# Finding UMVUE of difference of exponentals

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.

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

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

$$g(\theta)=e^{-\theta}-e^{-2\theta}=P_{\theta}(X_1>1)-P_{\theta}(X_1>2)=P_{\theta}(1

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

Therefore by Lehmann-Scheffé theorem, UMVUE of $$g(\theta)$$ is the conditional expectation $$E\left[I_{1 where $$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, 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_1a \\ 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 $$T$$.

• 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 $T$. May 18, 2021 at 8:44
• 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 $1$. May 18, 2021 at 9:49
• Ok. Following my previous comment and your second link, would the answer just be $\left(1 - \frac{2}{T}\right)^{n - 1}- \left(1 - \frac{1}{T}\right)^{n - 1}?$ May 18, 2021 at 12:49
• Not quite. The UMVUE depends on the range of $T$, if you follow the calculation closely. May 18, 2021 at 15:03
• 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. May 18, 2021 at 18:04