Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 51 1 a Solution Created 2026-10-03 Updated 2026-10-14
An order parameter distinguishes phases by a macroscopic quantity that changes at the transition. For an Ising model ferromagnet it is the magnetization per Ising spin, . At zero magnetic field the statistical Hamiltonian is invariant under spin inversion symmetry. The disordered phase preserves this symmetry, whereas a selected ordered phase has and a spin-reversed partner with . At finite volume with symmetric boundary conditions the expected value is zero even below the transition; spontaneous magnetization means taking the thermodynamic limit before sending a symmetry-breaking field to zero.
Landau-Ginzburg theory introduces a coarse-grained local order parameter and a symmetry-constrained free-energy functional. For a short-range system with scalar spin inversion symmetry, a useful expansion isThe coefficients vary smoothly with microscopic control parameters at the level of the Landau approximation; normally with . The gradient term penalizes spatial variation. The magnetic field is conjugate to the order parameter, and odd powers at are excluded by spin inversion symmetry. With , the sextic term may be omitted locally; with , a stabilizing positive is essential. The equilibrium uniform order parameter minimizes the local potential , givingThis equation of state follows from minimization, rather than being imposed as an independent assumption. Vector or tensor order parameters and their allowed invariants describe other symmetry classes.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 51 2 Solution Created 2026-10-03 Updated 2026-10-14
Use for inverse temperature, to distinguish it from the order-parameter critical exponent . In a translation-invariant pure thermodynamic phase, the connected two-point correlation function isFor finite correlation length, at it decays exponentially, possibly with an algebraic prefactor; is the asymptotic exponential decay rate. Below an ordered transition the pure phase is selected by taking the thermodynamic limit before or . A zero-field symmetric mixture of the two ordered phases has a nondecaying connected term and does not describe the usual pure-phase magnetic susceptibility.
Let denote extensive physical free energy and . Since the field contributes to the statistical Hamiltonian, differentiation givesExpanding the variance and using translation invariance proves the correlation-function susceptibility sum rule:The sum includes . If the source is instead defined as the dimensionless quantity , the corresponding derivative omits the displayed inverse-temperature factor.
A real-space renormalization group step of length factor uses a local blocking kernel satisfying . For example, a deterministic majority-spin rule assigns one coarse Ising spin to each block; tied blocks can be resolved with equal probabilities. Define the blocked statistical Hamiltonian byThe spacing becomes . All generated interactions, including multispin terms and the field-independent constant, must be retained for an exact transformation. Summing over uses normalization to prove . Iteration gives and .
The coarse probability law is the exact marginal law of these block variables. Indeed, for any coarse observable , its expected value is the microscopic expected value of . A blocking rule that retains the long-wavelength order-parameter mode, with source and observable transformations tracked, therefore retains long-distance response and correlation functions. Their field normalization may change, while their physical length scales and critical singularities do not. Partition-function equality alone would not justify this assertion for an arbitrary kernel that discards all slow variables. Truncating the generated operator family is likewise an approximation.
For the free-energy recursion use dimensionless free energy per site,in the thermodynamic limit. Exact partition-function invariance gives . If one step sends to , the coefficient records the identity-operator contribution from eliminated fluctuations and normalization. ConsequentlyRepeated substitution derives the additive free-energy recursion under blocking:In physical energy units, divide this convention by . The additive term is essential for absolute free energies even when it does not affect the leading critical powers.
A renormalization-group fixed point satisfies . Linearize in scaling coordinates so that . A relevant operator has , an irrelevant operator has , and a marginal operator has with behaviour decided by nonlinear terms. Thus the eigenvalues of the discrete linear map are ; the sign criterion applies to the generator exponents , not to the discrete multipliers. The critical surface is the stable manifold on which every relevant scaling field has been tuned away. Irrelevant coordinates flow to the renormalization-group fixed point along it. A repulsive renormalization-group trajectory leaves the renormalization-group fixed point along an unstable relevant direction, ultimately entering a noncritical phase or another scaling regime.
Local renormalization-group flow in the zero-field slice: attraction along the critical surface and repulsion in the thermal direction
. Details. To separate the singular free-energy density from its regular background, suppose the eliminated-mode contribution is analytic near the renormalization-group fixed point. Look for an analytic background satisfying . Then satisfies . In linear thermal and field coordinates, an analytic monomial has background coefficientprovided the denominator is nonzero. This explicitly explains when analytic subtraction of an inhomogeneous renormalization recursion works. The printed inhomogeneous equation may describe the free-energy function before this subtraction, or a singular part defined only up to analytic terms. The inhomogeneous sum need not be small compared with the full free energy: an analytic constant can dominate a vanishing singular contribution. It can be omitted in deriving leading singular powers after regular subtraction, provided no renormalization-group free-energy resonance, marginal logarithm or dangerously irrelevant coupling obstructs the homogeneous critical asymptotics. At a renormalization-group free-energy resonance , additive logarithms can remain, so simply dropping the term would miss them.
Suppose there are exactly two relevant scaling fields, thermal and magnetic , with positive exponents , and the remaining fields approach a regular limit. Physical reduced temperature and magnetic field agree with these coordinates to leading order, up to nonuniversal metric factors. The homogeneous scaling hypothesis for critical phenomena isas an asymptotic scaling equation. Choose to make the thermal argument or . Defining givesThe two functions describe respectively and : approaching the transition from disordered and ordered sides need not give the same function. In particular the ordered zero-field function has one-sided derivatives associated with spontaneous magnetization. Metric factors multiply this formula in different microscopic normalizations.
The correlation length in Bravais lattice units obeys . At , the same choice of gives . Two thermal derivatives of give . One and two field derivatives give, respectively, and , assuming the relevant branch amplitudes are nonzero. At , choose ; then , so . HenceSimple algebra now gives the Rushbrooke scaling relation and the hyperscaling relation . Also , giving the Widom scaling relation. These statements refer to leading singular contributions with the homogeneous two-field scaling assumptions. Above the upper critical dimension, the quartic coupling is dangerously irrelevant because it is still needed to stabilize the ordered phase; ordinary mean-field critical exponents then satisfy Rushbrooke and Widom but generally not hyperscaling. Marginal logarithms also require qualifications at the upper critical dimension.
For the connected-correlation scaling form , the function tends to a finite nonzero value as and decays exponentially at large , up to powers of , in a massive short-range phase. The scaling ansatz alone does not fix that prefactor. If the usual massive Gaussian field correlation tail applies, matching the powers gives , with a correlation-length-dependent amplitude in fixed by matching the critical scaling form.
In physical coordinates the lattice sum has site-density factor . Its long-distance singular part therefore giveswhere is the area of the unit sphere in dimensions. For , finite , an integrable exponential tail and a nonzero integral, the last factor tends to a finite nonzero constant as . Microscopic distances contribute a regular background. Thus , proving the Fisher scaling relationEquivalently the field has scaling dimension , so its conjugate field has , which gives the same identity from the exponent formulas above.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 59 4 Solution Created 2026-10-03 Updated 2026-10-14
For each hidden variable , a deterministic local hidden-variable model assigns values to all settings. The responses depend only on the local setting; their averaging distribution is setting independent, as required by measurement independence.
Let be the sum of the nonnegative modular representatives in the chained modular Bell inequality. Before reduction modulo , the chain telescopes:Reduction changes this integer only by multiples of . Since all reduced terms are nonnegative,The smallest possible nonnegative value in this residue class is . Averaging over the hidden variable provesAdding local random seeds to gives the same result for stochastic local hidden-variable theories.
For the coplanar qubit realization of the chained Bell inequality, shareAt angle in the plane, use the projective measurement of the observableHere are Pauli matrices, and the classical bit records eigenvalue . Since and both crossed expected values vanish in this Bell state,The one-party expected values vanish, so the complete joint probabilities areFor binary outcomes, is one exactly when , while is one exactly when . Consequently
For , set and choose Alice's angle and Bob's angle . Every adjacent pair or differs by , giving expectation . The closing pair has angle difference ; its agreement probability is also . Thus every one of the terms contributes equally:Indeed for , so this value is below . For an explicit two-setting instance, Alice measures and Bob measures . The four relevant expected values sum toviolating the local bound one. At there is no violation; the two terms always sum to one. As , the displayed quantum value is asymptotic to and approaches the algebraic lower bound zero.
The violation rules out a local hidden-variable theory with setting-independent preassigned local responses reproducing these quantum correlations. It does not permit faster-than-light communication: the joint formula gives each local outcome probability , independent of the remote setting, as required by the no-signalling theorem. The contradiction concerns the joint correlations and the Bell assumptions, not the ability to send a signal.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 60 4 a Solution Created 2026-10-03 Updated 2026-10-14
The terminal term is the objective reward. With the Hilbert-Schmidt inner product , a Hermitian target observable gives its final expected value. The vector is the density operator represented as a vector in operator space; it retains the same physical content as the density matrix, not a wavefunction of the system.
The term imposes the dynamical equation by a time-dependent costate . It is a Lagrange-multiplier operator in the same operator space, propagated backward from a terminal condition. It is not required to be a positive, trace-one density matrix. DefineThen the enforced state equation is . The uncontrolled coherent part is , the dissipative contribution is , and each describes a control coupling. For Quantum Hamiltonian control, .
The term is a quadratic field-energy or fluence penalty. For the positive scale and positive weights , larger make a given control amplitude more costly. Thus maximizing trades terminal performance against control fluence while enforcing the dynamics. The multiplier term vanishes on any dynamically feasible trajectory. The weights do not represent decoherence rates; they specify control-resource preferences.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 69 3 a Solution Created 2026-10-03 Updated 2026-10-14
The strain is an Ornstein-Uhlenbeck process. Multiplying its equation by the integrating factor and using the zero initial condition givesThe second expression is a stochastic integral against Brownian motion. A deterministic linear functional of Gaussian white noise is Gaussian, and its expected value is zero. For , the noise covariance givesFor , the initial-condition term is negligible. The stationary autocorrelation is thereforeIts normalized exponential decay has correlation time . The zero-start Ornstein-Uhlenbeck covariance also shows how stationarity is approached.
Past exam of the mathematics course of the University of Cambridge 2007 ia Paper 2 3F Solution Created 2026-09-24 Updated 2026-10-14
For , independence givesThus has probability density function , and its expected value isPointwise, and . By linearity of expectation, ; by independence, . Consequently the covariance of two uniform order statistics isThe order statistics are dependent despite the independence of the original observations.
Past exam of the mathematics course of the University of Cambridge 2007 ia Paper 2 9F Solution Created 2026-09-24 Updated 2026-10-14
For a nonnegative integer-valued random variable, its expected value is , allowing the value . Since , rearranging nonnegative summands gives the tail-sum formula for expectation:No integrability assumption is needed for the rearrangement. The same formula holds for an extended nonnegative integer-valued variable by the monotone convergence theorem, using .
For the first ascent in independent continuous observations, continuity makes ties probability zero. By order symmetry of independent random variables, the first observations have each of their strict orderings with equal probability. For , means exactly that the first observations are decreasing, soSubtracting consecutive tails giveswith probability zero for . The events decrease to ; continuity of probability under decreasing intersections gives . Finally,The first observation cannot itself end the run; this is why the tail includes the separate term .
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 35 5 Solution Created 2026-10-03 Updated 2026-10-14
For any , implies . Hence , and taking expected values gives the Chernoff boundAt this is the trivial bound one; an infinite exponential moment merely gives a noninformative bound. For positive at which all required moment-generating functions are finite, define . Independence of the source random variables impliesApplying the Chernoff bound to givesConsequentlyHere and, for a nontrivial target, .
The effective bandwidth is the capacity charge of one source for this exponential-moment tail criterion. Independent sources contribute additively. It incorporates variability as well as mean demand: the Jensen inequality gives , and, under exponential integrability near zero, as . For bounded traffic its large- limit is the essential peak, as described by the mean and peak limits of effective bandwidth. Thus larger gives more weight to rare high demands. This is a one-window demand bound; a queue's overflow probability over many windows requires additional temporal modeling, not just this single-window calculation.
For the normal distribution , completing the square in its density gives the moment-generating functionWrite and for the aggregate mean and variance. The exponential-moment capacity condition becomesFor , the last two terms are minimized at , where their sum is . Therefore Gaussian effective bandwidth yields the sufficient conditionEquivalently, optimizing the exponential bound directly when gives ; when the best nonnegative- bound is one. The square-root condition is sufficient rather than necessary because an exponential bound does not equal the exact normal tail.
For the exact condition, independence and multiplication of the normal moment-generating functions show that . For , let be the standard normal distribution function. The normal law is continuous, soSince is continuous and strictly increasing from zero to one, the exact Gaussian overflow constraint isThe constant depends only on the desired tail exponent , not on the source counts. It is positive when , zero when , and negative when . For example, at the exact normal quantile is about , whereas the Chernoff bound uses ; the latter reserves more capacity.
Finally, the positive-variance hypothesis must not be hidden. If , then almost surely, and the zero-variance boundary of an overflow constraint gives . For , the exact condition is then . In particular does not meet the target even though substituting into the printed non-strict square-root inequalities would allow it. The displayed normal-quantile equivalence applies to nondegenerate Gaussian traffic; the deterministic case requires this strict boundary convention.
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 45 1 Solution 2026-10-14
Condition on the claim count distribution. If and , then independence of the count and the independent and identically distributed random variables givesThe empty sum is zero. The law of total expectation and law of total variance therefore prove the random sum of independent claims formulasFor the variance calculation assume the count and severity have finite second moments; the expectation identity needs the corresponding integrability. Nonnegative claims also permit the expectation identity with extended nonnegative values.
Represent each policy payment as , where is a Bernoulli random variable of parameter , independent of its severity. Equivalently, specifies the payment distribution conditional on a claim. ThenIndependence of the policies makes their variances add. PutThe heterogeneous individual claims model consequently has
For the Poisson approximation, define the severity mixture distributionLet be the individual characteristic functions. Conditioning on each Poisson count and then multiplying transforms of the independent policy payments givesThis is Poisson superposition of insurance portfolios. By the uniqueness theorem for characteristic functions,with count and severities independent. Its compound Poisson distribution has distribution functionwhere is the convolution distribution of independent severities and is the distribution concentrated at zero. Characteristic functions avoid assuming that severity moment-generating functions exist. Since and , the random-sum formulas give
Use to distinguish the second, compound-binomial approximation from the Poisson one. It has with and the same severity law . Thus and . Applying the random sum of independent claims identities again givesHence
Both approximations match the exact expected value. The Cauchy-Schwarz inequality gives , proving the variance comparison for compound portfolio approximations:The excess variances are respectively and . The compound-binomial model is therefore closer in variance; equality with the exact variance occurs when all are equal. This variance comparison is not a general ordering of approximation errors or tail probabilities.
The Poisson replacement is useful when all claim probabilities are small: it replaces the individual restriction of at most one claim by an unrestricted Poisson count, and the variance error is of second order in those probabilities. The compound-binomial replacement retains a count bounded by , but pools both frequencies and severity laws and can lose policy-specific information. If all and all agree, it is exactly the original portfolio law. In a heterogeneous portfolio, agreement of means, or even variances, does not make either approximate law exact. Neither approximation is guaranteed accurate merely by having many policies.
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 80 3 b ii Solution Created 2026-10-03 Updated 2026-10-14
Inverse Fourier transform of the expected value givesindependent of both coordinates. Its squared modulus is the coherent scalar wave intensity , which must not be confused with the mean wave intensity.
For the latter use the stationary two-point statistics. Let be the uncentered spectral measure of a stationary random field of , normalized bySince , its total mass is . Under free paraxial propagation each spectral component is multiplied by ; in its wave-field correlation this wave phase multiplies its complex conjugate and cancels. The stationary intensity under free paraxial propagation is consequentlyStationarity is essential to this pointwise statement: it makes the spectral covariance diagonal. Conservation of integrated power alone would not show that a particular spatial point keeps the same scalar wave intensity.
The same result follows locally from , where . Taking expectations makes independent of , so . Thus coherent scalar wave intensity and mean wave intensity are both constant in propagation, with different values. The incoherent contribution has mean . Individual realizations can show focusing and changing scalar wave intensity even though the expected value remains constant. All these propagation statements concern the free parabolic wave equation from part (a).
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 80 3 b i Solution Created 2026-10-03 Updated 2026-10-14
Represent the thin layer by its wave phase jump at the exit and take the incident envelope to have amplitude one. Then . Stationarity makes the wave phase mean a constant ; it need not be zero, since the question specifies only the variance. For a normal distribution marginal, write with . If , integration by parts in the Gaussian density gives and , hence . Thus its characteristic function givesThe Fourier transform of this constant is , interpreted distributionally for the infinite plane wave. Applying the propagator from part (a),If the wave phase has zero mean, omit . A nonunit incident amplitude simply multiplies the expression. Only the one-point Gaussian distribution is needed for this expected value; no assumption about joint Gaussian distributions at separated points is needed.
Past exam of the mathematics course of the University of Cambridge 2008 ia Paper 2 4F Solution Created 2026-09-24 Updated 2026-10-14
Let be one score. It has the discrete uniform distribution on , so its expected value and second moment areThereforeFor independent throws, linearity of expectation and additivity of variance give and . The sample average consequently has expectation and variance .
The Chebyshev inequality follows by applying the Markov inequality to the nonnegative square of a centered random variable. It givesThis is at most if . Thus suffices, as does every larger integer. This is the smallest integer guaranteed by this particular Chebyshev inequality bound; it need not be the smallest integer satisfying the actual probability requirement.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 14 1 Solution Created 2026-10-03 Updated 2026-10-14
Choose a uniformly random permutation of and take its initial segments. This produces a Uniformly random maximal chain in a Boolean lattice. A fixed -set belongs to this chain with probability . An antichain meets the chain at most once, so taking the expected value of the number of meetings proves the LYM inequality:In particular the Sperner theorem bound follows by multiplying each summand by at most the largest binomial coefficient.
For the vector request, an ordering by ordinary subset need not work: distinct vectors may cancel. Instead, write and construct a separated block decomposition for vector subset sums. A block is separated if distinct labeled members have subset sums at distance at least one. We prove that the labeled subsets of can be partitioned into separated blocks, withblocks of size , for , with .
For the one subset forms one block. Suppose a block of sums has already been constructed for vectors, and let . Choose minimizing in the block. Replace the two copies of the block, with and without the new coordinate, by the following two blocks:The old labels are retained, so these partition the two copies even if equal sums occur in different blocks. Translation preserves separation in the first translated portion, and the second block is a subset of the old separated block. For the new cross pairs, Cauchy-Schwarz inequality givesThus both new blocks are separated. Their sizes are and , omitting an empty block. The resulting size-profile recurrence is , and Pascal's identity verifies the displayed binomial-difference formula, including the even middle boundary. The total number of blocks telescopes to .
A signed sum associated with is . If two signed sums lie in the open ball of radius one, their difference has norm less than two, so their subset sums have distance less than one. At most one can therefore lie in any separated block. Hence the Littlewood-Offord inequality isThe counting is by sign choices, not by distinct numerical values of the sums. The open-ball condition handles the equality-distance boundary correctly.
For the last request take real coefficients. Absorb their signs into the choices , so all coefficients are positive and at least one. A signed-sum real interval of length four corresponds to a subset-sum real interval of length two. A strict three-member inclusion chain would have , which cannot occur in that open real interval. Thus its set family is a k-Sperner family with . Every maximal chain in a Boolean lattice meets it at most twice, yielding . Write , so and . If are the two largest binomial coefficients and is the fraction at a largest level, then . This proves the two-level Littlewood-Offord bound:It is the greatest possible number uniformly over choices of coefficients and center. For even , take every coefficient equal to one and center : the real interval contains the sum levels , of multiplicities . For odd , take center zero, giving levels , each of multiplicity . These attain the bound. For the only sign choice gives the separate maximum one. Attainment here is at the exhibited centers; it need not hold at every prescribed center.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 34 3 b Solution Created 2026-10-03 Updated 2026-10-14
A filtration on a probability space is an increasing sequence of sigma-algebras contained in : . A real stochastic process is a martingale for this filtration if is -measurable, , andThe first condition is adaptedness; the tower property of conditional expectation then gives whenever .
The almost-sure martingale convergence theorem states that if a real martingale satisfies , then some integrable random variable satisfies almost surely. The more general almost sure submartingale convergence theorem needs only for a submartingale. In particular, a nonnegative martingale converges to a finite integrable limit almost surely, because its expected value is constant. Mere boundedness in the L1 norm does not imply convergence in L1. The stronger uniformly integrable martingale convergence theorem gives both almost-sure convergence and convergence in L1 when the martingale is uniformly integrable.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 34 4 a Solution Created 2026-10-03 Updated 2026-10-14
Put and take to be the trivial sigma-algebra. Each is measurable with respect to the natural filtration , andIndependence of from makes its conditional expectation equal to its expected value, which is zero. HenceThe measurability, integrability and conditional expectation conditions are all verified, so
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 34 4 c Solution Created 2026-10-03 Updated 2026-10-14
The hitting time is a stopping time for the natural filtration, because . The integer-valued increments are at most one. A path started at zero therefore cannot reach or exceed without first visiting , soDefine . It is nonnegative and integrable: . Moreover,The indicator function is -measurable, and is independent of with zero expected value. Taking conditional expectation therefore proves directly that is a nonnegative martingale.
The general result used here is that a nonnegative discrete-time martingale has a finite almost-sure limit; this is the martingale convergence theorem, since bounds its L1 norm. On it follows that converges finitely. A convergent sequence of integers is eventually constant, so would eventually be zero on this event.
However, the independent events all have the same strictly positive probability. The second Borel-Cantelli lemma says that they occur infinitely often almost surely. This contradicts eventual constancy on any positive-probability subset of . ThusNotice that no lower bound on the negative increments was needed.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 35 4 Solution Created 2026-10-03 Updated 2026-10-14
For every , the pointwise inequality holds: on the event the exponential is at least one, and off it the indicator is zero. Taking expected values and then the infimum proves the Chernoff boundAn infinite exponential moment merely gives a vacuous bound at that ; no existence assumption on moments is needed for this extended-value inequality.
For a positive at which the relevant moment-generating functions are finite, apply the Chernoff bound to . Independence of the sources givesThusClasses with make no contribution, so no moment assumption on those inactive classes is necessary. The effective bandwidth is the exponential-moment capacity demand of one class- source over this observation interval. Independent sources have additive effective bandwidths; reserving the margin below capacity certifies the desired overflow bound. If describes traffic during one unit of time, this is a bandwidth in traffic per unit time; for a window of length , divide by instead.
To establish monotonicity and bounds of effective bandwidth, write for a representative source. The Holder inequality gives, for in its moment domain,Equivalently, the cumulant-generating function is convex with , so its secant slope from zero is nondecreasing. Hence is nondecreasing, rather than necessarily strictly increasing: if is deterministic its effective bandwidth is constant.
Assuming the mean is defined and finite, the Jensen inequality gives , hence . LetFor finite , almost surely, so and . Combining these provesThe upper bound is vacuous if the essential supremum is infinite. For nonnegative traffic these comparisons also make sense with extended values; for an arbitrary real-valued source with undefined mean, the printed mean comparison itself requires an integrability qualification. The positive parameter domain is essential: is defined only by a limit when the appropriate moment assumptions hold, and the stated monotonicity comparisons concern .
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 42 3 Solution Created 2026-10-03 Updated 2026-10-14
Iterating the recurrence yieldsThus every finite collection of the is a linear transformation of independent standard normal random variables, and has a multivariate Gaussian distribution. The expected values are zero. Independence of the givesFor , the recurrence writes plus a sum of future innovations independent of . Hence . The required vector law isThe covariance matrix is positive definite: the lower-triangular transformation from has nonzero diagonal entries . This also identifies a stationary autoregressive process of order one, with marginal density .
With the standard normal density as the kernel for density estimation, the two characteristic functions are and . The Gaussian random variable has mean zero and variance , soThis expression is real. Substitution in the given integrated covariance expression, followed by the Gaussian integral for , gives
To bound the total Gaussian autoregressive kernel variance correction, apply the mean value theorem to . For , the whole interval between and is bounded below by . ThereforeSumming the geometric series gives the uniform boundIt follows from the supplied variance identity thatThe constant depends on the fixed , but not on or . In particular this applies to the requested shrinking-bandwidth regime; the covariance bound itself does not need .
All observations have the same marginal density , so dependence changes the variance but not the bias of a kernel density estimator: . Write and . For a smooth density with , the integrated variance of a kernel density estimator in the independent case is . A second-order Taylor expansion of the convolution gives in the L2 norm. Thus the bias-variance decomposition of mean squared error givesThe first two terms are the asymptotic mean integrated squared error. The additional dependence term is smaller than , so the leading optimal bandwidth and optimal rate are the same as for independent data:Here both and are the standard normal density. Since , direct Gaussian integrals giveConsequently the explicit leading bandwidth isAt this bandwidth the leading integrated variance is four times the leading integrated squared bias, which is also a check on the minimization constant.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 43 1 c Solution Created 2026-10-03 Updated 2026-10-14
Here , so iteration givesThe negative-binomial series , valid for , normalizes this toThis is the negative binomial distribution counting failures before the th success, with success probability . The convention is specified because a negative-binomial variable is sometimes defined to count trials instead. Its count expected value is , and the Panjer recursion begins at . For this reduces to the failure-count geometric distribution.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 83 2 Solution Created 2026-10-03 Updated 2026-10-14
Set and . Multiplying the Langevin equation by and integrating givesThe expected value of is zero. For the printed dot-product force covariance, write , so . Without any white-noise approximation the trace variance isThe intended short-correlation approximation replaces by , where is the Dirac delta. It requires the force correlation time to be much shorter than and the observation time. ThenAt thermal equilibrium, the equipartition theorem gives . Taking the long-time limit determines the vector noise normalization in inertial Langevin dynamics:This factor six corresponds to the dot-product covariance across three components. If were defined for one component instead, its integrated strength would be before dividing by , and the acceleration strength would be .
Finite rapid correlation is not literally a Dirac delta. For a stationary force and local drag, its exact limiting velocity trace isThis follows by splitting the double convolution into and . Thus equipartition fixes the weighted integral, not just , unless the short-correlation approximation is made. For example, gives . The remaining explicit formulae use the intended continuous-time Gaussian white noise limit.
Assume isotropic jointly Gaussian noise, as appropriate to an isotropic bath. The component covariance isThe scalar moment identities in the question must now be applied to a fixed Cartesian component , not to the magnitude of the vector. Its variance isIn the -fold integral for , the Isserlis theorem replaces the force product by a sum of complete pairings. Each pairing yields identical two-time integrals, hence . Pair the first labelled factor in ways, then the first unpaired factor in ways, and continue. The Gaussian even-moment pairing count isOdd force products vanish, soFor completeness, summing the scalar moment-generating function givesThe same Gaussian process argument for every linear combination of components gives the vector characteristic function . Inverting this Fourier transform yields the normalized multivariate Gaussian distributionThe initial condition is a velocity Dirac delta as , and the long-time limit is the thermal velocity density. The converted TeX's exponent is damaged; the PDF has the squared norm displayed here. The scalar pairing formula cannot be interpreted as a vector-norm formula: the radial moments of an isotropic Gaussian vector instead give in three dimensions.
Integrate the velocity once more. Interchanging the time integrals gives the integrated Ornstein-Uhlenbeck displacementFor the stated fixed initial position and velocity,Let and . Its component covariance comes from integrating the squared noise kernel:Therefore the total centered variance isThe mean-square displacement from additionally contains . Confusing this deterministic drift contribution with centered variance changes the short-time conclusion.
For times longer than the force correlation time but , the mean motion is and the centered trace is . Thus a nonzero fixed initial velocity produces a leading ballistic mean-square displacement ; conditioned on , the white-noise displacement variance instead starts as . If one averages over an independent equilibrium initial velocity, its covariance adds , givingThis is the usual thermal ballistic regime, before velocity memory has relaxed. At , the conditional mean approaches , while the centered trace is . Its leading behavior is diffusion with and , as successive displacement increments lose inertial memory. At times below a finite force correlation time the white-noise short-time expansion is not applicable; for a regular force covariance and fixed initial velocity the leading noise-induced displacement variance is of order .

