Expectation of the maximum of two exponential random variables [closed]

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, AweyganOct 13 '17 at 19:34

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

Hint:

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:$$P(Z>z)=P(X>z)+P(Y>z)-P(X>z)P(Y>z)$$

• $$P(Z>z)=P(X>z)+P(Y>z)-P(X>z \wedge Y>z)$$ $$P(Z>z)=P(X>z)+P(Y>z)-P(X>z)P(Y>z)$$ $$P(Z>z)=e^{-\lambda z}+e^{-\mu z}-e^{-(\lambda+\mu) z}$$ $$\mathbb{E}(Z)=\int_{0}^{\infty} e^{-\lambda z}+e^{-\mu z}-e^{-(\lambda+\mu) z} dz = \cfrac{1}{\lambda}+\cfrac{1}{\mu}-\cfrac{1}{\lambda+\mu}$$ – Uncle Iroh Oct 13 '17 at 11:18
• Well, that makes things easy, doesn't it? Excellent! – drhab Oct 13 '17 at 11:19
• 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. – drhab Oct 13 '17 at 11:22
• Thank you very much for the help! – Uncle Iroh Oct 13 '17 at 11:23
• You are welcome. If the answer meets your needs then you could accept it. – drhab Oct 13 '17 at 11:24