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.

  • $\begingroup$ 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? $\endgroup$
    – Math1000
    Oct 15, 2019 at 17:25
  • 1
    $\begingroup$ 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. $\endgroup$
    – jhon_wick
    Oct 15, 2019 at 17:40

1 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}

  • $\begingroup$ 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 $Y_3$. $\endgroup$
    – jhon_wick
    Oct 15, 2019 at 16:57
  • $\begingroup$ Ah, I missed that detail. I will need to rewrite my answer then. $\endgroup$
    – Math1000
    Oct 15, 2019 at 17:18
  • $\begingroup$ @jhon_wick I have rewritten my answer as according to what you were actually asking. It is possible it may have an error, but at least it is attempting to answer the correct question now :) $\endgroup$
    – Math1000
    Oct 15, 2019 at 19:38
  • $\begingroup$ thank you!. I have to write the solution by myself and check the steps. For now, this looks promising. Could you check this link please: math.stackexchange.com/questions/3387496/…. It's kind of similar. $\endgroup$
    – jhon_wick
    Oct 16, 2019 at 10:02

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