Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 28 / 2

For each fixed $t\in [0,1]$, the variables $1_{\{X_k\le t\}}$ are independent and identically distributed random variables with Bernoulli distribution of parameter $t$. Their expected value is $t$, and the strong law of large numbers gives $F_n(t)\to t$ almost surely.

For a finite set $F$, intersect these probability-one events over its finitely many elements. On the resulting single event, every one of the finitely many errors tends to zero, so their maximum tends to zero:If $F$ is empty, take the displayed maximum to be zero. The important point is simultaneous convergence on one event; pointwise convergence at an uncountable collection of arguments would not alone imply the uniform claim in part (ii).

For each positive integer $m$, apply part (i) to the grid $D_m=\{j/m:0\le j\le m\}$. Intersect the resulting probability-one events over all $m$, a countable intersection. Work on this common event.

For $x\in [j/m,(j+1)/m]$, monotonicity of the empirical distribution function gives $F_n(j/m)\le F_n(x)\le F_n((j+1)/m)$. ConsequentlyandThe endpoint $x=1$ is itself on the grid. ThereforeFor each fixed $m$, take the upper limit as $n\to \infty$; part (i) makes the maximum vanish. Then let $m\to \infty$. This provesThis is the uniform-sample case of the Glivenko-Cantelli theorem, proved by a finite-grid approximation rather than assumed.

Let $G$ be the cumulative distribution function of the arbitrary Borel probability measure $\mu$. It may have atoms or flat intervals. Use its quantile functionThe limits of $G$ at the two infinities make $Q(u)$ finite. Its monotonicity and right continuity imply the exact equivalence $Q(u)\le x⟺u\le G(x)$. For the nontrivial direction, if $Q(u)\le x$, points with $G\ge u$ lie arbitrarily close above $Q(u)$; if the infimum itself is $x$, right continuity gives $G(x)\ge u$. The other direction follows directly from the defining infimum.

Take independent uniform variables $U_k$ and set $Y_k^{\prime }=Q(U_k)$. Thenso $(Y_k^{\prime })$ has the same entire sequence law as the original iid sample $(Y_k)$. On this construction, the empirical distribution function satisfiesThus part (ii) givesThe supremum is measurable: since both distribution functions are right-continuous, it equals the supremum over rational $x$. The convergence event consequently depends measurably on the coordinate sequence, and equality of the sequence laws transfers its probability one to $(Y_k)$. HenceThis is the Quantile proof of the Glivenko-Cantelli theorem, including arbitrary distributions with atoms. The original PDF supplies the full definition and absolute-value signs in this part, which the TeX conversion incompletely records.

A real Gaussian process $(Z_t{)}_{t\in T}$ is a family of random variables such that every finite vector $(Z_{t_1},\dots ,Z_{t_d})$ has a multivariate normal distribution, allowing degenerate distributions. Equivalently every finite real linear combination is a Gaussian random variable. Its mean and covariance kernel are $m(t)=\mathbb{E}{Z}_t$ and $C(s,t)=\mathbb{E}[(Z_s-m(s))(Z_t-m(t))]$.

For any finite set of indices, these functions give the whole mean vector and covariance matrix. The characteristic function of the vector isThe standard uniqueness theorem for characteristic functions therefore determines every finite-dimensional distribution, including the singular cases. Cylinder events form a generating class for the product sigma-algebra on ${\mathbb{R}}^T$ and are closed under finite intersections, so agreement on all cylinder events gives agreement of the process laws by the uniqueness theorem for probability measures. Thus the mean and covariance determine the law on this coordinate sigma-algebra even when $T$ is uncountable. This statement does not claim uniqueness on an arbitrary finer sigma-algebra of path properties.

Let $(B_t{)}_{0\le t\le 1}$ be standard Brownian motion with its continuous version, and defineEvery finite vector is a linear transform of a Gaussian Brownian vector, so $Z$ is a centred Gaussian process. Using $\mathbb{E}[B_s{B}_t]=\min (s,t)$,For $s\le t$, this is $s(1-t)$, the prescribed Brownian bridge covariance kernel. Its paths are continuous on the probability-one event of Brownian continuity, because $t\mapsto t{B}_1$ is continuous on every sample. The construction also gives $Z_0=Z_1=0$. Thus it realizes a Brownian bridge with almost surely continuous paths, and part (i) identifies its process law.

Fix times $t_1,\dots ,t_d\in [0,1]$, and setThese are iid bounded centred random vectors, andTheir covariance matrix isWe prove the required central limit theorem calculation explicitly. For $u\in {\mathbb{R}}^d$, the scalar $u\cdot V_1$ is bounded, has mean zero and second moment $u^{T}Cu$. The Taylor expansion of the exponential, with its uniformly bounded cubic remainder, givesIndependence implies that the characteristic function of the sum vector is the $n$th power of this expression, so it tends to $\exp (-u^{T}Cu/2)$. This is the characteristic function of a centred multivariate normal distribution with covariance $C$. The Lévy continuity theorem states that convergence of characteristic functions to a characteristic function continuous at zero gives convergence in distribution; apply it here.

By part (ii), the corresponding Brownian bridge vector has exactly this covariance and mean. ThereforeRepeated times or the endpoint times may make $C$ singular; the characteristic-function argument still applies. This proves every finite-dimensional distribution convergence requested, the uniform case of the Finite-dimensional Brownian bridge limit of an empirical distribution process.