Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 8 / b

The central limit theorem states that if $W_1,W_2,\dots$ are independent and identically distributed with finite mean $\mu$ and variance $0<{\tau }^2<\infty$, thenHere $\stackrel{d}{\to }$ denotes convergence in distribution.

The algebraic identity ${\hat{\beta }}_0-{\beta }_0=n^{-1}{\sum }_i{ϵ}_i$ does not require a normal distribution for the errors. Applying the central limit theorem to their sample mean givesThus, for large $n$,Its mean ${\beta }_0$ and variance ${\sigma }^2/n$ remain exact, even though its normal distribution is now only an approximation. With the error probability distribution unspecified, the formula is retained as an estimator; it is not being asserted to remain a maximum likelihood estimation formula.

No, not from the stated assumptions alone. The centered slope is a weighted sum,and the weighted errors need not be identically distributed. The ordinary central limit theorem therefore does not apply directly.

For a concrete counterexample, for every $n\ge 3$ chooseThese satisfy all three centring and orthogonality assumptions, and both explanatory columns are nonzero. Let the errors be independent with Rademacher distribution, taking values $\pm 1$ with equal probability. Thenwhich takes values $-1,0,1$ with probabilities $1/4,1/2,1/4$ for every $n$. Increasing the sample size adds no information about this slope, and the standardized estimator does not approach a normal distribution.

A useful additional assumption for asymptotic normality of a weighted regression slope is that no single design value dominates:Under this condition and the given finite error variance, the Lindeberg-Feller central limit theorem yieldsjustifying the approximation $N({\beta }_1,{\sigma }^2/S_{xx})$ for the slope as well.