Deriving the UMVUE for Rayleigh scale parameter Let $X_1, \dots, X_n$ be i.i.d. with the PDF given by $f(x \mid \theta) = 2\theta^{-1} xe^{-\frac{x^2}{\theta}}$ for $x > 0$. My task is to find the UMVUE for $\theta$, and I’m given the following hint: “$U(X)=\sum_{i=1}^nX_i^2$ is sufficient for $\theta$, and $U/\theta$ is a multiple of $\chi_{2n}^2$”.
It’s my understanding that to find the UMVUE, you need to derive $E[T \mid U = u]$ where $T$ is an unbiased estimator and $U$ is a complete sufficient statistic. However, I’m not sure how to make sense of the given hint. Where does the Chi-squared come from?
 A: Joint density of $X_1,X_2,\ldots,X_n$ is 
\begin{align}
f_{\theta}(x_1,x_2,\ldots,x_n)&=\prod_{i=1}^n f(x_i\mid\theta)
\\&=\left(\frac{2}{\theta}\right)^n \left(\prod_{i=1}^n x_i\right) \exp\left(-\frac{1}{\theta}\sum_{i=1}^n x_i^2\right)\mathbf1_{x_1,\ldots,x_n>0}\quad,\,\theta>0
\end{align}
This pdf is a member of the one-parameter exponential family. 
So it follows that a complete sufficient statistic for $\theta$ is indeed
$$U(X_1,X_2,\ldots,X_n)=\sum_{i=1}^n X_i^2$$
Yes it is true that the UMVUE of $\theta$ if it exists is given by $E(T\mid U)$ where $T$ is any unbiased estimator of $\theta$. This is what the Lehmann-Scheffe theorem says. As a corollary, it also says that any unbiased estimator of $\theta$ based on a complete sufficient statistic has to be the UMVUE of $\theta$. Here this corollary comes in handy.
To make sense of the hint given, find the distribution of $Y=X^2$ where $X$ has the Rayleigh pdf you are given.
Via change of variables, the pdf of $Y$ is 
\begin{align}
f_Y(y)&=f(\sqrt y\mid\theta)\left|\frac{dx}{dy}\right|\mathbf1_{y>0}
\\&=\frac{1}{\theta}e^{-y/\theta}\mathbf1_{y>0}\quad,\,\theta>0
\end{align}
In other words, $X_i^2$ are i.i.d Exponential with mean $\theta$ for each $i=1,\ldots,n$.
Or, $$\frac{2}{\theta}X_i^2\stackrel{\text{ i.i.d }}\sim\text{Exp with mean }2\equiv \chi^2_2$$
Thus implying $$\frac{2}{\theta}\sum_{i=1}^n X_i^2=\frac{2U}{\theta} \sim \chi^2_{2n}$$
So, 
\begin{align}
E_{\theta}\left(\frac{2U}{\theta}\right)=2n\implies E_{\theta}\left(\frac{U}{n}\right)=\theta
\end{align}
Hence the UMVUE of $\theta$ is $$\boxed{\frac{U}{n}=\frac{1}{n}\sum_{i=1}^n X_i^2}$$
However, we did not require finding the distribution of $X_i^2$ since it is easy to show directly that $$E_{\theta}(U)=\sum_{i=1}^n \underbrace{E_{\theta}(X_i^2)}_{\theta}=n\theta$$
