If $X,Y$ are two non-negative integer RVs, show that $E[XY] = \sum[P(X>=n),(Y>=m)]$ I apologize for the missing latex commands but i'll try my best.
Show that $E[XY] = \sum\sum P(X \ge n, Y \ge m)$ 
The outer summation is for $n$ and the inner summation is for $m$ and the limits are as follows:
\begin{gather*}
1\le n < \infty\\
1\le m < \infty.
\end{gather*}
The problem before this asked me to prove that the expectation for non-negative integer RVs can be written as a distribution form instead of the conventional $xP(X=x)$ over all $x$ form. 
That is, $E[X]= \sum P(X \ge n)$ for $1\le n < \infty$
I did that successfully. However, I am slightly stuck here in proving joint expectation. In order for me to use my previous results, I have to somehow break up the joint PMF as a product of marginal PMFs which is only possible if the two RVs are independent. The problem does not state that the two RVs are independent. Any thoughts on how I could approach this? 
So I was thinking this \hellip. I don't know how much of it makes sense with my broken $\LaTeX$ but i'll try it anyway :).
Over all n ($\sum$ *P(X>=n, Y>=m)) * over all m ($\sum$ *P(X>=n, Y>=m
the first summation would be the expected value of Y and the expected value of X using the marginals... does it make sense? 
Thank you.
-SK
PS: yes, this is a homework assignment and I am not looking for the answer. Just a nudge in the right direction would be peachy! :) 
 A: Either you can prove directly, in a similar way as the previous one, or you can
use that result for conditional probabilities: introduce another random variable $Z_n$ for each fixed $n\in\Bbb N$ such that $P(Z_n=m)\ =\ \displaystyle\frac{P(X=n,\,Y=m)}{P(X=n)}$ (of course only if $P(X=n)\ne 0$). Then we have
$$E[XY]=\sum_{n,m}nm\,P(X=n,Y=m)=\sum_nn\,\left(\sum_mm\,P(X=n,Y=m)\right)= \\
= \sum_nn\,P(X=n)\,E(Z_n)=\sum_n \left(n\,P(X=n)\,\sum_m P(Z_n\ge m)\right)\,.$$
Then ovserve that $P(Z_n\ge m)=\displaystyle\sum_{k\ge m}P(Z_n=k)=\frac{P(X=n,\,Y\ge m)}{P(X=n)}$ as the events $(X=n,Y=k)$ are pairwise disjoint, so continuing
$$E[XY]= \sum_{n,m} n\,P(X=n,\,Y\ge m)$$
and then introduce one more family of random variables, $W_m$ such that $P(W_m=n)=\displaystyle\frac{P(X=n,Y\ge m)}{P(Y\ge m)}$.
A: I am finding this slightly easier to write this in a continuous setting, but indeed the proof is the same with summations. You can write that
$$
\mathbb{E}[XY] = \int_{x>0}\int_{y>0} xy \, p(x,y) \, dx \, dy
$$
and use the fact that $x = \int_{u>0} 1(u<x) \, du$ and $y = \int_{v>0} 1(v<y) \, dv$. After doing this and writing
$$
\mathbb{E}[XY] = \int_{x>0}\int_{x>0}\int_{u>0}\int_{v>0} \ldots
=
\int_{u>0}\int_{v>0} \int_{x>0}\int_{x>0} \ldots
$$
you will obtain exactly what you are looking for, namely
$$
\mathbb{E}[XY]
=
\int_{u>0}\int_{v>0} \int_{x>0}\int_{x>0} \mathbb{P}(X>u, Y > v) \, du \, dv.
$$
Indeed, this generalizes to any number of random variables.
