There is a result that $E(X)$ exists iff $E(|X|)<\infty$

Question

I have a question on an interesting property of $E(X)$

The if part is true because, in that case we get a series which is absolutely convergent , so the original series must converge. But what about the converse? Please dont refer to Lebesgue Theory, because i am a novice in it.

Not a result, but a part of the definition. – Did Mar 09 '17 at 06:56 — Did, Mar 09 '17 at 06:56

score 0 · Answer 1 · answered Mar 09 '17 at 08:29

You didn't want to get referred to with Lebesgue Theory, but one needs to talk about integrals at least.

Calculating $E(X)$ means that you calculate an integral (I am deliberately not going into detail here). This integral is calculated in a way that you essentally calculate the cases where $X>0$ and where $X<0$ separately and then add the results together.

Hence, if you can calculate $E(X)$, this means that you can calculate $E(X|X>0)$ and $E(X|X<0)$ separately, and thus also $E(|X|)$ can be calculated. This is what @Did meant with "it's part of the definition".

There is a result that $E(X)$ exists iff $E(|X|)<\infty$

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