Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 210 / 1

Writing $P_{n}g=n^{-1}{\sum }_{i=1}^{n}g(X_i)$ and $Pg=\mathbb{E}g(X)$, the class $\mathcal{G}$ satisfies a uniform law of large numbers whenin probability; a strong ULLN uses almost-sure convergence.

Fix $\epsilon >0$ and choose finitely many brackets $[g_j^L,g_j^U]$ of $L^1(P)$ width at most $\epsilon$. The weak law of large numbers, simultaneously for their finitely many endpoints, givesIf $g_j^L\le g\le g_j^U$, bracketing both $P_{n}g$ and $Pg$ showsTaking the supremum gives a limit superior at most $\epsilon$. Since $\epsilon$ is arbitrary, the ULLN follows.

The Glivenko-Cantelli theorem states that for the empirical distribution function $F_n$ of iid real observations with distribution function $F$,almost surely.

Apply part (a) to $\mathcal{G}=\{1_{(-\infty ,t]}:t\in \mathbb{R}\}$. For each $\epsilon >0$, choose finitely many quantile cutpoints so that the $P$-mass between consecutive cutpoints is at most $\epsilon$, treating atoms as cutpoints themselves. Indicators at adjacent cutpoints give finite $L^1(P)$ brackets of width at most $\epsilon$. Using the strong law for the finite bracket endpoints and then intersecting the probability-one events for $\epsilon =1/m$ gives the almost-sure conclusion.

No. For each realized sample let $A_n=\{X_1,\dots ,X_n\}$. This is a finite Borel set, and continuity of $F$ makes $P(A_n)=0$ almost surely, while $P_n(A_n)=1$. Thereforealmost surely for every $n$, so the class of all Borel-set indicators cannot satisfy a ULLN.