Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 201 / 6

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 $\psi (u)=-|u|$.