Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 18H (Statistics)

Write $\bar{X}=n^{-1}{\sum }_i{X}_i$ and $\bar{Y}=n^{-1}{\sum }_i{Y}_i$. Completing the square in each sample, the log likelihood is, up to a data-dependent constant,The maximum-likelihood estimators follow because it is strictly maximized over unrestricted means at $(\bar{X},\bar{Y})$. Under the null it is maximized at $(\bar{X},0)$. Thus the likelihood ratio is $\Lambda =\exp (-n{\bar{Y}}^2/2)$, and the likelihood-ratio test rejects for large $T=-2\log \Lambda =n{\bar{Y}}^2$.

Under every null value of ${\mu }_X$, $Z=\sqrt{n}\bar{Y}$ has the standard Gaussian distribution; hence $T$ has the chi-squared distribution with one degree of freedom. Let $z_p={\Phi }^{-1}(p)$, where $\Phi$ is the standard Gaussian cumulative distribution function. The exact size-$\alpha$ test isThe null rejection probability is $2[1-\Phi (z_{1-\alpha /2})]=\alpha$, independently of the nuisance mean ${\mu }_X$. No randomization is required because the statistic has a continuous distribution.

The unrestricted maximum-likelihood estimators remain $(\bar{X},\bar{Y})$. Subject to the null constraint, minimizing $({\mu }_X-\bar{X}{)}^2$ on $[A,\infty )$ gives ${\hat{\mu }}_X=\max (A,\bar{X})$, while ${\hat{\mu }}_Y=0$. ConsequentlyWe must calibrate this likelihood-ratio test over the entire composite null. Under that null let $W=\sqrt{n}(\bar{X}-A)$ and $Z=\sqrt{n}\bar{Y}$. Then $W\sim N(\delta ,1)$ with $\delta =\sqrt{n}({\mu }_X-A)\ge 0$, while $Z\sim N(0,1)$ independently, and $T=Z^2+(-W{)}_{+}^2$. Couple all null distributions by $W=W_0+\delta$, with a fixed standard Gaussian $W_0$. Increasing $\delta$ decreases $(-W_0-\delta {)}_{+}^2$ pointwise, so the largest rejection probability occurs at $\delta =0$, namely ${\mu }_X=A$.

At this boundary the sign of $W$ is independent of $W^2$, with each sign having probability one half. On $W\ge 0$, $T=Z^2\sim {\chi }_1^2$; on $W<0$, $T=Z^2+W^2\sim {\chi }_2^2$. Thus the least favourable null hypothesis law is the exact chi-bar-squared distribution $\frac{1}{2}{\chi }_1^2+\frac{1}{2}{\chi }_2^2$. Choose $c_{\alpha }>0$ byThe tail is a continuous function and strictly decreasing from one to zero, so this threshold is unique for $0<\alpha <1$. The null rejection probability is at most $\alpha$ and equals $\alpha$ at ${\mu }_X=A$. This proves the required size, rather than using an interior chi-squared approximation.

To compare, put $c_1=z_{1-\alpha /2}^2$, the threshold in (i). Since ${\chi }_2^2$ is stochastically larger than ${\chi }_1^2$ (add an independent squared standard Gaussian), the mixture threshold obeys $c_{\alpha }>c_1$. When $\bar{X}\ge A$, test (ii) therefore requires stronger evidence from $\bar{Y}$ than test (i). When $\bar{X}<A$, test (ii) also responds to violation of the mean constraint and can reject even with $\bar{Y}=0$. Neither critical region contains the other. In the coordinates $(W,Z)$, the acceptance region in (i) is the strip $|Z|\le \sqrt{c_1}$; in (ii) it is the right half-strip $W\ge 0$, $|Z|\le \sqrt{c_{\alpha }}$ together with the left half-disk $W<0$, $W^2+Z^2\le c_{\alpha }$. The extra constraint changes both the statistic and its correct null calibration.