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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45,701 | n observations from a random variable VS. 1 observation from n i.i.d random variables | I think it depends on the theoretical and mainstream approach. This approach could not be proved, but it is such an axiom. I looked in my books on math. stat., authors do not mention this question.
Some points about that:
Math. stat. theory approach based on tests and theory derived from i.i.d. condition. With this co... | n observations from a random variable VS. 1 observation from n i.i.d random variables | I think it depends on the theoretical and mainstream approach. This approach could not be proved, but it is such an axiom. I looked in my books on math. stat., authors do not mention this question.
So | n observations from a random variable VS. 1 observation from n i.i.d random variables
I think it depends on the theoretical and mainstream approach. This approach could not be proved, but it is such an axiom. I looked in my books on math. stat., authors do not mention this question.
Some points about that:
Math. stat.... | n observations from a random variable VS. 1 observation from n i.i.d random variables
I think it depends on the theoretical and mainstream approach. This approach could not be proved, but it is such an axiom. I looked in my books on math. stat., authors do not mention this question.
So |
45,702 | Maximum likelihood estimator that is not a function of a sufficient statistic | Nothing is wrong with what you said, just the statement that every maximum likelihood estimator has to be a function of any sufficient statistic, which is false as stated. A more correct form of putting this assertion is:
If $T$ is a sufficient statistic for $\theta$ and a unique MLE of
$\hat{\theta}$ exists, then $... | Maximum likelihood estimator that is not a function of a sufficient statistic | Nothing is wrong with what you said, just the statement that every maximum likelihood estimator has to be a function of any sufficient statistic, which is false as stated. A more correct form of putti | Maximum likelihood estimator that is not a function of a sufficient statistic
Nothing is wrong with what you said, just the statement that every maximum likelihood estimator has to be a function of any sufficient statistic, which is false as stated. A more correct form of putting this assertion is:
If $T$ is a suffici... | Maximum likelihood estimator that is not a function of a sufficient statistic
Nothing is wrong with what you said, just the statement that every maximum likelihood estimator has to be a function of any sufficient statistic, which is false as stated. A more correct form of putti |
45,703 | Maximum likelihood estimator that is not a function of a sufficient statistic | I think to preserve the theorem in cases like this one should define the MLE as the interval of MLEs. That is a function of the sufficient statistic.
This page takes a different point of view: For every sufficient statistic, there is at least one MLE that is a function of it. (So if there is only one MLE, then that one... | Maximum likelihood estimator that is not a function of a sufficient statistic | I think to preserve the theorem in cases like this one should define the MLE as the interval of MLEs. That is a function of the sufficient statistic.
This page takes a different point of view: For eve | Maximum likelihood estimator that is not a function of a sufficient statistic
I think to preserve the theorem in cases like this one should define the MLE as the interval of MLEs. That is a function of the sufficient statistic.
This page takes a different point of view: For every sufficient statistic, there is at least... | Maximum likelihood estimator that is not a function of a sufficient statistic
I think to preserve the theorem in cases like this one should define the MLE as the interval of MLEs. That is a function of the sufficient statistic.
This page takes a different point of view: For eve |
45,704 | Maximum likelihood estimator that is not a function of a sufficient statistic | Sufficient statistics only apply to exponential family distributions. Continuous uniform is not an exponential family distribution.
See [DeGroot, Morris H. Optimal Statistical Decisions, 1970, McGraw-Hill Book Company, New York] | Maximum likelihood estimator that is not a function of a sufficient statistic | Sufficient statistics only apply to exponential family distributions. Continuous uniform is not an exponential family distribution.
See [DeGroot, Morris H. Optimal Statistical Decisions, 1970, McGraw- | Maximum likelihood estimator that is not a function of a sufficient statistic
Sufficient statistics only apply to exponential family distributions. Continuous uniform is not an exponential family distribution.
See [DeGroot, Morris H. Optimal Statistical Decisions, 1970, McGraw-Hill Book Company, New York] | Maximum likelihood estimator that is not a function of a sufficient statistic
Sufficient statistics only apply to exponential family distributions. Continuous uniform is not an exponential family distribution.
See [DeGroot, Morris H. Optimal Statistical Decisions, 1970, McGraw- |
45,705 | Why periodically skip updating a parameter in MCMC? | This type of fine-tuned (Gibbs) MCMC is appropriate for cases when one conditional distribution is most "sticky" than other conditional distributions in the problem. For instance, updating only one [random] part of $\beta$ may be profitable when updating the whole vector results in high rejection rates or in very small... | Why periodically skip updating a parameter in MCMC? | This type of fine-tuned (Gibbs) MCMC is appropriate for cases when one conditional distribution is most "sticky" than other conditional distributions in the problem. For instance, updating only one [r | Why periodically skip updating a parameter in MCMC?
This type of fine-tuned (Gibbs) MCMC is appropriate for cases when one conditional distribution is most "sticky" than other conditional distributions in the problem. For instance, updating only one [random] part of $\beta$ may be profitable when updating the whole vec... | Why periodically skip updating a parameter in MCMC?
This type of fine-tuned (Gibbs) MCMC is appropriate for cases when one conditional distribution is most "sticky" than other conditional distributions in the problem. For instance, updating only one [r |
45,706 | Understanding max-pooling and loss of information | Max pooling loses information in a sense that it tells you whether a filtered feature was encountered or not, but forgets where in the data, how many times etc.
Suppose your filter is looking for vertical stripes in the image. Without max pooling it will output all stripes found. With max pooling, it will tell you whet... | Understanding max-pooling and loss of information | Max pooling loses information in a sense that it tells you whether a filtered feature was encountered or not, but forgets where in the data, how many times etc.
Suppose your filter is looking for vert | Understanding max-pooling and loss of information
Max pooling loses information in a sense that it tells you whether a filtered feature was encountered or not, but forgets where in the data, how many times etc.
Suppose your filter is looking for vertical stripes in the image. Without max pooling it will output all stri... | Understanding max-pooling and loss of information
Max pooling loses information in a sense that it tells you whether a filtered feature was encountered or not, but forgets where in the data, how many times etc.
Suppose your filter is looking for vert |
45,707 | Understanding max-pooling and loss of information | I'm not completely sure but I'm thinking that if any of the pixels are dark in a chunk of max pooling it will output that darkest pixel, no matter what other pixels are. Without max pooling weights can be applied on all the pixels of the previous layer so less data is lost. Even though the network will learn what infor... | Understanding max-pooling and loss of information | I'm not completely sure but I'm thinking that if any of the pixels are dark in a chunk of max pooling it will output that darkest pixel, no matter what other pixels are. Without max pooling weights ca | Understanding max-pooling and loss of information
I'm not completely sure but I'm thinking that if any of the pixels are dark in a chunk of max pooling it will output that darkest pixel, no matter what other pixels are. Without max pooling weights can be applied on all the pixels of the previous layer so less data is l... | Understanding max-pooling and loss of information
I'm not completely sure but I'm thinking that if any of the pixels are dark in a chunk of max pooling it will output that darkest pixel, no matter what other pixels are. Without max pooling weights ca |
45,708 | What are the implications of the curse of dimensionality for ordinary least squares linear regression? | Edit:
As @Richard Hardy pointed out, the linear model under squared loss and ordinary least squares (OLS) are different things. I revised my answer to discuss the linear regression model only, where we are trying to check if the curse of dimesionality (CoD) is present when solving the following optimization problem:
$$... | What are the implications of the curse of dimensionality for ordinary least squares linear regressio | Edit:
As @Richard Hardy pointed out, the linear model under squared loss and ordinary least squares (OLS) are different things. I revised my answer to discuss the linear regression model only, where w | What are the implications of the curse of dimensionality for ordinary least squares linear regression?
Edit:
As @Richard Hardy pointed out, the linear model under squared loss and ordinary least squares (OLS) are different things. I revised my answer to discuss the linear regression model only, where we are trying to c... | What are the implications of the curse of dimensionality for ordinary least squares linear regressio
Edit:
As @Richard Hardy pointed out, the linear model under squared loss and ordinary least squares (OLS) are different things. I revised my answer to discuss the linear regression model only, where w |
45,709 | What are the implications of the curse of dimensionality for ordinary least squares linear regression? | I think that everything that hxd1011 says is correct, however if one is interested in prediction rather than description, CoD can rear it's ugly head. For example if one is using Akaike I]information Criteria to decide on model accuracy, then the value is proportional to the number ,p, of variables. Since a lower AIC i... | What are the implications of the curse of dimensionality for ordinary least squares linear regressio | I think that everything that hxd1011 says is correct, however if one is interested in prediction rather than description, CoD can rear it's ugly head. For example if one is using Akaike I]information | What are the implications of the curse of dimensionality for ordinary least squares linear regression?
I think that everything that hxd1011 says is correct, however if one is interested in prediction rather than description, CoD can rear it's ugly head. For example if one is using Akaike I]information Criteria to decid... | What are the implications of the curse of dimensionality for ordinary least squares linear regressio
I think that everything that hxd1011 says is correct, however if one is interested in prediction rather than description, CoD can rear it's ugly head. For example if one is using Akaike I]information |
45,710 | Calculate standard deviation given mean and percentage | We can solve this problem almost instantly in our heads using the "68-95-99.7" rule. I will explain the process in detail because that is what matters. The answer is of little interest: the point to this question is to help us learn to think about probability distributions.
These numbers in the 68-95-99.7 rule are (a... | Calculate standard deviation given mean and percentage | We can solve this problem almost instantly in our heads using the "68-95-99.7" rule. I will explain the process in detail because that is what matters. The answer is of little interest: the point to | Calculate standard deviation given mean and percentage
We can solve this problem almost instantly in our heads using the "68-95-99.7" rule. I will explain the process in detail because that is what matters. The answer is of little interest: the point to this question is to help us learn to think about probability dis... | Calculate standard deviation given mean and percentage
We can solve this problem almost instantly in our heads using the "68-95-99.7" rule. I will explain the process in detail because that is what matters. The answer is of little interest: the point to |
45,711 | Calculate standard deviation given mean and percentage | So you can use R to get the answer:
target=function (sd){
b=pnorm(0.0278, mean = 0.0276, sd = sd)
a=pnorm(0.0275, mean = 0.0276, sd = sd)
return(abs(b-a-0.98))
}
sd=optim(1,target)
sd$par
This gives:
> sd$par
[1] 4.868167e-05
What we are doing is using numerical method to calculate $\sigma$ such that $$F(0.027... | Calculate standard deviation given mean and percentage | So you can use R to get the answer:
target=function (sd){
b=pnorm(0.0278, mean = 0.0276, sd = sd)
a=pnorm(0.0275, mean = 0.0276, sd = sd)
return(abs(b-a-0.98))
}
sd=optim(1,target)
sd$par
This | Calculate standard deviation given mean and percentage
So you can use R to get the answer:
target=function (sd){
b=pnorm(0.0278, mean = 0.0276, sd = sd)
a=pnorm(0.0275, mean = 0.0276, sd = sd)
return(abs(b-a-0.98))
}
sd=optim(1,target)
sd$par
This gives:
> sd$par
[1] 4.868167e-05
What we are doing is using num... | Calculate standard deviation given mean and percentage
So you can use R to get the answer:
target=function (sd){
b=pnorm(0.0278, mean = 0.0276, sd = sd)
a=pnorm(0.0275, mean = 0.0276, sd = sd)
return(abs(b-a-0.98))
}
sd=optim(1,target)
sd$par
This |
45,712 | Calculate standard deviation given mean and percentage | It has to be solved numerically. Here is a solution in R using a simple root finding algorithm. We simply solve the equation
$$
F_{\mu,\sigma}(b) - F_{\mu,\sigma}(a) - p = 0
$$
where $F_{\mu,\sigma}(\cdot)$ denotes the cumulative distribution function of the normal distribution with mean $\mu$ and standard deviation $\... | Calculate standard deviation given mean and percentage | It has to be solved numerically. Here is a solution in R using a simple root finding algorithm. We simply solve the equation
$$
F_{\mu,\sigma}(b) - F_{\mu,\sigma}(a) - p = 0
$$
where $F_{\mu,\sigma}(\ | Calculate standard deviation given mean and percentage
It has to be solved numerically. Here is a solution in R using a simple root finding algorithm. We simply solve the equation
$$
F_{\mu,\sigma}(b) - F_{\mu,\sigma}(a) - p = 0
$$
where $F_{\mu,\sigma}(\cdot)$ denotes the cumulative distribution function of the normal... | Calculate standard deviation given mean and percentage
It has to be solved numerically. Here is a solution in R using a simple root finding algorithm. We simply solve the equation
$$
F_{\mu,\sigma}(b) - F_{\mu,\sigma}(a) - p = 0
$$
where $F_{\mu,\sigma}(\ |
45,713 | Calculate standard deviation given mean and percentage | There is no simple way to calculate this, I believe. I'd suggest looking into numerical solutions for it.
Just to explain a bit, this is the normal distribution:
$f(x|\sigma, \mu)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
The percentage of values in the interval [a, b] is then given by:
$F(a, b)=\int... | Calculate standard deviation given mean and percentage | There is no simple way to calculate this, I believe. I'd suggest looking into numerical solutions for it.
Just to explain a bit, this is the normal distribution:
$f(x|\sigma, \mu)=\frac{1}{\sqrt{2\pi} | Calculate standard deviation given mean and percentage
There is no simple way to calculate this, I believe. I'd suggest looking into numerical solutions for it.
Just to explain a bit, this is the normal distribution:
$f(x|\sigma, \mu)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
The percentage of values... | Calculate standard deviation given mean and percentage
There is no simple way to calculate this, I believe. I'd suggest looking into numerical solutions for it.
Just to explain a bit, this is the normal distribution:
$f(x|\sigma, \mu)=\frac{1}{\sqrt{2\pi} |
45,714 | Calculate standard deviation given mean and percentage | I entered on WolframAlpha:
integral_0.0275^0.0278 (1/sqrt(2 π))/a exp(-(((x - 0.0276)/sqrt(2))/a)^2) dx = 0.98
Which got me
0.5 erf(0.0000707107/a) + 0.5 erf(0.000141421/a) = 0.98
Solving which assuming a is real.
gives a = 0.0000486817
Multiple expansions of error function are listed on its Wikipedia page and on ... | Calculate standard deviation given mean and percentage | I entered on WolframAlpha:
integral_0.0275^0.0278 (1/sqrt(2 π))/a exp(-(((x - 0.0276)/sqrt(2))/a)^2) dx = 0.98
Which got me
0.5 erf(0.0000707107/a) + 0.5 erf(0.000141421/a) = 0.98
Solving which a | Calculate standard deviation given mean and percentage
I entered on WolframAlpha:
integral_0.0275^0.0278 (1/sqrt(2 π))/a exp(-(((x - 0.0276)/sqrt(2))/a)^2) dx = 0.98
Which got me
0.5 erf(0.0000707107/a) + 0.5 erf(0.000141421/a) = 0.98
Solving which assuming a is real.
gives a = 0.0000486817
Multiple expansions of ... | Calculate standard deviation given mean and percentage
I entered on WolframAlpha:
integral_0.0275^0.0278 (1/sqrt(2 π))/a exp(-(((x - 0.0276)/sqrt(2))/a)^2) dx = 0.98
Which got me
0.5 erf(0.0000707107/a) + 0.5 erf(0.000141421/a) = 0.98
Solving which a |
45,715 | Inconsistent estimators in case of endogeneity | The mistake is, if $\text{plim}N^{-1}X'u = 0$, then we only need to assume $\text{plim}N^{-1}(X'X)^{-1}$ exists to prove consistency. But when $\text{plim}N^{-1}X'u \ne 0$, as is in the case when $x_k$ is endogenous (suppose $x_k$ is the last variable), we will need to check $\text{plim}N^{-1}(X'X)^{-1}$ to see if othe... | Inconsistent estimators in case of endogeneity | The mistake is, if $\text{plim}N^{-1}X'u = 0$, then we only need to assume $\text{plim}N^{-1}(X'X)^{-1}$ exists to prove consistency. But when $\text{plim}N^{-1}X'u \ne 0$, as is in the case when $x_k | Inconsistent estimators in case of endogeneity
The mistake is, if $\text{plim}N^{-1}X'u = 0$, then we only need to assume $\text{plim}N^{-1}(X'X)^{-1}$ exists to prove consistency. But when $\text{plim}N^{-1}X'u \ne 0$, as is in the case when $x_k$ is endogenous (suppose $x_k$ is the last variable), we will need to che... | Inconsistent estimators in case of endogeneity
The mistake is, if $\text{plim}N^{-1}X'u = 0$, then we only need to assume $\text{plim}N^{-1}(X'X)^{-1}$ exists to prove consistency. But when $\text{plim}N^{-1}X'u \ne 0$, as is in the case when $x_k |
45,716 | Inconsistent estimators in case of endogeneity | Elaborating on @Paul 's answer, let $\text{plim}(X'X)^{-1} = W$ with typical element $[w^{ij}]$. Then, the expression for, say the $\hat \beta_0$ element of the vector $\hat \beta$ is, when only $X_k$ is correlated with $u$,
$$\text{plim} \hat \beta_0 = \beta_0 + w^{1k}\cdot E(X_ku)$$
So only if $w^{1k} = 0$, correlati... | Inconsistent estimators in case of endogeneity | Elaborating on @Paul 's answer, let $\text{plim}(X'X)^{-1} = W$ with typical element $[w^{ij}]$. Then, the expression for, say the $\hat \beta_0$ element of the vector $\hat \beta$ is, when only $X_k$ | Inconsistent estimators in case of endogeneity
Elaborating on @Paul 's answer, let $\text{plim}(X'X)^{-1} = W$ with typical element $[w^{ij}]$. Then, the expression for, say the $\hat \beta_0$ element of the vector $\hat \beta$ is, when only $X_k$ is correlated with $u$,
$$\text{plim} \hat \beta_0 = \beta_0 + w^{1k}\cd... | Inconsistent estimators in case of endogeneity
Elaborating on @Paul 's answer, let $\text{plim}(X'X)^{-1} = W$ with typical element $[w^{ij}]$. Then, the expression for, say the $\hat \beta_0$ element of the vector $\hat \beta$ is, when only $X_k$ |
45,717 | Inconsistent estimators in case of endogeneity | Let's abstract from technicalities on Law of Large number and assume that for each $x_j$ for $j=1,...,k$ we have that $$\frac 1n\sum_{i=1}^{n} x_{i,j}u_i\to E(x_ju)$$ Now assme that for all but index $k$ we have that $E(x_ju)=0$. Let $E(x_ku)=\delta\neq0$. Let $X'X$ also have full rank, say you have an invertible limit... | Inconsistent estimators in case of endogeneity | Let's abstract from technicalities on Law of Large number and assume that for each $x_j$ for $j=1,...,k$ we have that $$\frac 1n\sum_{i=1}^{n} x_{i,j}u_i\to E(x_ju)$$ Now assme that for all but index | Inconsistent estimators in case of endogeneity
Let's abstract from technicalities on Law of Large number and assume that for each $x_j$ for $j=1,...,k$ we have that $$\frac 1n\sum_{i=1}^{n} x_{i,j}u_i\to E(x_ju)$$ Now assme that for all but index $k$ we have that $E(x_ju)=0$. Let $E(x_ku)=\delta\neq0$. Let $X'X$ also h... | Inconsistent estimators in case of endogeneity
Let's abstract from technicalities on Law of Large number and assume that for each $x_j$ for $j=1,...,k$ we have that $$\frac 1n\sum_{i=1}^{n} x_{i,j}u_i\to E(x_ju)$$ Now assme that for all but index |
45,718 | Can log-transformation, then z-scoring make a positive mean difference negative | While ordering of observations (and hence ordering of quantiles) are preserved through monotonic transformations -- so if medians or upper quartiles are ordered in one direction before taking logs and standardizing by a common location and scale they will be in that same direction afterward.
However, averages are not c... | Can log-transformation, then z-scoring make a positive mean difference negative | While ordering of observations (and hence ordering of quantiles) are preserved through monotonic transformations -- so if medians or upper quartiles are ordered in one direction before taking logs and | Can log-transformation, then z-scoring make a positive mean difference negative
While ordering of observations (and hence ordering of quantiles) are preserved through monotonic transformations -- so if medians or upper quartiles are ordered in one direction before taking logs and standardizing by a common location and ... | Can log-transformation, then z-scoring make a positive mean difference negative
While ordering of observations (and hence ordering of quantiles) are preserved through monotonic transformations -- so if medians or upper quartiles are ordered in one direction before taking logs and |
45,719 | Testing Measurement Invariance with Robust Estimators Yields Bizarre (Improved) Model Fit Indexes | The source of your problem is the 'robust' estimation of standard errors using the robust Satorra-Bentler Chi-square statistic. When testing for measurement invariance, we compare less constrained (configural invariance) to more constrained (metric or scalar invariance) models. The comparison that is usually applied is... | Testing Measurement Invariance with Robust Estimators Yields Bizarre (Improved) Model Fit Indexes | The source of your problem is the 'robust' estimation of standard errors using the robust Satorra-Bentler Chi-square statistic. When testing for measurement invariance, we compare less constrained (co | Testing Measurement Invariance with Robust Estimators Yields Bizarre (Improved) Model Fit Indexes
The source of your problem is the 'robust' estimation of standard errors using the robust Satorra-Bentler Chi-square statistic. When testing for measurement invariance, we compare less constrained (configural invariance) t... | Testing Measurement Invariance with Robust Estimators Yields Bizarre (Improved) Model Fit Indexes
The source of your problem is the 'robust' estimation of standard errors using the robust Satorra-Bentler Chi-square statistic. When testing for measurement invariance, we compare less constrained (co |
45,720 | Gaussian covariance matrix basic concept | Why they represent covariance with 4 separated matrices?
I emphasize this each notion as matrix. what happen if each notion become a matrix
In this case the vectors ${\boldsymbol Y}$ and ${\boldsymbol \mu}$ are really block vectors. In the case of an $n$-dimensional ${\boldsymbol Y}$ vector we could expand it as fo... | Gaussian covariance matrix basic concept | Why they represent covariance with 4 separated matrices?
I emphasize this each notion as matrix. what happen if each notion become a matrix
In this case the vectors ${\boldsymbol Y}$ and ${\boldsy | Gaussian covariance matrix basic concept
Why they represent covariance with 4 separated matrices?
I emphasize this each notion as matrix. what happen if each notion become a matrix
In this case the vectors ${\boldsymbol Y}$ and ${\boldsymbol \mu}$ are really block vectors. In the case of an $n$-dimensional ${\bolds... | Gaussian covariance matrix basic concept
Why they represent covariance with 4 separated matrices?
I emphasize this each notion as matrix. what happen if each notion become a matrix
In this case the vectors ${\boldsymbol Y}$ and ${\boldsy |
45,721 | Why do we use expectation in reinforcement learning? | Because life is uncertain. We don't know what the future might hold.
If we knew the future, we'd calculate the reward we'll receive for each possible action, and choose the best one -- but alas, no one can tell the future. Therefore, we can't be sure what the reward of each possible action will be, and we can't be su... | Why do we use expectation in reinforcement learning? | Because life is uncertain. We don't know what the future might hold.
If we knew the future, we'd calculate the reward we'll receive for each possible action, and choose the best one -- but alas, no o | Why do we use expectation in reinforcement learning?
Because life is uncertain. We don't know what the future might hold.
If we knew the future, we'd calculate the reward we'll receive for each possible action, and choose the best one -- but alas, no one can tell the future. Therefore, we can't be sure what the rewar... | Why do we use expectation in reinforcement learning?
Because life is uncertain. We don't know what the future might hold.
If we knew the future, we'd calculate the reward we'll receive for each possible action, and choose the best one -- but alas, no o |
45,722 | Why do we use expectation in reinforcement learning? | You can view this as a consequence of how we define "optimal" in most reinforcement learning applications: An optimal policy is that which maximizes expected discounted reward in a Markov decision process. MDPs are RL's core and longest-studied problem, making them a natural starting point.
Though natural, this definit... | Why do we use expectation in reinforcement learning? | You can view this as a consequence of how we define "optimal" in most reinforcement learning applications: An optimal policy is that which maximizes expected discounted reward in a Markov decision pro | Why do we use expectation in reinforcement learning?
You can view this as a consequence of how we define "optimal" in most reinforcement learning applications: An optimal policy is that which maximizes expected discounted reward in a Markov decision process. MDPs are RL's core and longest-studied problem, making them a... | Why do we use expectation in reinforcement learning?
You can view this as a consequence of how we define "optimal" in most reinforcement learning applications: An optimal policy is that which maximizes expected discounted reward in a Markov decision pro |
45,723 | Why do we use expectation in reinforcement learning? | We use expectations because we want to optimize the long-term performance of our algorithms. This is the weighted sum of all possible outcomes multiplied by their probabilities — the expected reward. | Why do we use expectation in reinforcement learning? | We use expectations because we want to optimize the long-term performance of our algorithms. This is the weighted sum of all possible outcomes multiplied by their probabilities — the expected reward. | Why do we use expectation in reinforcement learning?
We use expectations because we want to optimize the long-term performance of our algorithms. This is the weighted sum of all possible outcomes multiplied by their probabilities — the expected reward. | Why do we use expectation in reinforcement learning?
We use expectations because we want to optimize the long-term performance of our algorithms. This is the weighted sum of all possible outcomes multiplied by their probabilities — the expected reward. |
45,724 | Sampling from an Inverse Gamma distribution | This discrepancy arises because there are two different parameterizations of the Gamma distribution and each relate differently to the Inverse Gamma distribution.
On Wikipedia, the two parameterizations for the Gamma distribution are differentiated by using $(k,\theta)$ and $(\alpha, \beta)$.
$$\text{If } X \sim \text... | Sampling from an Inverse Gamma distribution | This discrepancy arises because there are two different parameterizations of the Gamma distribution and each relate differently to the Inverse Gamma distribution.
On Wikipedia, the two parameterizatio | Sampling from an Inverse Gamma distribution
This discrepancy arises because there are two different parameterizations of the Gamma distribution and each relate differently to the Inverse Gamma distribution.
On Wikipedia, the two parameterizations for the Gamma distribution are differentiated by using $(k,\theta)$ and $... | Sampling from an Inverse Gamma distribution
This discrepancy arises because there are two different parameterizations of the Gamma distribution and each relate differently to the Inverse Gamma distribution.
On Wikipedia, the two parameterizatio |
45,725 | How to select a range of dates in R? [closed] | Assuming you are using the Date class: if you are using a data.frame:
myData[myData$myDate >= "1970-01-01" & myData$myDate <= "2016-06-27",]
And if you are using a data.table:
myData[myDate >= "1970-01-01" & myDate <= "2016-06-27"] | How to select a range of dates in R? [closed] | Assuming you are using the Date class: if you are using a data.frame:
myData[myData$myDate >= "1970-01-01" & myData$myDate <= "2016-06-27",]
And if you are using a data.table:
myData[myDate >= "1970- | How to select a range of dates in R? [closed]
Assuming you are using the Date class: if you are using a data.frame:
myData[myData$myDate >= "1970-01-01" & myData$myDate <= "2016-06-27",]
And if you are using a data.table:
myData[myDate >= "1970-01-01" & myDate <= "2016-06-27"] | How to select a range of dates in R? [closed]
Assuming you are using the Date class: if you are using a data.frame:
myData[myData$myDate >= "1970-01-01" & myData$myDate <= "2016-06-27",]
And if you are using a data.table:
myData[myDate >= "1970- |
45,726 | What selection criteria to use and why? (AIC, RMSE, MAPE) - All possible model selection for time series forecasting | The short answer is that there is no silver bullet. The few selection criteria you have named are also by far not all there are (as I am sure you are aware of).
So let us start with the ones that are most commonly used for time series applications: the Bayesian-Schwarz Criterion (BIC), the Akaike Criterion (AIC), and ... | What selection criteria to use and why? (AIC, RMSE, MAPE) - All possible model selection for time se | The short answer is that there is no silver bullet. The few selection criteria you have named are also by far not all there are (as I am sure you are aware of).
So let us start with the ones that are | What selection criteria to use and why? (AIC, RMSE, MAPE) - All possible model selection for time series forecasting
The short answer is that there is no silver bullet. The few selection criteria you have named are also by far not all there are (as I am sure you are aware of).
So let us start with the ones that are mo... | What selection criteria to use and why? (AIC, RMSE, MAPE) - All possible model selection for time se
The short answer is that there is no silver bullet. The few selection criteria you have named are also by far not all there are (as I am sure you are aware of).
So let us start with the ones that are |
45,727 | What selection criteria to use and why? (AIC, RMSE, MAPE) - All possible model selection for time series forecasting | Whether it makes sense to create a short-list of just n models or to select a single model depends a lot on how much the data favors the "best" (according to a chosen criterion) model.
If a lot of models are "close together", then all of them are plausible models and selecting a single one or a subset of these when yo... | What selection criteria to use and why? (AIC, RMSE, MAPE) - All possible model selection for time se | Whether it makes sense to create a short-list of just n models or to select a single model depends a lot on how much the data favors the "best" (according to a chosen criterion) model.
If a lot of mo | What selection criteria to use and why? (AIC, RMSE, MAPE) - All possible model selection for time series forecasting
Whether it makes sense to create a short-list of just n models or to select a single model depends a lot on how much the data favors the "best" (according to a chosen criterion) model.
If a lot of model... | What selection criteria to use and why? (AIC, RMSE, MAPE) - All possible model selection for time se
Whether it makes sense to create a short-list of just n models or to select a single model depends a lot on how much the data favors the "best" (according to a chosen criterion) model.
If a lot of mo |
45,728 | When should the Pasting ensemble method be used instead of Bagging? | I am not an expert on the subject, but I think I have a sufficient answer:
Since pasting is without replacement, each subset of the sample can be used once at most, which means that you need a big dataset for it to work. As a matter of fact, pasting was originally designed for large data-sets, when computing power is l... | When should the Pasting ensemble method be used instead of Bagging? | I am not an expert on the subject, but I think I have a sufficient answer:
Since pasting is without replacement, each subset of the sample can be used once at most, which means that you need a big dat | When should the Pasting ensemble method be used instead of Bagging?
I am not an expert on the subject, but I think I have a sufficient answer:
Since pasting is without replacement, each subset of the sample can be used once at most, which means that you need a big dataset for it to work. As a matter of fact, pasting wa... | When should the Pasting ensemble method be used instead of Bagging?
I am not an expert on the subject, but I think I have a sufficient answer:
Since pasting is without replacement, each subset of the sample can be used once at most, which means that you need a big dat |
45,729 | When should the Pasting ensemble method be used instead of Bagging? | Bagging is to use the same training for every predictor, but to train them on different random subsets of the training set. When sampling is performed with replacement, this method is called bagging (short for bootstrap aggregating). When sampling is performed without replacement, it is called pasting.
In other words, ... | When should the Pasting ensemble method be used instead of Bagging? | Bagging is to use the same training for every predictor, but to train them on different random subsets of the training set. When sampling is performed with replacement, this method is called bagging ( | When should the Pasting ensemble method be used instead of Bagging?
Bagging is to use the same training for every predictor, but to train them on different random subsets of the training set. When sampling is performed with replacement, this method is called bagging (short for bootstrap aggregating). When sampling is p... | When should the Pasting ensemble method be used instead of Bagging?
Bagging is to use the same training for every predictor, but to train them on different random subsets of the training set. When sampling is performed with replacement, this method is called bagging ( |
45,730 | Does the sign of eigenvectors matter? [duplicate] | No, there is no difference. Notice that if $v$ is an eigenvector to $A$ with eigenvalue $\lambda$ and $\alpha$ is a scalar, then
$$ A \alpha v = \alpha A v = \lambda \alpha v $$
and thus $\alpha v$ is also an eigenvector with eigenvalue $\lambda$. Since $\alpha$ is any scalar, if you let $\alpha = -1$ then you see that... | Does the sign of eigenvectors matter? [duplicate] | No, there is no difference. Notice that if $v$ is an eigenvector to $A$ with eigenvalue $\lambda$ and $\alpha$ is a scalar, then
$$ A \alpha v = \alpha A v = \lambda \alpha v $$
and thus $\alpha v$ is | Does the sign of eigenvectors matter? [duplicate]
No, there is no difference. Notice that if $v$ is an eigenvector to $A$ with eigenvalue $\lambda$ and $\alpha$ is a scalar, then
$$ A \alpha v = \alpha A v = \lambda \alpha v $$
and thus $\alpha v$ is also an eigenvector with eigenvalue $\lambda$. Since $\alpha$ is any ... | Does the sign of eigenvectors matter? [duplicate]
No, there is no difference. Notice that if $v$ is an eigenvector to $A$ with eigenvalue $\lambda$ and $\alpha$ is a scalar, then
$$ A \alpha v = \alpha A v = \lambda \alpha v $$
and thus $\alpha v$ is |
45,731 | Verifying that a random generator outputs a uniform distribution | UPDATE 2: striked wrong points, and replaced some by stuff I think are correct.
UPDATE 1: added direct analysis of both your uniformity tests, and kept my old answer as a proposed superior uniformity test that I suggest.
Summary
Both you and your student have offered uniformity tests that work.
Both of your tests don'... | Verifying that a random generator outputs a uniform distribution | UPDATE 2: striked wrong points, and replaced some by stuff I think are correct.
UPDATE 1: added direct analysis of both your uniformity tests, and kept my old answer as a proposed superior uniformity | Verifying that a random generator outputs a uniform distribution
UPDATE 2: striked wrong points, and replaced some by stuff I think are correct.
UPDATE 1: added direct analysis of both your uniformity tests, and kept my old answer as a proposed superior uniformity test that I suggest.
Summary
Both you and your student... | Verifying that a random generator outputs a uniform distribution
UPDATE 2: striked wrong points, and replaced some by stuff I think are correct.
UPDATE 1: added direct analysis of both your uniformity tests, and kept my old answer as a proposed superior uniformity |
45,732 | Verifying that a random generator outputs a uniform distribution | Note that we cannot conclude on the basis of only a sample that data is drawn from a uniform, only that it is consistent with having come from a uniform (sufficiently small deviations will be undetectable at any given sample size), or that it is not consistent.x
Since the request is for a test rather than simply a meas... | Verifying that a random generator outputs a uniform distribution | Note that we cannot conclude on the basis of only a sample that data is drawn from a uniform, only that it is consistent with having come from a uniform (sufficiently small deviations will be undetect | Verifying that a random generator outputs a uniform distribution
Note that we cannot conclude on the basis of only a sample that data is drawn from a uniform, only that it is consistent with having come from a uniform (sufficiently small deviations will be undetectable at any given sample size), or that it is not consi... | Verifying that a random generator outputs a uniform distribution
Note that we cannot conclude on the basis of only a sample that data is drawn from a uniform, only that it is consistent with having come from a uniform (sufficiently small deviations will be undetect |
45,733 | Verifying that a random generator outputs a uniform distribution | It's like saying $a$ is close to be $b$ when $a-b$ is 0 or when $\frac{a}b$ is 1. I don't see why one should be better than the other. | Verifying that a random generator outputs a uniform distribution | It's like saying $a$ is close to be $b$ when $a-b$ is 0 or when $\frac{a}b$ is 1. I don't see why one should be better than the other. | Verifying that a random generator outputs a uniform distribution
It's like saying $a$ is close to be $b$ when $a-b$ is 0 or when $\frac{a}b$ is 1. I don't see why one should be better than the other. | Verifying that a random generator outputs a uniform distribution
It's like saying $a$ is close to be $b$ when $a-b$ is 0 or when $\frac{a}b$ is 1. I don't see why one should be better than the other. |
45,734 | Verifying that a random generator outputs a uniform distribution | A test that is actually used in testing the R random number generators is based on the cumulative distribution function (rather than histogram or density), with test statistic
$$d=\max_t\left\{\left|\frac{\sum_i (X_i\leq t)}{N}-F(t)\right|\right\}$$
This test takes advantage of Massart's inequality,
$$P(\sup | F_n - F ... | Verifying that a random generator outputs a uniform distribution | A test that is actually used in testing the R random number generators is based on the cumulative distribution function (rather than histogram or density), with test statistic
$$d=\max_t\left\{\left|\ | Verifying that a random generator outputs a uniform distribution
A test that is actually used in testing the R random number generators is based on the cumulative distribution function (rather than histogram or density), with test statistic
$$d=\max_t\left\{\left|\frac{\sum_i (X_i\leq t)}{N}-F(t)\right|\right\}$$
This ... | Verifying that a random generator outputs a uniform distribution
A test that is actually used in testing the R random number generators is based on the cumulative distribution function (rather than histogram or density), with test statistic
$$d=\max_t\left\{\left|\ |
45,735 | What can be inferred from this residual plot? | The person who produced that plot made a mistake.
Here's why. The setting is ordinary least squares regression (including an intercept term), which is where responses $y_i$ are estimated as linear combinations of regressor variables $x_{ij}$ in the form
$$\hat y_i = \hat\beta_0 + \hat \beta_1 x_{i1} + \hat\beta_2 x_{i... | What can be inferred from this residual plot? | The person who produced that plot made a mistake.
Here's why. The setting is ordinary least squares regression (including an intercept term), which is where responses $y_i$ are estimated as linear co | What can be inferred from this residual plot?
The person who produced that plot made a mistake.
Here's why. The setting is ordinary least squares regression (including an intercept term), which is where responses $y_i$ are estimated as linear combinations of regressor variables $x_{ij}$ in the form
$$\hat y_i = \hat\b... | What can be inferred from this residual plot?
The person who produced that plot made a mistake.
Here's why. The setting is ordinary least squares regression (including an intercept term), which is where responses $y_i$ are estimated as linear co |
45,736 | Downsides of inverse Wishart prior in hierarchical models | Here are some relevant resources (full disclosure: the first link is to a paper of mine):
http://newprairiepress.org/agstatconference/2014/proceedings/8
http://www.themattsimpson.com/2012/08/20/prior-distributions-for-covariance-matrices-the-scaled-inverse-wishart-prior/
http://andrewgelman.com/2012/08/29/more-on-scal... | Downsides of inverse Wishart prior in hierarchical models | Here are some relevant resources (full disclosure: the first link is to a paper of mine):
http://newprairiepress.org/agstatconference/2014/proceedings/8
http://www.themattsimpson.com/2012/08/20/prior | Downsides of inverse Wishart prior in hierarchical models
Here are some relevant resources (full disclosure: the first link is to a paper of mine):
http://newprairiepress.org/agstatconference/2014/proceedings/8
http://www.themattsimpson.com/2012/08/20/prior-distributions-for-covariance-matrices-the-scaled-inverse-wish... | Downsides of inverse Wishart prior in hierarchical models
Here are some relevant resources (full disclosure: the first link is to a paper of mine):
http://newprairiepress.org/agstatconference/2014/proceedings/8
http://www.themattsimpson.com/2012/08/20/prior |
45,737 | Maximum Likelihood Formulation for Linear Regression | In ordinary least squares regression the goal is to model the condition expectation;
$$
E[y_i|x_i] = x_i'\beta
$$
$y_i$ and $x_i$ are referred to as the dependent and independent variables respectively because we are literally conditioning $y_i$ on $x_i$.
Ordinary least squares is equivalent to maximum likelihood wh... | Maximum Likelihood Formulation for Linear Regression | In ordinary least squares regression the goal is to model the condition expectation;
$$
E[y_i|x_i] = x_i'\beta
$$
$y_i$ and $x_i$ are referred to as the dependent and independent variables respectivel | Maximum Likelihood Formulation for Linear Regression
In ordinary least squares regression the goal is to model the condition expectation;
$$
E[y_i|x_i] = x_i'\beta
$$
$y_i$ and $x_i$ are referred to as the dependent and independent variables respectively because we are literally conditioning $y_i$ on $x_i$.
Ordinary... | Maximum Likelihood Formulation for Linear Regression
In ordinary least squares regression the goal is to model the condition expectation;
$$
E[y_i|x_i] = x_i'\beta
$$
$y_i$ and $x_i$ are referred to as the dependent and independent variables respectivel |
45,738 | $p$-value for non-standard asymptotics | Assuming you know the $\lambda_i$, simulation is feasible.
Consider
library(MASS)
k <- 3
lambda <- c(.2,.3,.4) # pick your lambdas here
reps <- 100000
distr <- rep(NA,reps)
for (i in 1:reps){
distr[i] <- sum(lambda*rchisq(k,1))
}
distr <- sort(distr)
teststat <- 2 # pick your teststat here
pvalue <- which.min(ab... | $p$-value for non-standard asymptotics | Assuming you know the $\lambda_i$, simulation is feasible.
Consider
library(MASS)
k <- 3
lambda <- c(.2,.3,.4) # pick your lambdas here
reps <- 100000
distr <- rep(NA,reps)
for (i in 1:reps){
di | $p$-value for non-standard asymptotics
Assuming you know the $\lambda_i$, simulation is feasible.
Consider
library(MASS)
k <- 3
lambda <- c(.2,.3,.4) # pick your lambdas here
reps <- 100000
distr <- rep(NA,reps)
for (i in 1:reps){
distr[i] <- sum(lambda*rchisq(k,1))
}
distr <- sort(distr)
teststat <- 2 # pick yo... | $p$-value for non-standard asymptotics
Assuming you know the $\lambda_i$, simulation is feasible.
Consider
library(MASS)
k <- 3
lambda <- c(.2,.3,.4) # pick your lambdas here
reps <- 100000
distr <- rep(NA,reps)
for (i in 1:reps){
di |
45,739 | $p$-value for non-standard asymptotics | There are two useful approximations and at least three computations that would be exact with infinite precision arithmetic.
Let's call the distribution $Q(\lambda)$. And write $\bar\lambda$ for the mean of $\lambda$ and $\tau$ for the mean of $\lambda^2$.
The two approximations are implemented in pchisqsum in the R su... | $p$-value for non-standard asymptotics | There are two useful approximations and at least three computations that would be exact with infinite precision arithmetic.
Let's call the distribution $Q(\lambda)$. And write $\bar\lambda$ for the m | $p$-value for non-standard asymptotics
There are two useful approximations and at least three computations that would be exact with infinite precision arithmetic.
Let's call the distribution $Q(\lambda)$. And write $\bar\lambda$ for the mean of $\lambda$ and $\tau$ for the mean of $\lambda^2$.
The two approximations a... | $p$-value for non-standard asymptotics
There are two useful approximations and at least three computations that would be exact with infinite precision arithmetic.
Let's call the distribution $Q(\lambda)$. And write $\bar\lambda$ for the m |
45,740 | Why does a regression tree not split based on variance? | First off, let me just say that there are plenty of metrics that could be used to determine a split in the Regression Tree (that's an altogether different question) but the criterion you mentioned--minimizing the sum of squares--is definitely the most popular.
With respect to your second question (i.e. why we don't div... | Why does a regression tree not split based on variance? | First off, let me just say that there are plenty of metrics that could be used to determine a split in the Regression Tree (that's an altogether different question) but the criterion you mentioned--mi | Why does a regression tree not split based on variance?
First off, let me just say that there are plenty of metrics that could be used to determine a split in the Regression Tree (that's an altogether different question) but the criterion you mentioned--minimizing the sum of squares--is definitely the most popular.
Wit... | Why does a regression tree not split based on variance?
First off, let me just say that there are plenty of metrics that could be used to determine a split in the Regression Tree (that's an altogether different question) but the criterion you mentioned--mi |
45,741 | Why does a regression tree not split based on variance? | So the best split offers the best gain. If the loss function is sum of squares the gain function could look like:
$\sigma^2 = \frac{\sum{(y_i-\overline{y})^2}}{N} = SS/N$, (SS is sum of sqaures)
$gain = \frac{SS_{parent}}{n_{parent}}
- w_l \frac{SS_{left}}{n_{left}}
- w_r \frac{SS_{right}}{n_{right}}$
$w_l=n_{left} , w... | Why does a regression tree not split based on variance? | So the best split offers the best gain. If the loss function is sum of squares the gain function could look like:
$\sigma^2 = \frac{\sum{(y_i-\overline{y})^2}}{N} = SS/N$, (SS is sum of sqaures)
$gain | Why does a regression tree not split based on variance?
So the best split offers the best gain. If the loss function is sum of squares the gain function could look like:
$\sigma^2 = \frac{\sum{(y_i-\overline{y})^2}}{N} = SS/N$, (SS is sum of sqaures)
$gain = \frac{SS_{parent}}{n_{parent}}
- w_l \frac{SS_{left}}{n_{left... | Why does a regression tree not split based on variance?
So the best split offers the best gain. If the loss function is sum of squares the gain function could look like:
$\sigma^2 = \frac{\sum{(y_i-\overline{y})^2}}{N} = SS/N$, (SS is sum of sqaures)
$gain |
45,742 | Why does a regression tree not split based on variance? | In order to compare the behavior of the different splitting formulas, let's consider a simple example:
We have $N$ predictors with value $y=1$ in the left, $N$ predictors with value $y=-1$ in the right, and a single point at the right boundary with value $y^\ast$.
There are two reasonable split points in this scenari... | Why does a regression tree not split based on variance? | In order to compare the behavior of the different splitting formulas, let's consider a simple example:
We have $N$ predictors with value $y=1$ in the left, $N$ predictors with value $y=-1$ in the righ | Why does a regression tree not split based on variance?
In order to compare the behavior of the different splitting formulas, let's consider a simple example:
We have $N$ predictors with value $y=1$ in the left, $N$ predictors with value $y=-1$ in the right, and a single point at the right boundary with value $y^\ast$.... | Why does a regression tree not split based on variance?
In order to compare the behavior of the different splitting formulas, let's consider a simple example:
We have $N$ predictors with value $y=1$ in the left, $N$ predictors with value $y=-1$ in the righ |
45,743 | Why does a regression tree not split based on variance? | It's an old question but the reason I can think of is as follows:
We don't divide each term by the number of points in each region because while performing the split, we are not interested in average deviance instead we are interested in the absolute deviance; thus, a larger set of observations will usually have a larg... | Why does a regression tree not split based on variance? | It's an old question but the reason I can think of is as follows:
We don't divide each term by the number of points in each region because while performing the split, we are not interested in average | Why does a regression tree not split based on variance?
It's an old question but the reason I can think of is as follows:
We don't divide each term by the number of points in each region because while performing the split, we are not interested in average deviance instead we are interested in the absolute deviance; thu... | Why does a regression tree not split based on variance?
It's an old question but the reason I can think of is as follows:
We don't divide each term by the number of points in each region because while performing the split, we are not interested in average |
45,744 | Adding the probabilities of two events when one is a subset of the other | The inclusion-exclusion principle states
$P(A\cup B)=P(A)+P(B)-P(A\cap B).$ Therefore, you know that
$P(A\cup B)+P(A\cap B)=P(A)+P(B).$ Without further information/assumptions, it is not possible to uniquely identify $P(A)$ or $P(B).$
You use the word "conditional" in your title, but it's important to note that this i... | Adding the probabilities of two events when one is a subset of the other | The inclusion-exclusion principle states
$P(A\cup B)=P(A)+P(B)-P(A\cap B).$ Therefore, you know that
$P(A\cup B)+P(A\cap B)=P(A)+P(B).$ Without further information/assumptions, it is not possible to | Adding the probabilities of two events when one is a subset of the other
The inclusion-exclusion principle states
$P(A\cup B)=P(A)+P(B)-P(A\cap B).$ Therefore, you know that
$P(A\cup B)+P(A\cap B)=P(A)+P(B).$ Without further information/assumptions, it is not possible to uniquely identify $P(A)$ or $P(B).$
You use the... | Adding the probabilities of two events when one is a subset of the other
The inclusion-exclusion principle states
$P(A\cup B)=P(A)+P(B)-P(A\cap B).$ Therefore, you know that
$P(A\cup B)+P(A\cap B)=P(A)+P(B).$ Without further information/assumptions, it is not possible to |
45,745 | How to choose a constant for reject sampling | Let $\pi(x) = M f(x)$, where $M$ is the normalizing constant. In many situations, only $f(x)$ is known and $M$ is unknown.
To implement rejection sampling, you want $c$ such that, for all $x$,
$$\dfrac{\pi(x)}{h(x)} \leq c. $$
Then for all $x$,
$$\dfrac{f(x)}{h(x)} \leq \dfrac{c}{M} := c'. $$
You don't know $c$ or $M$,... | How to choose a constant for reject sampling | Let $\pi(x) = M f(x)$, where $M$ is the normalizing constant. In many situations, only $f(x)$ is known and $M$ is unknown.
To implement rejection sampling, you want $c$ such that, for all $x$,
$$\dfra | How to choose a constant for reject sampling
Let $\pi(x) = M f(x)$, where $M$ is the normalizing constant. In many situations, only $f(x)$ is known and $M$ is unknown.
To implement rejection sampling, you want $c$ such that, for all $x$,
$$\dfrac{\pi(x)}{h(x)} \leq c. $$
Then for all $x$,
$$\dfrac{f(x)}{h(x)} \leq \dfr... | How to choose a constant for reject sampling
Let $\pi(x) = M f(x)$, where $M$ is the normalizing constant. In many situations, only $f(x)$ is known and $M$ is unknown.
To implement rejection sampling, you want $c$ such that, for all $x$,
$$\dfra |
45,746 | How to choose a constant for reject sampling | More generally a principle to choose $M$ is
$$\inf_{\theta\in\Theta}\sup_{x\in\mathbb{R}}\frac{f(x)}{g_\theta(x)}$$
where $f$ is the normalized target and $g_\theta$ is the proposal density.
As an example, for $f(x)=(2\pi)^{-\frac{1}{2}}e^{-x^2/2}$ and $g(x)=\theta^{-1}e^{-\theta |x|}$
We will have $\sup_xf(x)/g_\thet... | How to choose a constant for reject sampling | More generally a principle to choose $M$ is
$$\inf_{\theta\in\Theta}\sup_{x\in\mathbb{R}}\frac{f(x)}{g_\theta(x)}$$
where $f$ is the normalized target and $g_\theta$ is the proposal density.
As an ex | How to choose a constant for reject sampling
More generally a principle to choose $M$ is
$$\inf_{\theta\in\Theta}\sup_{x\in\mathbb{R}}\frac{f(x)}{g_\theta(x)}$$
where $f$ is the normalized target and $g_\theta$ is the proposal density.
As an example, for $f(x)=(2\pi)^{-\frac{1}{2}}e^{-x^2/2}$ and $g(x)=\theta^{-1}e^{-... | How to choose a constant for reject sampling
More generally a principle to choose $M$ is
$$\inf_{\theta\in\Theta}\sup_{x\in\mathbb{R}}\frac{f(x)}{g_\theta(x)}$$
where $f$ is the normalized target and $g_\theta$ is the proposal density.
As an ex |
45,747 | How to choose a constant for reject sampling | Just as an example consider the density proportional to $e^{-(1+x^4)^\frac14},\quad-\infty<x<\infty \,.$
I definitely know what this function "looks like", since we have everything but the normalizing constant; I can draw a function proportion to the density albeit that I currently have no idea what the normalizing con... | How to choose a constant for reject sampling | Just as an example consider the density proportional to $e^{-(1+x^4)^\frac14},\quad-\infty<x<\infty \,.$
I definitely know what this function "looks like", since we have everything but the normalizing | How to choose a constant for reject sampling
Just as an example consider the density proportional to $e^{-(1+x^4)^\frac14},\quad-\infty<x<\infty \,.$
I definitely know what this function "looks like", since we have everything but the normalizing constant; I can draw a function proportion to the density albeit that I cu... | How to choose a constant for reject sampling
Just as an example consider the density proportional to $e^{-(1+x^4)^\frac14},\quad-\infty<x<\infty \,.$
I definitely know what this function "looks like", since we have everything but the normalizing |
45,748 | Difference in hypothesis testing using p-value and confidence interval | With large $N$, your test statistic will be distributed as a normal. Thus, we can refer to your test statistic as "$z$", and we can also use "$z$" to refer to the asymptotic sampling distribution. However, these are not the same $z$'s. When you form your $1-\alpha\%$ confidence interval, you need to multiply the sta... | Difference in hypothesis testing using p-value and confidence interval | With large $N$, your test statistic will be distributed as a normal. Thus, we can refer to your test statistic as "$z$", and we can also use "$z$" to refer to the asymptotic sampling distribution. H | Difference in hypothesis testing using p-value and confidence interval
With large $N$, your test statistic will be distributed as a normal. Thus, we can refer to your test statistic as "$z$", and we can also use "$z$" to refer to the asymptotic sampling distribution. However, these are not the same $z$'s. When you f... | Difference in hypothesis testing using p-value and confidence interval
With large $N$, your test statistic will be distributed as a normal. Thus, we can refer to your test statistic as "$z$", and we can also use "$z$" to refer to the asymptotic sampling distribution. H |
45,749 | Difference in hypothesis testing using p-value and confidence interval | I don't have rep to comment, but I see some flaws in that problem.
First, 6.3 sigmas are way more than 0.95 C.L. Just apply quickly Chebyshev theorem.
A confidence level says that "in N experiments, if at least 1 - p experiments you get the null hypothesis you can't reject it".
Then, a binomial distribution isn't symme... | Difference in hypothesis testing using p-value and confidence interval | I don't have rep to comment, but I see some flaws in that problem.
First, 6.3 sigmas are way more than 0.95 C.L. Just apply quickly Chebyshev theorem.
A confidence level says that "in N experiments, i | Difference in hypothesis testing using p-value and confidence interval
I don't have rep to comment, but I see some flaws in that problem.
First, 6.3 sigmas are way more than 0.95 C.L. Just apply quickly Chebyshev theorem.
A confidence level says that "in N experiments, if at least 1 - p experiments you get the null hyp... | Difference in hypothesis testing using p-value and confidence interval
I don't have rep to comment, but I see some flaws in that problem.
First, 6.3 sigmas are way more than 0.95 C.L. Just apply quickly Chebyshev theorem.
A confidence level says that "in N experiments, i |
45,750 | Difference in hypothesis testing using p-value and confidence interval | Just to grasp the concept, I've been working on a couple of plots to illustrate why the $z$ statistic calculates the standard error expected around the population or theoretical proportion (from the perspective of the Null hypothesis so to speak), and then figures out how many of these standard errors fit into the dist... | Difference in hypothesis testing using p-value and confidence interval | Just to grasp the concept, I've been working on a couple of plots to illustrate why the $z$ statistic calculates the standard error expected around the population or theoretical proportion (from the p | Difference in hypothesis testing using p-value and confidence interval
Just to grasp the concept, I've been working on a couple of plots to illustrate why the $z$ statistic calculates the standard error expected around the population or theoretical proportion (from the perspective of the Null hypothesis so to speak), a... | Difference in hypothesis testing using p-value and confidence interval
Just to grasp the concept, I've been working on a couple of plots to illustrate why the $z$ statistic calculates the standard error expected around the population or theoretical proportion (from the p |
45,751 | Why do you put all the exogenous variables into the first and second stage of 2SLS? | Technically, you are actually regressing $[X\;,\; W]$ on $[Z\;,\; W]$ so the resulting fitted values for the second stage regressors are $[\hat X\;,\; \hat W]=[\hat X \;,\;W]$.
$\hat W =W$ since the best prediction of $W$ available in the matrix $[Z\;,\; W]$ is obviously $W$ itself.
But the trivialities aside, $W$ i... | Why do you put all the exogenous variables into the first and second stage of 2SLS? | Technically, you are actually regressing $[X\;,\; W]$ on $[Z\;,\; W]$ so the resulting fitted values for the second stage regressors are $[\hat X\;,\; \hat W]=[\hat X \;,\;W]$.
$\hat W =W$ since the | Why do you put all the exogenous variables into the first and second stage of 2SLS?
Technically, you are actually regressing $[X\;,\; W]$ on $[Z\;,\; W]$ so the resulting fitted values for the second stage regressors are $[\hat X\;,\; \hat W]=[\hat X \;,\;W]$.
$\hat W =W$ since the best prediction of $W$ available in... | Why do you put all the exogenous variables into the first and second stage of 2SLS?
Technically, you are actually regressing $[X\;,\; W]$ on $[Z\;,\; W]$ so the resulting fitted values for the second stage regressors are $[\hat X\;,\; \hat W]=[\hat X \;,\;W]$.
$\hat W =W$ since the |
45,752 | Why do you put all the exogenous variables into the first and second stage of 2SLS? | This question is quite old --- however I haven't found a fully satisfactory answer online for this query, so I am adding my 2 cents. PLEASE add a comment if you notice errors!
In the case of simultaneous equations (one particular framework for thinking about endogeneity and IV), the stata documentation has a nice page ... | Why do you put all the exogenous variables into the first and second stage of 2SLS? | This question is quite old --- however I haven't found a fully satisfactory answer online for this query, so I am adding my 2 cents. PLEASE add a comment if you notice errors!
In the case of simultane | Why do you put all the exogenous variables into the first and second stage of 2SLS?
This question is quite old --- however I haven't found a fully satisfactory answer online for this query, so I am adding my 2 cents. PLEASE add a comment if you notice errors!
In the case of simultaneous equations (one particular framew... | Why do you put all the exogenous variables into the first and second stage of 2SLS?
This question is quite old --- however I haven't found a fully satisfactory answer online for this query, so I am adding my 2 cents. PLEASE add a comment if you notice errors!
In the case of simultane |
45,753 | Is there a closed form solution for L2-norm regularized linear regression (not ridge regression) | You will get the ridge regression solutions, but parametrised differently in terms of the penalty parameter $\lambda$. This holds more generally for convex loss functions.
If $L$ is a convex, differentiable function of $\beta$ let $\beta(\lambda)$ denote the unique minimiser of the strictly convex function
$$h(\beta) ... | Is there a closed form solution for L2-norm regularized linear regression (not ridge regression) | You will get the ridge regression solutions, but parametrised differently in terms of the penalty parameter $\lambda$. This holds more generally for convex loss functions.
If $L$ is a convex, differe | Is there a closed form solution for L2-norm regularized linear regression (not ridge regression)
You will get the ridge regression solutions, but parametrised differently in terms of the penalty parameter $\lambda$. This holds more generally for convex loss functions.
If $L$ is a convex, differentiable function of $\b... | Is there a closed form solution for L2-norm regularized linear regression (not ridge regression)
You will get the ridge regression solutions, but parametrised differently in terms of the penalty parameter $\lambda$. This holds more generally for convex loss functions.
If $L$ is a convex, differe |
45,754 | Notation for possible values of a random variable | There are sloppy ways and rigorous ways. The sloppy ways are shorthands, like "$X\in\{1,2,3\}$", that are either nonsensical or (in this example) just plain wrong when interpreted according to the correct conventional meanings of the symbols. (The second statement literally means $X$ is one of three specified integer... | Notation for possible values of a random variable | There are sloppy ways and rigorous ways. The sloppy ways are shorthands, like "$X\in\{1,2,3\}$", that are either nonsensical or (in this example) just plain wrong when interpreted according to the co | Notation for possible values of a random variable
There are sloppy ways and rigorous ways. The sloppy ways are shorthands, like "$X\in\{1,2,3\}$", that are either nonsensical or (in this example) just plain wrong when interpreted according to the correct conventional meanings of the symbols. (The second statement lit... | Notation for possible values of a random variable
There are sloppy ways and rigorous ways. The sloppy ways are shorthands, like "$X\in\{1,2,3\}$", that are either nonsensical or (in this example) just plain wrong when interpreted according to the co |
45,755 | Notation for possible values of a random variable | A full description of $X$ would include the pmf, yes. For example ...
$$
X \in \{1,2,3\} \\
\mathbb{P}(X=1)=\frac{1}{12} \\
\mathbb{P}(X=2)=\frac{7}{12} \\
\mathbb{P}(X=2)=\frac{1}{3}
$$
... which looks a bit clumsy, or do it in words: $X$ is a discrete random variable in $\{1,2,3\}$ with probabilities $p_1=\frac{1}{1... | Notation for possible values of a random variable | A full description of $X$ would include the pmf, yes. For example ...
$$
X \in \{1,2,3\} \\
\mathbb{P}(X=1)=\frac{1}{12} \\
\mathbb{P}(X=2)=\frac{7}{12} \\
\mathbb{P}(X=2)=\frac{1}{3}
$$
... which lo | Notation for possible values of a random variable
A full description of $X$ would include the pmf, yes. For example ...
$$
X \in \{1,2,3\} \\
\mathbb{P}(X=1)=\frac{1}{12} \\
\mathbb{P}(X=2)=\frac{7}{12} \\
\mathbb{P}(X=2)=\frac{1}{3}
$$
... which looks a bit clumsy, or do it in words: $X$ is a discrete random variable... | Notation for possible values of a random variable
A full description of $X$ would include the pmf, yes. For example ...
$$
X \in \{1,2,3\} \\
\mathbb{P}(X=1)=\frac{1}{12} \\
\mathbb{P}(X=2)=\frac{7}{12} \\
\mathbb{P}(X=2)=\frac{1}{3}
$$
... which lo |
45,756 | R: Test for correlation with a covariate? | Yes, you can use correlation with a covariate. This is called partial correlation. It produces a (partial) correlation coefficient that is normalized to the [-1, 1] range just like a regular correlation coefficient, except that the covariate is "controlled for" in the analysis -- a concept which is kind of subtle, but ... | R: Test for correlation with a covariate? | Yes, you can use correlation with a covariate. This is called partial correlation. It produces a (partial) correlation coefficient that is normalized to the [-1, 1] range just like a regular correlati | R: Test for correlation with a covariate?
Yes, you can use correlation with a covariate. This is called partial correlation. It produces a (partial) correlation coefficient that is normalized to the [-1, 1] range just like a regular correlation coefficient, except that the covariate is "controlled for" in the analysis ... | R: Test for correlation with a covariate?
Yes, you can use correlation with a covariate. This is called partial correlation. It produces a (partial) correlation coefficient that is normalized to the [-1, 1] range just like a regular correlati |
45,757 | How frequently am I going to lose this game? | We can ignore the rows and column, and just say that you have 40 positions and 40 cards that each point to a specific position.
For the first draw, you have 40 cards but only 1 card will make you lose the game (the card that points to the position you chose). This gives a probability of 39/40 to proceed to the next rou... | How frequently am I going to lose this game? | We can ignore the rows and column, and just say that you have 40 positions and 40 cards that each point to a specific position.
For the first draw, you have 40 cards but only 1 card will make you lose | How frequently am I going to lose this game?
We can ignore the rows and column, and just say that you have 40 positions and 40 cards that each point to a specific position.
For the first draw, you have 40 cards but only 1 card will make you lose the game (the card that points to the position you chose). This gives a pr... | How frequently am I going to lose this game?
We can ignore the rows and column, and just say that you have 40 positions and 40 cards that each point to a specific position.
For the first draw, you have 40 cards but only 1 card will make you lose |
45,758 | How frequently am I going to lose this game? | As @Jonas pointed out, in the whole game there's only one losing card, and it is the one you choose at first. If you ever flip that, you lose; if that is the last card you flip, you win.
Then a simpler approach to the question might be interpreting the game this way:
you name one card;
you shuffle your 40 cards deck;
... | How frequently am I going to lose this game? | As @Jonas pointed out, in the whole game there's only one losing card, and it is the one you choose at first. If you ever flip that, you lose; if that is the last card you flip, you win.
Then a simple | How frequently am I going to lose this game?
As @Jonas pointed out, in the whole game there's only one losing card, and it is the one you choose at first. If you ever flip that, you lose; if that is the last card you flip, you win.
Then a simpler approach to the question might be interpreting the game this way:
you na... | How frequently am I going to lose this game?
As @Jonas pointed out, in the whole game there's only one losing card, and it is the one you choose at first. If you ever flip that, you lose; if that is the last card you flip, you win.
Then a simple |
45,759 | Multivariate normal distributions density function: What is the value if the determinant is zero? | Return value for your function
If you want to write code for a function that computes a multivariate normal distribution, then, in case the determinant of the var-covar is zero, this return-value is undefined because the var-covar matrix is singular (the inverse does not exist).
Probably the best return value for your... | Multivariate normal distributions density function: What is the value if the determinant is zero? | Return value for your function
If you want to write code for a function that computes a multivariate normal distribution, then, in case the determinant of the var-covar is zero, this return-value is u | Multivariate normal distributions density function: What is the value if the determinant is zero?
Return value for your function
If you want to write code for a function that computes a multivariate normal distribution, then, in case the determinant of the var-covar is zero, this return-value is undefined because the v... | Multivariate normal distributions density function: What is the value if the determinant is zero?
Return value for your function
If you want to write code for a function that computes a multivariate normal distribution, then, in case the determinant of the var-covar is zero, this return-value is u |
45,760 | Multivariate normal distributions density function: What is the value if the determinant is zero? | Presume that the covariance matrix as specified is n by n, i.e., the Multivariate Normal random variable X is in n dimensions. Det(covariance matrix) = 0 if and only if the covariance matrix is singular. If the covariance matrix is singular, X does not have a density. There may exist a lower dimensional space (manif... | Multivariate normal distributions density function: What is the value if the determinant is zero? | Presume that the covariance matrix as specified is n by n, i.e., the Multivariate Normal random variable X is in n dimensions. Det(covariance matrix) = 0 if and only if the covariance matrix is singu | Multivariate normal distributions density function: What is the value if the determinant is zero?
Presume that the covariance matrix as specified is n by n, i.e., the Multivariate Normal random variable X is in n dimensions. Det(covariance matrix) = 0 if and only if the covariance matrix is singular. If the covarianc... | Multivariate normal distributions density function: What is the value if the determinant is zero?
Presume that the covariance matrix as specified is n by n, i.e., the Multivariate Normal random variable X is in n dimensions. Det(covariance matrix) = 0 if and only if the covariance matrix is singu |
45,761 | Multivariate normal distributions density function: What is the value if the determinant is zero? | Actually it can also be defined by the use of generalized inverse and "pseudo-determinant".
Solution is not unique though. | Multivariate normal distributions density function: What is the value if the determinant is zero? | Actually it can also be defined by the use of generalized inverse and "pseudo-determinant".
Solution is not unique though. | Multivariate normal distributions density function: What is the value if the determinant is zero?
Actually it can also be defined by the use of generalized inverse and "pseudo-determinant".
Solution is not unique though. | Multivariate normal distributions density function: What is the value if the determinant is zero?
Actually it can also be defined by the use of generalized inverse and "pseudo-determinant".
Solution is not unique though. |
45,762 | What is the name of this kind of visualization, with arrows showing count of different subsets? [duplicate] | It is called a Sankey Diagram. A notable example is Charles Joseph Minard's visualization of Napoleon's invasion of Russia.
(Edit) Also of interest may be these questions:
What's a good tool to create Sankey diagrams?
What is the proper name for a "river plot" visualisation | What is the name of this kind of visualization, with arrows showing count of different subsets? [dup | It is called a Sankey Diagram. A notable example is Charles Joseph Minard's visualization of Napoleon's invasion of Russia.
(Edit) Also of interest may be these questions:
What's a good tool to cr | What is the name of this kind of visualization, with arrows showing count of different subsets? [duplicate]
It is called a Sankey Diagram. A notable example is Charles Joseph Minard's visualization of Napoleon's invasion of Russia.
(Edit) Also of interest may be these questions:
What's a good tool to create Sankey ... | What is the name of this kind of visualization, with arrows showing count of different subsets? [dup
It is called a Sankey Diagram. A notable example is Charles Joseph Minard's visualization of Napoleon's invasion of Russia.
(Edit) Also of interest may be these questions:
What's a good tool to cr |
45,763 | How to prove Berkson's Fallacy? | $$
\begin{align}
P( A \mid A \cup B) &= \frac{P(A)}{P(A \cup B)} \\
&\geq P(A) .
\end{align}
$$ | How to prove Berkson's Fallacy? | $$
\begin{align}
P( A \mid A \cup B) &= \frac{P(A)}{P(A \cup B)} \\
&\geq P(A) .
\end{align}
$$ | How to prove Berkson's Fallacy?
$$
\begin{align}
P( A \mid A \cup B) &= \frac{P(A)}{P(A \cup B)} \\
&\geq P(A) .
\end{align}
$$ | How to prove Berkson's Fallacy?
$$
\begin{align}
P( A \mid A \cup B) &= \frac{P(A)}{P(A \cup B)} \\
&\geq P(A) .
\end{align}
$$ |
45,764 | How to prove Berkson's Fallacy? | I've mostly heard of it as Berkson's paradox and it refers to the spurious generation of associations when you are comparing an exposure and an outcome and you sample only individuals with either the exposure or the outcome. Suppose the population level association is:
$$ \begin{array}{c|ccc}
& D & \bar{D} & \\... | How to prove Berkson's Fallacy? | I've mostly heard of it as Berkson's paradox and it refers to the spurious generation of associations when you are comparing an exposure and an outcome and you sample only individuals with either the | How to prove Berkson's Fallacy?
I've mostly heard of it as Berkson's paradox and it refers to the spurious generation of associations when you are comparing an exposure and an outcome and you sample only individuals with either the exposure or the outcome. Suppose the population level association is:
$$ \begin{array}{c... | How to prove Berkson's Fallacy?
I've mostly heard of it as Berkson's paradox and it refers to the spurious generation of associations when you are comparing an exposure and an outcome and you sample only individuals with either the |
45,765 | How to prove Berkson's Fallacy? | Yes, it is true that $P(A|A \cup B) \ge P(A)$.
I find that helps to think of the two quantities as fractions. Then the whole thing follows from these two facts:
The numerator of both probabilities consists of the number of $A$s present in the population. This is trivial in the case of the unconditional probability $P(... | How to prove Berkson's Fallacy? | Yes, it is true that $P(A|A \cup B) \ge P(A)$.
I find that helps to think of the two quantities as fractions. Then the whole thing follows from these two facts:
The numerator of both probabilities co | How to prove Berkson's Fallacy?
Yes, it is true that $P(A|A \cup B) \ge P(A)$.
I find that helps to think of the two quantities as fractions. Then the whole thing follows from these two facts:
The numerator of both probabilities consists of the number of $A$s present in the population. This is trivial in the case of t... | How to prove Berkson's Fallacy?
Yes, it is true that $P(A|A \cup B) \ge P(A)$.
I find that helps to think of the two quantities as fractions. Then the whole thing follows from these two facts:
The numerator of both probabilities co |
45,766 | Should I adjust p-values when investigating an ANOVA interaction? | The next question I'd like to answer is "how many of the 30 subjects could reliably discriminate between the 3 headphone types"
Yes, using $p<0.05$ criterion will lead to several false positives expected by chance alone. You should either use some formal method of multiple testing adjustments, or perhaps simply lower ... | Should I adjust p-values when investigating an ANOVA interaction? | The next question I'd like to answer is "how many of the 30 subjects could reliably discriminate between the 3 headphone types"
Yes, using $p<0.05$ criterion will lead to several false positives expe | Should I adjust p-values when investigating an ANOVA interaction?
The next question I'd like to answer is "how many of the 30 subjects could reliably discriminate between the 3 headphone types"
Yes, using $p<0.05$ criterion will lead to several false positives expected by chance alone. You should either use some forma... | Should I adjust p-values when investigating an ANOVA interaction?
The next question I'd like to answer is "how many of the 30 subjects could reliably discriminate between the 3 headphone types"
Yes, using $p<0.05$ criterion will lead to several false positives expe |
45,767 | Should I adjust p-values when investigating an ANOVA interaction? | This may not fully answer your question, but you might look into mixed-effects models, which allow you to control for subject-level variation ("Joe always estimates high") without actually estimating coefficients for each subject. This lets you get at the original question you posed: "Do the headphones differ in subje... | Should I adjust p-values when investigating an ANOVA interaction? | This may not fully answer your question, but you might look into mixed-effects models, which allow you to control for subject-level variation ("Joe always estimates high") without actually estimating | Should I adjust p-values when investigating an ANOVA interaction?
This may not fully answer your question, but you might look into mixed-effects models, which allow you to control for subject-level variation ("Joe always estimates high") without actually estimating coefficients for each subject. This lets you get at t... | Should I adjust p-values when investigating an ANOVA interaction?
This may not fully answer your question, but you might look into mixed-effects models, which allow you to control for subject-level variation ("Joe always estimates high") without actually estimating |
45,768 | Should I adjust p-values when investigating an ANOVA interaction? | Normally subjects are kept in the Error term to reduce the effect of subject variability but your aim is to determine which subjects are significantly different from others. You can take help of graphics to identify such subjects. For each subject you can determine range of rating given (max rating - min rating) and pl... | Should I adjust p-values when investigating an ANOVA interaction? | Normally subjects are kept in the Error term to reduce the effect of subject variability but your aim is to determine which subjects are significantly different from others. You can take help of graph | Should I adjust p-values when investigating an ANOVA interaction?
Normally subjects are kept in the Error term to reduce the effect of subject variability but your aim is to determine which subjects are significantly different from others. You can take help of graphics to identify such subjects. For each subject you ca... | Should I adjust p-values when investigating an ANOVA interaction?
Normally subjects are kept in the Error term to reduce the effect of subject variability but your aim is to determine which subjects are significantly different from others. You can take help of graph |
45,769 | Define own noninformative prior in stan | You can define a proper or improper prior in the Stan language using the increment_log_prob() function, which will add its input to the accumulated log-posterior value that is used in the Metropolis step to decide whether to accept or reject a proposal for the parameters.
In your example, the model block would need to ... | Define own noninformative prior in stan | You can define a proper or improper prior in the Stan language using the increment_log_prob() function, which will add its input to the accumulated log-posterior value that is used in the Metropolis s | Define own noninformative prior in stan
You can define a proper or improper prior in the Stan language using the increment_log_prob() function, which will add its input to the accumulated log-posterior value that is used in the Metropolis step to decide whether to accept or reject a proposal for the parameters.
In your... | Define own noninformative prior in stan
You can define a proper or improper prior in the Stan language using the increment_log_prob() function, which will add its input to the accumulated log-posterior value that is used in the Metropolis s |
45,770 | svm functional margin and geometric margin | I think that the proper way to write the functional margin is
$$ \hat{\gamma}_i = y_i(w^Tx_i + b), $$
while the geometric margin is simply
$$ \gamma_i = \frac{\hat{\gamma}_i}{||w||}. $$
You can find the answer to your first question in here:
[...] the functional margin would give you a number but without a reference y... | svm functional margin and geometric margin | I think that the proper way to write the functional margin is
$$ \hat{\gamma}_i = y_i(w^Tx_i + b), $$
while the geometric margin is simply
$$ \gamma_i = \frac{\hat{\gamma}_i}{||w||}. $$
You can find t | svm functional margin and geometric margin
I think that the proper way to write the functional margin is
$$ \hat{\gamma}_i = y_i(w^Tx_i + b), $$
while the geometric margin is simply
$$ \gamma_i = \frac{\hat{\gamma}_i}{||w||}. $$
You can find the answer to your first question in here:
[...] the functional margin would ... | svm functional margin and geometric margin
I think that the proper way to write the functional margin is
$$ \hat{\gamma}_i = y_i(w^Tx_i + b), $$
while the geometric margin is simply
$$ \gamma_i = \frac{\hat{\gamma}_i}{||w||}. $$
You can find t |
45,771 | svm functional margin and geometric margin | but why it says that one should find the maximum geometrical margin?
Because the geometric margin is invariant to the scaling of the vector orthogonal to the hyperplane. Please see the answer here.
should it not be also to find the maximum of the functional margin
from the beginning?
Since scaling the paramet... | svm functional margin and geometric margin | but why it says that one should find the maximum geometrical margin?
Because the geometric margin is invariant to the scaling of the vector orthogonal to the hyperplane. Please see the answer her | svm functional margin and geometric margin
but why it says that one should find the maximum geometrical margin?
Because the geometric margin is invariant to the scaling of the vector orthogonal to the hyperplane. Please see the answer here.
should it not be also to find the maximum of the functional margin
fro... | svm functional margin and geometric margin
but why it says that one should find the maximum geometrical margin?
Because the geometric margin is invariant to the scaling of the vector orthogonal to the hyperplane. Please see the answer her |
45,772 | svm functional margin and geometric margin | Regarding following question
but why it says that one should find the maximum geometrical margin? should it not be also to find the maximum of the functional margin from the beginning?
You are very right our goal is to maximize both geometrical margin or functional margin but the confusion is that following formula ... | svm functional margin and geometric margin | Regarding following question
but why it says that one should find the maximum geometrical margin? should it not be also to find the maximum of the functional margin from the beginning?
You are very | svm functional margin and geometric margin
Regarding following question
but why it says that one should find the maximum geometrical margin? should it not be also to find the maximum of the functional margin from the beginning?
You are very right our goal is to maximize both geometrical margin or functional margin b... | svm functional margin and geometric margin
Regarding following question
but why it says that one should find the maximum geometrical margin? should it not be also to find the maximum of the functional margin from the beginning?
You are very |
45,773 | svm functional margin and geometric margin | Not going into unnecessary complications about this concept, but in the most simple terms here is how one can think of and relate functional and geometric margin.
Think of functional margin -- represented as 𝛾̂, as a measure of correctness of a classification for a data unit. For a data unit x with parameters w and b ... | svm functional margin and geometric margin | Not going into unnecessary complications about this concept, but in the most simple terms here is how one can think of and relate functional and geometric margin.
Think of functional margin -- represe | svm functional margin and geometric margin
Not going into unnecessary complications about this concept, but in the most simple terms here is how one can think of and relate functional and geometric margin.
Think of functional margin -- represented as 𝛾̂, as a measure of correctness of a classification for a data unit.... | svm functional margin and geometric margin
Not going into unnecessary complications about this concept, but in the most simple terms here is how one can think of and relate functional and geometric margin.
Think of functional margin -- represe |
45,774 | Homogeneous vs. Inhomogeneous Poisson point process | A homogeneous Poisson point process is also called complete spatial randomness described by a single parameter called the intensity (number of points per unit area). It distributes a random number of points completely randomly and uniformly in any given set. The number of points falling in two disjoint sets are indepen... | Homogeneous vs. Inhomogeneous Poisson point process | A homogeneous Poisson point process is also called complete spatial randomness described by a single parameter called the intensity (number of points per unit area). It distributes a random number of | Homogeneous vs. Inhomogeneous Poisson point process
A homogeneous Poisson point process is also called complete spatial randomness described by a single parameter called the intensity (number of points per unit area). It distributes a random number of points completely randomly and uniformly in any given set. The numbe... | Homogeneous vs. Inhomogeneous Poisson point process
A homogeneous Poisson point process is also called complete spatial randomness described by a single parameter called the intensity (number of points per unit area). It distributes a random number of |
45,775 | Why is the curse of dimensionality also called the empty space phenomenon? | I don't think the curse of dimensionality has anything to do with correlation, or at least not in my understanding. The curse is the notion that a local neighborhood of a point in a high dimensional space is not really so local - the number of data points it takes to uniformly "fill" a neighborhood of a point with a f... | Why is the curse of dimensionality also called the empty space phenomenon? | I don't think the curse of dimensionality has anything to do with correlation, or at least not in my understanding. The curse is the notion that a local neighborhood of a point in a high dimensional | Why is the curse of dimensionality also called the empty space phenomenon?
I don't think the curse of dimensionality has anything to do with correlation, or at least not in my understanding. The curse is the notion that a local neighborhood of a point in a high dimensional space is not really so local - the number of ... | Why is the curse of dimensionality also called the empty space phenomenon?
I don't think the curse of dimensionality has anything to do with correlation, or at least not in my understanding. The curse is the notion that a local neighborhood of a point in a high dimensional |
45,776 | Why do real-world high dimensional data often have much lower inherent dimensionality? | It's not that the dimensionality of the data (more precisely, the statistical process at work) is smaller than the coordinate space, it's that there is often not enough available data to get statistically significant results in every direction. This is a manifestation of the famed curse of dimensionality. PCA is an a... | Why do real-world high dimensional data often have much lower inherent dimensionality? | It's not that the dimensionality of the data (more precisely, the statistical process at work) is smaller than the coordinate space, it's that there is often not enough available data to get statistic | Why do real-world high dimensional data often have much lower inherent dimensionality?
It's not that the dimensionality of the data (more precisely, the statistical process at work) is smaller than the coordinate space, it's that there is often not enough available data to get statistically significant results in every... | Why do real-world high dimensional data often have much lower inherent dimensionality?
It's not that the dimensionality of the data (more precisely, the statistical process at work) is smaller than the coordinate space, it's that there is often not enough available data to get statistic |
45,777 | Why do real-world high dimensional data often have much lower inherent dimensionality? | The answer is relatively simple. Typically for a real world problem you don't know which features to use for the problem. Very often you end up throwing too many features and let the algorithm figure out which ones are discriminative.
Let's take MNIST digit classification as an example. You are given 28x28 black&white ... | Why do real-world high dimensional data often have much lower inherent dimensionality? | The answer is relatively simple. Typically for a real world problem you don't know which features to use for the problem. Very often you end up throwing too many features and let the algorithm figure | Why do real-world high dimensional data often have much lower inherent dimensionality?
The answer is relatively simple. Typically for a real world problem you don't know which features to use for the problem. Very often you end up throwing too many features and let the algorithm figure out which ones are discriminative... | Why do real-world high dimensional data often have much lower inherent dimensionality?
The answer is relatively simple. Typically for a real world problem you don't know which features to use for the problem. Very often you end up throwing too many features and let the algorithm figure |
45,778 | Why do real-world high dimensional data often have much lower inherent dimensionality? | What "the inherent dimensionality of data points will be smaller than the number of coordinates" means is this: If you have 2 dimensional data, you need a 2 dimensional coordinate system (x,y for example) in order to be able to show them. if you data has n dimensions, you need an n dimensional coordinate system, where ... | Why do real-world high dimensional data often have much lower inherent dimensionality? | What "the inherent dimensionality of data points will be smaller than the number of coordinates" means is this: If you have 2 dimensional data, you need a 2 dimensional coordinate system (x,y for exam | Why do real-world high dimensional data often have much lower inherent dimensionality?
What "the inherent dimensionality of data points will be smaller than the number of coordinates" means is this: If you have 2 dimensional data, you need a 2 dimensional coordinate system (x,y for example) in order to be able to show ... | Why do real-world high dimensional data often have much lower inherent dimensionality?
What "the inherent dimensionality of data points will be smaller than the number of coordinates" means is this: If you have 2 dimensional data, you need a 2 dimensional coordinate system (x,y for exam |
45,779 | Why do real-world high dimensional data often have much lower inherent dimensionality? | In real world everything's controlled by God's. That's why everything's dependent on God. So, everything's really in one dimension, the dimension of God's will. This is is the only true answer, but I doubt that your Prof will accept it. So, here's an easier one.
In real world we're probably not gathering some random da... | Why do real-world high dimensional data often have much lower inherent dimensionality? | In real world everything's controlled by God's. That's why everything's dependent on God. So, everything's really in one dimension, the dimension of God's will. This is is the only true answer, but I | Why do real-world high dimensional data often have much lower inherent dimensionality?
In real world everything's controlled by God's. That's why everything's dependent on God. So, everything's really in one dimension, the dimension of God's will. This is is the only true answer, but I doubt that your Prof will accept ... | Why do real-world high dimensional data often have much lower inherent dimensionality?
In real world everything's controlled by God's. That's why everything's dependent on God. So, everything's really in one dimension, the dimension of God's will. This is is the only true answer, but I |
45,780 | Why do real-world high dimensional data often have much lower inherent dimensionality? | The information in the high-dimensional space can often be captured by a smaller number of latent variables/dimensions because there tends to be dependence (e.g. multicollinearity) between variables in the high-dimensional space.
There tends to be dependence because many of the variables are probably going to form net... | Why do real-world high dimensional data often have much lower inherent dimensionality? | The information in the high-dimensional space can often be captured by a smaller number of latent variables/dimensions because there tends to be dependence (e.g. multicollinearity) between variables i | Why do real-world high dimensional data often have much lower inherent dimensionality?
The information in the high-dimensional space can often be captured by a smaller number of latent variables/dimensions because there tends to be dependence (e.g. multicollinearity) between variables in the high-dimensional space.
Th... | Why do real-world high dimensional data often have much lower inherent dimensionality?
The information in the high-dimensional space can often be captured by a smaller number of latent variables/dimensions because there tends to be dependence (e.g. multicollinearity) between variables i |
45,781 | Why do real-world high dimensional data often have much lower inherent dimensionality? | The statement in the exam question seems contentious. Take a simple example, like plotting life expectancy of a baby versus various factors including average daily salt consumption. The baby will (on average) die early if the salt consumption is near zero, will thrive it the salt consumption is intermediate, and will d... | Why do real-world high dimensional data often have much lower inherent dimensionality? | The statement in the exam question seems contentious. Take a simple example, like plotting life expectancy of a baby versus various factors including average daily salt consumption. The baby will (on | Why do real-world high dimensional data often have much lower inherent dimensionality?
The statement in the exam question seems contentious. Take a simple example, like plotting life expectancy of a baby versus various factors including average daily salt consumption. The baby will (on average) die early if the salt co... | Why do real-world high dimensional data often have much lower inherent dimensionality?
The statement in the exam question seems contentious. Take a simple example, like plotting life expectancy of a baby versus various factors including average daily salt consumption. The baby will (on |
45,782 | Interpretation of eigenvectors of Hessian inverse | "Are the eigenvectors/eigenvalues of the inverse Hessian related to those of the Hessian?"
Yes. The hessian is a symmetric matrix which can be diagonalized as
$H=Q\Lambda Q^{T}$
where $Q$ is an orthogonal matrix whose columns are eigenvectors of $H$ and $\Lambda$ is a diagonal matrix with the eigenvalues of $H$ on ... | Interpretation of eigenvectors of Hessian inverse | "Are the eigenvectors/eigenvalues of the inverse Hessian related to those of the Hessian?"
Yes. The hessian is a symmetric matrix which can be diagonalized as
$H=Q\Lambda Q^{T}$
where $Q$ is an or | Interpretation of eigenvectors of Hessian inverse
"Are the eigenvectors/eigenvalues of the inverse Hessian related to those of the Hessian?"
Yes. The hessian is a symmetric matrix which can be diagonalized as
$H=Q\Lambda Q^{T}$
where $Q$ is an orthogonal matrix whose columns are eigenvectors of $H$ and $\Lambda$ is... | Interpretation of eigenvectors of Hessian inverse
"Are the eigenvectors/eigenvalues of the inverse Hessian related to those of the Hessian?"
Yes. The hessian is a symmetric matrix which can be diagonalized as
$H=Q\Lambda Q^{T}$
where $Q$ is an or |
45,783 | How to compare two Pearson correlation coefficients | There are various tests you can apply. Biedenhofen & Musch (2015, PLoS ONE) give pointers and describe the cocor package for R, which implements these tests. You can also submit your correlations for testing to a web tool which internally uses the cocor package. | How to compare two Pearson correlation coefficients | There are various tests you can apply. Biedenhofen & Musch (2015, PLoS ONE) give pointers and describe the cocor package for R, which implements these tests. You can also submit your correlations for | How to compare two Pearson correlation coefficients
There are various tests you can apply. Biedenhofen & Musch (2015, PLoS ONE) give pointers and describe the cocor package for R, which implements these tests. You can also submit your correlations for testing to a web tool which internally uses the cocor package. | How to compare two Pearson correlation coefficients
There are various tests you can apply. Biedenhofen & Musch (2015, PLoS ONE) give pointers and describe the cocor package for R, which implements these tests. You can also submit your correlations for |
45,784 | How to compare two Pearson correlation coefficients | The cocor package seems to be a handy tool. I ran the cocor package with my parameters via the web tool as you suggested. The output of that calculation is the following:
Comparison between r1.jk = -0.747 and r2.hm = -0.885
Difference: r1.jk - r2.hm = 0.138
Group sizes: n1 = 159200, n2 = 2400
Null hypothesis: r1.jk is... | How to compare two Pearson correlation coefficients | The cocor package seems to be a handy tool. I ran the cocor package with my parameters via the web tool as you suggested. The output of that calculation is the following:
Comparison between r1.jk = - | How to compare two Pearson correlation coefficients
The cocor package seems to be a handy tool. I ran the cocor package with my parameters via the web tool as you suggested. The output of that calculation is the following:
Comparison between r1.jk = -0.747 and r2.hm = -0.885
Difference: r1.jk - r2.hm = 0.138
Group siz... | How to compare two Pearson correlation coefficients
The cocor package seems to be a handy tool. I ran the cocor package with my parameters via the web tool as you suggested. The output of that calculation is the following:
Comparison between r1.jk = - |
45,785 | What is the test statistics used for a conditional inference regression tree? | If both the regressor $X_{ji}$ and the response $Y_i$ are numeric, then both $g(\cdot)$ and $h(\cdot)$ are chosen to be the identiy by default. Thus, the linear test statistic $T_j$ is simply the sum of products $X_{ji} \cdot Y_i$. This corresponds essentially to the main ingredient of a covariance or correlation - and... | What is the test statistics used for a conditional inference regression tree? | If both the regressor $X_{ji}$ and the response $Y_i$ are numeric, then both $g(\cdot)$ and $h(\cdot)$ are chosen to be the identiy by default. Thus, the linear test statistic $T_j$ is simply the sum | What is the test statistics used for a conditional inference regression tree?
If both the regressor $X_{ji}$ and the response $Y_i$ are numeric, then both $g(\cdot)$ and $h(\cdot)$ are chosen to be the identiy by default. Thus, the linear test statistic $T_j$ is simply the sum of products $X_{ji} \cdot Y_i$. This corre... | What is the test statistics used for a conditional inference regression tree?
If both the regressor $X_{ji}$ and the response $Y_i$ are numeric, then both $g(\cdot)$ and $h(\cdot)$ are chosen to be the identiy by default. Thus, the linear test statistic $T_j$ is simply the sum |
45,786 | Non-Stationary: Larger-than-unit root [duplicate] | I think this is actually a quite good question, which is often neglected (as you have noticed) and which I myself haven't thought about much before. The main point, I would say, is that processes with larger-than-one roots (called explosive roots) are not as interesting. If you have something which is just slightly abo... | Non-Stationary: Larger-than-unit root [duplicate] | I think this is actually a quite good question, which is often neglected (as you have noticed) and which I myself haven't thought about much before. The main point, I would say, is that processes with | Non-Stationary: Larger-than-unit root [duplicate]
I think this is actually a quite good question, which is often neglected (as you have noticed) and which I myself haven't thought about much before. The main point, I would say, is that processes with larger-than-one roots (called explosive roots) are not as interesting... | Non-Stationary: Larger-than-unit root [duplicate]
I think this is actually a quite good question, which is often neglected (as you have noticed) and which I myself haven't thought about much before. The main point, I would say, is that processes with |
45,787 | Non-Stationary: Larger-than-unit root [duplicate] | There are several kinds of non-stationarities:
1) Series expected value is a function of time
2) Series variance depends on time and not just about lag
3) Etc
Series with linear trend is non-stationary but stationary around trend..
EDIT:
Your example of stochastic difference equation with explosive root is of course ... | Non-Stationary: Larger-than-unit root [duplicate] | There are several kinds of non-stationarities:
1) Series expected value is a function of time
2) Series variance depends on time and not just about lag
3) Etc
Series with linear trend is non-station | Non-Stationary: Larger-than-unit root [duplicate]
There are several kinds of non-stationarities:
1) Series expected value is a function of time
2) Series variance depends on time and not just about lag
3) Etc
Series with linear trend is non-stationary but stationary around trend..
EDIT:
Your example of stochastic dif... | Non-Stationary: Larger-than-unit root [duplicate]
There are several kinds of non-stationarities:
1) Series expected value is a function of time
2) Series variance depends on time and not just about lag
3) Etc
Series with linear trend is non-station |
45,788 | Binary outcome in randomized controlled trials -- OLS or logistic? | Edit for clarity: It looks like my responses here have led to some clarifying additions to the question or additional information in comments, which make parts of my answer now at least partially obsolete. However, I plan to leave my answer as is, partly for context and partly because I believe the points raised may b... | Binary outcome in randomized controlled trials -- OLS or logistic? | Edit for clarity: It looks like my responses here have led to some clarifying additions to the question or additional information in comments, which make parts of my answer now at least partially obs | Binary outcome in randomized controlled trials -- OLS or logistic?
Edit for clarity: It looks like my responses here have led to some clarifying additions to the question or additional information in comments, which make parts of my answer now at least partially obsolete. However, I plan to leave my answer as is, part... | Binary outcome in randomized controlled trials -- OLS or logistic?
Edit for clarity: It looks like my responses here have led to some clarifying additions to the question or additional information in comments, which make parts of my answer now at least partially obs |
45,789 | Binary outcome in randomized controlled trials -- OLS or logistic? | A lot of economists use linear probability models, arguing that LPM provides the linear approximation of the conditional expectation function, which is often considered "good enough." Consistent (in large samples) standard errors can be gotten by using ``robust'' variance-covariance matrix estimators.
This is an OK ... | Binary outcome in randomized controlled trials -- OLS or logistic? | A lot of economists use linear probability models, arguing that LPM provides the linear approximation of the conditional expectation function, which is often considered "good enough." Consistent (in | Binary outcome in randomized controlled trials -- OLS or logistic?
A lot of economists use linear probability models, arguing that LPM provides the linear approximation of the conditional expectation function, which is often considered "good enough." Consistent (in large samples) standard errors can be gotten by using... | Binary outcome in randomized controlled trials -- OLS or logistic?
A lot of economists use linear probability models, arguing that LPM provides the linear approximation of the conditional expectation function, which is often considered "good enough." Consistent (in |
45,790 | Cluster Sequences of data with different length | One way to do it (among many other ways) is to treat the element of your sequence as a word. In other words, if your assume your list is a sentence, then you can extract ngrams.
import nltk
from nltk import ngrams
a = [1, 15, 1, 1, 13, 14]
b = [1, 1, 1, 1, 12, 1, 7, 11, 9, 11, 7, 11, 7, 11, 7, 4, 7, 7, 14, 15, 9, 2]
c ... | Cluster Sequences of data with different length | One way to do it (among many other ways) is to treat the element of your sequence as a word. In other words, if your assume your list is a sentence, then you can extract ngrams.
import nltk
from nltk | Cluster Sequences of data with different length
One way to do it (among many other ways) is to treat the element of your sequence as a word. In other words, if your assume your list is a sentence, then you can extract ngrams.
import nltk
from nltk import ngrams
a = [1, 15, 1, 1, 13, 14]
b = [1, 1, 1, 1, 12, 1, 7, 11, 9... | Cluster Sequences of data with different length
One way to do it (among many other ways) is to treat the element of your sequence as a word. In other words, if your assume your list is a sentence, then you can extract ngrams.
import nltk
from nltk |
45,791 | Cluster Sequences of data with different length | K-means won't work on data of this type. To me, the strings you provided as examples lend themselves to information theoretic approaches to clustering based on MDL (minimum description length https://en.wikipedia.org/wiki/Minimum_description_length) or data compression. By compressing these strings to their unique sequ... | Cluster Sequences of data with different length | K-means won't work on data of this type. To me, the strings you provided as examples lend themselves to information theoretic approaches to clustering based on MDL (minimum description length https:// | Cluster Sequences of data with different length
K-means won't work on data of this type. To me, the strings you provided as examples lend themselves to information theoretic approaches to clustering based on MDL (minimum description length https://en.wikipedia.org/wiki/Minimum_description_length) or data compression. B... | Cluster Sequences of data with different length
K-means won't work on data of this type. To me, the strings you provided as examples lend themselves to information theoretic approaches to clustering based on MDL (minimum description length https:// |
45,792 | Cluster Sequences of data with different length | k-means must be able to compute means, so it won't work for you.
Consider using hierarchical clustering, with a Levenshtein or similar similarity metric. LCSS is also a good choice; any similarity designed for sequences. | Cluster Sequences of data with different length | k-means must be able to compute means, so it won't work for you.
Consider using hierarchical clustering, with a Levenshtein or similar similarity metric. LCSS is also a good choice; any similarity des | Cluster Sequences of data with different length
k-means must be able to compute means, so it won't work for you.
Consider using hierarchical clustering, with a Levenshtein or similar similarity metric. LCSS is also a good choice; any similarity designed for sequences. | Cluster Sequences of data with different length
k-means must be able to compute means, so it won't work for you.
Consider using hierarchical clustering, with a Levenshtein or similar similarity metric. LCSS is also a good choice; any similarity des |
45,793 | Sum of Gaussian is Gaussian? | No and this is a common fallacy. People tend to forget that the sum of two Gaussian is a Gaussian only if $X$ and $Y$ are independent or jointly normal.
Here is a nice explanation. | Sum of Gaussian is Gaussian? | No and this is a common fallacy. People tend to forget that the sum of two Gaussian is a Gaussian only if $X$ and $Y$ are independent or jointly normal.
Here is a nice explanation. | Sum of Gaussian is Gaussian?
No and this is a common fallacy. People tend to forget that the sum of two Gaussian is a Gaussian only if $X$ and $Y$ are independent or jointly normal.
Here is a nice explanation. | Sum of Gaussian is Gaussian?
No and this is a common fallacy. People tend to forget that the sum of two Gaussian is a Gaussian only if $X$ and $Y$ are independent or jointly normal.
Here is a nice explanation. |
45,794 | Sum of Gaussian is Gaussian? | You are saying in your second point “uncorrelated (hence independent)”. It’s not quite the case. You can have two Gaussian distributions both uncorrelated and dependent. Uncorrelated implies independence only if they are jointly Gaussian. | Sum of Gaussian is Gaussian? | You are saying in your second point “uncorrelated (hence independent)”. It’s not quite the case. You can have two Gaussian distributions both uncorrelated and dependent. Uncorrelated implies independe | Sum of Gaussian is Gaussian?
You are saying in your second point “uncorrelated (hence independent)”. It’s not quite the case. You can have two Gaussian distributions both uncorrelated and dependent. Uncorrelated implies independence only if they are jointly Gaussian. | Sum of Gaussian is Gaussian?
You are saying in your second point “uncorrelated (hence independent)”. It’s not quite the case. You can have two Gaussian distributions both uncorrelated and dependent. Uncorrelated implies independe |
45,795 | Independent replication experiments yielding contrasting results; how to combine them? | The p-value from each experiment should have a uniform distribution between 0 and 1 under the null hypothesis, so tests of the null hypothesis over all experiments can be based on this. Perhaps the most common test statistic is Fisher's: for p-values $p_j$ from $m$ independent experiments the negative log of each follo... | Independent replication experiments yielding contrasting results; how to combine them? | The p-value from each experiment should have a uniform distribution between 0 and 1 under the null hypothesis, so tests of the null hypothesis over all experiments can be based on this. Perhaps the mo | Independent replication experiments yielding contrasting results; how to combine them?
The p-value from each experiment should have a uniform distribution between 0 and 1 under the null hypothesis, so tests of the null hypothesis over all experiments can be based on this. Perhaps the most common test statistic is Fishe... | Independent replication experiments yielding contrasting results; how to combine them?
The p-value from each experiment should have a uniform distribution between 0 and 1 under the null hypothesis, so tests of the null hypothesis over all experiments can be based on this. Perhaps the mo |
45,796 | Is the sum of several Poisson processes a Poisson process? | If they're independent of each other, yes.
Indeed a more general result is that if there are $k$ independent Poisson processes with rate $\lambda_i, \, i=1,2,\ldots,k$, then the combined process (the superposition of the component processes) is a Poisson process with rate $\sum_i\lambda_i$.
It's really only necessary t... | Is the sum of several Poisson processes a Poisson process? | If they're independent of each other, yes.
Indeed a more general result is that if there are $k$ independent Poisson processes with rate $\lambda_i, \, i=1,2,\ldots,k$, then the combined process (the | Is the sum of several Poisson processes a Poisson process?
If they're independent of each other, yes.
Indeed a more general result is that if there are $k$ independent Poisson processes with rate $\lambda_i, \, i=1,2,\ldots,k$, then the combined process (the superposition of the component processes) is a Poisson proces... | Is the sum of several Poisson processes a Poisson process?
If they're independent of each other, yes.
Indeed a more general result is that if there are $k$ independent Poisson processes with rate $\lambda_i, \, i=1,2,\ldots,k$, then the combined process (the |
45,797 | Confusing Holt-Winters parameters | The small values for $\beta$ and $\gamma$ show that the trend and seasonality do not change much over time. They do not tell you that there is no trend or seasonality. | Confusing Holt-Winters parameters | The small values for $\beta$ and $\gamma$ show that the trend and seasonality do not change much over time. They do not tell you that there is no trend or seasonality. | Confusing Holt-Winters parameters
The small values for $\beta$ and $\gamma$ show that the trend and seasonality do not change much over time. They do not tell you that there is no trend or seasonality. | Confusing Holt-Winters parameters
The small values for $\beta$ and $\gamma$ show that the trend and seasonality do not change much over time. They do not tell you that there is no trend or seasonality. |
45,798 | Confusing Holt-Winters parameters | All parameters, $\alpha$, $\beta$ and $\gamma$, have values between 0 and 1. In broad terms, a simple exponential smoothing model looks like this (though the idea also works for double and triple exponential smoothing):
$$ \mathit{smoothed_t} = \color{blue}{\mathit{parameter}} \cdot \mathit{observation_t} + (\color{blu... | Confusing Holt-Winters parameters | All parameters, $\alpha$, $\beta$ and $\gamma$, have values between 0 and 1. In broad terms, a simple exponential smoothing model looks like this (though the idea also works for double and triple expo | Confusing Holt-Winters parameters
All parameters, $\alpha$, $\beta$ and $\gamma$, have values between 0 and 1. In broad terms, a simple exponential smoothing model looks like this (though the idea also works for double and triple exponential smoothing):
$$ \mathit{smoothed_t} = \color{blue}{\mathit{parameter}} \cdot \m... | Confusing Holt-Winters parameters
All parameters, $\alpha$, $\beta$ and $\gamma$, have values between 0 and 1. In broad terms, a simple exponential smoothing model looks like this (though the idea also works for double and triple expo |
45,799 | How to express "inequality" of a distribution in one number? | Perhaps the best known measure would be the Gini index.
The R package ineq (See here) implements the Herfindahl and Rosenbluth concentration measures (in function conc).
It also implements a number of inequality indexes (including the Gini) in function ineq -- the Gini coefficient, Ricci-Schutz coefficient (also called... | How to express "inequality" of a distribution in one number? | Perhaps the best known measure would be the Gini index.
The R package ineq (See here) implements the Herfindahl and Rosenbluth concentration measures (in function conc).
It also implements a number of | How to express "inequality" of a distribution in one number?
Perhaps the best known measure would be the Gini index.
The R package ineq (See here) implements the Herfindahl and Rosenbluth concentration measures (in function conc).
It also implements a number of inequality indexes (including the Gini) in function ineq -... | How to express "inequality" of a distribution in one number?
Perhaps the best known measure would be the Gini index.
The R package ineq (See here) implements the Herfindahl and Rosenbluth concentration measures (in function conc).
It also implements a number of |
45,800 | How to express "inequality" of a distribution in one number? | I've developed a method for quantifying "uniformity" that allows you to do what you are asking. Its helped out a couple other folks too.
See: https://math.stackexchange.com/questions/921084/how-to-calculate-peakiness-or-uniformity-in-histogram/921110#921110
Basically, you are just calculating the path length of the ass... | How to express "inequality" of a distribution in one number? | I've developed a method for quantifying "uniformity" that allows you to do what you are asking. Its helped out a couple other folks too.
See: https://math.stackexchange.com/questions/921084/how-to-cal | How to express "inequality" of a distribution in one number?
I've developed a method for quantifying "uniformity" that allows you to do what you are asking. Its helped out a couple other folks too.
See: https://math.stackexchange.com/questions/921084/how-to-calculate-peakiness-or-uniformity-in-histogram/921110#921110
B... | How to express "inequality" of a distribution in one number?
I've developed a method for quantifying "uniformity" that allows you to do what you are asking. Its helped out a couple other folks too.
See: https://math.stackexchange.com/questions/921084/how-to-cal |
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