Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 202 3 c Solution Created 2026-09-24 Updated 2026-10-03
True. The Dambis-Dubins-Schwarz theorem, with an independent continuation of the Brownian motion if is bounded, representsBecause is deterministic, every finite vector is a finite vector of a Brownian motion at deterministic times and therefore has a multivariate normal distribution. Hence is a Gaussian process. This is the deterministic quadratic variation characterizes a Gaussian continuous local martingale result.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 202 4 a Solution Created 2026-09-24 Updated 2026-09-25
On the Brownian path does not meet zero, so the power function is twice continuously differentiable along the path. The Itô formula givesSince , this becomeswhere is a standard Brownian motion by the Lévy characterization of Brownian motion. Thus the displayed equation in the paper is valid after the customary renaming of as .
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 202 5 b Solution Created 2026-09-24 Updated 2026-09-25
Work first under Wiener measure with coordinate Brownian motion . Boundedness of implies the Novikov condition, sohas expectation one. Define by . The Girsanov theorem makesa -Brownian motion, and hence is a weak solution of a stochastic differential equation.
For uniqueness in law, start with any weak solution under and apply the inverse change of measure with density . Boundedness again gives the Novikov condition, and under the resulting measure the process is Brownian. Reversing the density expresses the law of under as the same functional of a Wiener path. It is therefore independent of the chosen weak solution. This proves the Weak existence and uniqueness in law for an additive-noise SDE with bounded drift.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 202 6 a Solution Created 2026-09-24 Updated 2026-09-25
Fix and define for . The assumed smooth extension and the Neumann boundary condition at zero make a function. The heat equation givesThe Itô formula therefore makes a local martingale. Stop first when leaves a large compact interval. The exponential growth bound and the finite exponential moments of the maximum of Brownian motion on give uniform integrability, so localization and the dominated convergence theorem yieldThis is the Feynman-Kac formula for the Neumann heat problem.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 203 2 a Solution Created 2026-09-24 Updated 2026-09-25
For Schramm–Loewner evolution in , the Scaling invariance of SLE states that, for every ,has the same law as . The scaled Loewner driving function is . Since , the Brownian scaling identity proves the claim.
The Conformal Markov property of SLE states that, conditionally on the hull through time , the future hull mapped by is an independent in . More precisely,has driving function . The stationary increments and independent increments of Brownian motion show that is independent of and has the same law as . The deterministic correspondence between continuous drivers and Loewner chains completes the proof.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 203 2 b Solution Created 2026-09-24 Updated 2026-09-25
Put and defineThen . Differentiating with the Chordal Loewner equation givesUniqueness for this ordinary differential equation shows that . At ,which is the endpoint identity for the Reverse Loewner flow.
For , the pathwise identity is generally false. The left side is built from the reversed final driver segment , whereas the right side is built from the initial segment . By time reversal and symmetry of Brownian motion they have the same probability distribution, but they are not equal for the given Brownian path.
Past exam of the mathematics course of the University of Cambridge 2024 iii Paper 203 4 d Solution Created 2026-09-24 Updated 2026-09-25
Part (c), iterated over disjoint nested annuli, gives independent increments, and the law gives stationary increments. Every finite vector is jointly Gaussian by the definition of the Zero-boundary Gaussian free field, and a continuous version was assumed. Moreover , because the field has zero boundary values.
Thus is a continuous centered Gaussian process with stationary increments and independent increments. Its variance is a continuous additive function on the nonnegative real numbers, so for some . The Gaussian-process characterization of Brownian motion now givesfor standard Brownian motion . This is the Circle-average process of the Gaussian free field is Brownian motion.
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 201 1 a Solution Created 2026-09-24 Updated 2026-09-25
A standard Brownian motion is a real-valued process with , almost surely continuous paths, and independent increments satisfying for .
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 202 3 b Solution Created 2026-09-24 Updated 2026-09-25
The -dimensional Lévy characterization of Brownian motion says that a continuous local martingale with is a standard -dimensional Brownian motion exactly whenfor all and .
One direction follows directly from independent Gaussian increments. Conversely, fix . Applying Itô formula and the bracket assumption shows thatis a complex local martingale. After stopping on leaving large balls it is bounded, so optional sampling and then dominated convergence give, for ,This is the characteristic function of and is deterministic. Thus each increment is Gaussian with the required covariance and independent of the past. Together with continuity, these are precisely the defining properties of standard -dimensional Brownian motion.
Past exam of the mathematics course of the University of Cambridge 2025 ii Paper 4 29K b Solution Created 2026-09-24 Updated 2026-09-25
- almost surely;
- its sample paths are almost surely continuous;
- it has independent increments; and
- for ,
Equivalently, every future increment is independent of the past filtration and has the stated centred normal law.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 201 3 d Solution Created 2026-09-24 Updated 2026-10-03
Let be Brownian motion in started at a nonzero point. The function is a positive harmonic function on because its Laplacian vanishes there. Stopping on annuli and applying Itô formula shows that is a local martingale. Letting the inner boundary shrink to zero also shows that three-dimensional Brownian motion does not hit the origin.
A positive local martingale is a supermartingale, so is -bounded by . The upcrossing proof from part (c), which applies verbatim to a positive supermartingale, therefore gives a finite almost-sure limit
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 201 4 a Solution Created 2026-09-24 Updated 2026-10-03
A standard Brownian motion satisfies almost surely, has almost surely continuous sample paths, and has independent stationary increments such that
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 201 4 c Solution Created 2026-09-24 Updated 2026-09-25
Let be independent and identically distributed random variables with mean zero and variance one, and let . Define the linearly interpolated processThe Donsker invariance principle, also called the functional central limit theorem, states that converges weakly in the space with the uniform norm to standard Brownian motion.
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 202 1 a Solution Created 2026-09-24 Updated 2026-09-25
Apply Itô formula to . Its semimartingale decomposition isThe second term is a continuous finite-variation process. Since is assumed to be a local martingale, uniqueness of the semimartingale decomposition makes this term identically zero. Both and are nonzero, so the quadratic variation of is .
The Lévy characterization of Brownian motion now says that is a Brownian motion. Consequentlyis a constant multiple of an exponential Brownian martingale. It is therefore a true martingale for every .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 353 4 b i Solution Created 2026-09-24 Updated 2026-09-25
On , a standard Brownian motion has pathwise quadratic variation , whereas the time-rescaled process has quadratic variation . Therefore their path laws are mutually singular when .
For on with initial data and time-dependent Dirichlet boundary conditions , stop Brownian motion at its first hit of or . Applying Itô formula to yields
A Brownian motion starting outside a fixed affine subspace of codimension at least two almost surely never hits it. Choose a two-dimensional orthogonal projection annihilating the subspace's direction and with nonzero projected initial displacement. The projected process is planar Brownian motion starting away from zero, and a hit of the subspace would hit zero in that projection. The polar point for planar Brownian motion result excludes this. In three dimensions this shows avoidance of every fixed line when started outside it.
Realized absolute covariation 2026-09-24
For continuous local martingales and a sequence in which each term is a partition of an interval whose mesh tends to zero, the sumsconverge in the sense of uniform convergence on compacts in probability to a continuous increasing process. To identify the limit, put and choose Radon-Nikodym derivativesIf is a centered bivariate normal distribution with covariance matrix , thenLocalizing, representing the pair as stochastic integrals against a two-dimensional Brownian motion, and approximating the integrands by bounded predictable step processes proves the convergence. The step-process case follows from the weak law of large numbers for independent Gaussian increments; the Burkholder-Davis-Gundy inequality controls the approximation error. Since ,