Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 201 (Advanced Probability)

A standard Brownian motion is a real-valued process $(B_t{)}_{t\ge 0}$ with $B_0=0$, almost surely continuous paths, and independent increments satisfying $B_t-B_s\sim N(0,t-s)$ for $s<t$.

For a partition of $[a,b]$, every Riemann sum $S_n={\sum }_j{B}_{t_j}\Delta t_j$ is a centered Gaussian random variable, withPath continuity gives $S_n\to {\int }_a^b{B}_{s}ds$ almost surely, and the covariance bound $\mathbb{E}|B_s-B_t{|}^2=|s-t|$ also gives convergence in $L^2$. Hence the limit is Gaussian, centered, and passage to the limit in the displayed sums gives variance

Choose step functions $f_n\to f$ in $L^2[0,1]$. Part (i) shows that $\int B_s{f}_n(s)ds$ is centered Gaussian. Since the kernel $K(r,s)=\min (r,s)$ is bounded,The $L^2$ limit is therefore centered Gaussian, and its variance is

Writing $c_i(t)=⟨1_{[0,t]},f_i{⟩}_{L^2}$ gives $X_n(t)={\sum }_{i=1}^n{\alpha }_i{c}_i(t)$, henceBy Parseval identity, the variance tends to $\Vert 1_{[0,t]}{\Vert }_2^2=t$. Thus $X_n(t)\stackrel{d}{\to }N(0,t)$.

For fixed $t$, $(X_n(t))$ is a martingale with independent centered increments andThe $L^2$ Martingale convergence theorem gives convergence both almost surely and in $L^2$ to a random variable $X(t)$.

For real $u_1,\dots ,u_j$, the preceding series construction givesIt is centered Gaussian, and Parseval identity makes its varianceThis is the variance of ${\sum }_r{u}_r{B}_{t_r}$. The Cramer-Wold theorem proves equality of the two joint distributions.

The Strong Markov property says that for every almost surely finite stopping time $T$, the process $(B_{T+t}-B_T{)}_{t\ge 0}$ is a standard Brownian motion independent of ${\mathcal{F}}_T$.

The process $Y_t=B_{T-t}-B_T$ is centered Gaussian and continuous. Its covariance isas follows by expanding Brownian covariances, or by reading its increments backwards. The Gaussian-process characterization of Brownian motion therefore shows that $(Y_t{)}_{0\le t\le T}$ has the same law as $(B_t{)}_{0\le t\le T}$.

The event $\{B_T=I_T\}$ is the event that $B_{T-t}-B_T\ge 0$ for every $t\le T$. By part (ii), its probability equals the probability that Brownian motion started at zero remains nonnegative throughout $(0,T]$. The Brownian reflection principle implies $\mathbb{P}({\min }_{s\le T}{B}_s\ge 0)=0$. Hence $\mathbb{P}(B_T=I_T)=0$.

Continuity gives existence of a minimizer. If there were two, choose a rational $q$ strictly between them. Then the minima on $[0,q]$ and $[q,1]$ would coincide. Conditional on ${\mathcal{F}}_q$, the latter equals $B_q$ plus the minimum of an independent Brownian motion on $[0,1-q]$, whose distribution is continuous by the Brownian reflection principle. Thus equality has conditional probability zero. Taking the countable union over rational $q$ proves almost-sure uniqueness.

Almost surely, Brownian motion has a unique minimizer on every interval with rational endpoints, by rescaling part (iv). Every local minimum is the minimum on some rational interval contained in a witnessing neighbourhood. Each rational interval contributes at most one point, and there are countably many such intervals. The set of local minima is therefore countable.

The function $z\mapsto \log |z|$ is harmonic on the annulus $\epsilon <|z|<R$. Hence $\log |B_{t\wedge {\tau }_{\epsilon }\wedge {\tau }_R}|$ is a bounded martingale by Itô formula. Optional stopping givesSolving yields

For fixed $y\ne x$, part (i), translated by $y$, and then $\epsilon \downarrow 0$ show that planar Brownian motion has probability zero of ever hitting $y$ before leaving any fixed large disk. Letting the disk radius tend to infinity shows ${\mathbb{P}}_x(y\in B([0,1]))=0$. By Tonelli theorem,The nonnegative random area is therefore zero almost surely.

The conditional expectation $\mathbb{E}[X\mid \mathcal{G}]$ is the almost surely unique integrable, $\mathcal{G}$-measurable random variable $Z$ such that ${\int }_{A}Zd\mathbb{P}={\int }_{A}Xd\mathbb{P}$ for every $A\in \mathcal{G}$.

Conditional expectation is the orthogonal projection from $L^2(\mathcal{F})$ onto $L^2(\mathcal{G})$. The difference $X-\mathbb{E}[X\mid \mathcal{G}]$ is orthogonal to every $\mathcal{G}$-measurable square-integrable variable, including $\mathbb{E}[X\mid \mathcal{G}]-\mathbb{E}[X\mid \mathcal{H}]$. The Pythagorean theorem in an inner-product space applied togives the identity.

Put $M=\mathbb{E}[Y\mid \mathcal{G}]$. Expanding and using $\mathbb{E}[YM]=\mathbb{E}[M^2]$ givesIf $M\stackrel{d}{=}Y$, the right side is zero. Therefore $Y=M$ almost surely.

The conditional Jensen inequality states that for an integrable $X$ and convex $\varphi$, whenever the terms are integrable,For differentiable $\varphi$, the supporting-line inequality $\varphi (x)\ge \varphi (m)+{\varphi }^{\prime }(m)(x-m)$ with $m=\mathbb{E}[X\mid \mathcal{G}]$ gives the result after conditional expectation. Approximation by supporting affine functions proves the general convex case.

Let $M=\mathbb{E}[Y\mid \mathcal{G}]$ and choose the stated strictly convex differentiable $\varphi$ with $|\varphi (x)|\le |x|$. Conditional Jensen inequality gives $\varphi (M)\le \mathbb{E}[\varphi (Y)\mid \mathcal{G}]$. Since $M\stackrel{d}{=}Y$, both sides have the same expectation, so equality holds almost surely. Strictness of the supporting-line inequality on $\{Y\ne M\}$ then forces $Y=M$ almost surely.

The almost-sure Martingale convergence theorem states that a supermartingale $(X_n)$ satisfying ${\sup }_n\mathbb{E}[X_n^{-}]<\infty$ converges almost surely to a finite random variable.

For a nonnegative supermartingale, $X_n^{-}=0$, so the hypothesis of part (i) holds. Its almost-sure limit is finite.

Let independent increments $D_n$ equal $-1$ with probability $1-2^{-n}$ and $2^n-1$ with probability $2^{-n}$. Then $\mathbb{E}{D}_n=0$, so $X_n={\sum }_{j=1}^n{D}_j$ is a martingale. Since ${\sum }_n{2}^{-n}<\infty$, the Borel-Cantelli lemmas say that only finitely many positive jumps occur almost surely. Thereafter every increment is $-1$, and hence $X_n\to -\infty$ almost surely.

For each integer $m$, stop on first crossing below $-m$. Bounded increments ensure the stopped martingale is bounded below by $-m-C$; after adding $m+C$ it is a nonnegative supermartingale and therefore converges. Thus on the event that $(X_n)$ is bounded below, it converges finitely. Applying the same argument to $-X_n$ shows that boundedness above also forces convergence. Outside the finite-limit event the path is therefore unbounded in both directions, so its limsup is $+\infty$ and its liminf is $-\infty$. Hence $\mathbb{P}(A\cup B)=1$.

Let $Z$ be a fair Bernoulli variable measurable at time zero and let $(S_n)$ be an independent simple symmetric random walk. Then $X_n=Z{S}_n$ is a martingale with increments bounded by one. On $\{Z=0\}$ it converges to zero, while on $\{Z=1\}$ the recurrence of the simple symmetric random walk gives limsup $+\infty$ and liminf $-\infty$. Thus $\mathbb{P}(A)=\mathbb{P}(B)=1/2$.

Set $p_n=\mathbb{P}(B_n\mid {\mathcal{F}}_{n-1})$, $S_n={\sum }_{j\le n}{1}_{B_j}$, and $A_n={\sum }_{j\le n}{p}_j$. Then $M_n=S_n-A_n$ is a martingale with bounded increments and conditional variance at most $p_n$. On $\{A_{\infty }<\infty \}$, localization and the $L^2$ martingale convergence theorem make $M_n$ converge, so the integer-valued increasing sequence $S_n$ is finite. On $\{A_{\infty }=\infty \}$, applying martingale convergence toand Kronecker lemma gives $M_n/A_n\to 0$. Hence $S_n/A_n\to 1$ and $S_n\to \infty$. This is the Conditional Borel-Cantelli lemma, and proves the two events equal almost surely.

The sequence $X_n$ converges in distribution to $X$ when $\mathbb{E}f(X_n)\to \mathbb{E}f(X)$ for every bounded continuous function $f$.

Independence makes the joint law of $(X_n,Y_n)$ the product of its marginal laws. Weak convergence of both marginals implies convergence of these product measures to the product law of independent copies $(X,Y)$. The addition map is continuous, so the continuous mapping theorem gives $X_n+Y_n\stackrel{d}{\to }X+Y$.

Differentiability gives $\phi (u)=1+iau+o(u)$ at zero. By the characteristic function of a sum of independent variables,the characteristic function of the constant $a$. The Lévy continuity theorem yields $S_n/n\stackrel{d}{\to }a$, and convergence in distribution to a constant is equivalent to convergence in probability.

A family $({\mu }_n)$ of probability measures on a metric space is tight when for every $\eta >0$ there is a compact set $K$ such that ${\sup }_n{\mu }_n(K^c)<\eta$.

Choose $0<\alpha <\epsilon /p$. At dyadic level $m$, Markov inequality and a union bound giveuniformly in $n$. Summing over $m$ shows that, outside a set of probability at most $C_0{C}^{-p}$, all dyadic increments obey this bound. Chaining dyadic approximations and using continuity givesfor all $s,t$. Since $X_0^n=0$, the paths then lie in the stated compact Hölder set by the Arzela-Ascoli theorem. Taking $C$ large proves tightness.

Every evaluation map ${\pi }_t$ is continuous in the uniform metric, so $\mathcal{F}\subseteq \mathcal{B}$. Conversely, for $g\in C[0,1]$,with the harmless restriction to $m$ for which $r-1/m>0$. Thus every open ball belongs to $\mathcal{F}$. Separability makes every open set a countable union of balls, so $\mathcal{B}\subseteq \mathcal{F}$ and equality follows.

Let $d(x,A)$ be the distance to the closed set $A$. Continuity givesEvery variable in the countable infimum is ${\mathcal{F}}_t$-measurable by adaptedness, so the event belongs to ${\mathcal{F}}_t$. Hence $T_A$ is a stopping time.

Almost-sure convergence plus uniform integrability gives $X_n\to X$ in $L^1$. Conditional expectation is an $L^1$ contraction, so

Take $X_n=Y_n{Z}_n$ and $\mathcal{G}=\sigma (Y_1,Y_2,\dots )$. Then $\mathbb{P}(X_n=n)=n^{-2}$, so $(X_n)$ is uniformly integrable and $X_n\to 0$ almost surely by the Borel-Cantelli lemmas. Independence and $\mathbb{E}{Z}_n=1$ giveBut the independent events $\{Y_n=1\}$ have divergent probability sum, so the second Borel-Cantelli lemma makes them occur infinitely often. Thus these conditional expectations do not converge almost surely to zero.

The Skorokhod embedding theorem states that if $\mu$ is a centered probability law on $\mathbb{R}$ with finite second moment, there is a Brownian stopping time $T$ such that $B_T\sim \mu$, the stopped process $(B_{t\wedge T})$ is uniformly integrable, and $\mathbb{E}T=\int x^2\mu (dx)$.

Construct the times inductively. Suppose $(B_{T_0},\dots ,B_{T_n})$ has the law of $(S_0,\dots ,S_n)$. Conditional on the past, the martingale increment $S_{n+1}-S_n$ has mean zero and finite second moment. Apply the conditional form of the Skorokhod embedding theorem to this regular conditional law, using the fresh Brownian motion $B_{T_n+t}-B_{T_n}$ supplied by the Strong Markov property. This gives a stopping time increment ${\tau }_{n+1}$ and $T_{n+1}=T_n+{\tau }_{n+1}$ such that the next Brownian increment has the required conditional law. Induction provesfor every $k$.