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$X,Y$ are independent exponential RV with parameter $\lambda,\mu$. How to calculate

$ E[\min(X,Y) \mid X>Y+c] $

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    @MichaelHardy, the parameters is the rates and thank you for advise in latex~2012-12-19

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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}. $

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    There is no $E(A;B)$ in the picture. If $X$ is a random variable and $A$ an event, $E(X;A)$ is the expectation of $X$ restricted to $A$, that is, $E(X\mathbf 1_A)$. // The result is independent on the value of $c$ as long as $c\geqslant0$. The post says nothing about the case $c\lt0$.2012-12-19
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Assuming $c \ge 0$ and so $Y \lt X$, this is equivalent to $E[Y\mid X\gt Y+c]$ $= \frac{\int_{y=0}^{\infty} y \, \mu \exp(-\mu y)\, \int_{x=y+c}^\infty \lambda \exp(- \lambda x)\, dx \, dy}{\int_{y=0}^{\infty} \mu \exp(-\mu y)\, \int_{x=y+c}^\infty \lambda \exp(- \lambda x)\, dx \, dy}$