Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 201 (Advanced Probability)

The finite sigma-algebra ${\mathcal{F}}_n=\sigma (A_1,\dots ,A_n)$ is partitioned by the nonempty eventsThese are its atoms. Define the random variableand give it any finite value on the union of the null atoms. It is ${\mathcal{F}}_n$-measurable and integrable. Every $A\in {\mathcal{F}}_n$ is a union of atoms, soThis constructs the requested variable directly, without invoking the general existence theorem for conditional expectation.

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Uniqueness means almost sure equality. Suppose that two ${\mathcal{F}}_n$-measurable integrable random variables $Y$ and $Z$ satisfy the integral identity from part (a). The event $D=\{Y>Z\}$ belongs to ${\mathcal{F}}_n$, and henceThe integrand is nonnegative and is positive precisely on $D$, so $\mathbb{P}(D)=0$. Interchanging $Y$ and $Z$ gives $\mathbb{P}(Z>Y)=0$, and therefore $Y=Z$ almost surely.

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For $A\in {\mathcal{F}}_n$, the defining identity for $X_{n+1}$ givesPart (b) therefore identifies $X_n$ with $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]$, so $(X_n,{\mathcal{F}}_n)$ is a martingale.

The atom formula also provesLet $B_{n,K}=\{|X_n|>K\}$. Then $\mathbb{P}(B_{n,K})\le \mathbb{E}|X|/K$ by Markov inequality, whileThe uniform absolute continuity for a finite measure makes the right-hand side uniformly small as $K\to \infty$. Thus $(X_n)$ is uniformly integrable. The Martingale convergence theorem now supplies an integrable random variable $Y$ such that $X_n\to Y$ both almost surely and in $L^1$.

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Let $\mathcal{G}=\sigma (A_1,A_2,\dots )$. The limit $Y$ from part (c) is $\mathcal{G}$-measurable because each $X_n$ is. If $A\in {\bigcup }_n{\mathcal{F}}_n$, then $A\in {\mathcal{F}}_N$ for some $N$, and for every $n\ge N$,The $L^1$ convergence lets us pass to the limit and obtain $\mathbb{E}[Y{1}_A]=\mathbb{E}[X{1}_A]$.

The sets on which this identity holds form a Dynkin system, and ${\bigcup }_n{\mathcal{F}}_n$ is a generating pi-system. The Dynkin lemma therefore extends the identity to every $A\in \mathcal{G}$. Consequently $Y$ has both defining properties of $\mathbb{E}[X\mid \mathcal{G}]$, established here without appealing to the general existence theorem.

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For an integrable real random variable $X$, its cumulant-generating function is the extended-real convex functionwhere $\psi (\theta )=+\infty$ if the exponential moment diverges. Its Legendre transform of a cumulant-generating function is

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Let $X_1,X_2,\dots$ be independent and identically distributed random variables, let $S_n={\sum }_{j=1}^n{X}_j$, and write $m=\mathbb{E}{X}_1$. Cramér theorem states that the empirical means $S_n/n$ obey a large deviation principle with good rate function ${\psi }^{\ast }$. In particular, for $x\ge m$,with the usual extended-real interpretation; the analogous lower-tail formula holds for $x\le m$.

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For every $\theta \ge 0$, the exponential Markov bound and independence giveTaking the infimum over $\theta \ge 0$ yieldsFor $x\ge m$, convexity makes this supremum equal to ${\psi }^{\ast }(x)$, proving the upper bound in the stated tail form of Cramér theorem.

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Fix $a>m$ and then $x>a$. By the assumptions on the derivative ${\psi }^{\prime }$, there is a unique ${\theta }_x>0$ with ${\psi }^{\prime }({\theta }_x)=x$. Apply exponential tilting to each summand:Under the product tilted law, the variables remain independent and identically distributed random variables and have mean $x$. Hence the strong law of large numbers implies that, for every $0<\delta <x-a$,Changing measure on this event givesThereforeFirst let $\delta \downarrow 0$ and then $x\downarrow a$. The continuity of a convex function gives the lower bound $-{\psi }^{\ast }(a)$. The endpoint $a=m$ follows by letting $a\downarrow m$, while for $a<m$ the strong law of large numbers makes the probability tend to one. This proves the required lower bound.

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Let $(M_k,{\mathcal{F}}_k{)}_{k=0}^N$ be a martingale, let $a<b$, and let $U_N[a,b]$ count the completed upcrossings of $[a,b]$ by time $N$. Doob upcrossing inequality statesEquivalent conventions for a process with a prescribed initial holding add the corresponding endpoint term.

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Assume ${\sup }_n\mathbb{E}|M_n|<\infty$. For every pair of rationals $a<b$, Doob upcrossing inequality givesby monotone convergence theorem. Thus every rational interval is upcrossed only finitely often almost surely. If a real sequence has distinct limit inferior and limit superior, it completes infinitely many upcrossings of some rational interval between them. Hence $M_n$ converges in the extended real line almost surely.

Fatou lemma givesso the limit is finite almost surely and integrable. This is the $L^1$-bounded form of the Martingale convergence theorem.

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Let $(M_t{)}_{t\ge 0}$ be a right-continuous continuous-time martingale with ${\sup }_t\mathbb{E}|M_t|<\infty$. Restricting it to the nonnegative rational times gives a countable martingale. The argument of part (b), applied on successively finer rational grids, shows that it has a finite almost-sure limit as the rational time tends to infinity. Right-continuity and the upcrossing characterization prevent the values at arbitrary times from having a different limit. Thus $M_t$ converges almost surely to a finite integrable random variable as $t\to \infty$, which is the Continuous-time martingale convergence theorem.

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Let $(B_t)$ be Brownian motion in ${\mathbb{R}}^3$ started at a nonzero point. The function $h(x)=|x{|}^{-1}$ is a positive harmonic function on ${\mathbb{R}}^3\setminus \{0\}$ because its Laplacian vanishes there. Stopping on annuli and applying Itô formula shows that $h(B_t)$ is a local martingale. Letting the inner boundary shrink to zero also shows that three-dimensional Brownian motion does not hit the origin.

A positive local martingale is a supermartingale, so $h(B_t)$ is $L^1$-bounded by $h(B_0)$. The upcrossing proof from part (c), which applies verbatim to a positive supermartingale, therefore gives a finite almost-sure limit

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By Brownian scaling,where $Z$ has the standard three-dimensional multivariate normal distribution. The right-hand side tends to zero in probability, since $Z$ has no atom at the origin. Part (d) gives almost-sure convergence to $Y$, which also implies convergence in probability to $Y$. Uniqueness of a limit in probability therefore gives $Y=0$ almost surely. Hence $|B_t|\to \infty$ almost surely, proving the transience of Brownian motion in dimension at least three in dimension three.

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A standard Brownian motion $(B_t{)}_{t\ge 0}$ satisfies $B_0=0$ almost surely, has almost surely continuous sample paths, and has independent stationary increments such that

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The variable $Z={\int }_0^1{B}_{t}dt$ is the $L^2$ limit of Riemann sums of the Gaussian process $B$, so it is a Gaussian random variable. Its mean is zero, and Fubini's theorem with $\mathbb{E}[B_s{B}_t]=\min (s,t)$ givesTherefore $Z\sim N(0,1/3)$.

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Let $X_1,X_2,\dots$ be independent and identically distributed random variables with mean zero and variance one, and let $S_k={\sum }_{j=1}^k{X}_j$. Define the linearly interpolated processThe Donsker invariance principle, also called the functional central limit theorem, states that $W_n$ converges weakly in the space $C[0,1]$ with the uniform norm to standard Brownian motion.

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The evaluation map $f\mapsto f(1)$ is continuous on $C[0,1]$ in the uniform norm. The continuous mapping theorem applied to part (c) therefore givesThis recovers the central limit theorem from the functional version.

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Reversing the finite summations givesThis is a Riemann sum for the continuous functional $f\mapsto {\int }_0^{1}f(t)dt$ evaluated at the interpolated random walk; the interpolation error tends to zero in probability. The continuous mapping theorem and part (b) yield

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Let $D\subset {\mathbb{R}}^d$ be bounded, choose $R$ with $D\subset B(0,R)$, and let $T$ and ${\tau }_R$ be the respective exit times. Then $T\le {\tau }_R$. Sinceis a martingale, the optional sampling theorem for a supermartingale at ${\tau }_R\wedge n$ givesMonotone convergence theorem now gives

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For $u\in C^2(\overline{D})$, Itô formula shows thatis a martingale. Take expectations and let $t\to \infty$. The function $u$ is bounded on the compact set $\overline{D}$, while $\Delta u$ is bounded and ${\mathbb{E}}_{x}T<\infty$ by part (a). The dominated convergence theorem therefore gives Dynkin formula for Brownian motion

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Apply part (b) to $u_n$. Its zero boundary value givesThe assumptions say that $\Delta u_n(y)\to 0$ for every $y\in D$ and that $|\Delta u_n|\le C$ uniformly. Thus the integrand converges pointwise to zero and is dominated by $CT$, whose expectation is finite by part (a). The dominated convergence theorem gives $u_n(x)\to 0$ for every $x\in D$.

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A Lévy process $(X_t{)}_{t\ge 0}$ starts at zero, has independent and stationary increments, is stochastically continuous, and is taken with càdlàg sample paths. Thus for $0\le t_0<\cdots <t_n$, the increments $X_{t_j}-X_{t_{j-1}}$ are independent, and the law of $X_{s+t}-X_s$ depends only on $t$.

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Write $\mathbb{E}{X}_1=\mu$ and $\mathrm{Var}(X_1)={\sigma }^2<\infty$. For rational $t=m/n$, stationarity and independence of the $n$ increments over intervals of length $1/n$ giveusing variance additivity for independent random variables. Stochastic continuity extends both identities from rational to real $t$. In the centered case $\mu =0$, this becomes $\mathbb{E}{X}_t=0$ and $\mathbb{E}{X}_t^2=t{\sigma }^2$.

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For $s<t$, write $X_t=X_s+Y$, where the increment $Y=X_t-X_s$ is independent of the natural filtration at time $s$, has mean zero, and has variance $(t-s){\sigma }^2$. HenceIt follows that $M_t=X_t^2-t{\sigma }^2$ is a martingale, as asserted by the centered square-integrable Lévy martingale identity.

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Let $T=\inf \{t\ge 0:X_t\notin (-a,b)\}$, and suppose the jumps of $X$ have absolute value at most $c$. A nonconstant centered finite-variance Lévy process oscillates, so $T<\infty$ almost surely. Before $T$ the process lies in $(-a,b)$, and at $T$ its bounded overshoot gives $X_T\in [-a-c,b+c]$. Thus the variables $X_{T\wedge n}^2$ are uniformly bounded.

Apply the optional sampling theorem for a supermartingale to the martingale from part (c):Bounded convergence theorem on the left and monotone convergence theorem on the right yield

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Let $X_t=N_t^{+}-N_t^{-}$, where $N^{+}$ and $N^{-}$ are independent Poisson processes of rate $\lambda$. This symmetric Poisson difference process is centered, has jumps $\pm 1$, and has variance rate ${\sigma }^2=2\lambda$. If $a,b$ are positive integers, then $X_T\in \{-a,b\}$ exactly. Optional sampling of the martingale $X_t$ givesConsequentlyPart (d) now gives the mean exit time

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