# Probabilistic dominated convergence theorem

• Dec 13th 2009, 05:48 AM
kingwinner
Probabilistic dominated convergence theorem
Dominated convergence theorem: Suppose Xn->X almost surely and |Xn| ≤ W for all n, where E(W)<∞. Then E(Xn)->E(X).

1) Suppose Xn->X in probability and |Xn| ≤ W for all n, where E(W)<∞. Show that E(Xn)->E(X).

2) Suppose Xn->X in probability and (Xn)^2 ≤ W for all n, where E(W)<∞. Show that Xn->X in mean square. (i.e. E[(Xn-X)^2]->0)
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1) Let E(Xnk) be any subsequence of {E(Xn)}.
Then Xnk->X in probability => there exists a further subsequence Xnk' such that Xnk'->X almost surely.
By dominated convergence theorem, this implies that E(Xnk')->E(X) and so every subsequence of {E(Xn)} has a further subsequence of it which converges to E(X) => E(Xn)->E(X).

2) But now I am stuck with question 2. The assumption is replaced by (Xn)^2 ≤ W, and we need to prove more: Xn->X in mean square. I tried to modify the proof in question 1, but it doesn't seem to work here. How can we modify the proof? Or do we need something new?

Any help is much appreciated! :)
• Dec 13th 2009, 07:34 AM
Laurent
Quote:

Originally Posted by kingwinner
So every subsequence of {E(Xn)} has a further subsequence of it which converges to E(X) => E(Xn)->E(X).

This implication is true, but not obvious. Can you justify it? (or maybe it is from your lecture notes)

Quote:

2) But now I am stuck with question 2. The assumption is replaced by (Xn)^2 ≤ W, and we need to prove more: Xn->X in mean square. I tried to modify the proof in question 1, but it doesn't seem to work here. How can we modify the proof? Or do we need something new?
Nothing new is needed. Just prove that the sequence $Y_n=(X_n-X)^2$ satisfies the assumptions of question 1) (with 0 as the limit).

Hint: using the property about subsequences, you have $|X|^2\leq W$ as well.
• Dec 14th 2009, 02:40 AM
kingwinner
Quote:

Originally Posted by Laurent
This implication is true, but not obvious. Can you justify it? (or maybe it is from your lecture notes)

Yes, it's from my notes. I am not sure why it is true though.

Quote:

Nothing new is needed. Just prove that the sequence $Y_n=(X_n-X)^2$ satisfies the assumptions of question 1) (with 0 as the limit).
Hint: using the property about subsequences, you have $|X|^2\leq W$ as well.
Xn->X in probability
=>|Xn-X|->0 in probability
=>|Xn-X|^2 ->0 in probability
So the first assumption is satisfied.

But I don't know how to prove that the other assumptions are satisfied. Here is my attempt:
|(Xn-X)^2| = (Xn-X)^2 = (Xn)^2 - 2(Xn)(X) + X^2 ≤ |Xn|^2 + 2|Xn| |X| + |X|^2 ≤ 2W +2|Xn| |X|
But I have no idea how to prove E(2W +2|Xn| |X|) < ∞ :(

• Dec 14th 2009, 04:25 AM
Laurent
Quote:

Originally Posted by kingwinner
But I don't know how to prove that the other assumptions are satisfied. Here is my attempt:
|(Xn-X)^2| = (Xn-X)^2 = (Xn)^2 - 2(Xn)(X) + X^2 ≤ |Xn|^2 + 2|Xn| |X| + |X|^2 ≤ 2W +2|Xn| |X|

I told you you can prove $|X|^2\leq W$, so that $|X_n||X|\leq \sqrt{W}\sqrt{W}=W$. So you just have to prove what I claimed.
• Dec 14th 2009, 07:53 AM
kingwinner
Quote:

Originally Posted by Laurent
I told you you can prove $|X|^2\leq W$, so that $|X_n||X|\leq \sqrt{W}\sqrt{W}=W$. So you just have to prove what I claimed.

OK, so E(4W)< ∞, so it remains to prove that "if Xn->X in probability and (Xn)^2 ≤ W for all n, then X^2 ≤ W."

I wrote out all the given information and definitions, and searched for properties and theorems, but I don't see the connection. Which property about subsequence do I need?

Thanks!
• Dec 14th 2009, 10:52 AM
Laurent
Quote:

Originally Posted by kingwinner
Which property about subsequence do I need?

The fact that there exists a subsequence which converges a.s.. Can you see why?