Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

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 ?
share|improve this question
1  
The book "Statistics of Random Processes" Vol. 1 by Robert Lipster and Albert Shiryaev has a whole chapter devoted to various (abstract) forms of Bayes Law. If I remember correctly it is chapter 7. The eBook is available from SpringerLink if you have access. –  pbr142 Apr 3 at 11:27
    
thank you for the reference - seeing how it also answers my question - you can post it as an answer not only as a comment - the reference might be interesting to others –  Probilitator Apr 3 at 11:41

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
thanks - hopefully that also comprehensively clarifies your numeraire question –  Probilitator Apr 3 at 11:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.