Let $(\Omega,\mathcal F,P)$ be a probability space and $(Y_n)_{n\in\mathbb N}$ be independent random variables that are positive valued and with mean equal to $1$. Let $(X_n)_{n\in\mathbb > N}$ be the stochastic process $$X_n =\prod_{k=1}^n Y_k$$ Show that it is a martingale with respect to the filtration $(\mathcal F_n)_{n\in\mathbb N}$generated by $(Y_n)_{n\in\mathbb N}$
My attempt:
I need to show that $$\int_F X_{n+1} dP=\int_F X_n dP$$ for any $F\in\mathcal F_n$.
I started by writing down the definition of $X_{n+1}$.
$$\int_F \prod_{k=1}^{n+1} Y_k dP=\prod_{k=1}^{n+1}\int 1_F Y_k dP$$ (this first equality holds because of the independence of the $Y$'s.) $$=\bigg(\prod_{k=1}^{n}\int 1_F Y_k dP\bigg)\cdot\int 1_F Y_{n+1} dP$$
Now $\bigg(\prod_{k=1}^{n}\int 1_F Y_k dP\bigg)$ equals by definition $\int_F X_n dP$.
Now since the $Y$'s are independent, and the filtration is the natural filtration,we have that $\mathcal F_n$ and $Y_{n+1}$ are independent, and hence $1_F$ and $Y_{n+1}$ are independent (remember that $F$ is any $\mathcal F_n$-measurable set). But then I obtain
$$\int_F X_{n+1} dP=\bigg(\int_F X_n dP\bigg)\cdot \bigg(\int 1_F dP\bigg)\cdot\bigg(\int Y_{n+1}dP\bigg)$$
$$\int_F X_{n+1} dP=\bigg(\int_F X_n dP\bigg)\cdot P(F)$$ and hence we obtain a supermartingale instead.
Nevertheless I've seen another solution that uses a different approach:
$$E[X_{n+1}|\mathcal F_n]=E[X_{n}Y_{n+1}|\mathcal F_n]=X_nE[Y_{n+1}|\mathcal F_n]=X_nE[Y_{n+1}=X_n$$ By independence, and the fact that $Y_{n+1}$ is independent from the sigma algebra $\mathcal F_n$.
The approach seems quite similar, but I don't really see how to put them both together. Is there any mistake on my approach? Thanks in advance.
$\int (I_F X_n) Y_{N+1} dP=(\int X_n I_F dP) (\int Y_{n+1}dP)$ because $Y_{n+1}$ and $X_n I_F$ are independent. In your argument you are aassuming independence of the triple $X_n, I_F, Y_{n+1}$. You cannot separate $P(F)$ because $X_n$ and $I_F$ are not indepedent.