Conditional Expectation in Bayes' Rule Under a Change of Measure

Summary

The discussion examines steps in a proof about conditional expectations under two probability measures, P and Q. The question focuses on why a conditional expectation under Q can be treated as measurable with respect to a sigma-algebra G, and how that measurability permits taking it outside a conditional expectation under P. The answer also invokes the defining integral property of conditional expectation and the density relation between the measures.

The exchange sketches why these manipulations can be valid, while emphasizing measurability and the change-of-measure density as essential conditions. It does not reproduce the proposition or present a complete, carefully ordered derivation; the displayed notation is abbreviated and some claims depend on assumptions about integrability and the density. Readers applying the argument in pricing or arbitrage models should check those conditions against the original theorem rather than treating the brief explanation as a standalone proof.

Key ideas

• A conditional expectation given G is G-measurable, which permits pulling it outside another conditional expectation given G.
• The defining integral property of conditional expectation explains the equality of conditional averages and the original variable over G-measurable events.
• A Radon–Nikodym density connects expectations under the two probability measures.
• The exchange omits assumptions and a complete derivation, so the original proposition is needed for full conditions.

Tags

# Bayes Theorem with change of measure


# 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?

## Answer by Magic is in the chain (score 3)

https://quant.stackexchange.com/a/59371

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:

## Answer by ltrd (score 0)

https://quant.stackexchange.com/a/44237

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 $\int_{G} \mathbb{E}(L | G) dP = \int_{G} L dP$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.