Show that a conditional expectation is zero Consider two binary random variables $G,Z$ and a continuous random variable $\eta$. Assume that
$$
\begin{aligned}
& (A) \quad E(\eta|Z=1)=E(\eta|Z=0)=0\\
& (B) \quad  Pr(G=1| Z=1, \eta)=1 \quad \text{a.s.}
\end{aligned}
$$
Assuming also that the following expectations exist, which of them must be $0$?
\begin{aligned}
&(1)\quad E(\eta G| Z=1)\\
&(2) \quad E(\eta G Z)=0\\
&(3) \quad E(\eta Z| G=1)\\
& (4) \quad E(\eta| Z=1, G=1)\\
\end{aligned}

MY ATTEMPTS (The question has been revised thanks to the comments below)
For simplicity of notation, I assume that $\eta$ is discrete.
(1): Let $\mathcal{M}_1\equiv \{a\in \mathbb{R}: \Pr(\eta=a|Z=1)>0\}$.
We have that
$$
\begin{aligned}
E(\eta G |Z=1)& =\sum_{a\in \mathcal{M}_1} a \Pr(\eta=a, G=1|Z=1)\\
& =\sum_{a\in \mathcal{M}_1} a \underbrace{\Pr(G=1|\eta=a, Z=1)}_{=1}\times \underbrace{\Pr(\eta=a|Z=1)}_{\text{$>0$ because $a\in \mathcal{M}_1$}}\\
&=E(\eta|Z=1)=0
\end{aligned}
$$
(2):  We have that
$$
\begin{aligned}
E(\eta G Z)& = E(E(\eta G Z|Z))\\
&=E(\eta G Z|Z=1)\Pr(Z=1)+E(\eta G Z|Z=0)\Pr(Z=0)\\
&=1\times E(\eta G |Z=1)\Pr(Z=1)+0\times E(\eta G |Z=0)\Pr(Z=0)\\
&=E(\eta G |Z=1)\Pr(Z=1)\\
&\overbrace{=}^{(1)}0
\end{aligned}
$$
I'm not sure (3) and (4) hold.
 A: This answer is incorrect. I made the incorrect assumption that
$$
E[\eta Z \mid Z=1] = E[\eta]
$$
when actually
$$
E[\eta Z \mid Z=1] = E[\eta \mid Z=1]
$$
since $\eta$ and $Z$ are not necessarily independent. This assumption is also incorrectly made in other instances in this answer.

Using the law of total expectation
$$
\begin{align}
E[\eta Z] &= p(Z=0) \cdot E[\eta Z \mid Z=0] + p(Z=1) \cdot E[\eta Z \mid Z=1] \\
&= p(Z=0) \cdot E[0] + p(Z=1) \cdot E[\eta] \\
&= p(Z=1) \cdot E[\eta] \\
&= 0
\end{align}
$$
$$
\begin{align}
E[\eta G \mid Z=1] &= p(G=0 \mid Z=1) \cdot E[\eta G \mid Z=1,G=0] + p(G=1 \mid Z=1) \cdot E[\eta G \mid Z=1,G=1] \\
&= p(G=0 \mid Z=1) \cdot E[0 \mid Z=1] + p(G=1 \mid Z=1) \cdot E[\eta \mid Z=1] \\
&= p(G=1 \mid Z=1) \cdot E[\eta \mid Z=1] \\
&= \int_{\eta} \eta \cdot p(G=1 \mid Z=1) \cdot p(\eta \mid G=1,Z=1) \ \text{d} \eta
\end{align}
$$
Since
$$
p(\eta \mid G=1,Z=1) = \frac{p(G=1 \mid Z=1,\eta) \cdot p(\eta \mid Z=1) \cdot p(Z=1)}{p(G=1 \mid Z=1) \cdot p(Z=1)}
$$
And since
$$
p(G=1 \mid Z=1,\eta) = 1
$$
Then
$$
\begin{align}
E[\eta G \mid Z=1] &= \int_{\eta} \eta \cdot p(\eta \mid Z=1) \ \text{d} \eta \\
&= E[\eta \mid Z=1] \neq 0
\end{align}
$$
Similarly:
$$
\begin{align}
E[\eta Z \mid G=1] &= p(Z=0 \mid G=1) \cdot E[\eta Z \mid G=1,Z=0] + p(Z=1 \mid G=1) \cdot E[\eta Z \mid G=1,Z=1] \\
&= p(Z=0 \mid G=1) \cdot E[0 \mid G=1] + p(Z=1 \mid G=1) \cdot E[\eta \mid G=1] \\
&= p(Z=1 \mid G=1) \cdot E[\eta \mid G=1]
\end{align}
$$
It is safe to assume that this will also not be equal to 0, since it is already difficult to use the fact that
$$
p(G=1 \mid Z=1,\eta) = 1
$$
without making the expression more complicated.
