I am currently trying to solve this question when I noted that my knowledge about probability topics became a bit rusty.
Let x, y, and z be independently but not identically distributed on some common support [0,finite]. Additionally, let the distributions be non-atomic.
I'm struggling to figure out whether I am doing everything correctly when computing the probability of the relative order of the realizations of the random numbers. Say, what is Pr(X$>$Y and Y$>$Z and Z$<$a) where a is a non-random integer number$<$finite? Am I right that this would be
$$\int_{0}^{finite}\int_{0}^{a}\int_{0}^{Y}dF_{Z}(u)dF_{Y}(u)dF_{X}$$
The issue is that one of the integrals would than have Y as one point of evaluation with Y being a realization of some random number which confuses me a bit not only with notation.
To find $\mathbb P(X>Y>Z, Z<a)$ for any fixed $a$ varies in some bounds $(0,N)$ you need to integrate $dF_X(x)dF_Y(y)dF_Z(z)$ over the set $$ \{(x,y,z)\in\mathbb R^3_+: x>y>z,\ z<a\}. $$ Let outer integral be over $z$. The only restriction on $z$ is $0<z<a$. For any fixed $z$, $y$ should be greater than $z$. For any fixed $y$, $x$ should be geater than $y$. Therefore $$ \mathbb P(X>Y>Z, Z<a) = \int_{z=0}^{z=a} \int_{y=z}^{y=\infty} \int_{x=y}^{x=\infty} dF_X(x) dF_Y(y) dF_Z(z). $$ You can integrate in the other order: let the outer integral is over $x$. Then $0<x<\infty$. For any fixed $x$, $y$ should be less than $x$. And for any fixed $y$, $z$ should be less than $y$. And also $z$ should be less than $a$. This gives $z<\min{y,a}=y\wedge a$. Wrote it: $$ \mathbb P(X>Y>Z, Z<a) = \int_{x=0}^{x=\infty} \int_{y=0}^{y=x} \int_{z=0}^{z=y\wedge a} dF_Z(z) dF_Y(y) dF_X(x). $$ Note that there is no integration over $a$ since $a$ is a nonrandom parameter which is preliminary fixed. It cannot vary while we calculate probabililty. Therefore integrals over $(0,N)$ cannot arise here.