# Radon Nikodym Derivative for conditional expectation?

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 $$\mathcal{F}$$.

• From: probability: For basic questions about probability and the questions associated with the calculation of probability, expected value, variance, standard deviation, or similar statistical quantities. For questions about the theoretical footing of probability (especially using measure-theory), ask under probability-theory instead. Commented Jan 29, 2022 at 13:05
• ah oke! I will remember Commented Jan 29, 2022 at 13:07

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 $$G \in \mathcal G,$$
$$\mathbb E_{\mathbb Q}\left[ 1_G \frac{\mathbb{E}_{\mathbb{P}}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}Y|\mathcal{G}\right]}{\mathbb{E}_{\mathbb{P}}\left[\frac{d\mathbb{Q}}{d\mathbb{P}}|\mathcal{G}\right]}\right] = \mathbb E_{\mathbb Q}\left[1_G Y \right].$$