Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 201 (Advanced Probability)

The partial sums $S_n$ are a martingale, as are $S_n^2-n$. Since the increment has mean zero and nonzero variance, there are ${\delta }_{+},{\delta }_{-}>0$ with $\mathbb{P}(X_1\ge {\delta }_{+})>0$ and $\mathbb{P}(X_1\le -{\delta }_{-})>0$. From any point in $(-a,b)$, a sufficiently long run of either kind exits the interval. Independence in consecutive blocks therefore bounds $\mathbb{P}(T>km)$ by a geometric sequence. In particular, $T<\infty$ almost surely and $\mathbb{E}T<\infty$.

Apply the optional stopping theorem first to $T\wedge n$. Since the increments are bounded and $\mathbb{E}T<\infty$, the stopped variables are uniformly integrable and passage to the limit givesApplying the same argument to $S_n^2-n$, using $|S_T|\le \max \{a,b\}+c$, gives

On upper exit, $b\le S_T\le b+c$; on lower exit, $-a-c\le S_T\le -a$. With $p=\mathbb{P}(S_T\ge b)$ and $\mathbb{E}{S}_T=0$,Solving gives

Moreover $(S_T+a)(S_T-b)\ge 0$, so $\mathbb{E}{S}_T^2\ge ab$, which is stronger than the requested lower bound. Also $S_T\in [-a-c,b+c]$, and henceTaking expectations and using $\mathbb{E}{S}_T=0$ gives $\mathbb{E}{S}_T^2\le (a+c)(b+c)$, stronger than the requested upper bound. Sincethe two stated estimates follow from part a.

The strong law of large numbers states that for independent identically distributed integrable random variables,To prove it, set $Y_n=X_n{1}_{\{|X_n|\le n\}}$. The tail-sum formula gives ${\sum }_n\mathbb{P}(X_n\ne Y_n)<\infty$, so the Borel-Cantelli lemmas make the two sequences eventually equal. AlsoThe Kolmogorov convergence theorem implies that ${\sum }_n(Y_n-\mathbb{E}{Y}_n)/n$ converges almost surely, and Kronecker lemma yieldsFinally $\mathbb{E}{Y}_n\to \mu$, so the Cesaro mean of these expectations tends to $\mu$.

After replacing $X_j$ by $X_j-\mu$, the maximal inequality for independent averages givesThis follows from the Doob Lp maximal inequality by dyadically grouping the partial sums. The strong law makesalmost surely. The displayed maximal function is in $L^p$, so dominated convergence applied to its $p$th power proves convergence in $L^p$.

Prokhorov's theorem says that a sequence of Borel probability measures on $\mathbb{R}$ is relatively compact for weak convergence of probability measures exactly when it is tight: for every $ϵ>0$, some compact $K$ satisfies ${\sup }_n{\mu }_n(\mathbb{R}\setminus K)<ϵ$.

Put $z=y/\lambda$. Direct integration givesFor $|z|\ge 1$, the right side has a positive infimum $c_0$ because $\sin z/z<1$ there and it tends to one at infinity. Thus the claimed inequality holds with $C=c_0^{-1}$.

The masses ${\mu }_n(\mathbb{R})={\psi }_n(0)$ converge and are therefore bounded. Part b and Tonelli theorem giveThe integrands are uniformly bounded. By pointwise convergence and the continuity of $\psi$ at zero, the right side can be made uniformly small for all sufficiently large $n$ by taking $\lambda$ large; finitely many remaining measures are individually tight. Thus $({\mu }_n)$ is tight. Applying Prokhorov's theorem after normalizing the masses, or adjoining missing mass at one fixed point, gives a weakly convergent subsequence.

Yes. If a subsequence converges weakly to $\nu$, bounded continuity of $e^{iuy}$ givesThe uniqueness theorem for characteristic functions makes $\nu$ unique. Tightness implies that every subsequence has a further weakly convergent subsequence, and every such limit is $\nu$. This subsequence criterion proves that the entire sequence converges weakly to $\nu$.

If $t\le s$, then $\mathbb{E}(X_t\mid {\mathcal{F}}_s)=X_t$. If $s<t\le 1$, Gaussian conditional expectation for the Brownian bridge between $(s,X_s)$ and $(1,X_1)$ gives

Since $X_1-X_s$ is centered Gaussian with variance $1-s$,Fubini's theorem therefore shows that the defining integral for $A_1$ is absolutely finite almost surely.

For $s<r<1$, part a givesConditional Fubini then yieldsHence $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$, so $M$ is a martingale.

The process $M$ is continuous and Gaussian. Since $A$ has finite variation,The Lévy characterization of Brownian motion makes $M$ a Brownian motion in the enlarged filtration.

Moreover,Every finite vector from $M$ is jointly Gaussian with $X_1$, so zero covariance implies independence. Thus the whole process $M$ is independent of $X_1$.

The function $u(x)=ϵ/|x|$ is harmonic on $\{|x|>ϵ\}$, equals one on the target sphere, and tends to zero at infinity. Optional stopping at the first hit of radius $ϵ$ and the first exit from a ball of radius $R$ givesLetting $R\to \infty$ and taking $|x|=1$ yields ${\mathbb{P}}_x({\tau }_{ϵ}<\infty )=ϵ$.

Project Brownian motion modulo ${\mathbb{Z}}^3$ to the compact three-dimensional torus. The projected process has normalized volume as invariant probability measure and is irreducible, so it visits every nonempty open set infinitely often almost surely. The image of $S_3$ contains the radius-$ϵ$ ball about the origin. Hence the original Brownian motion hits $S_3$ at an unbounded set of times.

Yes. Quotient only the first coordinate modulo $\mathbb{Z}$. The resulting process lives on $\mathbb{T}\times {\mathbb{R}}^2$ and the image of $S_1$ is the radius-$ϵ$ ball about $(0,0,0)$. The two noncompact coordinates form planar Brownian motion, which is recurrent; during its infinitely many returns to a smaller disc, the independent circle coordinate has a fixed positive chance of lying in the required interval. The Strong Markov property then shows that the target ball is visited infinitely often. Lifting back proves that $S_1$ is hit at unbounded times.

A Poisson random measure $N$ of intensity $\mu$ assigns to disjoint measurable sets independent random variables, andwhenever $\mu (A)<\infty$, with the usual countable-additivity requirement.

A Lévy process starts at zero almost surely, has stationary independent increments, and is stochastically continuous; one normally takes its càdlàg modification.

Independent Poisson processes have stationary independent increments, so their weighted sum does too and is stochastically continuous. Moreover,where

Let $N(ds,dy)$ be a Poisson random measure on $(0,\infty {)}^2$ with intensity $dsK(dy)$ and defineThe assumption $\int yK(dy)<\infty$ makes this integral finite on compact time intervals. The exponential formula for a Poisson random measure givesso $X$ is a Lévy process with exponent $\psi$.

Choose finite-valued measurable functions $q_n\ge 0$ which vanish off $[1/n,n]$ and satisfyThis is possible by truncation followed by approximation by simple functions. PutThe measure of the support of $q_n$ is finite, and $q_n$ takes finitely many values, so $X^n$ is a simple pure-jump Lévy process. Under this common coupling,