Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 (Advanced Probability)

Put $q=1-p$. Sincethe martingale condition isThe root in $(0,1)$ is therefore

Let $\tau =T_a\wedge T_b$. The walk exits the finite interval $[a,b]$ almost surely, and the stopped martingale $\varphi (S_{n\wedge \tau })$ is bounded between ${\lambda }^b$ and ${\lambda }^a$. The optional stopping theorem and bounded convergence theorem giveSolving for the first probability gives the biased gambler's ruin probability

As $b\to \infty$, the events $\{T_a<T_b\}$ increase to $\{T_a<\infty \}$: every path that reaches $a$ has a finite maximum before doing so. Since ${\lambda }^b\to 0$,Similarly, as $a\to -\infty$, the events $\{T_b<T_a\}$ increase to $\{T_b<\infty \}$, andHence

The strong law of large numbers givesBecause $0<\lambda <1$, it follows that ${\lambda }^{S_n}\to 0$ almost surely. If this martingale were uniformly integrable, almost-sure convergence would imply convergence in $L^1$, and thereforeBut the martingale has constant expectation $\mathbb{E}[{\lambda }^{S_n}]=1$. This contradiction proves that

Write $v_m=\mathrm{Var}(X_m)={\sigma }_m^2$. Independence and $\mathbb{E}[X_{n+1}]=0$ giveThus

Put $V=\mathrm{Var}(S_n)={\sum }_{m=1}^n{v}_m$. If ${\max }_{m\le n}{S}_m\ge x$, thenThe Doob maximal inequality for a nonnegative submartingale therefore givesThe right side is minimized at $c=V/x$, and substitution gives

Let $V_n=\mathrm{Var}(S_n)={\sum }_{m=1}^n{v}_m$. ThenConsequently

LetBecause $|X_m|\le K$, one always has $|S_{\tau }|\le x+K$: this is clear if no crossing occurs, and at the first crossing the overshoot is at most one increment. Apply the optional stopping theorem to the martingale from part (c):On the event $\{{\max }_{m\le n}|S_m|\le x\}$ one has $\tau =n$, so $V_{\tau }=V_n$. Since $V_{\tau }\ge 0$ everywhere,Hence

Since $X_1$ is uniform on $\{-1,1\}$,For $|x|<1$, the supremum in the Legendre transform of a cumulant-generating function is attained whereSubstitution gives the Rademacher large-deviation rate functionfor $|x|\le 1$, with $0\log 0=0$; it is $+\infty$ for $|x|>1$.

By the definition of exponential tilting,On $A_n$ one has $S_n\le cn$, so for $\lambda \ge 0$,Therefore

Fix $a<b<c<1$ and chooseUnder ${\mathbb{P}}_{\lambda }$, the increments remain independent and identically distributed, with mean ${\psi }^{\prime }(\lambda )=b$. The strong law of large numbers therefore givesPart (b) impliesLet $c\downarrow b$ and then $b\downarrow a$. Since $\lambda b-\psi (\lambda )={\psi }^{\ast }(b)$ and ${\psi }^{\ast }$ is continuous on $[0,1)$,The same argument includes $a=0$ by taking $b\downarrow 0$.

An ${\mathbb{R}}^d$-valued process $(X_t{)}_{t\ge 0}$ is a Brownian motion started at $x$ when:

The paths of $U{X}_t$ are continuous, and linear transformation preserves independence of increments. Moreover,because $U$ is an orthogonal matrix. Thus all defining properties are preserved, proving the orthogonal invariance of Brownian motion:

Fix a closed ball $\overline{B}(x,r)\subset D$ and let $\tau$ be its first exit time. By the Strong Markov property, conditioning at $\tau$ givesThe orthogonal invariance of Brownian motion implies that $X_{\tau }$ is uniformly distributed on the sphere $\partial B(x,r)$. HenceThus $\varphi$ has the mean value property on every ball compactly contained in $D$. Since $0\le \varphi \le 1$, the mean-value characterization of harmonic functions yieldsThis function is the harmonic measure of $A$ viewed from $x$.

The required boundary values on the real axis are zero to the left of the origin and one to the right. The bounded harmonic function with those values is the upper-half-plane harmonic measure of the positive half-axis,Indeed, $\arg z$ is harmonic in the upper half-plane, and the displayed function tends to $1$ on the positive half-axis and to $0$ on the negative half-axis. The uniqueness of bounded solutions of the Dirichlet problem identifies it with the Brownian exit probability.

Planar Brownian motion is recurrent. More explicitly, the planar Brownian annulus hitting probability giveswhen $|x|>1$. Thus the unit disc is hit almost surely from every starting point. Applying the Strong Markov property after each departure and return shows that such returns occur after arbitrarily large times. Therefore

By the Tonelli theorem and the planar Brownian transition density,For $s\ge 1$,while for $0\le s\le 1$ it is at most $\Vert f{\Vert }_{\infty }$. Hencefor $t\ge 1$, and consequently

Because $f$ is a continuous probability density, there are a point $z$, a radius $r>0$, and $ϵ>0$ such thatBy the recurrence of planar Brownian motion, the smaller disc $B(z,r/2)$ is visited at arbitrarily large times. Starting anywhere in that smaller disc, Brownian continuity and compactness give a uniform probability $\delta >0$ of staying in $B(z,r)$ for a fixed time $u>0$.

Apply the Strong Markov property at successive visits separated by at least $u$. The conditional probability of each stay event is at least $\delta$, so the conditional Borel-Cantelli lemma gives infinitely many successful stays almost surely. Every success adds at least $ϵu$ to $A_t$. Since $A_t$ is nondecreasing,

The count $N_t=M(0,t]$ is a rate-$\lambda$ Poisson process. Over a time interval $(s,t]$, the incrementdepends only on the Poisson points and marks in that interval. Disjoint intervals give independent increments, and the distribution depends only on $t-s$. The paths are càdlàg step functions, $X_0=0$, andThus $X$ is stochastically continuous and

For $g\ge 0$, the Tonelli theorem and conditioning on $N_t$ giveSince $\mathbb{E}[N_t]=\lambda t$,with both sides allowed to be $+\infty$.

A martingale must be integrable. On the event $\{N_t=1\}$, $X_t=g(Y_1)$, so integrability of $X_t$ forcesFor $s<t$, independent increments giveTherefore the necessary and sufficient condition is

The joint process is a two-dimensional Compound Poisson process whose Lévy measure isTwo coordinates of a Lévy process are independent exactly when its Lévy measure charges only the coordinate axes and its Gaussian covariance has no cross term. Here there is no Gaussian part, so independence is equivalent toSince $g_1{g}_2$ is continuous, this is equivalent to pointwise vanishing. Conversely, when the product vanishes, the mark sets where $g_1$ and $g_2$ are nonzero are disjoint; independent thinning of the Poisson random measure gives independent coordinate processes. Hence