Assume $Z_1, Z_2, Z_3$ are independent exponential random variables with parameter $\lambda >0$ and $J$ denotes a bernoulli random variable with parameter $\theta \in [0,1]$.
$$ X = (1-\theta)Z_1 + JZ_3,$$ $$ Y = (1-\theta)Z_2 + JZ_3,$$
I'm trying to evaluate $\mathbb{P}(X > x, Y > y)$.
My question: what is the best approach for splitting up the sums like these?
I assume that we can always split the probabilities in the following way:
$$\begin{array}{ll}\mathbb{P}(X > x, Y > y) &= \mathbb{P}((1-\theta)Z_1 + JZ_3 > x, (1-\theta)Z_2 + JZ_3 > y) \\ &= \mathbb{P}(\min\{(1-\theta)Z_1, JZ_3\} > x, \min\{ (1-\theta)Z_2, JZ_3\} > y) ~+ \\ &~~~~ \mathbb{P}(\min\{(1-\theta)Z_1, JZ_3 \}> x, \max\{(1-\theta)Z_2, JZ_3\} > y) +~\\ &~~~~ \mathbb{P}(\max\{(1-\theta)Z_1, JZ_3 \}> x, \min\{(1-\theta)Z_2, JZ_3\} > y) +~\\ &~~~~ \mathbb{P}(\max\{(1-\theta)Z_1, JZ_3 \}> x, \max\{(1-\theta)Z_2, JZ_3\} > y)~\\ \end{array}$$ And from here proceed dissecting the $\min$'s and $\max$'s. In this particular case, it seems possible to write, i.e., $\mathbb{P}(\min\{(1-\theta)Z_1, JZ_3 \}> x)$ as $$\mathbb{P}((1-\theta)Z_1 > x, JZ_3 > x) = \mathbb{P}(Z_1 > \frac{x}{1-\theta}, JZ_3 > x),$$ however, what can $\mathbb{P}(\max\{(1-\theta)Z_1, JZ_3 \}> x)$ be reduced to? In this case it feels like a cleaner solution should exist, or if I'm missing something.
EDIT:
I will add that the solution I'm trying to arrive to is:
$$ P(X>y, Y>y) =\text{exp}\{{-\lambda \max\{x, y\}}\} + \frac{1-\theta}{1 + \theta}\text{exp}\left\{\frac{-\lambda}{1-\theta}(x+y)\right\}\left( 1 - \text{exp}\left\{\lambda \frac{1+\theta}{1-\theta}\min\{x,y\}\right\}\right) $$
Would he grateful for any hints or tricks!