Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 34 (Advanced Probability)

For a martingale and a stopping time $T\le N$, expand the stopped value using its martingale differences:The event $\{T\ge k\}=\{T>k-1\}$ belongs to the filtration ${\mathcal{F}}_{k-1}$. Each summand is integrable and its conditional expectation given ${\mathcal{F}}_{k-1}$ is zero. Taking expectations of this finite sum therefore proves $\mathbb{E}{M}_T=\mathbb{E}{M}_0$ for every bounded stopping time. This proves the forward direction of the bounded-stopping-time characterization of a martingale without assuming an optional stopping theorem.

For $A\in {\mathcal{F}}_n$, define the bounded stopping time $T=n+1$ on $A$ and $T=n$ on $A^c$. Before time $n$ its stopping events are empty, at time $n$ the event $\{T\le n\}=A^c$ belongs to ${\mathcal{F}}_n$, and subsequently it has stopped everywhere. Subtract the assumed mean identity for this $T$ from that for the constant stopping time $n$ to obtainSince $M_n$ is integrable and ${\mathcal{F}}_n$-measurable, this is precisely the defining property of conditional expectation giving $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Thus the adapted process is a martingale, completing the bounded-stopping-time characterization of a martingale.

The required condition is uniform integrability, written entirely in terms of the given probability laws:Together with the given almost sure convergence, this implies L1 convergence to an integrable $M_{\infty }$. Passing to the limit in $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$, using the contraction property of conditional expectation in L1 space, gives $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.

For every bounded stopping time $S$, the same identity holds with ${\mathcal{F}}_S$: split an event of its stopping-time sigma-algebra according to $\{S=k\}$ and use the deterministic-time identity. In particular $M_{T\wedge n}=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_{T\wedge n}]$. These conditional expectations are uniformly integrable: for an integrable $Y$ and $Z=\mathbb{E}[Y\mid \mathcal{G}]$, the event $A=\{|Z|>K\}$ satisfies $\mathbb{P}(A)\le \mathbb{E}|Y|/K$, andFirst choose $R$, then $K$, uniformly in $\mathcal{G}$. Define $M_T=M_{\infty }$ on $\{T=\infty \}$. Now $M_{T\wedge n}\to M_T$ almost surely and in L1 space, so the bounded-time mean identities imply $\mathbb{E}{M}_T=\mathbb{E}{M}_0$ even when $T$ can be infinite.

With the natural filtration ${\mathcal{F}}_n=\sigma (Z_1,\dots ,Z_n)$, the sums form a martingale. Put $v_n={\sum }_{k\le n}{a}_k^2$. Independence and the characteristic function of a standard normal random variable giveIf $M_n\to M_{\infty }$ almost surely with a finite limit, dominated convergence gives convergence of these characteristic functions to $\mathbb{E}{e}^{iu{M}_{\infty }}$. If $v_n\to \infty$, that limit would be $1$ at zero and $0$ at every nonzero $u$. A characteristic function is continuous at zero: apply dominated convergence to $e^{iu{M}_{\infty }}$ as $u\to 0$. This contradiction provesThis is also sufficient by the L2-bounded martingale convergence theorem, since $\mathbb{E}{M}_n^2=v_n$. Thus the almost sure convergence criterion for an independent Gaussian series is an equivalence.

The exact condition is again ${\sum }_k{a}_k^2<\infty$, with the terminal almost-sure limit used to define $M_T$ when $T=\infty$. In that case ${\sup }_n\mathbb{E}{M}_n^2<\infty$, andproves uniform integrability. The preceding unbounded optional stopping argument yields $\boxed{\mathbb{E}{M}_T=0}$.

For necessity, suppose $v_n\to \infty$. For every fixed $R>0$, the normal distribution of $M_n$ gives $\mathbb{P}(M_n\ge R)\to 1/2$. Hence $\mathbb{P}({\sup }_n{M}_n\ge R)\ge 1/2$, and intersecting over positive integer $R$ shows $\mathbb{P}({\sup }_n{M}_n=\infty )\ge 1/2$. Unboundedness above is unchanged by altering finitely many of the independent $Z_k$, because that changes all sufficiently late sums by one finite constant. It is therefore a tail event; the Kolmogorov zero-one law makes its probability one. Consequently $T<\infty$ almost surely and $M_T\ge 1$. Its expectation is at least one, possibly infinite, and cannot equal zero. This proves the necessity assertion in threshold stopping of an independent Gaussian series.

The almost sure submartingale convergence theorem states that a discrete-time submartingale $Y_n$ satisfying ${\sup }_n\mathbb{E}{Y}_n^{+}<\infty$ converges almost surely to a finite integrable limit. In particular, a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ converges almost surely to an integrable $M_{\infty }$, and $\mathbb{E}|M_{\infty }|\le {\sup }_n\mathbb{E}|M_n|$ by Fatou's lemma. This is the martingale convergence theorem. Uniform integrability gives the stronger conclusion of L1 convergence; it is not implied by bounded first moments alone.

Take $\Omega =[0,1)$ with its Borel sigma-algebra and Lebesgue measure of total mass one. Let ${\mathcal{F}}_n$ be the finite sigma-algebra generated by the half-open dyadic intervals of length $2^{-n}$. Set $X_0=f(1)-f(0)$ if the indexing begins at one. Each $X_n$ is adapted to this filtration, and if $L$ is a Lipschitz constant of $f$, then $|X_n|\le L$, so every coordinate is integrable.

On a parent interval $[u,u+2^{-n})$, conditional expectation averages the two equally likely child values. Their average isThus $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$, establishing the dyadic slope martingale. One can equivalently include $1$ as a separate null atom without changing any probability law.

The dyadic slope martingale is uniformly bounded by $L$. The martingale convergence theorem gives an almost-sure limit $\dot{f}$, and dominated convergence gives L1 convergence. Define $\dot{f}=0$ on the null exceptional set and at the endpoint. This gives a measurable representative with $|\dot{f}|\le L$ everywhere.

If $a,b$ are dyadic rationals, then for every sufficiently fine grid the integral telescopes:Pass to the limit by L1 convergence. For arbitrary endpoints choose dyadic $a_j\to a$, $b_j\to b$ with $a_j\le b_j$. The error in the integrals is at most $L(|a_j-a|+|b_j-b|)$, and Lipschitz continuity controls the error in the endpoint values by the same expression. ThereforeThis constructs the bounded integral density directly, without assuming in advance that a Lipschitz function has an almost everywhere derivative.

For each sample point put $A_n={\max }_{0\le k<2^n}|X_{(k+1)2^{-n}}-X_{k{2}^{-n}}|$. First take $s<t$ in the dyadic rationals, set $h=t-s$, and choose $N\ge 0$ with $2^{-N}\le h<2^{1-N}$. The left grid approximations $s_N=2^{-N}\lfloor 2^{N}s\rfloor$ and $t_N=2^{-N}\lfloor 2^{N}t\rfloor$ are at most two grid steps apart, so $|X_{t_N}-X_{s_N}|\le 2A_N$.

For either endpoint $r$, successive left approximations $r_n,r_{n+1}$ are equal or one step apart on the level-$(n+1)$ grid. Since $r$ is dyadic these approximations eventually equal $r$ exactly. The triangle inequality therefore givesThe second inequality is valid also for $\alpha =0$. Thus the required bound holds simultaneously for every dyadic pair whenever $K_{\alpha }$ is finite; if it is infinite the inequality holds in the extended sense. For $s=t$ the left side is zero. This is dyadic increment chaining, with the stated constant two.

For a maximum of finitely many nonnegative quantities, its $p$th power is at most the sum of their $p$th powers. Hence the increment assumption givesBy the triangle inequality in the Lp norm, the partial sums defining $K_{\alpha }$ have norms bounded by a geometric series. Monotone convergence of their $p$th powers givesIn particular $K_{\alpha }$ is finite almost surely, so the preceding pathwise estimate has a common probability-one domain for all dyadic pairs.

First construct a centred Gaussian process on the countable dyadic set with covariance function $\mathbb{E}(B_s{B}_t)=\min (s,t)$. These consistent finite-dimensional normal distributions exist because, for coefficients $a_j$,the Kolmogorov extension theorem realizes them on one probability space. Every increment has the normal distribution $N(0,|t-s|)$, so for every finite $p\ge 1$,Use the preceding estimate with $\beta =1/2$, $p>2$ and $0<\alpha <1/2-1/p$. It yields an almost surely Hölder continuous extension from the dense dyadic set to $[0,1]$. Approximate arbitrary times by dyadic times; continuity of the paths and of the Gaussian covariance function preserves all finite-dimensional laws. Disjoint increments are jointly Gaussian with zero cross-covariances, hence independent, with the required variances. Together with $B_0=0$ and continuity, this constructs Brownian motion.

For any $\alpha <1/2$, choose $p$ large enough. Taking a countable increasing sequence of exponents tending to $1/2$ gives simultaneous Hölder regularity of every order $\alpha <1/2$ on a common probability-one set; the same argument on each compact time interval gives local Brownian Hölder regularity. This method does not give the endpoint exponent. In fact that endpoint fails: if a path had a finite $1/2$-Hölder seminorm $K$, all its $2^n$ adjacent normalized increments at level $n$ would have magnitude at most $K$. These are independent standard normal random variables, so for a fixed integer $m$ the probability that their maximum is at most $m$ is $\mathbb{P}(|Z|\le m{)}^{2^n}\to 0$. The event of a finite $K$ is contained in the union over $m$ of events with this bound at every level, each of probability zero. Thus Brownian paths are not $1/2$-Hölder continuous on $[0,1]$ almost surely.

Write $u(z)=|z{|}^{2-d}$. For a radial function $r^{2-d}$, the Laplacian isThus $u$ is a harmonic function off zero. Fix $R>|x|$ and stop Brownian motion when it reaches either sphere of radius $\epsilon$ or $R$, at time $S_R$. This exit time is finite: stopping the martingale $|B_t{|}^2-dt$ at $S_R\wedge t$ yields $d\mathbb{E}(S_R\wedge t)\le R^2-|x{|}^2$, and monotone convergence applies. By Itô's formula, $u(B_{t\wedge S_R})$ is a bounded martingale. Optional stopping and dominated convergence therefore giveThe events $\{T_{\epsilon }<T_R\}$ increase to $\{T_{\epsilon }<\infty \}$ as $R\to \infty$: a continuous path up to any finite hitting time is bounded. Since $d\ge 3$, $R^{2-d}\to 0$, proving the Brownian hitting probability of a ball

To keep the normalization visible, denote the integral defining the printed constant by $A_d$. Substituting $r=|y{|}^2/(2s)$ gives, for $y\ne 0$,This is the Newtonian potential of the Brownian heat kernel. The singularity at zero is locally integrable in dimension $d$, since its radial integral behaves like ${\int }_0^1{r}^{2-d}{r}^{d-1}dr=1/2$. All integrands are nonnegative, so Tonelli's theorem permits both orders of integration.

Integrating in $s$ first givesIntegrating in $y$ first uses the Gaussian heat kernel convolution. Completing the square in $|y-x{|}^2/t+|y{|}^2/s$ givesConsequently $I={\int }_t^{\infty }p(r,0,x)dr$, and the actual identity isThe original PDF uses $A_d$ rather than $A_d^{-1}$ as the multiplier while defining its constant to equal $A_d$. That formula is false. For example $A_3=1/(2\pi )$, so the required multiplier in dimension three is $2\pi$. Equivalently, as $t\downarrow 0$ at $x\ne 0$, the left side tends to $|x{|}^{2-d}$, whereas the printed right side tends to $A_d^2|x{|}^{2-d}$. One can retain the printed form only by defining its multiplier as $c_d=A_d^{-1}$.

Continue with $A_d$ equal to the integral defined in the PDF and $u(z)=|z{|}^{2-d}$. Until the Brownian motion hits the radius-$\epsilon$ ball, $u(B_s)$ is a bounded local martingale, hence a true martingale. At the finite stopping time $t\wedge T_{\epsilon }$, optional stopping givesRearranging,The fixed random variable $u(B_t)$ is integrable by the Gaussian heat kernel and the local-integrability calculation above. Also ${\mathbb{P}}_x(T_{\epsilon }\le t)\le (\epsilon /|x|{)}^{d-2}\to 0$. For any $R>0$, the final expectation is at mostFirst send $\epsilon \downarrow 0$, then $R\to \infty$, proving that it vanishes. The identities from part (b) giveSubtracting yields the small-ball Brownian hitting asymptoticThus the constant in this printed limit requires exactly the same reciprocal correction as in part (b). For $d=3$ the multiplier is $2\pi$, not $1/(2\pi )$. The argument proves the corrected finite-time limit directly, rather than inferring it from the eventual hitting probability alone.

Let $R=|U|$. Uniform volume measure on the unit ball gives $\mathbb{P}(R\le r)=r^n$, $0\le r\le 1$, hence radius density $n{r}^{n-1}$. Conditional on $R=r>0$, independence leaves $W$ a standard Brownian motion. Its exit time from the radius-$r$ ball is finite: applying optional stopping to $|W_t{|}^2-nt$ stopped at that exit and then letting $t\to \infty$ gives $\mathbb{E}{\tau }_r=r^2/n$.

The probability law of the exit point is invariant under every orthogonal transformation, because both Brownian motion and the centred sphere are invariant under those transformations. The unique invariant probability on that sphere is normalized surface measure; for $n=1$ it is the equally weighted pair $\{-r,r\}$. This is also the conditional angular law of $U$ given $|U|=r$. Mixing these identical conditional laws against $n{r}^{n-1}dr$ proves the uniform ball sampling by Brownian stopping identityThe event $R=0$ has probability zero and causes no difficulty. Independence of $R$ permits its value to be included in the initial filtration, making $T$ a stopping time.

Allow $g_D$ to take the value $+\infty$. Fix a closed ball $\overline{B(z,r)}\subset D$, independently choose a radius $R$ with density $n{s}^{n-1}/r^n$ on $(0,r)$, and let $S$ be the first time the Brownian motion from $z$ reaches that radius. Conditional on $R=s$, the Strong Markov property at $S$ decomposes the remaining Brownian exit time from $D$. Since $S<T_D$ and the conditional exit time has mean $s^2/n$, Tonelli's theorem and part (a) give the Brownian exit-time ball averaging identityThis identity is valid in the extended nonnegative reals; no finiteness was assumed to derive it.

If $g_D(z)<\infty$, the integral over $B(z,r)$ is finite. For any $y\in B(z,r)$ choose $\rho >0$ with $\overline{B(y,\rho )}\subset B(z,r)$. The same identity at $y$ has a finite right side because its integral is over a subset of the integrable ball. Thus $g_D(y)<\infty$. The set $F=\{z\in D:g_D(z)<\infty \}$ is therefore open relative to $D$.

It is also relatively closed. If $z_j\in F$ converges to $y\in D$, choose $r>0$ with $\overline{B(y,3r)}\subset D$. For large $j$, $|z_j-y|<r$, and $\overline{B(z_j,2r)}\subset D$. The preceding argument shows that every point in $B(z_j,2r)$, including $y$, belongs to $F$. Hence $F$ is closed in $D$. Since $F$ is nonempty and $D$ is connected, $g_D(y)<\infty$ for every $y\in D$. This proves finiteness propagation of Brownian mean exit times without assuming beforehand that $g_D$ solves a smooth Poisson equation.

Start at $x=(x_1,\dots ,x_n)$ with all $x_i>0$, and let ${\tau }_i$ be the first zero of the $i$th coordinate. The orthant exit time is $T={\min }_i{\tau }_i$. The coordinate processes are independent, soHere the Brownian reflection principle gives ${\mathbb{P}}_a(\tau \le t)=2\mathbb{P}(B_t\le -a)$; $\Phi$ is the standard normal distribution function. Since $\Phi (z)-1/2=z/\sqrt{2\pi }+o(z)$ as $z\to 0$,The tail integral formula for expectation, ${\mathbb{E}}_{x}T={\int }_0^{\infty }{\mathbb{P}}_x(T>t)dt$, has a bounded integrand on $(0,1)$ and converges at infinity exactly when $n/2>1$. Therefore the Brownian exit-time expectation in an orthant givesThe divergence for $n=2$ is logarithmic; almost-sure finiteness of the exit time does not imply a finite expectation.

For $0<\delta <1$, define the compensated small-jump Poisson integralThis is an ordinary compensated finite-intensity sum. Independent Poisson random variables give mean zero and the L2 construction of compensated Poisson integrals gives, for $0<\eta <\delta$,Thus $Y_t^{\delta }$ is Cauchy in L2 space as $\delta \downarrow 0$, defining the first integral. It is not the difference of two separately convergent integrals. On a compact time interval $[0,L]$, the Doob L2 maximal inequality bounds the squared supremum of the same difference by $8cL(\delta -\eta )$. Hence the truncated martingales are Cauchy in the expected squared supremum norm. A sufficiently rapidly decreasing sequence of cutoffs converges uniformly almost surely; its limit has càdlàg paths, so the construction also supplies a consistent process.

For large jumps, $K(\{|y|>1\})=2c<\infty$. There are only finitely many such atoms before any fixed time, almost surely, and each atom has a finite mark. The uncompensated large-jump integral is therefore a finite random sum, even though its absolute first moment is infinite. The two terms are well-defined for these different reasons.

There is also a literal signed-integral defect in the PDF: $\int yK(dy)$ on either symmetric region is not $+\infty$. Its positive and negative parts both have infinite mass, so the ordinary signed Lebesgue integral is undefined. The correct divergence statement isA symmetric principal value may be zero, but is not an ordinary signed integral. Square-integrable compensation defines the small jumps; finite activity defines the large jumps.

For a finite-intensity Poisson random measure, independence of counts gives the exponential characteristic function formula. Apply it to the cutoff construction from part (a), including the deterministic compensation, and pass to the L2 space limit of the small jumps. The compensated integrand is $O(y^2)$ at zero and bounded on the large-jump region, so its integral converges absolutely:Symmetry makes the absolutely integrable imaginary part zero. Substituting $z=|u|y$ in the real part and using the specified normalization givesThus $\mathbb{E}{e}^{iu{X}_1}=e^{-|u|}$, and this construction is the standard Cauchy process. To calculate the density, its integrable characteristic function permits Fourier inversion:This is the standard Cauchy distribution. The normalization also gives $c=1/\pi$: integration by parts turns ${\int }_0^{\infty }(1-\cos z)z^{-2}dz$ into ${\int }_0^{\infty }\sin zdz/z=\pi /2$. For completeness, with exponential damping the latter integral equals $\arctan (1/\eta )$, because its derivative in frequency $b$ is ${\int }_0^{\infty }{e}^{-\eta z}\cos (bz)dz=\eta /({\eta }^2+b^2)$ and its value at $b=0$ is zero. Let $\eta \downarrow 0$; integration by parts bounds the undamped and damped tails uniformly by a constant divided by the lower cutoff, justifying the limit.

Disjoint time strips of the Poisson random measure are independent, and its intensity is translation invariant in time. The construction and its limits therefore give independent increments and stationary increments, withFor the scaling actually printed in the original PDF,For example, $\alpha =2$, $t=1$, $u=1$ gives $e^{-4}$ rather than $e^{-1}$. Thus the requested equality of process laws is false unless $\alpha =1$.

The intended self-similarity of a Cauchy process rescales space and time in opposite directions:Indeed, for $0\le s<t$,The transformed process retains independent increments. For any ordered times, the joint characteristic function of its successive increments is therefore the same product as for $X$; cumulative summation proves equality of all finite-dimensional distributions. Both constructions have càdlàg paths, so these distributions also determine their laws on the usual path space. This derives the corrected invariance and explicitly diagnoses the false printed assertion.