# Finding a function of sufficient statistic with the same expected value with $\psi(X)$

This problem is from a qualifying exam of mathematical statistics of Department of Mathematics, POSTECH, and I'm trying to solve this problem on reference of Casella & Berger's Statistical Inference.

Let $X$ be a random variable with pdf $f(x;\theta)$ and $T=T(X)$ be a sufficient statistic for $\theta$. A statistic $\psi$ is called a test function if $0\le \psi \le 1$. Show that for every test function $\psi(X)$, there exists a test function $\tilde{\psi}(T)$ such that $E_\theta\tilde{\psi}(T)=E_\theta\psi(X)$ for all $\theta$.

My attempt: $E_\theta\psi(X)=\int\psi(x)f(x;\theta)dx=\int\psi(x) \frac{f(x;\theta)}{g(t;\theta)}g(t;\theta)dx$

Here $g(t;\theta)$ is the pdf for $T=T(X)$. Now I stuck in manipulating the above integral formula. I'd like to exploit the sufficiency of $T$, but I can't find the way how to. Does anyone have ideas?

Any hints or advices are welcome! Thanks!

• Hi there and welcome. Please include an exact/full reference. Also, please inlcude the self-study tag. Best of luck. – Reviewer
– Jim
Jun 20, 2018 at 15:12
• Consider constructing the function $\tilde \psi$ directly from $\psi$ by exploiting the definition (not characterization!) of sufficient statistics.
– whuber
Jun 20, 2018 at 15:54
• @whuber Thanks! I finally figured out the answer. I'll post right here soon. Jun 21, 2018 at 5:59

By sufficiency of $T$, we get $f(x|t)f(t|\theta)=f(x|t,\theta)f(t|\theta)=f(x,t|\theta)$
Therefore, $E_\theta\psi(X)=\int\int\psi(x)f(x,t|\theta)dxdt=\int\int\psi(x)f(x|t)f(t|\theta)dxdt$
Letting $\tilde\psi(t)=\int\psi(x)f(x|t)dt$, we get the desired result.