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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 $\mathbb{E}(Z)$?

drhab
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Uncle Iroh
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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
  • @UncleIroh Great! You can edit your question and correct your question by clicking the small "edit"-button under your answer. – Bobson Dugnutt Oct 13 '17 at 11:10
  • @Lovsovs Actually, I am unsure as to the difference between $X\sim \lambda e^{-\lambda x} $ for $\lambda>0 $ and $X\sim e^{\lambda} $ – Uncle Iroh Oct 13 '17 at 11:22
  • @UncleIroh Well, they are completely different expressions. If $\lambda>0$ for both, $e^\lambda$ isn't even a valid PDF. – Bobson Dugnutt Oct 13 '17 at 11:30
  • @Lovsovs Ah yes, in my original posting, I had intended to mean that $X \sim exp(\lambda) $ is equivalent to $X$ is a random variable with an exponential distribution with parameter $\lambda$, sorry for the confusion. – Uncle Iroh Oct 13 '17 at 14:24
  • @UncleIroh No problem! That (especially wit the captital E in Exp, which I think you used) is actually also pretty standard notation, I don't know why I misinterpreted it, so I'm the one that should be apologizing. Cheers. – Bobson Dugnutt Oct 13 '17 at 16:20

1 Answers1

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

drhab
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  • $$P(Z>z)=P(X>z)+P(Y>z)-P(X>z \wedge Y>z) $$ $$P(Z>z)=P(X>z)+P(Y>z)-P(X>z)P(Y>z) $$ $$P(Z>z)=e^{-\lambda z}+e^{-\mu z}-e^{-(\lambda+\mu) z} $$ $$\mathbb{E}(Z)=\int_{0}^{\infty} e^{-\lambda z}+e^{-\mu z}-e^{-(\lambda+\mu) z} dz = \cfrac{1}{\lambda}+\cfrac{1}{\mu}-\cfrac{1}{\lambda+\mu} $$ – Uncle Iroh Oct 13 '17 at 11:18
  • Well, that makes things easy, doesn't it? Excellent! – drhab Oct 13 '17 at 11:19
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    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