Let $Z:=\text{max}(X,Y)$ where $X,Y$ are independent random variables having exponential distribution with parameters $\lambda$ and $\mu$ respectively.

My question is:

What is the expectation of $Z$, i.e. what is $\mathbb{E}(Z)$?


closed as off-topic by Did, Bobson Dugnutt, Namaste, Aqua, Aweygan Oct 13 '17 at 19:34

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  • $\begingroup$ Do you mean that $X\sim \lambda e^{-\lambda x}$ for $\lambda>0$? Also, this may be useful. $\endgroup$ – Bobson Dugnutt Oct 13 '17 at 10:48
  • $\begingroup$ You (most probably) forgot to mention that $X,Y$ are independent. $\endgroup$ – drhab Oct 13 '17 at 11:01
  • $\begingroup$ @Lovsovs, yes I did intend that. And that was very useful - thanks! $\endgroup$ – Uncle Iroh Oct 13 '17 at 11:08
  • $\begingroup$ @drhab Oops, yes I did forget to mention that. $\endgroup$ – Uncle Iroh Oct 13 '17 at 11:08
  • $\begingroup$ It is by far the most frequent lack by questions about probability. $\endgroup$ – drhab Oct 13 '17 at 11:09


Make use of: $$\mathbb EZ=\int_0^{\infty}P(Z>z)dz$$

and of course:$$P(Z>z)=P(X>z)+P(Y>z)-P(X>z\wedge Y>z)$$

By independence of $X,Y$ this results in:$$P(Z>z)=P(X>z)+P(Y>z)-P(X>z)P(Y>z)$$

  • $\begingroup$ $$P(Z>z)=P(X>z)+P(Y>z)-P(X>z \wedge Y>z) $$ $$P(Z>z)=P(X>z)+P(Y>z)-P(X>z)P(Y>z) $$ $$P(Z>z)=e^{-\lambda z}+e^{-\mu z}-e^{-(\lambda+\mu) z} $$ $$\mathbb{E}(Z)=\int_{0}^{\infty} e^{-\lambda z}+e^{-\mu z}-e^{-(\lambda+\mu) z} dz = \cfrac{1}{\lambda}+\cfrac{1}{\mu}-\cfrac{1}{\lambda+\mu} $$ $\endgroup$ – Uncle Iroh Oct 13 '17 at 11:18
  • $\begingroup$ Well, that makes things easy, doesn't it? Excellent! $\endgroup$ – drhab Oct 13 '17 at 11:19
  • $\begingroup$ Another way is exploiting $\max(X,Y)=X+Y-\min(X,Y)$. Then you can take advantage of the fact that a minimimum of expontially distributed variables is also expontially distributed. $\endgroup$ – drhab Oct 13 '17 at 11:22
  • $\begingroup$ Thank you very much for the help! $\endgroup$ – Uncle Iroh Oct 13 '17 at 11:23
  • $\begingroup$ You are welcome. If the answer meets your needs then you could accept it. $\endgroup$ – drhab Oct 13 '17 at 11:24

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