Let $X_1, \ldots, X_n$ be a sample from an exponential distribution with p.d.f. $f(x; \theta) = \theta e^{-\theta x}$ for $x > 0$ where $\theta > 0$ is an unknown parameter.

I would like to find the UMVUE of $e^{-\theta} - e^{-2\theta}$, but I've been struggling to do so.

I know that $T(X)$ is the UMVUE if $\text{Var}(T(X)) \leq \text{Var}(U(X))$ for any other unbiased estimator $U(X)$ of the expression. I've read through several examples in my textbook, but this is one problem that I am having some difficult with as I study. I am also aware of Lehmann-Scheffe's Theorem, which seemed to be useful in a couple of the examples I saw; however, my book only has two examples.

I've seen the example that $\sum_i X_i/n$ is the UMVUE for the parameter of a Poisson distribution, and I've seen that $(n + 1)X_{(n)}/n$ is the UMVUE for a uniform distribution with parameter $\theta$, but I'm not quite sure how to solve this other problem.

I thought that exponential family of distributions might be helpful here, but I'm not sure.

I would really appreciate any assistance with this problem. I've looked online for more examples but can't really find anything similar to this problem.

  • $\begingroup$ Is this a course related homework of some sort? $\endgroup$ May 18, 2021 at 18:06
  • $\begingroup$ It's related to the course in the sense that I'm studying for an exam. It's not an assignment that's due. $\endgroup$ May 18, 2021 at 18:06

1 Answer 1


Note that $P_{\theta}(X_1>x)=e^{-\theta x}$ for every $x>0$ and for every $\theta >0$, so you have


Hence an unbiased estimator of $g(\theta)$ is the indicator variable $I_{1<X_1<2}$.

Therefore by Lehmann-Scheffé theorem, UMVUE of $g(\theta)$ is the conditional expectation $E\left[I_{1<X_1<2}\mid T\right]$ where $T=\sum\limits_{i=1}^n X_i$ is a complete sufficient statistic.

Sufficiency of $T$ follows from Factorization theorem. Completeness of $T$ can be shown directly from definition or by arguing that the joint density of $X_1,\ldots,X_n$ is a member of a full-rank (regular) exponential family.

Now $E\left[I_{1<X_1<2}\mid T\right]=P\left[1<X_1<2\mid T\right]$, so to find this quantity you can refer to:

Using the fact that $\frac{X_1}T\sim \text{Beta}(1,n-1)$ is independent of $T$ (explained in linked posts), we find that for any $a>0$,

\begin{align} P(X_1<a\mid T)&=P\left(\frac{X_1}T<\frac aT\right) \\&=(n-1)\int_0^{\min\left\{\frac aT,1\right\}}(1-x)^{n-2}\,\mathrm dx \\&=\begin{cases} 1-\left(1-\frac aT\right)^{n-1} &,\text{ if } T>a \\ 1 &,\text{ if }0\le T\le a \end{cases} \end{align}

This suggests to me that $P(X_1<2\mid T)-P(X_1<1\mid T)$ is also a piecewise function of $T$.

  • $\begingroup$ I read the second answer, and I tried writing $P(1 < X_1 < 2 \mid T) = P(X_1 < 2 \mid T) - P(X_1 < 1 \mid T)$, but I'm still a little bit confused because they go from $T$ to conditioning on the actual value of the sum when I think that the final answer should be a function of $T$. $\endgroup$ May 18, 2021 at 8:44
  • $\begingroup$ The answer is a function of $T$. If $E[I_{1<X<2}\mid T=t]=h(t)$, then $E[I_{1<X<2}\mid T]=h(T)$ with probability $1$. $\endgroup$ May 18, 2021 at 9:49
  • $\begingroup$ Ok. Following my previous comment and your second link, would the answer just be $\left(1 - \frac{2}{T}\right)^{n - 1}- \left(1 - \frac{1}{T}\right)^{n - 1}?$ $\endgroup$ May 18, 2021 at 12:49
  • $\begingroup$ Not quite. The UMVUE depends on the range of $T$, if you follow the calculation closely. $\endgroup$ May 18, 2021 at 15:03
  • $\begingroup$ Thanks for your help so far. I've gone through the derivation probably 10 times, and I'm still a little bit confused as to what could be different in this case. $\endgroup$ May 18, 2021 at 18:04

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.