Find the value of P[X>YU] from a given joint distribution Suppose the joint distribution is given by some function $f(x, y, z, u)$. Then find the value of $P[X>Y<Z>U]$ assuming that $f$ is non-zero only for $x > 0, y>0, z>0, u>0$
My approach: Simply integrate the joint distribution function, taking care of appropriate limits for all the possible cases.
However, I keep getting some form of "circular limits" on the integral.
For example, one arrangement of the values for $x, y, z, u$ can be $y < u < x < z$
and the integral looks like $\int_{0}^{z}\int_{u}^{\infty}\int_{0}^{x}\int_{y}^{\infty}f(x, y, z, u)dx dy dz du$
Here, the limits for $z$ and $u$ are dependent on each other's value. Similarly for $x$ and $y$.
What is the way to deal with such integral limits? Forgive me if this a silly doubt, but I've been out of practice for a while. Please note that the RV are not independent.
Thanks
 A: For each integral you have to deal with (as you noted, there are several possible orderings of the variables), it may be easiest to integrate in order of the inequalities. So for instance in the one case mentioned in the original post ($y<u<x<z$), you could set up the integral as
$$\int_0^\infty \int_0^z \int_0^x \int_0^u f(x,y,z,u) dy du dx dz$$
Or you could do $$\int_0^\infty \int_y^\infty \int_u^\infty \int_x^\infty f(x,y,z,u) dzdxdudy$$
If you insist on the order $dxdydzdu$, I guess the integral would have to be
$$\int_0^\infty \int_u^\infty \int_0^u \int_u^z f(x,y,z,u) dxdydzdu$$
[Working from the outside, $u$ can be any positive real; $z>u$: $y<u$; $u<x<z$]
A: Writing the integrals in alternate form to make the manipulations easier to see:
$$\begin{align}&P[X>Y<Z>U]\\[2ex]
&=P[Y<\min(X,Z),\ \ U<Z]\\[2ex]
&=\int_0^\infty dz \int_0^\infty dx \int_0^{\min(x,z)} dy \int_0^z du\, f(x,y,z,u)\\[2ex]
&=\int_0^\infty dz \left\{\int_0^z dx +\int_z^\infty dx\right\}\int_0^{\min(x,z)} dy \int_0^z du\, f(x,y,z,u)\\[2ex]
&=\left\{\int_0^\infty dz \int_0^z dx \int_0^x dy \int_0^z du\, + \int_0^\infty dz \int_z^\infty dx \int_0^z dy \int_0^z du\right\}\,f(x,y,z,u)\\[2ex]
\end{align}$$
