Let $(X_1,...,X_n)$ be a random sample from an exponential distribution with pdf $f(x,\theta) = \frac{1}{\theta}\exp{\frac{-x}{\theta}}I_{(0,\infty)(x)}$ , which of the following estimators is the MVUE for $E(X^2)= 2\theta^2$

$$ T_1 = \frac{2(\sum X_i)^2}{n(n+1)}$$ $$ T_2 = \frac{\sum X_i^2}{n}$$ .

Based on the uniqueness of MVUE I should be able to find the MVUE without finding the variance of each estimator .

Now I can prove that the joint density of the random sample is a member of the exponential class of densities and $\sum X_i$ is a complete sufficient statistic so that based on Lehmann-Scheffe completeness theorem if there is an unbiased estimator for $q(\theta)$ and it is a function of the complete sufficient statistic $S$ such that $T=S(X)$ it will be the MVUE for $q(\theta)$ , but I can not decide which one of $T_1$ and $T_2$ is the MVUE for $2\theta^2$ since both are unbiased and functions of the complete sufficient statistic $\sum X_i$

  • 2
    $\begingroup$ Please show us how $\sum X_i^2$ can be written as a function of $\sum X_i$ when $n \gt 1$. $\endgroup$
    – whuber
    May 18, 2017 at 17:03
  • 1
    $\begingroup$ On a quick blush, $T_2$ is not a function of $\sum X_i$, it is a function of the sum of squares of $X_i$. $\endgroup$ May 18, 2017 at 17:04
  • $\begingroup$ @whuber thanks I thought that $$ T_2 = \frac{\sum X_i^2}{n} =\frac{(\sum X_i)^2 - 2\sum\sum X_i X_j}{n}$$ could be considered as function of $\sum X_i$ $\endgroup$ May 18, 2017 at 17:15
  • 1
    $\begingroup$ The notation might be confusing you. Consider writing down all the terms individually in the case $n=2$. Alternatively, consider level sets of $\sum X_i$ and $\sum X_i^2$: if one function can be expressed in terms of another, then their level sets will have the same shapes. Given that all $X_i$ are nonnegative, the level sets of $\sum X_i$ are line segments; the level sets of $\sum X_i^2$ are arcs of circles. $\endgroup$
    – whuber
    May 18, 2017 at 17:17
  • 1
    $\begingroup$ Another way I like to think about it is from the information content. Suppose the holder of the original data unfortunately got hit by a bus. If we only had access to the sufficient statistic (and $n$ of course), could we compute anything else we need? $\endgroup$ May 18, 2017 at 17:22


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.