Is $Z_n = X_n Y_n$ a martingale? Suppose $X_n$ and $Y_n$ are independent martingales with respect to filtration $\mathcal{F}_n$. Is $Z_n$ a martingale with respect to the same filtration, where $Z_n = X_n Y_n$ and we know that $\mathbb{E}\left[X_n^2\right] < \infty,\mathbb{E}\left[Y_n^2\right] < \infty$?
Well I did the first step and found that
$E|Z_n| = E|X_nY_n| = E\left[|X_n||Y_n|\right] \leq \sqrt{E|X_n|^2} \sqrt{E|Y_n|^2} < \infty$
I have a problem with the second part
$E\left[Z_n|\mathcal{F}_s\right] = E\left[X_nY_n|\mathcal{F}_s\right] = ? 
$
I wanted to do something like
$E\left[X_nY_n|\mathcal{F}_s\right] = E\left[X_n|\mathcal{F}_s\right]E\left[Y_n|\mathcal{F}_s\right] = X_sY_s = Z_s$,
but I'm not sure if I can do it only knowing that they are independent.
 A: I was somewhat surprised that the answer is negative.
(However, see the answer by John Dawkins, who notes that $Z_n$ is a Martingale with respect to   some filtration.)
Let $\{\xi_n\}$ and $\{\eta_n\}$ be i.i.d. variables of mean zero, taking values $\pm 1$. Consider the partial sums $X_n:=\sum_{k=1}^n \xi_k$ and
$Y_n:=\sum_{k=1}^n \eta_k$. Write $\gamma_j=\xi_j \eta_j$,  and observe that $\gamma_j$ is independent of $\xi_j$. Denote by $\mathcal F_n$ the $\sigma$-field generated by the $2n+1$ variables $\{\xi_j\}_{j=1}^{n} , 
\{\eta_j\}_{j=1}^{n}$ and $\gamma_{n+1}$.
Then
$$E[X_{n+1} -X_n |\mathcal F_n]=E[\xi_{n+1} |\mathcal F_n]=E[\xi_{n+1} |\gamma_{n+1}]=0 \,,$$
so  $\{X_n\}$ is a $\{\mathcal F_n\}$-martingale. Similarly,  $\{Y_n\}$ is a $\{\mathcal F_n\}$-martingale.
However, the products $Z_n=X_n Y_n$ satisfy
$$E[Z_{n+1} -Z_n |\mathcal F_n]
=E[\xi_{n+1}Y_n+\eta_{n+1}X_n+ \gamma_{n+1} |\mathcal F_n]=0+0+\gamma_{n+1} \,,$$
so  $\{Z_n\}$ is not a $\{\mathcal F_n\}$-martingale.
A: Partial answer: $(X_n)$ is a martingale with respect to its natural filtration $\mathcal F_n^X:=\sigma\{X_k: k\le n\}$, and likewise for  $(Y_n)$. Let $\mathcal G_n:=\mathcal F^X_n\vee\mathcal F^Y_n$. Using the independence of $(X_n)$ and $(Y_n)$ its eeasy to check that $(Z_n)$ is a martingale with respect to $\mathcal G_n$. [N.B. Because of the independence, and Fubini, the integrability of each $Z_n$ follows from that of $X_n$ and $Y_n$ — no square integrability is needed.] Of course $\mathcal G_n\subset\mathcal F_n$.
