Let $\mathbb{P},\mathbb{Q}$ be two equivalent probability measures on the space $(\Omega,\mathcal{F})$. And let $\frac{d\mathbb{Q}}{d\mathbb{P}}$ be the Radon-Nikodym derivative ($\mathbb{Q}$ w.r.t $\mathbb{P}$) and $Y$ an integrable random variable defined on $\Omega$. Then we have
$$\mathbb{E}_\mathbb{Q}[Y] = \mathbb{E}_{\mathbb{P}}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}Y\right] $$
Do we also have that
$$\mathbb{E}_\mathbb{Q}[Y|\mathcal{G}] = \mathbb{E}_{\mathbb{P}}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}Y|\mathcal{G}\right] $$
where $\mathcal{G}$ a sub sigma field of $\mathcal{F}$.
Best Answer
No, what we have is $$\mathbb{E}_\mathbb{Q}[Y|\mathcal{G}] = \frac{\mathbb{E}_{\mathbb{P}}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}Y|\mathcal{G}\right]}{\mathbb{E}_{\mathbb{P}}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}|\mathcal{G}\right]}. $$
In order to prove it you can use the definition of conditional expectation, i.e. show that for all $G \in \mathcal G,$
$$\mathbb E_{\mathbb Q}\left[ 1_G \frac{\mathbb{E}_{\mathbb{P}}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}Y|\mathcal{G}\right]}{\mathbb{E}_{\mathbb{P}}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}|\mathcal{G}\right]}\right] = \mathbb E_{\mathbb Q}\left[1_G Y \right].$$