# Deriving the UMVUE for Rayleigh scale parameter

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

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

• Read it this time. Thanks for the response. – David Nov 19 '18 at 16:02
• If I were to also find the UMVUE estimator of $\theta^{-1}$, how could I alter this process? – David Nov 20 '18 at 23:11
• @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. – StubbornAtom Nov 21 '18 at 5:27