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I'm told that it's possible that $Cov(x_i, \epsilon_i) = 0$ and $E[\epsilon] =0$ yet $E[\epsilon | X] \neq 0$ but I haven't been able to find an example. My professor proposed $\epsilon_i = x_i^2$ where $x_i \sim \text{uniform}(-1,1)$ but I realized that this fails $E[\epsilon_i] = 0$.

If we got rid of the assumption $E[\epsilon]=0$ than this satisfies $Cov(x_i, \epsilon_i) = 0$ yet $E[\epsilon | X] \neq 0$ but the task is to include $E[\epsilon]=0$.

The suggested pages do not provide a concrete example. @Whuber provides a nice image which appears to be of $\epsilon_i = x_i^3$ where $x_i \sim \text{uniform}(-2,2)$ but this doesn't work out:

Let $\epsilon_i = x_i^3$ where $x_i \sim \text{unif}(-2,2)$. We have: \begin{align} \begin{split} E[\epsilon_i] &= \int_{-2}^2 \frac{1}{4} x_i^3 dx \\ &= \frac{1}{16} x_i^4 \Big|_{-2}^2 \\ &= \frac{1}{16}((2)^4 - (-2)^4) \\ &= 0 \end{split} \end{align} But for the covariance, we have: \begin{align} \begin{split} Cov(\epsilon_i, x_i) &= E[\epsilon_i, x_i] + E[\epsilon_i]E[x_i]\\ &= E[(x_i^3) x_i] + 0E[x_i] \\ &= E[x_i^4] \\ &= \int_{-2}^2 \frac{1}{4}x^4 dx \\ &= \frac{1}{20}x^5 \Big|_{-2}^2 \\ &= \frac{1}{20}(2^5 - (-2)^5) \\ &\neq 0 \end{split} \end{align} Am I mistaken?

Professor gave me a solution which was just demeaning the variable. (Should have thought of doing that sooner.)

Let $\epsilon_i = x_i^2 - E[x_i^2]$

financial_physician
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  • read here: https://stats.stackexchange.com/questions/445883/can-zero-covariance-and-zero-expectation-imply-zero-conditional-expectation/517066#517066 – markowitz Apr 18 '21 at 21:16
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    Flagging as duplicate using above link: Does this answer your question? [Can zero covariance and zero expectation imply zero conditional expectation?](https://stats.stackexchange.com/questions/445883/can-zero-covariance-and-zero-expectation-imply-zero-conditional-expectation) – Arya McCarthy Apr 18 '21 at 21:18
  • I realized the function isn't exactly $y = x^3$ But something closer to $y = x(x-1)(x+1)$ but still can't get the result... Do you know what the function is? – financial_physician Apr 18 '21 at 22:15
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    @financial_physician whuber's picture is an example of samples drawn from a joint dist, and it's not $\epsilon=X^3$. There is even no deterministic relation between $\epsilon$ and $X$. But, $E[\epsilon|X]$ is clearly non-zero for some x. – gunes Apr 18 '21 at 22:33

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