How to calculate this expectation? $X,Y$ are independent exponential RV with parameter $\lambda,\mu$. How to calculate 
$$
E[\min(X,Y) \mid X>Y+c]
$$
 A: 
The result is independent on $c\geqslant0$.

To see this, recall that, for every measurable function $u$,
$$
\mathbb E(u(Y);X\gt Y+c)=\int_0^{+\infty} u(y)\mathbb P(X\gt y+c)\mathrm d\mathbb P_Y(y)=\mathbb E(u(Y)\mathrm e^{-\lambda(Y+c)}).
$$
Using this for $u:y\mapsto y$ and for $u:y\mapsto1$ and canceling the common factor $\mathrm e^{-\lambda c}$ yields
$$
\mathbb E(Y\mid X\gt Y+c)=\frac{\mathbb E(Y\mathrm e^{-\lambda Y})}{\mathbb E(\mathrm e^{-\lambda Y})}
$$
The denominator is 
$$
\mathbb E(\mathrm e^{-\lambda Y})=\int_0^{+\infty}\mathrm e^{-\lambda y}\mu\mathrm e^{-\mu y}\mathrm dy=\frac{\mu}{\mu+\lambda}.
$$
The numerator is minus the derivative of the denominator with respect to $\lambda$ hence the ratio is
$$
\mathbb E(Y\mid X\gt Y+c)=\frac{\mu/(\mu+\lambda)^2}{\mu/(\mu+\lambda)}=\frac1{\mu+\lambda}.
$$
A: Assuming $c \ge 0$ and so $Y \lt X$, this is equivalent to $$E[Y\mid X\gt Y+c]$$ $$= \frac{\int_{y=0}^{\infty} y \, \mu  \exp(-\mu y)\, \int_{x=y+c}^\infty   \lambda \exp(- \lambda x)\,  dx \, dy}{\int_{y=0}^{\infty}  \mu  \exp(-\mu y)\, \int_{x=y+c}^\infty  \lambda \exp(- \lambda x)\,  dx \, dy}$$
