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I'm trying to come up with an example showing that $E(\epsilon|z,\eta)=E(\epsilon|\eta)$ and $E(\epsilon)=0$ do not imply $E(\epsilon|z)=0$. The model is nonparametric IV model with the structural equation $y=m(x,z_1)+\epsilon$ and the reduced form $x=\pi(z)+\eta$, assuming that $E(\epsilon|z)=0$.

Note that for $E(\epsilon|z,\eta)=E(\epsilon|\eta)$ to hold, a sufficient condition is that $(\epsilon,\eta)$ and $z$ are independent.

How to construct the example?

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    $\begingroup$ It seems to me a trivial example would be $z = \eta$; if $E(\epsilon|\eta) \neq 0$, you're done. But maybe that's too trivial, or for some other reason doesn't work for you. $z$ equals any 1-1 transform of $\eta$ works too, obviously. $\endgroup$
    – jbowman
    Commented Apr 4 at 2:29
  • $\begingroup$ @jbowman But for $E(\epsilon|z,\eta)=E(\epsilon|\eta)$ to hold, we need $z$ and $\eta$ to be independent. If $z=\eta$, they can't be independent. Can they? $\endgroup$ Commented Apr 4 at 2:36
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    $\begingroup$ The independence is a sufficient condition, not a necessary one :) $\endgroup$
    – jbowman
    Commented Apr 4 at 15:10
  • $\begingroup$ got it. thank you! $\endgroup$ Commented Apr 4 at 17:03

1 Answer 1

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Continuous Example

If you are familiar with multivariate normal distribution properties, an example that satisfies all the constraints you listed is very easy to construct.

To begin with, let's set \begin{align*} \begin{bmatrix} \epsilon \\ z \\ \eta \end{bmatrix} \sim N_3\left( \begin{bmatrix} 0 \\ \mu_1 \\ \mu_2 \end{bmatrix}, \begin{bmatrix} 1 & \alpha & \beta\\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{bmatrix} \right). \end{align*}

Here the marginal mean of $\epsilon$ is set to $0$ to satisfy the constraint $E[\epsilon] = 0$. It then follows that \begin{align*} E[\epsilon | \eta] &= \beta(\eta - \mu_2), \tag{1}\label{1}\\ E[\epsilon | z, \eta] &= \begin{bmatrix} \alpha & \beta \end{bmatrix} \begin{bmatrix} 1 & \gamma \\ \gamma & 1 \end{bmatrix}^{-1} \begin{bmatrix} z - \mu_1 \\ \eta - \mu_2 \end{bmatrix} \\ \phantom{E[\epsilon | z, \eta] } &= \frac{1}{1 - \gamma^2}\left[(\alpha - \beta\gamma)(z - \mu_1) + (\beta - \alpha\gamma)(\eta - \mu_2)\right], \tag{2}\label{2} \\ E[\epsilon | z] &= \alpha(z - \mu_1). \tag{3}\label{3} \end{align*} A necessary and sufficient condition for $\eqref{1}$ and $\eqref{2}$ to be identical is thus clearly \begin{align*} \begin{cases} \alpha - \beta\gamma = 0, \\[1em] \beta(1 - \gamma^2) = \beta - \alpha\gamma, \end{cases} \end{align*} which is just $\alpha = \beta\gamma$. To ensure $\eqref{3}$ is non-zero, $\alpha$ must be non-zero. Therefore, any choice of $(\alpha, \beta, \gamma)$ such that $\alpha = \beta\gamma$, $\alpha \neq 0$ (as well as they together make a valid covariance matrix) would serve a valid example ($\mu_1, \mu_2$ can be chosen arbitrarily). For instance, $\mu_1 = \mu_2 = 0$, $\alpha = \frac{1}{2}$, $\beta = \gamma = \frac{1}{\sqrt{2}}$ give \begin{align*} & E[\epsilon] = 0, \\ & E[\epsilon |\eta] = E[\epsilon |z, \eta] = \frac{\sqrt{2}}{2}\eta, \\ & E[\epsilon | z] = \frac{1}{2}z \neq 0. \end{align*}

Discrete Example

Let sample space $\Omega = \{1, 2, 3, 4\}$, $\mathscr{F} = 2^{\Omega}$, $P(\{1\}) = P(\{2\}) = P(\{3\}) = P(\{4\}) = \frac{1}{4}$. Consider two partitions of $\Omega$: \begin{align*} & \mathscr{G}_1 = \{A_1, A_2\}, \text{ where } A_1 = \{1, 2\}, A_2 = \{3, 4\}, \\ & \mathscr{G}_2 = \{B_1, B_2, B_3, B_4\}, \text{ where } B_i = \{i\}, i = 1, 2, 3, 4. \end{align*} Clearly $\mathscr{G}_2$ is a refinement of $\mathscr{G}_1$. Now define \begin{align*} & \epsilon = I_{A_1} - I_{A_2}, \\ & z = I_{A_1} + 4I_{A_2}, \\ & \eta = \frac{1}{4}I_{B_1} + \frac{3}{4}I_{B_2} + I_{B_3} + 3I_{B_4}. \end{align*} It is then easy to verify that $\sigma(z) = \mathscr{G}_1 \subset \mathscr{G}_2 = \sigma(\eta)$ and $\sigma(z, \eta) = \sigma(\eta) = \mathscr{G}_2$, whence $E[\epsilon|z] = E[\epsilon |z, \eta] = E[\epsilon |\mathscr{G}_2]$. On the other hand, clearly we have $E[\epsilon] = P(A_1) - P(A_2) = 0$ and \begin{align*} E[\epsilon | z] = E[\epsilon|A_1]I_{A_1} + E[\epsilon|A_2]I_{A_2} = I_{A_1} - I_{A_2} \neq 0. \end{align*}

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