Show that the Radon-Nikodym density is $\mathcal{G}$ measurable iff $E_P[X|G] = E_Q[X|G]$

Question

Let $\alpha$ and $\beta$ be equivalent probability measures on $(\Omega, \mathcal{F})$, with Radon-Nikodym density of $\alpha$ wrt $\beta$ is $\eta$, i.e., for all $A \in \mathcal{F}, \beta(A) = \int_A\eta d\alpha$. Let $\mathcal{G}$ be a sub-$\sigma$-field of $\mathcal{F}$.

Show that $\eta$ is $\mathcal{G}$-measurable $\iff$ for any bounded random variable $X$, $E_{\alpha}[X|\mathcal{G}] = E_{\beta}[X|\mathcal{G}]$

I think I have the forward direction:

Let $\eta$ be $\mathcal{G}$ measurable. Using the General Baye's theorem: $$E_{\beta}[X|\mathcal{G}] = \frac{E_{\alpha}[\eta X|\mathcal{G}]}{E_{\alpha}[\eta|\mathcal{G}]}$$ However, $\eta$ is $\mathcal{G}$ measurable, so we have $E_{\alpha}[\eta X|\mathcal{G}] = \eta E_{\alpha}[X|\mathcal{G}]$ and $E_{\alpha}[\eta|\mathcal{G}] = \eta$, so $$E_{\beta}[X|\mathcal{G}] = \frac{E_{\alpha}[\eta X|\mathcal{G}]}{E_{\alpha}[\eta|\mathcal{G}]}$$ $$ = \frac{\eta E_{\alpha}[X|\mathcal{G}]}{\eta} = E_{\alpha}[X|\mathcal{G}]$$

I am just confused about the backward direction, I can't seem to get it. Also, I'm not sure I've done the forward direction in the best way. Thanks!

Answer 1

You didn't apply Bayes formula correctly; it should read

$$\mathbb{E}_{\color{red}{\beta}}(X \mid \mathcal{G}) = \frac{\mathbb{E}_{\color{red}{\alpha}}(\eta X \mid \mathcal{G})}{\mathbb{E}_{\color{red}{\alpha}}(\eta \mid \mathcal{G})}. \tag{1}$$

If you switch the roles of $\alpha$ and $\beta$ in your reasoning, then everything will work fine.

Regarding the backward direction:

1. Use Bayes formula $(1)$ and the tower property to show that $$\mathbb{E}_{\alpha} \bigg[X \cdot \mathbb{E}(\eta \mid \mathcal{G}) \mid \mathcal{G} \bigg] = \mathbb{E}_{\alpha}(\eta X \mid \mathcal{G})$$ and so $$\mathbb{E}_{\alpha} \bigg[ X \cdot \big(\mathbb{E}(\eta \mid \mathcal{G})-\eta\big) \mid \mathcal{G} \bigg] = 0. \tag{2}$$
2. By the definition of the conditional expectation, $(2)$ means that $$\int_G X \cdot \big(\mathbb{E}(\eta \mid \mathcal{G})-\eta\big) \, d\alpha = 0$$ for any $G \in \mathcal{G}$.
3. Set $X = \text{sgn}(\mathbb{E}(\eta \mid \mathcal{G})-\eta)$ to conclude that $$\int |\mathbb{E}(\eta \mid \mathcal{G})- \eta| \, d\alpha =0.$$