I have been thinking about this problem and have seen the answer but did not really understand the reasoning behind it. So im hoping someone could be so kindly to explain in detail what i am missing.
First the problem:
Let p$\in$(0,1), q=1-p, and define the following sequence of random variables:
$X_{n+1}= \begin{cases} p+qX_n & w.p.\,\, X_n\\ qX_n & w.p.\,\, 1-X_n \\ \end{cases}$
Show that this sequence is a martingale (with respect to itself).
Im only interested in one of the two conditions, namely $\mathbb{E}(X_{n+1}|\mathcal{F_n})=X_n$.
The solution just claimed that one would evaluate:
$\mathbb{E}(X_{n+1}|\mathcal{F_n})=(X_n)(p+qX_n)+(1-X_n)(qX_n)$ and that this sum would be $X_n$. That the sum is $X_n$ is fine but the first equality is not clear to me.
I have tried to evaluate this through the definition of conditional expectation w.r.t a sigma field but i do not get that $\mathbb{E}(X_{n+1}|\mathcal{F_n})=X_n$ almost surely.
So please if possible clarify the above part via the definition of conditional expectation or in a way that gives more insight if possible.
I have been trying to accomplish this "fact" with $\mathbb{E}(X_{2}|\mathcal{F_1})=X_1$ a.s. and would be really glad if someone could do it with sets (that is little omegas) so i could compare were my calculations are wrong.
Thanks for any help, very much appreciated!
I think you're overthinking this, especially when the problem doesn't appear to define the probability triplet explicitly. As far as $X_{n+1}$ is concerned, the only relevant bit of info from $\mathcal{F}_n$ is the value of $X_n$. Thus, \begin{aligned} \mathbb{E}(X_{n+1}|\mathcal{F}_n)&=\mathbb{E}(X_{n+1}|X_n)\\ &=\Pr(X_{n+1}=p+qX_n|X_n)\times(p+qX_n)+\Pr(X_{n+1}=qX_n|X_n)\times qX_n\\ &=X_n\times(p+qX_n)+(1-X_n)\times qX_n. \end{aligned}