Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 201 (Advanced Probability)

For the simple symmetric random walk, $S_n$ is a martingale and the square-minus-time martingale of a simple symmetric random walk isIndeed, conditioning on ${\mathcal{F}}_n$ and using $\mathbb{E}[X_{n+1}]=0$ and $X_{n+1}^2=1$ gives $\mathbb{E}[S_{n+1}^2\mid {\mathcal{F}}_n]=S_n^2+1$.

Apply the optional sampling theorem for a supermartingale to the bounded stopping time $T\wedge n$:Thus the stopped martingale $(S_{T\wedge n})$ is bounded in $L^2$. The L2 martingale convergence theorem gives convergence in $L^2$, and because $T<\infty$ almost surely its limit is $S_T$. Meanwhile the monotone convergence theorem gives $\mathbb{E}[T\wedge n]\to \mathbb{E}[T]$. Therefore

Finite mean is essential. Let $T=\inf \{n\ge 1:S_n=0\}$ be the first return to zero. The one-dimensional simple symmetric random walk is recurrent, so $T<\infty$ almost surely, but its first-return time has infinite mean. Since $S_T=0$,

The law of the iterated logarithm for a simple symmetric random walk states that almost surelySince $\sqrt{2\log \log n}$ is unbounded, $|S_n|/\sqrt{n}$ exceeds every fixed $c>0$ at some finite time. Hencealmost surely.

Suppose $c>1$ and $\mathbb{E}[T_c]<\infty$. Part (a) would giveBut $T_c\ge 1$ and its defining strict inequality gives $S_{T_c}^2>c^2{T}_c$ almost surely. Taking expectations would yieldwhich is impossible because $c^2>1$. Part (b) shows that $T_c$ is nevertheless finite almost surely. Consequently

Weak convergence of probability measures ${\mu }_n\to \mu$ on a metric space means thatfor every bounded continuous real function $f$. Weak convergence of random variables means weak convergence of their probability distributions; equivalently,for every bounded continuous $f$.

For a bounded measurable $f$, the dual estimate for total variation distance givesThe right-hand side tends to zero, so in particular the integrals converge for every bounded continuous $f$. Thus weak convergence of random variables follows.

The converse fails. On $\mathbb{R}$, let ${\mu }_n={\delta }_{1/n}$ and $\mu ={\delta }_0$. Continuity gives $f(1/n)\to f(0)$, so ${\mu }_n$ converges weakly to $\mu$. However, for $A=\{0\}$,for every $n$, so there is no convergence in total variation distance.

Assume first that $X_n\stackrel{d}{\to }X$. Because all variables take values in the compact interval $[-C,C]$, each monomial $x^k$ agrees there with a bounded continuous function on $\mathbb{R}$. The bounded moment criterion for weak convergence in this case begins withfor every natural number $k$.

Conversely, suppose all moments converge. Let $f$ be bounded and continuous. By the Weierstrass approximation theorem, for every $\epsilon >0$ there is a polynomial $P$ withMoment convergence implies $\mathbb{E}[P(X_n)]\to \mathbb{E}[P(X)]$, whileTaking the limit superior and then $\epsilon \downarrow 0$ proves $\mathbb{E}[f(X_n)]\to \mathbb{E}[f(X)]$, which is convergence in distribution.

Let $g:M^{\prime }\to \mathbb{R}$ be bounded and continuous. The composition $h=g\circ f$ is bounded and measurable, and every discontinuity point of $h$ is a discontinuity point of $f$. ThusThe Portmanteau theorem includes the null-discontinuity criterion: if $X_n\stackrel{d}{\to }X$ and a bounded measurable function is continuous at $X$ almost surely, then its expectations converge. HenceSince this holds for every bounded continuous $g$, it proves the continuous mapping theorem conclusion $f(X_n)\stackrel{d}{\to }f(X)$.

Let $\tau =T_r\wedge T_R$. The function $z\mapsto \log |z|$ is a harmonic function on the annulus $r<|z|<R$, so Itô formula shows thatis a bounded martingale. By the optional sampling theorem for a supermartingale applied to this martingale,where $p={\mathbb{P}}_x(T_r<T_R)$. Solving gives the planar Brownian annulus hitting probability

Each completed visit to radius $r_2$ begins a new radial excursion. By part (a), the conditional probability that the following excursion reaches radius $R$ before radius $r_1$ isThe Strong Markov property at the successive stopping times ${\tau }_{2k}$ makes these trials independent with the same success probability. Consequently $N(R)$ has a geometric distribution on $\{1,2,\dots \}$ with parameter $q_R$:As $R\to \infty$, $q_R\to 0$ andSlutsky theorem now gives

The one-dimensional Donsker invariance principle says that if $X_1,X_2,\dots$ are IID random variables with mean zero and variance one, then the linearly interpolated processconverges weakly in $C([0,1])$ with the uniform norm to standard Brownian motion.

The strong law for Brownian motion gives $B_t/t\to 0$ almost surely. Thereforealmost surely, and hence ${\tilde{B}}_t\to -\infty$ almost surely.

Let ${\sigma }_x=\inf \{t\ge 0:{\tilde{B}}_t=x\}$. Continuity gives $\{S\ge x\}=\{{\sigma }_x<\infty \}$. On this event, the Strong Markov property says thatis an independent Brownian motion with drift $-\mu$. It reaches level $y$ with probability $\mathbb{P}(S\ge y)$. Therefore

Under the Cameron-Martin theorem for a linear drift, the probability that Brownian motion with drift $-\mu$ reaches $x$ isUsing the supplied Laplace transform with $\lambda ={\mu }^2/2$ givesand henceThis is the survival function of the exponential distribution with rate $2\mu$.

Planar Brownian motion and the initial law ${\nu }_{0,R}$ are invariant under every rotation about the origin. The hitting time ${\tau }_{0,r}$ is also rotation invariant, so the law of $B_{{\tau }_{0,r}}$ is invariant under every rotation of the circle of radius $r$. Planar Brownian motion hits that circle almost surely by recurrence of planar Brownian motion. The unique rotation-invariant probability measure on the circle is its uniform measure, and therefore

Reflection in $L$ interchanges $x$ and $y$. Relabel the two points if necessary so that $x$ lies on the side of $L$ containing the center $a$; the absolute difference is symmetric in $x$ and $y$. Couple a Brownian motion $B$ started at $x$ to one $B^{\prime }$ started at $y$ by setting $B_t^{\prime }=\varphi (B_t)$ before $T_L$ and $B_t^{\prime }=B_t$ afterwards. The Brownian reflection coupling has the correct marginal laws because reflection is an isometry and the Strong Markov property applies at $T_L$. Once the paths meet on $L$, they agree forever.

If the path from $x$ does not hit the target circle before $T_L$, the coupled paths meet before the relevant uncoupled hitting outcomes can differ. The coupling inequality for total variation therefore gives

WriteThis is the harmonic measure of $A$ viewed from $z$, and is a bounded harmonic function outside the closed target disc. The disc is contained in $B(0,r_2)$. In the half-plane cut out by a line $L$ through the origin, the probability of reaching $B(0,r_2)$ before $L$, from a point of modulus $R$, is $O(r_2/R)$ uniformly in the direction. One sees this by mapping the half-plane outside the disc with $z\mapsto \log (z/r_2)$ to a half-strip and solving the corresponding Dirichlet problem by a sine series.

Part (b) consequently shows that the angular oscillation of $H_A$ on a large circle tends to zero, uniformly in the Borel set $A$. The same half-strip estimate in the annulus $R-|a|\le |z|\le R+|a|$ shows that shifting the center of that circle by the fixed vector $a$ changes the average of $H_A$ by $o(1)$, uniformly in $A$. Hence, for every $\epsilon >0$, all sufficiently large $R$ satisfyfor every Borel subset $A$ of the target disc.

By part (a), translated by $a$, starting with $B_0\sim {\nu }_{a,R}$ givesPart (c) says that the same hitting law from ${\nu }_{0,R}$ tends in total variation distance to ${\nu }_{a,r_1}$. On the other hand, every path from the circle of radius $R>r_2$ to the target disc must first hit the circle of radius $r_2$. Part (a) and the Strong Markov property show that its position there has law ${\nu }_{0,r_2}$ and thatindependently of $R$. Letting $R\to \infty$ in part (c) proveswhen $B_0\sim {\nu }_{0,r_2}$.

The Martingale convergence theorem states that a discrete-time martingale $(X_n)$ with uniform integrability has an integrable random variable $X_{\infty }$ such thatalmost surely and in $L^1$. Moreover, the martingale is closed by its limit:

For bounded stopping times $S\wedge n\le T\wedge n$, the optional sampling theorem for a supermartingale givesA stopped family drawn from a uniformly integrable martingale is uniformly integrable. Since $X_{T\wedge n}\to X_T$ and $X_{S\wedge n}\to X_S$ almost surely, uniform integrability upgrades both convergences to $L^1$. Passing to the limit yields

Put ${\mathcal{F}}_n=\sigma (X_0,\dots ,X_n)$ and $\tau =T_0\wedge T_y$. Before $\tau$, the integer-valued increment belongs to $\{-1,0,1\}$. The martingale property givesTheir sum is at least $1/2$, so each conditional probability is at least $1/4$. From any state in $\{1,\dots ,y-1\}$, a run of at most $y$ upward moves reaches $y$ and has conditional probability at least $4^{-y}$. Applied in successive blocks of $y$ steps, this givesso $\tau <\infty$ almost surely.

The stopped process $X_{n\wedge \tau }$ takes values in $[0,y]$, hence is a bounded martingale and has uniform integrability. The optional sampling theorem for a supermartingale givesbecause $X_{\tau }$ is zero or $y$. Therefore