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I want to find the expected value of $\text{max}\{X,Y\}$ where $X$ ist $\text{exp}(\lambda)$-distributed and $Y$ ist $\text{exp}(\eta)$-distributed. X and Y are independent. I figured out how to do this for the minimum of $n$ variables, but i struggle with doing it for 2 with the maximum.

(The context in which this was given is waiting for the later of two trains, with their arrival times being exp-distributed).

Thanks!

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3 Answers 3

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The minimum of two independent exponential random variables with parameters $\lambda$ and $\eta$ is also exponential with parameter $\lambda+\eta$.

Also $\mathbb E\big[\min(X_1,X_2)+\max(X_1,X_2)\big]=\mathbb E\big[X_1+X_2\big]=\frac{1}{\lambda}+\frac{1}{\eta}$. Because $\mathbb E\big[\min(X_1,X_2)\big]=\frac{1}{\lambda+\eta}$, we get $\mathbb E\big[\max(X_1,X_2)\big]=\frac{1}{\lambda}+\frac{1}{\eta}-\frac{1}{\lambda+\eta}.$

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Let $V=\max\{X,Y\}$. Then $$\mathbb{P}(V\leq t)=\mathbb{P}(X\leq t,Y\leq t)=\mathbb{P}(X\leq t)\mathbb{P}(Y\leq t).$$ Now find $f_V(t)$ and then $\int_{0}^{+\infty}tf_V(t)dt$, which should be $\frac{1}{\lambda}+\frac{1}{\eta}-\frac{1}{\lambda+\eta}$.

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The sample $(X,Y)$ have a density given by $f_X(x)f_Y(y)$ since $X$ and $Y$ are independent. You have to compute $$\iint_{\Bbb R^2}\max\{x,y\}f_X(x)f_Y(y)dxdy.$$ Cut this integral in two parts.

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  • $\begingroup$ This is good and mathematically more explicit than Julius'. But I can't be too critical of either. $\endgroup$ May 19, 2012 at 20:06
  • $\begingroup$ Can you please guide how to break this up in two integrals? $\endgroup$ Oct 18, 2016 at 7:17

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