Friday, February 19, 2016

probability - Baye's rule for conditional expectations (Proof review)



The Baye's rule for conditional expectations states


$$ E^Q[X|\mathcal{F}]E^P[f|\mathcal{F}]=E^P[Xf|\mathcal{F}] $$


With $f=dQ/dP$ - thus being the Radon-Nikodyn derivative and $X$ being some random variable and $\mathcal{F}$ being some sigma-algebrad.



For I wasn't able to find the proof in any of the books that I usually use I tried to prove it myself. This rule is often used in the context of the change of numeraire technique.


The proof uses the definition/characterization of conditional expectations. Thus one mainly needs to show


$$\int_A E^Q[X|\mathcal{F}]E^P[f|\mathcal{F}]dP=\int_AE^P[Xf|\mathcal{F}]dP $$ For all $A\in\mathcal{F}$


Again using the characterisation of conditional expectation the right side equals $\int_A Xf dP$ and with $f$ being the Radon-Nikodyn-derivative this is equal to $\int_A X dQ$ thus



$$\int_AE^P[Xf|\mathcal{F}]dP=\int_A X dQ $$


On the other side using measurability of $E^Q[X|\mathcal{F}]$ with respect to $\mathcal{F}$ the left side equals $$\int_A E^P\left[(E^Q[X|\mathcal{F}] f)\vert \mathcal{F}\right] dP$$ Once again using the characterisation of conditional expectation this is $$\int_A E^P\left[(E^Q[X|\mathcal{F}] f)\vert \mathcal{F}\right] dP=\int_A fE^Q[X|\mathcal{F}] dP$$ Finally with $f$ being the Radon-Nikodyn density one arrives at


$$\int_A fE^Q[X|\mathcal{F}] =\int_A E^Q[X|\mathcal{F}] dQ=\int_A X dQ$$ and thus $$\int_A E^Q[X|\mathcal{F}]E^P[f|\mathcal{F}]dP=\int_A X dQ$$


This concludes the proof.


Two question:



  • does anyone know of a source where I could cross-check that

  • is there an alternative way to proof the result ?



Answer




Is this the proof you are looking for?


enter image description here


-- from Shreve, S. E.'s book "Stochastic calculus for finance II, continuous-time Models", chapter 5.


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