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... 


Thanks again for your help!


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



a question on 0-1 valued stochastic process
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 $P(A^c) = 0$.


