# Maximums of two exponentials

The wikipedia page for the exponential distribution states that for $$X_1, X_2, \dots X_n$$ independent exponentially distributed with rate parameters $$\lambda_1,\lambda_2,\dots,\lambda_n$$, the index of the variable which achieves the minimum is distributed according to the law

$$\Pr(k|X_k = \min\{X_1, \dots, X_n\}) = \frac{\lambda_k}{\lambda_1+\dots+\lambda_n}.$$

Given the above, would I be correct in saying that in the special case $$n=2$$:

$$\Pr(1|X_1 = \max\{X_1, X_2\})=\Pr(X_1 >X_2) = \frac{\lambda_2}{\lambda_1+\lambda_2},$$

since when $$n=2$$ the variable that is not the minimum must be the maximum.

In other words, if $$U=I(X_1\geq X_2)$$, would $$U$$ have probability mass function:

$$\begin{equation} f(u) =\frac{1}{\lambda_1+\lambda_2} \begin{cases}\lambda_1, & u=0 \\ \lambda_2, & u = 1 \end{cases} \end{equation}$$

• The probabilities have to sum to unity: your conclusion follows immediately, since there are only two possible outcomes. – whuber Apr 15 at 13:05