3
$\begingroup$

I am supposed to provide a martingale proof of Kolmogorov's zero-one law.

Hint Let $X_n$ be independent random variables and let $\mathcal C_\infty$ be the corresponding tail $\sigma$-algebra. Let $C \in \mathcal C_\infty$ and $\mathcal F_n = \sigma(X_j; 0 \le j \le n)$.

Show that $E[1_C\mid \mathcal F_n] = P(C)$. Then show that $\lim_{n \to \infty} E[1_c \mid \mathcal F_n] = 1_C$ almost surely and deduce that $P(C) = 0$ or $1$

I have no idea how to approach this. I would like an answer but I would love an answer that provides insights about all this. I don't have any intuition about what $C_\infty$ is, and why should I follow the outlined proof, and what's the role of martingales.

EDIT 1

Let me try to provide a proof. Please point out any mistake, inaccuracy or redundancy that you can spot! I'll be most grateful.

Let us note that for every finite $N$, $\mathcal C_\infty$ is independent with $\mathcal F_N$. In fact $$\mathcal C_\infty = \bigcap_{n=1}^\infty \sigma\left(\bigcup_{m \ge n} F_m\right) = \bigcap_{n=N+1}^\infty \sigma\left(\bigcup_{m \ge n} F_m\right)$$ So $\mathcal C_\infty$ is the result of operations applied only on sigma algebras independent with $\mathcal F_N$

This implies $E[1_c \mid \mathcal F_n] = E[1_C] = P(C)$.

Let $\mathcal B_n = \sigma\left(\bigcup_{i \le n} \mathcal F_i\right)$. By the exact same reasoning, we find $E[1_c \mid \mathcal B_n] = P(C) \implies E[1_c \mid \mathcal B_n] = E[1_c \mid \mathcal F_n]$

This was necessary because now $\mathcal B_n$ is an increasing sequence of sigma algebra and I can apply Levy's zero one law: $$\lim_{n \to \infty} E[1_C \mid \mathcal F_n] = \lim_{n\to \infty} E[1_C \mid \mathcal B_n] = E[1_C \mid \mathcal B_\infty]\text{ a.s.}$$

Where $\mathcal B_\infty = \sigma\left(\bigcup_{n \ge 0} \mathcal B_n\right)$ Since $\mathcal C_\infty = \inf \mathcal B_\infty \implies \mathcal C_\infty \subset \mathcal B_\infty$ (is this correct?) the fact that $1_C$ is measurable with respect to $\mathcal C_\infty$ implies that $1_C$ is measurable with respect to $\mathcal B_\infty$, so

$$\lim_{n \to \infty} E[1_C \mid \mathcal F_n] = E[1_C \mid \mathcal B_\infty] = 1_C$$

Now $\lim_{n \to \infty} E[1_C \mid \mathcal F_n] = P(C) = 1_C$, hence $P(C) = 0$ or $1$.

Remarks

The proof seems somewhat convoluted and it is probably better done in another way. Also, I am not really sure is it correct. Moreover, I did not use directly martingale properties; I know Levy's zero one law is a consequence of the Martingale Convergence theorem, though.

EDIT 2

I deleted my answer and decided to put my attempt in the question, so I can award points to anyone who is kind enough to help me. I didn't do so previously because the wall of text may be a deterrent to read the whole question.

$\endgroup$
  • 2
    $\begingroup$ Don't you think that you get the best insights if you try to solve this on your own? What do you know about the tail-$\sigma$-algebra? What can you say about the relation of $\mathcal{F}_n$ and $C_{\infty}$ (keyword: independence)? $\endgroup$ – saz Feb 22 '15 at 7:41
  • $\begingroup$ @saz Thank you for your comment. I tried to write down a proof. I would love if you could take a look at it! :-) $\endgroup$ – Ant Feb 24 '15 at 11:58
  • $\begingroup$ Yes, there are some things in your proof which are a bit "convoluted", as you say. I'll write an answer (but not right now...) However, I'm glad that you did try to solve it on your own. $\endgroup$ – saz Feb 24 '15 at 13:07
  • $\begingroup$ @saz Thank you very much! I'm struggling a bit with this new topic! :) $\endgroup$ – Ant Feb 24 '15 at 13:25
1
$\begingroup$

First of all (and most importantly): The idea of your proof is correct. However, there are some things which can be improved:

$(\mathcal{F}_n)_{n \in \mathbb{N}}$ is a filtration, i.e. $\mathcal{F}_n$ is a $\sigma$-algebra for each $n \in \mathbb{N}$ and $\mathcal{F}_n \subseteq \mathcal{F}_{n+1}$. This follows directly from the definition of $\mathcal{F}_n$. This means in particular that you can apply Lévy's zero one law directly to the sequence

$$\mathbb{E}(1_C \mid \mathcal{F}_n);$$

there is no need for $\mathcal{B}_n$. (In fact, since the $\sigma$-algebras $\mathcal{F}_n$ are increasing, it holds that $\mathcal{B}_n = \mathcal{F}_n$.) This makes the proof much more easier.

Concerning martingale properties: You are right; Lévy's zero one law is a direct conseqeunce of the martingale convergence theorem. The martingale you are considering in this case is actually

$$X_n := \mathbb{E}(1_C \mid \mathcal{F}_{n}).$$

If it is not clear to you why this is a martingale, then try to prove it - it is a good exericse. :)

$\endgroup$
  • $\begingroup$ Thank you very much for your patience! I see why $\mathcal B_n$ were useless, I actually got confused :) Yes, $X_n$ is closed by a r.v. $\in L^1$, so it is a martingale, right?. (It is easy to prove with repeated conditional expectations). Thank you again! You've been most helpful :) $\endgroup$ – Ant Feb 24 '15 at 15:24
  • $\begingroup$ @Ant Thanks; you are welcome. (And yes, you are right about $X_n$ being a martingale.) $\endgroup$ – saz Feb 24 '15 at 15:27

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.