Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Since probabilities are always non-negative, the order of summation doesn't matter -- so you just need to convince yourself that there is a one-to-one correspondence between the terms on one side and the terms on the other. –  Henning Makholm Oct 9 '11 at 21:29
    
@DidierPiau: Yes, sorry. Edited. –  jean Oct 9 '11 at 21:32

4 Answers 4

up vote 4 down vote accepted

It doesn't matter what you sum (as long as the sums are convergent). The points in the $(r,v)$ plane that are being summed over can be illustrated by a diagram like this:

v

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

The two sides in the identity correspond to summing by columns first or rows first.

share|improve this answer

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.

share|improve this answer
    
This is equivalent to my answer. :) –  J. M. Oct 10 '11 at 3:11
    
Yes, it is, but I didn't use Iverson's brackets. Aren't all correct answers equivalent? :) –  Kirill Oct 10 '11 at 3:21
    
I mean you made the exact same inequality manipulations. ;) –  J. M. Oct 10 '11 at 3:24
    
@J.M.: and both of you have zero ;) –  Ilya Oct 10 '11 at 17:57
    
@Gortaur: I don't particularly mind. :) In any event: you might want to see page 3 of this... –  J. M. Oct 10 '11 at 18:02

Both say you're summing over $\{(x,r) : 0 \le r < x\}$.

share|improve this answer

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

$$\sum_x[x \geq 1] \sum_r [x-1 \geq r \geq 0]\mathbb P(X=x)=\sum_{x=1}^\infty\sum_{r=0}^{x-1}\mathbb P(X=x)$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.