Mistake in calculating $E\left[\min(X,Y)|X>Y+c\right]$ by a direct computation for independent exponential r.v.s $X$ and $Y$ Let $X,Y$ be independent exponential random variables with respective rates $\lambda$ and $\mu$, and let $c$ be a non-negative constant.
Find $E[\min(X,Y)|X>Y+c]$ by a direct computation.
I know that the answer is $\frac{1}{\lambda + \mu}$ but I can't approach this answer "directly".
I try to use
\begin{equation}
\int_{-\infty}^\infty\int_{-\infty}^\infty \min(x,y)f(x,y)\,dx\,dy
\end{equation}
where $f(x,y)$ is the joint probability distribution of $X,Y$ conditioned on $X>Y+c$.
Since $X$ is always greater than $Y$, $\min(x,y)$ is just $y$. The joint conditional probability distribution $f(x,y)$ for values of $X$ and $Y$, for those values where $X>Y+c$ is
\begin{split}
\frac{\lambda e^{-\lambda x}\mu e^{-\mu y}}{P(X>Y+c)} = 
\frac{\lambda e^{-\lambda x}\mu e^{-\mu y}}{e^{-\lambda(y+c)}}
\end{split}
This gets me
\begin{equation}
\int_0^\infty\int_{y+c}^\infty y\cdot \frac{\lambda e^{-\lambda x}\mu e^{-\mu y}}{e^{-\lambda(y+c)}}\,dx\,dy
\end{equation}
which I evaluate to something that isn't $\frac{1}{\lambda+\mu}$!
What am I doing wrong?
 A: There is seems to be an issue where you computed $P(X>Y+c)$. It should yield
$$P(X>Y+c)=\int_{[0,\infty)}\int_{[0,\infty)}\mathbb{I}_{\{x>y+c\}}(x,y)f_{X,Y}(x,y)dxdy= \\ = \lambda \mu\int_{[0,\infty)}\int_{(y+c,\infty)}e^{-\lambda x -\mu y}dxdy=\lambda \mu\int_{[0,\infty)}\bigg(\int_{(y+c,\infty)}e^{-\lambda x}dx\bigg) e^{-\mu y}dy= \\ =\lambda \mu\int_{[0,\infty)}\bigg(-\frac{1}{\lambda}\cdot 0+\frac{1}{\lambda}e^{-\lambda(y+c)}\bigg) e^{-\mu y}dy=\mu e^{-\lambda c}\int_{[0,\infty)}e^{-(\lambda + \mu)y}dy=\frac{\mu e^{-\lambda c}}{\lambda + \mu}$$
Furthermore, taking into account what you already said about $\min(x,y)$,
$$E[\min(X,Y)\mathbb{I}_{\{x>y+c\}}]=\lambda \mu\int_{[0,\infty)}\int_{(y+c,\infty)}\min(x,y)e^{-\lambda x -\mu y}dxdy = \\ =\lambda \mu\int_{[0,\infty)}\int_{(y+c,\infty)}ye^{-\lambda x -\mu y}dxdy=\mu e^{-\lambda c}\int_{[0,\infty)}ye^{-(\lambda+\mu) y}dy = \\ =\mu e^{-\lambda c}\frac{1}{(\lambda + \mu)^2}$$
So we can finally assert that
$$E[\min(X,Y)|X>Y+c]=\frac{1}{P(X>Y+c)}E[\min(X,Y)\mathbb{I}_{\{x>y+c\}}]=\frac{1}{\lambda + \mu}$$
