Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 220 / 2

For $u\ge 0$, let ${\mathcal{C}}_u$ be the collection of interval components of the superlevel set $\{r:g(r)\ge u\}$. Two times $s,t$ lie in the same member of ${\mathcal{C}}_u$ exactly when $u\le m_g(s,t)$. ThereforeFor arbitrary real ${\lambda }_1,\dots ,{\lambda }_n$, the Tonelli theorem now givesThus $(m_g(s_i,s_j){)}_{i,j}$ is a positive semidefinite matrix.

The head of the Brownian snake driven by $g$ is the centered Gaussian process $(Z_t{)}_{0\le t\le 1}$ with covariance function $\mathbb{E}[Z_s{Z}_t]=m_g(s,t)$. Part i shows that these finite-dimensional distributions exist consistently. Moreover,If $g$ has Hölder constant $L$, then $d_g(s,t)\le 2L|t-s{|}^{\alpha }$. The absolute moment formula for a centered normal distribution consequently gives, for every $p\ge 2$,The Kolmogorov continuity theorem, with $p$ arbitrarily large, produces a modification that is $\gamma$-Hölder continuous for every $\gamma <\alpha /2$. Taking $\gamma =\alpha /2-\epsilon$ proves the claim whenever $0<\epsilon <\alpha /2$; for larger $\epsilon$ the assertion is vacuous.

The set ${\mathcal{Q}}_n$ consists of rooted planar maps with $n$ faces, every face having degree four. The set ${\mathcal{Q}}_n^{\bullet }$ consists of these quadrangulations with an additional distinguished vertex.

For the trivial bijection between planar maps and quadrangulations, start from a rooted planar map with $n$ edges. Put a new vertex in every face and join it to the original vertex at every incident corner. Delete the original edges. The two endpoints of each deleted edge and the new vertices in its two adjacent faces bound a quadrangular face, so the result lies in ${\mathcal{Q}}_n$. Its bipartition distinguishes old from new vertices and reconstructs the original map. With the standard root convention this is a bijection, and hence

Consider an occurrence counted by $G_m$. Before the next counted occurrence, the exploration must first take a downward step of the simple symmetric random walk, which has probability $1/2$, and then choose the decrement $-1$ among the three equally likely values of $\xi$, which has probability $1/3$. Thus it terminates the visits to the current record value with probabilityThe Strong Markov property at successive counted occurrences makes these trials independent. Therefore $G_m$ has the geometric distribution on $\{1,2,\dots \}$ with parameter $1/6$, andfor every $j\ge 1$.

In the standard labelled encoding of a pointed planar quadrangulation, incidences at the distinguished vertex are represented by visits counted by the record variables $G_m$. Consequently its degree of a vertex is bounded by the largest such count encountered before the coding walk first reaches $-1$.

Before time $2n+1$ there are at most $2n+1$ possible record levels. The union bound and part i therefore implyConditioning on $\sigma =2n+1$ and using the supplied lower bound givesSet $r=C\log n$. If $C>5/(2\log (6/5))$, the right-hand side tends to zero. Hence for some constant $C>0$,