Past exam of the options II course of the University of Oxford / 2010 / O / O4

The maximum-likelihood estimator is a maximizerWriting ${\ell }_n=\log L_n$ and the score $S_n={\nabla }_{\theta }{\ell }_n$, the expected Fisher information matrix iswhere the equality uses the usual regularity conditions. For independent identically distributed observations, ${\mathcal{I}}_n=n{\mathcal{I}}_1$.

Under the standard identifiability, interior-parameter, differentiability and nonsingular-information assumptions, the large-sample distribution isor approximately $\hat{\theta }\sim N(\theta ,{\mathcal{I}}_n(\theta {)}^{-1})$. Boundary parameters need not obey this regular approximation.

The two totals are independent, with $n\sim \mathrm{Bin}(2N,q)$ and $m\sim \mathrm{Bin}(2M,p)$. Their joint likelihood isDifferentiating the separated log-likelihood givesThese formulas include all-zero or all-one totals as boundary maxima. For interior $p,q$, the Fisher information in the ordered parameter vector $(p,q)$ isThe off-diagonal entries vanish because the likelihood separates, and the two estimates are independent even at finite sample size. For large $M,N$, their joint distribution is approximately

A linear combination of independent normal variables is normal; the corresponding asymptotic statement follows from the joint central limit theorem and the continuous mapping theorem. ThusThe two variances add because of independence. This approximation assumes both sample sizes grow and the true probabilities are interior.

Under the null, replace both parameters by a common $r$. The likelihood becomesThe pooled maximum-likelihood estimate isUnder an interior null $p=q=r$, part (ii) gives a centered asymptotic normal difference with variance $r(1-r)(1/(2M)+1/(2N))$. The law of large numbers makes $\hat{r}\to r$ in probability. By Slutsky's theorem, replacing $r$ by this consistent estimate givesFor the two-sided alternative, reject when $|Z|>z_{1-\alpha /2}$, the corresponding standard-normal quantile. This has asymptotic significance level $\alpha$.

For a two-sided 5% test, the critical values are $\pm 1.96$. Since $|-1.89|<1.96$, do not reject the null at the 5% level.

For the prespecified directional alternative $q>p$, the numerator $\hat{p}-\hat{q}$ tends to be negative. Use the lower-tail critical region $Z<-1.64$ at the 5% level. Since $-1.89<-1.64$, reject in favor of $q>p$ in the one-sided test. The direction should be chosen from the additional scientific information rather than selected after inspecting the sign of this same statistic.