Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 28J / Solution

Let $F_n(x)=n^{-1}{\sum }_i{1}_{\{X_i\le x\}}$. The Kolmogorov-Smirnov statistic is $D_n={\sup }_x|F_n(x)-F(x)|$, often reported as $\sqrt{n}{D}_n$. Under the simple null, the probability integral transform makes $U_i=F(X_i)$ independent uniform variables on $[0,1]$. Strict monotonicity and continuity identify the empirical distribution under this change of coordinates, soIts null law is therefore independent of $F$.

For the composite null, the joint density is ${\theta }^{-n}{\prod }_i{1}_{\{0\le x_i\le \theta \}}$. Its parameter dependence is only through $M={\max }_i{X}_i$, proving sufficiency by the Fisher-Neyman factorization theorem; the larger statistic $(J,M)$ is sufficient too. Ties have probability zero. For fixed $j$, the density of $(J=j,M=\xi )$ is ${\xi }^{n-1}/{\theta }^n$, $0<\xi <\theta$, because the other coordinates range independently over $[0,\xi ]$. Dividing the joint density by this marginal gives conditional densityThus the remaining observations are conditionally independent $U[0,\xi ]$.

For $n\ge 2$, discard the maximum and apply the Kolmogorov-Smirnov test to the $n-1$ ratios $X_i/M$, $i\ne J$, against $U[0,1]$. Use the ordinary null critical values for sample size $n-1$. Its conditional, hence unconditional, level is independent of $\theta$. Keeping the fitted maximum and using the ordinary $n$-sample critical values would not have that justification.