# Conditional expectation of this stochastic process?

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.

• 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.
– whuber
Dec 30, 2021 at 15:42
• @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. Dec 30, 2021 at 15:45
• 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.
– whuber
Dec 30, 2021 at 15:52
• 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.
– whuber
Dec 30, 2021 at 16:20
• 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.
– whuber
Dec 30, 2021 at 16:37