Let $X$ and $Y$ be two exponentially distributed and independent random variables with mean $α$ and $β$, respectively. If $Z$ = min $(X, Y)$, then the mean of $Z$ is given by

Given n Random Var, the probability P(Y≤y)=P(min(X1…Xn)≤y) implies that at least one Xi be smaller than y. The probability that at least one Xi is smaller than y is equivalent to one minus the probability that all Xi are greater than y.

So, $\begin{align*}P\left \{Z > a \right \} &= P \left \{X > a \right \} P \left \{Y > a \right \} \left(\because \text{X and Y are independent events and } \\Z > \min \left(X, Y \right) \right)\\&=e^ {-\frac{1}{\alpha} a} e^ {-\frac{1}{\beta} a} \\&=e^{-\left(\frac{1}{\alpha} + \frac{1}{\beta} \right)a} \\&=e^{-\left(\frac{\alpha + \beta} {\alpha \beta} \right)a}\end{align*}$

This shows that $Z$ is also exponentially distributed with parameter $\frac{\alpha + \beta} {\alpha \beta}$ and mean $\frac{\alpha \beta} {\alpha + \beta}$.