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.


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 $1$.


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.


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.

  • 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

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. :)

  • $\begingroup$ 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

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