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I asked this question here. Unfortunately there was not a satisfying answer. So I hope here is someone who could help me.

I'm solving some exercises and I have a question about this one:

Let $(X_i)$ be a sequence of random variables in $ L^2 $ and a filtration $ (\mathcal{F}_i)$ such that $X_i$ is $\mathcal{F}_i$ measurable. Define $ M_n := \sum_{i=1}^n \left(X_i-E(X_i|\mathcal{F}_{i-1})\right) $

I should show the following:

  1. $M_n $ is a martingale.
  2. $M_n $ is square integrable.
  3. $M_n $ converges a.s. to $ M^*$ if $ M_\infty := \sum_{i=1}^\infty E\left((X_i-E(X_i|\mathcal{F}_{i-1}))^2|\mathcal{F}_{i-1}\right)<\infty$ .
  4. If $\sum_{i=1}^\infty E(X_i^2) <\infty \Rightarrow 3)$

I was able to show 1 and with Davide Giraudo's comment 2. is clear too. But I got stuck at 3. and 4. So I'm very thankful for any help!

hulik

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    What you mean by bracket? Do you mean quadratic variation? Why should it have this form? Looking at the formal definition on Wikipedia it looks different. Sorry I'm not very familiar with this. If it is the quadratic variation, is there a theorem, that if it's finite then it converge a.s.?2011-12-14

2 Answers 2

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For 3, compute $E[M_n^2]$. Having done this conclude that $E[M_n^2]\le M_\infty$ for all $n$. This means that $\{M_n\}$ is an $L^2$-bounded martingale, to which the martingale convergence theorem may be applied.

For 4, the $i$th term in the sum defining $M_\infty$ is equal to $E[(X_i-E[X_i|\mathcal{F}_{i-1}])^2]$, which in turn is equal to $E[X_i^2] -E[(E[X_i|\mathcal{F}_{i-1}])^2]\le E[X_i^2]$.

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    and why is $ E(X_i)^2 -2E(X_iE(X_i|\mathcal{F}_{i-1})) + E((E(X_i|\mathcal{F}_{i-1}))^2) = E(X_i^2)-E((E(X_i|\mathcal{F}_{i-1}))^2) $? I would agree if $E(X_i|\mathcal{F}_{i-1})$ is constant, otherwise I do not see why this sould be true.2012-01-05
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I will use the following reference: Klenke

Unfortunately the book is in German, nevertheless here is my answer, based on it:

On p. 208 Klenke shows ("Satz 10.4) that for a square integrable martingale $\{X\}$ the so called quadratic variation process is given by

$ \langle X \rangle_n = \sum_{i=1}^n E((X_i-X_{i-1})^2|\mathcal{F}_{i-1}).$

Since you have already proved, that $ M_n$ is a square integrable martingale, we find:

$ \langle M \rangle_n = \sum_{i=1}^n E((M_i-M_{i-1})^2|\mathcal{F}_{i-1})$

since $M_n$ is defined through a sum, this is equal

$\langle M \rangle_n = \sum_{i=1}^n E((X_i-E(X_i|\mathcal{F}_{i-1}))^2|\mathcal{F}_{i-1}).$

So far nothing happens, but it's important for the things, which follow:

The main Theorem can be found on page 225, "Korollar 11.11", I cite (and translate):

If $ X = \{X_n\}$ is a square integrable martingale with quadratic variation process $\langle X \rangle $ the following are equivalent:

  1. $\sup_n E(X_n^2) < \infty $
  2. $\lim_n E(\langle X\rangle_n) < \infty$
  3. $X$converges in $ L^2$.
  4. $X$ converges in $ L^2$ and almost surely.

What we will use is the equivalence of $2.\iff 4.$ First let me point out, that I do not see how to prove your 3. without knowing that $ M_\infty \le c < \infty$. Though it's not hard to prove your 4.

Assume that $ \sum_{i=1}^\infty E(X_i^2) < \infty $ then we want to show, that $M_n$ converges a.s.

As mentioned, I will use the equivalence of $2.\iff 4.$: $\lim E(\langle M\rangle_n) = E(\langle M\rangle_\infty)$ using monotone convergence once more,

$E(\langle M\rangle_\infty)=\sum_{i=1}^\infty E(E((X_i-E(X_i|\mathcal{F}_{i-1}))^2|\mathcal{F}_{i-1}))$

by basic properties of conditional expectation:

$\sum_{i=1}^\infty E(E((X_i-E(X_i|\mathcal{F}_{i-1}))^2|\mathcal{F}_{i-1})) =\sum_{i=1}^\infty E(E((X_i-E(X_i|\mathcal{F}_{i-1}))^2).$

Now observe:

$0\le(a-b)^2=a^2-2ab+b^2=|a^2-2ab+b^2|\le a^2+2|ab|+b^2$

and by basic analysis, $ 2|ab| \le a^2 +b^2 \Rightarrow (a-b)^2\le 2a^2+2b^2$, this leads to:

$E(E((X_i-E(X_i|\mathcal{F}_{i-1}))^2)\le 2(E(X_i^2) + E(E(X_i|\mathcal{F}_{i-1})^2)).$

Now use Jensen for conditional expectation $ E(E(X_i|\mathcal{F}_{i-1})^2) \le E(E(X_i^2|\mathcal{F}_{i-1}))= E(X_i^2)$, hence

$ \sum_{i=1}^\infty E(E((X_i-E(X_i|\mathcal{F}_{i-1}))^2)\le 4\sum_{i=1}^\infty E(X_i^2) < \infty$

by assumption, and therefore $ M_n $ converges a.e. and even more, we see that $ \langle M \rangle_\infty < \infty$ a.s.

As I said, I don't know how to prove 3. but I decided to post an answer, because maybe my calculation helps someone to prove 3. and second, it's too long for a comment.

cheers

math

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    @ math: Thank you for your answer! Now I do understand much better the answer of John Dawkins, too. I accepted your answer since 4 was quite important for me and you explanations were clear. Perhaps there's a mistake in 3. I will again check this.2012-01-11