Estimation of $\mathbb{P}(X > t)$ Let $X_1, \ldots, X_n \sim X$ be iid random variables. The goal is to find an estimator of $\theta = \mathbb{P}(X > t)$ for a given $t > 0$. 
1) Show that $\hat{\theta}_1 = 1 - F_n(t)$ is an unbiased estimator of $\theta$ and find its MSE ($F_n$ is the empirical distribution function). 
2) Let now $X_1, \ldots, X_n$ be exponential($\lambda$). Find the UMVUE of $\theta$. (Hint: $X_1$ and $X_1/\sum_{i=1}^n X_i$ are independent and the pdf of $X_1/\sum_{i=1}^n X_i$ is : $(n-1)(1-x)^{n-2} \mathbb{I}_{ x \in (0,1) }$)
Some thoughts 
1) If I'm not wrong:
$$
\hat{\theta}_1 = 1 - F_n(t) = \frac{1}{n} \sum_{i=1}^{n} \mathbb{I}_{\{X_i > t \}}
$$
Therefore $n \hat{\theta}_1 \sim \mathrm{Binomial}(n, 1 - F_X(x))$ and then:
$$
\mathbb{E}(\hat{\theta}_1) = \frac{1}{n}\mathbb{E}(n\hat{\theta}_1) = 1 - F_X(x) = \mathbb{P}(X > x) = \theta \qquad \mathrm{(unbiased)}
$$
$$
\mathrm{MSE}(\hat{\theta}_1) = \mathbb{V}(\hat{\theta}_1) = \frac{1}{n^2}\mathbb{V}(n\hat{\theta}_1) = \frac{\theta(1-\theta)}{n}
$$
2) We know that $\hat{\theta}_1$ is an unbiased estimator of $\theta = e^{-t/ \lambda}$ and $S = \sum_{i=1}^n X_i$ is a complete sufficient statistic of $\lambda$. Therefore $U = \mathbb{E}(\hat{\theta}_1 \, | \, S)$ is the UMVUE of $\theta$. The problem is that I'm unable to "compute" $U$ or find an alternative solution. 
Edit: Alternative solution
I think I've found an alternative (and rather tedious) solution which doesn't make use of the hint. I was trying to write it as a comment, but it was too long. Mr. Hardy's answer is clearer and better in any sense, so I apologize in advance.
From Michael Hardy's answer:

Notice that $W=\mathbb{I}_{\{X_1>t\}}$ is an unbiased estimator of
  $\theta$, so the Rao-Blackwell estimator is $\mathbb{E}(W\mid S)$. 
  Because of completeness, [...] all unbiased estimators
  based on the sufficient statistic $S$ will be the same. So we seek $\mathbb{E}(W\mid S) = > \Pr(X_1>t\mid S)$, and this must be equal to $\mathbb{E}(\hat{\theta}_1 \mid S)$.

The conditional pdf of $X_1$ given $S = s$ is:
$$f_{X_1 | S = s}(x) = \frac{f_{X_1,S}(x,s)}{f_S(s)} $$
We know that $S$ is a Gamma r.v. because it is the sum of $n$ Exponential r.v. :
$$f_S(s) = \frac{\lambda^{-n} s^{n-1} e^{-s/\lambda}}{(n-1)!} $$
The joint distribution can be rewritten as $f_{X_1,S}(x,s) = f_{X_1}(x)f_{S | X_1 = x}(s)$. 
It should also be noted that:
$$
\begin{align}
\mathbb{P}(S \leq s | X_1 = x) &= \mathbb{P} \left(\left. x + \sum_{i=2}^n X_i \leq s \; \right|\; X_1 = x \right) \\ &= \mathbb{P} \left(\left. \sum_{i=2}^n X_i \leq s -x \; \right|\; X_1 = x\right) \\ &= \mathbb{P} \left(\sum_{i=2}^n X_i \leq s -x\right) 
\end{align}
$$
Where $\sum_{i=2}^n X_i$ is a Gamma r.v..
Thus:
$$
f_{S | X_1 = x}(s) = f_{\sum_{i=2}^n X_i}(s-x) = \frac{\lambda^{-(n-1)} (s-x)^{n-2}e^{-(s-x)/\lambda}}{(n-2)!}
$$
If $S = s$, then $0 \leq X_1 \leq s$. If we simplify it turns out that:
$$
f_{X_1 | S = s}(x) = (n-1) \frac{(s-x)^{n-2}}{s^{n-1}} \,  \, \mathbb{I}_{\{0 \leq x \leq s\}}
$$
Finally:
$$
\mathbb{P}(X_1 > t | S = s) = \int_{t=1}^S (n-1)\frac{(s-x)^{n-2}}{s^{n-1}} \, \mathrm{d}x = \frac{n-1}{s^{n-1}}\left[- \frac{(s-x)^{n-1}}{n-1} \right]^{s}_{t} = \left(1 - \frac{t}{s}\right)^{n-1}$$
Which is exactly the same thing. 
 A: The "hint" seems like more than just a hint; I'd have phrased it as "You may use the fact that...." etc. instead of just "Hint".
Notice that $W=\mathbb{I}_{\{X_1>t\}}$ is an unbiased estimator of $\theta$, so the Rao-Blackwell estimator is $\mathbb{E}(W\mid S)$.  Because of completeness, which you mentioned, all unbiased estimators based on the sufficient statistic $S$ will be the same.  So we seek $\mathbb{E}(W\mid S) = \Pr(X_1>t\mid S)$, and this must be equal to the conditional expectation that you say you didn't know how to find.  One thing that's clear is that $X_1>t$ is impossible if $t>S$, so $\Pr(X_1>t\mid S)=0$ if $t>S$.
The "hint" says in effect $Y=X_1/S\sim\operatorname{Beta}(1,n-1)$ and this is independent of $X_1$.  So
$$
\begin{align}
& \Pr(X_1>t\mid S)=\Pr\left(\left.\frac{X_1}{S} >\frac{t}{S}\; \right|\; S\right) \\[10pt]
= {} & \Pr\left(\left.Y>\frac{t}{S} \;\right|\; S\right) = \int_{t/S}^1 (n-1)(1-x)^{n-2} \;dx. \\[10pt]
= {} & \left[-(1-x)^{n-1}\right]_{x=t/S}^{x=1} = \left(1-\frac{t}{S}\right)^{n-1}.
\end{align}
$$
You might want to check my details, and I will too, but if I have this right then the best unbiased estimator that you seek is
$$
\begin{cases} \left(1-\frac{t}{S}\right)^{n-1} & \text{if }S>t, \\[10pt] 0 & \text{if } S<t. \end{cases}
$$
Maybe this qualifies as another instance in which unbiasedness doesn't make a lot of sense and maximum likelihood might.
Notice that the MLE for $\lambda$ is $S/n$, so that of $\theta=e^{-t/\lambda}$ is $e^{-nt/S}$.  Clearly this is biased, since it's a function of a complete sufficient statistic and it differs from the best unbiased estimator.
