# strong and weakly orthogonality for martingales

If I have two RCLL martingales $X,Y$, both bounded in $L^2$, hence uniformly integrable. Then we call $X,Y$ weakly orthogonal if $E[X_\infty Y_\infty]=0$ and we call $X,Y$ strongly orthogonal if $XY$ is a martingale. Now I have some question about this:

1. Why is the product $XY$ uniformly integrable?
2. If I know that $X^\tau Y^\tau$ is a martingale for every stopping time $\tau$, then $XY$ is martingale. Is this just because I can take $\tau:=\infty$ which is a stopping time, since it is constant to obtain the result?

1. Why would the product we uniformly integrable, and under which conditions? 2. It seems yes (if $X_t^{\tau}$ denotes $X_{\min\{t,\tau\}}$. – Davide Giraudo Jul 23 '12 at 17:11
@DavideGiraudo Sorry, I was not very precisely! About your comment: Yes, $X^\tau:=X_{\tau\wedge t}$. I want to show that $XY$ is a martingale. Actually I know that for every stopping time $\tau$ we have $X^\tau Y$ is a martingale. Hence I want to use the stopping theorem, which I can just apply for uniformly integrable martingales or bounded stopping times. I guess, instead of using uniformly integrability, I have to use bounded stopping times: if $\tau =+\infty$ then I already know that $XY$ is a martingale (since $X^\tau =X$)..... – math Jul 24 '12 at 9:36