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.
 A: Note that $P_{\theta}(X_1>x)=e^{-\theta x}$ for every $x>0$ and for every $\theta >0$, so you have
$$g(\theta)=e^{-\theta}-e^{-2\theta}=P_{\theta}(X_1>1)-P_{\theta}(X_1>2)=P_{\theta}(1<X_1<2)$$
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:

*

*Find the joint distribution of $X_1$ and $\sum_{i=1}^n X_i$


*UMVUE for Exponential probabilties

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 $T$.
