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I'm just beginning to learn about stochastic processes and encountered this very elementary problem that confused me a bit:

We toss a coin that lands on Head with probability $p$ and Tail with $q=1-p$. This probability never changes. We define $W_n$ to be $-1$ when the $n$-th toss is Tail and $1$ if Head. The problem was to determine whether or not $(W_n)_n$ is a martingale. While solving, the lecturer writes this: $$\mathbb{E}(W_{n+1}\vert\mathcal{F}_n)=\mathbb{E}(W_{n+1})=p-q$$ Here $\mathcal{F}_n=\sigma(W_1,\dots,W_n)$. He argues that since the $n$-th toss and the $n+1$-th toss are independent, $W_{n+1}$ and $\mathcal{F}_n$ are independent, which is why the above holds. I get this.

But I also think that $W_n$'s are defined on the space $(\{H,T\},\{\emptyset,\{H\},\{T\},\{H,T\}\},\mathbb{P})$ where $\mathbb{P}(H)=p$ and $\mathbb{P}(T)=q$. So $\sigma(W_1,\dots,W_n)$ is the entire $\sigma$-algebra $\{\emptyset,\{H\},\{T\},\{H,T\}\}$. But then $$\mathbb{E}(W_{n+1}\vert\mathcal{F}_n)=W_{n+1}$$ Could someone please explain to me what I'm doing wrong? This will really help me learn about the topic.

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    $\begingroup$ The entire martingale is defined on the space of all countable sequences of possible outcomes. You cannot make sense of, say, $\mathcal{F}_2$ otherwise. $\endgroup$
    – whuber
    Dec 30, 2021 at 15:42
  • $\begingroup$ @whuber Thank you for the response. I am not sure I understand what you mean. For instance, I think the conclusion to the problem is that $(W_n)$ is NOT a martingale. $\endgroup$
    – Not Euler
    Dec 30, 2021 at 15:45
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    $\begingroup$ Sorry, I meant "stochastic process" rather than "martingale." What I am inviting you to consider is how you could possibly even define both $W_1$ and $W_2$ on the tiny space you have described in a way that makes them independent. $\endgroup$
    – whuber
    Dec 30, 2021 at 15:52
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    $\begingroup$ The space can be some general, abstract one. But a standard model--the smallest possible--is the set $\Omega=\{(\omega_1,\omega_2,\ldots)\mid \omega_i \in \{H,T\}\}.$ The problem is that if you have one copy of your "tiny space" and a separate, distinct copy, you still haven't any way to compare a random variable defined on one with a random variable defined on the other. You need a space on which all the random variables are simultaneously defined. See stats.stackexchange.com/a/123754/919 for details of how this is done for just $W_1$ and $W_2$ alone. $\endgroup$
    – whuber
    Dec 30, 2021 at 16:20
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    $\begingroup$ You define the probability measure on a basis of the measurable sets. That basis naturally corresponds to all finite prefixes of sequences. See stats.stackexchange.com/a/164995/919 for the details. $\endgroup$
    – whuber
    Dec 30, 2021 at 16:37

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