Compute $P(X<Y<Z)$ for $(X,Y,Z)$ with joint density $e^{-x-y-z}$

Question

Suppose that X,Y and Z have a joint density function given by
$$f(x,y,z) = \begin{cases}e^{-x-y-z}&\text{if }x,y,z>0\\ 0&\text{otherwise}\end{cases}$$ Compute $P(X<Y<Z)$

I think I should try get the marginal density for X, Y and Z. and then after that integrated them with this restriction, $x,y,z>0$ and $x<y<z$. But I think it is wrong for I got wrong answer, right answer is 1/6.

Answer 1

Write the probability in terms of the vector $(X,Y,Z)$, i.e. $$ P(X<Y<Z)=P((X,Y,Z)\in A)=\int f(x,y,z)\mathbf{1}_A(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz $$ where $A=\{(x,y,z)\in\mathbb{R}^3\mid x<y<z\}$. Then note that $$ f(x,y,z)\mathbf{1}_A(x,y,z)=e^{-x-y-z} $$ if $0\leq z$, $0\leq y<z$ and $0\leq x<y$ and it is zero otherwise, and hence

$$ P(X<Y<Z)=\int_0^\infty \int_0^z\int_0^y e^{-x-y-z}\,\mathrm dx\,\mathrm dy\,\mathrm dz. $$

Answer 2

Forget the exact formula for the joint density. By symmetry, $P(X\lt Y\lt Z)=\frac16$.

This holds for every independent random variables $(X,Y,Z)$ with the same continuous distribution. Then the probability of ties is zero and the six possible orderings of $X$, $Y$ and $Z$ are an equiprobable partition of the sample space.