Marginal pdf with nonnegative random variables

Question

Suppose that $X_1$ and $X_2$ are two nonnegative random variables that satisfy

$$P(X_1 > x_1, X_2 > x_2) = \exp \left[−\mu x_1 − \nu x_2 − \lambda \max(x_1, x_2)\right] ,\text{ for }x_1 \ge 0 \text{ and }x_2 \ge 0\,,$$

where $\mu, \nu$, and $\lambda$ are positive parameters.

Find the marginal p.d.f.s of $X_1$ respectively.

(My Answer)

1. x1≥x2 , max(x1,x2)=x1 $$ f(x_1) = \int_0^\infty exp(−µx_1 − νx_2 − λ max(x_1, x_2) dx_2 $$ $$ = \int_0^\infty exp(−µx_1 − νx_2 − λ x_1) dx_2 $$ $$ =exp(−µx_1− λ x_1)\int_0^\infty exp(− νx_2 ) dx_2 $$ $$ =exp(−µx_1− λ x_1) (-\frac{1}{v})exp(− νx_2 )|_0^\infty$$ $$ = \frac{exp(-(µ+λ)x_1)}{v}$$

2. x1<x2 , max(x1,x2)=x2 In the same way as x1≥x2, $$ f(x_2) = \int_0^\infty exp(−µx_1 − νx_2 − λ max(x_1, x_2) dx_2 $$ $$ = \int_0^\infty exp(−µx_1 − νx_2 − λ x_2) dx_2 $$ $$ =exp(-µx_1)\int_0^\infty exp(− νx_2 − λ x_2) dx_2 $$ $$ =exp(-µx_1) (−\frac{1}{ν+λ})exp(-(ν+λ)x_2)|_0^\infty$$ $$ = \frac{exp(-µx_1)}{ν+λ}$$

Is it right?

Answer 1

Just out $x_2=0$ to get $P(X_1>x_1)=e^{-(\lambda+\mu)x_1}$ for $x_1 \geq 0$. Hence the marginal density of $X_1$ is $(\lambda+\mu) e^{-(\lambda+\mu)x_1}$ for $x_1 \geq 0$. Similarly write down the density of $X_2$.