How to find $\mathbb{E}[X\mid\min(X,Y)]$?

Question

Say I have two independent variables $X$ and $Y$ that are exponentially distributed with respective rates $\lambda_X$ and $\lambda_Y$. How do I compute $\mathbb{E}[X\mid \min\{X,Y\}]$?

Answer 1

(Answering a query made in the comments.)

Recall that, for every integrable random variable $X$, $E(X\mid Z)$ is defined as the almost surely unique random variable $u(Z)$ such that $E(Xv(Z))=E(u(Z)v(Z))$ for every bounded measurable function $v$.

In words, $E(X\mid Z)$ is at the same time measurable with respect to $Z$ and similar to $X$ in that $E(X\mid Z)$ and $X$ give the same expectation when integrated against any function of $Z$.

In the present case, $Z=\min\{X,Y\}$ hence one needs the distribution of $Z$ to compute $E(u(Z)v(Z))$. Next, the task is to compute $E(Xv(Z))$. And finally one must equate these, that is, find the unique $u$ such that these coincide for every $v$...

The final result is $$ E(X\mid\min\{X,Y\})=\min\{X,Y\}+\frac{\mu}{\lambda+\mu}\,\frac1{\lambda}, $$ which has some nice interpretations in terms of marked Poisson processes or other similar notions.