Sum and product of Martingale processes

Given two Martingale processes $(X_t)$ and $(Y_t)$, are their sum $(X_t+Y_t)$ and their product $(X_t \times Y_t)$ also Martingale?

If not, will the two $(X_t)$ and $(Y_t)$ being independent grant their sum and product Martingale? Thanks!

The product of two independent martingales is a martingale--or rather it is or it is not, depending on the precise formulation of the hypothesis! When it is, one says that the martingales are orthogonal. This is explained, for example, by Alexander Cherny in the chapter Some Particular Problems of Martingale Theory of the Shiryaev Festschrift.

And yes, the sum of two independent martingales is a martingale but, here again, it might be wise to state the result with some care and, first of all, as mentioned by steveO in a comment, to specify the filtration(s) one is considering.

The trivial version is that if $X$ and $Y$ are two martingales (independent or not) with respect to a given filtration $\mathcal{G}$, then the sum $X+Y$ is also a martingale with respect to $\mathcal{G}$. But what happens if one assumes that $X$ is a martingale with respect to its own filtration $\mathcal{F}^X$ and that $Y$ is a martingale with respect to its own filtration $\mathcal{F}^Y$?

(Recall that the filtration $\mathcal{F}^Z$ of a process $Z$ is defined by $\mathcal{F}^Z_n=\sigma(\{Z_k;k\le n\})$ for every $n$.)

Then, if $X$ and $Y$ are independent, $X+Y$ is a martingale with respect to its own filtration $\mathcal{F}^{X+Y}$ but, first, the proof, while not terribly difficult, requires to be careful (and uses more than the linearity of conditional expectations), and, second, for non independent martingales, this becomes horribly wrong.

To get an idea of the problem, consider a given integrable random variable $\xi$ and two $\sigma$-algebras $\mathcal{A}$ and $\mathcal{B}$ and try to find conditions guaranteeing that $E(\xi|\mathcal{A}\vee\mathcal{B})=E(\xi|\mathcal{A})$. Is $\xi$ independent of $\mathcal{B}$ enough? No, one has to assume that $\mathcal{B}$ is independent of $\sigma(X)\vee\mathcal{A}$.