10
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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 $\endgroup$ – Yemon Choi Apr 7 '15 at 20:16
  • 2
  • 2
    $\begingroup$ 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). $\endgroup$ – Todd Trimble Apr 9 '15 at 10:56
  • $\begingroup$ What is the quantifier corresponding to $\epsilon$? $\endgroup$ – R W Apr 9 '15 at 21:41
6
$\begingroup$

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 $P(A^c) = 0$.

$\endgroup$
  • $\begingroup$ Dear Nate, thank you so much for your prompt response and your excellent solution! We are really happy. This is an important step in our construction. We were stuck for quite a while. $\endgroup$ – Arkadi Predtetchinski Apr 10 '15 at 11:47
  • $\begingroup$ @ArkadiPredtetchinski: You're welcome! You can mark this answer as "accepted" by clicking the check mark to the left. $\endgroup$ – Nate Eldredge Apr 10 '15 at 18:21

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.