Conditional expectation conditional on exponential random variable Given $X_1$, $X_2$ are independent EXP(1) random variables, how do I compute the following expectations:


*

*$E[X_1 | X_1 + X_2]$

*$P(X_1 > 3 | X_1 + X_2)$

*$E[X_1 | min(X_1, t)]$
 A: I'll do the first one. Please, try to find similar solutions to the others. We know that $X_1$ and $X_2$ are IID (their specific distributions don't matter yet). Hence, by symmetry we have
$$
  \textrm{E}[X_1\mid X_1+X_2] = \textrm{E}[X_2\mid X_1+X_2] \quad \textrm{a.s.}
$$
Therefore, using the properties of the conditional expectation, we find
$$
  \textrm{E}[X_1\mid X_1 + X_2] = \frac{1}{2} \Bigg( \textrm{E}[X_1\mid X_1+X_2] + \textrm{E}[X_2\mid X_1+X_2] \Bigg)
$$
$$
  = \frac{1}{2} \textrm{E}[X_1 + X_2 \mid X_1+X_2] =\frac{X_1+X_2}{2} \quad \textrm{a.s.}
$$
Since $X_1$ and $X_2$ are independent $\textrm{Exp}(1)\sim\textrm{Gamma}(1,1)$, we know that
$$
  X_1+X_2\sim\textrm{Gamma}(2,1) \, ,
$$
yielding
$$
  \textrm{E}[X_1\mid X_1+X_2] \sim \textrm{Gamma}(2,1/2) \, .
$$
It's always good to make a quick verification. By the tower property, we know that it must be the case that
$$
\textrm{E}[\textrm{E}[X_1\mid X_1+X_2]]=\textrm{E}[X_1] \, ,
$$
but from what we've just found, we have $\textrm{E}[\textrm{E}[X_1\mid X_1+X_2]]=1$, and also $\textrm{E}[X_1]=1$.
So the expectations match as expected, which is not bad.
