Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 8

For known ${\sigma }^2>0$, maximizing the normal likelihood over $\mu$ is equivalent to minimizing ${\sum }_i(X_i-\mu {)}^2$. The identityshows that the unique minimum occurs at $\boxed{\hat{\mu }=\overline{X}}$. Equivalently, the log-likelihood has second derivative $-n/{\sigma }^2<0$.

The sample mean is exactly $N(\mu ,{\sigma }^2/n)$, so $(\overline{X}-\mu )/(\sigma /\sqrt{n})$ has the standard normal distribution. The given quantile gives probability $0.95$ to the interval $[-1.96,1.96]$. Inverting this pivot yields the exact $95\%$ confidence interval

Put $D_j=X_j-\overline{X}$. The supplied linear-transformation result shows that $(\overline{X},D_2,\dots ,D_n)$ is jointly multivariate normal. Its marginal distributions areIndeed, $\mathrm{Cov}(X_j,\overline{X})={\sigma }^2/n$, soFor a jointly Gaussian vector, zero cross-covariance between blocks implies their independence. Thus $\overline{X}$ is independent of the entire vector $(D_2,\dots ,D_n)$, not merely of each coordinate separately. The residual coordinates themselves need not be independent: for $j\ne k$, their covariance is $-{\sigma }^2/n$.

Each residual has mean zero and variance ${\sigma }^2(1-1/n)$. By linearity of expectation, without needing independence of residuals,Hence $S^2$ is an unbiased estimator of ${\sigma }^2$.

Since ${\sum }_{i=1}^n{D}_i=0$, we have $D_1=-{\sum }_{i=2}^n{D}_i$. Squaring and adding the other residual squares givesTherefore $S^2$ is a measurable function of $(D_2,\dots ,D_n)$. By the block independence proved in part i, it is independent of $\overline{X}$. This is the independence of the normal sample mean and sample variance; pairwise independence alone would not justify passing to an arbitrary function of several residuals.