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Math Help - 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  (\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!
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  2. #2
    Member mabruka's Avatar
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    Any thoughts on this ?
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