Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4 / 19C / ii / Solution

Let $\mathrm{RSS}={\sum }_i(Y_i-\hat{\alpha }-\hat{\beta }{x}_i{)}^2$. Differentiating the log-likelihood in $v={\sigma }^2$ givesFor $\mathrm{RSS}>0$ this changes from positive to negative at $v=\mathrm{RSS}/n$, soTo find its bias, substitute the model and use the orthogonal fit directions $1$ and $\mathbf{x}$, obtainingThe three expectations are $n{\sigma }^2$, ${\sigma }^2$, and ${\sigma }^2$, respectively: $\mathrm{Var}(\bar{\epsilon })={\sigma }^2/n$ and $\mathrm{Var}({\sum }_i{x}_i{\epsilon }_i)={\sigma }^2{S}_{xx}$. Hence $\mathbb{E}\mathrm{RSS}=(n-2){\sigma }^2$, giving the unbiased estimatorWith $n>2$, a zero residual sum of squares is a probability-zero event in the model. For that exceptional dataset the likelihood is unbounded as ${\sigma }^2\downarrow 0$, rather than having a maximum at a positive variance. In the intercept-only case $S_{xx}=0$, the corresponding unbiased multiplier is $n/(n-1)$ for $n>1$.