Conditional expectation and Radon-Nikodym

Question

Let $P,Q$ be two probability measures on $(\Omega,\mathcal{F})$ such that $Q<<P$, let $\mathcal{G}\subset{\mathcal{F}}$ be a sub $\sigma$-algebra. Say $P_{\mathcal{G}},Q_{\mathcal{G}}$ are the restriction of $P$ and $Q$ to $\mathcal{G}$ and denote with $Z=dQ/dP, Z_{\mathcal{G}}=dQ_{\mathcal{G}}/dP_{\mathcal{G}}$ the Radon-Nikodym derivatives. I need to show that, given $X\in{L^1(\Omega,\mathcal{F},Q)}$ and $\forall{G}\in{\mathcal{G}}$ we have: $$\int_{\Omega}1_{{G}}\mathbb{E}_Q[X|\mathcal{G}]dQ=\int_{\Omega}1_{G}\mathbb{E}[X|\mathcal{G}]Z_\mathcal{G}dP.$$ In particular I don't get how we turn an integral in $dQ$ to an integral in $dP$.

Answer 1

The important fact here, is that for any $f\in L^1(\Omega,\mathcal F,P)$, we know that $\int_\Omega fdQ=\int_\Omega fZdP$ (this is true if we replace $\mathcal F$ with $\mathcal G$ and the restricted measures replacing the original). You can prove this by noting that this is true for simple functions simply by the definition of the Radon-Nikodym derivative, and then using the usual approximation and limiting argument. We also know that $\int_G\mathbb E_Q[X|\mathcal G]dQ=\int_G\mathbb E_Q[X|\mathcal G]dQ_{\mathcal G}$ (see here). Thus, combining these and using the definition of conditional expectation: $$\int_G\mathbb E_Q[X|\mathcal G]dQ=\int_G\mathbb E_Q[X|\mathcal G]dQ_{\mathcal G}=\int_\Omega\mathbb 1_G\mathbb E_Q[X|\mathcal G]Z_{\mathcal G}dP_{\mathcal G}=\int_\Omega\mathbb 1_G\mathbb E_Q[X|\mathcal G]Z_{\mathcal G}dP.$$ Now, finally, using standard properties of conditional expectation and recalling that $Z_\mathcal G$ is $\mathcal G$ measurable, we have $$\int_\Omega\mathbb 1_G\mathbb E_Q[X|\mathcal G]Z_{\mathcal G}dP_{\mathcal G}=\int_\Omega\mathbb 1_G\mathbb E_Q[X|\mathcal G]\mathbb E_P[Z_{\mathcal G}|\mathcal G]dP_{\mathcal G}=\int_\Omega\mathbb 1_G\mathbb E_P[XZ_{\mathcal G}|\mathcal G]dP_{\mathcal G}=\int_\Omega\mathbb 1_G\mathbb E_P[X|\mathcal G]Z_{\mathcal G}dP_{\mathcal G}$$ where the second last equality follows from the so called "abstract Bayes formula". Thus, because $\mathbb 1_G\mathbb E_P[X|\mathcal G]Z_{\mathcal G}$ is $\mathcal G$ measurable, we conclude: $$\int_G\mathbb E_Q[X|\mathcal G]dQ=\int_\Omega\mathbb 1_G\mathbb E_P[X|\mathcal G]Z_{\mathcal G}dP$$.