The sampling distribution of variance: the general case
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Suppose I take many samples of size $n$ from a known probability distribution. And for each sample, I calculate the sample variance of the $n$ observations. How will my sample variances be distributed?
(Is it possible to determine the exact distribution of sample variances without needing to assume my known probability distribution is Gaussian? In other words, I realize that when the known probability distribution is Gaussian, the sampling distribution of variance is Chi-squared. But what about the cases where we do not sample from Gaussian the distribution?)
probability-theory
asked Dec 30 '18 at 10:38
user120911user120911
229111
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add a comment |
$begingroup$
Suppose I take many samples of size $n$ from a known probability distribution. And for each sample, I calculate the sample variance of the $n$ observations. How will my sample variances be distributed?
(Is it possible to determine the exact distribution of sample variances without needing to assume my known probability distribution is Gaussian? In other words, I realize that when the known probability distribution is Gaussian, the sampling distribution of variance is Chi-squared. But what about the cases where we do not sample from Gaussian the distribution?)
probability-theory
asked Dec 30 '18 at 10:38
user120911user120911
229111
$endgroup$
$begingroup$
Or do you mean the distribution of the sample estimator $s^2$ of the variance? This will depend on the the underlying distribution: saying $(n−1)s^2/σ^2$ has a chi-square distribution with $n-1$ degrees of freedom is based on the underlying distribution being normal
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– Henry
Dec 30 '18 at 11:37
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@Henry Please see my clarification in the posted question.
$endgroup$
– user120911
Dec 30 '18 at 11:50
1
$begingroup$
Then the answer is no, if you do not know the underlying distribution then you do not know the distribution of the sample variance. Experiment yourself with a standard normal (variance $1$) and an exponential random variable (rate and variance $1$): you will find that the sample variance for the exponential is more widely dispersed than the sampling variance for the normal
$endgroup$
– Henry
Dec 30 '18 at 14:54
add a comment |
$begingroup$
Suppose I take many samples of size $n$ from a known probability distribution. And for each sample, I calculate the sample variance of the $n$ observations. How will my sample variances be distributed?
(Is it possible to determine the exact distribution of sample variances without needing to assume my known probability distribution is Gaussian? In other words, I realize that when the known probability distribution is Gaussian, the sampling distribution of variance is Chi-squared. But what about the cases where we do not sample from Gaussian the distribution?)
probability-theory
asked Dec 30 '18 at 10:38
user120911user120911
229111
$endgroup$
Suppose I take many samples of size $n$ from a known probability distribution. And for each sample, I calculate the sample variance of the $n$ observations. How will my sample variances be distributed?
(Is it possible to determine the exact distribution of sample variances without needing to assume my known probability distribution is Gaussian? In other words, I realize that when the known probability distribution is Gaussian, the sampling distribution of variance is Chi-squared. But what about the cases where we do not sample from Gaussian the distribution?)
probability-theory
probability-theory
asked Dec 30 '18 at 10:38
user120911user120911
229111
asked Dec 30 '18 at 10:38
user120911user120911
229111
edited Dec 30 '18 at 12:04
user120911
asked Dec 30 '18 at 10:38
user120911user120911
229111
asked Dec 30 '18 at 10:38
user120911user120911
229111
asked Dec 30 '18 at 10:38
user120911user120911
229111
229111
$begingroup$
Or do you mean the distribution of the sample estimator $s^2$ of the variance? This will depend on the the underlying distribution: saying $(n−1)s^2/σ^2$ has a chi-square distribution with $n-1$ degrees of freedom is based on the underlying distribution being normal
$endgroup$
– Henry
Dec 30 '18 at 11:37
$begingroup$
@Henry Please see my clarification in the posted question.
$endgroup$
– user120911
Dec 30 '18 at 11:50
1
$begingroup$
Then the answer is no, if you do not know the underlying distribution then you do not know the distribution of the sample variance. Experiment yourself with a standard normal (variance $1$) and an exponential random variable (rate and variance $1$): you will find that the sample variance for the exponential is more widely dispersed than the sampling variance for the normal
$endgroup$
– Henry
Dec 30 '18 at 14:54
add a comment |
$begingroup$
Or do you mean the distribution of the sample estimator $s^2$ of the variance? This will depend on the the underlying distribution: saying $(n−1)s^2/σ^2$ has a chi-square distribution with $n-1$ degrees of freedom is based on the underlying distribution being normal
$endgroup$
– Henry
Dec 30 '18 at 11:37
$begingroup$
@Henry Please see my clarification in the posted question.
$endgroup$
– user120911
Dec 30 '18 at 11:50
1
$begingroup$
Then the answer is no, if you do not know the underlying distribution then you do not know the distribution of the sample variance. Experiment yourself with a standard normal (variance $1$) and an exponential random variable (rate and variance $1$): you will find that the sample variance for the exponential is more widely dispersed than the sampling variance for the normal
$endgroup$
– Henry
Dec 30 '18 at 14:54
$begingroup$
Or do you mean the distribution of the sample estimator $s^2$ of the variance? This will depend on the the underlying distribution: saying $(n−1)s^2/σ^2$ has a chi-square distribution with $n-1$ degrees of freedom is based on the underlying distribution being normal
$endgroup$
– Henry
Dec 30 '18 at 11:37
$begingroup$
Or do you mean the distribution of the sample estimator $s^2$ of the variance? This will depend on the the underlying distribution: saying $(n−1)s^2/σ^2$ has a chi-square distribution with $n-1$ degrees of freedom is based on the underlying distribution being normal
$endgroup$
– Henry
Dec 30 '18 at 11:37
$begingroup$
@Henry Please see my clarification in the posted question.
$endgroup$
– user120911
Dec 30 '18 at 11:50
$begingroup$
@Henry Please see my clarification in the posted question.
$endgroup$
– user120911
Dec 30 '18 at 11:50
1
1
$begingroup$
Then the answer is no, if you do not know the underlying distribution then you do not know the distribution of the sample variance. Experiment yourself with a standard normal (variance $1$) and an exponential random variable (rate and variance $1$): you will find that the sample variance for the exponential is more widely dispersed than the sampling variance for the normal
$endgroup$
– Henry
Dec 30 '18 at 14:54
$begingroup$
Then the answer is no, if you do not know the underlying distribution then you do not know the distribution of the sample variance. Experiment yourself with a standard normal (variance $1$) and an exponential random variable (rate and variance $1$): you will find that the sample variance for the exponential is more widely dispersed than the sampling variance for the normal
$endgroup$
– Henry
Dec 30 '18 at 14:54
add a comment |
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$begingroup$
Or do you mean the distribution of the sample estimator $s^2$ of the variance? This will depend on the the underlying distribution: saying $(n−1)s^2/σ^2$ has a chi-square distribution with $n-1$ degrees of freedom is based on the underlying distribution being normal
$endgroup$
– Henry
Dec 30 '18 at 11:37
$begingroup$
@Henry Please see my clarification in the posted question.
$endgroup$
– user120911
Dec 30 '18 at 11:50
1
$begingroup$
Then the answer is no, if you do not know the underlying distribution then you do not know the distribution of the sample variance. Experiment yourself with a standard normal (variance $1$) and an exponential random variable (rate and variance $1$): you will find that the sample variance for the exponential is more widely dispersed than the sampling variance for the normal
$endgroup$
– Henry
Dec 30 '18 at 14:54