# Conditional expectations with different measures

Let $\mathbb Q \sim \mathbb P$ on ($\Omega, \mathcal F$) and $\mathcal G \subset \mathcal F$. We also have a Radon-Nykodym derivative $$\rho = \frac{d \mathbb Q}{d\mathbb P} \bigg|_{\mathcal F}$$ For $X\in \mathcal L^1(\Omega, \mathcal F, \mathbb Q)$ I want to show that $$E_{\mathbb Q}[X \mid \mathcal G] = \frac{E_{\mathbb P}[\rho X \mid \mathcal G]}{E_{\mathbb P}[\rho \mid \mathcal G]}.$$ I start with using the definition of conditional expectation so that we have to show that $$E_{\mathbb Q}[1_G \text{RHS}] = E_{\mathbb Q}[1_G X]$$ for $G \in \mathcal G$. I have tried converting the $\mathbb P$-expectations to $\mathbb Q$-expectations by $$E_{\mathbb Q} \left[1_G \frac{E_{\mathbb P}[\rho X \mid \mathcal G]}{E_{\mathbb P}[\rho \mid \mathcal G]}\right] =E_{\mathbb Q} \left[1_G \frac{E_{\mathbb Q}[\hat \rho\rho X \mid \mathcal G]}{E_{\mathbb Q}[\hat \rho\rho \mid \mathcal G]}\right]$$ where $\hat \rho = E_\mathbb P[\rho \mid \mathcal G]$, and then try different manipulations but without much luck.

## 1 Answer

If $V$ denotes the right-hand side, then $$\mathrm{E}_{\mathbb{P}}[\rho V\mid \mathcal{G}]=\mathrm{E}_{\mathbb{P}}[\rho X\mid\mathcal{G}]$$ and hence $$\mathrm{E}_{\mathbb{P}}[\mathbf{1}_G\rho V]=\mathrm{E}_{\mathbb{P}}[\mathbf{1}_G \mathrm{E}_{\mathbb{P}}[\rho V\mid \mathcal{G}]]=\mathrm{E}_{\mathbb{P}}[\mathbf{1}_G \mathrm{E}_{\mathbb{P}}[\rho X\mid\mathcal{G}]]=\mathrm{E}_{\mathbb{P}}[\mathbf{1}_G\rho X]$$ holds for all $G\in\mathcal{G}$. Thus, $$\mathrm{E}_{\mathbb{Q}}[\mathbf{1}_GV]=\mathrm{E}_{\mathbb{P}}[\mathbf{1}_G\rho V]=\mathrm{E}_{\mathbb{P}}[\mathbf{1}_G\rho X]=\mathrm{E}_{\mathbb{Q}}[\mathbf{1}_GX]$$ for all $G\in\mathcal{G}$.