Find UMVUE of ${\rm e}^{-\theta \tau}$ in which $\theta$ is parameter of ${\rm Exp}(\theta)$ Suppose $X_1,... ,X_n\ {\rm i.i.d.\sim Exp}(\theta)$, i.e. $X_i \sim f(x) = \theta{\rm e}^{-\theta x}I_{(x>0)}$. Find UMVUE of ${\rm e}^{-\theta \tau}$ in which $\tau > 0$ is given.
Hint: ${\rm e}^{-\theta\tau} = P_\theta(X_1 > \tau)$
I find complete and sufficient statistic is $T = \sum_{i=1}^n X_i \sim p(x) = \frac{\theta^n t^{n-1}}{(n-1)!}{\rm e}^{-\theta t}$, and $P_{\theta}(X_1 > \tau) = EI_{(X_1 > \tau)}$.
Hence I want to find a function $I_{(X_1>\tau)} = g(T)$. Any hint is appreciated.
 A: As mentioned in my comment, $g(T) := E[I_{X_1 > \tau} \mid T]$ is the UMVUE.
The conditional density of $X_1$ given $T=t$ is
\begin{align}
\frac{f_{X_1, T}(x_1, t)}{f_T(t)}
&= \frac{f_{X_1}(x_1) f_{X_2 + \cdots + X_n}(t-x_1)}{f_T(t)}
\\
&= \frac{\theta e^{-\theta x_1} \cdot \theta^{n-1} (t-x_1)^{n-2} e^{-\theta (t-x_1)} / (n-2)!}{\theta^n t^{n-1} e^{-\theta t} / (n-1)!}
\\
&= (n-1)\frac{(t-x_1)^{n-2}}{t^{n-1}}
\\
&= \frac{n-1}{t} \left(1-\frac{x_1}{t}\right)^{n-2}
\end{align}
for $0 < x_1 < t$, and zero otherwise.
Thus,
\begin{align}
g(t)
&= P(X_1 > \tau \mid T=t)
\\
&= \frac{n-1}{t} \int_\tau^t (1-x_1/t)^{n-2} \, dx_1
\\
&= (n-1) \int_{\tau/t}^1 (1-u)^{n-2} \, du
\\
&= \left[-(1-u)^{n-1}\right]_{u=\tau/t}^1
\\
&= (1-\tau/t)^{n-1}
\end{align}
for $0 < \tau < t$, and zero otherwise. Thus the UMVUE is
$$\left(1 - \frac{\tau}{T}\right)^{n-1} \mathbf{1}_{T > \tau}.$$

Empirical sanity check that this statistic is unbiased:
theta <- 1
n <- 10
tau <- 1.5

GenerateUmvue <- function(theta, tau, n) {
  t <- rgamma(1, shape=n, rate=theta)
  if (t > tau) {return((1 - tau / t)^(n-1))}
  else {return(0)}
}

umvue_dist <- numeric(0) 
for (i in 1:1e4) {
  umvue_dist <- c(umvue_dist, GenerateUmvue(theta, tau, n))
}
mean(umvue_dist)
exp(- theta * tau)

> mean(umvue_dist)
[1] 0.223106

> exp(- theta * tau)
[1] 0.2231302

