Axioms of expectation:

1. X≥0 => E(X)≥0

2. E(1)=1

3. E(aX+bY) = aE(X) + bE(Y)

4. If X_n is a nondecreasing sequence of numbers and lim(X_n) = X, then E(X)=lim E(X_n) [montone convergence theorem]

Definition:P(A)=E[I(A)]

Using the above, prove that if X=0 almost surely [i.e. P(X=0)=1 ], then E(X)=0.

Proof:

X=0 almost surely <=> |X|=0 almost surely

[note: I is the indicator/dummy variable

I(A)=1 if event A occurs

I(A)=0 otherwise]

|X| = |X| I(|X|=0) + |X| I(|X|>0)

=> E(|X|) = E(0) + E[|X| I(|X|>0)]

=E(0*1) + E[|X| I(|X|>0)]

=0E(1) + E[|X| I(|X|>0)] (axiom 3)

=0 + E[|X| I(|X|>0)] (axiom 2)

=E[|X| I(|X|>0)]

=E[lim |X| * I(0<|X|≤N)] (lim here means the limit as N->∞)

=lim E[|X| * I(0<|X|≤N)] (axiom 4)

≤lim E[N * I(0<|X|≤N)]

=lim N * E[I(0<|X|≤N)]

=lim N * P(0<|X|≤N) (by definition)

=lim (0) since P(X=0)=1 => P(0<|X|≤N)=0

=0

=>E(X)=0

=======================================

Now, I don't understand the parts in red.

1) The following is a proof of the claim E[|X| I(|X|=0)] =0.

"If X=0, then |X|=0, so |X| I(|X|=0) = 0

If X≠0, then I(|X|=0) =0, so |X| I(|X|=0) = 0

Therefore, |X| I(|X|=0) is always 0 and E[|X| I(|X|=0)] =0. "

But in the assumptions, we are given that X=0 almost surely [i.e. P(X=0)=1]. We can say that X=0with certainty. It is impossible for X to not be 0. Then, why are we even considering the case X≠0? There should only be one possible case, X=0, right?

2) |X| * I(0<|X|≤N) ≤ N * I(0<|X|≤N)

=> E[|X| * I(0<|X|≤N)] ≤ E[N * I(0<|X|≤N)] (I am OK with this step)

But does this imply

lim E[|X| * I(0<|X|≤N)] ≤ lim E[N * I(0<|X|≤N)] ???????????

(note: lim=the limit as N->∞)

I was flipping through my calculus textbooks, but I couldn't find a theorem that applies and justifies the last step about the limit.

Any help is greatly appreciated!

[also under discussion in talk stats forum and S.O.S. math cyberboard, yet nobody has provided an answer to the two follow-up questions above]