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?

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    $\begingroup$ Do you know that $E^Q(X/\Gamma)$ is $\Gamma$ measurable? $\endgroup$ – Gordon Feb 22 '19 at 23:34
  • $\begingroup$ @Gordon thanks i think i get the drift, edited the question now. $\endgroup$ – dayum Feb 23 '19 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:

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

  • $\begingroup$ could you explain what you mean by 'defining relation of conditional expectation' $\endgroup$ – dayum Feb 23 '19 at 19:17
  • $\begingroup$ The last equation is a defining relation. It is just an equation (almost surely) that is in the definition of conditional expectation. $\endgroup$ – Lipa_FNTE Feb 24 '19 at 9:25
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    $\begingroup$ @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 $A$. $\endgroup$ – fpessoa Jan 23 '20 at 23:10

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