Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 35 (Stochastic Calculus and Applications)

Relative to a filtration satisfying the usual conditions, a local martingale is an adapted càdlàg process $X$ for which there exist increasing stopping times $S_n\uparrow \infty$ almost surely such that every stopped process $X_t^{S_n}=X_{t\wedge S_n}$ is an integrable martingale. Thus, for $s\le t$,The sequence is a localizing sequence. A continuous local martingale additionally has continuous sample paths. The usual definition entails integrability of the initial value; local martingales need not be integrable at later times without stopping.

Let $V_t$ be the total variation of $X$ on $[0,t]$, and set ${\tau }_n=\inf \{t:V_t\ge n\}$. Adaptation and continuity of $V$ make these stopping times; finite variation on every compact interval gives ${\tau }_n\uparrow \infty$. The stopped process $U=X^{{\tau }_n}$ satisfies $|U_t|\le n$ and has total variation at most $n$. It is a martingale by the bounded local martingale criterion, and is square-integrable.

Fix $t$ and take deterministic partitions $0=t_0<\cdots <t_m=t$ with mesh tending to zero. The martingale-difference orthogonality of the increments givesPathwise, however,by uniform continuity on $[0,t]$. These sums are bounded by $n^2$, so the dominated convergence theorem gives $\mathbb{E}{U}_t^2=0$. Hence $U_t=0$ almost surely. Take all rational $t$ on a common probability-one event and use continuity to obtain $U\equiv 0$. Then let $n\to \infty$. This proves the continuous finite-variation local martingale is constant result in its zero-starting form:The proof does not assume the uniqueness of quadratic variation that is to be established next.

For two candidates $A,C$, their difference is a continuous adapted finite-variation process. It is also a local martingale, sinceApply part (b) after subtracting its initial value. ConsequentlyThis proves the uniqueness of an increasing square compensator when the initial value is prescribed, in particular under the standard quadratic variation normalization $A_0=C_0=0$.

The printed assertion omits that normalization and is literally false. For standard Brownian motion $Y$, both $A_t=t$ and $C_t=t+1$ are continuous adapted increasing processes; both $Y_t^2-t$ and $Y_t^2-t-1$ are martingales. Thus uniqueness holds after fixing the initial value; otherwise it holds only up to an initial constant.

For a simple predictable process, define its stochastic integral byIt is constant after $t_n$, so the symbol $(H\cdot M{)}_{\infty }$ means its value there. Put ${\Delta }_{k}M=M_{t_{k+1}}-M_{t_k}$. If $k<\ell$, then $Z_k{\Delta }_{k}M{Z}_{\ell }$ is ${\mathcal{F}}_{t_{\ell }}$-measurable, and the conditional expectation of ${\Delta }_{\ell }M$ given that sigma-algebra is zero. Boundedness of the coefficients and square-integrability of $M$ justify the expectations, so all cross terms vanish:For each diagonal term, the martingale increment identity also givesIndeed, the omitted term $2Z_k^2{M}_{t_k}{\Delta }_{k}M$ has expectation zero. Apply the assumed martingale property of $M^2-[M]$ to replace this difference of squares by the corresponding quadratic variation increment. ThusThe intervals $(t_k,t_{k+1}]$ in the Lebesgue–Stieltjes integral exactly match these increments, including possible jumps of $M$.

Put $M_t={\int }_0^t{H}_{s}d{B}_s$ and $A_t={\int }_0^t{H}_s^{2}ds=[M{]}_t$. The assumed path properties make $A$ continuous, strictly increasing and unbounded. Therefore $T$ is a finite stopping time and $A_T={\sigma }^2$, even though its definition uses a strict inequality.

For $u\in \mathbb{R}$, the Itô formula givesIts real and imaginary parts are local martingales, and $|E_t|\le e^{u^2{\sigma }^2/2}$. Thus $E$ is a bounded true martingale. Continuity of $M$, finiteness of $T$, and dominated convergence yield $\mathbb{E}{E}_T=1$. Rearranging givesThe characteristic function uniquely identifies the distribution, proving the Gaussian terminal value at a deterministic bracket level:This proof does not use the time-change theorem that the later part asks us to prove.

The Dambis-Dubins-Schwarz theorem states: let $M$ be a continuous local martingale with $M_0=0$ and $[M{]}_{\infty }=\infty$ almost surely. SetThen $W_u=M_{{\tau }_u}$ is standard Brownian motion relative to $({\mathcal{G}}_u)$ andStrict increase of the bracket is not required in the general theorem: $M$ is constant on intervals on which its bracket is constant. Continuity of $M$ is required. A nonzero initial value is handled by applying the theorem to $M-M_0$ and adding it back. If the terminal bracket is finite, the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem supplies the Brownian motion up to that bracket time and, if necessary, an independent continuation on an enlarged probability space.

Use $A_t={\int }_0^t{H}_s^{2}ds$ and its inverse ${\tau }_u$ as above. Strict increase and continuity make $u\mapsto {\tau }_u$ continuous, with $A_{{\tau }_u}=u$ and ${\tau }_{A_t}=t$. Thus $W_u=M_{{\tau }_u}$ is continuous and adapted to ${\mathcal{G}}_u={\mathcal{F}}_{{\tau }_u}$, with $W_0=0$.

The essential remaining point is independence of increments, which does not follow merely from the Gaussian marginal in part (a). Fix $0\le u<v$ and $\lambda \in \mathbb{R}$. The complex exponentialis bounded by $e^{{\lambda }^{2}v/2}$ in modulus, hence is a uniformly integrable martingale. The optional sampling theorem at the finite, possibly unbounded, times ${\tau }_u\le {\tau }_v$ givesDivide by the nonzero right-hand exponential. ThenThis deterministic conditional characteristic function says that $W_v-W_u$ is $N(0,v-u)$ and independent of ${\mathcal{G}}_u$. Successively conditioning proves independence of every finite family of increments. Together with continuity and the zero initial value, this proves that $W$ is Brownian motion. The inverse-clock identity gives $W_{A_t}=M_{{\tau }_{A_t}}=M_t$, completing the requested special-case proof of the Dubins-Schwarz theorem.

Because $W$ is continuous and adapted, the Borel function $\mathrm{sgn}(W)$ is predictable. Its value at zero is $-1$, so its square is exactly one everywhere. Consequently $A$ is a zero-starting continuous local martingale withThe Lévy characterization of Brownian motion therefore makes $A$ a Brownian motion in the same filtration. Here that characterization is the deterministic-clock case of the conditional-exponential argument in question 2.

Apply the Itô formula to $V=W^2$:Since $\sqrt{V_t}=|W_t|$ and $|W_t|\mathrm{sgn}(W_t)=W_t$, including at zero, substitution yieldsThus $W^2$ is a squared Bessel process of dimension one.

Use the common filtration in which the two driving processes are independent Brownian motions and the solutions are adapted. Let $Z=X+Y$, and define predictable bounded weightswith the ratios defined only on $\{Z>0\}$. The nonnegative solutions have $X=Y=0$ when $Z=0$. SetIndependent Brownian drivers have zero cross-variation. Since $h_1^2+h_2^2=1$ even on the zero set, $[B{]}_t=t$, and the Lévy characterization of Brownian motion makes $B$ a Brownian motion. Moreover $\sqrt{Z}{h}_1=\sqrt{X}$ and $\sqrt{Z}{h}_2=\sqrt{Y}$ everywhere. Adding the two original equations therefore proves additivity of independently driven squared Bessel processes:The unit-vector fill on $\{Z=0\}$ is essential: simply dividing by $\sqrt{Z}$ there would leave the Brownian construction undefined.

Localize to compact subintervals of $(0,\infty )$, so that the Itô formula can be applied to $f(z)=\sqrt{z}$. Its derivatives are $f^{\prime }(z)=1/(2\sqrt{z})$ and $f^{\prime \prime }(z)=-1/(4z^{3/2})$, while $d[Z{]}_t=4Z_{t}dt$. Thus, before $\zeta$,Hence the Bessel process equation isThe inverse-power drift is interpreted only while $R>0$; its boundary behaviour is justified next, not assumed in this calculation.

Uniqueness in law means that, for a fixed initial value, every two weak stochastic solutions of the stochastic differential equation have the same distribution as entire solution paths, even if their filtered probability spaces and driving Brownian motions differ. For an equation on $(0,\infty )$, initially consider the stopped path and lifetime at the boundary.

We first justify that the boundary cannot intervene when $\gamma \ge 2$. For any solution of the displayed Bessel equation, let ${\tau }_{\epsilon ,L}$ be its first exit from $(\epsilon ,L)$, where $0<\epsilon <r<L$. Applying Itô to $R^2$ givesThus the annulus exit has finite expectation and is almost surely finite. The one-dimensional generator is $\frac{1}{2}{\partial }_{rr}+\frac{\gamma -1}{2r}{\partial }_r$. Its harmonic scale functions are $\log r$ for $\gamma =2$ and $r^{2-\gamma }$ for $\gamma >2$, as direct differentiation verifies. They are bounded on the stopped annulus, so optional sampling givesBoth tend to zero as $\epsilon \downarrow 0$. A continuous path hitting zero before ${\tau }_L$ would hit every inner level first, so that event has probability zero. Exhausting the upper levels proves that zero is not hit at any finite time. Nor can an upper explosion occur: localization of $R^2$ gives $\mathbb{P}({\tau }_L\le t)\le (r^2+\gamma t)/L^2\to 0$. These are the Hitting-zero classification for a Bessel process arguments needed for the dimension-at-least-two case.

Now suppose $\gamma$ is an integer and take a $\gamma$-dimensional Brownian motion $U$ from any deterministic vector $u$ with $|u|=r$. For $\rho =|U|>0$, differentiation of the Euclidean norm givesDefine the radial noise $\beta ={\sum }_i\int U_i/\rho d{W}_i$, filling the integrand by a fixed unit vector on the zero set, as in part (b). It has bracket $t$ and is a Brownian motion. the Itô formula, initially before a possible zero, givesThe boundary argument just given makes this a global positive weak stochastic solution. The assumed uniqueness in law therefore identifies every solution's path law with that of $|U|$. Orthogonal invariance of Brownian motion makes this radial law independent of the selected point $u$ on the sphere. ConsequentlyThis proves equality of process laws, not merely equality of one-time marginal distributions.

Let $X,Y$ be two solutions with the same Brownian driver and the same initial value. Write $D=X-Y$, and take a common Lipschitz constant $L$ for both coefficients. Stop when $|X|+|Y|$ reaches $n$. Before this stopping time ${\tau }_n$, all difference integrands below are bounded. The Itô formula givesThe stopped stochastic integral is a true martingale on every finite horizon. Lipschitz continuity therefore yieldsSince $D_0=0$, the Gronwall inequality makes the left side zero. Let $n\to \infty$; the continuous global solutions are bounded on each compact time interval, so their stopping times exhaust it. Equality first at every rational time and then by continuity gives indistinguishability. This proves pathwise uniqueness for globally Lipschitz drift and diffusion coefficients. The argument does not require the initial value to have a global second moment: after stopping, the difference is bounded and initially zero.

The Itô product rule with a deterministic integrating factor givesIntegrating from zero provesBoth coefficients are globally Lipschitz, so part (a) proves that this is the pathwise unique solution. Under a measure where $B$ is a Brownian motion, the solution is centered Gaussian with variance $e^{2t}{\int }_0^t{e}^{-2s}ds=(e^{2t}-1)/2$. The positive drift sign gives an unstable linear diffusion rather than a mean-reverting one.

Use $\tilde{\mathbb{P}}$ for the reference measure under which $X$ is a Brownian motion. Pathwise, $B_t=X_t-{\int }_0^t{X}_{s}ds$. Put $h_s=X_s{1}_{\{s\le T\}}$ and defineThe stopped integrand has absolute value at most one. The Novikov condition holds on every finite horizon, so $Z$ is a density martingale. The Itô formula also gives the more useful identityThus $Z$ is uniformly integrable over the entire time axis, not just a true martingale on each finite horizon. Under the reference measure, $\tilde{\mathbb{E}}(t\wedge T)=\tilde{\mathbb{E}}{X}_{t\wedge T}^2\le 1$, so $T<\infty$ almost surely. Hence $Z_t\to Z_T>0$ and $\tilde{\mathbb{E}}{Z}_T=1$.

Define the probability measure on the original sigma-algebra byIts restriction to ${\mathcal{F}}_t$ has density $Z_t$. The Girsanov theorem states that subtracting the integrated density integrand from the reference Brownian motion gives a Brownian motion under the new measure. Thereforeis a Brownian motion under $\mathbb{P}$, and $B_{t\wedge T}=W_{t\wedge T}^{\mathbb{P}}$. This is exactly the requested Brownian property until $T$. The bounded Girsanov density for exit of an unstable linear diffusion also proves global absolute continuity, so a mere collection of unspecified finite-horizon measures is unnecessary.

At exit, continuity gives $X_T^2=1$. The preceding density identity becomesSince $|X_s|\le 1$ before exit, ${\int }_0^T{X}_s^{2}ds\le T$. On $\{T\le t\}$ it follows that $Z_T\ge e^{1/2-T}\ge e^{1/2-t}$. The definition of the changed measure now yieldsThe positive sign of the stochastic integral in the density is what introduces the positive drift $X_{t}dt$; reversing that sign would give a different measure and would not prove this inequality.

Use one $d$-dimensional Brownian motion $W$ to couple the diffusion processesThe global existence theorem for stochastic differential equations with Lipschitz coefficients supplies a unique nonexplosive strong stochastic solution. Let $x$ be the deterministic drift trajectory. Subtraction and the Gronwall inequality give, on every finite horizon $t$,This also follows directly from the integral equations with continuous forcing, so the coupling can be chosen simultaneously for all $\epsilon$.

For completeness derive the needed Feynman-Kac formula with its potential sign. For fixed $t$, apply the Itô formula and the Itô product rule toIts drift is the exponential factor times $-u_t^{\epsilon }+({\epsilon }^2\Delta /2+b\cdot \nabla )u^{\epsilon }+c{u}^{\epsilon }$, which vanishes by the partial differential equation. It is a local martingale. Boundedness of $u^{\epsilon }$ on the finite time slab and boundedness of $c$ make $Y$ bounded there, so it is a true martingale. Its endpoint expectations giveThe deterministic trajectory on $[0,t]$ is compact. Uniform pathwise convergence puts all sufficiently small-noise paths in a fixed compact neighbourhood of that trajectory. Continuity of $c$ therefore implies ${\sup }_{s\le t}|c(X_s^{\epsilon })-c(x_s)|\to 0$; global uniform continuity of $c$ is not required. Continuity of $f$ gives convergence of the terminal factor. Finally,The dominated convergence theorem proves the zero-noise limit with a bounded potential:At $t=0$ this is the given initial condition. All boundedness arguments concern a fixed finite horizon; no uniform bound over infinite time is needed.

A Markov jump process, considered up to its explosion time, is an adapted càdlàg piecewise-constant process whose conditional future law given ${\mathcal{F}}_t$ depends only on its current state. Describe its rates by an off-diagonal kernel $q(x,dy)$ with finite total rate $q(x,E)$. In state $x$ the holding time has the exponential distribution with this rate, followed by a destination distributed as $q(x,dy)/q(x,E)$; a zero-rate state is absorbing. Its Markov jump-process generator acts byThe Markov property must hold relative to the specified filtration, not merely the natural filtration: future-revealing information could change the jump compensator.

Let $\mu$ count jumps, marked by their destination states. For this jump measure of a Markov jump process, the predictable compensator isThe compensated random-measure theorem says that a predictable $H(s,y)$ with $\mathbb{E}{\int }_0^t\int |H|d\nu <\infty$ gives a true martingaleThe same statement holds locally when that expected integrability is obtained after localization. The compensator identity applied to $1_A{1}_{(s,t]}H$, for $A\in {\mathcal{F}}_s$, shows that each increment has zero conditional expectation. Taking $H(s,y)=f(y)-f(X_{s-})$ gives the generator local martingale $f(X_t)-f(X_0)-{\int }_0^{t}Qf(X_s)ds$. This states the integrability conditions in the compensated jump-measure local martingale result explicitly.

For the birth-process claim first take the customary fixed initial state $X_0=i$. Write $N_t=X_t-i$, ${\Lambda }_t={\int }_0^t\lambda (X_s)ds$ and $a=e^{\theta }-1$. The compensator of $N$ is $\Lambda$, whose integrand may equivalently use $X_{s-}$ because jump times have zero Lebesgue measure. A birth multiplies $e^{\theta X}$ by $e^{\theta }$, so the jump product rule givesFor an explicit localization, stop at ${\tau }_n=\inf \{t:X_t\ge n\}\wedge n$ with $n>i$. Up to that time the statewise rates are bounded by the maximum of the finitely many rates below $n$, the jump count is bounded, and $M$ and its compensated-integral integrand are bounded on the finite stopped horizon. The compensated-measure theorem therefore makes each stopped process a true martingale. The times increase to the explosion lifetime (or infinity if there is no explosion), proving $M$ is a local martingale on $[0,\zeta )$ for every real $\theta$.

Now suppose $\lambda (j)\le C$ uniformly. For $C>0$, propose candidate births at the times of a rate-$C$ Poisson process, accepting each with probability $\lambda (X_{s-})/C$ using fresh independent uniform marks. While the state is fixed, Poisson thinning gives a holding time with the required exponential distribution of rate $\lambda$; accepted births have exactly the prescribed law. Thus there can be no explosion, and the true birth count $N_T$ is stochastically dominated by a Poisson random variable of mean $CT$. In particular,The case $C=0$ is constant and immediate. For every fixed horizon $T$ and all $s\le T$, including localization times,This is one integrable dominating variable for the entire stopped family. Conditional dominated convergence removes the localization and provesso the exponential martingale of a pure birth process is a genuine martingale for uniformly bounded rates. Only uniform integrability on each finite horizon is asserted; it need not be uniformly integrable over infinite time.

For a random initial state, the same proof requires $\mathbb{E}{e}^{\theta X_0}<\infty$. The source does not explicitly specify the initial distribution, so that qualification cannot be dropped: take constant rate one and an independent initial law $\mathbb{P}(X_0=n)=6/({\pi }^2{n}^2)$. For $\theta >0$, already $\mathbb{E}{M}_0=\infty$, precluding even the integrability in the definition of a martingale. The fixed-initial-state interpretation above supplies the intended complete proof; bounded rates alone do not remedy an arbitrary heavy-tailed initial law.