Question

Let $X_1,\dots ,X_n$be independent and identically distributed random variables. Find

$E\left[X_1\mid X_1+\cdots +X_n=x\right]$

Short answer

$E\left(X_1\mid X_1+\cdots +X_n=x\right)=\frac{x}{n}$

Step-by-step solution

Given information

Given in the question that, let$X_1,\dots ,X_n$be independent and identically distributed random variables.

Explanation

Given,

and since all the variables are equally distributed, we have the symmetry

for all$j=1,\dots ,n$. Using the linearity of the conditional expectation, we have that

$=\sum _j E\left(X_j\mid X_1+\cdots +X_n=x\right)$

$=\sum _j E\left(X_1\mid X_1+\cdots +X_n=x\right)$

$=nE\left(X_1\mid X_1+\cdots +X_n=x\right)$

Which implies that,

$E\left(X_1\mid X_1+\cdots +X_n=x\right)=\frac{x}{n}$

Final answer

$E\left(X_1\mid X_1+\cdots +X_n=x\right)=\frac{x}{n}$