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!


  • $\begingroup$ Since $X_i$ and $E(X_i\mid \mathcal F_{i-1})$ are $L^2$, $M_n$ is a $L^2$ martingale. In fact, $M_{\infty}$ is the sum for $\left[M,M\right]_n$ where $\left[M,M\right]_n$ is the bracket of $M$ (such that $\{M_n^2-\left[M,M\right]_n\}$ is a $(\mathcal F_n)$-martingale). $\endgroup$ – Davide Giraudo Dec 10 '11 at 23:53
  • $\begingroup$ Hm...ok that was not to hard. But what's about 3? I have seriously no idea how to prove this. Some help would be appreciated. $\endgroup$ – user20869 Dec 11 '11 at 8:21
  • $\begingroup$ 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.? $\endgroup$ – user20869 Dec 14 '11 at 14:31

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

  • $\begingroup$ @ John Dawkins: Thanks for your answer, but there are some question around before accepting your answer. First question: How should I compute $E(M_n^2)$ ? This looks complicated, since I square the whole sum, and why is $ E(M_n^2) \le M_\infty$ for all $n$. Second question: Why is $ E((X_i-E(X_i|\mathcal{F}_{i-1}))^2|\mathcal{F}_{i-1})=E((X_i-E(X_i|\mathcal{F}_{i-1}))^2)= E(X_i^2)-E(E(X_i|\mathcal{F}_{i-1})^2)$ ? $\endgroup$ – user20869 Jan 4 '12 at 17:53
  • $\begingroup$ @hulik, concerning 3: Unfortunately, I misread your definition of $M_\infty$, mentally adding an extra expectation in front. It is true that $E[M_n^2]\le E[M_\infty]$, so if $E[M_\infty]$ is finite, then $\{M_n\}$ is $L^2$-bounded, as claimed. Under your condition that $M_\infty<\infty$ (almost surely), the proof that $M_n$ converges is more involved; the only argument that comes to mind requires a supplementary condition such as $|X_i-E(X_i|{\mathcal F}_{i-1})|\le Z$ for all $i$, and some square-integrable r.v. $Z$. $\endgroup$ – John Dawkins Jan 4 '12 at 21:42
  • $\begingroup$ @hulik, concerning 4: What is true here is that $E[M_\infty] =\sum_{i=1}^\infty E[(X_i-E[X_i|{\mathcal F}_{i-1}])^2]$, and the $i$th term of this sum is equal to $E[X_i^2] -2E[X_iE[X_i|{\mathcal F}_{i-1}]]+E[(E[X_i|{\mathcal F}_{i-1}])^2] =E[X_i^2] -E[(E[X_i|{\mathcal F}_{i-1}])^2]$, by properties of conditional expectation. Since this latter expectation is at most $E[X_i^2]$, the expectation of $M_\infty$ is at most $\sum_{i=1}^\infty E[X_i^2]$. If this sum is finite, then $E[M_\infty]<\infty$, and consequently $M_\infty<\infty$ almost surely. $\endgroup$ – John Dawkins Jan 4 '12 at 21:54
  • $\begingroup$ @ John Dawkins: Thanks a lot for your patience, but there still some arguements, which I do not understand: 1. In your second last comment do you mean $ E(M_n^2) \le M_\infty $ instead of $E(M_n^2) \le E(M_\infty)$? It would be appreciated a lot, if you would update your answer and show how you get this. 2. Unfortunately there's no assumption as $|X_i-E(X_i|\mathcal{F}_{i-1})|\le Z$. $\endgroup$ – user20869 Jan 5 '12 at 8:15
  • $\begingroup$ 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. $\endgroup$ – user20869 Jan 5 '12 at 8:26

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.



  • $\begingroup$ @ 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. $\endgroup$ – user20869 Jan 11 '12 at 8:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy