Radon Nikodym computation, step of a proof

**mabruka** · March 4th 2010, 12:31 PM

Let $(\Omega,\mathcal F)$ be a measureable space (in fact Polish, so we can guarantee the existence of regular conditional probabilities), $\lambda, \mu$ two probability measures on $\mathcal F$ and $\mathcal F_1 \subset \mathcal F_2 \subset \mathcal F$ two sub $\sigma$ -algebras.

Let $\lambda_\omega$ and $\mu_\omega$ be the regular conditional probabilities of $\lambda$ and $\mu$ given $\mathcal F_1$ .

If $\mu <<\lambda$ on $\mathcal F_2$ and $\mu_\omega << \lambda_\omega$ $\lambda$ -almost everywhere

then

$\frac{d\mu}{d\lambda} _{\mathcal F_2}(\omega)= \frac{d\mu}{d\lambda} _{\mathcal F_1}(\omega) \frac{d\mu_\omega}{d\lambda_\omega} _{\mathcal F_2} (\omega)$ (i think it should say $\lambda$ -almost everywhere too, but maybe it is implicit)

Any help regarding this step? What would be a proper approach?

I have tried to use the next properties:

-Conditional expectation (with respect to $\lambda$ )

$\mathbb E \left [ \frac{d\mu}{d\lambda} _{\mathcal F_2}| \mathcal F_1 \right ]= \frac{d\mu}{d\lambda} _{\mathcal F_1}$

$\mathbb E \left [ \frac{d\mu}{d\lambda} _{\mathcal F_2}| \mathcal F_1 \right ](\omega)= \int_{\Omega} \frac{d\mu}{d\lambda} _{\mathcal F_2} d\lambda_\omega$

Just in case $\frac{d\mu}{d\lambda} _{\mathcal F_2}$ means the radon-nikodym derivative on $\mathcal F_2$

Any help would be greatly appreciated!

**mabruka** · March 5th 2010, 10:09 AM

Any thoughts on this

?

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