Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 4 / 17H (Statistics)

Write the linear mean model asThe errors are independent but heteroscedastic. Minimizing the unweighted residual sum of squaresgives the normal equationThus, provided ${\sum }_i{x}_i^2>0$,Since $\mathbb{E}{Y}_i=\theta x_i$,Hence it is an unbiased estimator. This is the least-squares part of the estimators for a Poisson exposure model.

Ignoring terms independent of $\theta$, the log-likelihood isIts score function isSince ${\ell }^{\prime \prime }(\theta )=-({\sum }_i{Y}_i)/{\theta }^2\le 0$, the interior critical point is the maximum, givingThe sum of independent Poisson distributions is Poisson with mean $\theta {\sum }_i{x}_i$, soThus the maximum-likelihood estimator is also unbiased.

PutIndependence and $\mathrm{var}(Y_i)=\theta x_i$ givewhereas ${\sum }_i{Y}_i\sim \mathrm{Poisson}(\theta S_1)$ givesThe Cauchy-Schwarz inequality, applied to $(x_i^{1/2})$ and $(x_i^{3/2})$, yieldsConsequentlyEquality holds exactly when all positive exposures $x_i$ are equal. This is the variance comparison for Poisson exposure estimators.

Both alternatives have larger $\theta$, so use upper-tail rejection regions. Let $z_{0.95}={\Phi }^{-1}(0.95)\simeq 1.645$ be the 95th quantile of the standard normal distribution.

For the likelihood estimator,When $S_1$ is large, the normal approximation to the Poisson distribution gives, under $H_0$,An approximate size-$0.05$ test therefore rejects $H_0$ when

When every $x_i$ is large, $Y_i\approx N(\theta x_i,\theta x_i)$ independently. A linear combination of independent normal variables is normal, so under $H_0$,The corresponding approximate size-$0.05$ test rejects whenIn both cases the null rejection probability is approximately $0.05$, while values near the alternative mean $2$ increasingly fall in the rejection region as the total exposure grows. These are the normal-approximation tests for a Poisson exposure model.