Probability distribution of Maximum of the minimum of exponential random variables

Question

We have six exponentially distributed random variables: $X_1, X_2, X_3, Y_1, Y_2, Y_3$ with mean $\alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3$. We want make pairwise selection: $X_i,Y_i$, where $i = \{ 1,2,3\}$. Thus, we can either choose ($X_1$ and $Y_1$) or ($X_2$ and $Y_2$) or ($X_3$ and $Y_3$). We want to make sure that the pair selected has maximum of the minimum values.

Selection criteria: Select the maximum of the $\min(X_i, Y_i)$.

I would like to find the probability distribution $P(X_i > \tau, Y_i > \tau)$. Thank you.

Answer 1

In general, if $X\sim\mathrm{Exp}(\lambda)$ and $Y\sim\mathrm{Exp}(\mu)$ are independent, then for any $t>0$ we have $$ \mathbb P(X\wedge Y>t) = \mathbb P(X>t)\mathbb P(Y>t) = e^{-\lambda t}e^{-\mu t} = e^{-(\lambda+\mu)t}, $$ so that $X\wedge Y$ has $\mathrm{Exp}(\lambda+\mu)$ distribution. It follows that $$ Z_i:=X_i\wedge Y_i\sim \mathrm{Exp}(\alpha_i+\beta_i),\ i=1,2,3.\\ $$ By symmetry we have \begin{align} \mathbb P(\min\{Z_1,Z_2,Z_3\}=Z_i) = \frac{\alpha_i+\beta_i}{\sum_{j=1}^3(\alpha_j+\beta_j)}. \end{align} Conditioned on $\{\min\{Z_1,Z_2,Z_3\}=Z_i\}$, we have $$ \mathbb P(X_i>\tau,Y_i>\tau) = \mathbb P(X_i>\tau)\mathbb P(Y_i>\tau) = e^{-(\alpha_i+\beta_i)\tau}. $$ It follows then that \begin{align} \mathbb P(\min\{Z_1,Z_2,Z_3\}=Z_i, X_i>\tau,Y_i>\tau) &= \mathbb P(X_i>\tau,Y_i>\tau\mid \min\{Z_1,Z_2,Z_3\}=Z_i)\mathbb P(\min\{Z_1,Z_2,Z_3\}=Z_i)\\ &= \frac{e^{-(\alpha_i+\beta_i)\tau}(\alpha_i+\beta_i)}{\sum_{j=1}^3(\alpha_j+\beta_j)} \end{align}