UMVUE for Exponential probabilties Let $X_i$ be i.i.d $exp(\lambda)$ and take any $a > 0$.
I want to find the UMVUE of $P(X_i < a ) = 1-\exp(-\lambda a)$.
My attempt
By properties of the exponential family we know $\sum_i X_i$ is a complete sufficient statistic for $\lambda$.
Hence we only need to:


*

*Find an unbiased estimator of $P(X_i < a)$

*Condition it on $T(X) = \sum_i X_i$ and apply Lehmann-Scheff Theorem
An obvious candidate for an unbiased estimator is using indicator functions. For example
$$\delta (X) = 1(X_1 < a)$$ 
Is an unbiased estimator. Hence the estimator
$$T(X) = E[\delta(X) | T(X) = t]$$
Is the UMVUE of $P(X < a)$. However, I am having trouble deriving this. The problem I am having is the fact I have conditioned a continuous random variable on a discrete condition $T(X) = t$. How can  I proceed?
 A: Going to post a late answer as it's quite simple but useful trick. The UMVUE is indeed given by
$$P(X_i < a|T)$$
The trick is to use Renyi's representation which says that
$$\frac{X_1}{\sum_{i=1}^n X_i} =_d U_{(1)}$$
Where $U_{(1)}$ is the 1st order statistic from a uniform(0,1) random sample
$$U_1, U_2, \dots U_{n-1}$$
$$P(X_1 < a | \sum_i X_i = t) = P \left(\frac{X_1}{\sum_i X_i} < \frac{a}{t}|\sum_i X_i=t \right) = P\left(\frac{X_1}{\sum_i X_i} < \frac{a}{t}\right)$$
The last inequality follows from Basu's theorem which says every complete sufficient statistic is independent of every ancillary statistic. Since $X_i$ come from a scale family, it follows $X_i/\sum_i X_i$ is ancillary. Hence the probability is independent of the complete sufficient statistic $T(X) = \sum_i X_i$.
Hence by the above Renyi's representation, using the fact that the minimum of a Uniform(0,1) sample (of size $n-1$) has distribution $P(U_{(1)} \leq u ) = 1- (1-u)^{n-1}$, it follows that the UMVUE is given by
$$P(X_1 < a | \sum_i X_i = t) = P\left(U_{(1)} < \frac{a}{t} \right) = 1 - \left(1-\frac{a}{t} \right)^{n-1}$$
ie.
$\delta (X) = 1 - (1 - \frac{a}{\sum_i X_i})^{n-1}$ is UMVUE
