I want to show the following:
$\mathbb{E}_{P}\left(\left(\frac{d Q}{d P}\right)_{\mathcal{F}} | \mathcal{G}\right)=\left(\frac{d Q}{d P}\right)_{\mathcal{G}}$
where P and Q are probability measures such that $Q \ll P$. And $\mathcal{G}\subset\mathcal{F}$ are $\sigma$-algebras.
What I have done so far:
just check the definition of RHS, we have that $\forall g\in L^{1}(\Omega, \mathscr{G}, P)$, we have $$\int g d Q=\int g \left(\frac{d Q}{d P}\right)_{\mathcal{G}} d P$$ (I also wonder can I write it as $\mathbb{E}_{Q}(g)=\mathbb{E}_{P}\left(g \left(\frac{d Q}{d P}\right)_{\mathcal{G}}\right)$ ?).
Then, for LHS, we know by definition that $\mathbb{E}_{P}\left(\left(\frac{d Q}{d P}\right)_{\mathcal{F}} | \mathcal{G}\right)$ is the unique random variable in $ L^{1}(\Omega, \mathscr{G}, P)$ such that $\mathbb{E}_{P}\left(\left(\frac{d Q}{d P}\right)_{\mathcal{F}}Y\right)=\mathbb{E}_{P}\left(\mathbb{E}_{P}\left(\left(\frac{d Q}{d P}\right)_{\mathcal{F}} | \mathcal{G}\right)Y\right)$ for every bounded, $\mathcal{G}$-measurable random variable $Y$.
But this is getting messy and I am not sure about how to proceed, also, do I need to check the derivative is in $L^{1}$?
Best Answer
You are making things complicated. We already know that RHS is measurable w.r.t. $\mathcal G$. What is needed is to show that $\int_A (\frac {dQ} {dP})_{\mathcal F} dP= \int_A (\frac {dQ} {dP})_{\mathcal G}dP$ if $A \in\mathcal G$ . But both sides are equal to $P(A)$ because $A $ belongs to both $\mathcal G$ and $\mathcal F$