Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 2 / 29J (Principles of Statistics)

The Kolmogorov-Smirnov theorem states that, for continuous $F$,where $B$ is a Brownian bridge; the finite-sample law is also independent of continuous $F$. To test $H_0:F=F_0$, computeand reject when $\sqrt{n}{D}_n$ exceeds the appropriate null quantile.

Split $Z_1,\dots ,Z_{mn}$ into $m$ blocks of size $n$. Compute the same Kolmogorov-Smirnov statistic against $F_0$ for the observed $X$ sample and for every $Z$ block. Under $H_0$ the $m+1$ statistics are exchangeable and, because the distribution is continuous, ties have probability zero. Reject exactly when the statistic from the $X$ sample is the largest. Its null rejection probability is exactly

Let $h=n^{-1/3}$. Away from the boundary, the estimator isIts variance is at mostand the bias of a kernel density estimator is bounded by $h\Vert f^{\prime }{\Vert }_{\infty }/2$, so its squared bias is at most $h^2/4$. This gives the claimed bound, with room to spare, whenever the full window lies in the support.

At $x=0$, however, the assertion is false as printed. For examplesatisfies $\sup (|f|+|f^{\prime }|)=3/4<1$, butwhereas $f(0)=1/2$. The mean-square error therefore tends to $1/16$, contradicting a bound tending to zero. The intended statement needs $x\ge h/2$, a boundary correction, or a condition such as $f(0)=0$.