How can I prove that the minimum of two exponential random variables is another exponential random variable, i.e. Z = min(X,Y)
$\endgroup$2 Answers
$\begingroup$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 $\lambda+\mu$.
$\endgroup$ 2 $\begingroup$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.
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