Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 27 (Stochastic Calculus and Applications)

Put $V_t=⟨X{⟩}_t$ and $S_t={\sup }_{s\le t}|X_s|$. Suppose first that $|X|\le K$. The Itô formula for $|x{|}^p$, which is twice continuously differentiable for $p\ge 2$, givesFor $p=2$, the second derivative is interpreted as the constant $2$. The stochastic term has mean zero: its integrand is bounded, and $\mathbb{E}{V}_t<\infty$ makes it a square-integrable martingale. ConsequentlyThis is the first required estimate.

A bounded local martingale is a true martingale. Apply the allowed Doob Lp maximal inequality, and then Hölder's inequality with conjugate exponents $p/(p-2)$ and $p/2$ for $p>2$. With $D_p=(p/(p-1){)}^p$, this yieldsIf the left side is zero there is nothing to prove; otherwise divide by its indicated power and raise to $p/2$. For $p=2$, the same conclusion follows directly from $\mathbb{E}{X}_t^2=\mathbb{E}{V}_t$ and Doob's inequality. A usable constant is thereforeIn particular $C_2=4$. This is the upper maximal moment bound for a continuous local martingale.

For an unbounded continuous local martingale, stop at ${\tau }_n=\inf \{s:|X_s|\ge n\}$. Its stopped path is bounded, and its bracket at $t$ is $V_{t\wedge {\tau }_n}\le V_t$. The proved inequality gives a bound by $C_p\mathbb{E}{V}_t^{p/2}$, independent of $n$. Continuity makes ${\tau }_n\uparrow \infty$, and the stopped maxima increase to $S_t$. Monotone convergence proves the same inequality for the original unbounded process, with the same constant. This localization step also shows that the assumed bracket moments supply all the needed maximal moments.

Use $V=⟨X⟩$. Applying the Itô formula and the Itô product rule givesThe finite-variation terms cancel in the prescribed combination, leavingThus $Y$ is initially a continuous local martingale. It is a true martingale, not merely local. On each fixed horizon $T$,Part (a) with $p=4$, the assumed bracket moments and Cauchy-Schwarz inequality make this bound integrable. The integrable-supremum martingale criterion now applies, by localization and dominated conditional expectations. The same reasoning makes $X^2-V$ a martingale, so $\mathbb{E}{V}_t=\mathbb{E}{X}_t^2=t$.

The zero covariance now means $\mathbb{E}(X_t^2{V}_t)=t^2$. Since $Y_0=0$ and $\mathbb{E}{Y}_t=0$,This is the fourth-moment deficit and bracket variance identity. If equality holds for every $t$, then $V_t=t$ almost surely for each $t$. Take a single probability-one event for all rational times and use continuity to obtain $V_t=t$ simultaneously for all times. The Lévy characterization of Brownian motion then proves $X$ is Brownian motion in its given filtration. A proof of that characterization by conditional characteristic functions is included in Question 2(a).

Let $V_t=⟨M{⟩}_t$, and first suppose $V_{\infty }=\infty$ almost surely in addition to strict increase. The Dambis-Dubins-Schwarz theorem says thatdefine a Brownian motion $B$ in the time-changed filtration and giveEach $T_u$ is a stopping time and is finite. Continuity and strict increase of $V$ make $u\mapsto T_u$ continuous, $V_{T_u}=u$ and $T_{V_t}=t$.

Here are the martingale details behind this inverse-clock proof of the Dambis-Dubins-Schwarz theorem. Stopping a continuous local martingale when its bracket reaches $n$ makes it an L2-bounded continuous martingale. This follows from the stopped Itô isometry or from the $p=2$ estimate proved in Question 1(a), applied after localization. In particular it is uniformly integrable, and optional sampling is valid even at an unbounded stopping time by taking limits. Applying this to $M$ stopped at $T_n$ shows that $(M_{T_{u\wedge n}}{)}_{u\ge 0}$ is a martingale. Thus $B$ is a continuous local martingale. Time-changing $M^2-V$ in the same way shows $B_u^2-u$ is a local martingale, so $⟨B{⟩}_u=u$.

For completeness, the Lévy characterization of Brownian motion follows directly from the Itô formula. If a continuous local martingale $N$, starting at zero, has bracket $u$, thenis a complex local martingale. Its modulus is bounded on each deterministic finite horizon, so it is a true martingale there. ConsequentlyConditional characteristic functions give Gaussian increments independent of the past. Iterating this identity gives independent increments, and continuity completes the Brownian characterization. The identical vector argument proves the Lévy characterization of multidimensional Brownian motion when the bracket matrix is $uI$.

The printed strict-increase hypothesis does not imply $V_{\infty }=\infty$. For example, $M_t={\int }_0^t{e}^{-s}d{W}_s$ has strictly increasing bracket $(1-e^{-2t})/2$. To state the theorem under exactly the printed hypothesis, allow an independent enlargement of the probability space if the terminal clock $L=V_{\infty }$ can be finite.

On $\{L<\infty \}$ the martingale $M$ has a finite terminal limit. Indeed, stopping at each bracket level $n$ gives an L2-bounded continuous martingale which converges; on $\{V_{\infty }<n\}$ the stopped process is the original one. This proves the finite-bracket convergence lemma. Set $T_u=\infty$ when $u\ge L$ and continue $N_u=M_{T_u}$ by that terminal limit. The optional-sampling argument just given makes $N$ a continuous local martingale with bracket $u\wedge L$. Moreover $L$ is a stopping time in ${\mathcal{G}}_u={\mathcal{F}}_{T_u}$.

On a product extension add an independent Brownian motion $\beta$ in clock time, and putThe two summands have zero quadratic covariation, and their brackets are $u\wedge L$ and $(u-L{)}^{+}$. Thus $⟨\tilde{B}{⟩}_u=u$; the proved characterization makes $\tilde{B}$ Brownian. Since $V_t<L$ at every finite $t$ when $L$ is finite, $M_t={\tilde{B}}_{V_t}$ still holds. This is the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem. An infinite clock gives Brownian motion on the original space; a finite clock may require the independent extension.

Define the radial martingale and its clock byThe Itô formula gives $R_t^2=2H_t+2t$. Independence of the coordinate Brownian motions makes their cross variation zero, soThe last identity is orthogonality of the radial martingale and planar Brownian area.

The clock is adapted and continuous, and it is strictly increasing almost surely. Otherwise the two coordinate paths would both vanish throughout a nontrivial interval. Such an interval contains a rational subinterval, while a Brownian increment over each fixed rational subinterval is a nondegenerate Gaussian and cannot be zero with positive probability.

Also $A(\infty )=\infty$ almost surely. If it were finite, the finite-bracket convergence lemma would make $H_t$ converge to a finite limit. Then $R_t^2=2t+O(1)$ on that event, forcing ${\int }_0^{\infty }{R}_t^{2}dt=\infty$, a contradiction. Thus no finite-lifetime extension is needed here.

Use the same inverse clock $T_u=\inf \{t:A(t)>u\}$ for both martingales, and set $W_u=H_{T_u}$, $B_u=Z_{T_u}$. Their bracket matrix isThe vector characterization proved in part (a) makes $(W,B)$ a two-dimensional Brownian motion; in particular its two coordinate processes are independent. Reversing the common clock givesThis is a common-clock Brownian representation of radius and area. Independence follows from the joint time change and identity bracket matrix; no independence of either Brownian motion from $A$ is asserted.

For a finite list of real coefficients $(a_j{)}_{j\le m}$,A deterministic Itô integral is centered Gaussian: first verify it for step functions as a linear combination of independent Gaussian Brownian increments, then pass to an L2 approximation using the Itô isometry and characteristic functions. Its variance is the squared L2 norm of the integrand, which is ${\sum }_j{a}_j^2$ by the supplied orthonormality. ThereforeThis factors as the joint characteristic function of independent standard normals. Since every finite subfamily has this law, the entire sequence is independent and each ${\xi }_n$ has law N(0,1). This is the Gaussian coordinates of deterministic orthonormal Wiener integrands principle.

Put ${\omega }_n=(n-\frac{1}{2})\pi$. The deterministic derivative is $g_n^{\prime }=-{\omega }_n{h}_n$, and $g_n(1)=0$, while $W_0=0$. The Itô product rule with a deterministic smooth function givesAlmost every Brownian path is continuous, hence belongs to $L^2[0,1]$. Its Fourier coefficient in the given orthonormal basis $(h_n)$ is therefore ${\xi }_n/{\omega }_n$. The Parseval identity for a Hilbertian basis gives, pathwise on a probability-one event,This is the squared-norm consequence of the Brownian half-integer sine expansion; completeness, rather than pointwise convergence of a Fourier series, is all that is needed.

The nonnegative series in part (b) has independent squared-standard-normal terms. If $G$ is standard normal, direct Gaussian integration gives $\mathbb{E}{e}^{-a{G}^2}=(1+2a{)}^{-1/2}$ for $a\ge 0$. Consequently, by independence and dominated convergence applied to the exponentials of increasing partial sums,Use the permitted product identity with $x=\sqrt{2\lambda }$. The answer isThis Laplace transform of the integrated square of Brownian motion equals $1$ at zero. Its first derivative there gives mean $1/2$, in agreement with ${\int }_0^1\mathbb{E}{W}_t^{2}dt$.

Since the density martingale is uniformly integrable, $Z_t={\mathbb{E}}_{\mathbb{P}}[Z_{\infty }\mid {\mathcal{F}}_t]$. Equivalence of measures, as stipulated, means $Z_{\infty }>0$ almost surely. Write $L=\int Z^{-1}dZ$ for the stochastic logarithm. The Itô formula gives $\log Z=L-⟨L⟩/2$, and hence the quadratic covariation in the drift correction isIt involves the local-martingale part of $\log Z$; its continuous finite-variation part has zero covariation.

With $Y=X-C$, the Itô product rule now gives an exact cancellation:Thus $ZY$ is a P-local martingale. To transfer this conclusion rigorously, set ${\tau }_n=\inf \{t:|Y_t|\ge n\}$. The same computation for the stopped $Y$ givesThis is again a P-local martingale. Moreover its absolute value is at most $n{Z}_t$. Uniform integrability of $Z$ makes the family over bounded stopping times uniformly integrable, so the product is a true martingale. This is the bounded-process density-product criterion.

The Bayes formula for conditional expectation therefore gives, for $s\le t$,Continuity gives ${\tau }_n\uparrow \infty$, proving $X-⟨\log Z,X⟩$ is a Q-local martingale. This proves the needed Girsanov theorem rather than invoking it.

For the Brownian-filtration conclusion, use this precise Brownian martingale representation theorem: every continuous local martingale in the usual augmentation of the natural Brownian filtration is an Itô integral with a predictable integrand locally square integrable in time. In particularDefineOn each finite horizon a strictly positive continuous $Z$ has a positive pathwise minimum. Thus ${\int }_0^t{\alpha }_s^{2}ds<\infty$ almost surely, andThe already proved measure-change result makes $\hat{W}=W-\int \alpha ds$ a continuous Q-local martingale. Its quadratic variation is $t$, unchanged by its finite-variation correction. The Lévy characterization of Brownian motion provesNo additional exponential-integrability condition is needed, since the equivalent uniformly integrable density process is already given.

Strict positivity lets us define the continuous local martingaleThe integrand is locally bounded because a positive continuous path has positive minimum on every compact time interval. The Itô formula for $\log M$ givesThis is the stochastic exponential representation of a positive continuous local martingale.

If $⟨X{⟩}_{\infty }$ were finite on an event of positive probability, the finite-bracket convergence lemma proved in Question 2(a) would make $X_t$ converge to a finite limit there. The exponential would then have a strictly positive limit, contradicting the assumed $M_t\to 0$. ThereforeThis is the divergent logarithmic clock for a positive local martingale tending to zero.

Fix $a>1$. Continuity ensures $M_{T_a}=a$ when $T_a<\infty$, even though the defining inequality is strict. Before that infimum the process is at most $a$. Thus $M^{T_a}$ is a bounded nonnegative local martingale and hence a true martingale, with expectation one. For deterministic $t$,The stopped process converges to $a$ on $\{T_a<\infty \}$ and to zero on its complement. It is bounded by $a$, so dominated convergence gives $1=a\mathbb{P}(T_a<\infty )$. The crossing event is exactly $\{{\sup }_{t\ge 0}{M}_t>a\}$. ThereforeThis is the maximal identity for a continuous nonnegative local martingale tending to zero. It also shows there is no atom at a level greater than one. The tail tends to one as $a\downarrow 1$, so the overall maximum has no atom at its lower endpoint either.

Let $S$ denote this maximum and put $\tau =\tau (-b)$. Brownian motion reaches $-b$ almost surely: the Brownian reflection principle gives crossing probability $2\mathbb{P}(W_t<-b)\to 1$. Continuity gives $W_{\tau }=-b$, with the strict-crossing infimum interpreted as in part (b). The processis a continuous nonnegative local martingale starting at one and tending to zero. Its maximum is $1+S/b$. Apply part (b) at $a=1+x/b$, for $x>0$:Differentiating gives the maximum before a lower Brownian barrier densityThe tail tends to one as $x\downarrow 0$, so there is no atom at zero. The density integrates to one.

Use the positive exponential Brownian martingaleThe strong law for Brownian motion, $W_t/t\to 0$, makes its exponent tend to $-\infty$ and hence $M_t\to 0$. If $S={\sup }_{t\ge 0}(W_t-\lambda t)$, then $\sup M=e^{2\lambda S}$. Part (b) gives, for $x>0$,Thus the maximum is exponentially distributed with rate $2\lambda$:There is no atom at zero, by letting $x\downarrow 0$ in the tail. This is the infinite-horizon crossing probability for Brownian motion with negative drift.

Define the diffusion generator $L=b{\partial }_x+{\sigma }^2{\partial }_{xx}/2$. Fix a horizon $t$ and start the strong solution from the deterministic state $x$. The Itô formula applied to the time-reversed test function givesThe drift vanishes by the Kolmogorov backward equation. This is initially a local martingale; localization on compact state/time sets justifies the stochastic integral without a global derivative bound. Since $u$ itself is bounded, this local martingale is a true martingale on $[0,t]$. Its two endpoint expectations giveThis is the bounded backward-equation stochastic representation. Starting the strong solution at deterministic $x$ is the precise meaning of the conditional notation at $X_0=x$. Also $f=u(0,\cdot )$ is bounded, even though boundedness was not separately imposed on $f$.

Put $q(x)=\sigma (x{)}^{2}p(x)$. The explicit density formula and local integrability assumption giveHence its Fokker-Planck probability current $bp-q^{\prime }/2$ is zero. The stationary adjoint equation is $L^{\ast }p=-(bp{)}^{\prime }+q^{\prime \prime }/2=0$. We prove the required integral identity with cutoffs, avoiding an unstated boundary condition on derivatives of $u$.

Choose smooth $0\le {\chi }_R\le 1$, equal to one on $[-R,R]$, supported in $[-2R,2R]$, with $|{\chi }_R^{\prime }|\le C_1/R$ and $|{\chi }_R^{\prime \prime }|\le C_2/R^2$. Since $pLu=(q{u}_x{)}^{\prime }/2$, integrate the backward equation in time and integrate by parts twice in space:These integrations have compact support, as permitted in the question. Bounded coefficients and normalized $p$ giveThe right side is bounded in absolute value bywhich tends to zero. Dominated convergence on the left therefore provesThis is the cutoff proof of invariance for a zero-flux diffusion density. In particular, if an independent initial state has density $p$, part (a) and Fubini give a constant expectation of this test function at every time. This consequence alone is sufficient for the final coupling argument.

Use the same Brownian motion for both solutions: this is a synchronous coupling. For $D_t=X_t-Y_t$, the Itô formula givesOn any fixed finite horizon the stochastic integral has mean zero. Indeed, boundedness of $\sigma$ and the assumed second-moment bounds make the expectation of its squared integrand integrable in time. With $h(t)=\mathbb{E}{D}_t^2$, the contraction condition yields, for every $s\le t$,The allowed Gronwall inequality gives the mean-square contraction of synchronously coupled diffusionsThe identical estimate can also be obtained by localizing the nonnegative local supermartingale $e^{kt}{D}_t^2$ and using Fatou's lemma, a formulation useful when the drift has linear growth.

There is a genuine compatibility issue in the printed global assumptions. If $|b|\le B$, then for $d=|x-y|$,It cannot be at most $-k{d}^2$ for all $d>4B/k$. Thus a globally bounded drift cannot satisfy the stated strict contraction on all of the real line. The stochastic estimate above is the requested conditional calculation. For a non-vacuous application, global boundedness of the drift must be relaxed while retaining suitable existence and moment hypotheses; it is not silently changed here.

Enlarge the space if needed to choose $Y_0$ with density $p$, independent of the driving Brownian motion, and solve the same equation with the same driver as $X_0=x$. The second moment of $p$ makes the initial difference square integrable. By parts (a) and (b),while part (c) givesWe must not assume that the smooth bounded $f$ is globally Lipschitz. The hypotheses give $C_Y={\sup }_t\mathbb{E}{Y}_t^2<\infty$, hence uniform tightness of $(Y_t)$. For any $R$ and $0<\delta \le 1$, define the modulus of continuity of $f$ on $[-R-1,R+1]$ by ${\omega }_R(\delta )$. Splitting according to $|Y_t|\le R$ and $|X_t-Y_t|\le \delta$, and applying Markov inequality, givesFirst choose $R$ large, then $\delta$ small using uniform continuity on that compact interval, and finally let $t\to \infty$. The right side can be made arbitrarily small. Together with part (a), this provesThis is convergence by synchronous coupling for bounded continuous test functions. It only needs the constant expectation from part (b) and uniform second moments; it does not add an unstated global bound on $f^{\prime }$ or require a separate stationarity theorem. As explained in part (c), the literal bounded-drift plus strict-contraction assumptions have no global example; the calculation records the intended consequence under compatible dissipative-drift hypotheses as well.