Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 26 (Advanced Probability)

Use the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$. The first exit time is a stopping time. If $r\le x$, it is zero, so its expectation is already finite. Now assume $0<x<r$.

The positive-increment hypothesis gives a number $h>0$ with $\mathbb{P}(X_1>h)>0$. Choose an integer $m$ so large that $r/m<h$, and put $p=\mathbb{P}(X_1>r/m)>0$. A block of $m$ increments all exceeding $r/m$ has probability $p^m$. From any point still in $(0,r)$, that block forces an upper exit before the block ends.

By independence, conditional on ${\mathcal{F}}_{km}$ and survival to time $km$, the next block has this same probability. Thus the geometric tail bound from a uniform escape probability givesUsing the tail-sum formula for the expected value of a nonnegative integer-valued random variable,This is the random-walk exit bound from a positive-increment block. In particular, the exit occurs with probability one. The mean-zero assumption is not needed for this first bound.

For $0<x<r$, the exit time is at least one. Since $\{\eta \ge n\}=\{\eta >n-1\}\in {\mathcal{F}}_{n-1}$, this event is independent of $X_n$. The Tonelli theorem givesTherefore the integrability of a stopped random-walk increment bound isIt is the survival event, not the exit-at-$n$ event, that is independent of the next increment. The selected exit increment need not have the same distribution or mean as $X_1$.

For $r\le x$, the printed variable $X_{\eta }=X_0$ is undefined because the increment sequence starts at one. Either restrict this part to $0<x<r$, or make the harmless additional convention $X_0=0$. With that convention the conclusion also holds in the immediate-exit case. The integrability assertion needs this indexing qualification.

The random walk $S_n$ is a martingale, since its integrable increments are independent of the past and have mean zero. For the bounded stopping time $n\wedge \eta$, the bounded optional stopping theorem, proved in the next question, gives $\mathbb{E}{S}_{n\wedge \eta }=x$.

If $0<x<r$, the value immediately before exit lies in $(0,r)$, soBefore exit the stopped value has absolute value less than $r$, and after exit it equals $S_{\eta }$. Consequently $|S_{n\wedge \eta }|\le r+|X_{\eta }|$ for every $n$. This is an integrable dominating random variable by the preceding part. Since $\eta <\infty$ with probability one, the dominated convergence theorem givesFor $r\le x$, $S_{\eta }=S_0=x$ directly, without any convention about $X_0$. Thus the expected stopped position is well-defined and has the stated value for every $r>0$.

The conditional expectation $\mathbb{E}[X\mid \mathcal{G}]$ is a $\mathcal{G}$-measurable integrable random variable $Z$ such thatIt is defined up to almost sure equality. Measurability with respect to the sub-sigma-algebra and equality of these integrals are both essential: the first expresses that only the information in $\mathcal{G}$ is retained, and the second preserves all averages visible through that information. Existence follows from the Radon-Nikodym theorem; uniqueness is up to sets of probability zero.

Let $T\le N$ for a deterministic integer $N$. The stopped martingale has the finite-sum representationEach summand is integrable, and $\{T\ge k\}=\{T>k-1\}\in {\mathcal{F}}_{k-1}$ by the stopping time property. Therefore the conditional expectation identity for a martingale givesTaking expectations in the finite sum proves the bounded optional stopping theorem:No limiting argument or uniform-integrability assumption is needed for a bounded stopping time.

For the real-valued variables here, take a $\mathcal{G}$-measurable version of $X=\mathbb{E}[Y\mid \mathcal{G}]$. Since the variables are bounded, the conditional expectation defining identity extends to the bounded $\mathcal{G}$-measurable multiplier $X$, givingThe equality case for conditional second moments now givesA nonnegative random variable with zero expectation vanishes with probability one. Thus $X=Y$ as an almost sure equality. The same proof works for square-integrable variables; boundedness is more than is needed.

The bounded optional stopping theorem gives $\mathbb{E}{X}_{T\wedge n}=\mathbb{E}{X}_0$. Decompose the difference from the terminal value:ConsequentlyThe first term tends to zero by hypothesis. The second tends to zero by the dominated convergence theorem, because $X_T$ is integrable and $T$ is finite with probability one. Thus the stopped martingale converges to $X_T$ with convergence in L1, which permits passage of expectations to the limit:The explicit tail condition supplies exactly the missing control for an unbounded stopping time.

First prove the strong law for Brownian motion, $B_t/t\to 0$ with probability one. For each $\epsilon >0$, the Gaussian tail bound givesBy stationary increments and the Brownian reflection principle, followed by the same tail bound,Both bounds are summable in $n$. The First Borel-Cantelli lemma therefore implies that, eventually, both quantities inside these probability events are at most $\epsilon n$. For $n\le t\le n+1$ this gives $|B_t|/t\le 2\epsilon$. Intersect over a sequence of positive rational $\epsilon$ tending to zero to obtain the asserted continuous-time limit.

On this one almost sure event,so $B_t+at\to +\infty$ for $a>0$ and to $-\infty$ for $a<0$. For each real $x$, the path is eventually strictly on the corresponding side of $x$. HenceThe same event works simultaneously for all levels $x$, giving the requested transience of Brownian motion with drift.

For $t>0$, write $\Phi$ for the standard normal distribution function. Since the Brownian running maximum is nonnegative, $a\le 0$ gives simply $\mathbb{P}(B_t\le b)=\Phi (b/\sqrt{t})$.

For $a>0$ and $b<a$, reflect the Brownian motion after its first hit of $a$. This is a stopping time, and the Strong Markov property together with symmetry of Brownian increments shows that reflection preserves its law. On paths that hit $a$, it sends the endpoint $B_t$ to $2a-B_t$. Thus the event with endpoint at most $b$ maps to endpoints at least $2a-b$, all of which necessarily hit $a$. This provesThe same reflection gives $\mathbb{P}(M_t\ge a)=2(1-\Phi (a/\sqrt{t}))$. If $b\ge a$, every path with $B_t>b$ has hit $a$, so subtracting that endpoint tail gives the complete answer:The expressions agree at $b=a$. This is the joint distribution of Brownian motion and its running maximum. In particular $M_t$ has the distribution of $|B_t|$ and has no atoms for $t>0$. If $t=0$, the pair is $(0,0)$ deterministically, so the requested probability is $1_{\{a\le 0,b\ge 0\}}$.

Continuity ensures that the maximum on $[0,1]$ is attained, so its first attainment time $\tau$ lies in $[0,1]$. To exclude the endpoint, use Brownian time reversal on a finite interval:This is a standard Brownian motion on this interval. Indeed, its increments are increments of $B$ on disjoint intervals taken in reverse order, hence are independent centered normal variables with the required variances, and its paths are continuous.

If $B_1$ is the maximum, then $R_s\ge 0$ for every $s$. But by the Brownian reflection principle and symmetry,the running maximum of $-R$ has the distribution of $|R_1|$, whose probability of being zero is zero. In particular the event $\tau =1$ has probability zero. Thus $\tau <1$ with probability one. Uniqueness of the maximizing time is not needed for this proof.

Put $Y_t=B_{\tau +t}-B_{\tau }$. Because $B_{\tau }$ is the maximum on $[0,1]$,The preceding part makes this a strictly positive random interval with probability one. A standard Brownian motion cannot have this path property. For each deterministic $h>0$, the Brownian reflection principle gives $\mathbb{P}({\sup }_{0\le t\le h}{B}_t\le 0)=0$. Taking the countable union over $h=1/k$ shows that the probability of staying nonpositive on any initial interval of positive length is zero.

Therefore $Y$ is not a Brownian motion. This is the obstruction for Brownian motion shifted at its finite-horizon maximum. It concerns its path law, irrespective of any proposed filtration for the shifted process.

Fix $0<t<1$. The event $\{\tau \le t\}$ means that the future path through time one never exceeds the maximum already attained by time $t$. Given ${\mathcal{F}}_t$, future increments form an independent Brownian motion $W$, soHere the Brownian running maximum distribution is the one derived above. On $\{B_t<0\}$, an event of probability $1/2$, the finite quantity $M_t-B_t$ is strictly positive, so this conditional probability lies strictly between zero and one.

If $\tau$ were a stopping time, $\{\tau \le t\}$ would be ${\mathcal{F}}_t$-measurable, and the displayed conditional expectation of its indicator would equal that indicator, taking only zero and one. The contradiction proves $\tau$ is not an $({\mathcal{F}}_t)$-stopping time. Thus the Strong Markov property cannot be invoked at $\tau$ to assert a Brownian shifted process; the direct argument here does not assume that property at this random time.

Use the Kolmogorov continuity theorem in its one-parameter form: if on a compact interval a process satisfies $\mathbb{E}|X_t-X_s{|}^p\le C|t-s{|}^{1+\beta }$ for some $p,\beta >0$, it has a modification whose paths are Hölder continuous of every order $\gamma <\beta /p$ on that interval.

For Brownian motion, normal increments give, for every $p>0$,Every such normal moment is finite. Taking $p>2$ yields $\beta =p/2-1$, hence any order below $1/2-1/p$. For a prescribed $0<\alpha <1/2$, choose $p$ with $1/2-1/p>\alpha$.

The continuous modification and the given continuous Brownian motion agree at all rational times on one almost sure event; continuity makes them agree everywhere on the interval. To obtain all exponents and all compact intervals simultaneously, apply the theorem to integer intervals $[0,N]$ and a countable sequence of positive exponents increasing to $1/2$, then intersect these almost sure events. A bound at exponent $\gamma >\alpha$ implies a bound at $\alpha$ on a compact interval. Consequently the Brownian Hölder regularity conclusion iswith finite random constants on one common event of probability one. This uses the usual positive-exponent meaning of Hölder continuity.

A finite right derivative at zero would make the difference quotients $B_{1/n}/(1/n)$ eventually bounded. Therefore that event is contained inFor fixed $K,N$, its probability is at most $\mathbb{P}(|B_{1/n}|\le K/n)$ for each $n\ge N$. By the normal distribution of the Brownian increment,Thus every event in this countable union has probability zero. With probability one, Brownian motion has no finite right derivative at zero, which is the appropriate derivative for its time domain. No independence of the quotients is assumed or needed. This proves the required endpoint case of nowhere differentiability of Brownian motion.

Fix rational $0<p<q$. Divide $[p,q]$ into $N$ equal intervals. Its Brownian increments are independent centered normal variables, so the probability that all $N$ are nonnegative is $2^{-N}$. A nondecreasing path would force this event for every $N$, hence its probability is zero. The same argument with nonpositive increments excludes a nonincreasing path.

There are countably many rational pairs $p<q$, so with probability one none of these intervals supports a monotone path. Every real interval $0<a<b<\infty$ contains such a rational subinterval. Monotonicity on the larger interval would imply monotonicity on that subinterval, a contradiction. Thus the nowhere monotonicity of Brownian motion assertion holds simultaneously:Both nondecreasing and nonincreasing behavior, including a constant path segment, are excluded.

Use the Donsker invariance principle: for independent and identically distributed random variables of mean zero and variance one, the linearly interpolated processes $W_n$ with $W_n(k/n)=S_k/\sqrt{n}$ converge in distribution to standard Brownian motion in $C[0,1]$ with the uniform topology. No moment beyond the finite second moment is required.

The maximum functional $F(f)={\max }_{0\le t\le 1}f(t)$ is continuous, since $|F(f)-F(g)|\le \Vert f-g{\Vert }_{\infty }$. A linear function on each interpolation interval has its maximum at an endpoint, so $F(W_n)=M_n/\sqrt{n}$. The continuous mapping theorem therefore givesBy the Brownian reflection principle, the limiting random variable has the distribution of $|Z|$, $Z\sim N(0,1)$. Its distribution function is continuous and has no atom at any $x\ge 0$, so convergence in distribution permits passage to these tail probabilities. HenceAt $x=0$ the left side is exactly one for every $n$, because $S_0=0$ is included in the maximum, and the right side is also one.

A real Lévy process starts at zero with probability one, has stationary increments and independent increments, is stochastically continuous, and is taken in its càdlàg version. Stationarity means $X_{t+s}-X_s$ has the law of $X_t$; independence means increments over disjoint ordered time intervals are independent. Stochastic continuity means $X_s\to X_t$ in probability as $s\to t$. One can equivalently impose starting at zero, stationary independent increments and stochastic continuity first, and then take a càdlàg modification.

A Poisson random measure with sigma-finite intensity $\nu$ on a measurable space $(E,\mathcal{E})$ is a countably additive integer-valued random measure $N$ such that $N(A)$ has Poisson distribution with parameter $\nu (A)$ whenever $\nu (A)<\infty$, and counts on disjoint measurable sets are independent. A set of infinite intensity has infinite count with probability one. For jump processes the space is often time times a mark space, with intensity $dt\nu (dz)$. These conditions specify both the marginal count laws and their joint independence.

Let $Z=X+Y$, where the two Lévy processes are independent as processes. For any disjoint ordered time intervals, the increment vectors of $X$ and $Y$ are independent of one another, and each vector has independent coordinates. Thus the pairs of corresponding increments are independent across intervals, and so are their sums. The law of each summed increment is the convolution of the two increment laws, depending only on the interval length. This proves independent increments and stationary increments for $Z$.

Also $Z_0=0$, and for every $\epsilon >0$,Thus $Z$ is stochastically continuous. The sum of two càdlàg functions is càdlàg. All defining properties hold, so $X+Y$ is a Lévy process. Independence of the entire two processes, not merely equality of some one-time laws, is what supplies independent summed increments.

For $t>0$, use the continuous logarithm of the characteristic function, normalized to zero at $u=0$, to writeThus the deterministic drift is $+c$. In particular the last contribution is not the negative drift of a compensated unit-jump process.

Take a Poisson random measure $N$ on $(0,\infty )\times \{+,-,p\}$ with intensityDefine the jump marks $j(+)=b$, $j(-)=-b$, and $j(p)=1$. A realization with the required process law is the Poisson stochastic integral with finite intensitywhere the three counts are independent Poisson processes of rates $a/2,a/2,c$. Their characteristic functions multiply towhich is exactly the given expression. The constructed process is a Lévy process, and stationary independent increments make its entire finite-dimensional law determined by these one-time characteristic functions. This gives a representation in law of the specified process.

Equivalently, using its nonzero-jump measure, the Lévy measure isand the unmarked representation is $X_t=ct+{\int }_{(0,t]\times (\mathbb{R}\setminus \{0\})}zN(ds,dz)$ with intensity $ds\nu (dz)$. The finite-jump case of the Lévy–Itô decomposition realizes this pathwise using the jump measure of a version of $X$; the integral is an uncompensated finite sum.

The atomic compound Poisson process with drift has càdlàg paths with finitely many nonzero jumps on every bounded time interval, linear slope $c$ between jumps, and no Brownian component. When $b>0$, jumps of sizes $b,-b,1$ occur at the stated rates; when $b=1$, positive unit-jump rates combine to $a/2+c$. When $b=0$, the two symmetric marks have zero effect and are omitted from the Lévy measure; the process reduces to $ct+N_t^p$, so $a$ has no effect. If $c=0$ the paths are piecewise constant, and if also $ab=0$ they are identically zero. All paths have finite variation on bounded intervals. As a check on the drift sign,