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?

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