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We know one of the Gauss-Markov conditions is $\mathrm{Cor}[e_i, e_j]=0$ for $i \neq j.$

Also, we assume the errors are normally distributed, i.e., $e_i \sim \mathcal{N}(0, \sigma^2).$

My teacher said that the normality assumption implies that the errors are independent since they are uncorrelated.

However, what I know is only multivariate normal distribution with zero correlation can imply independent. I do not know why we can draw the conclusion that $e_i$ are independent.

Thanks!

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  • $\begingroup$ Saying the errors are individually (i.e. marginally normal) is insufficient. They have to be jointly normal for the implication to hold, as you say. This is covered in a one or two previous posts on site; I'll try to dig one up. $\endgroup$
    – Glen_b
    Oct 6, 2019 at 7:13
  • $\begingroup$ For the bivariate case a counterexample is mentioned in the answer here: stackoverflow.com/c/moderators/questions/1316/1584#1584 ... that's not the post I was looking for though $\endgroup$
    – Glen_b
    Oct 6, 2019 at 7:29

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You're right: it is when the distribution is jointly Gaussian that a lack of correlation implies independence. Our late friend BruceET gives a good visual demonstration of that here.

Consequently, if your teacher only mentioned marginal normality, the claim is not quite true. If you teacher mentioned multivariate normality, however, the claim is true.

In defense of your teacher, it is not so unusual to write something like $\epsilon\sim N(0,\sigma^2I)$ to indicate joint normality of the errors, so it is reasonable to think that you teacher left implicit the joint error normality.

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