UMVUE for $P(X_1>t)$ for some fixed $t>\mu$ when $X_i \sim \operatorname{Exp}(\sigma, \mu)$ $X_1,\ldots,X_n$ is a sample from the density
$$f_X(x)=\frac{1}{\sigma}e^{-\frac{x-\mu}{\sigma}}\quad,\,x>\mu$$
I know form a previous step that the UMVUE for $\sigma$ and $\mu$ is $\frac{1}{n-1}(\sum_iX_i-nX_{(1)})$ and $\frac{n}{n-1}X_{(1)}-\frac{1}{n-1}\bar{X}$. Also, I know that $P(X_1>t) = e^{-\frac{(x-\mu)}{\sigma}}$.
My question is if there is a way I can use this information to find the UMVUE for $P(X_1>t)$?
I tried to use the Rao-Blackwell theorem, with $h(X_1) = 1 \text{ if } X_1>t$ 0 OW unbiased estimator for $e^{-\frac{(x-\mu)}{\sigma}}$. However, the computations for the estimator became tedious and couldn't get to a final answer.
 A: Suppose $Y=T-nX_{(1)}$ where $T=\sum\limits_{i=1}^n X_i$.
Then a complete sufficient statistic for $\theta=(\mu,\sigma)$ based on the sample $X_1,\ldots,X_n$ is $U=(X_{(1)},Y)$. The UMVUE of $P(X_1>t)$ for a fixed $t>\mu$ is given by $$E_{\theta}[I_{X_1>t}\mid U]=P_{\theta}(X_1>t\mid U)$$
Clearly $P_{\theta}(X_1>t\mid U)=1$ for $X_{(1)}>t$.
For $X_{(1)}\le t$, noting that $U$ and $(X_1-X_{(1)})/Y$ are independent by Basu's theorem,
\begin{align}
P_{\theta}(X_1>t\mid U=u)&=P\left(\frac{X_1-X_{(1)}}{Y}>\frac{t-x_{(1)}}{y}\,\Big|\, U=u\right)
\\&=P\left(\frac{X_1-X_{(1)}}{Y}>\frac{t-x_{(1)}}{y}\right)
\\&=\sum_{j=1}^n P\left(\frac{X_1-X_{(1)}}{Y}>\frac{t-x_{(1)}}{y}\,\Big|\,X_j=X_{(1)}\right)P(X_j=X_{(1)})\tag{1}
\end{align}
Now $P(X_j=X_{(1)})=\frac1n$ for every $j=1,\ldots,n$ as the $X_i$'s are i.i.d continuous.
But $P\left(\frac{X_1-X_{(1)}}{Y}>\frac{t-x_{(1)}}{y}\,\Big|\,X_1=X_{(1)}\right)=0$ and the conditional probabilities given $X_j=X_{(1)}$ are all equal for $j=2,\ldots,n$. So $(1)$ reduces to
\begin{align}
P_{\theta}(X_1>t\mid U=u)&=\frac{n-1}{n} P\left(\frac{X_1-X_{(1)}}{Y}>\frac{t-x_{(1)}}{y}\,\Big|\,X_n=X_{(1)}\right)
\\&=\left(1-\frac1n\right) P\left(\frac{X_1-X_{(1)}}{\sum_{i=1}^{n-1}(X_i-X_{(1)})}>\frac{t-x_{(1)}}{y}\,\Big|\,X_n=X_{(1)}\right)
\end{align}
Now consider  that given $X_n=X_{(1)}$, the variables $X_i-X_{(1)}$ are i.i.d exponential with mean $\sigma$ for $i=1,\ldots,n-1$ (see related post) which would imply that $(X_1-X_{(1)})/\sum_{i=1}^{n-1}(X_i-X_{(1)})$ conditioned on $X_n=X_{(1)}$ has a certain Beta distribution.
