# Bayes Theorem with change of measure

Tomas bjork- arbitrage theory in continuous time. Appendix B, proposition B41 says:

The proof is not clear to me.

Thanks to Gordon's comment below of $$E^Q (X/G)$$ being $$G$$ measurable, I think the part where Bjork seems to imply that

$$E^Q (X/G) . E^P (L/G) = E^P[(L.E^Q(X/G))/G]$$

is valid since $$E(x.y/\tau) = yE(x/\tau)$$ if $$y$$ is $$\tau$$ measurable.

However in the next step, Bjork seems to say

$$E^P[(L.E^Q(X/G))/G] = L.E^Q(X/G)$$

Why would this be valid?

Moreover the RHS seems to imply

$$E^P[(L.X)/G] = L.X$$

Why is this valid?

• Do you know that $E^Q(X/\Gamma)$ is $\Gamma$ measurable? Feb 22, 2019 at 23:34
• @Gordon thanks i think i get the drift, edited the question now. Feb 23, 2019 at 0:14

Nothing new in this answer - I have just consolidated what others have said in the answer and the comments, and put the explanation next to each step. I have to move the original equations so as to have one equation on each line:

The last one is valid since it is a defining relation of conditional expectation. Ane also we have $$\frac{dQ}{dP} = Z$$ and it implies that $$dP \thinspace Z = dQ$$. And this is the last equation.

Let's consider the first equation: $$\mathbb{E}^P[L \thinspace \mathbb{E}^Q(X | G) \thinspace | \thinspace G] = L \thinspace \mathbb{E}^Q(X | G)$$

As it was said before, $$\mathbb{E}^Q(X | G)$$ is G-measurable, so we can take this expression before the whole conditional expectation and again we use defining relation of the conditional expectation $$\int_{G} \mathbb{E}(L | G) dP = \int_{G} L dP$$

• could you explain what you mean by 'defining relation of conditional expectation' Feb 23, 2019 at 19:17
• The last equation is a defining relation. It is just an equation (almost surely) that is in the definition of conditional expectation.
– ltrd
Feb 24, 2019 at 9:25
• @dayum skipping the formalism, the conditional expectation $E(X|G)$ is defined by two properties: (1) Measurability - $E(X|G)$ is G-measurable and (2) Partial Averaging - for all sets $A \in G$, the integral of $E(X|G)$ in $A$ equals de integral of $X$ in $A$. Jan 23, 2020 at 23:10