Mixing
Sufficient statistic for $N(mu,1)$
$begingroup$
Let $X_1,ldots,X_n$ be a random sample from $N(mu,1)$, where $mu$ is an unknown parameter. Show that $(overline{x}/{S^2}, S^2)$ is a sufficient statistic for $mu$, where $S^2$ is the sample variance.
My Approach
$$
sum_{i=1}^n (x_i-mu)^2 = n(bar x-mu)^2 + sum_{i=1}^n (x_i-bar x)^2.
$$
By Neyman Fisher Factorisation,
begin{align}
f_{X_1,ldots,X_n}(x_1,ldots,x_n) propto {} & prod_{i=1}^n frac 1 {(sqrt2pi)^n} expleft( frac{-1} 2 left( {x_i-mu} right)^2 right) = frac 1 {(sqrt2pi)^n} expleft( frac{-1}{2} sum_{i=1}^n (x_i-mu)^2 right) \[10pt]
= {} & (sqrt2pi)^{-n} expleft( frac{-1}{2} left( n(bar x - mu)^2 + sum_{i=1}^n (x_i - bar x)^2 right) right) \[10pt]
= {} & (sqrt2pi)^{-n} expleft( - frac n {2} ((bar x - mu)^2 + s^2) right).
end{align}
How do I proceed from here?
statistics statistical-inference
asked Jan 30 at 20:58
LadyLady
1298
$endgroup$
add a comment |
$begingroup$
Let $X_1,ldots,X_n$ be a random sample from $N(mu,1)$, where $mu$ is an unknown parameter. Show that $(overline{x}/{S^2}, S^2)$ is a sufficient statistic for $mu$, where $S^2$ is the sample variance.
My Approach
$$
sum_{i=1}^n (x_i-mu)^2 = n(bar x-mu)^2 + sum_{i=1}^n (x_i-bar x)^2.
$$
By Neyman Fisher Factorisation,
begin{align}
f_{X_1,ldots,X_n}(x_1,ldots,x_n) propto {} & prod_{i=1}^n frac 1 {(sqrt2pi)^n} expleft( frac{-1} 2 left( {x_i-mu} right)^2 right) = frac 1 {(sqrt2pi)^n} expleft( frac{-1}{2} sum_{i=1}^n (x_i-mu)^2 right) \[10pt]
= {} & (sqrt2pi)^{-n} expleft( frac{-1}{2} left( n(bar x - mu)^2 + sum_{i=1}^n (x_i - bar x)^2 right) right) \[10pt]
= {} & (sqrt2pi)^{-n} expleft( - frac n {2} ((bar x - mu)^2 + s^2) right).
end{align}
How do I proceed from here?
statistics statistical-inference
asked Jan 30 at 20:58
LadyLady
1298
$endgroup$
$begingroup$
If $T$ is sufficient for $mu$, then so is $(T,T')$ for any other statistic $T'$.
$endgroup$
– StubbornAtom
Jan 31 at 7:07
add a comment |
$begingroup$
Let $X_1,ldots,X_n$ be a random sample from $N(mu,1)$, where $mu$ is an unknown parameter. Show that $(overline{x}/{S^2}, S^2)$ is a sufficient statistic for $mu$, where $S^2$ is the sample variance.
My Approach
$$
sum_{i=1}^n (x_i-mu)^2 = n(bar x-mu)^2 + sum_{i=1}^n (x_i-bar x)^2.
$$
By Neyman Fisher Factorisation,
begin{align}
f_{X_1,ldots,X_n}(x_1,ldots,x_n) propto {} & prod_{i=1}^n frac 1 {(sqrt2pi)^n} expleft( frac{-1} 2 left( {x_i-mu} right)^2 right) = frac 1 {(sqrt2pi)^n} expleft( frac{-1}{2} sum_{i=1}^n (x_i-mu)^2 right) \[10pt]
= {} & (sqrt2pi)^{-n} expleft( frac{-1}{2} left( n(bar x - mu)^2 + sum_{i=1}^n (x_i - bar x)^2 right) right) \[10pt]
= {} & (sqrt2pi)^{-n} expleft( - frac n {2} ((bar x - mu)^2 + s^2) right).
end{align}
How do I proceed from here?
statistics statistical-inference
asked Jan 30 at 20:58
LadyLady
1298
$endgroup$
Let $X_1,ldots,X_n$ be a random sample from $N(mu,1)$, where $mu$ is an unknown parameter. Show that $(overline{x}/{S^2}, S^2)$ is a sufficient statistic for $mu$, where $S^2$ is the sample variance.
My Approach
$$
sum_{i=1}^n (x_i-mu)^2 = n(bar x-mu)^2 + sum_{i=1}^n (x_i-bar x)^2.
$$
By Neyman Fisher Factorisation,
begin{align}
f_{X_1,ldots,X_n}(x_1,ldots,x_n) propto {} & prod_{i=1}^n frac 1 {(sqrt2pi)^n} expleft( frac{-1} 2 left( {x_i-mu} right)^2 right) = frac 1 {(sqrt2pi)^n} expleft( frac{-1}{2} sum_{i=1}^n (x_i-mu)^2 right) \[10pt]
= {} & (sqrt2pi)^{-n} expleft( frac{-1}{2} left( n(bar x - mu)^2 + sum_{i=1}^n (x_i - bar x)^2 right) right) \[10pt]
= {} & (sqrt2pi)^{-n} expleft( - frac n {2} ((bar x - mu)^2 + s^2) right).
end{align}
How do I proceed from here?
statistics statistical-inference
statistics statistical-inference
asked Jan 30 at 20:58
LadyLady
1298
asked Jan 30 at 20:58
LadyLady
1298
asked Jan 30 at 20:58
LadyLady
1298
asked Jan 30 at 20:58
LadyLady
1298
asked Jan 30 at 20:58
LadyLady
1298
1298
$begingroup$
If $T$ is sufficient for $mu$, then so is $(T,T')$ for any other statistic $T'$.
$endgroup$
– StubbornAtom
Jan 31 at 7:07
add a comment |
$begingroup$
If $T$ is sufficient for $mu$, then so is $(T,T')$ for any other statistic $T'$.
$endgroup$
– StubbornAtom
Jan 31 at 7:07
$begingroup$
If $T$ is sufficient for $mu$, then so is $(T,T')$ for any other statistic $T'$.
$endgroup$
– StubbornAtom
Jan 31 at 7:07
$begingroup$
If $T$ is sufficient for $mu$, then so is $(T,T')$ for any other statistic $T'$.
$endgroup$
– StubbornAtom
Jan 31 at 7:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since the variance in this problem is a known constant, it is actually the case that $bar{x}$ is sufficient for $mu$, and since $bar{x}$ can be obtained from the stated statistic (as the product of the two elements), that statistic is sufficient also sufficient for $mu$. To see that $bar{x}$ is sufficient for $mu$ we can apply the Fisher-Neyman factorisation theorem just as you have attempted to do. In this problem your likelihood function is given by:$^dagger$
$$begin{equation} begin{aligned}
L_mathbf{x}(mu)
&= (2 pi)^{-n/2} cdot exp Big( - frac{1}{2} (x_i-mu)^2 Big) \[6pt]
&= (2 pi)^{-n/2} cdot exp Big( - frac{n}{2} cdot (bar{x}-mu)^2 - frac{n-1}{2} cdot s^2 Big) \[6pt]
&= underbrace{(2 pi)^{-n/2} cdot exp Big( - frac{n-1}{2} cdot s^2 Big)}_{h(mathbf{x})} cdot underbrace{exp Big( - frac{n}{2} cdot (bar{x}-mu)^2 Big)}_{g_mu(bar{x},n)}. \[6pt]
end{aligned} end{equation}$$
It follows that $bar{x}$ is sufficient for $mu$. With some more work it is possible to show that it is minimal sufficient, but that is unnecessary for the result you want in this case.
$^dagger$ I have used the standard notation where the sample variance uses Bessel's correction) so my result is slightly different to yours.
answered Jan 31 at 3:31
BenBen
1,900215
$endgroup$
$begingroup$
how do I express $(overline{x}/{S^2}, S^2)$ as a sufficient statistic for $mu$?
$endgroup$
– Lady
Jan 31 at 11:08
$begingroup$
It is already sufficient. Since $mu$ is sufficient, then $mu / S^2 times S^2$ is sufficient, so the vector of both elements is sufficient (but not minimal sufficient).
$endgroup$
– Ben
Feb 1 at 23:34
add a comment |
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1 Answer
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$begingroup$
Since the variance in this problem is a known constant, it is actually the case that $bar{x}$ is sufficient for $mu$, and since $bar{x}$ can be obtained from the stated statistic (as the product of the two elements), that statistic is sufficient also sufficient for $mu$. To see that $bar{x}$ is sufficient for $mu$ we can apply the Fisher-Neyman factorisation theorem just as you have attempted to do. In this problem your likelihood function is given by:$^dagger$
$$begin{equation} begin{aligned}
L_mathbf{x}(mu)
&= (2 pi)^{-n/2} cdot exp Big( - frac{1}{2} (x_i-mu)^2 Big) \[6pt]
&= (2 pi)^{-n/2} cdot exp Big( - frac{n}{2} cdot (bar{x}-mu)^2 - frac{n-1}{2} cdot s^2 Big) \[6pt]
&= underbrace{(2 pi)^{-n/2} cdot exp Big( - frac{n-1}{2} cdot s^2 Big)}_{h(mathbf{x})} cdot underbrace{exp Big( - frac{n}{2} cdot (bar{x}-mu)^2 Big)}_{g_mu(bar{x},n)}. \[6pt]
end{aligned} end{equation}$$
It follows that $bar{x}$ is sufficient for $mu$. With some more work it is possible to show that it is minimal sufficient, but that is unnecessary for the result you want in this case.
$^dagger$ I have used the standard notation where the sample variance uses Bessel's correction) so my result is slightly different to yours.
answered Jan 31 at 3:31
BenBen
1,900215
$endgroup$
$begingroup$
how do I express $(overline{x}/{S^2}, S^2)$ as a sufficient statistic for $mu$?
$endgroup$
– Lady
Jan 31 at 11:08
$begingroup$
It is already sufficient. Since $mu$ is sufficient, then $mu / S^2 times S^2$ is sufficient, so the vector of both elements is sufficient (but not minimal sufficient).
$endgroup$
– Ben
Feb 1 at 23:34
add a comment |
$begingroup$
Since the variance in this problem is a known constant, it is actually the case that $bar{x}$ is sufficient for $mu$, and since $bar{x}$ can be obtained from the stated statistic (as the product of the two elements), that statistic is sufficient also sufficient for $mu$. To see that $bar{x}$ is sufficient for $mu$ we can apply the Fisher-Neyman factorisation theorem just as you have attempted to do. In this problem your likelihood function is given by:$^dagger$
$$begin{equation} begin{aligned}
L_mathbf{x}(mu)
&= (2 pi)^{-n/2} cdot exp Big( - frac{1}{2} (x_i-mu)^2 Big) \[6pt]
&= (2 pi)^{-n/2} cdot exp Big( - frac{n}{2} cdot (bar{x}-mu)^2 - frac{n-1}{2} cdot s^2 Big) \[6pt]
&= underbrace{(2 pi)^{-n/2} cdot exp Big( - frac{n-1}{2} cdot s^2 Big)}_{h(mathbf{x})} cdot underbrace{exp Big( - frac{n}{2} cdot (bar{x}-mu)^2 Big)}_{g_mu(bar{x},n)}. \[6pt]
end{aligned} end{equation}$$
It follows that $bar{x}$ is sufficient for $mu$. With some more work it is possible to show that it is minimal sufficient, but that is unnecessary for the result you want in this case.
$^dagger$ I have used the standard notation where the sample variance uses Bessel's correction) so my result is slightly different to yours.
answered Jan 31 at 3:31
BenBen
1,900215
$endgroup$
$begingroup$
how do I express $(overline{x}/{S^2}, S^2)$ as a sufficient statistic for $mu$?
$endgroup$
– Lady
Jan 31 at 11:08
$begingroup$
It is already sufficient. Since $mu$ is sufficient, then $mu / S^2 times S^2$ is sufficient, so the vector of both elements is sufficient (but not minimal sufficient).
$endgroup$
– Ben
Feb 1 at 23:34
add a comment |
$begingroup$
Since the variance in this problem is a known constant, it is actually the case that $bar{x}$ is sufficient for $mu$, and since $bar{x}$ can be obtained from the stated statistic (as the product of the two elements), that statistic is sufficient also sufficient for $mu$. To see that $bar{x}$ is sufficient for $mu$ we can apply the Fisher-Neyman factorisation theorem just as you have attempted to do. In this problem your likelihood function is given by:$^dagger$
$$begin{equation} begin{aligned}
L_mathbf{x}(mu)
&= (2 pi)^{-n/2} cdot exp Big( - frac{1}{2} (x_i-mu)^2 Big) \[6pt]
&= (2 pi)^{-n/2} cdot exp Big( - frac{n}{2} cdot (bar{x}-mu)^2 - frac{n-1}{2} cdot s^2 Big) \[6pt]
&= underbrace{(2 pi)^{-n/2} cdot exp Big( - frac{n-1}{2} cdot s^2 Big)}_{h(mathbf{x})} cdot underbrace{exp Big( - frac{n}{2} cdot (bar{x}-mu)^2 Big)}_{g_mu(bar{x},n)}. \[6pt]
end{aligned} end{equation}$$
It follows that $bar{x}$ is sufficient for $mu$. With some more work it is possible to show that it is minimal sufficient, but that is unnecessary for the result you want in this case.
$^dagger$ I have used the standard notation where the sample variance uses Bessel's correction) so my result is slightly different to yours.
answered Jan 31 at 3:31
BenBen
1,900215
$endgroup$
Since the variance in this problem is a known constant, it is actually the case that $bar{x}$ is sufficient for $mu$, and since $bar{x}$ can be obtained from the stated statistic (as the product of the two elements), that statistic is sufficient also sufficient for $mu$. To see that $bar{x}$ is sufficient for $mu$ we can apply the Fisher-Neyman factorisation theorem just as you have attempted to do. In this problem your likelihood function is given by:$^dagger$
$$begin{equation} begin{aligned}
L_mathbf{x}(mu)
&= (2 pi)^{-n/2} cdot exp Big( - frac{1}{2} (x_i-mu)^2 Big) \[6pt]
&= (2 pi)^{-n/2} cdot exp Big( - frac{n}{2} cdot (bar{x}-mu)^2 - frac{n-1}{2} cdot s^2 Big) \[6pt]
&= underbrace{(2 pi)^{-n/2} cdot exp Big( - frac{n-1}{2} cdot s^2 Big)}_{h(mathbf{x})} cdot underbrace{exp Big( - frac{n}{2} cdot (bar{x}-mu)^2 Big)}_{g_mu(bar{x},n)}. \[6pt]
end{aligned} end{equation}$$
It follows that $bar{x}$ is sufficient for $mu$. With some more work it is possible to show that it is minimal sufficient, but that is unnecessary for the result you want in this case.
$^dagger$ I have used the standard notation where the sample variance uses Bessel's correction) so my result is slightly different to yours.
answered Jan 31 at 3:31
BenBen
1,900215
answered Jan 31 at 3:31
BenBen
1,900215
answered Jan 31 at 3:31
BenBen
1,900215
answered Jan 31 at 3:31
BenBen
1,900215
1,900215
$begingroup$
how do I express $(overline{x}/{S^2}, S^2)$ as a sufficient statistic for $mu$?
$endgroup$
– Lady
Jan 31 at 11:08
$begingroup$
It is already sufficient. Since $mu$ is sufficient, then $mu / S^2 times S^2$ is sufficient, so the vector of both elements is sufficient (but not minimal sufficient).
$endgroup$
– Ben
Feb 1 at 23:34
add a comment |
$begingroup$
how do I express $(overline{x}/{S^2}, S^2)$ as a sufficient statistic for $mu$?
$endgroup$
– Lady
Jan 31 at 11:08
$begingroup$
It is already sufficient. Since $mu$ is sufficient, then $mu / S^2 times S^2$ is sufficient, so the vector of both elements is sufficient (but not minimal sufficient).
$endgroup$
– Ben
Feb 1 at 23:34
$begingroup$
how do I express $(overline{x}/{S^2}, S^2)$ as a sufficient statistic for $mu$?
$endgroup$
– Lady
Jan 31 at 11:08
$begingroup$
how do I express $(overline{x}/{S^2}, S^2)$ as a sufficient statistic for $mu$?
$endgroup$
– Lady
Jan 31 at 11:08
$begingroup$
It is already sufficient. Since $mu$ is sufficient, then $mu / S^2 times S^2$ is sufficient, so the vector of both elements is sufficient (but not minimal sufficient).
$endgroup$
– Ben
Feb 1 at 23:34
$begingroup$
It is already sufficient. Since $mu$ is sufficient, then $mu / S^2 times S^2$ is sufficient, so the vector of both elements is sufficient (but not minimal sufficient).
$endgroup$
– Ben
Feb 1 at 23:34
add a comment |
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$begingroup$
If $T$ is sufficient for $mu$, then so is $(T,T')$ for any other statistic $T'$.
$endgroup$
– StubbornAtom
Jan 31 at 7:07