When deriving the OLS estimators, we get to a point where we need to analyze asymptotically the following term:

$n^{-1}\sqrt{n} \sum_{i=1}^n (X_i - \mu_x)u_i$

We'll normally have the weaker assumption that $E(X_iu_i) = 0$, which will imply that $E((X_i - \mu_x)u_i) = 0$.

However, to apply the Central Limit Theorem, we need that the sequence $((X_i - \mu_x)u_i)_{i=1}^n$ be independent and identically distributed (i.i.d).

So my question is, given that the sequence $(u_i, X_i)_{i=1}^n$ is i.i.d by assumption, how can we grant $(u_i, X_i)_{i=1}^n$ is i.i.d?

$E(X_iu_i) = 0$ will only grant uncorrelatedness in this case.

  • $\begingroup$ it is the sampling procedure that grants you that ...You extract the pair $(Y_i,X_i)$ that come from a bivariate (or multivariate in general) distribution. The error is just a linear combination of $Y$ and $X$. If the sampling procedure is done correctly then the observations $(Y_1,X_1), (Y_2,X_2), ... , (Y_n,X_n)$ are i.i.d and also a function of those is formed by independent observation since they are defined on the same sample space (that is a product space) $\endgroup$ – Hard Core Mar 16 '18 at 21:39
  • $\begingroup$ Thank you @HardCore. Do you have a source for that information? Also, the result that any function of those is i.i.d holds? I mean, for example, $(X_i)^8$ + $ln(Y_i)$ will be i.i.d across i? $\endgroup$ – jpugliese Mar 16 '18 at 22:00
  • $\begingroup$ try this .. it explains my brief comment .. I have my lecture notes on measure theory as source, but I think @whuber explanation is more than enough stats.stackexchange.com/questions/94872/… $\endgroup$ – Hard Core Mar 16 '18 at 22:32

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