# Poisson process problem; joint and marginal distributions of time till first event

There are two companies, Company 1 and Company 2. Stock market crashes occur according to a poisson process with rate $\lambda_0$ and destroy both. A type 1 event (type 2 event) also follows a poisson process with rate $\lambda_1$($\lambda_2$) and destroys only Company 1 (Company 2). All three poisson processes are independent. Let $X_1$ and $X_2$ be how long Company 1 and Company 2 will survive.

(a) Find the marginal distribution of $X_1$ and $X_2$.

(b) Find $P(X_1 > x_1 , X_2 > x_2)$, and use this to find the joint CDF of $X_1$ and $X_2$.

My approach:

(a) The time till the first event in each poisson process occurs is exponentially distributed with rate $\lambda_0$, $\lambda_1$ and $\lambda_2$. Thus, using the fact that the minimum of two independent exponential random variables is also an exponential random variable,

$$X_1 \sim \operatorname{Expo}(\lambda_0+\lambda_1) \quad \text{and} \quad X_2 \sim \operatorname{Expo}(\lambda_0+\lambda_2).$$

(b)

\begin{align} P(X_1 >x_1, X_2>x_2) &= P(\text{no crash before } max(x_1,x_2)) \\ &\times P(\text{no type 1 event before } x_1) \\ &\times P(\text{no type 2 event before } x_2) \end{align}

But here I don't know how to move on.

When $x_1{}>{}x_2$, $$\begin{eqnarray*} f(x_1,x_2){}={}\dfrac{\partial^2}{\partial x_1\partial x_2}P\left(X_1>x_1,X_2>x_2\right){}={}\dfrac{\partial^2}{\partial x_1\partial x_2}\left(e^{-\left(\lambda_1 x_1{}+{}\lambda_2 x_2{}+{}x_1\lambda_0\right)}\right)\,, \end{eqnarray*}$$
and, when $x_1{}<{}x_2$,
$$\begin{eqnarray*} f(x_1,x_2){}={}\dfrac{\partial^2}{\partial x_1\partial x_2}P\left(X_1>x_1,X_2>x_2\right){}={}\dfrac{\partial^2}{\partial x_1\partial x_2}\left(e^{-\left(\lambda_1 x_1{}+{}\lambda_2 x_2{}+{}x_2\lambda_0\right)}\right)\,. \end{eqnarray*}$$
The case $\{x_1{}={}x_2\}$ is a set of lebesgue measure zero, so unimportant.