Properties about a certain martingale

Question

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

Answer 1

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

Answer 2

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