Find $E(X_{(1)}\mid T)$ where $T=\sum_{i=1}^n X_i$

Question

Let $X_1,X_2,\ldots,X_n$ be a random sample with $n\geq 2$ from an exponential distribution. $X_{(1)}=\min(X_1,X_2,\ldots,X_n)$. Find $E(X_{(1)}\mid T)$ where $T=\sum_{i=1}^n X_i$.

I was able to find the Cdf of $X_{(1)}$ which is \begin{align} F_{X_{(1)}}(x)&= \begin{cases} 0, & x<0 \\ 1-e^{-n \lambda x}, & x\geq 0 \end{cases}\label{6} \end{align} But I am unable to understand how to proceed from here. Please help.

Answer 1

Assuming $X_1,X_2,\ldots,X_n$ are i.i.d Exponential with mean $1/\lambda$, the conditional expectation can be found indirectly using the theory of minimum variance unbiased estimation.

As $T=\sum\limits_{i=1}^n X_i$ is a complete sufficient statistic for this family of distributions, $E\left[X_{(1)}\mid T\right]$ is simply the uniformly minimum variance unbiased estimator (UMVUE) of $E\left[X_{(1)}\right]$ by Lehmann-Scheffé theorem.

But note that $X_{(1)}$ is Exponential with mean $\frac{1}{n\lambda}$ and

$$E\left[\frac1{n^2}\sum_{i=1}^n X_i\right]=\frac1{n^2}\sum_{i=1}^n \frac1{\lambda}=\frac1{n\lambda}$$

By Lehmann-Scheffé, this again suggests that $\frac1{n^2}\sum\limits_{i=1}^n X_i$ is the UMVUE of $E\left[X_{(1)}\right]=\frac1{n\lambda}$.

As UMVUE is unique whenever it exists, it must be that

$$E\left[X_{(1)}\mid T\right]=\frac{T}{n^2} \quad,\text{ a.e.}$$

Answer 2

Sketch for a solution: surely you knows that $\operatorname{E}[X|\sigma (Y)](\omega)=\operatorname{E}[X|Y=Y(\omega)]$ and $$ \operatorname{E}[X|Y=t]=\int_{\mathbb{R}}x\,\frac{f_{X,Y}(x,t)}{f_{Y}(t)}\mathop{}\!d x\quad \text{ a.e. } $$ when $(X,Y)$ have a density. Then if you find $g(t):=\operatorname{E}[X|Y=t]$ you will have that $\operatorname{E}[X|\sigma (Y)]=g\circ Y$. It seems that you don't have an explicit definition of the $X_k$ so probably you are just interested in $g$, from the identity above you can get it, you only need to compute $f_{X,Y}$ and $f_Y$.