Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 201 (Advanced Probability)

Relative to a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ and a filtration $({\mathcal{F}}_n)$, the real stochastic process $(M_n)$ is a martingale if it is adapted, each $M_n$ is an integrable random variable, andHere adapted means $M_n$ is ${\mathcal{F}}_n$-measurable, and integrability means $\mathbb{E}|M_n|<\infty$. By the tower property of conditional expectation, the equivalent multi-step condition is $\mathbb{E}[M_j\mid {\mathcal{F}}_i]=M_i$ for all $i\le j$. The filtration is part of the definition; specifying only the marginal means is insufficient.

The discrete-time Doob L2 maximal inequality states that a square-integrable martingale satisfiesEquivalently, the Lp norm of the maximum at $p=2$ is at most $2\Vert M_n{\Vert }_2$. The same assertion holds for a nonnegative square-integrable submartingale. No assumption $M_0=0$ is required; for a bound in terms of fluctuations from the initial value, apply it to $(M_k-M_0)$.

Use the natural filtration ${\mathcal{F}}_k=\sigma (X_1,\dots ,X_k)$ and define $M_k=S_k-mk$. The increments $X_k-m$ are centered, independent, and square integrable, so their conditional expectations given the past are zero. Thus $(M_k)$ is a square-integrable martingale with $M_0=0$. Independence givesApplying the Doob L2 maximal inequality yieldsThis includes ${\sigma }^2=0$, when every centered increment vanishes almost surely.

Put $M_k=S_k-mk$ as above. Centering the linear interpolation also interpolates the centered values: for $u\in [0,1]$,The absolute value of this convex combination is at most the larger endpoint absolute value. For every fixed $A>0$, part (c) therefore givesHence there is uniform convergence on compacts in probability to the deterministic path $h(t)=mt$, by Markov inequality. Equip $C([0,\infty ),\mathbb{R})$ with its standard compact-open topology, metrized byFor each finite number of terms their suprema converge to zero in probability, and the remaining tail is bounded deterministically by ${\sum }_{j>J}{2}^{-j}$. Thus $d(S^{(N)},h)\to 0$ in probability. Convergence in probability to a deterministic point implies weak convergence of probability measures, so the fluid limit isHere ${\delta }_h$ is the Dirac measure concentrated on that continuous path. The topology is locally uniform convergence; no assertion of uniform convergence over the entire unbounded half-line is needed. The interpolation is intended for integers $k\ge 0$, including the initial interval $[0,1]$. If the PDF's ${\mathbb{Z}}^{+}$ is interpreted as strictly positive integers, that initial piece is omitted from the displayed definition and must be supplied by the same formula using $S_0=0$.

For the forward implication, let $(M_n)$ be a martingale and let the stopping time $T$ satisfy $0\le T\le N$ for a deterministic integer $N$. The stopped variable is integrable because it uses only finitely many integrable values. The pathwise identityexpresses it through martingale increments. Since $\{T\ge j\}=\{T>j-1\}\in {\mathcal{F}}_{j-1}$, each summand has expectation zero by conditional expectation. ConsequentlyThis proves the bounded-time case of the optional stopping theorem directly; no assumptions about an unbounded stopping time are being used.

For the converse in the characterization of a martingale by stopped expectations, fix $n\ge 0$ and $A\in {\mathcal{F}}_n$. Define the bounded stopping timeIndeed, its only nontrivial sublevel event is $\{T_A\le n\}=A^c\in {\mathcal{F}}_n$. Applying the assumed stopped-expectation equality to $T_A$ and to the deterministic stopping time $n$ givesThis holds for every $A\in {\mathcal{F}}_n$. Since $M_n$ is adapted and both variables are integrable, the defining test-event property of conditional expectation yieldsTogether with the given adaptation and integrability, this is exactly the martingale property.

For $m\ge 1$, the increments have exponential momentThus $Z_n=2^{S_n}$ is an exponential martingale of a random walk relative to the natural filtration. We justify its stopping limit. From any surviving state in $(-m,m)$, a block of $2m$ successive increments all equal to $-1$ forces an exit. The block has probability $q=(6/7{)}^{2m}>0$, independently of the preceding increments. The geometric tail bound from a uniform escape probability givesso $T$ is finite almost surely and has finite expectation.

At the lower exit $S_T=-m$ exactly; at the upper exit $S_T$ is either $m$ or $m+1$. Before exit the same overall bound $-m\le S_{n\wedge T}\le m+1$ holds. Hence $Z_{n\wedge T}$ is bounded by $2^{m+1}$, uniformly in $n$. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{n\wedge T}=Z_0=1$, and the dominated convergence theorem now yieldsAllowing the upper overshoot by one is essential; the terminal state need not equal $m$ on upper exit.

The mean increment is $\mathbb{E}{X}_1=-6/7+2/7=-4/7$, so $M_n=S_n+(4/7)n$ is a martingale. Applying the bounded optional stopping theorem at $n\wedge T$ givesThe terminal sums are bounded by the exit-state bounds from (b), while $n\wedge T$ increases to the integrable stopping time $T$. The dominated convergence theorem on the left and the monotone convergence theorem on the right therefore proveThis is also Wald's equation here. In fact, the stopped identity already implies $\mathbb{E}(n\wedge T)\le 7m/4$ from $S_{n\wedge T}\ge -m$, giving another direct finite-expectation justification.

Write $p_m=\mathbb{P}(S_T=m)$, $p_{m+1}=\mathbb{P}(S_T=m+1)$, and $q_m=p_m+p_{m+1}$. Part (b) givesso $q_m\le 2^{-m}$. The exit-state decomposition also givesCombining this with part (c), including the upper overshoot, yieldsFor example, the same calculation supplies the quantitative boundThe asymptotic linear growth is determined by the negative mean increment; the exponentially unlikely upper exit gives a vanishing correction.

The martingale convergence theorem in its $L^1$-bounded form says that a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ has an integrable limit $M_{\infty }$ andThe norm bound on the limit follows from Fatou lemma. Boundedness in $L^1$ by itself does not imply convergence in L1. The uniformly integrable martingale convergence theorem gives the stronger conclusion: if $(M_n)$ is uniformly integrable, then $M_n\to M_{\infty }$ both almost surely and in L1 norm, and $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. Conversely, convergence in L1 implies uniform integrability.

For the distinction, the fair-coin doubling martingale $M_n=2^n{1}_{\{\text{first }n\text{ tosses are heads}\}}$ has expectation one for every $n$ but converges almost surely to zero. Its L1 norm remains one, so its convergence is not in L1 norm. These two formulations specify exactly which hypothesis is needed in part (d).

For $1<p<\infty$, the Lp martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|M_n{|}^p<\infty$ has a limit $M_{\infty }\in L^p$ such thatMoreover, $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$ and $\mathbb{E}|M_{\infty }{|}^p\le {\sup }_n\mathbb{E}|M_n{|}^p$. To see why $p>1$ matters, the Doob Lp maximal inequality gives an integrable dominating variable $({\sup }_n|M_n|{)}^p$. The martingale convergence theorem first supplies the almost-sure limit, then the dominated convergence theorem supplies convergence in the Lp norm. This maximal estimate is unavailable at $p=1$ in the required form.

Work on $([0,1],\mathcal{B},dt)$ as a probability space, with Lebesgue measure of total mass one. Let ${\mathcal{F}}_n$ be the filtration generated by the level-$n$ half-open dyadic cells together with the separate null cell $\{1\}$. Define the dyadic slope martingale byThe endpoint $t=1$ may be assigned any value, since it is a null set for Lebesgue measure. The two child slopes average to their parent slope, by telescoping the two increments. Therefore $\mathbb{E}[G_{n+1}\mid {\mathcal{F}}_n]=G_n$ almost everywhere, so this is a martingale. The Lipschitz condition gives $|G_n|\le K$ everywhere except possibly at the freely chosen endpoint, where we take zero.

Apply the Lp martingale convergence theorem with $p=2$. Its limit $g$ has $|g|\le K$ almost everywhere and $G_n\to g$ in L1 norm as well, by Cauchy-Schwarz inequality. Choose a measurable representative of $g$ and set it to zero on any exceptional null set; it is then a bounded measurable function on the entire interval.

Set $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$. Telescoping at the grid points shows that $f_n$ is the linear interpolation of $f$ on the dyadic grid. The Lipschitz condition gives $\Vert f_n-f{\Vert }_{\infty }\le K{2}^{-n}\to 0$. AlsoThe two uniform limits coincide, giving the absolutely continuous function representationThe chosen bound holds for every $t$ after the null-set modification; the integral identity holds for every $x$ simultaneously.

Use the same dyadic slope martingale $G_n$ and dyadic filtration as in (c). Each $G_n$ is integrable, since it takes finitely many finite values. On each dyadic cell, the absolute slope is $2^n$ times the absolute endpoint increment. Consequently the hypothesis in the PDF is exactlyThus $(G_n)$ is uniformly integrable. It is also bounded in L1 norm: choose a finite ${\lambda }_0>0$ at which the supremum of the tails is finite, and use $\Vert G_n{\Vert }_1\le {\lambda }_0+{\sup }_j{V}_j(f,{\lambda }_0)$. The uniformly integrable martingale convergence theorem supplies $g\in L^1([0,1])$ with $G_n\to g$ in L1 norm.

The functions $f_n(x)=f(0)+{\int }_0^x{G}_n(t)dt$ are again the dyadic linear interpolations of $f$. Since $f$ is continuous on a compact interval, it is uniformly continuous, and $\Vert f_n-f{\Vert }_{\infty }\le {\omega }_f(2^{-n})\to 0$, where ${\omega }_f$ is its modulus of continuity. On the other hand, the integral of $G_n-g$ is uniformly bounded in absolute value by $\Vert G_n-g{\Vert }_1\to 0$. Hence the dyadic slope-tail criterion for absolute continuity givesNo boundedness of $g$ is asserted here; the tail condition permits integrable densities that are unbounded.

Write $A_n={\max }_{0\le k<2^n}|{\xi }_{(k+1)2^{-n}}-{\xi }_{k{2}^{-n}}|$. The maximum of nonnegative numbers to the power $p$ is bounded by their sum, so the increment hypothesis givesTherefore $\Vert A_n{\Vert }_p\le C{2}^{-n(\beta -1/p)}$. The Minkowski inequality and completeness of $L^p$ imply convergence of the defining series wheneverFor a fully explicit argument, the norm of every tail is at most $2C{\sum }_{n\ge N}{2}^{-n(\beta -1/p-\alpha )}$, which tends to zero. Since the summands are nonnegative, their pointwise increasing sum equals this finite $L^p$ limit and is finite almost surely. This is a guaranteed range; no endpoint convergence follows from the displayed summability estimate. The assumptions control increments only and do not require ${\xi }_0\in L^p$.

Fix an exponent in the range from (a), and work on the single probability-one event where $K_{\alpha }<\infty$ and all dyadic values are finite. We give the dyadic increment chaining argument. For dyadic $s<t$, put $\delta =t-s$ and choose $n\ge 0$ with $2^{-n}\le \delta <2^{1-n}$. Let $s_j=2^{-j}\lfloor 2^{j}s\rfloor$ and $t_j=2^{-j}\lfloor 2^{j}t\rfloor$. The level-$n$ approximations are at most two grid steps apart, so their difference is bounded by $2A_n$. At each subsequent level an approximation either stays fixed or moves by one adjacent level increment. Because $s,t$ are dyadic, these approximations eventually equal $s,t$. ThusThe dyadic rationals are dense in $[0,1]$, so this uniform bound gives a unique continuous extension $X$ to the entire interval, withAll these equalities hold on that one event, not merely one event per dyadic time. Define $X$ to be the zero path on its null complement. Each $X_t$ is measurable as the limit of the random variables at deterministic left dyadic approximations. Alternatively, their measurable linear interpolations converge uniformly to $X$, which also shows measurability as a random element of $C([0,1])$. This is the continuous extension version of a continuous modification.

The Gaussian process assumption and the covariance formula give a centered Gaussian increment with varianceFor every finite $p>1$, a standard normal variable $Z$ therefore givesFor any $0<\alpha <1/2$, choose $p>\max (2,(1/2-\alpha {)}^{-1})$. Then $\alpha <1/2-1/p$, so parts (a) and (b) give a Hölder continuous function of exponent $\alpha$ as an extension. To obtain one extension with every required exponent, take a countable sequence ${\alpha }_j\uparrow 1/2$ and apply the construction with suitable $p_j$. Intersect the probability-one events. Their continuous extensions agree on the dense set $D$ and hence agree everywhere. For each smaller exponent choose $j$ with $\alpha <{\alpha }_j$; since $|t-s|\le 1$, the bound with ${\alpha }_j$ also implies the bound with $\alpha$. ThusThe exponent constants may depend on $\alpha$ and the sample path. This proves Brownian Hölder regularity for the continuous Gaussian extension without making an uncountable intersection of unrelated full-probability events.

Let ${\tau }_r=\inf \{t\ge 0:|B_t|=r\}$, ${\tau }_R=\inf \{t\ge 0:|B_t|=R\}$, and $\tau ={\tau }_r\wedge {\tau }_R$. The function $u(y)=|y{|}^{-1}$ is harmonic away from the origin: its radial Laplacian in dimension three isThe Itô formula therefore makes $u(B_{t\wedge \tau })$ a local martingale, bounded between $1/R$ and $1/r$, hence a true martingale. The annulus exit time is finite almost surely. For instance, exit from the containing open ball is finite because from every point in the open ball a sufficiently large Gaussian vector increment has a uniform positive probability of leaving it in unit time; iterate the geometric tail bound from a uniform escape probability.

The optional stopping theorem at $t\wedge \tau$, followed by the dominated convergence theorem, gives $\mathbb{E}u(B_{\tau })=u(x)$. By continuity, there is no radial overshoot and the two exit spheres cannot be hit simultaneously. Writing $p={\mathbb{P}}_x({\tau }_r<{\tau }_R)$ givesAll harmonicity and boundedness statements apply only in the annulus, which stays away from the singularity at zero.

Take a sequence $R\uparrow \infty$ with $R>|x|$. The events $\{{\tau }_r<{\tau }_R\}$ increase, and their union is $\{{\tau }_r<\infty \}$. Indeed, if the sphere of radius $r$ is reached at a finite time, continuity makes the path bounded up to that time, so a sufficiently large outer sphere has not yet been hit. Conversely, each event in the union includes a finite inner hit.

Taking the limit of part (a)'s Brownian sphere-hitting probability in dimension three yields the sharper formulaThis establishes the required escape probability without presupposing the transience of Brownian motion in dimension at least three.

First suppose $|x+z|\le r$ for some lattice point $-z$. If $|x+z|=r$, the desired hit is already at time zero. If $|x+z|<r$, finite exit from that bounded open ball and continuity give the required sphere hit almost surely.

It remains to consider $|x+z|>r$ for every $z\in {\mathbb{Z}}^3$. At each integer time $n$, choose a nearest lattice point $-z_n$ to $B_n$ by rounding its three coordinates, with a fixed rule at ties. Then $|B_n+z_n|\le \sqrt{3}/2$. Put $\delta =\min (r/2,1/4)>0$. Conditional on the past, $B_{n+1}-B_n$ is a standard three-dimensional Gaussian vector. Its probability density function is bounded below on the open ball of radius $\delta$ centered at $-(B_n+z_n)$ byMultiplying by the volume of that open ball gives a constant $\epsilon >0$, independent of $n$ and the past, such thatThe probability of avoiding all lattice-centered radius-$r$ open balls at the first $N$ positive integer times is therefore at most $(1-\epsilon {)}^N$, by iterated conditional expectation. Eventually the path enters one such open ball almost surely. Since the initial point was outside that particular open ball, continuity forces an earlier crossing of its sphere. This proves Brownian hitting of lattice spheres:A uniform chance of hitting some member of the periodic family replaces recurrence to any one prescribed sphere; there is no contradiction with (b).

For $a>0$, the Brownian reflection principle gives, for $t>0$,where $\Phi$ is the standard normal distribution function. Hence $H_a<\infty$ almost surely; for $a=0$, $H_0=0$.

For $u\ge 0$, use the Exponential martingale for Brownian motion $Z_t=\exp (u{B}_t-u^{2}t/2)$. At $H_a\wedge t$, it is bounded by $e^{ua}$ because $B_s<a$ before the first hit and $B_{H_a}=a$ by continuity. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{H_a\wedge t}=1$. Its limit is $e^{ua-u^2{H}_a/2}$, and the dominated convergence theorem yields $\mathbb{E}{e}^{ua-u^2{H}_a/2}=1$. Equivalently, the Brownian first-passage Laplace transform isTaking $\lambda =u^2/2$ proves the displayed exponential identity. The cases $u=0$ and $a=0$ are included. The restriction $u\ge 0$ is what gives the upper bound on the stopped exponential.

Fix a deterministic $a\ge 0$. By the Strong Markov property at the finite Brownian first-passage time $H_a$, the process ${\tilde{B}}_s=B_{H_a+s}-a$ is a standard Brownian motion started at zero. Such a process takes positive values arbitrarily soon almost surely: for every $\epsilon >0$, the Brownian reflection principle makes its maximum on $[0,\epsilon ]$ have the distribution of $|B_{\epsilon }|$, so the probability that the maximum is zero is zero. Intersecting these events for $\epsilon =1/n$ gives an infimum of positive-crossing times equal to zero. ThereforeThe infimum defining the strict crossing time need not itself be a time when $B_t>a$; continuity gives $B_{T_a}=a$. The quantifier here is fixed-level almost sure equality, not indistinguishability of stochastic processes as the level varies.

Let $A={\max }_{0\le t\le 1}{B}_t$, a random level. The Brownian reflection principle gives $A\stackrel{d}{=}|B_1|$, so $A>0$ almost surely. Also $A-B_1\stackrel{d}{=}|B_1|$: this follows by applying reflection to the reversed increment process $(B_1-B_{1-s}{)}_{0\le s\le 1}$, which has the law of a standard Brownian motion. Thus $B_1<A$ almost surely as well.

By continuity and compactness, $A$ is attained at some time strictly between zero and one, so $H_A<1$. It is not exceeded anywhere on $[0,1]$; since $B_1<A$, continuity even excludes an exceedance in a positive interval just after time one. Consequently $T_A>1$. Both conclusions hold on a single probability-one event, giving the fixed-level versus simultaneous Brownian passage-time equality distinction:Thus the simultaneous assertion is false. Equality for each deterministic level, and even simultaneously for all rational levels, cannot be extended to all real levels by an uncountable intersection.

Use the usual completed right-continuous Brownian filtration. The strict passage process $(T_a{)}_{a\ge 0}$ is finite at every level on one probability-one event: finiteness at all integer levels from (a) suffices by monotonicity. Also $T_0=0$ almost surely by (b).

For each fixed $a$, $T_a$ is a stopping time with $B_{T_a}=a$. The Strong Markov property shows that the passage times above this level depend on a new independent standard Brownian motion. Thus for $0\le a<b$, $T_b-T_a$ is independent of the past at $T_a$ and has the same law as $T_{b-a}$. Iteration gives independent increments and stationary increments in the level parameter. From (a) and fixed-level equality,For $\lambda ,\eta >0$, this also proves continuity in probability at zero, sinceFinally, with $M_t={\sup }_{s\le t}{B}_s$, we have $T_a=\inf \{t:M_t>a\}$. This strict generalized inverse of a nondecreasing function is right-continuous: if $a_n\downarrow a$, then for any $t>T_a$ we have $M_t>a$, and eventually $M_t>a_n$, which forces $T_{a_n}\le t$. Monotonicity gives the reverse bound. Monotonicity and local finiteness also give finite left limits. Hence $T$ is the Brownian first-passage subordinator, with càdlàg paths.

Now condition on this clock, which is independent of $W$. For a deterministic partition $0=a_0<a_1<\cdots <a_k$, write $\Delta T_j=T_{a_j}-T_{a_{j-1}}$ and $\Delta X_j=W_{T_{a_j}}-W_{T_{a_{j-1}}}$. Conditional on the clock these are independent centered Gaussian increments, with respective variances $\Delta T_j$. ConsequentlyFactorization and dependence only on interval lengths prove independent increments and stationary increments for $X$. For small $h$, $T_h\to 0$ in probability, and independence and continuity of $W$ imply $W_{T_h}\to 0$ in probability. For example, bound its deviation probability by $\mathbb{P}(T_h>\eta )+\mathbb{P}({\sup }_{t\le \eta }|W_t|>\epsilon )$ and then let $h\downarrow 0$ and $\eta \downarrow 0$. Stationary increments give stochastic continuity at every deterministic level. Composition of the continuous path of $W$ with the nondecreasing càdlàg clock gives càdlàg paths for $X$, and $X_0=0$. ThusThis is subordination of a Lévy process. Using strict passage times ensures the required right-continuous path choice, despite their fixed-level equality with the non-strict times.

Specify the sign convention for the Lévy characteristic exponent byConditioning on $T_a$ and using the Brownian first-passage Laplace transform givesThereforeThe absolute value is essential for negative $u$. The process is the standard symmetric Cauchy process; for $a>0$, $X_a$ has the Cauchy distribution with location zero and scale $a$. If the exponent convention instead uses $\mathbb{E}{e}^{iu{X}_a}=e^{a\psi (u)}$, the answer is $\psi (u)=-|u|$.