The problem here is that you are adding together two values for a specific case of x and then adding together the result for all values of x (is that clear? pro

The problem here is that you are adding together two values for a specific case of x and then adding together the result for all values of x (is that clear? probably not). It's analogous to another age old problem: what is the value of 1 + (-1) + 1 + (-1) + 1 + (-1) ... You could argue that (1 + -1) + (1 + -1) ... = 0 + 0 ... = 0. You could also argue that 1 + (-1 + 1) + (-1 + 1) = 1 + 0 ... = 1. There are actually way to group the numbers in that series to get any integer answer you want :D Since we haven't actually defined a distribution over x the problem is not well-defined anyway. But lets pick, say, P(x=2^i) = 1/(2^i) for all i>1. Then the expected value of A is E(A) = (1/2 * $2) + (1/4 * $4) + (1/8 * $8) ... = +infinity. Similary E(B) = +infinity. Now for E(B-A) E(B-A) = (1/2 * $1) + (1/2 * -$1) + (1/4 * $2) + (1/4 * -$2) ... Like the example above, we can add this up in different ways: E(B-A) = ((1/2 * $1) + (1/2 * -$1)) + ((1/4 * $2) + (1/4 * -$2)) ... = $0 + $0 ... = $0 Or E(B-A) = (1/2 * $1) + ((1/2 * -$1) + (1/4 * $2)) + ((1/4 * -$2) ... = $0.5 + $0 + $0 ... = $0.5 When adding up infinite numbers of things you have to be very careful. I go into more detail in this thread - https://news.ycombinator.com/item?id=6387344 https://news.ycombinator.com/item?id=6387344 Cookie please :D