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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.

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  • $\begingroup$ Is this a self-study question? In other words, would you like help in terms of hints and checking your approach? $\endgroup$
    – AdamO
    Commented May 14, 2021 at 17:36
  • $\begingroup$ No, it is not. I would like to get the full answers. Thanks $\endgroup$
    – Star
    Commented May 14, 2021 at 17:49
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    $\begingroup$ I don't agree that the expectation is necessarily 0 for (2). Suppose that $\eta$ is normally distributed with mean 0, $G$ is always 1, and $Z$ is 1 if $\eta > 0$ and 0 otherwise. Then the expectation of (2) is positive, because conditioning on $Z = 1$, we know that $\eta > 0$. $\endgroup$
    – fblundun
    Commented May 17, 2021 at 17:04
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    $\begingroup$ @TEX I suggest you check each equality in your derivation until you find one which does not hold for my counterexample. That's the wrong step. $\endgroup$
    – fblundun
    Commented May 17, 2021 at 23:10
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    $\begingroup$ @TEX both directions hold in my counterexample, since I said "$Z$ is 1 if $\eta > 0$ and 0 otherwise". BTW if you don't mention my username with the "@fblundun" syntax in your replies here I don't get automatically notified of them. $\endgroup$
    – fblundun
    Commented May 18, 2021 at 9:19

1 Answer 1

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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.

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  • $\begingroup$ Thanks. When proving $E(\eta Z)=0$, we have $$p(Z=0)E(0)+p(Z=1) E(\eta Z|Z=1)=p(Z=1) E(\eta |Z=1)$$ How do you replace $E(\eta|Z=1)$ with $E(\eta)$? $\endgroup$
    – Star
    Commented May 17, 2021 at 17:40
  • $\begingroup$ With regards to $E(\eta G |Z=1)$ and $E(\eta Z |G=1)$: you don't show that they are equal to zero. Are they or are they not? $\endgroup$
    – Star
    Commented May 17, 2021 at 17:42
  • $\begingroup$ I've edited my answer to clarify your second question. As for your first question, what I am doing is replacing $E[\eta Z \mid Z=1]$ with $E[\eta]$, since it is given in the expectation that $Z=1$. $\endgroup$
    – mhdadk
    Commented May 17, 2021 at 18:57
  • $\begingroup$ @mhdadk that replacement is wrong, because we don't know that $\eta$ and $Z$ are independent. $\endgroup$
    – fblundun
    Commented May 17, 2021 at 19:00
  • $\begingroup$ What are you talking about? You used this replacement when you wrote $E(\eta G |Z=1,\eta=a) = a E( G |Z=1,\eta=a)$ in your derivation. In fact, $E( G |Z=1,\eta=a) = E( G |Z=1)$, since the value of $\eta$ is given. $\endgroup$
    – mhdadk
    Commented May 17, 2021 at 19:03

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