Assume $X$ and $Y$ are non-negative integers. They are random variables.
Show that $E[X]=\sum_{n>o} P[X\geqslant n]$ Show also that $E[X*Y]=\sum_{n=1}^{\inf}\sum_{m=1}^{\inf}P[(X\geqslant n) \cap (Y \geqslant m)]$
I am having trouble because I thought these were just the formal definitions of the expectation.
EDIT: $E[X]=\sum n*P[X=n]$ $=(P[X=1]+P[X=2]+P[X=3]+\dots)+(P[X=2]+P[X=1])+P[X=1]$ $=\sum_{n=1}P[X=n]+\sum_{n=2}P[X=n]+\sum_{n=3}P[X=n]+\dots$ $=P[X>0]+P[X>1]+P[X>2]+\dots$ $=\sum_{n>0}P[X \geqslant n]$
Here is my attempt for the first equivalency. I think that makes sense, and it's what I wish to show. Can someone confirm?
For the second, I'm starting with the definition that
$E[XY]=Cov(X,Y)+E[X]E[Y]$, but I'm not sure how this will turn into the right hand of the expression.
No, they are not the formal definitions of expectation.
The formal definition is: $\mathsf E(X) = \sum\limits_{n=0}^\infty n\,\mathsf P(X=n)$
You've been asked to show: $\mathsf E(X) = \sum\limits_{n=1}^\infty \mathsf P(X\geq n)$
Can you see the difference? Can you make an attempt?
Hint: $\; n = \sum\limits_{k=1}^n 1\;$ and $\;\mathsf P(X\geq k) = \sum\limits_{k=n}^\infty \mathsf P(X=k)$