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How can I prove that the minimum of two exponential random variables is another exponential random variable, i.e. Z = min(X,Y)

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up vote 8 down vote accepted

Note that you must assume that $X$ and $Y$ are independent, otherwise the result is easily seen to be false.

There is a constant $\lambda$ such that $P(X \geq t)=e^{-\lambda t}$ for every $t>0$.

There is a constant $\mu$ such that $P(Y \geq t)=e^{-\mu t}$ for every $t>0$.

Then for every $t>0$ we have

$$ P(Z \geq t)=P(X\geq t,Y\geq t)=P(X\geq t)P(Y\geq t)=e^{-(\lambda+\mu)t} $$

So $Z$ is an exponential random variable with parameter $\lambda+\mu$.

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So I like this explanation, but I think it could be strengthened by adding an explanation of why $P(Z ≥ t) = P(X ≥ t, Y ≥ t)$ when $Z = min(X,Y)$. I'm assuming the reasoning you were presuming is that the probability that Z ≥ t is the probability that both X ≥ t and Y ≥ t, but it requires an additional step to understand why the distribution should be defined over joint ccdf's of the component RVs and not in terms of the cdfs. Even linking to would point to why joint ccdfs ($P( X ≥ t , ∀X ∈ \{X_1, X_2, \ldots\} )$) are used for finding minima. – mpacer Dec 15 '15 at 3:51

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