Past exam of the options II course of the University of Oxford / 2012 / O (Statistics)

The sample mean has expectation $\theta$ and variance ${\theta }^2/n$. For $g(z)=\log z$, $g^{\prime }(\theta )=1/\theta$. The first-order delta-method expansion isTaking its mean and variance givesThese are large-sample approximations. Retaining the quadratic Taylor term would add the leading mean correction $-1/(2n)$, but it does not change the requested first-order expression.

For nonnegative integers $j,k$, repeated integration by parts gives the beta integral ${\int }_0^1{y}^j(1-y{)}^{k}dy=j!k!/(j+k+1)!$. Multiplying the stated density by $y$ and using this formula yieldsCancellation gives

The same beta integral with one additional power of $y$ givesSubtracting the square of the mean yieldsAt $r=1$ and $r=n$, therefore,Their equality also follows by reflecting every uniform observation via $y\mapsto 1-y$.

For the shifted exponential model, $F(x)=1-e^{-\lambda (x-1)}$ for $x\ge 1$. It is increasing, so the probability integral transform preserves order: $F(X_{(r)})=Y_{(r)}$. The expected uniform plotting positions are $r/(n+1)$ by part (b). Replacing the random transformed order statistic by this representative position givesThus plotting $x_{(r)}$ against $t_r=-\log (1-r/(n+1))$ should give an approximate line with intercept one and slope $\boxed{1/\lambda }$.

The replacement is a quantile-plot approximation, not an assertion that expectation commutes exactly with the nonlinear inverse distribution function. The random points need not lie exactly on the line.

For the Pareto model, integration gives $F(x)=1-x^{-\lambda }$ on $x\ge 1$. The same plotting positions suggest $1-x_{(r)}^{-\lambda }\approx r/(n+1)$. Taking logarithms givesTherefore plot $\log x_{(r)}$ vertically against $t_r=-\log (1-r/(n+1))$ horizontally. A reasonable fit should be approximately a straight line through the origin, with slope $\boxed{1/\lambda }$. This logarithmic transformation distinguishes the Pareto tail model from the shifted exponential plot in part (i).

Independence and normality give $\overline{X}\sim N(\mu ,1/n)$. Consequently $Z=\sqrt{n}(\overline{X}-\mu )$ is standard normal, andRearranging the event gives $\mu >\overline{X}-z_{\alpha }/\sqrt{n}$. Thus the random interval $(\overline{X}-z_{\alpha }/\sqrt{n},\infty )$ covers the fixed parameter with probability $1-\alpha$. After observing the sample, its realized confidence interval isThe coverage refers to repeated samples, not to a posterior probability for the fixed parameter.

The Neyman-Pearson lemma concerns testing two simple hypotheses with densities $f_0$ and $f_1$ relative to a common measure. A test that rejects when $f_1(x)>k{f}_0(x)$, accepts when $f_1(x)<k{f}_0(x)$, and randomizes on equality if necessary to achieve size $\alpha$, is most powerful among all tests with size at most $\alpha$:Here $\phi$ is the described likelihood-ratio test and test functions take values in $[0,1]$. In the continuous normal setting below the boundary has probability zero, so randomization is unnecessary.

The likelihood ratio between the two normal means isSince ${\mu }_1>{\mu }_0$, it is strictly increasing in $\overline{x}$. The Neyman–Pearson test therefore rejects for large sample means. Under the null, $\sqrt{n}(\overline{X}-{\mu }_0)\sim N(0,1)$, so choosing a right-tail probability $\alpha$ givesThis has size exactly $\alpha$ because the sampling distribution is continuous.

For every particular alternative mean ${\mu }_1>{\mu }_0$, the likelihood ratio is increasing in the same statistic $\overline{x}$. Its size-$\alpha$ threshold is fixed solely by the null distribution, so it is always the same $c_{\alpha }$ found in part (c), independent of ${\mu }_1$.

The single rejection region is therefore most powerful against every member of the composite alternative. Hence the test is uniformly most powerful for $H_1:\mu >{\mu }_0$.

For the observed mean, the one-sided $p$-value is the null probability of a sample mean at least as large:Standardization under the null gives

The standard normal distribution function is strictly increasing. Thus $p>\alpha$ impliesRearranging givesThis is the confidence-interval/test duality for the matching one-sided procedures. The strict inequalities agree with the printed open interval.