Discrete distribution taking only non-negative integer values

[rickywaldron]

I can't seem to wrap my head around the types of sums used in probability theory, and here is a classic example. Section 6.1 of this article:
http://en.wikipedia.org/wiki/Expect...ution_taking_only_non-negative_integer_values

The first line of the proof, what is going on here? I know how summation works, except I can't see the relation between the LHS and the RHS

Then the last step, I can't see how the second summation goes away and is just replaced by a j!
I always get confused by this notation but when I understand it intuitively I am much more comfortable.

Thanks

 

[CompuChip]

In the first line of the proof, they are replacing
$$P(X \ge i) = \sum_{j = i}^\infty P(X = j)$$
which is basically just writing out what the inequality sign means (if X is greater than or equal to i, then it is equal to i, or i + 1, or i + 2, etc).

Then they interchange the order of the summation and note that nothing in the summation depends on the summation variable (i) anymore, and they use
$$\sum_{i = 1}^j 1 = 1 + 1 + \cdots + 1 \text{ (j times)} = j$$

 

[rickywaldron]

Thanks, really good response

 

[CompuChip]

In fact, looking at the proof, I would probably put it the other way around, starting from the definition $E[X] = \sum j P(X = j)$. This makes the proof a bit less readable perhaps, the tricks described above seem to come even more from thin air as they do now, but that is quite typical in mathematical proofs, I think.

Also, if you want to be very rigorous: when dealing with infinite sums it is generally not allowed to interchange the order of the summation without additional restrictions on the summand. So when handing this in as an exercise for a class you would want to elaborate a bit on that, I guess.

 

[ankitnaudy]

@CompuChip Could you please explain how the summation interchanged. I could not get how limits of i changed from infinity to till j. Thanks.