Summing up discrete probabilities - trivial?

38 Views Asked by At

$$ \sum_{i=1}^\infty \sum_{k=1}^i p_X(i) = \sum_{i=1}^\infty i p_X(i). $$ This is from a problem solution in introductory probability theory course (problem 3a here, solution here) . X is a random variable that takes nonnegative integer values. The expression implies that sum from 1 to i of probabilities of X being i is equal to i (inner sum collapses into i). But I'm not sure why that is the case.

I understand that 1) probabilities for ALL i's must add up to 1, and 2) if we take 1 i times (the summation) we should get i. But it doesn't connect... Perhaps not fully getting notation here?

Any help would be much appreciated.

1

There are 1 best solutions below

0
On BEST ANSWER

This is not about probabilities adding up to $1.$ Suppose, for example, that $i=4.$ Then $$ \sum_{k=1}^i p_X(i) = \sum_{k=1}^4 p_X(4) = \underset{\Large\underset{k=1}\uparrow}{p_X(4)} + \underset{\Large\underset{k=2}\uparrow}{p_X(4)} + \underset{\Large\underset{k=3}\uparrow}{p_X(4)} + \underset{\Large\underset{k=4}\uparrow}{p_X(4)} = 4p_X(4) = ip_X(i). $$

Note that in the expression $p_X(i),$ the index $k$ that goes from $1$ to $4$ does not appear. That is why all four terms are the same as each other.