We have $Z_t$ a random variable, and $S_t$ is a sigma algebra. The subscript is essentially time. I want to understand the relationship given below, given another random variable $V$, and a constant $A$. All random variables are on the same probability space.
$E[(A - V) \mathbb{1}_{\{Z_t > A\}}| S_t] = E[A - V | Z_t > A, S_t] P(Z_t > A | S_t)$
Here is my reasoning. For any $\mathcal{H} \subset S_t$ we have, $\int_\mathcal{H} (A-V) \mathbb{1}_{\{Z_t > A\}} | H] dP = \int_\mathcal{H} (A - V) \mathbb{1}_{\{Z_t > A\}} dP$
$=\int(A-V) \mathbb{1}_{\{Z_t > A, \mathcal{H}\}} dP$
I am not sure where to go from here. The result above is really used in the paper by Glosten and Milgrom 1985. "BID, ASK Transaction Prices in a specialist market with heterogeneous informed traders"
Thank you in advance for your responses.
Let $(\Omega,\mathcal{A},\mu)$ be a probability space and $X$ a random variable. Then fix some event $H \in \mathcal{A}$, we have that $$\nu(A)=\frac{\mu(A\cap H)}{\mu(H)} \ \ \ A \in \mathcal{A}$$ Is a probability measure. Indeed: $$\nu(\emptyset)=\frac{\mu(\emptyset\cap H)}{\mu(H)}=0 \ \ \ \ \ \ \ \ \ \ \nu(\Omega)=\frac{\mu(\Omega\cap H)}{\mu(H)}=1$$ and for pairwise disjoint $(A_n)_{n \in \mathbb{N}}\subset \mathcal{A}$ we also have $$\nu(\cup_{n \in \mathbb{N}}A_n)=\frac{\mu(\cup_{n \in \mathbb{N}}A_n\cap H)}{\mu(H)}=\frac{\mu(\cup_{n \in \mathbb{N}}(A_n\cap H))}{\mu(H)}=\sum_{n \in \mathbb{N}}\frac{\mu(A_n\cap H)}{\mu(H)}=\sum_{n \in \mathbb{N}}\nu(A_n)$$ This is the conditional probability on the event $H$. The conditional expectation is $$E[X|H]:=\int_\Omega X(\omega) \nu(d\omega)=\int_\Omega X(\omega) \frac{\mu(d\omega\cap H)}{\mu(H)}$$ Thus $$E[X|H]\mu(H)=\int_\Omega X(\omega) \mu(d\omega\cap H)=\int_HX(\omega) \mu(d\omega)=E[X\mathbb{I}_H]$$