra | How to compare rates of occurence in consecutive time series count data?
To keep things really simple, you could consider using a simple mean/standard deviation inspired ratio, a bit like a z-score?
If you assume that the counts for two days, $X_1$ and $X_2$ are Poisson random samples with $\lambda_1$ and $\lambda_2$ r... | How to compare rates of occurence in consecutive time series count data?
To keep things really simple, you could consider using a simple mean/standard deviation inspired ratio, a bit like a z-score?
If you assume that the counts for two days, $X_1$ and $X_2$ are Poisson ra |
50,653 | Laplace distribution and, generally, interpreting an undefined moment | I was incorrectly using the moment generating function which led to my misunderstanding of the Laplace distribution.
The moment generating function is $M_X(\theta) = \text{E}(e^{\theta X})$.
When you use that to find the $n^{\text{th}}$ moment, you take the $n^{\text{th}}$ derivative at $\theta=0$:
$$\frac{d^{n}(M_... | Laplace distribution and, generally, interpreting an undefined moment | I was incorrectly using the moment generating function which led to my misunderstanding of the Laplace distribution.
The moment generating function is $M_X(\theta) = \text{E}(e^{\theta X})$.
When y | Laplace distribution and, generally, interpreting an undefined moment
I was incorrectly using the moment generating function which led to my misunderstanding of the Laplace distribution.
The moment generating function is $M_X(\theta) = \text{E}(e^{\theta X})$.
When you use that to find the $n^{\text{th}}$ moment, yo... | Laplace distribution and, generally, interpreting an undefined moment
I was incorrectly using the moment generating function which led to my misunderstanding of the Laplace distribution.
The moment generating function is $M_X(\theta) = \text{E}(e^{\theta X})$.
When y |
50,654 | How to interpret variation explained by principal coordinates? | In preparing a workshop on ordination techniques, I realized I was having the same difficulty in interpreting the eigenvalues of principal coordinate analysis for the same reasons that have puzzled you (@Paul Igor Costea), so I started digging around for some answers.
I have a few books on multivariate statistics that... | How to interpret variation explained by principal coordinates? | In preparing a workshop on ordination techniques, I realized I was having the same difficulty in interpreting the eigenvalues of principal coordinate analysis for the same reasons that have puzzled yo | How to interpret variation explained by principal coordinates?
In preparing a workshop on ordination techniques, I realized I was having the same difficulty in interpreting the eigenvalues of principal coordinate analysis for the same reasons that have puzzled you (@Paul Igor Costea), so I started digging around for so... | How to interpret variation explained by principal coordinates?
In preparing a workshop on ordination techniques, I realized I was having the same difficulty in interpreting the eigenvalues of principal coordinate analysis for the same reasons that have puzzled yo |
50,655 | The gamma distribution and Poisson processes | Interval between events in a Poisson process is exponentially distributed. If you skip intervals, and only count k-th event, then you get convolution of exponential r.v.'s which gives you Erlang distribution (which is a special case of Gamma distribution). Is this what you meant in the first paragraph?
If your growth p... | The gamma distribution and Poisson processes | Interval between events in a Poisson process is exponentially distributed. If you skip intervals, and only count k-th event, then you get convolution of exponential r.v.'s which gives you Erlang distr | The gamma distribution and Poisson processes
Interval between events in a Poisson process is exponentially distributed. If you skip intervals, and only count k-th event, then you get convolution of exponential r.v.'s which gives you Erlang distribution (which is a special case of Gamma distribution). Is this what you m... | The gamma distribution and Poisson processes
Interval between events in a Poisson process is exponentially distributed. If you skip intervals, and only count k-th event, then you get convolution of exponential r.v.'s which gives you Erlang distr |
50,656 | Bootstrap residuals: Wild vs Block Bootstrap? | Have you read this paper:
Cameron, A. C., Gelbach, J. B., & Miller, D. L. (2008). Bootstrap-Based Improvements for Inference with Clustered Errors. Review of Economics and Statistics, 90(3), 414–427. https://doi.org/10.1162/rest.90.3.414
For wild bootstrapping when you suspect clustering, this is probably the most comp... | Bootstrap residuals: Wild vs Block Bootstrap? | Have you read this paper:
Cameron, A. C., Gelbach, J. B., & Miller, D. L. (2008). Bootstrap-Based Improvements for Inference with Clustered Errors. Review of Economics and Statistics, 90(3), 414–427. | Bootstrap residuals: Wild vs Block Bootstrap?
Have you read this paper:
Cameron, A. C., Gelbach, J. B., & Miller, D. L. (2008). Bootstrap-Based Improvements for Inference with Clustered Errors. Review of Economics and Statistics, 90(3), 414–427. https://doi.org/10.1162/rest.90.3.414
For wild bootstrapping when you susp... | Bootstrap residuals: Wild vs Block Bootstrap?
Have you read this paper:
Cameron, A. C., Gelbach, J. B., & Miller, D. L. (2008). Bootstrap-Based Improvements for Inference with Clustered Errors. Review of Economics and Statistics, 90(3), 414–427. |
50,657 | Seemingly unrelated regression and multivariate Regression | As mentioned in the Wikipedia: SUR is equivalent to the equation-by-equation OLS under the following two conditions:
i. When there are no cross-equation correlations between the error terms
ii. When each equation contains exactly the same set of regressors.
That being said, if your model and data don't satisfy abov... | Seemingly unrelated regression and multivariate Regression | As mentioned in the Wikipedia: SUR is equivalent to the equation-by-equation OLS under the following two conditions:
i. When there are no cross-equation correlations between the error terms
ii. When | Seemingly unrelated regression and multivariate Regression
As mentioned in the Wikipedia: SUR is equivalent to the equation-by-equation OLS under the following two conditions:
i. When there are no cross-equation correlations between the error terms
ii. When each equation contains exactly the same set of regressors.
... | Seemingly unrelated regression and multivariate Regression
As mentioned in the Wikipedia: SUR is equivalent to the equation-by-equation OLS under the following two conditions:
i. When there are no cross-equation correlations between the error terms
ii. When |
50,658 | What is the standard procedure for evaluating a user-based CF algorithm with a dataset offline? | Regarding "The three sets"
training set: The users in the training folds restricted to the ratings before the split date. This data is used to build the model.
validation set: The users in the validation fold restricted to the ratings after the split date. This data is used to evaluate the model.
testing set: The user... | What is the standard procedure for evaluating a user-based CF algorithm with a dataset offline? | Regarding "The three sets"
training set: The users in the training folds restricted to the ratings before the split date. This data is used to build the model.
validation set: The users in the valida | What is the standard procedure for evaluating a user-based CF algorithm with a dataset offline?
Regarding "The three sets"
training set: The users in the training folds restricted to the ratings before the split date. This data is used to build the model.
validation set: The users in the validation fold restricted to ... | What is the standard procedure for evaluating a user-based CF algorithm with a dataset offline?
Regarding "The three sets"
training set: The users in the training folds restricted to the ratings before the split date. This data is used to build the model.
validation set: The users in the valida |
50,659 | Partitioning Around Medoids (PAM) with Gower distance matrix | If the binary variable is not very useful, try putting less weight on it.
There is nothing wrong with having a domain expert manually assign weights to different attributes to help the algorithm find new information. That the binary attribute splits the data into two is a correct result, now you want to find something ... | Partitioning Around Medoids (PAM) with Gower distance matrix | If the binary variable is not very useful, try putting less weight on it.
There is nothing wrong with having a domain expert manually assign weights to different attributes to help the algorithm find | Partitioning Around Medoids (PAM) with Gower distance matrix
If the binary variable is not very useful, try putting less weight on it.
There is nothing wrong with having a domain expert manually assign weights to different attributes to help the algorithm find new information. That the binary attribute splits the data ... | Partitioning Around Medoids (PAM) with Gower distance matrix
If the binary variable is not very useful, try putting less weight on it.
There is nothing wrong with having a domain expert manually assign weights to different attributes to help the algorithm find |
50,660 | Partitioning Around Medoids (PAM) with Gower distance matrix | To better justify the chosen number of clusters (k) you can use other partition quality indices than Silhouette width. Example indices based on arbitrary dissimilarity are: Caliński & Harabasz index (chosen as the best in Milligan and Cooper study 1985 and as 4th best in Dimitriadou et al. 2002) generalized for dissimi... | Partitioning Around Medoids (PAM) with Gower distance matrix | To better justify the chosen number of clusters (k) you can use other partition quality indices than Silhouette width. Example indices based on arbitrary dissimilarity are: Caliński & Harabasz index ( | Partitioning Around Medoids (PAM) with Gower distance matrix
To better justify the chosen number of clusters (k) you can use other partition quality indices than Silhouette width. Example indices based on arbitrary dissimilarity are: Caliński & Harabasz index (chosen as the best in Milligan and Cooper study 1985 and as... | Partitioning Around Medoids (PAM) with Gower distance matrix
To better justify the chosen number of clusters (k) you can use other partition quality indices than Silhouette width. Example indices based on arbitrary dissimilarity are: Caliński & Harabasz index ( |
50,661 | How to get from $R$ to $R^2$ the hard way | You are missing square roots for $S_X$, $S_Y$ and $SS_{reg}$ should be $\sum(\hat{y}-\overline{y})^2$ but I assume that these are just typos.
Let $y=\alpha+\beta x+\varepsilon$ and let us note that
$$r=\beta\frac{S_X}{S_Y}$$
$$r^2=\beta^2\frac{\overline{x^2}-\overline{x}^2}{\overline{y^2}-\overline{y}^2}$$
On the othe... | How to get from $R$ to $R^2$ the hard way | You are missing square roots for $S_X$, $S_Y$ and $SS_{reg}$ should be $\sum(\hat{y}-\overline{y})^2$ but I assume that these are just typos.
Let $y=\alpha+\beta x+\varepsilon$ and let us note that
$ | How to get from $R$ to $R^2$ the hard way
You are missing square roots for $S_X$, $S_Y$ and $SS_{reg}$ should be $\sum(\hat{y}-\overline{y})^2$ but I assume that these are just typos.
Let $y=\alpha+\beta x+\varepsilon$ and let us note that
$$r=\beta\frac{S_X}{S_Y}$$
$$r^2=\beta^2\frac{\overline{x^2}-\overline{x}^2}{\o... | How to get from $R$ to $R^2$ the hard way
You are missing square roots for $S_X$, $S_Y$ and $SS_{reg}$ should be $\sum(\hat{y}-\overline{y})^2$ but I assume that these are just typos.
Let $y=\alpha+\beta x+\varepsilon$ and let us note that
$ |
50,662 | Normalize binary features for logistic regression | Normalization means that you put your data in a particular range, often $0$ to $1$. If you have coded your binary variable with $0$ and $1$, you already have this property and do not need to do anything. If you use a different type of coding of your binary variable, such as $\pm 1$, then the usual $\dfrac{x_i - \min(x)... | Normalize binary features for logistic regression | Normalization means that you put your data in a particular range, often $0$ to $1$. If you have coded your binary variable with $0$ and $1$, you already have this property and do not need to do anythi | Normalize binary features for logistic regression
Normalization means that you put your data in a particular range, often $0$ to $1$. If you have coded your binary variable with $0$ and $1$, you already have this property and do not need to do anything. If you use a different type of coding of your binary variable, suc... | Normalize binary features for logistic regression
Normalization means that you put your data in a particular range, often $0$ to $1$. If you have coded your binary variable with $0$ and $1$, you already have this property and do not need to do anythi |
50,663 | Within, between or overall R-square for random effects in Stata | Random effect estimator (GLS estimator) is a weighted average of between and within estimators. In Stata, the default is random effect and you need to use R-squared: overall. As specified here, R-sq: within is not correct for fixed effect and there are alternatives to correct that in Stata. For example you need to use... | Within, between or overall R-square for random effects in Stata | Random effect estimator (GLS estimator) is a weighted average of between and within estimators. In Stata, the default is random effect and you need to use R-squared: overall. As specified here, R-sq: | Within, between or overall R-square for random effects in Stata
Random effect estimator (GLS estimator) is a weighted average of between and within estimators. In Stata, the default is random effect and you need to use R-squared: overall. As specified here, R-sq: within is not correct for fixed effect and there are al... | Within, between or overall R-square for random effects in Stata
Random effect estimator (GLS estimator) is a weighted average of between and within estimators. In Stata, the default is random effect and you need to use R-squared: overall. As specified here, R-sq: |
50,664 | What is the probability regression coefficient is larger than its OLS estimate | The OLS (or any other) estimator, $\hat \beta$, is a random variable. Namely, it is a real-valued function. It takes as input the sample data and produces a real number. This real number is the sample-specific estimate. The habit of using the same symbol to denote the function and a specific value of it can become conf... | What is the probability regression coefficient is larger than its OLS estimate | The OLS (or any other) estimator, $\hat \beta$, is a random variable. Namely, it is a real-valued function. It takes as input the sample data and produces a real number. This real number is the sample | What is the probability regression coefficient is larger than its OLS estimate
The OLS (or any other) estimator, $\hat \beta$, is a random variable. Namely, it is a real-valued function. It takes as input the sample data and produces a real number. This real number is the sample-specific estimate. The habit of using th... | What is the probability regression coefficient is larger than its OLS estimate
The OLS (or any other) estimator, $\hat \beta$, is a random variable. Namely, it is a real-valued function. It takes as input the sample data and produces a real number. This real number is the sample |
50,665 | Standard error of a weighted mean when observations are not independent | You say you have known sampling error. If you have multinormal data $(X_1, \dots, X_n)=X$ with commom mean $\mu$ and known covariance matrix $\Sigma$, then you can use the result that if $A$ is a matrix of constants, and $X \sim MN(\mu 1_n, \Sigma)$ then $AX \sim MN(A\mu, A \Sigma A^T)$. Just use the result with
$A= ... | Standard error of a weighted mean when observations are not independent | You say you have known sampling error. If you have multinormal data $(X_1, \dots, X_n)=X$ with commom mean $\mu$ and known covariance matrix $\Sigma$, then you can use the result that if $A$ is a mat | Standard error of a weighted mean when observations are not independent
You say you have known sampling error. If you have multinormal data $(X_1, \dots, X_n)=X$ with commom mean $\mu$ and known covariance matrix $\Sigma$, then you can use the result that if $A$ is a matrix of constants, and $X \sim MN(\mu 1_n, \Sigma... | Standard error of a weighted mean when observations are not independent
You say you have known sampling error. If you have multinormal data $(X_1, \dots, X_n)=X$ with commom mean $\mu$ and known covariance matrix $\Sigma$, then you can use the result that if $A$ is a mat |
50,666 | What can I do if the confidence intervals of the predicted mean are small but the predicted intervals are large | Very little can be said in general terms about this sort of situation. We certainly can't conclude that this is the best model and just submit it. There may be other variables that are better than this one, or that complement it well by explaining the remaining randomness. To answer the question "what should I do he... | What can I do if the confidence intervals of the predicted mean are small but the predicted interval | Very little can be said in general terms about this sort of situation. We certainly can't conclude that this is the best model and just submit it. There may be other variables that are better than t | What can I do if the confidence intervals of the predicted mean are small but the predicted intervals are large
Very little can be said in general terms about this sort of situation. We certainly can't conclude that this is the best model and just submit it. There may be other variables that are better than this one,... | What can I do if the confidence intervals of the predicted mean are small but the predicted interval
Very little can be said in general terms about this sort of situation. We certainly can't conclude that this is the best model and just submit it. There may be other variables that are better than t |
50,667 | What can I do if the confidence intervals of the predicted mean are small but the predicted intervals are large | To add some information to my own question, I could imagine that the predictor $X$ is length, and response $Y$ is weight. In this case, length perfectly predicts the average weight. However, the random error at each length is high.
We could imagine a second predictor Country {Asia, Europe, America...} and it satisfies... | What can I do if the confidence intervals of the predicted mean are small but the predicted interval | To add some information to my own question, I could imagine that the predictor $X$ is length, and response $Y$ is weight. In this case, length perfectly predicts the average weight. However, the rando | What can I do if the confidence intervals of the predicted mean are small but the predicted intervals are large
To add some information to my own question, I could imagine that the predictor $X$ is length, and response $Y$ is weight. In this case, length perfectly predicts the average weight. However, the random error ... | What can I do if the confidence intervals of the predicted mean are small but the predicted interval
To add some information to my own question, I could imagine that the predictor $X$ is length, and response $Y$ is weight. In this case, length perfectly predicts the average weight. However, the rando |
50,668 | How many counts are required to estimate a mean to a certain precision? | Your question seems to be assuming the rate of bubbles is constant across images.
If we do that, then you want that a $100(1-\alpha)\%$ CI for $\mu$ is $< 0.05\mu$.
$\text{se}(\hat{\mu}) = \sqrt{\mu/n}$
For a 95% interval, approximately $1.96 \sqrt{\mu/n} = .05 \mu$ ;
approximating again: $2 \sqrt{\mu/n} < .05 \mu$
i.... | How many counts are required to estimate a mean to a certain precision? | Your question seems to be assuming the rate of bubbles is constant across images.
If we do that, then you want that a $100(1-\alpha)\%$ CI for $\mu$ is $< 0.05\mu$.
$\text{se}(\hat{\mu}) = \sqrt{\mu/ | How many counts are required to estimate a mean to a certain precision?
Your question seems to be assuming the rate of bubbles is constant across images.
If we do that, then you want that a $100(1-\alpha)\%$ CI for $\mu$ is $< 0.05\mu$.
$\text{se}(\hat{\mu}) = \sqrt{\mu/n}$
For a 95% interval, approximately $1.96 \sqr... | How many counts are required to estimate a mean to a certain precision?
Your question seems to be assuming the rate of bubbles is constant across images.
If we do that, then you want that a $100(1-\alpha)\%$ CI for $\mu$ is $< 0.05\mu$.
$\text{se}(\hat{\mu}) = \sqrt{\mu/ |
50,669 | Testing for non-random overlap of polygons | In this case you should use spatial pattern analysis to identify relations between human activity and distribution of rocks. Rocks have to be represented as a point shp-file. Adding another level of abstract to distribution of rocks (making them polygons instead of points) is highly questionable approach that doesn't s... | Testing for non-random overlap of polygons | In this case you should use spatial pattern analysis to identify relations between human activity and distribution of rocks. Rocks have to be represented as a point shp-file. Adding another level of a | Testing for non-random overlap of polygons
In this case you should use spatial pattern analysis to identify relations between human activity and distribution of rocks. Rocks have to be represented as a point shp-file. Adding another level of abstract to distribution of rocks (making them polygons instead of points) is ... | Testing for non-random overlap of polygons
In this case you should use spatial pattern analysis to identify relations between human activity and distribution of rocks. Rocks have to be represented as a point shp-file. Adding another level of a |
50,670 | Fisher's method of combining p-values when one of the p-values is zero | Irrespective of the discussion in the comments about how these $p$-values of $0$ arose there are methods for combining $p$-values which can be calculated if $p=0$.
As the OP indicated neither Fisher's method nor Stouffer's works.
The method of Edgington based on the sum of $p$, the closely related mean $p$ method, the ... | Fisher's method of combining p-values when one of the p-values is zero | Irrespective of the discussion in the comments about how these $p$-values of $0$ arose there are methods for combining $p$-values which can be calculated if $p=0$.
As the OP indicated neither Fisher's | Fisher's method of combining p-values when one of the p-values is zero
Irrespective of the discussion in the comments about how these $p$-values of $0$ arose there are methods for combining $p$-values which can be calculated if $p=0$.
As the OP indicated neither Fisher's method nor Stouffer's works.
The method of Edgin... | Fisher's method of combining p-values when one of the p-values is zero
Irrespective of the discussion in the comments about how these $p$-values of $0$ arose there are methods for combining $p$-values which can be calculated if $p=0$.
As the OP indicated neither Fisher's |
50,671 | Bias term in support vector machine | This is a paper you probably should read:
Poggio, T., S. Mukherjee, R. Rifkin, A. Rakhlin and A. Verri. b, CBCL Paper #198/AI Memo #2001-011, Massachusetts Institute of Technology, Cambridge, MA, July 2001. (PostScript)
I rather doubt there is a paper with a shorter title! | Bias term in support vector machine | This is a paper you probably should read:
Poggio, T., S. Mukherjee, R. Rifkin, A. Rakhlin and A. Verri. b, CBCL Paper #198/AI Memo #2001-011, Massachusetts Institute of Technology, Cambridge, MA, July | Bias term in support vector machine
This is a paper you probably should read:
Poggio, T., S. Mukherjee, R. Rifkin, A. Rakhlin and A. Verri. b, CBCL Paper #198/AI Memo #2001-011, Massachusetts Institute of Technology, Cambridge, MA, July 2001. (PostScript)
I rather doubt there is a paper with a shorter title! | Bias term in support vector machine
This is a paper you probably should read:
Poggio, T., S. Mukherjee, R. Rifkin, A. Rakhlin and A. Verri. b, CBCL Paper #198/AI Memo #2001-011, Massachusetts Institute of Technology, Cambridge, MA, July |
50,672 | Is the p-postulate (equal p-values provide equal evidence against the null) true? | The Royall paper begins with two quotes that providing apparently contradictory interpretations of the p-value. Both rely on interpreting the p-value in light of the sample size and as such both are flawed interpretations of the p-value.
A p-value tells us one thing and one thing only--the probability of observing a... | Is the p-postulate (equal p-values provide equal evidence against the null) true? | The Royall paper begins with two quotes that providing apparently contradictory interpretations of the p-value. Both rely on interpreting the p-value in light of the sample size and as such both are | Is the p-postulate (equal p-values provide equal evidence against the null) true?
The Royall paper begins with two quotes that providing apparently contradictory interpretations of the p-value. Both rely on interpreting the p-value in light of the sample size and as such both are flawed interpretations of the p-value.... | Is the p-postulate (equal p-values provide equal evidence against the null) true?
The Royall paper begins with two quotes that providing apparently contradictory interpretations of the p-value. Both rely on interpreting the p-value in light of the sample size and as such both are |
50,673 | MLE for simulated case of binomial p with constant labelling rate | No, it is not possible to estimate the overcounting error rate $e$. This experiment behaves essentially like a Binomial experiment in which the erroneous "heads" cannot be distinguished from true heads and therefore the proportion of erroneous heads cannot be estimated.
Alternative interpretations
There is an ambigui... | MLE for simulated case of binomial p with constant labelling rate | No, it is not possible to estimate the overcounting error rate $e$. This experiment behaves essentially like a Binomial experiment in which the erroneous "heads" cannot be distinguished from true hea | MLE for simulated case of binomial p with constant labelling rate
No, it is not possible to estimate the overcounting error rate $e$. This experiment behaves essentially like a Binomial experiment in which the erroneous "heads" cannot be distinguished from true heads and therefore the proportion of erroneous heads can... | MLE for simulated case of binomial p with constant labelling rate
No, it is not possible to estimate the overcounting error rate $e$. This experiment behaves essentially like a Binomial experiment in which the erroneous "heads" cannot be distinguished from true hea |
50,674 | Tests for spatial stationarity (homogeneity)? | Three comments based on a mixture of experience and prejudice:
What should be important here is that the researcher's substantive knowledge (or that of a collaborator), which may make the question obvious at some level. That is, it may be foolish to apply models assuming stationarity if there are known to be gross tr... | Tests for spatial stationarity (homogeneity)? | Three comments based on a mixture of experience and prejudice:
What should be important here is that the researcher's substantive knowledge (or that of a collaborator), which may make the question o | Tests for spatial stationarity (homogeneity)?
Three comments based on a mixture of experience and prejudice:
What should be important here is that the researcher's substantive knowledge (or that of a collaborator), which may make the question obvious at some level. That is, it may be foolish to apply models assuming ... | Tests for spatial stationarity (homogeneity)?
Three comments based on a mixture of experience and prejudice:
What should be important here is that the researcher's substantive knowledge (or that of a collaborator), which may make the question o |
50,675 | Tests for spatial stationarity (homogeneity)? | Leung, Mei, and Zhang have developed two tests for whether GWR is a better fit than OLS regression. Their paper is here, but it's behind a paywall if you don't have academic access. As for variograms, etc. I know that Bivand, et al. cover tools and mechanisms In their book. I know a pdf of this exists because I have it... | Tests for spatial stationarity (homogeneity)? | Leung, Mei, and Zhang have developed two tests for whether GWR is a better fit than OLS regression. Their paper is here, but it's behind a paywall if you don't have academic access. As for variograms, | Tests for spatial stationarity (homogeneity)?
Leung, Mei, and Zhang have developed two tests for whether GWR is a better fit than OLS regression. Their paper is here, but it's behind a paywall if you don't have academic access. As for variograms, etc. I know that Bivand, et al. cover tools and mechanisms In their book.... | Tests for spatial stationarity (homogeneity)?
Leung, Mei, and Zhang have developed two tests for whether GWR is a better fit than OLS regression. Their paper is here, but it's behind a paywall if you don't have academic access. As for variograms, |
50,676 | Speed up web a/b tests with sample size checkpoints | This approach doesn't have the properties you would have if you fixed the sample size ahead of time.
The situation where you look for a particular result while your experiment continues and have some 'stopping rule' (halt your experiment early if a particular situation is achieved) is a version of sequential analysis;... | Speed up web a/b tests with sample size checkpoints | This approach doesn't have the properties you would have if you fixed the sample size ahead of time.
The situation where you look for a particular result while your experiment continues and have some | Speed up web a/b tests with sample size checkpoints
This approach doesn't have the properties you would have if you fixed the sample size ahead of time.
The situation where you look for a particular result while your experiment continues and have some 'stopping rule' (halt your experiment early if a particular situati... | Speed up web a/b tests with sample size checkpoints
This approach doesn't have the properties you would have if you fixed the sample size ahead of time.
The situation where you look for a particular result while your experiment continues and have some |
50,677 | Understanding the construction of Dirichlet process | 1) infinite does not mean continuous.
2) The total mass (measure) of a probabilistic measure F is always 1. In the former case, F is partitioned by k bins (discrete). However F is not required to be discrete. If F is continuous, it can be partitioned by k regions (T1...Tk) | Understanding the construction of Dirichlet process | 1) infinite does not mean continuous.
2) The total mass (measure) of a probabilistic measure F is always 1. In the former case, F is partitioned by k bins (discrete). However F is not required to be | Understanding the construction of Dirichlet process
1) infinite does not mean continuous.
2) The total mass (measure) of a probabilistic measure F is always 1. In the former case, F is partitioned by k bins (discrete). However F is not required to be discrete. If F is continuous, it can be partitioned by k regions (T1... | Understanding the construction of Dirichlet process
1) infinite does not mean continuous.
2) The total mass (measure) of a probabilistic measure F is always 1. In the former case, F is partitioned by k bins (discrete). However F is not required to be |
50,678 | Understanding the construction of Dirichlet process | Not an answer but rather a long comment regarding your point 2 (I am not an expert so take my explanation with caution)
It was not clear to me in what sense you have used the terms infinite and finite in your statement 2. Below is my take on how you can argue that indeed "In the finite case, $F$ is indexed by $1,\ldot... | Understanding the construction of Dirichlet process | Not an answer but rather a long comment regarding your point 2 (I am not an expert so take my explanation with caution)
It was not clear to me in what sense you have used the terms infinite and finit | Understanding the construction of Dirichlet process
Not an answer but rather a long comment regarding your point 2 (I am not an expert so take my explanation with caution)
It was not clear to me in what sense you have used the terms infinite and finite in your statement 2. Below is my take on how you can argue that in... | Understanding the construction of Dirichlet process
Not an answer but rather a long comment regarding your point 2 (I am not an expert so take my explanation with caution)
It was not clear to me in what sense you have used the terms infinite and finit |
50,679 | Why don't we look at $R^2$ when fitting an autoregressive model? | IMHO, using the R2 is irrelevant since it would just push you to use a larger regression order $k$ which would generally give you a smaller R2. The idea of fitting an AR (or any GLP) is to reproduce the underlying process with a model that is as simple as possible (since the idea is also to extract meaning out of the d... | Why don't we look at $R^2$ when fitting an autoregressive model? | IMHO, using the R2 is irrelevant since it would just push you to use a larger regression order $k$ which would generally give you a smaller R2. The idea of fitting an AR (or any GLP) is to reproduce t | Why don't we look at $R^2$ when fitting an autoregressive model?
IMHO, using the R2 is irrelevant since it would just push you to use a larger regression order $k$ which would generally give you a smaller R2. The idea of fitting an AR (or any GLP) is to reproduce the underlying process with a model that is as simple as... | Why don't we look at $R^2$ when fitting an autoregressive model?
IMHO, using the R2 is irrelevant since it would just push you to use a larger regression order $k$ which would generally give you a smaller R2. The idea of fitting an AR (or any GLP) is to reproduce t |
50,680 | Standard error for a statistic obtained via simulation | I do not know what you mean with "certain conditions" but if you want to calculate the uncertainty in the regression coefficients via simulation, you have to take account of the uncertainty about the residual standard deviation in general (because Var$(\,\hat{\beta}\,|\,X\,) = \sigma^2\,(X^TX)^{-1}$, $\sigma^2$ should... | Standard error for a statistic obtained via simulation | I do not know what you mean with "certain conditions" but if you want to calculate the uncertainty in the regression coefficients via simulation, you have to take account of the uncertainty about the | Standard error for a statistic obtained via simulation
I do not know what you mean with "certain conditions" but if you want to calculate the uncertainty in the regression coefficients via simulation, you have to take account of the uncertainty about the residual standard deviation in general (because Var$(\,\hat{\beta... | Standard error for a statistic obtained via simulation
I do not know what you mean with "certain conditions" but if you want to calculate the uncertainty in the regression coefficients via simulation, you have to take account of the uncertainty about the |
50,681 | Estimating parameters of a normal distribution from noisy observation of samples | I think I've figured it out now. As stated in the comments, I think there is indeed a problem with the estimators.
I found another answer on using a linear estimator for a similar problem. If I didn't make any mistakes, the estimator for $\mu$ in my case should be $$\hat{\mu} = \frac{ \sum_{i=1..n} z_i/(C+\Sigma_i)}{ ... | Estimating parameters of a normal distribution from noisy observation of samples | I think I've figured it out now. As stated in the comments, I think there is indeed a problem with the estimators.
I found another answer on using a linear estimator for a similar problem. If I didn't | Estimating parameters of a normal distribution from noisy observation of samples
I think I've figured it out now. As stated in the comments, I think there is indeed a problem with the estimators.
I found another answer on using a linear estimator for a similar problem. If I didn't make any mistakes, the estimator for $... | Estimating parameters of a normal distribution from noisy observation of samples
I think I've figured it out now. As stated in the comments, I think there is indeed a problem with the estimators.
I found another answer on using a linear estimator for a similar problem. If I didn't |
50,682 | Robust parameter estimation for shifted log normal distribution | In case anyone is still interested, I have managed to implement Aristizabal's formulae in Java. This is more proof-of-concept than the requested "robust" code, but it is a starting point.
/**
* Computes the point estimate of the shift offset (gamma) from the given sample. The sample array will be sorted by this method... | Robust parameter estimation for shifted log normal distribution | In case anyone is still interested, I have managed to implement Aristizabal's formulae in Java. This is more proof-of-concept than the requested "robust" code, but it is a starting point.
/**
* Compu | Robust parameter estimation for shifted log normal distribution
In case anyone is still interested, I have managed to implement Aristizabal's formulae in Java. This is more proof-of-concept than the requested "robust" code, but it is a starting point.
/**
* Computes the point estimate of the shift offset (gamma) from ... | Robust parameter estimation for shifted log normal distribution
In case anyone is still interested, I have managed to implement Aristizabal's formulae in Java. This is more proof-of-concept than the requested "robust" code, but it is a starting point.
/**
* Compu |
50,683 | predict() - multinomial logistic regression | When the estimates do not have the expected sign, the usual suspect is multicollinearity. Below is a passage from the Wikipedia page.
The usual interpretation of a regression coefficient is that it provides an estimate of the effect of a one unit change in an independent variable, $X_1$, holding the other variables co... | predict() - multinomial logistic regression | When the estimates do not have the expected sign, the usual suspect is multicollinearity. Below is a passage from the Wikipedia page.
The usual interpretation of a regression coefficient is that it p | predict() - multinomial logistic regression
When the estimates do not have the expected sign, the usual suspect is multicollinearity. Below is a passage from the Wikipedia page.
The usual interpretation of a regression coefficient is that it provides an estimate of the effect of a one unit change in an independent var... | predict() - multinomial logistic regression
When the estimates do not have the expected sign, the usual suspect is multicollinearity. Below is a passage from the Wikipedia page.
The usual interpretation of a regression coefficient is that it p |
50,684 | Creating fuzzy values for binary data | I have never seen this done, and I doubt other people have either. One usually gets informed answers on this site within a couple of hours of posting something. It's been a day, and no joy.
My thinking is this: if you want to tell the model that some values are more trustworthy than others, use weights. If you downweig... | Creating fuzzy values for binary data | I have never seen this done, and I doubt other people have either. One usually gets informed answers on this site within a couple of hours of posting something. It's been a day, and no joy.
My thinkin | Creating fuzzy values for binary data
I have never seen this done, and I doubt other people have either. One usually gets informed answers on this site within a couple of hours of posting something. It's been a day, and no joy.
My thinking is this: if you want to tell the model that some values are more trustworthy tha... | Creating fuzzy values for binary data
I have never seen this done, and I doubt other people have either. One usually gets informed answers on this site within a couple of hours of posting something. It's been a day, and no joy.
My thinkin |
50,685 | Preliminary estimates of ARIMA in R? | In base R's arima() take a look at the method= argument. From the help docs:
Fitting method: maximum likelihood or minimize conditional
sum-of-squares. The default (unless there are missing values) is to
use conditional-sum-of-squares to find starting values, then maximum
likelihood.
From a little further down... | Preliminary estimates of ARIMA in R? | In base R's arima() take a look at the method= argument. From the help docs:
Fitting method: maximum likelihood or minimize conditional
sum-of-squares. The default (unless there are missing values | Preliminary estimates of ARIMA in R?
In base R's arima() take a look at the method= argument. From the help docs:
Fitting method: maximum likelihood or minimize conditional
sum-of-squares. The default (unless there are missing values) is to
use conditional-sum-of-squares to find starting values, then maximum
li... | Preliminary estimates of ARIMA in R?
In base R's arima() take a look at the method= argument. From the help docs:
Fitting method: maximum likelihood or minimize conditional
sum-of-squares. The default (unless there are missing values |
50,686 | What is the relationship between correlation coefficients and regression coefficients in multiple regression? | Let's assume that the variable $x_1$ and $x_2$ are centered, it will make things easier (nothing prevents you to do that before doing your regression). Then, it is straightforward to see that:
$r_{y, x_1}\sigma_y=\beta_1\sigma_{x_1} + \beta_2r_{x_1, x_2} \sigma_{x_2}$
and
$r_{y, x_2}\sigma_y=\beta_2\sigma_{x_2} + \bet... | What is the relationship between correlation coefficients and regression coefficients in multiple re | Let's assume that the variable $x_1$ and $x_2$ are centered, it will make things easier (nothing prevents you to do that before doing your regression). Then, it is straightforward to see that:
$r_{y, | What is the relationship between correlation coefficients and regression coefficients in multiple regression?
Let's assume that the variable $x_1$ and $x_2$ are centered, it will make things easier (nothing prevents you to do that before doing your regression). Then, it is straightforward to see that:
$r_{y, x_1}\sigma... | What is the relationship between correlation coefficients and regression coefficients in multiple re
Let's assume that the variable $x_1$ and $x_2$ are centered, it will make things easier (nothing prevents you to do that before doing your regression). Then, it is straightforward to see that:
$r_{y, |
50,687 | How to compare coefficients of a negative binomial regression for determining relative importance? | First you'd have to figure out what change in one variable is "equal" to a what change in another. The usual standardization uses the standard deviation, but that may or may not be ideal. It may not be possible to figure this out - particularly if the IVs are related to each other, in which case a change in one would ... | How to compare coefficients of a negative binomial regression for determining relative importance? | First you'd have to figure out what change in one variable is "equal" to a what change in another. The usual standardization uses the standard deviation, but that may or may not be ideal. It may not | How to compare coefficients of a negative binomial regression for determining relative importance?
First you'd have to figure out what change in one variable is "equal" to a what change in another. The usual standardization uses the standard deviation, but that may or may not be ideal. It may not be possible to figure... | How to compare coefficients of a negative binomial regression for determining relative importance?
First you'd have to figure out what change in one variable is "equal" to a what change in another. The usual standardization uses the standard deviation, but that may or may not be ideal. It may not |
50,688 | How to compare coefficients of a negative binomial regression for determining relative importance? | For a quick way to get at the standardized beta coefficients directly from any lm (or glm) model in R, try using lm.beta(model). In the example provided, this would be:
library("MASS")
nb = glm.nb(responseCountVar ~ predictor1 + predictor2 +
predictor3, data=myData, control=glm.control(maxit=125))
summary(nb)
libr... | How to compare coefficients of a negative binomial regression for determining relative importance? | For a quick way to get at the standardized beta coefficients directly from any lm (or glm) model in R, try using lm.beta(model). In the example provided, this would be:
library("MASS")
nb = glm.nb(re | How to compare coefficients of a negative binomial regression for determining relative importance?
For a quick way to get at the standardized beta coefficients directly from any lm (or glm) model in R, try using lm.beta(model). In the example provided, this would be:
library("MASS")
nb = glm.nb(responseCountVar ~ pred... | How to compare coefficients of a negative binomial regression for determining relative importance?
For a quick way to get at the standardized beta coefficients directly from any lm (or glm) model in R, try using lm.beta(model). In the example provided, this would be:
library("MASS")
nb = glm.nb(re |
50,689 | Satisfaction of detailed balance equation in Metropolis-Hastings algorithms? | I'll address question 2.
If you fix a distribution supported on a finite set, the Markov chains which have that distribution as a stable distribution form a polytope. You can interpolate between any two by following one rule with probability $p$ and the other with probability $1-p$ and the convex combination will also... | Satisfaction of detailed balance equation in Metropolis-Hastings algorithms? | I'll address question 2.
If you fix a distribution supported on a finite set, the Markov chains which have that distribution as a stable distribution form a polytope. You can interpolate between any | Satisfaction of detailed balance equation in Metropolis-Hastings algorithms?
I'll address question 2.
If you fix a distribution supported on a finite set, the Markov chains which have that distribution as a stable distribution form a polytope. You can interpolate between any two by following one rule with probability ... | Satisfaction of detailed balance equation in Metropolis-Hastings algorithms?
I'll address question 2.
If you fix a distribution supported on a finite set, the Markov chains which have that distribution as a stable distribution form a polytope. You can interpolate between any |
50,690 | Is a sequence of random variables indexed by a homogeneous Poisson process process strictly stationary? | If the process $(N_t)$ and the sequence $(X_k)$ are independent, then indeed the identity $Y_t=X_{N_t}$ defines a strictly stationary process $(Y_t)$.
To show this, consider $Y^s_t=Y_{s+t}$, then, for every nonnegative time $s$, conditionally on $N_s=m$, the process $(Y^s_t)_{t\geqslant0}$ is such that, for every $t\... | Is a sequence of random variables indexed by a homogeneous Poisson process process strictly stationa | If the process $(N_t)$ and the sequence $(X_k)$ are independent, then indeed the identity $Y_t=X_{N_t}$ defines a strictly stationary process $(Y_t)$.
To show this, consider $Y^s_t=Y_{s+t}$, then, f | Is a sequence of random variables indexed by a homogeneous Poisson process process strictly stationary?
If the process $(N_t)$ and the sequence $(X_k)$ are independent, then indeed the identity $Y_t=X_{N_t}$ defines a strictly stationary process $(Y_t)$.
To show this, consider $Y^s_t=Y_{s+t}$, then, for every nonnega... | Is a sequence of random variables indexed by a homogeneous Poisson process process strictly stationa
If the process $(N_t)$ and the sequence $(X_k)$ are independent, then indeed the identity $Y_t=X_{N_t}$ defines a strictly stationary process $(Y_t)$.
To show this, consider $Y^s_t=Y_{s+t}$, then, f |
50,691 | The relationship between expectation-maximization and majorization-minimization | The general idea of Minorant Maximization algorithms is:
(a) Approximate target function with a dominated function.
(b) Climb in approximating function.
(c) Return to (a), approximating at new coordinate.
The "flavor" of the MM is given by the approximating method, and the climbing method.
EM is a particular instanc... | The relationship between expectation-maximization and majorization-minimization | The general idea of Minorant Maximization algorithms is:
(a) Approximate target function with a dominated function.
(b) Climb in approximating function.
(c) Return to (a), approximating at new coor | The relationship between expectation-maximization and majorization-minimization
The general idea of Minorant Maximization algorithms is:
(a) Approximate target function with a dominated function.
(b) Climb in approximating function.
(c) Return to (a), approximating at new coordinate.
The "flavor" of the MM is given ... | The relationship between expectation-maximization and majorization-minimization
The general idea of Minorant Maximization algorithms is:
(a) Approximate target function with a dominated function.
(b) Climb in approximating function.
(c) Return to (a), approximating at new coor |
50,692 | What's the null hypothesis in a one-sided Kolmogorov-Smirnov test? | I think most of the tables providing p-values for the K-S statistic are based on a two-sided test. The null hypothesis assumed by the values in the table is that the two samples are drawn from the same distribution (ie, that $C_x=C_y$). So really the table is only concerned with the absolute value of the difference bet... | What's the null hypothesis in a one-sided Kolmogorov-Smirnov test? | I think most of the tables providing p-values for the K-S statistic are based on a two-sided test. The null hypothesis assumed by the values in the table is that the two samples are drawn from the sam | What's the null hypothesis in a one-sided Kolmogorov-Smirnov test?
I think most of the tables providing p-values for the K-S statistic are based on a two-sided test. The null hypothesis assumed by the values in the table is that the two samples are drawn from the same distribution (ie, that $C_x=C_y$). So really the ta... | What's the null hypothesis in a one-sided Kolmogorov-Smirnov test?
I think most of the tables providing p-values for the K-S statistic are based on a two-sided test. The null hypothesis assumed by the values in the table is that the two samples are drawn from the sam |
50,693 | Assumption for an M/M/1 queue | I think the main advantage to modeling the arrival distribution as Poisson is the memory-less property. It greatly simplifies the subsequent calculations to be able to assume that the number of arrivals in a particular time interval depends only on the length of the interval, rather than when the interval occurs, how m... | Assumption for an M/M/1 queue | I think the main advantage to modeling the arrival distribution as Poisson is the memory-less property. It greatly simplifies the subsequent calculations to be able to assume that the number of arriva | Assumption for an M/M/1 queue
I think the main advantage to modeling the arrival distribution as Poisson is the memory-less property. It greatly simplifies the subsequent calculations to be able to assume that the number of arrivals in a particular time interval depends only on the length of the interval, rather than w... | Assumption for an M/M/1 queue
I think the main advantage to modeling the arrival distribution as Poisson is the memory-less property. It greatly simplifies the subsequent calculations to be able to assume that the number of arriva |
50,694 | Assumption for an M/M/1 queue | There's a theorem in renewal theory related to that. It is roughly equivalent to the Central Limit Theorem for sums of independent random variables. It says that, under certain general conditions, the superposition of a large number of independent arrival processes converges to a Poisson process. That explains why the ... | Assumption for an M/M/1 queue | There's a theorem in renewal theory related to that. It is roughly equivalent to the Central Limit Theorem for sums of independent random variables. It says that, under certain general conditions, the | Assumption for an M/M/1 queue
There's a theorem in renewal theory related to that. It is roughly equivalent to the Central Limit Theorem for sums of independent random variables. It says that, under certain general conditions, the superposition of a large number of independent arrival processes converges to a Poisson p... | Assumption for an M/M/1 queue
There's a theorem in renewal theory related to that. It is roughly equivalent to the Central Limit Theorem for sums of independent random variables. It says that, under certain general conditions, the |
50,695 | Assumption for an M/M/1 queue | To use an M/M/I queuing model, we must assume Poisson arrivals and exponential service distribution. Then, to derive characteristics for this types of waiting line, we must consider some other assumptions. Most importantly, there must be only one service channel, which arrivals enter one at a time. More so, it is assum... | Assumption for an M/M/1 queue | To use an M/M/I queuing model, we must assume Poisson arrivals and exponential service distribution. Then, to derive characteristics for this types of waiting line, we must consider some other assumpt | Assumption for an M/M/1 queue
To use an M/M/I queuing model, we must assume Poisson arrivals and exponential service distribution. Then, to derive characteristics for this types of waiting line, we must consider some other assumptions. Most importantly, there must be only one service channel, which arrivals enter one a... | Assumption for an M/M/1 queue
To use an M/M/I queuing model, we must assume Poisson arrivals and exponential service distribution. Then, to derive characteristics for this types of waiting line, we must consider some other assumpt |
50,696 | With a binary Y, why are R's lowess fits so often flat? | It works consistently fine for me.
y <- rbinom(100, 1, (0:99)/100)
x <- 1:100
m <- loess(y~x)
plot(y ~ x)
lines(predict(m)) # also illustrating the newer loess function that has different defaults
lines(lowess(x,y), col = 'blue')
If you run that a few times you'll notice the blue line does tend to stick closer to the ... | With a binary Y, why are R's lowess fits so often flat? | It works consistently fine for me.
y <- rbinom(100, 1, (0:99)/100)
x <- 1:100
m <- loess(y~x)
plot(y ~ x)
lines(predict(m)) # also illustrating the newer loess function that has different defaults
lin | With a binary Y, why are R's lowess fits so often flat?
It works consistently fine for me.
y <- rbinom(100, 1, (0:99)/100)
x <- 1:100
m <- loess(y~x)
plot(y ~ x)
lines(predict(m)) # also illustrating the newer loess function that has different defaults
lines(lowess(x,y), col = 'blue')
If you run that a few times you'l... | With a binary Y, why are R's lowess fits so often flat?
It works consistently fine for me.
y <- rbinom(100, 1, (0:99)/100)
x <- 1:100
m <- loess(y~x)
plot(y ~ x)
lines(predict(m)) # also illustrating the newer loess function that has different defaults
lin |
50,697 | Bayesian hypothesis testing and Bayes factors | Let $p(\alpha,\beta|M_1)=\mathrm{Ga}(\alpha,\beta)$ be your first prior,
and $p(\alpha,\beta|M_2)=\mathrm{Ga}(\alpha+1,\beta)$ be your second prior.
Furthermore, let $D$ be your observed data.
Your posterior distribution can then be expressed as:
$$p(\alpha,\beta|D,M_i) = \frac{L(D|\alpha,\beta)p(\alpha,\beta|M_i)}{p(D... | Bayesian hypothesis testing and Bayes factors | Let $p(\alpha,\beta|M_1)=\mathrm{Ga}(\alpha,\beta)$ be your first prior,
and $p(\alpha,\beta|M_2)=\mathrm{Ga}(\alpha+1,\beta)$ be your second prior.
Furthermore, let $D$ be your observed data.
Your po | Bayesian hypothesis testing and Bayes factors
Let $p(\alpha,\beta|M_1)=\mathrm{Ga}(\alpha,\beta)$ be your first prior,
and $p(\alpha,\beta|M_2)=\mathrm{Ga}(\alpha+1,\beta)$ be your second prior.
Furthermore, let $D$ be your observed data.
Your posterior distribution can then be expressed as:
$$p(\alpha,\beta|D,M_i) = \... | Bayesian hypothesis testing and Bayes factors
Let $p(\alpha,\beta|M_1)=\mathrm{Ga}(\alpha,\beta)$ be your first prior,
and $p(\alpha,\beta|M_2)=\mathrm{Ga}(\alpha+1,\beta)$ be your second prior.
Furthermore, let $D$ be your observed data.
Your po |
50,698 | Bayesian hypothesis testing and Bayes factors | How would a Bayes factor for the two prior distributions be formed?
The Bayes factor $K$ for competing models $M_1$ and $M_2$ is the quotient of the data likelihoods of the two models. Applying this definition to the problem you describe gives:
$$
K = \frac{p(\mathbf{x} |M_1)}{p(\mathbf{x}|M_2)} =
\frac{\int_{\lambda... | Bayesian hypothesis testing and Bayes factors | How would a Bayes factor for the two prior distributions be formed?
The Bayes factor $K$ for competing models $M_1$ and $M_2$ is the quotient of the data likelihoods of the two models. Applying this | Bayesian hypothesis testing and Bayes factors
How would a Bayes factor for the two prior distributions be formed?
The Bayes factor $K$ for competing models $M_1$ and $M_2$ is the quotient of the data likelihoods of the two models. Applying this definition to the problem you describe gives:
$$
K = \frac{p(\mathbf{x} |M... | Bayesian hypothesis testing and Bayes factors
How would a Bayes factor for the two prior distributions be formed?
The Bayes factor $K$ for competing models $M_1$ and $M_2$ is the quotient of the data likelihoods of the two models. Applying this |
50,699 | Computing confidence region for Gaussian mixture model | In general, it is possible for instance to compute what is the probability contained in a ball $\mathcal{B}(c,r)$. I suppose that your gaussian mixture writes
$$p(x) = \sum_{j=1}^K \mathcal{N}(x;\mu_j,\Sigma_j)\mathbb{P}(J=j)$$
There exist elementary pieces of code to make an algorithm to compute $F(c,r)=\int_{\mathcal... | Computing confidence region for Gaussian mixture model | In general, it is possible for instance to compute what is the probability contained in a ball $\mathcal{B}(c,r)$. I suppose that your gaussian mixture writes
$$p(x) = \sum_{j=1}^K \mathcal{N}(x;\mu_j | Computing confidence region for Gaussian mixture model
In general, it is possible for instance to compute what is the probability contained in a ball $\mathcal{B}(c,r)$. I suppose that your gaussian mixture writes
$$p(x) = \sum_{j=1}^K \mathcal{N}(x;\mu_j,\Sigma_j)\mathbb{P}(J=j)$$
There exist elementary pieces of code... | Computing confidence region for Gaussian mixture model
In general, it is possible for instance to compute what is the probability contained in a ball $\mathcal{B}(c,r)$. I suppose that your gaussian mixture writes
$$p(x) = \sum_{j=1}^K \mathcal{N}(x;\mu_j |
50,700 | Expectation notations | It might be helpful to isolate all the moving parts in a particular expectation. In doing so, recall that an expectation is a Lebesgue-Stieltjes integral.
I will adapt the notation slightly and use $X$ to denote a random variable, $x$ to denote a fixed value taken by that random variable in its support, that is, $x\in... | Expectation notations | It might be helpful to isolate all the moving parts in a particular expectation. In doing so, recall that an expectation is a Lebesgue-Stieltjes integral.
I will adapt the notation slightly and use $ | Expectation notations
It might be helpful to isolate all the moving parts in a particular expectation. In doing so, recall that an expectation is a Lebesgue-Stieltjes integral.
I will adapt the notation slightly and use $X$ to denote a random variable, $x$ to denote a fixed value taken by that random variable in its s... | Expectation notations
It might be helpful to isolate all the moving parts in a particular expectation. In doing so, recall that an expectation is a Lebesgue-Stieltjes integral.
I will adapt the notation slightly and use $ |
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