Let $X_1, X_2, … , X_n$ be a random sample form an exponential distribution $E(\theta)$, $\theta>0$. Obtain minimum-variance unbiased estimator of a function $g(\theta)=\frac{1}{\theta^2}$ ($E(X) = \frac{1}{\theta}$ and
$Var(X) = \frac{1}{\theta^2}$).
This is what I did [update]:
Since pdf of the exponential distribution is:
$f(x) = \theta e^{-\theta x}$
The pdf for $X = (X_1, X_2, … , X_n)$ is:
$f(X) = \theta^n e^{-\theta \sum_{i=1}^{n} X_i}$
So according to the factorization criterium:
$g(T(X)) = e^{-\theta \sum_{i=1}^{n} X_i}$
and
$h(x) = 1$
$T(x) = \sum_{i=1}^{n} X_i$ is a sufficient statistic of the parameter $\theta$.
$E(T(X))
= nE(X)
= \frac{n}{\theta}$ and $E(\overline{X})
= \frac{1}{\theta}$
$\overline{X}$ is the MVUE for $\frac{1}{\theta}$ so my intuition was to use $E(\overline{X}^2)$
$E(\overline{X}^2) = Var(\overline{X}) + [E(\overline{X})]^2
=\frac{1}{n\theta^2} + \frac{1}{\theta^2}
=\frac{n + 1}{n\theta^2}$
Above given estimator is biased but:
$(\frac{n}{n + 1})\overline{X}^2$
is an unbiased estimator of $\frac{1}{\theta^2}$ and a function of a sufficient statistic, therefore it is an MVUE of $\frac{1}{\theta^2}$.
Is such a solution correct?
Best Answer
Yes, this solution is almost correct, and it's a nice solution. The only slight technical error is that you should argue that $T$ is a complete sufficient statistic. If a sufficient statistic isn't complete, different unbiased estimators based on it may have different variance.