# Expected Value of the maximum of two exponentially distributed random variables

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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Are these two variable independent? – Davide Giraudo May 19 '12 at 8:36
Yes they are. Forgot to add that, sorry. – ifubic May 19 '12 at 8:41
You can find the cumulative distribution function, and for non-negative random variables there is a formula which links the CDF with the expectation. – Davide Giraudo May 19 '12 at 8:43

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

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Thanks! That worked out nicely! – ifubic May 19 '12 at 9:59
That would be appropriate to accept the answer then. – Julius May 19 '12 at 10:21

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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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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This is good and mathematically more explicit than Julius'. But I can't be too critical of either. – Michael Chernick May 19 '12 at 20:06