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Suppose we have two probability measures $\mathbb{P}$ and $\mathbb{Q}$ on $(\Omega,F)$ and $\frac{d\mathbb{Q}}{d\mathbb{P}}=Z$. Let $\mathbb{P}_{n}$ and $\mathbb{Q}_{n}$ be the restrictions to $F_{n}$.

It is now given that $\frac{d\mathbb{Q}_{n}}{d\mathbb{P}_{n}}=\mathbb{E}_{p}[Z|F_{n}]$

Could anyone help me understand why?

1 Answers 1

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You have to use the definition of the conditional expectation $E^P[Z\mid \mathcal{F}_n]$, namely that $ \int_A E^P[Z\mid\mathcal{F}_n]\,\mathrm{d}P=\int_A Z\,\mathrm{d}P,\quad A\in\mathcal{F}_n. $ That $P_n$ is the restriction of $P$ to $\mathcal{F}_n$ means that $P_n(A)=P(A)$ for every $A\in\mathcal{F}_n$. Now if $A\in \mathcal{F}_n$, then $ Q_n(A)=Q(A)=\int_A Z\,\mathrm{d}P=\int_A E^P[Z\mid\mathcal{F}_n]\,\mathrm{d}P =\int_A E^P[Z\mid\mathcal{F}_n]\,\mathrm{d}P_n, $ which is what you need.

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    Can't believe I missed that.. Thank you so much for your help, really appreciate it.2012-10-30