I try to solve this:
Let $X,Y$ be two independent exponential r.v. with parameters $\mu,\lambda>0$. > Let $T:=\min(X,Y)$ Compute $\mathbb{E}(T\vert X)$
Now there is a hint to compute $\mathbb{E}(Tf(X))$ for some measurable function $f:\mathbb{R}\to \mathbb{R}$ but what confuses me, is that $\min(x,y)$ has two components and is not from $\mathbb{R}\to \mathbb{R}$. So I tried to rewrite $$\mathbb{E}(Tf(X))=\mathbb{E}(\mathbb{E}(Tf(X)\vert X))=\mathbb{E}(f(X)\mathbb{E}(T\vert X))$$ but I don't see whether this is useful or not.
My second attempt was to work with the density. I computed the CDF for $T$ which is $F(t)=1-e^{-t(\lambda + \mu)}$ and the PDF $$f(t)=(\lambda + \mu)e^{-t(\lambda + \mu)}$$
But I'm not sure, if this is useful here.
Best Answer
For this question, you can also work it out directly as follows: \begin{align*} E(\min(X, Y) \mid X) &=E(X1_{Y\ge X} \mid X) + E(Y1_{X\ge Y} \mid X)\\ &=X \int_X^{\infty}\lambda e^{-\lambda y} dy + \int_0^{X}\lambda y e^{-\lambda y} dy\\ &=\frac{1}{\lambda}\left(1-e^{-\lambda X} \right). \end{align*}