# Bayes' 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 ?
• 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. Apr 3, 2014 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 Apr 3, 2014 at 11:41

Is this the proof you are looking for? -- from Shreve, S. E.'s book "Stochastic calculus for finance II, continuous-time Models", chapter 5.

• thanks - hopefully that also comprehensively clarifies your numeraire question Apr 3, 2014 at 11:42
• athos, is this correct? quant.stackexchange.com/questions/22268
– BCLC
Jan 31, 2021 at 8:51

$$\def\Filtr{\mathcal{F}} \def\EF{E^\Filtr}$$ Let $$f=dQ/dP$$, and denote by $$E$$, $$E_Q$$ the expectation with respect to the measure $$P$$, $$Q$$, respectively. Let us also write $$\EF$$, $$\EF_Q$$ instead of $$E(\cdot|\Filtr)$$, $$E_Q(\cdot|\Filtr)$$.

Assume that all random variables listed below are integrable, in particular, that $$E|\xi|$$, $$E|f\xi|$$, $$E|f^2\xi|<\infty$$. Let $$\Filtr$$ be any $$\sigma$$-field.

Thanks to self-adjointness property of conditional expectation ($$E(\xi\EF\eta)=E(\eta\EF\xi)$$), we have for every $$A\in\Filtr$$: \begin{align*} \newcommand{\eqby}{\stackrel{\text{#1}}{=}} E(\xi f\EF(f1_A)) &= E(f1_A\EF(\xi f)),\\ E_Q(\xi\EF(f1_A)) &= E_Q(\EF(\xi f)1_A),\\ E_Q(\xi(\EF f)1_A) &= E_Q(\EF(\xi f)1_A),\\ \EF_Q(\xi\EF f) &\eqby{a.s.} \EF\xi f,\\ (\EF f)(\EF_Q\xi) &\eqby{a.s.} \EF\xi f. \end{align*} Second equality follows from the definition of $$f$$, third from the pull-out property ($$\EF\xi\eta=\xi\EF\eta$$, if $$\xi$$ is $$\Filtr$$-measurable) and from $$\Filtr$$-measurability of $$1_A$$, fourth from the definition of conditional expectation $$\EF_Q$$, and the last one by the pull-out property, as $$\EF f$$ is already $$\Filtr$$-measurable.