Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 24 (Advanced Probability)

The Martingale convergence theorem says that a discrete-time martingale $(M_n,{\mathcal{F}}_n)$ satisfyinghas an almost sure convergence limit $M_{\infty }$, finite almost surely and in $L^1$. The theorem asserts that the limit is integrable; it does not assert convergence in L1. More generally, the almost sure submartingale convergence theorem applies to a submartingale with ${\sup }_n\mathbb{E}{M}_n^{+}<\infty$. Uniform integrability is the additional condition that upgrades a martingale's convergence to convergence in L1.

For the requested distinction, let $({\xi }_k)$ be independent fair Bernoulli variables and use their natural filtration. The coin-doubling martingaleis a nonnegative martingale: conditionally on ${\mathcal{F}}_n$, the next factor is $2{\xi }_{n+1}$ with mean one, so $\mathbb{E}(M_{n+1}\mid {\mathcal{F}}_n)=M_n$. Also $\mathbb{E}|M_n|=\mathbb{E}{M}_n=1$ for every $n$, giving the required uniform $L^1$ bound. The probability that all the Bernoulli variables equal one is ${\lim }_n{2}^{-n}=0$. Therefore a zero is eventually encountered almost surely, after which $M_n$ stays zero. ThusThere can be no other $L^1$ limit, since convergence in L1 implies convergence in probability, whose limit must agree with the almost sure limit. This martingale satisfies the almost sure theorem but does not converge in $L^1$.

Put $v=\mathbb{E}{X}_n^2$. For $c\ge 0$, conditional Jensen inequality applied to the convex function $x\mapsto (x+c{)}^2$ shows thatis a nonnegative submartingale. Its integrability follows from the square integrability of $X_k$. If $X_k\ge \lambda$, then $Z_k\ge (\lambda +c{)}^2$, since $c\ge 0$. The Doob maximal inequality for a nonnegative submartingale yieldswhere zero mean removes the cross term. For completeness, the maximal inequality follows by stopping at the first crossing: on the event of a crossing at $k$, the submartingale property gives $\mathbb{E}[Z_n{1}_{\{T=k\}}]\ge \mathbb{E}[Z_k{1}_{\{T=k\}}]$. Sum over $k\le n$, and use nonnegativity on the event of no crossing.

The derivative of the last ratio isFor $v>0$, the minimum over $c\ge 0$ is attained at $c=v/\lambda$. Substitution gives the one-sided maximal inequality for a centered square-integrable martingale:If $v=0$, $X_n=0$ almost surely and $X_k=\mathbb{E}(X_n\mid {\mathcal{F}}_k)=0$ almost surely for every $k\le n$, so the bound also holds. The optimization is the same one underlying the Cantelli inequality, but the submartingale argument controls the entire finite-time maximum.

A real Lévy process is a real-valued stochastic process $(X_t{)}_{t\ge 0}$ with the following properties:

One convention also includes càdlàg paths in the definition. Equivalently, under the intrinsic definition above one chooses the càdlàg modification, which exists for such a stochastic process. Thus the usual working version of a Lévy process has right-continuous paths with left limits. There is no assumption of finite moments or continuous paths.

Fix $u\in \mathbb{R}$ and let $s\to t$. For any $\delta >0$, use $|e^{ia}-e^{ib}|\le \min (2,|a-b|)$ to obtainThe probability tends to zero by stochastic continuity. Taking the limit superior and then letting $\delta \downarrow 0$ provesThis proves continuity of Lévy characteristic functions from the elementary estimate, without requiring moments or replacing convergence in probability by an unjustified almost sure limit. At $t=0$, time approaches from the right. If stochastic continuity is formulated only at zero, stationary increments give the same argument at every $t$: the absolute value of $X_s-X_t$ has the law of $|X_{|s-t|}|$. For $u=0$ the characteristic function is identically one.

For fixed $u$, write $h(t)={\phi }_{X_t}(u)$. Independent increments and stationary increments giveThe preceding part gives continuity. Also $h$ never vanishes: if $h(t_0)=0$ with $t_0>0$, then $h(t_0/m{)}^m=0$ for every positive integer $m$, contradicting $h(t_0/m)\to 1$.

Here is a direct proof of the exponential form of Lévy characteristic functions. Choose $a>0$ small enough that $|h(t)-1|<1/2$ on $[0,a]$. The principal complex logarithm gives a continuous $L(t)=\log h(t)$ there, with $L(0)=0$. For $s,t\ge 0$ with $s+t\le a$, the multiplicative identity impliesThis difference is continuous on the connected triangle of allowed $(s,t)$ and equals zero at $(0,0)$, so it is identically zero. Thus $L$ satisfies the additive Cauchy functional equation locally. Subdivision gives $L(a/m)=L(a)/m$ and $L(ka/m)=kL(a)/m$; continuity then gives $L(t)=tL(a)/a$ for every $t\in [0,a]$.

Define $\eta (u)=L(a)/a$. For any $t\ge 0$, choose an integer $m$ with $t/m\le a$; thenThe coefficient is unique: if two coefficients give the same exponential for every $t\ge 0$, their derivatives at zero agree. In particular $\eta (0)=0$, and $\mathrm{Re}\eta (u)\le 0$ follows from $|h(t)|\le 1$. The characteristic exponent of a Lévy process has therefore been obtained from first principles, without invoking the Lévy–Khintchine formula.

Take two independent rate-one Poisson processes $N^{+}$ and $N^{-}$, and letThe difference of independent Poisson processes starts at zero and has stationary increments and independent increments. Its paths are càdlàg. For an interval of length $h$, the probability of any jump is $1-e^{-2h}\to 0$, proving stochastic continuity. Thus $X$ is a Lévy process. The characteristic function of a Poisson distribution with mean $t$ is $\exp (t(e^{iu}-1))$, so independence givesThe sample paths are integer-valued step functions with jumps $+1$ or $-1$. On every bounded interval there are only finitely many jumps, and independent Poisson arrival times coincide with probability zero. The combined arrival rate is two: holding times are independent exponentials of rate two, and each jump direction has probability $1/2$, independently of the holding times. This is equivalently a Compound Poisson process of rate two with Rademacher distribution jump sizes. Its paths have finite variation on compact time intervals, although there are infinitely many jumps over the whole half-line almost surely.

Use the independent increments of Brownian motion, rather than merely checking that an Itô formula drift vanishes. For $0\le s<t$, put $h=t-s$ and write $B_t=B_s+Z$, where $Z$ is independent of ${\mathcal{F}}_s$ and has normal distribution $N(0,h)$. Its first four moments are $0,h,0,3h^2$. ThereforeFor the cubic expression, the coefficient of $B_s$ after conditioning is $3(t-s)+\alpha (t)$, so $\alpha (t)=-3t$ makes it $\alpha (s)$. For the quartic expression, choose $\beta (t)=-6t$. Its conditioned coefficient of $B_s^2$ is then $6(t-s)-6t=-6s$. The constant term becomeswhich equals $3s^2$ when $\gamma (t)=3t^2$. Thus a standard choice isThe resulting stochastic processes are Hermite polynomial martingales $H_3(B_t,t)$ and $H_4(B_t,t)$. They are genuine integrable martingales, since Gaussian moments are finite at every finite time and the displayed conditional identities establish the martingale property directly.

The choice is not unique. Constants $c,d,e\in \mathbb{R}$ give the valid family $\alpha (t)=c-3t$, $\beta (t)=d-6t$, and $\gamma (t)=3t^2-dt+e$: these add $c{B}_t$ to the cubic martingale and $d(B_t^2-t)+e$ to the quartic one. The boxed choice sets these harmless additions to zero.

Write $\tau ={\tau }_x$ and $T_t=t\wedge \tau$. Path continuity gives $|B_{T_t}|\le x$. Stop the martingale $B_s^2-s$ at the bounded stopping time $T_t$ and apply the optional stopping theorem:By the monotone convergence theorem, $\mathbb{E}\tau \le x^2$, so $\tau <\infty$ almost surely. Path continuity then gives $B_{\tau }\in \{-x,x\}$. The dominated convergence theorem for the bounded variables $B_{T_t}^2$ shows

Next stop the quartic Hermite polynomial martingale from part (a), again only at $T_t$. Its expectation is zero, soMonotone convergence proves $\mathbb{E}{\tau }^2\le 2x^4$, establishing the needed second-moment integrability before the final passage to the limit. Since $T_t{B}_{T_t}^2\le x^2\tau$ and $B_{T_t}^4\le x^4$, dominated convergence givesConsequently the Brownian symmetric interval-exit moments are

To obtain the Laplace transform of symmetric Brownian interval-exit time, put $a=\sqrt{2\lambda }$. The Exponential martingale for Brownian motion shows thatis a martingale with $H_0=1$. Bounded-time stopping gives $\mathbb{E}{H}_{T_t}=1$. Its stopped values are bounded by $\cosh (ax)$, so dominated convergence applies as $t\to \infty$. Since $\cosh (a{B}_{\tau })=\cosh (ax)$, it yieldsEvery use of stopping at $\tau$ has thus been justified through bounded stopping and an explicit integrability or domination argument.

A version of a stochastic process means a stochastic process on the same probability space such that, for every fixed $t$,The exceptional null set may depend on $t$. Indistinguishability of stochastic processes means that there is one null set outside which $X_t=Y_t$ for all $t\ge 0$ simultaneously.

For an example separating the definitions, let $U$ have uniform distribution on $(0,1)$, and setFor every fixed $t$, $\mathbb{P}(U=t)=0$, so $Y$ is a version of a stochastic process with original stochastic process $X$. But for every sample outcome the stochastic processes differ at its time $t=U$. HenceThe spike path of $Y$ is not right-continuous at $U$, which explains why the next part's regularity assumption rules out this example.

For each nonnegative rational $q$, equality of the two versions gives $\mathbb{P}(X_q=Y_q)=1$. Intersect these countably many full-probability events with the full-probability event on which both paths are càdlàg. Call the resulting event $A$; then $\mathbb{P}(A)=1$.

Fix $\omega \in A$ and any real $t\ge 0$. Choose rational numbers $q_n>t$ decreasing to $t$. Right continuity givesThe same event $A$ works for every $t$, because the argument is pathwise after $\omega$ is fixed. The stochastic processes are therefore indistinguishable. This proves that càdlàg versions are indistinguishable; the left limits are not needed for this implication, since right continuity alone suffices.

Relative to a filtration $({\mathcal{F}}_t)$, progressive measurability means that, for every $T<\infty$, the mapis measurable for the product sigma-algebra $\mathcal{B}([0,T])\otimes {\mathcal{F}}_T$ and the Borel sigma-algebra on $\mathbb{R}$.

Fix $T>0$ and divide $[0,T]$ into $2^m$ equal subintervals with mesh $h_m=T/2^m$. DefineSince $X$ is adapted, every random variable $X_{k{h}_m}$ is ${\mathcal{F}}_{k{h}_m}$-measurable, hence ${\mathcal{F}}_T$-measurable. Each approximation is consequently $\mathcal{B}([0,T])\otimes {\mathcal{F}}_T$-measurable. For $t<T$, its sampling time lies strictly to the right of $t$, tends to $t$, and never exceeds $T$. Right continuity implies $X_t^{(m)}\to X_t$; at $T$ equality is exact. Thus $X$ is the pointwise limit of measurable functions on this product space. As $T$ was arbitrary, $X$ is progressively measurable. This is the theorem that right-continuous adapted processes are progressively measurable.

The right-endpoint approximations need not themselves be adapted at their intermediate times. What the proof requires is their joint measurability with respect to the single terminal sigma-algebra ${\mathcal{F}}_T$. The proof uses the pathwise càdlàg convention. If path regularity is assumed only almost surely, under a completed filtration setting the stochastic process to zero on its common exceptional null event gives an indistinguishable progressively measurable version. Arbitrary values on that null event need not make the original stochastic process progressively measurable: even with a complete filtration, a null sample point may be assigned a non-Borel time function. This is why almost sure path regularity does not ensure progressive measurability.

Fix $t$ and put $\sigma =t\wedge \tau$. The stopping time property implies that $\sigma$ is an ${\mathcal{F}}_t$-measurable random variable with values in $[0,t]$: for $s<t$,and for $s\ge t$ the event is the whole space. By part (c), the restriction of $X$ to $[0,t]\times \Omega$ is $\mathcal{B}([0,t])\otimes {\mathcal{F}}_t$-measurable.

The evaluation map $\omega \mapsto (\sigma (\omega ),\omega )$ is measurable from $(\Omega ,{\mathcal{F}}_t)$ to this product space: the inverse image of a measurable rectangle $A\times C$ is $\{\sigma \in A\}\cap C$. Composing it with the jointly measurable stochastic process gives an ${\mathcal{F}}_t$-measurable random variableSince this holds for every fixed $t$, the stopped process is adapted. If path regularity holds only almost surely, first apply the proof to its pathwise regular representative; with a completed filtration, the original stopped variable differs only on a null event and is also ${\mathcal{F}}_t$-measurable. The adaptedness of a stopped right-continuous process requires neither boundedness of $\tau$ nor a martingale assumption; $\tau =\infty$ is harmless because $t\wedge \tau \le t$.

The Skorokhod embedding of a centered random walk states that a random walk with independent identically distributed centered steps of finite variance ${\sigma }^2$ can be realized on an appropriate probability space aswhere $B$ is a standard Brownian motion and the $T_n$ are finite stopping times. More precisely, the stopped positions have the same joint law as the given random walk, and the pairsmay be chosen independent and identically distributed. Their spatial component has the step law, and $\mathbb{E}(T_n-T_{n-1})={\sigma }^2$. Repeating the one-step Skorokhod embedding theorem with the Strong Markov property gives this formulation. In the present normalization, the mean time increment is one, and the strong law of large numbers gives $T_n/n\to 1$ almost surely.

The Donsker invariance principle states that the linearly interpolated diffusively rescaled random walkconverges weakly as a random element of $C[0,1]$, equipped with the uniform norm, to standard Brownian motion restricted to $[0,1]$. At $t=1$ the fractional term is zero. The only step assumptions needed here are zero mean, unit variance, and independent identical distributions; a higher moment or bounded support is not required. This is a functional central limit theorem, concerning the entire interpolated path rather than only its endpoint.

Let $W_n$ be the interpolation in part (a). The operationis continuous, since $|I(f)-I(g)|\le \Vert f-g{\Vert }_{\infty }$. By the Donsker invariance principle and the continuous mapping theorem,The exact trapezoidal integral of the linear interpolation isTherefore the statistic in question differs from $I(W_n)$ by $S_n/(2n^{3/2})$. Using independence, zero means, and unit variances givesThis error tends to zero in $L^2$, hence in probability. The Slutsky theorem now proves the integrated random-walk limit:The limiting law can also be made explicit. The time integral is a Gaussian random variable, as a mean-square limit of linear combinations of a Gaussian process. It is centered, and the covariance identity $\mathbb{E}(B_s{B}_t)=\min (s,t)$ givesThus the terminal value of integrated Brownian motion here has law $N(0,1/3)$.

Sample Brownian motion at integer times and put $Z_n=B_n/\sqrt{n}$. Every $Z_n$ has normal distribution $N(0,1)$, although the $Z_n$ are correlated. LetFor every fixed integer $m$,because $B_m/\sqrt{n}\to 0$ almost surely. The right-hand expression depends only on the future independent increments $B_j-B_{j-1}$ with $j>m$. Thus $\{L\ge a\}$ is, up to a null set, a tail event of those independent increments, and its probability is zero or one by the Kolmogorov zero-one law.

For any finite real $a$, every $Z_n$ exceeds $a$ with the same positive probability $p_a=\mathbb{P}(N(0,1)\ge a)$. For each $N$,Taking the decreasing intersection over $N$ shows that $Z_n\ge a$ infinitely often with probability at least $p_a$. This event implies $L\ge a$, so $\mathbb{P}(L\ge a)>0$. The zero-one law makes it one. Intersecting over positive integers $a$ gives $L=+\infty$ almost surely. The continuous-time limit superior is at least the one along integers, henceThis proves that Brownian fluctuations exceed the square-root scale without needing the lower bound in the law of the iterated logarithm.

For $t>e$, put $\Phi (t)=\sqrt{2t\log \log t}$. Fix $a>1$ and $\epsilon >0$, and setfor $n$ large enough that $a^n>e$. The Brownian reflection principle and the Gaussian tail estimate giveSince $(1+\epsilon {)}^2>1$, the sum of these probabilities is finite. The Borel-Cantelli first lemma implies that, almost surely, for all sufficiently large $n$,Now take $t\in [a^{n-1},a^n]$. The function $\Phi$ is increasing for $t>e$, soHere the positive upper bound for $B_t$ can first be divided by $\Phi (t)$, and the denominator can then be bounded below by $\Phi (a^{n-1})$; this remains valid even when $B_t<0$. Moreover,Thus, for each fixed pair $(a,\epsilon )$,Use the countable choices $a_m=1+1/m$ and ${\epsilon }_m=1/m$, intersect their probability-one events, and let $m\to \infty$. This proves the Brownian upper law of the iterated logarithm:Only large times are involved. The hint's related monotonicity assertion for $\Phi (t)/t$ is also valid eventually: the derivative of $2\log \log t/t$ is $2(1/\log t-\log \log t)/t^2$, which is negative for $t\ge e^e$. No monotonicity at the small-time edge of the logarithmic expression is required.