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.
P.S.- I also posted this on Cross Validated. I hope I am not violating any community guidlines by doing this.

When you define the $W_n$ stochastic process you're not just defining it on the space $\{H,T\}$, because that space would only contain information on one coin toss. You define it on the space of all possible outcomes of all coin tosses. Your space is $\Omega = \{\omega = (\omega_1, \omega_2, \dots) | \;\; \omega_i \in \{H, T\}\}$. Therefore $\sigma(W_1, \dots, W_n)$ is not the entire $\sigma$-algebra.
