Consider an uncountably infinite space, an infinite coin-tossing.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be the probability space. If a set $A\in\mathcal{F}$ satisfies $\mathbb{P(A)=1},$ then we say that event A occurs almost surely.

My Question:

  1. Why is it the case every individual coin-toss sequence has probability zero in this uncountable probability space? In Shreve's explanation, he defines two sets: $$A_H=\{\omega\in\Omega_\infty:\omega_1=H\}\\A_T=\{\omega\in\Omega_\infty:\omega_1=T\}.$$

He sets $\mathbb{P}(A_H)=p,\mathbb{P}(A_T)=1-p=q.$ Hence, instead of assigning probability measure to any single sequence, he looks at a set of sequences where it kicks off with either H or T. Why does this set-up makes sense compared to how we deal with coin-tossing in a finite sample space. Usually, when we deal with two coins, we are given an assumption that the coin is fair, so $P(H)=\frac{1}{2}$. However, in the uncountably infinite coin-tossing experiment, any single infinity sequence gets measure $0$ while the set of sequences with certain characteristics (e.g. first flip $H$) gets a strictly positive probability.

  1. What exactly does $\mathbb{P}(A)=1$ almost surely mean? According to Shreve:

"Whenever an event is said to be almost sure, we mean it has probability one, even though it may not include every possible outcome. The outcome or set of outocme not included, taken all together, has probability zero."

Does this mean the event that a coin toss gets at least one tail will happen with certainty? Conversely, when we say $\mathbb{P}(\{\omega\in\Omega_\infty:(\omega_i)_i=H\space\forall\space i\in\mathbb{N}\}=0,$ does this mean something with probability zero CANNOT happen, so it is impossible to obtain an infinite coin-toss sequence with all Heads?

Shreve, Steven E. $\textit{Stochastic Calculus for Finance II : Continuous-Time Models}$. Springer, 2008.


1 Answer 1

  1. Imagine a finite sequence of 1 billion billion billion coin flips instead, and you want the query the probability of 1 specific sequence. You will get 1/billion billion billion, approximately 0. The idea is similar here.

  2. Almost sure means P(A) = 1. Recall $P(\Omega)=1.$ A is not necessarily $\Omega$. All almost sure means the set of omegas that don't satisfy this event is "insignificant" or negligible. That is P(All the other little omegas) = 0. I made up an example just now, but I'm not 100% this delivers the intended idea. I suspect it'll be better than nothing. Let $\Omega=\{Head, Tail, Edge\}$. The event {H,T} occurs almost surely. That is P({H,T})=1. P({Edge})=0.


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