0
$\begingroup$

I am trying to solve the exercise at the end of this page, the framework is that of measure theory where we are tossing a coin infinitely often so we are working with a probability triple $( \Omega, F, P)$, where $w$ is a point of $\Omega$, $\Omega =[H,T]^N, \qquad w =(w_1, w_2, \dots ), \quad w_n \in \{H,T \},$ and

$$F = \sigma(\{ w \in \Omega : w_n = W \}: n \in N, W \in \{H,T\})$$

The formal definition (2.3,b) that is cited in the text below is

$$F^* = \left \{ w : \frac{\text{number of } (k \le n : w_k = H)}{n} \rightarrow \frac{1}{2} \right \} $$

enter image description here

I think that for any $w$ an $\alpha$ s.t. $w \notin F_\alpha$ is $\alpha(k) = 1$, correct?

Moreover I am missing the importance of this discussion, why is the fact that the intersection is the empty set important?

$\endgroup$
  • 2
    $\begingroup$ No, you need $\alpha(1)<\alpha(2)<\dots$. As to why it matters, I bet the discussion in the book continues after the exercise. $\endgroup$ – David C. Ullrich Jul 13 '15 at 20:29
  • 1
    $\begingroup$ Could you please let us know the source of your excerpt? $\endgroup$ – zoli Jul 13 '15 at 21:09
  • 1
    $\begingroup$ Why $\alpha(1)<\alpha(2)<\dots$? Two reasons: (i) It says so right there in the statement of the problem! (ii) The idea is that $(\omega_{\alpha(k)})$ is supposed to be a subsequence of $(\omega_1,\omega_2,\dots)$. That's part of the definition of "subsequence". $\endgroup$ – David C. Ullrich Jul 13 '15 at 22:32
  • 1
    $\begingroup$ Regarding why it matters, I posted an "answer" just now. Read it quick before it gets nuked for imprecision... $\endgroup$ – David C. Ullrich Jul 13 '15 at 22:45
  • 1
    $\begingroup$ "It matters" because it shows, on an explicit (and, supposedly, easy to grasp) example, that the statement that, for every alpha, the property F_alpha holds almost surely cannot be replaced by the statement that the property F_alpha holds for every alpha, almost surely. In turn, as Williams explains, this example shows the need for rigor in the treatment of "events that they never happen", which the framework of measure theory provides. $\endgroup$ – Did Jul 14 '15 at 9:26
1
$\begingroup$

This is going to be a little fuzzy. It's a fuzzy question on a fuzzy topic... the whole point is why axiomatic measure-theoretic probability theory was needed to unfuzzify things.

A person might attempt to define probability in terms of those limits: By "definition", saying that $P(H)=1/2$ means that that limit is $1/2$.

Then when someone else points out that the sequence $HHH\dots$ is possible the reply is that that's not a "random" sequence. Hard to argue with that since we haven't defined random...

So. We've attempted to define $P=1/2$ by saying that it means that we get asymptotically half heads in any "random" sequence of tosses. But now wait, we have a problem. If we make a "random" sequence of tosses then the results from the odd-numbered tosses should certainly still count as "random", so half the odd-numbered tosses must also be heads. And the same for any subsequence of tosses.

The exercise shows this can't work, by showing that it simply can't happen that half of any subsequence of tosses is heads. No matter what sequence we toss, there's some subsequence that violates this. So our definition leads to a contradiction. Bad.

$\endgroup$
  • $\begingroup$ Wow thanks this was very helpful, now I still need to find the $\alpha$ that proves that the intersection of the exercise is empty, could I have some help? And why is "almost surely" a better way of doing things? $\endgroup$ – Monolite Jul 13 '15 at 22:47
  • 1
    $\begingroup$ Given $\omega$ you need to find a subsequence which does not have half heads. Hmm. A subsequence consisting of nothing but heads would do. Or a subsequence with no heads. "Almost surely" is better because we know the definition $\endgroup$ – David C. Ullrich Jul 13 '15 at 22:51
  • $\begingroup$ Could I choose $\alpha(k) = m$ s.t. $ w_m = H$? is this allowed? $\endgroup$ – Monolite Jul 13 '15 at 23:18
  • 1
    $\begingroup$ I'm not sure what you mean. What you wrote looks like $\alpha(1)=\alpha(2)$ again. What I suspect you meant is this: Let $\alpha(1)$ be the smallest $m$ with $\omega_m=H$. Let $\alpha_2$ be the second-smallest $m$ with $\omega_m=H$, etc. That works fine, if infinitely many of the $\omega_j=H$. If only finitely many $\omega_j=H$ then... $\endgroup$ – David C. Ullrich Jul 13 '15 at 23:36
  • 1
    $\begingroup$ @Monolite The axiom of choice to guarantee the existence of a minimum of every nonempty subset of N... Sure about that? $\endgroup$ – Did Jul 14 '15 at 9:28

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.