4
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Is $G$ Bernoulli ? $\endgroup$ Commented Nov 13, 2022 at 20:36
  • $\begingroup$ Yes, it is binary $\endgroup$
    – Star
    Commented Nov 13, 2022 at 21:05

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

Using the antecedent conditions in your post you have:

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

$\endgroup$
5
  • $\begingroup$ $E(G\eta | G= g, Z = 1) = E(g\eta|Z = 1)$ seems questionable. $\endgroup$
    – Zhanxiong
    Commented Nov 13, 2022 at 23:41
  • $\begingroup$ I agree with the comment. We need to go ahead with $E(1\times \eta|G=1, Z=1)$ in the proof. $\endgroup$
    – Star
    Commented Nov 13, 2022 at 23:42
  • $\begingroup$ @Zhanxiong: Answer edited to correct. $\endgroup$
    – Ben
    Commented Nov 13, 2022 at 23:59
  • $\begingroup$ @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. $\endgroup$
    – Zhanxiong
    Commented Nov 14, 2022 at 0:07
  • $\begingroup$ @Zhanxiong: Oops, forgot to replace sum with integral --- now corrected! (I'm on fire today!) $\endgroup$
    – Ben
    Commented Nov 14, 2022 at 0:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.