Re: decomposition of r.v.

Hello,

I don't quite understand your question... You're quoting decompositions of martingales, but is X,Y or Z a martingale ?

If not, your question is trivial, since a F-measurable rv will be G-measurable : consider Z=X and Y= 0.

Re: decomposition of r.v.

$\displaystyle X,Y,Z$ aren't martingales ($\displaystyle \mathcal{F}$ and $\displaystyle \mathcal{G}$ are no filtrations).

Sorry, I've forgotten to mention, that $\displaystyle Y,Z$ are not measurable with respect to $\displaystyle \mathcal{F}$.

Thank you.

Re: decomposition of r.v.

Okay, here is a guess (I didn't entirely check if it works)

Consider $\displaystyle \mathcal H$ to be the set of all elements belonging to $\displaystyle \mathcal G$ and not belonging to $\displaystyle \mathcal F$. It can be proved that $\displaystyle \mathcal H$ is a sigma-algebra. Hence we can consider the conditional expectation of X with respect $\displaystyle \mathcal H$.

So if you take $\displaystyle Y=E[X|\mathcal H]$ and $\displaystyle Z=X-E[X|\mathcal H]$, it should work.