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


2 Answers 2


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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    $\begingroup$ 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 stats.stackexchange.com/a/10072/2800 would point to why joint ccdfs ($P( X ≥ t , ∀X ∈ \{X_1, X_2, \ldots\} )$) are used for finding minima. $\endgroup$
    – mpacer
    Dec 15, 2015 at 3:51
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    $\begingroup$ Note that this extends easily to $n$ exponential variables (with the resulting minimum simply summing the rate parameters) $\endgroup$ Dec 14, 2017 at 9:41

It might be more intuitive to work with the CDF in this case.

$$F_Z(z) = P(Z < z) = P(\min(X,Y) < z)$$

What is the probability that the minimum of $X$ and $Y$ is below $z$? This will happen if at least one of $X$ and $Y$ is below $z$. The easiest way to deal with probability questions that include the phrase at least one is to find the complementary probability and subtract it from one.

$$P(\text{at least one of $X$ and $Y$ ≤ $z$}) = 1 - P(\text{each of $X$ and $Y$ > z})$$

By independence of $X$ and $Y$ this becomes $1 - P(X > z)P(Y > z)$.

We find $P(X > z) = 1 - F_X(z) = 1 - (1 - e^{-\lambda_X z}) = e^{-\lambda_X z}$ and similarly $P(Y > z) = e^{-\lambda_Y z}$.

Therefore $F_Z(z) = 1 - e^{-\lambda_X z}e^{-\lambda_Y z} = 1 - e^{-(\lambda_X + \lambda_Y) z}$ which is the CDF of an exponential variable with parameter $\lambda_X + \lambda_Y$.

This generalises easily to the case with more than two independent exponential variables.

  • $\begingroup$ Why is it xz instead of just z? I am referring to general formula of CDF. $\endgroup$
    – puffles
    Oct 10, 2018 at 13:08
  • $\begingroup$ In the CDF $P(X \leq z)$, the random variable is $Z$ but the parameter is $\lambda_X$ (the rate parameter for $X$) so it might look like there's an "xz". Does that answer your question, or is there a typo in my answer I didn't notice? $\endgroup$
    – Silverfish
    Oct 11, 2018 at 1:22
  • $\begingroup$ makes sense. Thanks1 $\endgroup$
    – puffles
    Oct 13, 2018 at 12:00

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