# Order statistics of a max-min problem.

$X_{A,1}, X_{B,1}, X_{A,2}, X_{B,2}$ are independent and identically distributed (i.i.d.) exponential random variables with parameter $\lambda=1$, i.e., the means of the four random variables are all equal to 1. Let $$q_l=\min\{X_{A,l},X_{B,l}\}\tag{1}$$ for $l=1, 2$, and $$l^*=\arg\max_{1\leq l \leq 2} q_l.\tag{2}$$ What is the cumulative distribution function (CDF) of $X_{A,l^*}$?

The following is my try: First, from the fact that the minimum of two independent exponential random variables is again an exponential random variable with a rate parameter equals to the sum of the two parameters, i.e., $q_1, q_2 \sim \exp(2)$, the CDF of $q_l$ is $$F_{q_l}(r)=1-e^{-2r}, r\geq0.\tag{3}$$ Furthermore, we can find the CDF of $q_{l^*}=\max\{q_1,q_2\}$, which is \begin{align*} F_{q_{l^*}}(r) &=P[q_{l^*}\leq r]\\ &=P[\max\{q_1, q_2\}\leq r]\\ &=P[q_1\leq r\ \text{and}\ q_2\leq r]\\ &=P[\{q_1\leq r\}\cap \{q_2\leq r\}]\\ &\stackrel{\text{(a)}}{=}P[\{q_1\leq r\}] \cdot P[\{q_2\leq r\}]\\ &\stackrel{\text{(b)}}{=}\left( P[\{q_1\leq r\}] \right)^2 \\ &=\left[ F_{q_l}(r) \right]^2 \\ &\stackrel{\text{(c)}}{=}(1-e^{-2r})^2, r\geq0, \end{align*}\tag{4} where (a) results from the fact that $q_1$ and $q_2$ are independent, (b) results from the fact that $q_1$ and $q_2$ have the same distribution, and (c) results from (3). Then I don't know how to continue. Any answers or comments are welcome.