We now relax the OLS distributional assumption, assumption 6. This is the assumption that the errors $u$ are normally distributed that we first used to obtain a distribution of $\hat\beta$ expressed as the sampling error. We only make use of the “classic” assumptions of the OLS:
$E(u_t) = 0, \quad E(u_t^2) = \sigma^2 < \infty.$ i.e. vector of errors $E(\mathbf{uu’}) =\sigma^2I_n$
Assume $u_t \sim \text{IID}$. We know $\widehat{\beta} =(X’X)^{-1} X’ y = \beta + (X’X)^{-1}X’u$
$\widehat\beta$ collapses onto a single point as $ n \rightarrow \infty$. Therefore investigate asymptotic normality (X non-stochastic):
$$ \therefore \sqrt{n}(\hat\beta – \beta) = \sqrt{n}(X’X)^{-1}{X’u}$$
$$ = \dfrac{n}{\sqrt{n}}(X’X)^{-1}{X’u}$$
$$=\biggl(\underbrace{\dfrac{1}{n} X’ X}_{2 \xrightarrow{p}Q} \biggr) \biggl(\underbrace{\dfrac{1}{\sqrt{n}} X’ u}_{1 \xrightarrow{d} N( 0, \sigma^2 Q)} \biggr) $$
\begin{align*}\dfrac{1}{\sqrt{n}}\sum_{t=1}^n x_t u_t \xrightarrow{d} N( 0, \sigma^2 Q)\Rightarrow \mathsf{E}(x_t u_t) \\
= \mathsf{E}(x_t) \mathsf{E}(u_t) = 0\equiv \mu_t\end{align*}
2. Assumption #9: Regressors are “well-behaved”
\begin{align*}& \lim_{n \to \infty} \biggl(\frac{1}{n} X’ X \biggr) = Q\\
\Omega &= \lim_{n \to \infty} \frac{1}{n}\sum_{t=1}^n \text{var} (x_t u_t) – \mu_t\\
&=\lim_{n \to \infty} \dfrac{1}{n}\sum_{t=1}^n \biggl(\sigma^2 \mathbf{x_t x_t’}\biggr)-0= \sigma^2Q
\end{align*}
if X is stochastic, we have plims above (instead of lims) and
\begin{align*}
\Rightarrow E(\mathbf{x_t x_t’}) = Q,\; E(\mathbf{x_t x_t’}) E(\mathbf{uu’})= \sigma^2 Q\\
\therefore \sqrt{n}(\hat\beta – \beta) \xrightarrow{d} Q^{-1} N( 0, \sigma^2 Q) \end{align*}
by the Cramer Linear Transformation Theorem.
$$\therefore \sqrt{n}(\hat\beta – \beta) \xrightarrow{d} N( 0, \sigma^2 Q^{-1} Q Q^{-1}) $$