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A microstate specifies the microscopic state: classically the positions and momenta of all particles, or quantum mechanically an allowed many-body state. A macrostate specifies a few coarse variables such as energy, volume and particle number. Many microstates can give the same macrostate. A statistical ensemble is a probability assignment over hypothetical copies prepared with the same macroscopic constraints; it does not require a sample to be literally divided into many physical copies. An observable's ensemble mean is $⟨A⟩={\sum }_i{p}_i{A}_i$, with an integral replacing the sum in classical phase space.

For an isolated system at fixed $U,V,N$, the microcanonical ensemble assigns equal probability to accessible microstates in a narrow energy shell. If their number is $\Omega (U,V,N)$, the Boltzmann entropy is $S=k_B\ln \Omega$. Thermal contact between two initially isolated systems allows energy to be shared while their sum stays fixed. The most probable division maximizes $S_1(U_1)+S_2(U-U_1)$, so equilibrium requires equal $\partial S/\partial U$. This gives the statistical temperature definitionFor positive temperature, energy tends to flow from lower entropy slope to higher entropy slope until those slopes agree. The narrow energy-shell width represents macroscopic resolution; it must not exclude the physically accessible states.

For a small subsystem exchanging energy with a much larger bath, the bath has energy $U_{\mathrm{tot}}-E_i$ when subsystem microstate $i$ has energy $E_i$. Expanding its entropy gives $S_b(U_{\mathrm{tot}}-E_i)=S_b(U_{\mathrm{tot}})-E_i/T+\cdots$. Counting compatible bath states therefore yields the Boltzmann distribution and canonical ensemble:The exponential weights individual microstates. An energy level of degeneracy $g_j$ has total probability $g_j{e}^{-\beta E_j}/Z$; a more degenerate excited level can outweigh one ground-state level even though each of its microstates is less probable. The Gibbs entropy $-k_B{\sum }_i{p}_i\ln p_i$ gives $S=k_B\ln Z+U/T$. The equilibrium energy variance is $⟨(\Delta E{)}^2⟩=k_B{T}^2{c}_V$, linking measurable fluctuations to a macroscopic response.

If particle exchange is also allowed, a reservoir fixes $T$ and chemical potential $\mu$. The grand canonical ensemble weights a state by $e^{-\beta (E_i-\mu N_i)}/\Xi$, with $\Xi ={\sum }_{N,i}{e}^{-\beta (E_i-\mu N)}$. Mean particle number is $k_{B}T\partial \ln \Xi /\partial \mu$, while volume remains a constraint in this version. Thus the ensemble follows the physical exchange conditions: fixed energy, thermal contact, or thermal and particle contact.

In ordinary large short-range systems, relative fluctuations often decrease as $N^{-1/2}$ and the ensembles give the same bulk thermodynamics away from singular limits. Finite systems, critical fluctuations and long-range interactions require care. Interpreting a time average as an ensemble average additionally requires appropriate sampling or mixing dynamics; it is not guaranteed solely by naming an ensemble. thermal equilibrium probability assignments do not describe every driven or relaxing state.

For a classical dilute ideal gas in thermal equilibrium, the probability of a single-particle velocity is proportional to the Boltzmann factor for kinetic energy, $e^{-m{v}^2/(2k_{B}T)}$. Velocities with speed between $v$ and $v+dv$ fill a spherical shell of velocity-space volume $4\pi v^{2}dv$. The normalized Maxwell speed distribution is thereforeThe factor $v^2$ counts states, rather than representing an extra Boltzmann energy factor.

Set $\alpha =m/(2k_{B}T)$ and use $I_j={\int }_0^{\infty }{v}^j{e}^{-\alpha v^2}dv$. Normalization cancels in ratios, so the given recurrence givesConsequentlyThe root mean square speed is larger than the mean speed, consistent with the positive variance of the speed distribution.

Temperature as an equilibrium relation. Classical thermodynamics begins with the zeroth law of thermodynamics: thermal equilibrium is transitive. A thermometer can therefore assign a common empirical temperature to systems in mutual thermal equilibrium. Different thermometer properties initially give different monotone empirical scales; a resistance reading by itself does not define the absolute scale.

The Second law of thermodynamics provides a substance-independent thermodynamic scale. A reversible engine operating between reservoirs has a heat ratio depending only on their temperatures. One chooses absolute temperature so that $Q_h/Q_c=T_h/T_c$, with $Q_h,Q_c$ the positive heat magnitudes. Its reversible efficiency is $1-T_c/T_h$. Equivalently, reversible heat defines entropy through $dS=\delta Q_{\mathrm{rev}}/T$. For an equilibrium simple compressible system,Classical thermodynamics uses this relation without needing a microscopic account of entropy. A dilute-gas thermometer realizes an empirical scale approaching the thermodynamic scale after corrections for nonideal behavior.

Statistical meaning of the same derivative. In the microcanonical description, an equilibrium energy shell containing $\Omega (U,V,N)$ accessible states has $S=k_B\ln \Omega$. Its statistical temperature is consequentlyTo see why this is an equilibrium temperature, let two weakly coupled systems exchange a small energy while their total energy stays fixed. The combined multiplicity is ${\Omega }_A(U_A){\Omega }_B(U_{\mathrm{tot}}-U_A)$. Maximizing its logarithm giveswhich is equality of the two inverse temperatures. Stability requires the total entropy maximum, not only stationarity. The overwhelmingly dominant macrostate in a large system therefore realizes the thermodynamic equilibrium condition.

A large reservoir gives a small subsystem the canonical probabilities $p_i\propto e^{-\beta E_i}$, with $\beta =1/(k_{B}T)$. In that ensemble, $F=-k_{B}T\ln Z$ and $S=-k_B{\sum }_i{p}_i\ln p_i$ agree with macroscopic thermodynamic relations in their usual equilibrium regime. The equipartition identity for a classical dilute monatomic gas, $⟨p^2/(2m)⟩=3k_{B}T/2$, is a useful operational connection between microscopic motion and temperature, but is not a universal definition: quantum modes need not obey equipartition, and a system away from equilibrium need not have one temperature at all.

For finite systems one must specify the entropy convention and energy resolution; surface and fluctuation corrections can distinguish definitions that agree in the ordinary thermodynamic limit. Bounded-energy population-inverted systems can also have a negative entropy derivative and hence a negative statistical temperature. That possibility lies outside an ordinary positive-temperature ideal-gas thermometer; it does not mean temperature is merely an arbitrary average kinetic energy.

Obtaining the energy–temperature conversion. The Boltzmann constant connects entropy measured in $\mathrm{J}{\mathrm{K}}^{-1}$ with the dimensionless logarithm of a state count. Historically it could be determined from the ideal gas law $pV=N{k}_{B}T$: measure the molar gas constant $R$ and independently determine the Avogadro constant $N_A$, then obtain $k_B=R/N_A$. The independent particle-number determination matters; using $R$ alone would not fix energy per particle.

An electrical method uses Johnson-Nyquist noise. A resistor of resistance $R_e$ at known temperature has, in the classical range $hf\ll k_{B}T$, a one-sided open-circuit voltage-noise spectrum $S_V(f)=4k_{B}T{R}_e$. Over a calibrated flat bandwidth $\Delta f$,A practical measurement uses an equilibrium resistor, a calibrated low-noise amplifier and a known transfer function, subtracts or independently characterizes amplifier noise, and replaces a nominal bandwidth by the actual effective noise bandwidth. Varying temperature and fitting the mean-square voltage is more informative than treating one amplified voltage sample as the thermal variance. Other primary methods relate temperature to sound speed in a dilute gas or to calibrated particle diffusion and mobility.

There is a historically relevant unit-definition qualification: by the date of this exam, the revised SI had already taken effect on 20 May 2019. The kelvin is now defined by fixingThe same experiments now realize or compare thermodynamic temperatures using this fixed constant, whereas historically they measured the conversion constant on an independently defined temperature scale. The physical distinction between equilibrium temperature and microscopic state counting is unchanged.^[ref]

For the same massless spring of spring constant $k$, Hooke's law gives the vertical simple harmonic motion about the displaced equilibrium position:Gravity shifts the equilibrium extension but does not change this frequency. Use the measured sample means in the ratio estimate. For small measurement errors, logarithmic differentiation givesAssuming the two frequency estimates have independent errors, their variances add. Writing $ϵ=\delta \nu /\overline{\nu }$ for their common fractional standard uncertainty, monomial uncertainty propagation yieldsThis is a first-order uncertainty formula, rather than an exact identity for an arbitrary noisy ratio. It assumes that errors in the two measured frequencies have zero covariance.

There is a statistical distinction in the wording: the standard deviation of individual repetitions is not the standard error of their sample mean. If the supplied $\sigma ({\nu }_i)$ describes individual trials, the frequency means have errors $\sigma ({\nu }_i)/\sqrt{n}$, and the ratio formed from those means instead satisfiesThus the printed formula describes either a single-pair ratio or a convention in which $\delta \nu$ already denotes the uncertainty of each averaged frequency. Under either consistent interpretation, independent repeated measurements reduce statistical error in proportion to $n^{-1/2}$. Halving it requires $4n$ trials in total for each mass, or $3n$ additional trials. Repetition alone does not reduce a fixed systematic error.

Rearrange the supplied form of Kepler's third law to make the gravitational constant the subject:Convert the semi-major axis and orbital period to SI units before substitution:The central value is thereforeFor monomial uncertainty propagation, logarithmic differentiation givesIf the quoted errors are independent standard uncertainties, the variances of these contributions add:Thus ${\sigma }_G=1.61335\times 10^{-12}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, and the measured result under that statistical convention isThe semi-major axis measurement supplies the largest individual contribution to the relative uncertainty.

The paper does not specify whether its errors are statistical uncertainties or bounds. If they are maximum input errors, the first-order worst-case error propagation estimate adds magnitudes:Correlation between input errors would instead require the covariance terms in the general propagation of uncertainty formula; they are not specified.

With microstate energies held fixed when differentiating,Since $d\beta /dT=-1/(k_B{T}^2)$,The canonical ensemble fixes volume and particle number, and any other external Hamiltonian parameters. Thus the derivative is $C_V$; pressure-fixed fluctuations would naturally involve enthalpy in a different ensemble.

The translational and two-level energy variance follows because translation and internal states factorize for this dilute classical gas:Differentiating twice yieldsThe terms are independent translational and two-level binomial variances. At $k_{B}T\ll ϵ$, excitation freezes out exponentially and translation dominates. At $k_{B}T\gg ϵ$, the internal variance tends to $N{ϵ}^2/4$ while the translational variance grows as $T^2$; the internal heat capacity consequently tends to zero. Fractional total-energy fluctuations are of order $N^{-1/2}$ for large $N$. The low-temperature statement presumes translation is still nondegenerate; at strictly zero temperature the classical-gas model fails.

Let $\mu =\mathbb{E}[X_1]$ and ${\sigma }^2=\mathrm{Var}(X_1)\in (0,\infty )$. For independent and identically distributed random variables, the central limit theorem statesThe limit is the standard normal distribution. If the variance is zero, then $X_i=\mu$ almost surely; the centered sums are zero, and division by $\sigma$ is not defined.

By part (c) with $\lambda =1$, let $T_n=E_1+\cdots +E_n\sim \mathrm{Gamma}(n,1)$ for independent and identically distributed exponential random variables. The requested integral is $\mathbb{P}(T_n\le n)$. Each summand has expectation one and variance one, so the central limit theorem givesZero is a continuity point of the standard normal distribution, and consequently

Write $A_k=N_{2k}-N_k$ and $B_k=N_k$. The stationary increments and independent increments of the Poisson process make $A_k$ and $B_k$ independent random variables, each with Poisson distribution of parameter $k$. ThusMultiplication of the two moment-generating functions givesThe odd terms cancel in Taylor's theorem, so $\log M_k(t)=t^2+O(k^{-1})$. The limit $e^{t^2}$ is the moment-generating function of a normal distribution with mean zero and variance two. The same continuity theorem for moment-generating functions therefore yieldsIn particular, the two equal-length increments must be separated before using independence: $N_{2k}$ and $N_k$ themselves are dependent.

The Poisson process gives $N_k\sim \mathrm{Pois}(k)$, with expected value and variance both equal to $k$. Consequently $Y_k=(N_k-k)/\sqrt{k}$. Scaling and translating the argument in the moment-generating function of a Poisson distribution gives, for every real $t$,The centering term must remain inside the exponent: omitting it would leave a divergent first-order contribution.

The strong law of large numbers says that, for real independent and identically distributed random variables $Z_1,Z_2,\dots$ with $\mathbb{E}|Z_1|<\infty$,The assertion is almost sure convergence of the entire sequence of sample means. Finite first absolute moment suffices; a finite variance is not required.

Differentiating the moment-generating function at zero givesThe second derivative is the second raw moment, so the variance is $M_X^{\prime \prime }(0)-(M_X^{\prime }(0){)}^2$. The assumed exponential integrability justifies these derivatives.

Let $Y_1,Y_2,\dots$ be real independent and identically distributed random variables with finite expected value $\mu$ and variance $0<{\sigma }^2<\infty$. The central limit theorem statesEquivalently, the cumulative distribution functions of these normalized sums converge at every real point to the standard normal distribution function. The theorem itself does not require a finite moment-generating function.

The given expected value and variance imply $M_X(0)=1$, $M_X^{\prime }(0)=0$ and $M_X^{\prime \prime }(0)=1$. The Taylor expansion therefore givesSince $\log (1+u)=u+o(u)$ as $u\to 0$, it follows thatThis is the quadratic expansion of a cumulant-generating function for a centered variable of variance one.

The unit-time increments $D_j=N_j-N_{j-1}$ are independent and identically distributed random variables, by independent increments and stationary increments of the Poisson process. Each has the Poisson distribution of parameter one, hence expected value one and variance one. The central limit theorem therefore gives

Set $T_0=0$ and $E_j=T_j-T_{j-1}$. The Poisson interarrival times $E_j$ are independent and identically distributed random variables with the rate-one exponential distribution. In particular, $\mathbb{E}{E}_j=1$ and $\mathrm{Var}(E_j)=1$, so the finite, positive variance hypothesis of the central limit theorem holds. Since $T_n={\sum }_{j=1}^n{E}_j$,

Direct integration of the uniform distribution yields $\mathbb{E}{U}_j=b/2$ and $\mathbb{E}{U}_j^2=b^2/3$. The expectation of a product of independent random variables givesSubtracting the square of the expected value gives the variance:Since $W_n\ge 0$, Markov's inequality implies, for every $\epsilon >0$,Hence $\boxed{W_n\to 0\text{ in probability for }0<b<2.}$ The first moment suffices: the variance need not tend to zero throughout this range.

Fix $0<c<1$ and choose $R_c$, independently of $X$, with the normal distributionThe linear combination of independent normal random variables $cX+R_c$ is normal. Its expected value is $c\mu +(1-c)\mu =\mu$ and its variance is $c^2{\sigma }^2+(1-c^2){\sigma }^2={\sigma }^2$. Hence $cX+R_c\stackrel{d}{=}X$ for every $c$, proving that every specified normal distribution is a self-decomposable distribution.

For independent and identically distributed random variables $X_j\sim \mathcal{N}(0,{\sigma }^2)$, the linear combination of independent normal random variables $n^{-1/2}{\sum }_{j=1}^n{X}_j$ has expected value zero and variance $n^{-1}n{\sigma }^2={\sigma }^2$. ThusThis is exactly the requested $2$-stability. The paper's definition is a strictly stable distribution, since it allows no centering term.

Minimizing the squared residual norm gives the normal equationsFull column rank makes $X^{T}X$ invertible. With columns corresponding to $1,x,x^2-4$, the design matrix isSymmetry gives zero cross products between the odd column and either even column, and $\sum (x^2-4)=28-28=0$ gives the remaining orthogonality. Their squared norms are seven, 28 and 84, so $X^{T}X=\mathrm{diag}(7,28,84)$. ThereforeSince the error covariance is ${\sigma }^{2}I$, least squares has covariance ${\sigma }^2(X^{T}X{)}^{-1}$. ThusThe estimators have zero covariances. independence would require an additional distributional assumption such as Gaussian errors; the stated mean and variance assumptions alone give only the covariance calculation.