Let $X_1,...,X_n$ be iid with the pdf given by $f(x|\theta)=2\theta^{-1}xe^{-x^2/\theta}$ 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 $E[T|U=u]$ 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?


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 $$\frac{2}{\theta}\sum_{i=1}^n X_i^2=\frac{2U}{\theta} \sim \chi^2_{2n}$$


\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 $$E_{\theta}(U)=\sum_{i=1}^n \underbrace{E_{\theta}(X_i^2)}_{\theta}=n\theta$$

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    $\begingroup$ Read it this time. Thanks for the response. $\endgroup$ – David Nov 19 '18 at 16:02
  • $\begingroup$ If I were to also find the UMVUE estimator of $\theta^{-1}$, how could I alter this process? $\endgroup$ – David Nov 20 '18 at 23:11
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    $\begingroup$ @DavidS The procedure is the same, except you would have to find an unbiased estimator of $\theta^{-1}$ based on the complete sufficient statistic $U$. It is reasonable to start with $E(1/U)$ and see whether it results in the form $k/\theta$. Then $1/(kU)$ is your UMVUE. $\endgroup$ – StubbornAtom Nov 21 '18 at 5:27

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