Regression (in a wide sense), homoskedasticity and independence of error term

Question

I suspect this might have a simple answer, but I have been stuck on it for a while. It is a simple true or false question. I suspect it to be false, but I haven't come up with a counterexample. At this point, I feel more confused than I did before I started the problem.

If $Y = X'\beta + e$, $\mathbb{E}[e|X] = 0$ and $\mathbb{E}[e^{2}|X] = \sigma^{2}$, is it true that $e$ is independent of $X$?

I have tried to find an example satisfying the above assumptions with $\mathbb{E}[Xe] \neq 0$, since that would imply dependence. I have also tried to think about an error term that is some function of $X$, so not independent of $X$, but still satisfying the above assumptions. That has also been unsuccessful. For anyone wondering, the problem is from Bruce Hansen's textbook Econometrics (exercise 2.14), the august 2021 edition. Note that there are no further assumptions, such as for example $\beta$ being the OLS coefficient.

Thanks in advance to anyone that took the time to read this request. Any answers will be much appreciated.

Edit:

I suppose on could try the following: Define $e = XU$, where $U$ is a random variable satisfying $\mathbb{E}[U|X] = 0$ and $\mathbb{E}[U^{2}|X] = \frac{1}{X^{2}}\sigma^{2}$. Then the assumptions of the question are satisfied, but I would think that for most definitions of $U$ then $e$ depends on $X$. The question still remains as to whether such a $U$ exists.

Answer 1

Welcome here, nice question!

The assumption $\mathbb{E}[e|X] = \mathbb{E}[e] = 0$ is called mean independence.

Independent variables are always mean independent. The reverse does however not hold in general, see this Cross Validated question here. There are also examples mentioned in which case variables are mean independent but not independent.

Mean independence in turn implies uncorrelatedness, see here, but again not the reverse.

Answer 2

This might be an answer to my question. Consider the following (artificial) setup: Simplify to the case where $X$ is only a 1-dimensional random vector. Let $\Omega$ denote the sample space of $X$. If $X\in A$ then $e$ takes the values $1$ or $-1$ with equal probability. If $X\in \Omega\setminus A$ then $e\sim N(0,1)$. We then have $\mathbb{E}[e|X] = 0$ and $\mathbb{E}[e^{2}|X] = 1$. However, $e$ is not independent of $X$. It follows that the statement in the original post is false.

I am grateful to Arne Jonas Warnke for his reply. Should my own answer be correct, I still want to award him points by accepting this answer if this is not, for some reason, considered wrong. All feedback on this answer will be much appreciated. Thank you in advance to anyone that spends their time on a comment.