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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?

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    $\begingroup$ It is not clear in what sense your work is a "solution." What is $T(X)$? I would expect an unbiased estimator to be of the form $W=w(X_1, \ldots, X_n)$, and satisfy $E[W]=1/\theta^2$ while $E[(W-1/\theta^2)^2]$ is hopefully small (and going to $0$ as $n\rightarrow\infty$). In particular, you should show how your samples $\{X_1, \ldots, X_n\}$ can be used to form an estimate. $\endgroup$
    – Michael
    Sep 7, 2015 at 20:55
  • $\begingroup$ I updated the post, could you please review it again? $\endgroup$
    – kaksat
    Sep 8, 2015 at 12:39

1 Answer 1

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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.

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  • $\begingroup$ Thank you! So using the pdf function for $X$ and writing it as a formula for an exponnetial family I obtain: $f(X) = exp(nln\theta - \theta\sum_{i=1}^{n} x_i)$ and: $h(x) = 1$, $T(X) = \sum_{i=1}^{n} x_i$, $C(\theta) = -\theta$, and $B(\theta) = -nln\theta$. Therefore T(X) is a complete sufficient statistic of the parameter $\theta$. I therefore assume that instead of using the factorization criterium I should have writtten the pdf function as the exponential family? $\endgroup$
    – kaksat
    Sep 9, 2015 at 9:42
  • $\begingroup$ @kaksat: Well, I wouldn't say "instead" -- the exponential form is also factorized, and essentially the same as what you wrote for applying the factorization criterion -- factorization and thus sufficiency drop out as a byproduct when you write it as a member of the exponential family -- but additionally you get completeness. $\endgroup$
    – joriki
    Sep 9, 2015 at 10:51

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