Double summation limits Is there a way to "see" that $$\sum\limits_{r=0}^\infty \sum\limits_{x=r+1}^\infty \mathbb P(X=x)=\sum\limits_{x=1}^\infty\sum\limits_{r=0}^{x-1}\mathbb P(X=x)\; ?$$ Thanks.
 A: It doesn't matter what you sum (as long as the sums are convergent). The points in the $(r,x)$ plane that are being summed over can be illustrated by a diagram like this:
x

5 *****
4 ****
3 ***
2 **
1 *
  012345  r

The two sides in the identity correspond to summing by columns first or rows first.
A: Alternatively, you can use Iversonian brackets to prove the claim. Recall that an Iversonian bracket works like so:
$$[p]=\begin{cases}1&\text{if }p\text{ is true}\\0&\text{if }p\text{ is false}\end{cases}$$
and has the property $[p\text{ and }q]=[p][q]$. With these in mind, consider the following:
$$\sum_{r=0}^\infty \sum_{x=r+1}^\infty \mathbb P(X=x)=\sum_r[r \geq 0]\sum_x [x \geq r+1]\mathbb P(X=x)$$
This is equivalent to your original sum, in that the Iversonian brackets zero out all the other terms that do not belong to your original series. Using the multiplication property of the Iversonian brackets, we can turn this into
$$\sum_r[r \geq 0]\sum_x [x-1 \geq r]\mathbb P(X=x)=\sum_r\sum_x [x-1 \geq r \geq 0]\mathbb P(X=x)$$
which can be turned into
$$\sum_x[x \geq 1] \sum_r [x-1 \geq r \geq 0]\mathbb P(X=x)=\sum_{x=1}^\infty\sum_{r=0}^{x-1}\mathbb P(X=x)$$
A: The summation range is given by the double inequality $$0 \leq r < x.$$ So long as all sums converge absolutely and uniformly, it does not matter in which order the inequality is interpreted. So the following are both equivalent to the inequality above:
$$ 0\leq r, \qquad r < x$$
and
$$ 0 < x, \qquad 0\leq r < x.$$
This is precisely what happens when that order of summation is interchanged.
A: Both say you're summing over $\{(x,r) : 0 \le r < x\}$.
