Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 30 (Stochastic Calculus and Applications)

Let $N$ be the zero-starting continuous local martingale, and let $V_t$ be the total variation of its path on $[0,t]$. Stop when $|N|$ or $V$ reaches $n$, and also at time $n$. The stopped process is bounded, hence a true square-integrable martingale by the bounded local martingale criterion, and its total variation is bounded by $n$.

For a deterministic partition $0=t_0<\cdots <t_m=t$, martingale increments are orthogonal in $L^2$, soThe sum inside the expectation is at most the largest absolute increment times the stopped total variation. As the mesh tends to zero, path continuity makes the largest increment tend to zero. The product is bounded by $2n^2$, so the dominated convergence theorem makes the displayed expectation zero. Thus the stopped process vanishes at every fixed time. Use rational times, continuity and ${\tau }_n\uparrow \infty$ to concludeThis proves that a continuous finite-variation local martingale is constant.

Write a normalized semimartingale decomposition as $X=X_0+N+A$, with both $N_0=A_0=0$, $N$ a continuous local martingale and $A$ of finite variation. The difference of two such martingale parts is also the difference of two finite-variation parts, and hence vanishes by the result just proved. The other parts coincide too. The normalized continuous semimartingale decomposition is unique. Fixing the initial constants is necessary: without normalization a constant can be transferred between the two parts.

Fix a sequence of deterministic partitions with mesh tending to zero and define the completed-increment sumsFor a continuous semimartingale, these sums converge to $[X{]}_t$ in uniform convergence on compacts in probability, the standard path-increment formula for quadratic variation. The sums use only the path of $X$, and do not mention a filtration. If the same process is a semimartingale for two filtrations, the same sums have both proposed quadratic variations as limits. Uniqueness of limits in probability, followed by continuity, makes these limits indistinguishable.

For the change of measure, let $R=d\mathbb{Q}/d\mathbb{P}$. If $\mathbb{P}(A_n)\to 0$, thentruncate $R$ at a large constant, use $\mathbb{E}[R{1}_{\{R>K\}}]\to 0$, and then let $\mathbb{P}(A_n)\to 0$. Apply this argument to the event ${\sup }_{s\le t}|S_s^{(n)}-[X{]}_s^{\mathbb{P}}|>\epsilon$. The same sums converge to the P-version of the bracket in Q-probability, and the Q-semimartingale formula makes them converge to the Q-version. ThereforeThis is quadratic variation under an absolutely continuous measure change. With only $\mathbb{Q}\ll \mathbb{P}$, the equality is asserted under Q: a Q-version is unconstrained on a Q-null set that might have positive P-probability.

For each fixed horizon $t$, the stipulated property is the Class D process condition on $[0,t]$, and imposing it for every finite horizon gives a Class DL process.

First suppose $M$ is a true martingale. For every stopping time $T\le t$, the bounded-time optional sampling theorem givesThe conditional expectations of one integrable variable form a uniformly integrable family. To see the uniformity explicitly, put $V=\mathbb{E}(U\mid \mathcal{H})$. For any $r,K>0$, conditional Jensen and Markov inequalities giveChoose $r$ large and then $K$ large. The bound does not depend on the conditioning sigma-algebra. This proves the required uniform integrability. Bounded-time optional sampling here can also be justified directly: approximate $T$ from above by finite-grid stopping times, use discrete optional sampling with terminal value $M_t$, and use this same conditional-expectation uniform integrability and right continuity to pass to the limit.

Conversely, take a localizing sequence ${\tau }_n$ for the local martingale. For fixed $s\le t$, both families $M_{t\wedge {\tau }_n}$ and $M_{s\wedge {\tau }_n}$ are uniformly integrable, because their times are bounded by $t$. They converge almost surely to $M_t$ and $M_s$, and therefore converge in $L^1$. The stopped martingale identitypasses to the limit by $L^1$ contractivity of conditional expectation. Thus $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$, and all time values are integrable. Hence a local martingale is a true martingale exactly when it is of Class DL. This is a finite-horizon criterion, not a claim of uniform integrability over all times simultaneously.

For continuous semimartingales $U,V$, the semimartingale integration by parts formula isIndeed each partition increment satisfiesSumming telescopes the left side. The first two sums converge to the stochastic integrals by left-endpoint approximation, and the last sum converges to the quadratic covariation by polarization of the quadratic-increment formula. All limits are in uniform convergence on compacts in probability. Hence the equality holds at every time outside one null set, using continuity. In differential notation it is the Itô product rule $d(UV)=UdV+VdU+d[U,V]$.

For a vector of continuous semimartingales $X=(X^1,\dots ,X^d)$ and $F\in C^{1,2}$, the multidimensional Itô formula isOne route from the product formula is to apply it repeatedly to powers and products of the coordinates, obtaining this expression for polynomials; polarization identifies the mixed second derivatives. Time is another coordinate of finite variation, so it has no quadratic-covariation contribution. Localize to compact ranges, approximate a $C^{1,2}$ function and its indicated derivatives by smooth polynomial approximations, and pass to the limit using stochastic-integral continuity and the variation bounds on the finite-variation terms. Equivalently, second-order Taylor expansion along partitions gives the same limit, with integration by parts accounting for every quadratic term.

Write $X_t=x+W_t+\mu t$ with $W$ a zero-starting Brownian motion. A unit-time increment exceeds $2b$ with probabilityIf the process has not exited before an integer time $k$, it is in $(-b,b)$; an increment exceeding $2b$ forces it outside at time $k+1$ and hence forces an exit by then. Independence of Brownian increments gives $\mathbb{P}(T>k)\le (1-p{)}^k$. Thus the geometric tail bound from a uniform escape probability yieldsIn particular $T$ is finite almost surely.

The diffusion generator is $\mathcal{L}=\frac{1}{2}{\partial }_{xx}+\mu {\partial }_x$. The function $f(z)=e^{-2\mu z}$ satisfies $\mathcal{L}f=0$, so the Itô formula gives $df(X_t)=-2\mu f(X_t)d{W}_t$. It is a true martingale, since it is $e^{-2\mu x}$ times the Gaussian exponential martingale $e^{-2\mu W_t-2{\mu }^{2}t}$. The stopped values $f(X_{t\wedge T})$ are uniformly bounded because the stopped path stays in $[-b,b]$. Optional sampling and bounded convergence givewhere continuity ensures that $X_T$ is exactly one of the two endpoints. Solving,This drifted Brownian interval-exit probability tends to $(b-x)/(2b)$ as $\mu \to 0$, as expected.

The deterministic clock is $a(t)=e^{2\lambda t}$, so $a(0)=1$. For $s\le t$, the martingale property of Brownian motion givesIntegrability follows from the Gaussian second moment. Thus $X$ is a martingale in the time-changed filtration.

Its increments have variance $a(t)-a(s)$ and are independent of ${\mathcal{G}}_s$. Equivalently, $X_t^2-a(t)$ is a martingale. Since a quadratic variation starts at zero, its compensator must be normalized at $t=0$:The initial random variable $X_0=B_1$ does not contribute to quadratic variation. The expression $e^{2\lambda t}$ alone would violate $[X{]}_0=0$.

Define the stochastic integralIt is adapted to ${\mathcal{G}}_t$, starts at zero and has continuous paths. Integrals over disjoint clock intervals are centered Gaussian and independent, and an increment from $s$ to $t$ is independent of ${\mathcal{G}}_s$. The Itô isometry gives its varianceThese are precisely the defining properties of a ${\mathcal{G}}_t$-Brownian motion; in particular its future increments are independent of ${\mathcal{G}}_0={\mathcal{F}}_1$.

Deterministic substitution in the stochastic integral, first for step integrands and then by the Itô isometry, gives $d{X}_s=\sqrt{2\lambda }{e}^{\lambda s}d{W}_s$. HenceThe initial term is essential: a stochastic integral from time zero alone cannot equal $X$, since $X_0=B_1$. This is the unscaled part of the exponential Brownian time change to a stationary Ornstein-Uhlenbeck process.

Use the Itô product rule with the deterministic factor $e^{-\lambda t}$. Its quadratic covariation with $X$ is zero, givingMultiplication by the integrating factor $e^{\lambda t}$ solves the Ornstein-Uhlenbeck stochastic differential equation:The initial value $B_1$ is standard normal and independent of the future driving Brownian motion. Thus $Y$ is a centered Gaussian process with variance one. For $s\le t$, its independent future noise gives $\mathbb{E}(Y_s{Y}_t)=e^{-\lambda (t-s)}$. Its Gaussian finite-dimensional distributions consequently depend only on time differences, so it is a stationary process. This verifies the exponential Brownian time change to a stationary Ornstein-Uhlenbeck process directly.

The printed pathwise conclusion is false without specifying the initial value. We first prove exactly what the hypotheses do imply, by stationary drift saturation.

Stationarity gives $\mathbb{E}{Z}_s^2=1$. The Itô formula yieldsThe stochastic integral is a true square-integrable martingale on every finite horizon, since $\mathbb{E}{\int }_0^t{Z}_s^{2}ds=t$. Also $\mathbb{E}|Z_s{g}_s|\le \lambda$ by the Cauchy-Schwarz inequality. Taking expectations therefore givesEvery integrand is nonnegative, so $\mathbb{E}[Z_s{g}_s]=-\lambda$ for almost every $s$. For those times,It follows that $g_s=-\lambda Z_s$ for almost every time and outcome. Equality of these integrable drift terms gives the exact solutionThus $Z$ solves the same equation, but stationarity has not identified its initial random variable with $B_1$.

For an explicit counterexample, takeThis is adapted to ${\mathcal{G}}_t$. Its independent normal initial value and Brownian noise make it a centered stationary Gaussian process with covariance $e^{-\lambda |t-s|}$, exactly as for $Y$. Hence $\mathbb{E}{Z}_0^2=1$ and $\mathbb{E}{g}_t^2={\lambda }^2$, satisfying every printed hypothesis. Nevertheless $Z_t-Y_t=-2e^{-\lambda t}{B}_1$, which is nonzero almost surely for every finite $t$.

The corrected result is indistinguishability if one adds $Z_0=Y_0$ almost surely. Without that addition, there is equality in law of the stationary Ornstein-Uhlenbeck processes, not necessarily equality of their coupled paths. To see that the initial law is forced, independence of Brownian increments from ${\mathcal{G}}_0$ and stationarity give, for its characteristic function $\phi$,Letting $t\to \infty$ proves $\phi (u)=e^{-u^2/2}$. This is the distinction captured by same-noise stationary Ornstein-Uhlenbeck processes.

Set $F_t=e^{\alpha B_t-{\alpha }^{2}t/2}$, a positive stochastic exponential with $d{F}_t=\alpha F_{t}d{B}_t$. The Itô formula givesFor $Y=X/F$, the Itô product rule and $d[X,F^{-1}{]}_t=-{\alpha }^2{X}_t{F}_t^{-1}dt$ cancel both the Brownian terms and the extra drift. ConsequentlyThis is an ordinary Bernoulli differential equation with continuous random coefficients. For $\delta \ne 1$, separate variables to obtainTherefore the Bernoulli stochastic differential equation has the explicit maximal positive solutionvalid while its bracket remains strictly positive. When $\delta =1$, the reduced equation is $d{Y}_t=Y_{t}dt$, andSubstituting these expressions back through the product rule verifies the original stochastic differential equation, not just the reduced equation.

For $0<\delta <1$, the bracket in the explicit solution is positive and increasing. Its integral is finite on every finite interval because $F$ is continuous and strictly positive there. The same is immediate from the exponential formula for $\delta =1$. Neither solution can reach zero on a finite interval. Hence both cases have global existence almost surely.

For $\delta >1$, put $q=\delta -1>0$ and $K=x_0^{-q}/q$. The explosion threshold for a Bernoulli stochastic differential equation isIf the threshold is reached at a finite time, $F_{\tau }$ is positive and finite, and $X_t\to \infty$ as $t\uparrow \tau$.

When $\alpha =0$, $F=1$ and $\tau =K$ deterministically. When $\alpha \ne 0$, choose a constant $c$ large enough that $e^{qc}>K$. The eventhas positive probability: the first part is a Gaussian tail event, the second has positive probability by the Brownian reflection principle, and they are independent. On this event $\alpha B_s-{\alpha }^{2}s/2>c$ throughout $[1,2]$, so ${\int }_1^2{F}_s^{q}ds>e^{qc}>K$. Thus $\mathbb{P}(\tau \le 2)>0$ and global existence is not almost sure. We concludeThis argument asserts positive explosion probability for $\alpha \ne 0$, not certain explosion. Indeed $B_t/t\to 0$ almost surely makes ${\int }_0^{\infty }{F}_s^{q}ds$ finite in that case; that observation does not remove the positive chance that its value exceeds the threshold.

Denote the density process by $L_t=\exp (M_t-[M{]}_t/2)$. The given identities for all $A\in {\mathcal{F}}_s$ implyThus $L$ is a strictly positive true martingale, not merely a local martingale. The Itô formula gives $dL=LdM$, and henceFirst work with the zero-starting corrected increments $U=X-X_0-[X,M]$. Integration by parts givesThe finite-variation bracket has zero quadratic covariation with $L$, so no other term appears. Thus $LU$ is a P-local martingale.

To justify the change of measure, stop $U$ at ${\tau }_n=\inf \{t:|U_t|\ge n\}$, with optional further localization if needed. The product $L_t{U}_{t\wedge {\tau }_n}$ remains a P-local martingale: before the stopping time the cancellation above applies, and afterwards only its bounded constant factor multiplies $dL$. Its absolute value is bounded by $n{L}_t$. On a finite horizon the stopped values of $L$ are uniformly integrable, so Q1(c) upgrades this product to a true martingale. The Bayes formula for conditional expectation now givesFinite-horizon equivalence of the measures transfers the localizing sequence to Q. Therefore $U$ is a continuous Q-local martingale. If $X_0\in L^1(\mathbb{Q})$, adding its constant-in-time process provesThis proves the Girsanov theorem assertion under the usual zero or deterministic initial-value convention. More generally the normalization $M_0=0$ makes $L_0=1$, so P and Q agree on ${\mathcal{F}}_0$ and the required initial integrability is automatic.

There is a literal initial-integrability qualification if neither convention is imposed. On positive integers let $P(n)=2^{-n}$, $Q(n)=c/n^2$, with $c=({\sum }_{n\ge 1}{n}^{-2}{)}^{-1}$, and let every filtration equal the full sigma-algebra. Constant processes $M_t=\log (c{2}^n/n^2)$ and $X_t=n$ are P-martingales, and $e^{M_t}$ is precisely the prescribed density. But ${\mathbb{E}}_Q|X_0|=\infty$ and $[X,M]=0$, so the asserted Q-local martingale cannot have an integrable initial value. This is why initial integrability under a density change is part of the precise formulation.

If $X$ is a P-Brownian motion, its starting value is deterministic, so Q5(a) applies. The correction $[X,M]$ has finite variation and does not alter quadratic variation; Q1(b) gives $[\tilde{X}{]}_t=t$ under Q. The Lévy characterization of Brownian motion therefore identifies $\tilde{X}$ as a Q-Brownian motion with the same starting value.

For the likelihood, use Wiener measure ${\mathbb{Q}}^x$ as the reference measure and write its coordinate as ${\omega }_t=x+W_t$. Setand let $L=\mathcal{E}(M)$ be its stochastic exponential. We must prove it is a true martingale; unverified large-horizon Novikov estimates would not suffice.

Fix $T$ and stop at ${\tau }_n=\inf \{s:|{\omega }_s|\ge n\}$, with $n>|x|$. The bounded integrand satisfies the Novikov condition, so $L_{T\wedge {\tau }_n}$ defines a probability $P_n$ on ${\mathcal{F}}_T$. By the result just established,is Brownian under $P_n$ on this horizon. Therefore $d{\omega }_s=d{W}_s^{(n)}-\mu 1_{\{s\le {\tau }_n\}}{\omega }_{s}ds$. The Gronwall inequality givesThe Doob L2 maximal inequality then bounds its squared expectation by $C_T=2e^{2|\mu |T}(x^2+4T)$, uniformly in $n$. In particular $P_n({\tau }_n\le T)\le C_T/n^2$. On $\{{\tau }_n>T\}$ the stopped and unstopped densities agree, soContinuous Brownian paths are bounded on $[0,T]$, and the indicators increase to one. The monotone convergence theorem proves $\mathbb{E}{L}_T=1$. Since a nonnegative local martingale is a supermartingale, constant expectation at every time makes $L$ a true martingale.

Under its tilted measure, ${\hat{W}}_t=W_t+\mu {\int }_0^t{\omega }_{s}ds$ is Brownian and $d{\omega }_t=d{\hat{W}}_t-\mu {\omega }_{t}dt$. The explicit linear solution gives uniqueness of this law, identifying it with ${\mathbb{P}}^x$. Consequently the Ornstein-Uhlenbeck likelihood relative to Wiener measure isThis is a finite-horizon density statement; it does not assert absolute continuity of the full infinite-horizon path laws.

The differential operator without its potential is the diffusion generator of the d-dimensional Ornstein-Uhlenbeck processIts explicit solution $X_s=e^{-\lambda s}x+{\int }_0^s{e}^{-\lambda (s-r)}d{B}_r$ defines its path law ${\mathbb{P}}^x$ on $C([0,t];{\mathbb{R}}^d)$. For the potential $V(x)={\lambda }^2|x{|}^2/2$ and initial function one, the Feynman-Kac formula gives the nonnegative path integralThe sign in the exponential is positive, because the PDE contains $+Vu$. One can first stop the diffusion in bounded balls, where the usual localized Feynman-Kac argument applies, and then examine finiteness of the resulting expectation. The critical quadratic potential for the Ornstein-Uhlenbeck generator below supplies that finiteness check.

Apply Q5(b) in each independent coordinate, with $\mu =\lambda$, to change from the Ornstein-Uhlenbeck path law to d-dimensional Wiener measure ${\mathbb{Q}}^x$. Its density isMultiplication by the positive potential factor cancels the second term exactly. ThereforeThe integral in its exponent is an Itô integral under Wiener measure, not an ordinary pathwise line integral. The Itô formula for $|\omega {|}^2$ givesHence an equivalent endpoint-only Wiener integral is

Under Wiener measure, the endpoint has normal distribution $N(x,t{I}_d)$. For $t>0$ and $1+\lambda t>0$, the Gaussian integral isComplete the square:Evaluating the centered Gaussian gives $(1+\lambda t{)}^{-d/2}\exp [-\lambda |x{|}^2/(2(1+\lambda t))]$. With the preceding prefactor, the critical quadratic potential for the Ornstein-Uhlenbeck generator solution isThe initial value is one. As a direct PDE check, write $u=a(t)e^{b(t)|x{|}^2/2}$, whereSince $\nabla u=bxu$ and $\Delta u=(db+b^2|x{|}^2)u$, these identities give precisely the required time derivative and spatial operator.

For $\lambda \ge 0$, this is finite for every $t\ge 0$; at $\lambda =0$ it reduces to $u=1$. If the parameter is allowed to be negative, the expectation is finite only for $t<-1/\lambda$. At and beyond that time the quadratic coefficient in the endpoint Gaussian integral is nonnegative, so the integral diverges. A nonnegative classical solution on the entire half-line cannot then exist: stopping the Feynman-Kac martingale in balls and discarding its nonnegative boundary term bounds any such solution below by the truncated path expectations; their limit is infinite. Thus the printed global positive-solution premise implicitly requires the nonnegative parameter convention.