Let
$$p(x|\theta)=\frac{\left(\ln\theta\right)^x}{\theta x!}I_{\{0,1,2,\dots\}}(x)\; $$
I have to find UMVUE for $h(\theta)=\ln\theta$ and for $h(\theta)=(\ln\theta)^2$
I know $T(X_1,\dots,X_n)=\sum\limits_{i=1}^n X_i$ is complete and sufficient for $\theta$
By CRLB I found $\overline{X}$ as UMVUE for $h(\theta)=\ln\theta$ but now I don't know how to find the second one. Can you help me please?
Best Answer
The parent distribution is just $\mathsf{Poisson}(\ln\theta)$ where I assume $\theta>1$.
So assuming $X_1,\ldots,X_n$ are i.i.d $\mathsf{Poisson}(\ln\theta)$, you have $T=\sum\limits_{i=1}^n X_i\sim \mathsf{Poisson}(n\ln\theta)$.
Now,
$$E_{\theta}(T^2)=\operatorname{Var}_{\theta}(T)+(E_{\theta}(T))^2=n\ln\theta+n^2(\ln\theta)^2\quad,\,\forall\,\theta$$
Therefore, $$E_{\theta}(T^2-T)=E_{\theta}(T^2)-E_{\theta}(T)=n^2(\ln\theta)^2\quad,\,\forall\,\theta$$
Hence the UMVUE of $(\ln\theta)^2$ is $$\widehat{(\ln\theta)^2}=\frac{T(T-1)}{n^2}$$