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Let $X,Y$ be independent and both $\text{Exp}(\lambda)$ distributed. How does one calculate $$E(\max (X,Y)\mid X)\,?$$

By independence $\max (X,Y)$ is $\text{Exp}(2\lambda) $ distributed but I do not see how to continue. I am aware of memorylessness of exponential distribution. Help is welcome!

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    $\begingroup$ By independence $\min(X,Y)$ has exponential distribution. Not $\max(X,Y)$. $\endgroup$ – drhab Jul 12 '18 at 12:16
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For a fixed $x\geq0$ we find:

$$\begin{aligned}\mathbb{E}\max\left(x,Y\right) & =\mathbb{E}\max\left(x,Y\right)\mathbf{1}_{Y\leq x}+\mathbb{E}\max\left(x,Y\right)\mathbf{1}_{Y>x}\\ & =xP\left(X\leq x\right)+\mathbb{E}\left(Y\mid Y>x\right)P\left(Y>x\right)\\ & =x\left(1-e^{-\lambda x}\right)+\left(x+\lambda^{-1}\right)e^{-\lambda x}\\ & =x+\lambda^{-1} e^{-\lambda x} \end{aligned} $$

Here the equality $\mathbb{E}\left(Y\mid Y>x\right)=x+\lambda^{-1}$ is based on the lack of memory of exponential distribution.

Because $X$ and $Y$ are independent we find:

$$\mathbb{E}\left[\max\left(X,Y\right)\mid X=x\right]=\mathbb{E}\left[\max\left(x,Y\right)\mid X=x\right]=\mathbb{E}\max\left(x,Y\right)=x+\lambda^{-1} e^{-\lambda x}$$

allowing us to conclude that:

$$\mathbb{E}\left[\max\left(X,Y\right)\mid X\right]=X+\lambda^{-1} e^{-\lambda X}$$

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  • $\begingroup$ thank you, and what would be if consider the minimum instead of the maximum? How could one handle this? $\endgroup$ – user563311 Jul 12 '18 at 14:56
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    $\begingroup$ Write $\min(X,Y)=X+Y-\max(X,Y)$ and apply the result in my answer concerning $max(X,Y)$. $\endgroup$ – drhab Jul 12 '18 at 15:13

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