Maximum of the Sum of Exponential Random variables

Question

We have six independent exponentially distributed random variables: $X_1, X_2, X_3, Z_1, Z_2, Z_3$, with mean values $\alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3$, respectively. Also, $\alpha_1 \neq \alpha_2 \neq \alpha_3 \neq \beta_1 \neq \beta_2 \neq \beta_3$.

We want to make pairwise selection: ($X_i, Z_i$), where $i = \{1, 2, 3 \}$. Thus, we can either select ($X_1, Z_1$) or ($X_2, Z_2$) or ($X_3, Z_3$).

Selection Criteria: Let, $Y_i = X_i + Z_i$. Select $\max_i (X_i + Z_i)$, where $i = 1, 2, 3$.

Given that we choose $X_i$ and $Z_i$ based on the given criteria, what will be the probability distribution $P(X_i > \tau, Z_i > \tau)$?

I started this as: First, we need to find the probability distribution of summation of $X_i$ and ${Z}_i$, where $i = \{ 1, 2, 3\}$.

$F_{Y_1}(y) = P(X_1 + Z_1) = \int \int_{x_1 + z_1 \leq y} f_{X_1}(x_1) f_{Z_1}(z_1) \text{ } dx_1 dz_1 $

$ = \int_{-\infty}^\infty \int_{-\infty}^{y-x_1} f_{X_1}(x_1) f_{{z}_1}({z}_1) \text{ } dx_1 d{z}_1 $

Since we are interested in the event that $X_1 > \tau$ and ${Z}_1 > \tau$, we can write this as

$F_{Y_1 \mid X_1 > \tau, {Z}_1 > \tau}(y \mid X_1 > \tau, {Z}_1 > \tau) = $

$\frac{\int_{\tau}^\infty \left( \int_{\tau}^{y-x_1} f_{Z_1}(z_1) \text{ } dz_1 \right) f_{X_1}(x_1) \text{ } dx_1} {P(X_1 > \tau) P(Z_1 > \tau) } $

Also, how should I proceed to find the expectation?

Answer 1

Now that you have the density, you'll want to find the CDF (which isn't too hard in this case).

In general, the approach for saying something about the maximum of independent variables $X_1, X_2, \dots, X_n$ is to consider the CDF of the max. Note that $$\max\{X_1, \dots, X_n\} \leq k \iff X_1 \leq k, X_2 \leq k, \dots, X_n \leq k$$ so if $G$ is the CDF for $\max\{X_1, \dots, X_n\}$ and $F_i$ is the CDF for $X_i$, $$G(k) = F_1(k) F_2(k) \dots F_n(k).$$ Once you have the CDF, you can recover the density, and once you have that, you're home free for pretty much anything else you want to do.