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Thread: Radon Nikodym computation, step of a proof

  1. #1
    Member mabruka's Avatar
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    Radon Nikodym computation, step of a proof

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

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


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

    then

    $\displaystyle \frac{d\mu}{d\lambda} _{\mathcal F_2}(\omega)= \frac{d\mu}{d\lambda} _{\mathcal F_1}(\omega) \frac{d\mu_\omega}{d\lambda_\omega} _{\mathcal F_2} (\omega)$ (i think it should say $\displaystyle \lambda$-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 $\displaystyle \lambda$)

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


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




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

    Any help would be greatly appreciated!
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  2. #2
    Member mabruka's Avatar
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    Any thoughts on this ?
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