Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 29 (Advanced Probability)

Write $Y=\mathbb{E}[X\mid \mathcal{G}]$. The conditional expectation is an $L^2$ contraction by the conditional Jensen inequality, so $X,Y$ are square-integrable. The defining integral identity for conditional expectation, first with bounded measurable truncations of $Y$ and then with their $L^2$ limit, givesFor clarity, if $Y^{(K)}=\max (-K,\min (Y,K))$, then $\mathbb{E}[X{Y}^{(K)}]=\mathbb{E}[Y{Y}^{(K)}]$, and the Cauchy-Schwarz inequality permits the limit $K\to \infty$. Expanding the square now proves the equality case for conditional second moments:A nonnegative random variable of zero expected value vanishes almost surely. Therefore $\boxed{X=Y\text{ almost surely}.}$

Equality of the probability distributions implies equality of all integrals for which they are finite, in particular $\mathbb{E}[(\mathbb{E}[X\mid \mathcal{G}]{)}^2]=\mathbb{E}[X^2]<\infty$. Part (a), applied to this conditional expectation, therefore gives $\boxed{X=\mathbb{E}[X\mid \mathcal{G}]\text{ almost surely}.}$ The second-moment assumption justifies the squared-norm argument; the next part removes it.

Again put $Y=\mathbb{E}[X\mid \mathcal{G}]$. For any fixed real $c$, the random variableis nonnegative and integrable. The sign function of $Y-c$ is bounded and $\mathcal{G}$-measurable, so conditioning and equality of the probability distributions giveThus $D_c=0$ almost surely. On $\{Y>c\}$ this forces $X\ge c$; on $\{Y<c\}$ it forces $X\le c$; on $\{Y=c\}$ it forces $X=c$. In particular $\{Y=c\}\subseteq \{X=c\}$ up to a null set. These two events have equal probability, again by equality of the probability distributions, so they agree up to a null set. ConsequentlyTaking $c=0$ establishes the suggested sign assertion, but this alone would not exclude unequal positive values. Apply the same argument simultaneously to the countable set of rational $c$. Outside the union of their null sets, unequal $X$ and $Y$ would have a rational strictly between them, giving opposite signs. This proves the conditional expectation preserving a distribution result:Only integrability was used; no finite second moment or hidden square-integrability assumption is needed.

A convenient precise version is the almost sure submartingale convergence theorem: if $(S_n)$ is a real submartingale and ${\sup }_n\mathbb{E}{S}_n^{+}<\infty$, then $S_n$ converges almost surely to a finite integrable random variable. In particular the martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|S_n|<\infty$. It asserts neither convergence in L1 nor preservation of the initial expected value without additional uniform integrability.

Here is a proof including the crossing estimate. Fix real $a<b$, and let $U_n[a,b]$ count completed upcrossings by time $n$. Use a predictable indicator $H_k$ which holds one unit during each crossing attempt: enter when $S_k\le a$, exit when $S_k\ge b$. Witheach completed trade gains at least $b-a$, while the possible unfinished trade loses no more than $(S_n-a{)}^{-}$. Hence $G_n\ge (b-a)U_n[a,b]-(S_n-a{)}^{-}$. The submartingale property and $0\le H_k\le 1$ give $\mathbb{E}{G}_n\le \mathbb{E}(S_n-S_0)$: the complementary predictable gains have nonnegative expected value. ThereforeThe right side is uniformly bounded, since $(S_n-a{)}^{+}\le S_n^{+}+|a|$. By the monotone convergence theorem, every rational interval has finitely many upcrossings almost surely. A real sequence whose lower and upper limits differ crosses some rational interval infinitely often, so the countable intersection of these events gives an extended-real limit. Furthermoreso ${\sup }_n\mathbb{E}|S_n|<\infty$. The Fatou lemma makes the limit finite and integrable, completing the proof. Applying this theorem to $-Z_n$ gives the almost sure supermartingale convergence theorem for nonnegative supermartingales.

For the given adapted multiplicative drift, putThe next denominator $P_{n+1}$ is ${\mathcal{F}}_n$-measurable, and $P_n\ge 1$ makes $Z_n$ integrable. ThusThe nonnegative supermartingale $Z_n$ has a finite almost sure convergence limit $Z_{\infty }$. On the probability-one event where the specified series is finite,Consequently $P_n\uparrow P_{\infty }\in [1,\infty )$ and the multiplicative correction for summable adapted drift givesThe argument does not require a finite expected value for the infinite product or for the total drift.

The Brownian first-passage time $T_a$ is finite almost surely. Indeed the Brownian reflection principle gives $\mathbb{P}(T_a\le t)=2\mathbb{P}(B_t\ge a)\to 1$ as $t\to \infty$. Continuity of the paths means the first passage to $x+y$ occurs after the first passage to $x$. By the Strong Markov property, the process $B_{T_x+s}-x$ is an independent fresh Brownian motion. Thus $T_{x+y}-T_x$ has the same probability distribution as $T_y$ and is independent of $T_x$. Taking the Laplace transforms of nonnegative random variables givesFor fixed $\lambda \ge 0$, these transforms lie in $(0,1]$. Put $g(a)=-\log {\phi }_a(\lambda )$. It is additive on positive arguments and nondecreasing, because the hitting times increase with the level. Writing $c(\lambda )=g(1)$, additivity gives $g(r)=rc(\lambda )$ for positive rational $r$, and rational upper and lower approximations then give $g(a)=ac(\lambda )$ for every $a>0$. Hence ${\phi }_a(\lambda )=e^{-ac(\lambda )}$, with $c(\lambda )\ge 0$.

The Brownian scaling identity $T_a\stackrel{d}{=}{a}^2{T}_1$ givesFor $\lambda >0$, take $a=\sqrt{\lambda }$ in this relation with the transform parameter one. It follows that $c(\lambda )=C\sqrt{\lambda }$, where $C=c(1)\ge 0$. At zero, finiteness of the hitting time gives $c(0)=0$. Thus $\boxed{{\phi }_a(\lambda )=e^{-aC\sqrt{\lambda }}}$, with the constant determined in part (b).

Fix $\lambda >0$ and choose the positive rootUse the Exponential martingale for Brownian motion $M_t=\exp (\theta B_t-{\theta }^{2}t/2)$. Before the linear-boundary stopping time $\tau$, path continuity gives $B_s<a+bs$. At the boundary there is equality. ConsequentlyThe optional stopping theorem at the bounded stopping time $t\wedge \tau$ yields $\mathbb{E}{M}_{t\wedge \tau }=1$. Separate the two events:The last term is at most $e^{\theta a-\lambda t}$ and tends to zero. The first term converges by the monotone convergence theorem. Taking $e^{-\lambda \infty }=0$ therefore provesThis bounded-stopping argument does not assume in advance that $\tau$ is finite. For $b\le 0$, letting $\lambda \downarrow 0$ makes the right side tend to one, so it also proves $\mathbb{P}(\tau <\infty )=1$. Setting $b=0$ gives the Brownian first-passage Laplace transformThe zero-parameter value follows from the just-proved almost-sure finiteness.

For probability measures on a metric space, weak convergence of probability measures ${\mu }_n\Rightarrow \mu$ means $\int fd{\mu }_n\to \int fd\mu$ for every bounded continuous real function $f$. On the real line it is equivalently convergence of the cumulative distribution functions at every continuity point of the limiting cumulative distribution function. Bounded continuous upper and lower approximations to interval indicators give one direction; a partition at continuity points, uniform approximation on a compact interval, and control of the tails give the reverse direction.

For a probability measure $\mu$ on $\mathbb{R}$, its characteristic function is ${\phi }_{\mu }(t)=\int e^{itx}\mu (dx)$. The Lévy continuity theorem states: ${\mu }_n\Rightarrow \mu$ implies ${\phi }_{{\mu }_n}(t)\to {\phi }_{\mu }(t)$ for every real $t$; conversely, if these characteristic functions have a pointwise limit $\phi$ which is continuous at zero, there is a unique probability measure $\mu$ with characteristic function $\phi$, and ${\mu }_n\Rightarrow \mu$. In particular, for a specified limiting probability measure, weak convergence is equivalent to pointwise convergence of its characteristic functions.

The forward implication follows by testing cosine and sine. The dominated convergence theorem shows that every characteristic function is continuous at zero and has value one there. For the converse, first establish tightness of probability measures. By Fubini's theorem, for $h>0$,The integrand on the right is at least $1/2$ when $|x|\ge 2/h$. This proves the characteristic-function tightness boundThe integrals converge for each $h$ by dominated convergence, and continuity of $\phi$ at zero makes the limit as small as desired by choosing $h$ small. Thus all sufficiently late measures have a common tail bound; the finitely many earlier measures also have small tails once the radius is increased. This is uniform tightness of probability measures.

We give the required compactness argument explicitly. For any subsequence, choose a further diagonal subsequence whose cumulative distribution functions $F_n$ converge at every rational $q$, to $f(q)$. DefineThis function is nondecreasing and right-continuous: given a rational $q>x$ with $f(q)$ close to $F(x)$, the same bound holds for $F(y)$ when $x<y<q$. Tightness gives $F(-\infty )=0$ and $F(+\infty )=1$, so $F$ is a cumulative distribution function of a probability measure $\nu$. For rationals $r<x<s$, $F_n(r)\le F_n(x)\le F_n(s)$. The limits bracket $F(x-)$ and $F(x)$; at a continuity point of $F$ they coincide. Thus $F_n(x)\to F(x)$ at all such points. To obtain weak convergence, choose a compact interval with small tails and partition it using continuity points of $F$. A bounded continuous function is uniformly continuous there, so step-function approximations and convergence of interval probabilities prove convergence of its integral. This proves Helly selection for distribution functions and supplies a weak subsequential limit $\nu$.

Its characteristic function is $\phi$ by the already proved forward implication. We also prove uniqueness rather than assume it. If two probability measures $\nu ,\rho$ have the same characteristic function, convolve each with a centered normal distribution of variance ${\epsilon }^2>0$. Their densities are, by the Gaussian integral and Fubini's theorem,and likewise for $\rho$. The integrable Gaussian factor justifies exchanging the integrals, so these densities agree. For every bounded continuous $f$, letting $\epsilon \downarrow 0$ in $\int \mathbb{E}[f(x+\epsilon Z)]\nu (dx)$, with $Z$ a standard normal random variable, gives $\int fd\nu$ by bounded convergence theorem. Hence the two measures have equal integrals against all these test functions, and approximating interval indicators shows $\nu =\rho$. This proves the uniqueness theorem for characteristic functions.

Every subsequence therefore has a further subsequence weakly converging to the same unique $\mu$. If the original sequence failed weak convergence, some bounded continuous test function would have a subsequence of integrals staying a fixed distance from its integral under $\mu$, contradicting that further convergence. This completes both directions of the theorem. Continuity at zero is essential: the laws $N(0,n)$ have pointwise limiting characteristic function zero away from zero and one at zero, and are not tight.

For the finite probability distribution, put ${\xi }_j=(X_j+1)/2$. These are independent fair zero-one variables, andThe binary integer ${\sum }_{j=1}^n{2}^{n-j}{\xi }_j$ is uniform on $\{0,\dots ,2^n-1\}$. HenceThese are the midpoints of $2^n$ equal subintervals of $[-1,1]$. For every bounded continuous $f$, the Riemann sums giveThus $\boxed{Y_n\Rightarrow \mathrm{Unif}[-1,1]}$ in the sense of weak convergence of probability measures.

For the almost-sure and $L^2$ assertions, the Dyadic Rademacher series $Y={\sum }_{j\ge 1}{2}^{-j}{X}_j$ converges absolutely on every sign sequence, since $|X_j|=1$. Its tail satisfies $|Y-Y_n|\le 2^{-n}$, proving almost sure convergence and convergence in L2. More precisely, zero means and independence eliminate the cross terms:This identity follows first for finite tails and then by dominated convergence. Bounded continuous test functions and bounded convergence theorem identify the probability distribution of $Y$ with the weak limit above, namely the uniform distribution on $[-1,1]$.

For the requested product identity, each Rademacher random variable has characteristic function $\mathbb{E}{e}^{it{X}_j}=\cos t$. The characteristic function of a sum of independent variables therefore yieldsBy part (a), or directly by bounded convergence theorem, this converges to the characteristic function of a uniform distribution, $\frac{1}{2}{\int }_{-1}^1{e}^{itx}dx=\sin t/t$. Consequently the dyadic cosine product isBoth sides are one at zero. At zeros of the sinc function the identity is understood as the limit of the finite products, so no logarithm or division by a vanishing factor is needed.

Let $B$ be a standard planar Brownian motion started at $z$. Neighbourhood recurrence means that, with probability one, every nonempty open subset of ${\mathbb{R}}^2$ is visited at arbitrarily large times. Not hitting points means that for each fixed $w\ne z$, ${\mathbb{P}}_z(\exists t\ge 0:B_t=w)=0$. If $w=z$, time zero must be excluded, and there is still no return at any positive time. This assertion concerns each prescribed point; it does not say that the path contains no points.

We prove both claims from the planar Brownian annulus hitting probability. Fix a center $w$ and radii $0<r<\rho =|z-w|<R$, and define the stopping times ${\tau }_r=\inf \{t:|B_t-w|=r\}$ and ${\tau }_R=\inf \{t:|B_t-w|=R\}$. Their minimum $\sigma$ is finite almost surely: a coordinate of Brownian motion cannot stay forever in a bounded interval, by the Brownian reflection principle, so exit from the outer disk is finite. On the annulus,The Itô formula therefore makes $\log |B_{t\wedge \sigma }-w|$ a local martingale. It is bounded between $\log r$ and $\log R$, hence is a genuine martingale. The optional stopping theorem at $t\wedge \sigma$, followed by the bounded convergence theorem, givesThusThe exit can only be at one of the two distinct boundaries by path continuity.

For a fixed $r$, let $R\to \infty$. These events increase to $\{{\tau }_r<\infty \}$: any path up to a finite hitting time is bounded and so avoids some sufficiently large outer circle. The ratio tends to one, proving that any closed disk of positive radius is hit almost surely from any starting point outside it. An initial point inside it already counts as a hit.

To prove the stronger recurrence statement, take a closed disk contained in any prescribed nonempty open set. At each integer time $n$, the Markov property and the preceding disk-hitting result imply that some time $t\ge n$ visits this disk with probability one. Intersecting these events over all $n$ proves visits at arbitrarily large times. A countable basis of disks with rational centers and rational radii makes this simultaneous for all nonempty open sets. This is recurrence of planar Brownian motion, or equivalently neighbourhood recurrence.

For point avoidance, keep $R>\rho$ fixed and let $r\downarrow 0$. Hitting $w$ before ${\tau }_R$ requires hitting every inner circle first. The displayed ratio tends to zero, so this point-hitting event has probability zero. Any finite hit of $w$ would occur before exit from some outer circle with integer radius, again because the preceding path is bounded. A countable union now gives $\boxed{{\mathbb{P}}_z(T_w<\infty )=0(z\ne w)}$, the polar point for planar Brownian motion assertion.

Finally, even if the process starts at $w$, for each deterministic $s>0$ its normal distribution has a density, so $\mathbb{P}(B_s=w)=0$. The Markov property at $s$ and the result from other starting points exclude any visit to $w$ at or after $s$. Taking $s=1/n$ excludes all positive-time returns. Thus planar Brownian motion repeatedly approaches every point, while almost surely never hitting any prescribed point at positive time.

Interpret the comparison at $N$ on $\{N<\infty \}$; values at an infinite stopping time are not required. The stopping time events make $Y_n$ adapted, and $|Y_n|\le |X_n^1|+|X_n^2|$ gives integrability. On $\{N\le n\}$ the next value is $X_{n+1}^2$. On $\{N>n\}$ it is $X_{n+1}^1$ unless $N=n+1$, and on that latter event switching only decreases the value. Consequently the pointwise inequalityholds even though $\{N=n+1\}$ need not be known at time $n$. Both indicators displayed here are ${\mathcal{F}}_n$-measurable. Taking conditional expectations and using the two supermartingale inequalities therefore givesThus $\boxed{Y\text{ is a supermartingale}.}$ This proves pasting supermartingales with downward jumps without incorrectly treating the future switching event as predictable.

Put $q=b/a>1$. Each successive crossing time $N_j$, $j\ge 1$, is a stopping time, since in discrete time the event that a threshold is first met after the previous stopping time can be written as a finite union of events known at the candidate time. Use $\inf \varnothing =\infty$, with all later crossing times then infinite as well.

For the supermartingale assertion, build a process by finitely many switches. Begin with the constant supermartingale one. At $N_1$ switch to $X_n/a$; the new value at that time is at most one. At $N_2$ switch to the constant $q$; the old value there is $X_{N_2}/a\ge q$. At $N_3$ switch to $q{X}_n/a$, whose value is at most $q$, and at $N_4$ switch to the constant $q^2$, whose value is at most the previous process. More generally, the switches at $N_{2\ell -1}$ and $N_{2\ell }$ are respectively from cash $q^{\ell -1}$ to $q^{\ell -1}{X}_n/a$ and from that stock value to cash $q^{\ell }$. Every switch is downward, including possible overshoots of either threshold.

Each candidate continuation is a supermartingale, being either constant or a fixed positive multiple of $X$. Part (a) therefore proves by induction that the process $M^{(j)}$ obtained after the first $j$ switches is a supermartingale. It agrees with $Y$ through $N_j$. For fixed $j,n$, its values are bounded in absolute value by a deterministic constant depending on $j$ times $1+X_n$, which also verifies the required integrability rather than assuming it for an infinite sequence of pastings.

Stopping a discrete supermartingale preserves the property. Indeed its stopped increment is $1_{\{N_j>n\}}(M_{n+1}^{(j)}-M_n^{(j)})$, with an ${\mathcal{F}}_n$-measurable indicator; integrability at any fixed time follows from a finite sum of the absolute values at earlier times. HenceThis is the multiplicative upcrossing supermartingale construction. The printed $j=0$ case uses $N_0=-1$, while $Y$ was only defined from time zero. With the natural pre-start convention $Y_{-1}=1$, it is the constant process one and the assertion also holds for $j=0$. Without such an extension, that single displayed stopped process is undefined.

For the probability bound, the actual initial value isbecause $N_1=0$ exactly when $X_0\le a$. For an integer $k\ge 1$, stop after the $k$th completed upcrossing, at $N_{2k}$. The nonnegative stopped supermartingale satisfiessince $Y_{N_{2k}}=q^k$ on the event in the left side. Letting $n\to \infty$, and identifying $\{U\ge k\}=\{N_{2k}<\infty \}$, proves the Dubins upcrossing inequality:Neither finiteness of $N_{2k}$ nor uniform integrability is needed. The bound is for positive integers $k$; extending it to $k=0$ would generally be false, because its left side would be one while $\mathbb{E}\min (X_0/a,1)$ can be less than one.