# Show that $\Pr(G=1|Z=1, \eta)=1$ and $\mathbb{E}(\eta|Z=1)=0$ imply $\mathbb{E}(G\times \eta|Z=1)=0$

Take two binary random variables $$G,Z$$ and a continuous random variable $$\eta$$.

Assume $$\Pr(G=1|Z=1, \eta)=1 \text{ almost surely}$$ and $$\mathbb{E}(\eta|Z=1)=0$$

Could you help me to show that this implies $$\mathbb{E}(G\times \eta|Z=1)=0 \quad?$$

If this implication is wrong, could you explain why?

• Is $G$ Bernoulli ? Commented Nov 13, 2022 at 20:36
• Yes, it is binary
– Star
Commented Nov 13, 2022 at 21:05

Yes, it is correct.

To put the discussion under the measure-theoretic conditional expectation/probability framework, let's first clarify the exact meaning of the notation "$$P(G = 1 | Z = 1, \eta)$$" (where the conditioning is a hybrid of a random variable and an event), which seems uncommon in rigorous probability textbooks. As $$P(G = 1 | Z, \eta) = E[G | Z, \eta]$$ is $$\sigma(Z, \eta)$$-measurable, we can denote it by $$f(Z, \eta)$$, where $$f$$ is a Borel function from $$(\mathbb{R}^2, \mathscr{R}^2)$$ to $$(\mathbb{R}^1, \mathscr{R}^1)$$. Then $$P(G = 1 | Z = 1, \eta)$$ should be interpreted as a $$\sigma(\eta)$$-measurable random variable $$f(1, \eta)$$. Hence $$P(G = 1 | Z = 1, \eta) = 1$$ a.s. means that $$f(1, \eta) = 1$$ a.s.

As we know that when $$A$$ is an event with $$P(A) > 0$$, $$E[X|A]$$ is defined as $$E[X|A] = \frac{E[XI_A]}{P(A)}$$. So to prove $$E[G\eta|Z = 1] = 0$$ is equivalent to prove $$E[G\eta I_{\{Z = 1\}}] = 0$$. It follows that \begin{align*} & E[G\eta I_{\{Z = 1\}}] \\ =& E[E[G\eta I_{\{Z = 1\}} | Z, \eta]] \tag{Law of iterative expectations} \\ =& E[\eta I_{\{Z = 1\}}E[G | Z, \eta]] \tag{Pull out known factors} \\ =& E[\eta I_{\{Z = 1\}}f(Z, \eta)] \\ =& E[\eta I_{\{Z = 1\}}f(1, \eta)] \tag{Notation definition} \\ =& E[\eta I_{\{Z = 1\}}] \tag{Condition 1} \\ =& 0. \tag{Condition 2} \end{align*} This completes the proof.

Using the antecedent conditions in your post you have:

\begin{align} \mathbb{E}(G \times \eta|Z=1) &= \int \eta \cdot p(\eta,G=1|Z=1) \ d\eta \\[6pt] &= \int \eta \cdot \mathbb{P}(G=1|\eta,Z=1) \cdot p(\eta|Z=1) \ d\eta \\[6pt] &= \int \eta \cdot p(\eta|Z=1) \ d\eta \\[10pt] &= {E}(\eta|Z=1) \\[14pt] &= 0. \\[6pt] \end{align}

• $E(G\eta | G= g, Z = 1) = E(g\eta|Z = 1)$ seems questionable. Commented Nov 13, 2022 at 23:41
• I agree with the comment. We need to go ahead with $E(1\times \eta|G=1, Z=1)$ in the proof.
– Star
Commented Nov 13, 2022 at 23:42
• @Zhanxiong: Answer edited to correct.
– Ben
Commented Nov 13, 2022 at 23:59
• @Ben Hmm, in this case, why the first equality holds? OP stated that $\eta$ is a continuous r.v. -- even let's say $\eta$ is discrete, wouldn't the first equality be $E(G\eta | Z = 1) = \sum_g\sum_y gyP(G = g, \eta = y | Z = 1)$? Given that $\eta$ is continuous, the second equality is then also questionable. Commented Nov 14, 2022 at 0:07
• @Zhanxiong: Oops, forgot to replace sum with integral --- now corrected! (I'm on fire today!)
– Ben
Commented Nov 14, 2022 at 0:09