Mixing
UMVU for $sigma ^ p$ normal distribution.
$begingroup$
I am having some trouble with the following problem:
let $X_1, ldots, X_n$ independent from a Normal distribution with unknown mean $mu$ and variance $sigma^2$. Find the UMVU estimator for $sigma^p$ where $p>0$ is real.
I have found the maximum likelihood estimator but was not able to find and correct its mean, so I do not know how to proceed. (I have also tried to use Rao-Blacwell theorem but I did not know which unbiased estimator to use).
Thank you for your help.
statistics
asked Jan 22 at 7:44
franzkfranzk
1435
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add a comment |
$begingroup$
I am having some trouble with the following problem:
let $X_1, ldots, X_n$ independent from a Normal distribution with unknown mean $mu$ and variance $sigma^2$. Find the UMVU estimator for $sigma^p$ where $p>0$ is real.
I have found the maximum likelihood estimator but was not able to find and correct its mean, so I do not know how to proceed. (I have also tried to use Rao-Blacwell theorem but I did not know which unbiased estimator to use).
Thank you for your help.
statistics
asked Jan 22 at 7:44
franzkfranzk
1435
$endgroup$
$begingroup$
If I find the MLE then I know it is a function of complete sufficient statistics (property of MVE), then if I can trasnsform it to make it unbiased I will have an unbiased estimator which is a function of complete sufficient statistic and therefore it is umvue because of the lehman-scheffe theorem
$endgroup$
– franzk
Jan 22 at 10:31
add a comment |
$begingroup$
I am having some trouble with the following problem:
let $X_1, ldots, X_n$ independent from a Normal distribution with unknown mean $mu$ and variance $sigma^2$. Find the UMVU estimator for $sigma^p$ where $p>0$ is real.
I have found the maximum likelihood estimator but was not able to find and correct its mean, so I do not know how to proceed. (I have also tried to use Rao-Blacwell theorem but I did not know which unbiased estimator to use).
Thank you for your help.
statistics
asked Jan 22 at 7:44
franzkfranzk
1435
$endgroup$
I am having some trouble with the following problem:
let $X_1, ldots, X_n$ independent from a Normal distribution with unknown mean $mu$ and variance $sigma^2$. Find the UMVU estimator for $sigma^p$ where $p>0$ is real.
I have found the maximum likelihood estimator but was not able to find and correct its mean, so I do not know how to proceed. (I have also tried to use Rao-Blacwell theorem but I did not know which unbiased estimator to use).
Thank you for your help.
statistics
statistics
asked Jan 22 at 7:44
franzkfranzk
1435
asked Jan 22 at 7:44
franzkfranzk
1435
asked Jan 22 at 7:44
franzkfranzk
1435
asked Jan 22 at 7:44
franzkfranzk
1435
asked Jan 22 at 7:44
franzkfranzk
1435
1435
$begingroup$
If I find the MLE then I know it is a function of complete sufficient statistics (property of MVE), then if I can trasnsform it to make it unbiased I will have an unbiased estimator which is a function of complete sufficient statistic and therefore it is umvue because of the lehman-scheffe theorem
$endgroup$
– franzk
Jan 22 at 10:31
add a comment |
$begingroup$
If I find the MLE then I know it is a function of complete sufficient statistics (property of MVE), then if I can trasnsform it to make it unbiased I will have an unbiased estimator which is a function of complete sufficient statistic and therefore it is umvue because of the lehman-scheffe theorem
$endgroup$
– franzk
Jan 22 at 10:31
$begingroup$
If I find the MLE then I know it is a function of complete sufficient statistics (property of MVE), then if I can trasnsform it to make it unbiased I will have an unbiased estimator which is a function of complete sufficient statistic and therefore it is umvue because of the lehman-scheffe theorem
$endgroup$
– franzk
Jan 22 at 10:31
$begingroup$
If I find the MLE then I know it is a function of complete sufficient statistics (property of MVE), then if I can trasnsform it to make it unbiased I will have an unbiased estimator which is a function of complete sufficient statistic and therefore it is umvue because of the lehman-scheffe theorem
$endgroup$
– franzk
Jan 22 at 10:31
add a comment |
1 Answer
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$begingroup$
Define $overline X=frac{1}{n}sumlimits_{k=1}^n X_k$ and $S^2=frac{1}{n-1}sumlimits_{k=1}^n (X_k-overline X)^2$.
Then a complete sufficient statistic for $(mu,sigma^2)$ is given by $(overline X, S^2)$.
By Lehmann-Scheffe theorem, any unbiased estimator of $sigma^p$ based on $(overline X, S^2)$ will be the UMVUE of $sigma^p$.
Recall that $$frac{(n-1)S^2}{sigma^2}sim chi^2_{n-1}$$
So,
begin{align}
Eleft[frac{(n-1)S^2}{sigma^2}right]^{p/2}&=frac{1}{2^{frac{n-1}{2}}Gammaleft(frac{n-1}{2}right)}int_0^infty t^{p/2},e^{-t/2},t^{frac{n-1}{2}-1},mathrm{d}t
end{align}
Simplifying both sides of the above equation you will finally arrive at $$Eleft[cS^pright]=sigma^p$$
for some function $c(.)$ of $n$ and $p$.
Thus the UMVUE of $sigma^p$ is $c(n,p) S^p$.
answered Jan 22 at 10:29
StubbornAtomStubbornAtom
6,14311339
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1 Answer
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$begingroup$
Define $overline X=frac{1}{n}sumlimits_{k=1}^n X_k$ and $S^2=frac{1}{n-1}sumlimits_{k=1}^n (X_k-overline X)^2$.
Then a complete sufficient statistic for $(mu,sigma^2)$ is given by $(overline X, S^2)$.
By Lehmann-Scheffe theorem, any unbiased estimator of $sigma^p$ based on $(overline X, S^2)$ will be the UMVUE of $sigma^p$.
Recall that $$frac{(n-1)S^2}{sigma^2}sim chi^2_{n-1}$$
So,
begin{align}
Eleft[frac{(n-1)S^2}{sigma^2}right]^{p/2}&=frac{1}{2^{frac{n-1}{2}}Gammaleft(frac{n-1}{2}right)}int_0^infty t^{p/2},e^{-t/2},t^{frac{n-1}{2}-1},mathrm{d}t
end{align}
Simplifying both sides of the above equation you will finally arrive at $$Eleft[cS^pright]=sigma^p$$
for some function $c(.)$ of $n$ and $p$.
Thus the UMVUE of $sigma^p$ is $c(n,p) S^p$.
answered Jan 22 at 10:29
StubbornAtomStubbornAtom
6,14311339
$endgroup$
add a comment |
$begingroup$
Define $overline X=frac{1}{n}sumlimits_{k=1}^n X_k$ and $S^2=frac{1}{n-1}sumlimits_{k=1}^n (X_k-overline X)^2$.
Then a complete sufficient statistic for $(mu,sigma^2)$ is given by $(overline X, S^2)$.
By Lehmann-Scheffe theorem, any unbiased estimator of $sigma^p$ based on $(overline X, S^2)$ will be the UMVUE of $sigma^p$.
Recall that $$frac{(n-1)S^2}{sigma^2}sim chi^2_{n-1}$$
So,
begin{align}
Eleft[frac{(n-1)S^2}{sigma^2}right]^{p/2}&=frac{1}{2^{frac{n-1}{2}}Gammaleft(frac{n-1}{2}right)}int_0^infty t^{p/2},e^{-t/2},t^{frac{n-1}{2}-1},mathrm{d}t
end{align}
Simplifying both sides of the above equation you will finally arrive at $$Eleft[cS^pright]=sigma^p$$
for some function $c(.)$ of $n$ and $p$.
Thus the UMVUE of $sigma^p$ is $c(n,p) S^p$.
answered Jan 22 at 10:29
StubbornAtomStubbornAtom
6,14311339
$endgroup$
add a comment |
$begingroup$
Define $overline X=frac{1}{n}sumlimits_{k=1}^n X_k$ and $S^2=frac{1}{n-1}sumlimits_{k=1}^n (X_k-overline X)^2$.
Then a complete sufficient statistic for $(mu,sigma^2)$ is given by $(overline X, S^2)$.
By Lehmann-Scheffe theorem, any unbiased estimator of $sigma^p$ based on $(overline X, S^2)$ will be the UMVUE of $sigma^p$.
Recall that $$frac{(n-1)S^2}{sigma^2}sim chi^2_{n-1}$$
So,
begin{align}
Eleft[frac{(n-1)S^2}{sigma^2}right]^{p/2}&=frac{1}{2^{frac{n-1}{2}}Gammaleft(frac{n-1}{2}right)}int_0^infty t^{p/2},e^{-t/2},t^{frac{n-1}{2}-1},mathrm{d}t
end{align}
Simplifying both sides of the above equation you will finally arrive at $$Eleft[cS^pright]=sigma^p$$
for some function $c(.)$ of $n$ and $p$.
Thus the UMVUE of $sigma^p$ is $c(n,p) S^p$.
answered Jan 22 at 10:29
StubbornAtomStubbornAtom
6,14311339
$endgroup$
Define $overline X=frac{1}{n}sumlimits_{k=1}^n X_k$ and $S^2=frac{1}{n-1}sumlimits_{k=1}^n (X_k-overline X)^2$.
Then a complete sufficient statistic for $(mu,sigma^2)$ is given by $(overline X, S^2)$.
By Lehmann-Scheffe theorem, any unbiased estimator of $sigma^p$ based on $(overline X, S^2)$ will be the UMVUE of $sigma^p$.
Recall that $$frac{(n-1)S^2}{sigma^2}sim chi^2_{n-1}$$
So,
begin{align}
Eleft[frac{(n-1)S^2}{sigma^2}right]^{p/2}&=frac{1}{2^{frac{n-1}{2}}Gammaleft(frac{n-1}{2}right)}int_0^infty t^{p/2},e^{-t/2},t^{frac{n-1}{2}-1},mathrm{d}t
end{align}
Simplifying both sides of the above equation you will finally arrive at $$Eleft[cS^pright]=sigma^p$$
for some function $c(.)$ of $n$ and $p$.
Thus the UMVUE of $sigma^p$ is $c(n,p) S^p$.
answered Jan 22 at 10:29
StubbornAtomStubbornAtom
6,14311339
answered Jan 22 at 10:29
StubbornAtomStubbornAtom
6,14311339
answered Jan 22 at 10:29
StubbornAtomStubbornAtom
6,14311339
answered Jan 22 at 10:29
StubbornAtomStubbornAtom
6,14311339
6,14311339
add a comment |
add a comment |
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$begingroup$
If I find the MLE then I know it is a function of complete sufficient statistics (property of MVE), then if I can trasnsform it to make it unbiased I will have an unbiased estimator which is a function of complete sufficient statistic and therefore it is umvue because of the lehman-scheffe theorem
$endgroup$
– franzk
Jan 22 at 10:31