UMVUE of $\theta$ when $X_i$'s are i.i.d with pdf $f(x;\theta)=\theta x^{-(1+\theta)}I_{x>1}$

Question

I am trying to find the Unique minimum variance unbiased estimator for $\theta$ where $\{X_i\}_{i=i}^{n}\sim f(x;\theta)=\theta x^{-(1+\theta)}$ where $x>1$ and $\theta\in(1,\infty)$.

I start by showing that $f(x;\theta)=\theta x^{-(1+\theta)}$ is in the exponential:

$$f(x;\theta)=e^{\ln(\theta)-(1+\theta)\ln(x)}I_{x>1}$$

Since $f(x;\theta)$ is a member of the exponential family of full rank since the parameter space contains an open interval. then $\sum_{i=1}^{n}\ln(x_i)$ is a complete and minimally sufficient statistic. Since, $g(x)=e^x$ is a one to one transformation then, $\prod_{i=1}^{n}x_i$ is also a minimally sufficient statistic. By a similar argument one can conclude that $S(X)=\sum_{i=1}^{n}x_i$ is also minimally sufficient and complete. Note:

$$\int_{1}^{\infty}x\theta x^{-(1+\theta)}dx=\theta\int_{1}^{\infty}x^{-\theta}dx=\frac{\theta}{1-\theta}$$

Since, $\theta>1$ and $x>1$. Then $E[\sum_{i=1}^{n}X_i]=\frac{n\theta}{1-\theta}$.

Note that can only achieve CR Lower bound if in exponential family and estimating a linear function of the minimum sufficient statistic. Note: $E[a+bS(X)]=a+\frac{bn\theta}{1-\theta}$ so no linear combination of $S(X)$ can achieve an unbiased estimator of $\theta$ so there does not exist a UMVUE for $\theta$. IS my logic correct?

Answer 1

Your reasoning is incorrect in two places. Firstly, you are right that $\prod_{i=1}^{n}X_i$ or $\sum_{i=1}^{n}\ln X_i$ is minumal sufficient statistics. And the $S(X)=\sum_{i=1}^{n}X_i$ is not sufficient at all unless $n\neq 1$. No similar argument will give it. The second wrong place is that if CR lower bound is not attained then UMVUE does not exists. This is exactly the example where the unbiased estimate of $\theta$ is a function of complete and minimal sufficient statistic (an then UMVUE) but does not attain CR lower bound of variance.

You can find that $\ln X_1$ has exponential distribution with expectation $\frac1\theta$, then $\sum_{i=1}^n \ln X_i$ is gamma distributed, then calculate expectation of $\frac{1}{\sum_{i=1}^n \ln X_i}$ and finally obtain that $$\mathbb E\left[\frac{n-1}{\sum_{i=1}^n\ln X_i}\right]=\theta$$ so $\theta^*=\frac{n-1}{\sum_{i=1}^n\ln X_i}$ is the unbiased estimate of $\theta$. And it is UMVUE being a function of complete and sufficient statistic.