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Incoming links: Normal distribution
Past exam of the physics A course of the University of Cambridge 2017 Paper 1 Section A 5 Solution 2026-10-18
Use the sample mean to estimate each magnetic susceptibility, and the unbiased sample variance with denominator to estimate the repeatability of individual readings. In the numerical scale supplied by the paper,If the repeated measurements have independent random errors, their mean estimates have standard errorsThe difference and its estimated standard uncertainty are thereforeThe separation is about standard errors. Under an approximately normal distribution model with no experimenter-dependent bias, Welch's t-test gives about statistical degrees of freedom and a two-sided p-value of . Thus there is modest evidence against equal underlying susceptibilities at a conventional significance level; overlap of the ranges of individual readings is not an adequate comparison of the means. Five readings per sample provide limited information about the error distribution.
Crucially, material and student exhibit complete confounding: each student measured only one sample. An experimenter-dependent systematic error could mimic the difference. Paramagnetism also depends on conditions such as temperature, so equal material alone does not guarantee equal readings without controlling those conditions. The defensible physical conclusion is the data suggest a difference, but do not establish different materials. Both students should measure both samples in randomized or alternating order, with calibration and temperature controlled, to separate a sample effect from an operator effect. No unit or scale factor absent from the original table has been invented.
A chi-squared test compares squared deviations with the size expected from the assumed random errors. For independent observations with a normal distribution with known standard deviations and predictions , useFor a correct fixed model, the standardized independent residuals are unit-variance normal distributions, so has a chi-squared distribution with . Fitting independent linear parameters removes residual directions, giving . This distribution is exact for a full-rank linear model with known errors with a normal distribution; it is an approximation for many nonlinear fits. With known correlated errors with a multivariate normal distribution, replace the sum by , where is the covariance matrix and its rank and fitted constraints determine the available statistical degrees of freedom.
Under the null hypothesis, and . The reduced statistic should therefore be near one, with standard deviation , rather than equal to one exactly. An upper-tail p-value isCompare it with a significance threshold chosen in advance. A small value indicates tension with the model together with its error assumptions; it can arise from a wrong mean model, underestimated uncertainties or neglected correlations. An unusually small statistic can suggest overestimated uncertainties, correlations or overfitting. A p-value is not the probability that the model is true, and failure to reject is not proof of the model.
For counts in categories, the Pearson form is . Its chi-squared distribution is an asymptotic approximation requiring adequate expected counts. For a fixed-total multinomial sample there is one normalization constraint, so the usual statistical degrees of freedom are after fitting other parameters. Independent Poisson bins instead start with statistical degrees of freedom before parameters are fitted. Sparse counts require an appropriate count likelihood or a calibrated exact or simulation-based test; plugging tiny expected counts into the asymptotic formula is unreliable. The error model, binning, fitted parameters and chosen test must therefore be reported with the result.
Complete the square in the integral against the standard normal density:The remaining integral is one because its integrand is the probability density function of the normal distribution . Hence the moment-generating function of a standard normal variable is finite on the whole real line.
Use the deterministic weights . Their squared sum and maximum areTo apply the quadratic expansion of a cumulant-generating function rigorously, write , with . Fix and letAll arguments lie in this shrinking interval. ConsequentlyTaking logs of the moment-generating function and using independence of random variables now givesThe case is immediate. Since is the moment-generating function of a normal distribution of variance , the convergence theorem for moment-generating functions givesThis is a weighted central limit theorem with negligible maximum weight. Its limit variance also agrees with the exact value from variance additivity for independent random variables.
Fix and choose , independently of , with the normal distributionThe linear combination of independent normal random variables is normal. Its expected value is and its variance is . Hence for every , proving that every specified normal distribution is a self-decomposable distribution.
Past exam of the quantum condensed matter course of the University of Cambridge 2004 Section B B7 c Solution Created 2026-10-18 Updated 2026-10-20
This relates an thermodynamic equilibrium variance to the local stiffness of an appropriate thermodynamic free energy. At fixed bath temperature, suppose the effective thermodynamic free energy includes the constraints and the measure for a chosen coordinate. Its thermodynamic equilibrium probability density function is proportional to . Near a stable minimum , writeThe resulting normal distribution gives the Gaussian fluctuation approximationThis is exact for a quadratic thermodynamic free energy with the stated flat coordinate measure; in general it is a small-fluctuation result. A smaller curvature means a softer restoring response and larger fluctuations. Near vanishing curvature, higher-order terms become important, so the quadratic approximation cannot alone predict an actual infinite variance.
For example optical trap Langevin calibration uses the Brownian displacement of a trapped bead: gives . A camera or position detector records the fluctuations. Finite exposure averages rapid motion and can bias the measured variance; the measurement protocol matters as well as the thermodynamic equilibrium identity.
For a normalized state the position probability density is . With a real potential, the Time-dependent Schrödinger equation and its conjugate giveThis is the continuity equation . Integrating over the line yields under the stated decay. A real-valued wave function gives , so its instantaneous current vanishes; the same is true up to a spatially constant phase.
Put . Complete the square in the overlap:The Gaussian integral is , also for complex displacement by analytic continuation. HenceThe product of square roots fixes the branch convention implicit in the printed square root of . At the norm squared is , which becomes one for the stated . In particularThe centre is and its position variance is .
When differentiating in , keep fixed on the chosen constant-radius orbit. ThenSubstitution gives . Since , the chain rule gives . Continue continuously along the time orbit; its sign after a full orbit supplies the usual oscillator zero-point phase. The case of zero displacement is obtained as a limit, since the separate printed and expressions are singular at .
Finally . For real nonzero this isThe printed final current omits the Gaussian normalization factor . Retaining that factor is required both by the normalized state and by the continuity equation. This current is the Gaussian probability density times the velocity of its moving centre, .
Past exam of the random matrix theory course of the University of Oxford 2025 1 a i Solution 2026-10-25
Taking the transpose gives , so is a symmetric matrix. For , has the standard normal distribution, whereas has the normal distribution . Distinct upper-triangular entries depend on disjoint sets of the original independent random variables, so they remain independent.
Consequently the joint probability density function, relative to , isHere because is a symmetric matrix. This is the Gaussian orthogonal ensemble, with off-diagonal variance one and diagonal variance two; the rescaled matrix has the corresponding off-diagonal variance.
We will also use its entry covariance. Expanding in the independent entries of , for which , givesThe symbols are Kronecker deltas. This formula includes the diagonal case, where the two terms both contribute.
Substitute into the transformed partial differential equation. Comparing coefficients gives , , whenceThe stated Gaussian Fourier transform pair, with its inverse factor , givesThis is a normal distribution density with mean zero and variance . Its total integral is one. As it converges to in the sense of distributions, while as it tends to the standard normal distribution density. This is the transition density from zero of the Ornstein-Uhlenbeck process with drift and diffusion coefficient .
An importance weight has expected value . From the ratio in (a),The supplied normal distribution transform applies with quadratic coefficient , giving each factor . ThusFor , the base exceeds one because : importance weight variance increases exponentially with dimension. For , the importance weight is identically one and its variance is zero. This calculation concerns the importance weight, rather than the weighted integrand.
The integrand is positive, so the minimum-variance importance distribution is its target-weighted probability density function, normalized by :By completing the square,Thus is the multivariate normal distribution , where is the first standard basis vector. Its importance weight and weighted integrand areA variance cannot be negative, and this proposal makes every summand constant, proving optimality directly. It is admissible under the arbitrary-proposal condition even though its mean vector is not zero. Although optimal proposals usually involve an unknown normalizing integral, here both that integral and the resulting normal distribution are explicit.
At equal sample count, rejection sampling eventually has smaller variance for fixed ; at equal expected proposal cost, importance sampling eventually wins. The distinction matters because a sample of the rejection sampling estimator means an accepted draw.
For the importance sampling estimator, all coordinates contribute to the importance weight although only one enters the integrand. The same normal distribution transform as in (c), now with linear coefficient for coordinate one, givesBoth estimators are unbiased estimators of , andFor fixed , the first variance grows exponentially while the second stays constant. This is the benefit of returning exact target draws in rejection sampling. However, obtaining those draws costs proposals each on average.
To compare computational work, multiply each variance by its expected proposal count. For importance sampling that count is ; for rejection sampling it is . The ratio of the resulting cost-normalized variances isbecause . Thus importance sampling eventually gives better precision for the same expected number of proposals, assuming comparable work per proposal. This is an asymptotic cost comparison, not a guarantee that a fixed small computing budget produces any accepted draws. Both methods still deteriorate exponentially in this comparison. For they coincide and every proposal is accepted.
Sample-count and expected-proposal-cost comparisons for the one-coordinate integrand at sigma squared equal to two
. Details. 
A practical improvement is to exploit the integrand's one-coordinate structure directly: average for standard normal distribution draws , without generating unused coordinates. The optimal change of proposal distribution in (iii) improves further.
Let and , evaluated from the given polylogarithm formulas. Generate a large sample of independent and identically distributed random variables with
r_sample_f, and compute its sample mean and sample variance using mean and var. The central limit theorem gives the first standardized check:For the second check, the correct asymptotic variance of the sample variance involves the fourth central moment :The denominator is not unless the probability distribution has the requisite fourth central moment, as does the normal distribution. Here the probability density function decays exponentially, so all required moments are finite.To compute , expand the denominator of the probability density function as a geometric series and integrate each nonnegative term. The gamma function integral gives the raw momentsThus and . The following R code implements both checks for chosen interior parameters and large
n:p <- 0.4
beta <- 1.3
n <- 100000
raw_m <- vapply(1:4, function(j) {
-factorial(j) * polylog(j + 1, 1 - p) / (beta^j * log(p))
}, numeric(1))
m <- raw_m[1]
v <- raw_m[2] - m^2
mu4 <- raw_m[4] - 4 * m * raw_m[3] +
6 * m^2 * raw_m[2] - 3 * m^4
x <- r_sample_f(n, p, beta)
c(mean_check = sqrt(n) * (mean(x) - m) / sqrt(v),
variance_check = sqrt(n) * (var(x) - v) / sqrt(mu4 - v^2))For each check separately, an absolute value below is consistent with its approximate confidence interval. Repeat with many independent samples: the two standardized statistics should each resemble a standard normal distribution, with about outside that range. Also vary , and
n; a systematic discrepancy that grows with suggests a bias. The two checks from the same sample are generally dependent, so passing both is not itself a joint statement, nor can a finite test prove that a generator is correct.The second central limit theorem follows from the central limit theorem for the sample variance: differs by a quantity tending to zero in probability from .
Assume and a current state with , . The unnormalized posterior density isThus the log-posterior, up to an additive constant, is the log-likelihood plus the beta distribution and exponential distribution log prior densities. The exponential distribution rate is , since is its mean. [ref]
https://stat.ethz.ch/R-manual/R-devel/library/stats/html/Beta.html. Official R documentation: dbeta arguments and logarithmic density output
[ref]https://stat.ethz.ch/R-manual/R-devel/library/stats/html/Exponential.html. Official R documentation: dexp rate parameter and mean equal to inverse rate
Choose a symmetric joint random-walk Metropolis algorithm proposal: and , for independent standard normal distribution draws. Its scale depends only on fixed hyperparameters, so forward and reverse proposal probability density functions agree. Reject immediately outside ; otherwise accept with Metropolis–Hastings acceptance probabilityThe following function makes exactly one proposal, with rejection retaining both old parameters:
MHstep <- function(x, p, beta, a, b, mu) {
p_new <- rnorm(1, mean = p, sd = 0.1)
beta_new <- rnorm(1, mean = beta, sd = mu / 2)
if (p_new <= 0 || p_new >= 1 || beta_new <= 0)
return(list(p = p, theta = beta))
logpost <- function(pp, bb) {
ll_f(pp, bb, x) + dbeta(pp, a, b, log = TRUE) +
dexp(bb, rate = 1 / mu, log = TRUE)
}
logA <- logpost(p_new, beta_new) - logpost(p, beta)
if (log(runif(1)) <= min(0, logA)) {
p <- p_new
beta <- beta_new
}
list(p = p, theta = beta)
}The returned field
theta contains the updated value of beta. The original PDF really asks for p and theta, although no parameter is defined in this question. This choice preserves its literal interface while making the intended second parameter explicit. A caller therefore advances with state <- MHstep(x, p, beta, a, b, mu), p <- state$p, and beta <- state$theta.The proposal is an unrestricted symmetric normal distribution followed by rejection outside the support. Repeatedly drawing until a valid proposal appears would instead truncate and renormalize the proposal, requiring a proposal-density correction. The code uses immediate rejection and needs no such correction. The displayed scales are valid fixed choices; practical efficiency can be improved using a separate tuning phase.
The central limit theorem gives , so, with the cumulative distribution function of the standard normal distribution,The error has two tails. Using the supplied quantile, whose two-sided coverage is approximately , givesThe supplied quantile would be appropriate for a one-sided calculation; it would give only approximately two-sided coverage. Using the exact quantile instead of would give draws; we consistently use the paper's rounded value below.
For routine Monte Carlo integration, the central limit theorem calculation is preferable as a practical budget: it uses about one fifteenth as many draws as the Chebyshev's inequality bound. However, its coverage is approximate. Finite variance alone supplies no universal rate of convergence to the normal distribution; an integrand with large skewness or a heavy-tailed distribution can make the approximation poor even at a large sample count. If an unconditional finite-sample guarantee is required under only the stated assumptions, use the Chebyshev's inequality budget.
With a fixed budget of independent proposal trials, the accepted count has binomial distribution with success probability . HenceFor , the central limit theorem gives the one-sided approximationUse , since now only the lower tail of the accepted count matters. Requiring and solving the resulting quadratic equation in gives the one-sided normal approximation for a rejection proposal budgetEquivalently, before rounding, this is . For the value in (a)(iii),When , every proposal is accepted and exactly. Otherwise the positive safety margin above allows for random fluctuations in the number accepted. The average cost in (b)(ii) describes stopping at the th acceptance; this answer chooses a deterministic budget that reaches that acceptance with approximately probability.
The approximation requires a sufficiently accurate normal distribution approximation to the binomial distribution. An exact budget can instead be found from its tail probabilities. Also, a chance of enough accepted draws and approximate precision conditional on obtaining them are separate claims: their intersection is not automatically a event.
Use the squared three-dimensional endpoint displacements, not their arithmetic mean. The mean squared displacement estimate and corresponding diffusion coefficient areThe factor six is for three-dimensional Brownian motion. For an isolated no-slip sphere, the Stokes–Einstein relation gives the Brownian radius estimate from endpoint displacements:Here room temperature is taken as ; choosing gives , a much smaller difference than the sampling uncertainty. Under independent increments with a normal distribution, for . The sampling uncertainty of an endpoint diffusivity estimate therefore makes this a rough radius estimate. It assumes negligible drift and hydrodynamic coupling between probes, the supplied bulk viscosity, and actual three-dimensional tracking. Localization noise adds to the squared displacements and would bias this inferred radius downward if uncorrected.
For fixed , the likelihood ratio is strictly decreasing as increases. A likelihood-ratio test rejects for small , so its rejection rule is equivalently large . Under the null hypothesis, the standard normal distribution and chi-squared distribution results give independent quantitiesTheir ratio therefore has Student's t-distribution with statistical degrees of freedom. The exact level- rule isThis is precisely the two-sided Student's t-test; choosing the corresponding cutoff for yields the same rejection region and the same p-value.
Define and , with . The unrestricted maximum-likelihood estimates are and . Under the null hypothesis, maximizing the normal distribution's likelihood over the remaining variance givesbecause the cross term vanishes when . At either fitted variance, the exponential factor in the likelihood function is , so it cancels in the ratio:The standardized test statistic isThe sample variance uses divisor , whereas the maximum-likelihood estimate uses divisor ; this explains the in the final expression. Under a nondegenerate normal distribution, with probability one.
For the normal distribution, the density at its centre is . The sample mean has exact variance , whereas part (b) gives the sample median the leading variance . Thus the requested asymptotic relative efficiency isAt a fixed confidence level, a confidence interval based on the sample mean has a leading half-width smaller by the factor . With known , also has exact normal coverage. Prefer the sample mean under the normal model.
For the Laplace distribution, symmetry puts both the mean and median at . Its density there is , andConsequently the sample mean has variance , while the sample median has leading variance . HenceA large-sample confidence interval based on the sample median has leading half-width smaller by ; with known scale it is . Prefer the sample median under the Laplace model. Consistent estimation of an unknown scale preserves these leading comparisons. The reason for the reversal is quantitative: the median's precision depends on the central density , while the sample mean's precision depends on the whole population variance.
