# A question about a proof of Neyman's factorization theorem

This question comes from the proof of Neyman's factorization theorem in Robert V. Hogg, Joseph W. McKean, Allen T. Craig, "Introduction to Mathematical Statistics", 6th edition, pp 376-377.

In the proof, a one-to-one transformation is used which is indicated by the red line. But I could not understand why such a one-to-one transformation surely exists. Can you tell me?

Thank you for any help!

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$y_2,\dots,y_n$ can be chosen to be anything so that the transformation $(x_1,\dots,x_n)\mapsto (y_1,\dots,y_n)$ is bijective. The choice is not going to afect the proof. – Ashok Jun 21 '12 at 7:12
I can't get it. Could you please explain in a bit more detail? For example, if $u_1$ is not bijective, how to construct such a one-to-one transformation? – Zhou Heng Jun 21 '12 at 7:49
Yes, you seem to be right. – Ashok Jun 21 '12 at 8:48
The transformation of x1, x2, ...xn to y1 is the sufficient statistic given in the problem. But I am not sure how the other yis are constructed to get the 1-1 transformation from the xs to the ys. – Michael Chernick Jun 22 '12 at 22:34
This proof seems to be handwavy on purpose. For example, $y_{1}$ might not be differentiable and then you would not be able to get the Jacobian. The important claim is that if T is sufficient and T(x) = t then the density of t is proportional to the density of x''. There are some problems with that... the density of $t$ with respect to what? Is $T$ necessarily continuous? etc... I can show this rigorously using Measure Theory but I don't know if that would be good for you. – madprob Jun 26 '12 at 3:52

I wrote this stuff as my personal notes in some class as an adaptation of the proof in Theory of Statistics by Mark Schervish.

We use this definition of Sufficiency:

$\textbf{Definition 1.1}:$ A statistic $T$ is $\textit{sufficient}$ for $\Theta$ if $\forall A \in \sigma(X)$, $\exists$ a version of $P_{\theta}(A|T)$ functionally independent of $\theta$. In this case, we call $P_{\theta}(A|T)$ by $P(A|T)$.

This is the more abstract version of $P_{\theta}(X=x|T=t)$ not depending on $\theta$.

Next, we need these Lemmas (I can write the proof if it helps):

$\textbf{Lemma 1.1}$: If $\forall \theta \in \Omega$, $P_{\theta} << \nu$ for a $\sigma$-finite $\nu$, then $\exists (c_{i})_{i=1}^{\infty}$ in $[0,1]$ s.t. $\sum_{i=1}^{\infty}{c_{i}} = 1$ and $(\theta_{i})_{i=1}^{\infty}$ in $\Omega$ s.t. $\nu^{*} := \sum_{i=1}^{\infty}{c_{i}P_{\theta_{i}}}$ is a probability measure and $\forall \theta \in \Omega$, $P_{\theta} << \nu^{*} << \nu$.

$\textbf{Lemma 1.2}$: Suppose $\exists$ $\sigma$-finite $\nu$ on $(\chi,\beta_{X})$ s.t. $\forall \theta \in \Omega, P_{\theta} << \nu$ and that $T$ is sufficient for $\Theta$. Take $\nu^{*}$ as in Lemma $1.1$. $\forall \theta \in \Omega$, $\nu^{*}(A|T)$ is a $\nu^{*}$-version of $P_{\theta}(A|T)$.

$\textbf{Lemma 1.3}$: Suppose $\exists$ $\sigma$-finite $\nu$ on $(\chi,\beta_{X})$ s.t. $\forall \theta \in \Omega, P_{\theta} << \nu$ and take $\nu^{*}$ as in Lemma $1.1$. There $\exists$ functions $m_{1} \& m_{2,\theta}$ s.t. $\forall \theta \in \Omega$, a $\nu$-version of $\frac{dP_{\theta}}{d\nu}$ has $\forall x \in \chi$, $\frac{dP_{\theta}}{d\nu} = m_{1}m_{2,\theta}$, $m_{1}$ functionally independent of $\theta$ and $m_{2}$ $\sigma(T)$-measurable iif $\forall \theta \in \Omega$, there exists a $\sigma(T)$-measurable $\nu^{*}$-version of $\frac{dP_{\theta}}{d\nu^{*}}$. Whenever there is no confusion, we will also call this version by $\frac{dP_{\theta}}{d\nu^{*}}$.

$\textbf{Proof}$: $\textbf{Stage}$ 1: if''

$\$

First, observe that:

$$\frac{d\nu^{*}}{d\nu} = \sum_{j}{c_{j}\frac{dP_{\theta_{j}}}{d\nu}} =^{\nu} m_{1}\sum_{j}{c_{j}m_{2,\theta_{j}}}$$

Next, observe that, since $\nu^{*} << \nu$:

$$m_{1}m_{2,\theta} =^{\nu^{*}} \frac{dP_{\theta}}{d\nu^{*}} = \frac{dP_{\theta}}{d\nu} \frac{d\nu^{*}}{d\nu} =^{\nu^{*}} \frac{dP_{\theta}}{d\nu} m_{1}\sum_{j}{c_{j}m_{2,\theta_{j}}}$$

Observe that $A = \{x \in \chi: m_{1}\sum_{j}{c_{j}m_{2,\theta_{j}}} = 0\}$ is such that $\nu^{*}(A) = 0$, see\footnote{$\nu^{*}(A) = \int_{A}{d\nu^{*}} = \int_{A}{\frac{d\nu^{*}}{d\nu}d\nu} = \int_{A}{m_{1}\sum_{j}{c_{j}m_{2,\theta_{j}}}d\nu} = 0$}. Hence,

$$\frac{dP_{\theta}}{d\nu^{*}} =^{\nu^{*}} \frac{m_{1}m_{2,\theta}}{m_{1}\sum_{j}{c_{j}m_{2,\theta_{j}}}} = \frac{m_{2,\theta}}{\sum_{j}{c_{j}m_{2,\theta_{j}}}}$$

Since $\frac{m_{2,\theta}}{\sum_{j}{c_{j}m_{2,\theta_{j}}}}$ is $\sigma(T)$-measurable, the proof is complete.

$\$

$\textbf{Stage}$ 2: only if''

$\$

We wish to prove that there exist $m_{1}$ and $m_{2,\theta}$ s.t. $\forall A \in \sigma(X)$,

$$\int_{A}{\frac{dP_{\theta}}{d\nu}d\nu} = \int_{A}{m_{1}m_{2,\theta}d\nu}$$

$$\int_{A}{\frac{dP_{\theta}}{d\nu}d\nu} =^{R.N} \int_{A}{dP_{\theta}} =^{R.N} \int_{A}{\frac{dP_{\theta}}{d\nu^{*}}d\nu^{*}} =^{R.N.} \int_{A}{\frac{dP_{\theta}}{d\nu^{*}}\frac{d\nu^{*}}{d\nu}d\nu}$$

Taking $m_{1} = \frac{d\nu^{*}}{d\nu}$ and $m_{2,\theta} = \frac{dP_{\theta}}{d\nu^{*}}$, the proof is complete.

$\$

$\$

$\textbf{Theorem 1.1}$ (Fisher-Neyman Factorization): Suppose $\exists$ $\sigma$-finite $\nu$ on $(\chi,\beta_{X})$ s.t. $\forall \theta \in \Omega, P_{\theta} << \nu$. Then $T$ is sufficient for $\Theta$ iif $\exists$ functions $m_{1} \& m_{2,\theta}$ s.t. $\forall \theta \in \Omega$, a $\nu$-version of $\frac{dP_{\theta}}{d\nu}$ has $\forall x \in \chi$, $\frac{dP_{\theta}}{d\nu} = m_{1}m_{2,\theta}$, $m_{1}$ functionally independent of $\theta$ and $m_{2}$ is $\sigma(T)$-measurable.

$\$

$\textbf{Proof}:$ $\textbf{Stage}$ 1: only if''

$\$

We wish to find that $m_{1}$ and $m_{2}$ as in the Theorem such that, $\forall A \in \sigma(X)$, $\forall \theta \in \Omega$:

$$\int_{A}{\frac{dP_{\theta}}{d\nu}d\nu} = \int_{A}{m_{1}m_{2,\theta}d\nu}$$

Take an arbitrary $A \in \sigma(X)$ and consider $\nu^{*}$ as in Lemma $1.1$.

$$\int_{A}{\frac{dP_{\theta}}{d\nu}d\nu} =^{R.N} E_{P_{\theta}}(I_{A}) = E_{P_{\theta}}(P_{\theta}(A|T)) = E_{\nu^{*}}(P_{\theta}(A|T) \frac{dP_{\theta}}{d\nu^{*}}) =^ {L.2} E_{\nu^{*}}(\nu^{*}(A|T) \frac{dP_{\theta}}{d\nu^{*}}) =^{T.L.}$$

$$= E_{\nu^{*}}(\nu^{*}(A|T) E(\frac{dP_{\theta}}{d\nu^{*}}|T)) := E_{\nu^{*}}(\nu^{*}(A|T) m_{2,\theta}) = E_{\nu^{*}}(E_{\nu^{*}}(I_{A}m_{2,\theta}|T)) =^{T.L.} E_{\nu^{*}}(I_{A}m_{2,\theta}) =^{R.N.}$$

$$= E_{\nu}(I_{A}\frac{d\nu^{*}}{d\nu}m_{2,\theta}) = \int_{A}{\frac{d\nu^{*}}{d\nu}m_{2,\theta}d\nu} := \int_{A}{m_{1}m_{2,\theta}d\nu}$$

$\textbf{Stage}$ 2: if''

$\$

We will show that $\forall \theta \in \Omega$, $\nu_{*}(A|T)$ is a $P_{\theta}$-version of $P_{\theta}(A|T)$. Since $\nu^{*}(A|T)$ is functionally independent of $\theta$, the proof will be complete. Take an arbitrary $\theta$, since $\nu_{*}(A|T)$ and $P_{\theta}(A|T)$ are $\sigma(T)$-measurable, we wish to show that:

$$\forall B \in \sigma(T), \int_{B}{P_{\theta}(A|T)dP_{\theta}} = \int_{B}{\nu^{*}(A|T)dP_{\theta}}$$

Take an arbitrary $B \in \sigma(T)$:

$$\int_{B}{P_{\theta}(A|T)dP_{\theta}} = E_{P_{\theta}}(I_{A}I_{B}) =^{R.N.} E_{\nu^{*}}(I_{A}I_{B}\frac{dP_{\theta}}{d\nu^{*}}) =^{T.L.} E_{\nu^{*}}(E_{\nu^{*}}(I_{A}\frac{dP_{\theta}}{d\nu^{*}}|T)I_{B}) =^{L.3}$$

$$= E_{\nu^{*}}(E_{\nu^{*}}(I_{A}|T)\frac{dP_{\theta}}{d\nu^{*}}I_{B}) = E_{\nu^{*}}(\nu^{*}(A|T)I_{B}\frac{dP_{\theta}}{d\nu^{*}}) =^{R.N.} E_{P_{\theta}}(\nu^{*}(A|T)I_{B}) = \int_{B}{\nu^{*}(A|T)dP_{\theta}}$$

Which completes the proof of Theorem $1.1$.

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Thank you very much! I'll try my best to understand it. Thank you again! – Zhou Heng Jun 28 '12 at 0:15
Good Luck ;) and feel free to ask for any clarifications. – madprob Jun 28 '12 at 0:48

The proof is essentially the same as the one in the discrete case. Let $X = X_{1},\ldots,X_{n}$ be random variables with a discrete distribution.

$T$ is sufficient for $\theta$ if $P_{\theta}(X=x|T=t)$ does not depend on $\theta$.

If $T$ is sufficient,

$$P_{\theta}(X=x) = P_{\theta}(X=x,T=t) = P_{\theta}(T=t)P_{\theta}(X=x|T=t) = g_{\theta}(t)f(x)$$

If $P_{\theta}(X=x) = g_{\theta}(t)f(x)$,

$$P_{\theta}(T=t) = \sum_{x: T(x)=t}{P_{\theta}(X=x)} = g_{\theta}(t)\sum_{x: T(x)=t}{f(x)}$$

Hence,

$$P_{\theta}(X=x|T=t) = \frac{P_{\theta}(X=x,T=t)}{P_{\theta}(T=t)} = \frac{P_{\theta}(X=x)}{P_{\theta}(T=t)} = \frac{g_{\theta}(t)f(x)}{g_{\theta}(t)\sum_{y: T(y)=t}{f(y)}} = \frac{f(x)}{\sum_{y: T(y)=t}{f(y)}}$$

which does not depend on $\theta$.

This idea is also more or less the one the book you posted was trying to follow.

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