Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 4 14C Solution Created 2026-09-23 Updated 2026-10-03
This is a birth-death process with birth and death ratesIts transition diagram has the two arrowsThe gain into state comes from a death in state or a birth in state , while the loss is the sum of both rates out of state . The birth-death master equation is consequently
Multiplying by , summing over the nonnegative integers, and shifting the summation indices shows that each birth contributes and each death contributes . Thus the first-moment equation of a birth-death process givesAt a stationary state, use the variance identity . Solving the resulting quadratic equation givesThe minus root is inadmissible whenever it is negative, namely when . Equality would give zero mean, which is also incompatible with a positive immigration rate , so for the minus branch requires even to be a possible mean.
For a continuum approximation, write and . Applying the Kramers-Moyal expansion to the two gain terms and retaining derivatives through second order gives the Fokker-Planck equationwhere the negative drift and infinitesimal jump variance areThis truncation requires the typical population and the scale on which varies to be much larger than the unit jump size, with the rates varying smoothly across that scale.
The positive zero of isso in particular as . Put . The linear noise approximation useswhereand, because ,In a stationary state with zero Fokker-Planck probability current,Its normalized solution is the normal distributionIt follows that
These estimates agree with the exact stationary first-moment relation on its plus branch to leading order: inserting gives . Moreover,Thus the stationary mass lies far from the boundary , is narrow relative to its mean, yet changes across many lattice sites. These are precisely the large-population and slow-variation conditions needed for the diffusion approximation.
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 201 1 b i Solution Created 2026-09-24 Updated 2026-10-03
For a partition of , every Riemann sum is a centered Gaussian random variable, withPath continuity gives almost surely, and the covariance bound also gives convergence in . Hence the limit is Gaussian, centered, and passage to the limit in the displayed sums gives variance
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 219 1 a Solution Created 2026-09-24 Updated 2026-10-03
An apparent magnitude is the sum of absolute magnitude and distance modulus. ThusThe independent unbiased distance-modulus estimate is , so closure of independent normal distributions under linear combinations givesHence and .
Past exam of the mathematics course of the University of Cambridge 2025 iii Paper 335 1 i Solution Created 2026-09-24 Updated 2026-09-25
The phase screen adds the phase accumulated across its thickness, so the reduced field just after the screen isFor a zero-mean random variable with a normal distribution and variance , its characteristic function givesThe same expression is approximately valid for a non-Gaussian weak fluctuation: the cumulant expansion begins with , while higher cumulants give higher-order corrections.
For , every realization obeys the parabolic wave equationLinearity permits ensemble averaging, soThe initial mean is independent of , hence diffraction does not change it and for every .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 201 4 b Solution Created 2026-09-24 Updated 2026-10-03
The variable is the limit of Riemann sums of the Gaussian process , so it is a Gaussian random variable. Its mean is zero, and Fubini's theorem with givesTherefore .
Past exam of the mathematics course of the University of Cambridge 2026 iii Paper 226 3 c Solution Created 2026-09-24 Updated 2026-09-25
The one-arm event depends on the field values in the finite ball . Replace by for . In the new variables the event has threshold zero, while the product normal distribution density is . Differentiating this finite-dimensional integral under the integral sign gives the Gaussian shift identityThis is also an instance of Gaussian integration by parts. In particular, the asserted inequality holds, in fact with equality.
No, not from the stated assumptions alone. The centered slope is a weighted sum,and the weighted errors need not be identically distributed. The ordinary central limit theorem therefore does not apply directly.
For a concrete counterexample, for every chooseThese satisfy all three centring and orthogonality assumptions, and both explanatory columns are nonzero. Let the errors be independent with Rademacher distribution, taking values with equal probability. Thenwhich takes values with probabilities for every . Increasing the sample size adds no information about this slope, and the standardized estimator does not approach a normal distribution.
A useful additional assumption for asymptotic normality of a weighted regression slope is that no single design value dominates:Under this condition and the given finite error variance, the Lindeberg-Feller central limit theorem yieldsjustifying the approximation for the slope as well.
The algebraic identity does not require a normal distribution for the errors. Applying the central limit theorem to their sample mean givesThus, for large ,Its mean and variance remain exact, even though its normal distribution is now only an approximation. With the error probability distribution unspecified, the formula is retained as an estimator; it is not being asserted to remain a maximum likelihood estimation formula.
The least-squares estimators choose the intercept and slope to minimize the sum of squared regression residuals:After observation, replace by to obtain the numerical estimates. Their definition does not require a normal distribution for the errors. A unique slope requires some variation in the fixed predictors; under , this is , assumed below. If all , the slope is unidentifiable.
They need not be independent. For a counterexample take , , , and independent errors taking with equal probability. ThenEach centered estimator is zero with probability , but they are never both zero. Their zero events therefore violate independence.
A sufficient additional assumption is that the errors have a common normal distribution. The two estimators are then Jointly Gaussian, being linear combinations of the independent Gaussian random variables. Their zero covariance, established in iii, implies independence of uncorrelated jointly normal variables. Normal errors are a sufficient condition here, not a claim that no other error probability distribution could ever work.
The least-squares estimators depend only on the observed responses and design, so the formulas in i still hold. Their variances and covariance need only the error covariance matrix, now . ThereforeFor example the first formula follows from , and the second from . The formulas remain the same as before with ; this calculation does not establish independence or a normal distribution for the estimators.
At sample size 100, keep essentially the same numerical confidence interval, but treat its calibration as approximate. For independent heights with finite, nonzero variance, the central limit theorem makes the sample mean approximately normally distributed, and the sample variance consistently estimates the population variance. Hence the studentized sample mean is approximately standard normal even without a normal distribution for individual heights.
Using the permitted cutoff 2 givesA Student t-distribution cutoff gives a similar approximation for large sample size, but its justification at fixed sample size is no longer exact. Extreme skewness or influential outliers could make 100 observations insufficient for accurate coverage probability; abandoning normality alone supplies no specific numerical correction. The sampling with replacement assumption still precludes a finite population correction.
The independent uniform choices from the student list describe sampling with replacement. Under the stated independent normal distribution model for the heights, the population variance is unknown and the given cm is the sample standard deviation . The exact pivotal quantity isThus the exact 95% Student t confidence interval isusing . The supplied approximation instead gives a normal approximation for large sample sizeThis latter confidence interval is approximate even under normality because it substitutes the sample standard deviation for the unknown population value. There is no finite population correction under the specified independent selections. Interpreting the finite student list literally as a discrete population makes normality an idealized model; it does not justify changing the stated sampling design.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2002 Paper 2 4 Solution Created 2026-10-18 Updated 2026-10-20
The number of successes has a binomial distribution. Each arrangement of successes and failures has probability by independence, and the binomial coefficient counts the arrangements. ThereforeThe final equality is the binomial theorem. Writing as a sum of independent Bernoulli random variables gives its expectation and variance: , . Thus the matching normal distribution hasThe normal approximation is appropriate when both expected counts and are sufficiently large; large alone is insufficient if is extremely close to zero or one.
For the guessing student, , , . Therefore the expected number of correct answers is , and , . The inclusive integer range becomes after the continuity correction. Standardizing the normal distribution givesHere , and the last step uses the supplied bound. The approximate probability is . The normal approximation provides the requested conclusion; as a separate check, summing the five exact binomial probabilities gives approximately , also exceeding one half.
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2008 Paper 1 15 b i Solution Created 2026-10-18 Updated 2026-10-20
The density is the symmetric sumFor small , the normal distributions are narrow and well separated, giving peaks near each nonzero integer. The largest two peaks lie near and ; those near have approximately one quarter of their height, and the envelope falls as . There is no component centered at zero, and the density there is exponentially small. The left panel shows the resulting smooth comb.
Treat each observed arrival label as a constraint on an unobserved walking duration. Departures are treated as fixed design information. Writing for the minutes available until work starts, the derived late-label thresholds have frequencies ; the early-label thresholds have frequencies . These are censoring thresholds, not measured journey times. A late label means , and an early label means ; equality has probability zero under the continuous model.
The paper specifies no prior distribution. One explicit proper model iswith independent priors; all times are in minutes. These ranges are a transparent modelling choice, not information deduced from the observations. In particular, this is a prior on the standard deviation , not a uniform prior on the variance . The normal distribution permits negative durations; it is the mathematical approximation requested here, rather than a claim that such durations are physically possible. A truncated model would be a different model.
The following Python implements hard conditioning by rejection sampling.
None marks a rejected trace; an accepted Boolean answers whether all five future arrivals are late. The same parameters are shared across past and future days.import math
import random
LATE = {20: 1, 21: 3, 25: 3, 27: 1, 32: 2}
EARLY = {29: 1, 34: 1, 36: 3, 37: 1, 38: 1, 39: 1, 43: 1, 44: 1}
RECORDS = [(c, True) for c, n in LATE.items() for _ in range(n)]
RECORDS += [(c, False) for c, n in EARLY.items() for _ in range(n)]
def journey_trace(rng):
mu = rng.uniform(0.0, 60.0)
sigma = rng.uniform(1.0, 20.0)
# Twenty latent past durations and five future durations.
past = [rng.gauss(mu, sigma) for _ in RECORDS]
future = [rng.gauss(mu, sigma) for _ in range(5)]
if not all(t > c if late else t < c
for t, (c, late) in zip(past, RECORDS)):
return None
return all(t > 30.0 for t in future)
def accepted_average(trace, n, rng):
if n <= 0:
raise ValueError("n must be positive")
successes = accepted = 0
while accepted < n:
result = trace(rng)
if result is not None:
accepted += 1
successes += int(result)
return successes / nFor example,
accepted_average(journey_trace, n, random.Random(2021)) estimates the desired conditional probability from accepted traces. Accepting False is essential: treating a false answer as a rejected trace would condition on the desired conclusion. The model deliberately emphasizes clear semantics; rejection can be inefficient because all twenty past-label constraints must hold together.A confidence interval procedure produces random endpoints such that , under the model. In everyday terms, repeating the same sampling and interval construction many times would give intervals containing the true mean in about of repetitions.
For , define the sample mean and the sample variance . For independent observations from a normal distribution with , the standard sampling results areThe first follows from the independent normal sample mean and sample variance. Let be the positive Student quantile. Symmetry gives the central confidence intervalIts width is , so
The independent identically distributed central limit theorem states that for variables of common finite mean and finite positive variance ,For the exponential distribution, and . Thus the event in question is exactlyThe limiting normal distribution has a continuous distribution function at every real , so convergence in distribution givesHere , including the strict-inequality convention because the limiting distribution is continuous.
For each fixed , the Taylor expansion of the exponential yieldsThe limiting moment-generating function is that of the standard normal distribution and is finite around zero. The moment-generating function continuity theorem therefore gives
A psychometric function specifies the probability of a measured response as a physical stimulus variable changes. The response might be “present,” “clockwise,” or correct in a two-alternative forced-choice task. These are different dependent variables. Understanding the function's shape is necessary to distinguish sensory changes from decision, attention and measurement effects. A reported threshold without its task, response definition and performance criterion is incomplete. For example, a visual experiment could use calibrated display visual contrasts, timed two-interval presentations, fixation control and button choices; the observer’s choice supplies the dependent variable.
A smooth curve need not mean that a receptor has a smoothly varying conscious threshold. Consider a noisy internal response , with “yes” reported when . Here is a decision criterion, not necessarily a biological detection limit. If is the noise cumulative distribution function,With Gaussian noise of standard deviation , symmetry of the normal distribution gives a Gaussian psychometric functionThe sigmoid is therefore a distribution tail swept by a changing mean response. At low means, few responses exceed the criterion; near it, many trials switch category; at high means, almost all do. The derivative is , so slope depends on both sensory gain and variability. Moving shifts the curve without necessarily improving sensitivity. If noise vanished and the mean crossed the criterion sharply, this simple model would approach a step rather than a graded transition.
There is no universal sigmoid in every stimulus coordinate. A nonlinear transduction function changes the curve, and a logarithmic intensity axis changes its appearance. As a counterexample, a toy detector responding whenever at least one event occurs in a Poisson distribution with mean has . This is concave on a linear positive-intensity axis. It follows immediately by subtracting the zero-event probability from one. It is a mathematical example, not evidence that human detection universally uses a one-event rule. Variation in attention or decision criterion across trials can broaden a measured curve even if the within-state sensory noise is unchanged.
Task conventions determine the endpoints. A useful psychometric function with guessing and lapses isHere is the lower asymptote and the upper error asymptote. In balanced two-alternative forced-choice performance, is commonly fixed at . A yes/no response probability does not automatically have this lower bound. In this particular cumulative-Gaussian parameterization, ; thus gives 75% correct only for and . Other function families use different scale parameters. Nor is automatically the fraction of inattentive trials: if a fraction of trials involves random two-choice guessing, those trials contribute an upper error probability .
Original theoretical noisy-decision and psychometric-curve examples; the right panel is a phenomenological family, not observed participant data
. Details. The relation to signal detection theory is equally important. Under independent equal-variance Gaussian signal and noise samples, define sensitivity index . In an ideal two-interval comparison, the response difference has mean and variance , givingThis Gaussian forced-choice sensitivity relation follows by standardizing the Gaussian difference. It is a specified decision model, not an identity for every forced-choice experiment. In a yes/no task, hits and false alarms are both needed: under the corresponding equal-variance model, . A criterion change can increase both rates while leaving unchanged. The phenomenological fit above need not have parameters identical to these internal-response parameters.
These distinctions affect experiments. Felix A. Wichmann and N. Jeremy Hill used simulation and statistical modeling to show that unmodeled stimulus-independent errors can bias threshold and slope estimates, even when lapses are uncommon.[ref]
The psychometric function: I. Fitting, sampling, and goodness of fit by Felix A. Wichmann and N. Jeremy Hill. Notation, lapse simulations, and summary; Percept Psychophys 63, pp. 1293–1313
Ignoring high-intensity errors can make a function appear shallower and falsely suggest worse sensory precision. A broad group-average curve can likewise reflect a mixture of individually sharp functions with different locations.At predetermined stimulus levels, independent trials give counts . The parameter-dependent log-likelihood is . Fitting this binomial likelihood, checking goodness of fit and quantifying uncertainty are preferable to drawing a curve through noisy proportions. Easy trials help constrain lapses; levels near the transition constrain location and slope. Randomization, stable calibration and checks for learning or fatigue help distinguish a stationary sensory relation from drift. Adaptive procedures require inference appropriate to their sampling design.
Thus the function is both a description and a testable model. Its shape determines what a threshold means, how an experiment should sample, and which claims about sensory mechanisms are warranted.

