So the question is there are three random variables $X$, $Y$ and $\epsilon$, which the relation can be shown by:

$Y = \beta_0 + \beta_1 X + \epsilon$

Hence see if the following is right and prove why if it is wrong or right

[1] If $E(X\epsilon) = 0$, then $E(X^2\epsilon) = 0$

[2] If $E(\epsilon|X) = 0$, then $E(X^2\epsilon) = 0$

[3] If $E(\epsilon|X) = 0$, then $X$ and $\epsilon$ are independent

[4] If $E(X\epsilon) = 0$, then $E(\epsilon|X) = 0$

[5] If $E(\epsilon|X) = 0$ and $E(\epsilon^2|X) = \sigma^2$, then $X$ and $\epsilon$ are independent

My professor did not really went through error terms in lecture but he sent this as task... The only thing that was written on the lecture note was that

Y and X can be described $Y = E(Y|X) + \epsilon$ ,where

1.$\epsilon$ is mean-independent of X that is $E(\epsilon|X) = 0$

  1. $\epsilon$ is uncorrelated with any function of X

I have no idea what this means at all, but I tried by:

for [1]

$Cov(X,X\epsilon) = E(X^2\epsilon) - E(X)E(X\epsilon)$

Since $E(X\epsilon) = 0$....... Really got stuck here


This one I think I kind of proved it

$E(X^2\epsilon) = E(E(X^2\epsilon|X))$

$E(X^2E(\epsilon|X)) = 0$ ??

Hence, this is true but I am kind of stuck.. here. I don't even understand the concept. I need some help.

  • $\begingroup$ Are you supposed to assume points 1 and 2 when answering [1]-[5]? For example, 1 obviously implies [4] but in general I would have thought [4] need not be true. $\endgroup$
    – Henry
    Commented Apr 21, 2022 at 15:26
  • $\begingroup$ @Henry no those points are independent. each is individual question. I think I know 2 and 5 are true, also based on what I heard from colleagues, but I do not know how to prove it $\endgroup$ Commented Apr 21, 2022 at 16:09


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