Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 28 (Advanced Probability)

Let $(Z_k,{\mathcal{F}}_k{)}_{0\le k\le n}$ be an integrable real submartingale, and put $Z_n^{\ast }={\max }_{0\le k\le n}{Z}_k$. The refined Doob maximal inequality isIn particular, for a nonnegative submartingale the right-hand side is at most $\mathbb{E}{Z}_n$, giving the usual weak maximal bound.

For the proof, partition $A=\{Z_n^{\ast }\ge \lambda \}$ by its first crossing time: $A_k=\{Z_j<\lambda (j<k),Z_k\ge \lambda \}\in {\mathcal{F}}_k$. The submartingale property gives $\mathbb{E}[Z_n\mid {\mathcal{F}}_k]\ge Z_k$, soSumming over the disjoint crossing events proves the refined inequality, including a possible crossing at time zero. No independence assumption is needed.

The strong form is the Doob Lp maximal inequality. For $p>1$ and a nonnegative submartingale with $Z_n\in L^p$,Indeed, integrate the refined inequality against $p{\lambda }^{p-2}$ up to $K$. The tail integral formula for moments and the Tonelli theorem giveApply the Holder inequality, divide by $\Vert Z_n^{\ast }\wedge K{\Vert }_p^{p-1}$ when it is nonzero, and let $K\to \infty$ by the monotone convergence theorem. For a martingale $M$, the process $|M_k|$ is a submartingale by the conditional Jensen inequality, giving the corresponding bound for $\max |M_k|$. On a bounded continuous-time interval the nonnegative, right-continuous version follows by applying these inequalities on refining finite time grids and taking limits.

Use ${\mathcal{F}}_m=\sigma ({\xi }_1,\dots ,{\xi }_m)$ and $S_0=0$. Independence and zero means imply $\mathbb{E}[S_{m+1}\mid {\mathcal{F}}_m]=S_m$, so the partial sums form a square-integrable martingale. More explicitly,Thus $S_m^2$ is a nonnegative submartingale. The Doob maximal inequality at threshold $x^2$ now proves the Kolmogorov maximal inequality:Finally, expanding the terminal square and using independence and the zero means gives $\mathbb{E}{S}_n^2={\sum }_{j=1}^n\mathbb{E}{\xi }_j^2$. Identical distributions are not required.

Put $V=\mathbb{E}{X}_n^2$. For $c\ge 0$, the process $(X_m+c{)}^2$ is a nonnegative submartingale, by the conditional Jensen inequality. A crossing of $x$ by $X$ implies a crossing of $(x+c{)}^2$ by this squared process. Since $\mathbb{E}{X}_n=\mathbb{E}{X}_0=0$, the Doob maximal inequality givesFor $V>0$, differentiation of the last expression gives $2(cx-V)/(x+c{)}^3$, so its minimum over $c\ge 0$ occurs at $c=V/x$. Substitution proves the one-sided martingale maximal inequalityIf $V=0$, then $X_n=0$ almost surely and $X_m=\mathbb{E}[X_n\mid {\mathcal{F}}_m]=0$ for every $m\le n$, so the same conclusion holds. The constant is sharp: at one step take $X_1=x$ with probability $V/(V+x^2)$ and $X_1=-V/x$ otherwise. This has mean zero, variance $V$, and achieves equality.

Finite second moment implies $\mathbb{E}X<\infty$. For $\lambda$ near any positive ${\lambda }_0$, the derivative of $e^{-\lambda X}$ is bounded in absolute value by $X$, provided the neighbourhood stays in $(0,\infty )$. The dominated convergence theorem therefore permits differentiation under the expectation:For the expansion, put $r(z)=e^{-z}-1+z-z^2/2$. Taylor expansion gives $r(z)/z^2\to 0$ as $z\downarrow 0$. Also, for $z\ge 0$,The inequalities follow by integrating $0\le 1-e^{-u}\le u$ from zero to $z$. Consequently $r(\lambda X)/{\lambda }^2\to 0$ pointwise and its absolute value is bounded by $X^2/2$. Another application of the dominated convergence theorem gives the second-moment expansion of a Laplace transform:No third moment is needed for the remainder estimate.

First the exit time has sufficiently many finite moments. Over any interval of length $a^2$, a Brownian motion increment has probability $q=\mathbb{P}(|B_{a^2}|>2a)>0$ of absolute size larger than $2a$. If the process is still inside $(-a,a)$, such an increment forces it to exit before the next interval endpoint. Independence of increments at deterministic times therefore givesThus $T<\infty$ almost surely and all its positive moments are finite. This justifies extracting moments from its transform rather than presupposing them.

For $\lambda >0$, put $\theta =\sqrt{2\lambda }$. The processis a martingale, being the average of two exponential martingales for Brownian motion. Apply the optional stopping theorem at the bounded time $T\wedge t$. Since $|B_{T\wedge t}|\le a$, the stopped values are bounded by $\cosh (\theta a)$. Continuity gives $B_T\in \{-a,a\}$. Letting $t\to \infty$ by the dominated convergence theorem yieldsHence the Laplace transform of symmetric Brownian interval-exit time isThe value at zero follows from almost-sure finiteness. Expanding the reciprocal givesCompare with part (a), now applicable because of the geometric tail bound. The Brownian symmetric interval-exit moments are

Use the natural filtration ${\mathcal{G}}_n=\sigma ({\xi }_1,\dots ,{\xi }_n)$. Define the deterministic tail $r_n={\sum }_{j=n}^{\infty }{\delta }_j$ and put $Z_n={\xi }_n+r_n$. This is nonnegative and integrable. Since $r_n={\delta }_n+r_{n+1}$,Thus the deterministic summable drift correction produces a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives a finite limit $Z_{\infty }$ almost surely. Its finiteness is also reflected in the bound $\mathbb{E}{Z}_n\le \mathbb{E}{Z}_1$ and Fatou's lemma, which gives $\mathbb{E}{Z}_{\infty }\le \mathbb{E}{Z}_1<\infty$. The convergence theorem follows from the upcrossing argument: bounded expected negative parts of $-Z$ rule out infinitely many crossings of rational intervals, and the expectation bound excludes an infinite limit.

Since $r_n\to 0$, the required conclusion is $\boxed{{\xi }_n⟶Z_{\infty }\text{ almost surely, with }{Z}_{\infty }<\infty }$. The original sequence need not itself be a supermartingale; subtracting the vanishing deterministic tail recovers it from one that is.

Set $\inf \varnothing =\infty$. Since each $Y_j$ is ${\mathcal{F}}_j$-measurable, the eventis in ${\mathcal{F}}_n$, so ${\tau }_a$ is a stopping time. The stopped predictable budget for adapted increments is $A_n=a-U_{n\wedge {\tau }_a}$, for $n\ge 1$.

If ${\tau }_a=k<\infty$, the sum before the overshooting increment is $U_k={\sum }_{j<k}{Y}_j\le a$. If ${\tau }_a>n$, the sum through time $n$ is at most $a$, and in particular $U_n\le a$. ThusThe process is adapted, and boundedness gives integrability even when the increments have no finite expectations. Directly from the one-step-lagged sum,This decrement is ${\mathcal{F}}_n$-measurable, and $0\le Y_n{1}_{\{{\tau }_a>n\}}\le a$. ThereforeHence $A$ is a nonnegative supermartingale; in fact its paths are nonincreasing. At time zero one may set $U_0=0$ and $A_0=a$; then $A_1=a$ too. Stopping the sum including $Y_{{\tau }_a}$ would lose nonnegativity, which is why the shifted definition of $U$ matters.

Work from time one; a finite initial prefix does not affect convergence. Fix a positive integer $a$, and use the stopping time ${\tau }_a$ and budget $A_n$ from part (b). DefineIt is nonnegative. For each finite $n$, the stopped variable $X_{n\wedge {\tau }_a}$ is integrable: it is bounded by the finite sum $X_1+\cdots +X_n$, and the budget is bounded by $a$.

The stopped increment of $X$ isTaking conditional expectations, the assumed drift bound givesThe last term is bounded by $a$; no integrability assumption on the unlocalized $Y_n$ has been used. Its contribution is exactly cancelled by the budget decrement from part (b), givingThus $Z^{(a)}$ is an integrable nonnegative supermartingale. By the almost sure supermartingale convergence theorem it has a finite almost-sure limit. The bounded decreasing budget $A_n$ has a finite pathwise limit as well, so $X_{n\wedge {\tau }_a}=Z_n^{(a)}-A_n$ converges to a finite limit almost surely.

Take the intersection of these probability-one convergence events over all positive integers $a$. On the further probability-one event ${\sum }_{j\ge 1}{Y}_j<\infty$, choose an integer $a$ strictly larger than that total. Then ${\tau }_a=\infty$, so the stopped sequence is the original one. ConsequentlyThis proves supermartingale convergence with summable adapted drift under almost-sure summability alone; it does not require $\mathbb{E}{\sum }_j{Y}_j<\infty$.

Here is the one-parameter Kolmogorov continuity theorem. Suppose a real-valued stochastic process on $[0,1]$ satisfies, for some $p>0$, $q>0$ and finite $C$,Then it has a continuous modification of a stochastic process whose paths are Hölder continuous of every positive exponent $\gamma <q/p$. A modification means equality almost surely at each fixed time, not an a priori simultaneous equality at all times. Other compact time intervals follow by rescaling; the vector-valued version uses a norm in place of absolute value.

For the proof fix $0<\gamma <q/p$ and use the dyadic grids $D_n=\{j{2}^{-n}:0\le j\le 2^n\}$. Let $D_n^{\ast }$ be the largest adjacent increment on this grid. The union bound and Markov inequality giveThe right-hand side is summable. The Borel-Cantelli first lemma shows that, outside a null set, an almost-surely finite random constant $K$ can absorb the finitely many early grids so that

We now use dyadic increment chaining. For two distinct dyadic points $s,t$, choose $n$ with $2^{-(n+1)}<|t-s|\le 2^{-n}$. Approximate them from below by $s_j,t_j\in D_j$, for $j\ge n$. Each refinement moves by zero or one grid step, while $s_n,t_n$ are at most two grid steps apart. The sequences eventually equal the original dyadic points. Telescoping giveswhere a deterministic finite $C_{\gamma }$ absorbs the geometric series and $2^{-n}<2|t-s|$. Thus the sample function on the dense union of dyadic grids has a unique Hölder-continuous extension to $[0,1]$. Define this extension to be $\tilde{X}$ on the probability-one event, and set it to zero on the exceptional null set.

For a fixed $t$, take dyadic $t_j\to t$. The moment condition gives $X_{t_j}\to X_t$ in probability, while the constructed extension gives $X_{t_j}\to {\tilde{X}}_t$ almost surely. Uniqueness of a probability limit yields ${\tilde{X}}_t=X_t$ almost surely. This proves the required modification property. Performing the argument for a countable increasing sequence of exponents tending to $q/p$ gives one version with all smaller positive Hölder exponents, since its extensions agree on the dyadic points.

For Brownian motion, Gaussian increments give, for any integer $r\ge 2$,Use $p=2r$ and $q=r-1$. The theorem gives every exponent $\alpha <(r-1)/(2r)=1/2-1/(2r)$. For any $0<\alpha <1/2$, choose $r$ large enough. Intersect probability-one events over integer $r$ and compact intervals $[0,N]$ to obtainThe usual continuous Brownian version has the same property: its continuous paths agree with the constructed version on a countable dense set and hence everywhere, outside one null set. The Hölder constant depends on the compact interval. This is the precise local meaning of Brownian Hölder regularity; it is not a uniform Hölder bound on the entire unbounded time axis.

The force from one location is defined almost surely, since the uniform distribution assigns probability zero to the origin. Symmetry of that uniform distribution makes its characteristic function real:Near zero the integrand is bounded by $2$. For large $x$ it is at most $m^2{t}^2/(2x^4)$. The monotone convergence theorem therefore gives the finite integralIndependence of the locations gives the characteristic function of the total force as $q_n(t{)}^n$. For fixed $t$, $I_n(t)$ is bounded, so Taylor expansion of the logarithm givesThe logarithm is legitimate for all sufficiently large $n$.

For $t\ne 0$, substitute $u=m|t|/x^2$. The cosine integral for a symmetric stable exponent, evaluated directly in part (b), givesAt $t=0$ both sides vanish. The limiting characteristic function is continuous at zero, so the Lévy continuity theorem proves weak convergence of random variables to the symmetric stable distribution of index $1/2$ withThe coefficient reflects the spatial density $n/(2n)=1/2$. Symmetry was used in the characteristic function, without assuming that the singular force has a finite mean.

Interpret the signed inverse-power force as $m\mathrm{sign}(x)/|x{|}^p$, so noninteger $p$ is meaningful on either side of the origin. Write $\alpha =1/p\in (0,2)$. Exactly as in part (a), the one-star characteristic function is $1-I_{n,p}(t)/n$, whereThe integrand is bounded near zero and is at most $m^2{t}^2/(2x^{2p})$ at infinity. The condition $2p>1$ thus makes its integral finite. Independence and the logarithm calculation in part (a) give the limiting characteristic function $\exp (-I_p(t))$, withHere we substituted $u=m|t|/x^p$. The integral for $J_{\alpha }$ is finite because its integrand is $O(u^{1-\alpha })$ at zero and $O(u^{-1-\alpha })$ at infinity.

To determine the constant rather than leave it as an unevaluated integral, use the Gamma function representationThe integrand multiplied by $1-\cos u$ is nonnegative, so the Tonelli theorem permits exchanging the integrals. For $v>0$ the inner integral isConsequently, substituting $z=v^2$ and using the Euler beta function and Gamma reflection formula,Since $\Gamma (1+\alpha )=\alpha \Gamma (\alpha )$, this givesEvery factor in $c_p$ is positive in the stated range. The Lévy continuity theorem gives weak convergence of random variables because this characteristic function is continuous at zero. The limit is strictly stable: if $Z_1,Z_2$ are independent copies of it, the characteristic function of $a{Z}_1+b{Z}_2$ iswhich is the characteristic function of $(|a{|}^{\alpha }+|b{|}^{\alpha }{)}^{1/\alpha }{Z}_1$. This proves the asserted index, as well as identifying the symmetric stable distribution. In particular $p=2$ gives $c_2=\sqrt{\pi m/2}$, while $p=1$ gives $c_1=\pi m/2$, the centered Cauchy distribution with that scale.

Condition on the number $M$ of stars. The independence assumptions give $\mathbb{E}[e^{it{F}_n}\mid M]=q_n(t{)}^M$, with the inverse-square one-star characteristic function $q_n$ from part (a). Averaging over the Poisson distribution therefore gives the compound Poisson distribution formula directly:Since $I_n(t)\to \sqrt{\pi m/2}|t{|}^{1/2}$, the Lévy continuity theorem gives exactly the same weak convergence of random variables as in part (a):The random number of stars introduces no additional limiting term: its effect is already contained in the exact exponential formula. This is the symmetric inverse-power Poisson field on a line at spatial intensity $1/2$ and exponent $p=2$.

For $\epsilon >0$, the triangle inequality and the union bound giveThus $\boxed{X_n+Y_n⟶X+Y\text{ in probability}}$. This proves closure of convergence in probability under addition. No independence is needed; the random variables are compared on their given common probability space.

Take a random variable $Z$ with the Rademacher distribution, and setBoth marginal distributions equal the law of $Z$ for every $n$. Hence $X_n$ and $Y_n$ each have weak convergence of random variables to $Z$; one may take $X=Y=Z$ on this probability space. But their sums alternate:Indeed, at $t=\pi /2$ their characteristic functions are $-1$ for even $n$ and $1$ for odd $n$, so the sums have no weak limit at all. Separate marginal weak convergence does not imply convergence of the sum. The marginal weak convergence does not control sums counterexample changes the dependence while leaving the marginals fixed. Unlike convergence in probability, separate marginal weak convergence of random variables contains no information about the joint coupling.

Write the constant limit as $c$. First, convergence in distribution to a constant implies convergence in probability: the Portmanteau theorem applied to the closed set $\{y:|y-c|\ge \epsilon \}$ givesNow compare characteristic functions without imposing independence. For fixed $t\in \mathbb{R}$,The second bound uses $|e^{iu}-e^{iv}|\le \min \{2,|u-v|\}$. By convergence in probability, the limiting upper bound is $|t|\epsilon$; then let $\epsilon \downarrow 0$. Meanwhile $X_n$ has weak convergence of random variables to $X$, so $\mathbb{E}{e}^{it{X}_n}\to \mathbb{E}{e}^{itX}$. Thus the characteristic function of the sum tends to $e^{itc}\mathbb{E}{e}^{itX}$, the characteristic function of $X+c$, which is continuous at zero. The Lévy continuity theorem provesThis is the additive case of the Slutsky theorem, with its proof supplied. The constant limiting marginal removes the dependence obstruction in part (b).