# 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 ? Nov 13, 2022 at 20:36
• Yes, it is binary
– TEX
Nov 13, 2022 at 21:05

Yes, it is correct.

To put the discussion under the measure-theoretic conditional expectation/probability framework, first let's clarify the exact meaning of the notation $$P(G = 1 | Z = 1, \eta)$$. As $$P(G = 1 | Z, \eta)$$ 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 $$(\mathbb{R}^1, \mathscr{R}^1)$$. $$P(G = 1 | Z = 1, \eta) = 1$$ almost surely then means $$f(Z(\omega), \eta(\omega)) = 1$$ for $$\omega \in A := \{\omega \in \Omega: Z(\omega) = 1\}$$ with probability $$1$$. Accordingly, $$P(G = 1 | Z, \eta) = E(G | Z, \eta)$$ can be written as \begin{align} E(G | Z, \eta) = P(G = 1|Z, \eta)I_A + P(G = 1|Z, \eta)I_{A^c} = I_A + I_{A^c}f(Z, \eta). \end{align}

Now applying the property of conditional expectation: $$E[X|\mathscr{F}_1] = E[E[X|\mathscr{F_2}]|\mathscr{F}_1]$$ if $$\mathscr{F}_1 \subset \mathscr{F}_2$$ yields \begin{align} & E[G\eta|Z] = E[E[G\eta|Z, \eta]|Z] \\ =& E[\eta(I_A + I_{A^c}f(Z, \eta))|Z] = I_AE[\eta|Z] + I_{A^c}E[\eta f(Z, \eta)|Z]. \tag{1} \end{align}

Again, by noticing that $$E[G\eta|Z = 1]$$ means the (common) value of $$E[G\eta|Z]$$ on set $$A$$, it follows by $$(1)$$ and $$E[\eta|Z = 1] = 0$$ that for all $$\omega \in A$$ but a null-set that \begin{align} E[G\eta|Z](\omega) = 1 \times E[\eta|Z](\omega) + 0 \times E[\eta f(Z, \eta)|Z](\omega) = E[\eta | Z](\omega) = 0. \end{align}

This shows $$E[G\eta | Z = 1] = 0$$.

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. 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.
– TEX
Nov 13, 2022 at 23:42
• @Zhanxiong: Answer edited to correct.
– Ben
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. Nov 14, 2022 at 0:07
• @Zhanxiong: Oops, forgot to replace sum with integral --- now corrected! (I'm on fire today!)
– Ben
Nov 14, 2022 at 0:09