Minimal sufficient statistic criterion
$begingroup$
(Note: This is not about Bayesian inference, but about classical inference)
Let ${P_theta}_{thetain Theta}$ be a family of probability measures on $mathbb{R}^n$ with density functions $f_theta$.
Let $T:mathbb{R}^n rightarrow mathscr{T}$ be a statistic.
Define $$D(x):={yin mathbb{R}^n: text{ There exists a positive function } h text{ such that } f_{theta}(x)=f_{theta}(y)h(x,y) text{ for all } theta }$$.
(Note that $D(x)$ forms a partition of $mathbb{R}^n$)
Assume that $T(x)=T(y)$ if and only if $xin D(y)$.
Many textbooks and even wikipedia asserts that, with the above assumption, one can show that $T$ is minimal sufficient.
However, I do not get this Here is a proof that textbooks I have seen have:
Let $alpha:range(T)rightarrow mathbb{R}^n$ be a representative function for $T$. (That is, $T(alpha(t))=t$)
If we pick $xinmathbb{R}^n$, from our assumption,$f_theta(x)=f_theta(alpha(T(x)))h(x,alpha(T(x))$ holds for all $theta$ for some positive function $h>0$.
If we take $g_theta:=f_thetacirc alpha$, then by Fisher-Neymann theorem, $T$ is sufficient.
However, since $alpha$ need not be measurable, $g_theta$ need not be measurable. So we cannot apply Fisher-Neymann theorem. (As you can see here, $alpha$ need mot be measurable)
Is $g_theta$ measurable in anyways? Or is there a correct proof for this?
Thank you in advance.
statistics statistical-inference
asked Dec 14 '18 at 0:41
RubertosRubertos
5,6902824
$endgroup$
add a comment |
$begingroup$
(Note: This is not about Bayesian inference, but about classical inference)
Let ${P_theta}_{thetain Theta}$ be a family of probability measures on $mathbb{R}^n$ with density functions $f_theta$.
Let $T:mathbb{R}^n rightarrow mathscr{T}$ be a statistic.
Define $$D(x):={yin mathbb{R}^n: text{ There exists a positive function } h text{ such that } f_{theta}(x)=f_{theta}(y)h(x,y) text{ for all } theta }$$.
(Note that $D(x)$ forms a partition of $mathbb{R}^n$)
Assume that $T(x)=T(y)$ if and only if $xin D(y)$.
Many textbooks and even wikipedia asserts that, with the above assumption, one can show that $T$ is minimal sufficient.
However, I do not get this Here is a proof that textbooks I have seen have:
Let $alpha:range(T)rightarrow mathbb{R}^n$ be a representative function for $T$. (That is, $T(alpha(t))=t$)
If we pick $xinmathbb{R}^n$, from our assumption,$f_theta(x)=f_theta(alpha(T(x)))h(x,alpha(T(x))$ holds for all $theta$ for some positive function $h>0$.
If we take $g_theta:=f_thetacirc alpha$, then by Fisher-Neymann theorem, $T$ is sufficient.
However, since $alpha$ need not be measurable, $g_theta$ need not be measurable. So we cannot apply Fisher-Neymann theorem. (As you can see here, $alpha$ need mot be measurable)
Is $g_theta$ measurable in anyways? Or is there a correct proof for this?
Thank you in advance.
statistics statistical-inference
asked Dec 14 '18 at 0:41
RubertosRubertos
5,6902824
$endgroup$
add a comment |
$begingroup$
(Note: This is not about Bayesian inference, but about classical inference)
Let ${P_theta}_{thetain Theta}$ be a family of probability measures on $mathbb{R}^n$ with density functions $f_theta$.
Let $T:mathbb{R}^n rightarrow mathscr{T}$ be a statistic.
Define $$D(x):={yin mathbb{R}^n: text{ There exists a positive function } h text{ such that } f_{theta}(x)=f_{theta}(y)h(x,y) text{ for all } theta }$$.
(Note that $D(x)$ forms a partition of $mathbb{R}^n$)
Assume that $T(x)=T(y)$ if and only if $xin D(y)$.
Many textbooks and even wikipedia asserts that, with the above assumption, one can show that $T$ is minimal sufficient.
However, I do not get this Here is a proof that textbooks I have seen have:
Let $alpha:range(T)rightarrow mathbb{R}^n$ be a representative function for $T$. (That is, $T(alpha(t))=t$)
If we pick $xinmathbb{R}^n$, from our assumption,$f_theta(x)=f_theta(alpha(T(x)))h(x,alpha(T(x))$ holds for all $theta$ for some positive function $h>0$.
If we take $g_theta:=f_thetacirc alpha$, then by Fisher-Neymann theorem, $T$ is sufficient.
However, since $alpha$ need not be measurable, $g_theta$ need not be measurable. So we cannot apply Fisher-Neymann theorem. (As you can see here, $alpha$ need mot be measurable)
Is $g_theta$ measurable in anyways? Or is there a correct proof for this?
Thank you in advance.
statistics statistical-inference
asked Dec 14 '18 at 0:41
RubertosRubertos
5,6902824
$endgroup$
(Note: This is not about Bayesian inference, but about classical inference)
Let ${P_theta}_{thetain Theta}$ be a family of probability measures on $mathbb{R}^n$ with density functions $f_theta$.
Let $T:mathbb{R}^n rightarrow mathscr{T}$ be a statistic.
Define $$D(x):={yin mathbb{R}^n: text{ There exists a positive function } h text{ such that } f_{theta}(x)=f_{theta}(y)h(x,y) text{ for all } theta }$$.
(Note that $D(x)$ forms a partition of $mathbb{R}^n$)
Assume that $T(x)=T(y)$ if and only if $xin D(y)$.
Many textbooks and even wikipedia asserts that, with the above assumption, one can show that $T$ is minimal sufficient.
However, I do not get this Here is a proof that textbooks I have seen have:
Let $alpha:range(T)rightarrow mathbb{R}^n$ be a representative function for $T$. (That is, $T(alpha(t))=t$)
If we pick $xinmathbb{R}^n$, from our assumption,$f_theta(x)=f_theta(alpha(T(x)))h(x,alpha(T(x))$ holds for all $theta$ for some positive function $h>0$.
If we take $g_theta:=f_thetacirc alpha$, then by Fisher-Neymann theorem, $T$ is sufficient.
However, since $alpha$ need not be measurable, $g_theta$ need not be measurable. So we cannot apply Fisher-Neymann theorem. (As you can see here, $alpha$ need mot be measurable)
Is $g_theta$ measurable in anyways? Or is there a correct proof for this?
Thank you in advance.
statistics statistical-inference
statistics statistical-inference
asked Dec 14 '18 at 0:41
RubertosRubertos
5,6902824
asked Dec 14 '18 at 0:41
RubertosRubertos
5,6902824
edited Dec 14 '18 at 1:03
Rubertos
asked Dec 14 '18 at 0:41
RubertosRubertos
5,6902824
asked Dec 14 '18 at 0:41
RubertosRubertos
5,6902824
asked Dec 14 '18 at 0:41
RubertosRubertos
5,6902824
5,6902824
add a comment |
add a comment |
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