showing that $S=\sum^n_{i=1} \ln(X_i)$ is a complete sufficient statistic

Question

We have a random sample $X_1,X_2,\ldots,X_n$ from a probability with density:

$$ f(x)=\theta x^{-(\theta +1)} $$ given that $x>1$ and $0$ else.

now the question is:

Show that $S=\sum^n_{i=1} \ln(X_i)$ is a complete sufficient statistic and use this result to derive the UMVUE for $\frac{1}{\theta}$.

Now I made the joint density function like so: $$ \begin{align} f_\bar{x}(x_1,\ldots,x_n;\theta) &= \theta^n\prod_{i=1}^n x_i^{-(\theta +1)}I_{(1,\infty)}(x_i)\\ &= \theta^n\prod_{i=1}^n [I_{(1,\infty)}(x_i) ] e^{\sum(-\theta-1)\log(x_i)}\\ &=\theta^n\prod_{i=1}^n I_{(1,\infty)}(x_i)e^{(-\theta-1)\sum \log(x_i)} \end{align} $$ Now im doing this blindy, but im just following my book. what am I doing? what does being a complete sufficient statistic tell me? This a kink in my brain, I cant do this sort of thing without knowing what it is. Also how do i go from here? how to get the UMVUE?

Answer 1

You can write $\prod_i I_{(1,\infty)}(x_i)$ as $I_{(1,\infty)} (\min\{ x_1,\ldots,x_n\}) \cdot \prod_i x_i$. That matters in contexts where instead of $(1,\infty)$ you have $(\kappa,\infty)$ and $\kappa$ itself is to be estimated. It means that the minimum observation is itself one component of a sufficient tuple.

One thing to be careful about is that your last displayed expression is really $$ \theta^n\left(\prod_i I_{(1,\infty)}(x_i) \right)\left(e^{(-\theta-1)\sum_i\log x_i}\right) $$ (where, of course $\log$ is the same thing as $\ln$; I mention this since you used both notations in your posted question).

Fisher's factorization criterion tells you that that sum of logarithms is indeed sufficient.

What that means is: The conditional probability distribution of $X_1,\ldots,X_n$ given the value of $\sum_{i=1}^n\log X_i$ does not depend on $\theta$.

Next, what does "complete" mean? A complete statistic is one that admits no unbiased estimator of $0$ except the trivial one. That means there is no function $g$ such that $$ E\left(g\left(\sum_{i=1}^n \log X_i \right)\right) $$ remains equal to $0$ as $\theta$ changes (where of course one must not allow $g$ to vary as $\theta$ changes).

The Lehmann-Scheffe theorem now says you can get the UMVUE by starting with any crude unbiased estimator of $1/\theta$ --- call this estimator $T=T(X_1,\ldots,X_n)$ --- and finding $E(T\mid \sum_{i=1}^n \log X_i)$. Because of sufficiency this will be a function of the data that does not depend on $\theta$ ---- hence a statistic. It will be the UMVUE.

It may be a good idea to seek your crude unbiased estimator of $1/\theta$ among functions of $X_1$ alone rather than all of the $X$s, simply because it's easier to find.

Then you have to find the conditional expecation, which may take some work.