# X=0 almost surely => E(X)=0

• Jul 10th 2009, 01:25 PM
kingwinner
X=0 almost surely => E(X)=0
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=0 with 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]
• Jul 10th 2009, 06:37 PM
halbard
Suppose the random variable $X$ has the standard normal distribution, $X\sim\mathrm N(0,1)$. Pick any real number, say $1.5$. What is the probability of the event $\{X=1.5\}$? The answer is $0$, i.e. $\mathrm P(X=1.5)=0$, as any high school student will tell you. So the event $\{X\neq1.5\}$ has probability $1$, that is $X\neq1.5$ almost surely.

Do you think this means that it is impossible for $X$ to be exactly $1.5$? You may say yes, and you may be right, but then the same is true for any other value. You are left wondering how it is possible for $X$ to take any value whatsoever, given that it's impossible for $X$ to take any particular value.

Do you begin to see the difference between the terms surely and almost surely?
• Jul 10th 2009, 10:47 PM
kingwinner
Quote:

Originally Posted by halbard
Suppose the random variable $X$ has the standard normal distribution, $X\sim\mathrm N(0,1)$. Pick any real number, say $1.5$. What is the probability of the event $\{X=1.5\}$? The answer is $0$, i.e. $\mathrm P(X=1.5)=0$, as any high school student will tell you. So the event $\{X\neq1.5\}$ has probability $1$, that is $X\neq1.5$ almost surely.

Do you think this means that it is impossible for $X$ to be exactly $1.5$? You may say yes, and you may be right, but then the same is true for any other value. You are left wondering how it is possible for $X$ to take any value whatsoever, given that it's impossible for $X$ to take any particular value.

Do you begin to see the difference between the terms surely and almost surely?

1) But since high school, I was taught that an event with probability 1 is a certain event and an event with probability 0 is an impossible event.
P(Ω)=1
P(empty set)=0

I still don't quite see the difference between "X=0" and "X=0 almost surely".

Also, about your example P(X=1.5)=0, it is actually a big puzzle to me...I am never able to fully understand why P(X=1.5) would be exactly 0. What I actually beleive is that P(X=1.5) would be very close to 0, instead of exactly 0.
• Jul 11th 2009, 12:01 AM
CaptainBlack
Quote:

Originally Posted by kingwinner
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.

You need to rewrite this so that you can see what is happening:

$f_1(N) \le f_2(N)$

Then:

$\lim_{N\to N_0} f_1(N) \le \lim_{N \to N_0} f_2(N)$

This should be familiar, since it is equivalent to:

$g(N)=f_2(N)-f_1(N)\ge 0$

Then:

$\lim_{N \to N_0}g(N) \ge 0$

(I have missed out the caveats about what is meant if any of the limits do not exist)

CB
• Jul 11th 2009, 12:10 AM
kingwinner
Quote:

Originally Posted by CaptainBlack
You need to rewrite this so that you can see what is happening:

$f_1(N) \le f_2(N)$

Then:

$\lim_{N\to N_0} f_1(N) \le \lim_{N \to N_0} f_2(N)$

This should be familiar, since it is equivalent to:

$g(N)=f_2(N)-f_1(N)\ge 0$

Then:

$\lim_{N \to N_0}g(N) \ge 0$

(I have missed out the caveats about what is meant if any of the limits do not exist)

CB

Just to clarify a bit, by writing f(N)≤g(N), you mean "f(N)≤g(N) for ALL N", am I right?

Is it true that f(N)≤g(N) ALWAYS implies
lim f(N)≤ lim g(N) ?
N->∞ N->∞
And we can safely take the limit of both sides while still preserving the same inequality sign??
(I flipped through my calculus textbooks again, but still I am unable to find the exact statement of the theorem that guarantees the above, so I am not sure whether it is a correct statement or not.)