# Radon-Nikodym with restrictions to filtration

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?

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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.
Thank you for your reply. I've actually tried to use this in the following way for $A\in F_{n}$ $$\mathbb{Q}_{n}(A)=\mathbb{Q}(A)=\int_{A} Z d\mathbb{P}=\int_{\Omega}1_{A}Zd\mathbb{P}=\int_{\Omega}1_{A}Zd\mathbb{P}_{n}$$ If $1_{A}Z$ is $F_{n}$ measurable i can then find $\mathbb{E}_{p}[1_{A}Z|F_{n}]=1_{A}Z$ and filling this into the integral and youre there. My problem is proving that $1_{A}Z$ is $F_{n}$ measurable.. Am I at least on the right track with this? – Math Girl Oct 30 '12 at 18:54
$1_A Z$ is not necessarily $\mathcal{F}_n$-measurable because $Z$ is not necessarily $\mathcal{F}_n$-measurable, so you can not conclude that $E^P[1_A Z\mid \mathcal{F}_n]=1_A Z$. But what you do know is that $$\int_\Omega 1_A Z\,\mathrm{d}P = \int_\Omega 1_A E[Z\mid\mathcal{F}_n]\,\mathrm{d}P$$. This is the very definition of the conditional expectation. – Stefan Hansen Oct 30 '12 at 19:19