Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 33 (Advanced Probability)

Use the filtered probability space $([0,1),\mathcal{B}([0,1)),({\mathcal{F}}_n),\lambda )$, with its half-open dyadic intervals. Complete the initial definition by putting $X_0=\mu ([0,1))$. This is the same cell-average formula at level zero, where the only cell is $[0,1)$.

The function $X_n$ is ${\mathcal{F}}_n$-measurable and nonnegative. Its expectation isEvery parent cell is the disjoint union of its two half-open child cells. On $D_k^n$, the conditional expectation of $X_{n+1}$ is its cell average:Thus $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$, including at $n=0$. This proves that $X$ is a nonnegative martingale, specifically the dyadic density martingale. The half-open endpoints are essential to the disjoint partition for an arbitrary measure, which may have atoms.

The martingale convergence theorem in the form needed here states that a nonnegative supermartingale with uniformly bounded expectations converges almost surely to a finite integrable limit. Here $\mathbb{E}{X}_n=\mu ([0,1))$ for all $n$, so it applies to $X_n$. The ordinary Fatou lemma givesFor fixed $n$, apply the Conditional Fatou lemma to the nonnegative variables $X_m$, $m\ge n$. The martingale property makes each $\mathbb{E}[X_m\mid {\mathcal{F}}_n]$ equal to $X_n$, henceThe inequality is simultaneous for all $n$ after removing a countable union of null sets. Equality of expectations in the limit has not been assumed; it can fail for a measure with a singular component.

Write $\eta (A)={\int }_A{X}_{\infty }d\lambda$. This is a finite positive measure. For $A\in {\mathcal{F}}_n$, the conditional expectation inequality givesThe union $\mathcal{A}={\bigcup }_n{\mathcal{F}}_n$ is an algebra of sets generating $\mathcal{B}([0,1))$, since half-open dyadic intervals generate all open sets and hence all Borel sets. The class $\{A:\eta (A)\le \mu (A)\}$ is closed under increasing and decreasing limits, by continuity of these finite measures. The Monotone class theorem extends the inequality from $\mathcal{A}$ to every Borel set. Approximation by nonnegative simple functions and monotone convergence then giveThe difference $\nu =\mu -\eta$ is therefore a finite nonnegative measure, not merely a signed measure. For $A\in {\mathcal{F}}_n$,Consequently its restricted Radon-Nikodym derivative isChoose the cell-average versions of all the conditional expectations. Then $Y_n$ is constant on each level-$n$ cell and equals $2^n\nu (D_k^n)$ there, so these versions are nonnegative at every point.

Put $Z_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$. It is a conditional-expectation martingale. We justify why its limit is $X_{\infty }$, rather than just quoting convergence to an unspecified random variable.

First the family is uniformly integrable. For $A=\{Z_n>K\}\in {\mathcal{F}}_n$, nonnegativity and the conditional expectation identity giveThus its tail expectations are at most twice the integrable tail of $X_{\infty }$, uniformly in $n$. The uniformly integrable martingale convergence theorem gives a limit $Z$ almost surely and in $L^1$. For any $A\in {\mathcal{F}}_m$ and $n\ge m$,Pass to the $L^1$ limit and extend from $\mathcal{A}$ to all Borel sets by the Monotone class theorem. The equality of these integrals forces $Z=X_{\infty }$ almost surely. Combining this with the preceding convergence of $X_n$ proves

The event $A_{n,\epsilon }=\{Y_n\le \epsilon \}$ belongs to ${\mathcal{F}}_n$. Its $\nu$-measure can therefore be computed using the finite-level density:Let $H=\{{\mathrm{limsup}}_n{Y}_n=0\}$. Because the chosen versions satisfy $Y_n\ge 0$ everywhere, every point of $H$ eventually lies in all the $A_{n,\epsilon }$. For each $N$, the set ${\bigcap }_{n\ge N}{A}_{n,\epsilon }$ has $\nu$-measure at most $\epsilon$; these sets increase as $N$ increases. Continuity from below givesLetting $\epsilon \downarrow 0$ yields $\nu (H)=0$. In contrast, the preceding part says $\lambda (H)=1$. Thus $\nu$ is concentrated on the $\lambda$-null set $H^c$. This proves that $\nu$ and $\lambda$ are mutually singular measures, and completes the martingale construction of Lebesgue decomposition:The key step is that the density identity is available on each ${\mathcal{F}}_n$ even though $\nu$ need not have a density on the full Borel sigma-algebra.

Write the residual process asFor $s<t<1$, the filtration tells us whether each $U_i$ has arrived by time $s$, and its exact arrival time if it has. On $\{U_i>s\}$, its conditional law is uniform on $(s,1)$. Independence of the original $U_i$ means that the histories of the other arrivals do not alter this law. HenceThis conditional identity can also be checked against products of bounded functions of the censored arrival times and extended to ${\mathcal{F}}_s$ by the Monotone class theorem. Summing the identity gives $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$.

Each $M_t$ is adapted and integrable, withIts paths are smooth between the finitely many arrival times and have downward jumps of size ${\theta }_i/(1-U_i)$ at arrivals before time one. The indicators use $U_i>t$, so the value at a jump is the post-jump value: paths are right continuous with finite left limits on $[0,1)$. Thus $M$ is a càdlàg martingale, specifically the uniform-arrival residual mass martingale.

Since there are finitely many arrival times and each is strictly below one almost surely, $U_{\ast }:={\max }_i{U}_i<1$ almost surely. For every $t\ge U_{\ast }$ with $t<1$, all weights have arrived and $M_t=0$. Therefore $M_t\to 0$ almost surely as $t\uparrow 1$.

Nevertheless $\mathbb{E}{M}_t=S>0$ for every $t<1$. If the family were uniformly integrable, its almost-sure convergence along any deterministic sequence tending to one would imply convergence in $L^1$, and hence convergence of expectations to zero. This is impossible. Thus the martingale is not uniformly integrable. Its constant expected mass is carried by increasingly rare, late arrivals.

If $S=1$, the strict inequality in question already fails at time zero, almost surely, since $X_0=0$. Its probability is therefore zero, as required. Assume for the rest that $0<S<1$, so $M_0=S<1$.

Before time one the boundary inequality is equivalent to $M_t<1$. The process $M$ increases continuously between arrivals and has only downward jumps. Its first time $T<1$ of reaching level one must consequently be a continuous crossing, except possibly if an arrival occurs at the very same instant. All possible first crossing times belong to the finite deterministic setA uniform arrival hits none of these times almost surely. Thus $M_T=M_{T-}=1$ on $\{T<1\}$. This finite candidate set also proves the stopping-time property directly: determining whether a crossing has occurred by time $t$ only uses the arrival history up to $t$. The boundary set always contains time one because $X_1=S$, so $T\le 1$.

For a deterministic $r<1$, the martingale is bounded on $[0,r]$ by $S/(1-r)$. The optional stopping theorem for a bounded stopping time therefore givesMoreover $0\le M_{T\wedge r}\le 1$: if $T\le r$ its value is one, and otherwise no crossing has yet occurred. As $r\uparrow 1$, the stopped value tends to one on $\{T<1\}$ and to zero on $\{T=1\}$, because after the last arrival the unstopped process is zero. Dominated convergence now givesThis is the finite-jump boundary crossing identity. It uses uniform integrability of the bounded stopped values, not uniform integrability of the original family, which fails.

If ${\sigma }_1>n$, all the integer-time positions through $n$ lie in $(-1,1)$, so every increment over an intervening unit interval has absolute value at most two. The unit-time increments of Brownian motion are independent standard normal random variables. PutThe inclusion of events and independence giveIn particular ${\sigma }_1$ is finite almost surely. For real $t\ge 0$, its tail is at most ${\rho }^{\lfloor t\rfloor }$. The tail integral formula for moments therefore gives, for every $p>0$,An exponential geometric factor dominates the polynomial factors in this sum. This proves the requested moment finiteness for $p>1$, and in fact for every positive $p$.

The process $W_t=a^{-1}{B}_{a^{2}t}$ is standard Brownian motion, by Brownian scaling. Its first exit from $(-1,1)$ occurs at ${\sigma }_a/a^2$, soThe first part ensures integrability. Set $C=\mathbb{E}{\sigma }_1$; it is finite and strictly positive because path continuity at zero gives ${\sigma }_1>0$ almost surely. ThereforeWe keep $C$ unspecified until the comparison of limits requested below.

At each finite stopping time ${\sigma }_a^k$, the Strong Markov property says that $B_{{\sigma }_a^k+t}-B_{{\sigma }_a^k}$ is a standard Brownian motion independent of the past sigma-algebra. Its exit time from $(-a,a)$ is exactly ${\sigma }_a^{k+1}-{\sigma }_a^k$. By induction the stopping times are finite almost surely, and these durations are independent with common law ${\sigma }_a$, not merely identically distributed. This is the Brownian exit-time skeleton.

Put $v=\mathrm{Var}({\sigma }_1)<\infty$, which follows from the moment estimate. Brownian scaling gives $\mathrm{Var}({\sigma }_a)=a^{4}v$. Forthere are $2^{2n}$ independent durations of mean $C{2}^{-2n}$ and variance $v{2}^{-4n}$. ConsequentlyFor each $\epsilon >0$, Chebyshev's inequality givesApply the Borel-Cantelli first lemma for a countable sequence of positive errors tending to zero. We obtain the Brownian exit-skeleton clock convergence:No independence between the clocks for different values of $n$ is required.

Let ${\xi }_j=B_{{\sigma }_1^j}-B_{{\sigma }_1^{j-1}}$. Continuity makes each displacement equal to $+1$ or $-1$. Reflection symmetry of a fresh Brownian motion gives probability one half to each sign, and the Strong Markov property makes these signs independent across successive exits. Thus $B_{{\sigma }_1^n}={\sum }_{j=1}^n{\xi }_j$.

For the normalized position, the characteristic function issince $\log \cos z=-z^2/2+O(z^4)$ near zero. By the Lévy continuity theorem, this provesThis is the Gaussian limit of a Brownian exit skeleton.

Compare along the subsequence $n=2^{2m}$ with ${\tau }_m={\sigma }_{2^{-m}}^{2^{2m}}$. Joint Brownian scaling of the successive exits givesThe right-hand side converges weakly to $N(0,1)$. The preceding part and path continuity give $B_{{\tau }_m}\to B_C$ almost surely on the original Brownian probability space. Hence its weak limit is also the law $N(0,C)$. Uniqueness of the weak limit forces $C=1$, so finally $\mathbb{E}{\sigma }_a=a^2$. Independence of the position and its random clock was not assumed in this comparison.

For $x>0$, reflect a Brownian path after its first hitting of $x$. The Strong Markov property and symmetry preserve its law and exchange the events $\{S_1\ge x,B_1<x\}$ and $\{B_1>x\}$. The terminal equality event has probability zero, so the Brownian reflection principle givesContinuity of this distribution also gives no atom at zero. ThereforeA useful global bound follows from the maximum of the standard normal distribution density:

Monotonicity of $f$ gives, for each $n\ge 1$,Hence ${\sum }_{n\ge 1}f(2^{-n})<\infty$; the additional $n=0$ term is finite by continuity at one. Also $f(t)\to 0$ as $t\downarrow 0$, since a positive limiting lower bound would make the integral diverge.

By Brownian scaling, $S_{2^{-n-1}}$ has law $2^{-(n+1)/2}{S}_1$. Apply the small-value bound for the Brownian running maximum to obtainSumming yields the requested finite probability sum. The argument needs no independence of these nested-time events.

The Borel-Cantelli first lemma applied to that finite sum says that almost surely, for all sufficiently large $n$,For $2^{-n-1}\le t\le 2^{-n}$, both the Brownian running maximum and the positive function $\sqrt{t}f(t)$ are nondecreasing. Thus eventuallyThis proves the lower limit bound one, with interpolation across all small real times rather than only the dyadic sequence.

For every positive integer $m$, the function $mf$ satisfies the same assumptions and integrability condition. Repeating the argument gives an eventual lower bound $S_t\ge m\sqrt{t}f(t)$ on a probability-one event. Take the countable intersection of these events. On it, the ratio is eventually at least any prescribed positive integer, proving the integral lower envelope for the Brownian maximum:

Let $R=\{B_t:0\le t\le 1\}$ and $h=1/2$. For planar Brownian motion, the process $h^{-1/2}{B}_{ht}$ is again standard Brownian motion. Thus its time-$h$ range has law $\sqrt{h}R$. Two-dimensional Lebesgue measure scales by the square of the spatial factor, givingThe increments $B_{h+t}-B_h$, $0\le t\le h$, form another standard Brownian motion. Its range has the same law as $A_1$, and translation by $B_h$ does not change its area. HenceThe equality concerns the laws of the areas. No independence between these two original areas is needed here.

Path continuity makes each time-segment range compact and therefore Borel measurable. The membership event is jointly measurable in the sample and spatial variable: distance to the range equals the infimum of $|B_t-y|$ over a countable dense set of times. We may consequently apply the Tonelli theorem to its indicator.

Put $m=\mathbb{E}\lambda (R)<\infty$, using the allowed integrability hypothesis. Since $R=A_1\cup A_2$, finite-measure inclusion-exclusion and the preceding expectation identities giveFinally Tonelli theorem givesFiniteness of $m$ is what justifies the subtraction of expectations; an infinite-minus-infinite calculation would not establish this conclusion.

Set $W_t^{-}=B_{h-t}-B_h$ and $W_t^{+}=B_{h+t}-B_h$, $0\le t\le h$, with $h=1/2$. Both start at zero and are continuous. Increments of $W^{-}$ over consecutive disjoint time intervals are negatives of Brownian increments over corresponding disjoint past intervals, read in reverse order. They are independent centered normal random variables with covariance equal to the interval length times the identity. The same properties hold for $W^{+}$ using future increments. Every past increment is independent of all the future increments after deterministic time $h$, so the two whole processes are independent, by finite-dimensional independence and generation by rational-time evaluations. They are the independent Brownian arms at a deterministic time.

Let $R^{-}$ and $R^{+}$ be their ranges. Pathwise,Although $W^{-}$ need not be independent of $B_h$, this last equality uses only translation invariance of Lebesgue measure. Put $p(y)=\mathbb{P}(T_y\le h)$ for a standard Brownian motion started at zero. Independence of the centered processes and Tonelli theorem giveSince $p\ge 0$, it follows that $p(y)=0$ for Lebesgue-almost every $y$. ThereforeThe scaling relation in part (a) gives $\mathbb{E}\lambda (R)=2\mathbb{E}\lambda (A_1)=0$. A nonnegative random variable of expectation zero vanishes almost surely, proving the area of a planar Brownian path resultThe integral argument establishes the needed almost-everywhere spatial assertion; it does not require a prior theorem that individual points are polar.

Let $E_i=\{M(I_i)=0\}$. For any subset $K\subseteq \{1,\dots ,n\}$, simultaneous voidness is voidness of the union. Since the intervals are disjoint, the assumed void probability functional givesThis includes the empty subset. To check all event patterns explicitly, apply the inclusion-exclusion principle to the complements in a chosen pattern:Thus the void events are mutually independent, and their complementary occupied-cell indicators are mutually independent too. This is joint independence, not merely pairwise independence.

At level $n$, the $N_n$ dyadic cells fully contained in $J$ are disjoint. Their occupied indicators are independent by the preceding part, and every cell has length $2^{-n}$, so each has success probabilityTheir sum therefore satisfiesIf there are no fully contained cells, $N_n=0$ and this means the deterministic zero count. The binomial law follows directly from independence of occupation indicators, without assuming that the full counts $M(D_k^n)$ are already Poisson.

Refining an occupied cell replaces it by two child cells, at least one of which is occupied. All children of a cell contained in $J$ remain contained in $J$, and additional cells can enter the sum. Hence $M_n(J)$ is nondecreasing, pathwise, and is at most $M(J)$.

We must also address boundary points for a fixed interval $J$. For every fixed deterministic $x$, take bounded intervals $I_r$ containing $x$ with lengths tending to zero. The assumed void law givesThus the fixed endpoints of $J$ are almost surely not atoms. This is the exponential void probabilities forbid fixed atoms observation; it is not a simultaneous assertion for all possible real endpoints.

In a fixed bounded interval containing $J$, local finiteness gives only finitely many atoms. Simplicity makes their locations distinct, so there is a positive minimum distance between any two, unless there is at most one. Every atom in the interior of $J$ also has positive distance from its endpoints. For sufficiently fine dyadic cells each such atom lies alone in a cell fully contained in $J$, including if the atom itself is at a dyadic boundary because the cells are half-open. Thus the dyadic occupancy approximation of a simple point measure eventually equals $M(J)$, almost surely. In particular,

The union of fully contained level-$n$ cells misses at most two boundary portions of total length at most $2^{1-n}$. Consequently $N_n{2}^{-n}\to \lambda (J)$, and $p_n/2^{-n}\to 1$. For $0\le z\le 1$, the probability generating function from the binomial calculation givesThe convergence follows by expanding the logarithm, since $N_n{p}_n\to \lambda (J)$ and $N_n{p}_n^2\to 0$. Since the counts converge almost surely, bounded convergence identifies the same expression as $\mathbb{E}[z^{M(J)}]$. It is the generating function of the Poisson distribution, whose power-series coefficients determine its law. Thus

For finitely many disjoint bounded intervals $J_1,\dots ,J_r$, all dyadic cells used in their level-$n$ sums are disjoint. The independent occupied indicators proved in part (a) form independent groups, so $M_n(J_1),\dots ,M_n(J_r)$ are independent. Their almost-sure limits are also independent: for $0\le z_j\le 1$, bounded convergence givesThis joint probability generating function determines the joint law as that of independent Poisson counts. If an interval is unbounded, truncate all intervals by $[0,R)$ and let $R\uparrow \infty$ in the joint Laplace transforms. An unbounded interval has infinite count almost surely, since its Poisson means tend to infinity; this extends the independence assertion to such intervals as well.

For an interval-step function $s={\sum }_j{a}_j{1}_{J_j}$ with $a_j\ge 0$ and disjoint bounded intervals, independent Poisson distributions yieldWe now extend this to arbitrary nonnegative Borel functions, rather than stopping at interval-step functions. The mean measure $m(A)=\mathbb{E}M(A)$ is countably additive by Tonelli theorem, is finite on bounded intervals, and agrees with $\lambda$ on intervals by the preceding Poisson count calculation. Uniqueness of finite measures on each bounded Borel space, then exhaustion by bounded spaces, gives $m=\lambda$ on every Borel set. Simple-function approximation therefore givesFor bounded $f$ supported in a bounded interval, choose nonnegative interval-step functions $s_k$ with $\int |s_k-f|d\lambda \to 0$. Their existence follows by approximating measurable level sets in finite Lebesgue measure by finite unions of intervals, then approximating $f$ by simple functions. The mean-measure identity impliesThe function $e^{-u}$ is one-Lipschitz on $[0,\infty )$, so the expectations of the exponentials converge. The deterministic integrals of $1-e^{-s_k}$ converge by the same Lipschitz bound. Hence the Laplace formula holds for this bounded Borel $f$.

For general $f\ge 0$, put $f_k=(f\wedge k)1_{[0,k)}$. These functions increase to $f$, giving $M(f_k)\uparrow M(f)$ and $\int (1-e^{-f_k})d\lambda \uparrow \int (1-e^{-f})d\lambda$. Bounded convergence for the exponentials and monotone convergence for the integrals prove the Laplace functional of a Poisson random measure:with $e^{-\infty }=0$.

Finally take $f={\sum }_j{u}_j{1}_{A_j}$ for disjoint Borel sets of finite Lebesgue measure. The formula gives the product joint Laplace transform of independent Poisson counts with means $\lambda (A_j)$. Uniqueness of that transform proves the required count laws on all Borel sets. For infinite-measure sets the formula gives $\mathbb{E}{e}^{-uM(A)}=0$ for $u>0$, so their counts are infinite almost surely. Together with the given local finiteness, these facts show that $M$ is a Poisson random measure with intensity $\lambda$. This completes the exponential void probabilities characterize a Poisson random measure argument; simplicity was used to recover counts from fine-cell occupancy.