How do I find the UMVUE of $(1-\theta)^2$?

Question

Let $X_1,X_2,\ldots,X_n$ have the pdf $f(x;\theta) = \frac {\theta}{x^{\theta+1}},$ if $x>1$ and $f(x;\theta) = 0$ otherwise.

Let $\theta>0$. I have to find the UMVUE of $(1-\theta)^2$.

I’ve found the UMVUE of $\theta$ to be $\frac{n-1}{T}$ where $T=\sum_{i=1}^n \log X_i$ is a complete and sufficient statistic for the parameter $\theta$. $T$ has a $\Gamma(\theta,n)$ distribution and I found out the expectation of $\frac{1}{T}$ which came out to be $\frac{\theta}{n-1}$, and adjusted accordingly to get the UMVUE for $\theta$.

Now, I’m not sure how to proceed further. Is there any theorem similar to the invariance property in case of MLEs which would relate UMVUE’s of $\theta$ to get the UMVUE of $(1-\theta)^2$. If not, how else can I proceed?

Answer 1

For a polynomial in $\theta$, it suffices to find UMVUE of the individual powers of $\theta$.

Verify that for $n>2$,

\begin{align} E_{\theta}\left[\frac{(n-1)(n-2)}{T^2}\right]=\theta^2\quad,\forall\,\theta>0 \end{align}

You already have $$E_{\theta}\left[\frac{n-1}{T}\right]=\theta\quad,\forall\,\theta>0$$

So whenever $n>2$, the UMVUE of $\theta^2-2\theta+1$ must be $$h(T)=\frac{(n-1)(n-2)}{T^2}-\frac{2(n-1)}{T}+1$$