Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 20H / b / Solution

Write the one-column design as $x\in {\mathbb{R}}^n\setminus \{0\}$ and $S=x^{T}x$. Maximizing the normal linear model likelihood first gives $\hat{\beta }=x^{T}Y/S$. Its minimized residual sum of squares is $R=\Vert Y-x\hat{\beta }{\Vert }^2$. For variance $v>0$, the remaining log likelihood is $-\frac{n}{2}\log v-R/(2v)$ plus a constant, whose maximum is atHere $R>0$ almost surely because $n>1$ and the true variance is positive. In the exceptional data case $R=0$, the likelihood is unbounded as $v\downarrow 0$, so there is no positive-variance maximum.

Let $P=x{x}^T/S$, the orthogonal projection matrix onto the design column. Then $\hat{\beta }-\beta =x^T\epsilon /S$ and $Y-x\hat{\beta }=(I-P)\epsilon$. These are orthogonal linear projections of an isotropic multivariate normal distribution, hence independent. An orthonormal coordinate system adapted to $x$ makes the residual squared length a sum of $n-1$ independent squared standard normals. Thus the joint distribution of least-squares and variance estimators is the product lawIn particular the maximum likelihood variance estimator is biased: $\mathbb{E}{\hat{\sigma }}^2=(n-1){\sigma }^2/n$; the unbiased residual estimator uses denominator $n-1$.