# Probability distribution of Maximum of the minimum of exponential random variables

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.

• What you have asked: "Select the maximum of the $\min(X_i,Y_i)$" is different from "I would like to find the probability distribution $\mathbb P(X_i>\tau, Y_i>\tau)$." Can you clarify exactly what you are asking? – Math1000 Oct 15 at 17:25
• Say, you roll two dices together three times. you get the values: $(x_1, y_1)$, $(x_2, y_2)$ and $(x_3, y_3)$ at first, second and third throw, respectively. Here, $x_i$ is the value from one dice and $y_i$ is the value from the other dice. Now you want to find $\max \{ \min (x_i, y_i) \}$ first, where $i = \{1,2, 3\}$. After determining which throw has the maximum of minimum values, you want to check the probability that both $x_i$ and $y_i$ is above a certain value. – jhon_wick Oct 15 at 17:40

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}
• Not sure why $\tilde{X}:= \min \{ X_1, X_2, X_3\}$? This is a pairwise selection, where we can pick either $X_1$ and $Y_1$ or $X_2$ and $Y_2$ or $X_3$ and $Y_3$. – jhon_wick Oct 15 at 16:57