Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Infinite coin flips, etc.

  1. Feb 13, 2009 #1
    Hi, I'm new here. I'm trying to teach myself measure theory and probability and recently wanted to find an example of a probability space (X, A, P) where X is uncountable and the sigma-algebra A is the entire power set of X. Here was my idea: let X be the set of all strings c1c2c3... for [tex]c_{i} \in \{ 0, 1\}[/tex], thought of as an infinite sequence of coin flips (with 1 corresponding to heads and 0 to tails), and A = X's power set. For x in X, define the n-restriction xn of x as the string of the first n digits of x. Furthermore, for S in A, define Sn := {xn : x is in S}. Finally, define [tex]P (S) := lim_{n\rightarrow\infty} card(S_{n})/2^{n}[/tex], i.e., the probability of S is the long-term proportion of the initial segments of the members of S to all possible initial segments.

    I wanted to prove that (X, A, P) is a probability space. I thought I found a (rather torturous) proof of this claim, but got some initial results that led me to think my proof must have gone wrong somewhere. But before I type out my whole line of reasoning, does anyone already know for sure if the above triplet satisfies the criteria of a probability space? Thanks in advance!
    Last edited: Feb 13, 2009
  2. jcsd
  3. Feb 13, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    The construction of the Vitali set shows that there's no measure m defined on all subsets of R satisfying sigma-additivity, m(T(S)) = m(S) for any translation T, and m([0,1]) = 1. I think you can mimic that construction for your set to show that it too won't work.

    Regard X as functions from N to {0,1}. Define a relation ~ on X by:

    x~y iff x and y disagree on a finite subset

    It's clear that ~ is an equivalence relation. Let [x] denote the equivalence class of x. Invoking the Axiom of Choice, there exists a set R consisting of exactly one representative from each equivalence class. If S is a finite subset of N, x in X, define xS as follows:

    xS(n) = x(n) if n is not in S
    xS(n) = 1 - x(n) if n is in S

    Define RS = {xS : x in R}

    It's clear that if S and T are distinct finite subsets of N, then RS and RT are disjoint, and moreover

    [tex]\bigcup _{S \in [\mathbb{N}]^{<\omega}}R_S = X[/tex]

    where [itex][\mathbb{N}]^{<\omega}[/itex] is the set of finite subsets of N. The union above is a countable union since there are countably many finite subsets of N, and it's a union of disjoint sets. So

    [tex]1 = P(X) = P\left (\bigcup _{S \in [\mathbb{N}]^{<\omega}}R_S\right ) = \sum _{S \in [\mathbb{N}]^{<\omega}} P(R_S)[/tex]

    It's not hard to see that P(RS) = P(R) for all finite S. So the right hand side is P(R) added to itself countably many times. If P(R) is 0, then the right side is 0, contradicting the fact that the left side is 1. If P(R) is non-zero, then the right side is infinity, contradicting the fact that the left side is 1.

    So P isn't defined at R.
  4. Feb 13, 2009 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Aha, here's the problem.

    The key feature of this probability measure is that it's insensitive to what happens 'at infinity'. So, let's do something weird there.

    Let S be the set of all eventually constant sequences.

    What is P(S)? P(X-S)?
  5. Feb 14, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    Actually, the problem you're trying to solve, mag487 is not remotely easy. Look up measurable cardinals. I'm just starting to understand this area of set theory, but the existence of "a probability space (X, A, P) where X is uncountable and the sigma-algebra A is the entire power set of X" is something that's either:

    i) not known to be provable from ZFC and not known to be disprovable from ZFC,
    ii) known to be neither provable nor disprovable from ZFC, or
    iii) something like the above two possibilities.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook