# decomposition of r.v.

• Nov 5th 2011, 12:04 AM
Juju
decomposition of r.v.
Hallo everybody,

does anyone know, if there exists a decomposition of the random variable $\displaystyle X$ such that
$\displaystyle X=Z+Y$
where
$\displaystyle X$ is $\displaystyle \mathcal{F}-$measurable
$\displaystyle Z, Y$ are $\displaystyle \mathcal{G}-$measurable
$\displaystyle \mathcal{G}$ and $\displaystyle \mathcal{F}$ are two $\displaystyle \sigma-$algebras with $\displaystyle \mathcal{F} \subset \mathcal{G}$

I just know the decomposition of stochastic processes (continuous in time), such that a (local) martingale remains a (local) martingale in the enlarged filtration (initial enlargement of filtration). The Doob-Meyer Decomposition doesn't work here either.

• Nov 5th 2011, 02:17 AM
Moo
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.
• Nov 5th 2011, 11:41 AM
Juju
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.
• Nov 5th 2011, 03:48 PM
Moo
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.