# a question on 0-1 valued stochastic process

Here's a question on probability theory from a layman (I'm a game theorist). It is very likely that the question will be a straightforward matter for someone who is a probability theorist. I guess I'm missing a very standard technique that could be used to address the problem. Any ideas would be very helpful! Thank you all very much in advance.

Consider a stochastic process $X_{t}$ taking values in the set $\{0,1\}$ according to the probability measure $\mu$. Let $$Y_{t} = \mu\left(\limsup_{T \rightarrow \infty}\frac{1}{T}\sum_{t = 0}^{T - 1}X_{t} \geq \frac{1}{2}\mid X_{0},\dots,X_{t - 1}\right).$$ We know that $$\mu(X_{t} = 1 \mid Y_{t} < \epsilon) > 1 - \epsilon.$$ We believe (but unable to prove) that $$Y_{t} = 1\text{ for every }t,\,\mu\text{-almost surely.}$$

This might have to do with Levy's zero-one law...

P.S. My apologies for an earlier post that did not meet the standards of the forum.

• Why all the votes to close? Maybe I am missing something, since I am not a probabilist: but I don't see why the problem is obvious or off-topic Apr 7, 2015 at 20:16
• Apr 9, 2015 at 1:34
• Arkadi, you may be aware of the threads linked to in my previous comment. Sometimes the phrase "show that" makes a question look like a homework problem, which we don't want (and also MO users don't like imperative forms of address). When you have a chance, please read through the material in the help center on how to ask questions that will be well-received, as people here can be picky (I include myself). Apr 9, 2015 at 10:56
• What is the quantifier corresponding to $\epsilon$?
– R W
Apr 9, 2015 at 21:41

Let $A$ be the event $A = \left\{ \limsup_{T \to \infty}\frac{1}{T}\sum_{t = 0}^{T - 1}X_{t} \ge \frac{1}{2}\right\}$, so that $Y_t = P(A \mid X_1, \dots, X_t)$. We will show that $P(A) = 1$ and hence $Y_t = 1$ almost surely for all $t$.
First, Lévy's zero-one law asserts that $Y_t \to 1_A$ almost surely. Now assume there exists $\epsilon < 1/2$ such that for all $t$, we have $P(X_t = 1 \mid Y_t < \epsilon) > 1-\epsilon$ for all $t$ (I'm guessing that's what you want the quantifiers to be). We can rewrite this as $$P(X_t = 1, Y_t < \epsilon) > (1-\epsilon) P(Y_t < \epsilon)$$ or equivalently, letting $Z_t = 1_{\{Y_t < \epsilon\}}$, $$E[X_t Z_t] > (1-\epsilon) E[Z_t].$$ Since $Y_t \to 1_A$ we have $Z_t \to 1_{A^c}$ almost surely, so by dominated convergence the right side converges to $(1-\epsilon)P(A^c)$. On the left side, let us add and subtract $E[X_t 1_{A^c}]$: $$E[X_t(Z_t - 1_{A^c})] + E[X_t 1_{A^c}] > (1-\epsilon)E[Z_t]$$ Since $Z_t \to 1_{A^c}$ almost surely, and $|X_t| \le 1$, the first term on the left goes to 0. Now let us average over $T$: $$\frac{1}{T} \sum_{t=1}^T E[X_t(Z_t - 1_{A^c})] + E\left[\frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right] > (1-\epsilon) \frac{1}{T} \sum_{t=1}^T E[Z_t].$$ The first term on the left side goes to 0 and the right side goes to $(1-\epsilon) P(A^c)$, so we have $$(1-\epsilon) P(A^c) \le \limsup_{T \to \infty} E\left[\frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right].$$ The quantity inside the expectation is bounded above by 1, so Fatou's lemma gives $$\limsup_{T \to \infty} E\left[\frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right] \le E\left[\limsup_{T \to \infty} \frac{1}{T} \sum_{t=1}^T X_t 1_{A^c}\right].$$ But by definition of $A$, on the event $A^c$ we have $\limsup_{T \to \infty} \frac{1}{T} \sum_{t=1}^T X_t \le \frac{1}{2}$. Therefore $$(1-\epsilon) P(A^c) \le \frac{1}{2} P(A^c).$$ Since $\epsilon < 1/2$ this implies $P(A^c) = 0$.