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Availability and total entropy. The reservoir changes by $d{U}_R=-dU$ and $d{V}_R=-dV$. Its particle number and the system's particle number are fixed. The reservoir's first law of thermodynamics therefore givesAdd the system's entropy change. Because $T_R,p_R$ are constant,Thus define the availabilityAn additive constant in $A$ is immaterial; its value above the equilibrium reference is the exergy. The Second law of thermodynamics gives $\Delta S_{\mathrm{tot}}\ge 0$, hence $\Delta A\le 0$. A stable equilibrium maximizes total entropy over the allowed variations and therefore minimizes $A$. The environmental $T_R,p_R$ enter this definition even when the system has not yet reached their values. Because both thermal and mechanical exchanges are allowed, equilibrium here includes mechanical balance; equality of temperatures alone does not ensure the minimum.

Probability and curvature. For a fixed value of a macroscopic coordinate $x$, let $S(x)$ count the compatible system microstates. The product of the system and reservoir multiplicities isAll factors independent of $x$ enter a normalization constant, givingHere $A(x)$ is the constrained or effective availability for this coordinate, including the multiplicity of any eliminated variables. A nonuniform coordinate measure must likewise be retained or incorporated in that effective availability.

At a stable interior minimum, $A^{\prime }(x_0)=0$ and $A^{\prime \prime }(x_0)>0$. Its Taylor expansion givesKeeping the quadratic term yields a normal distribution,This is availability curvature and equilibrium fluctuations. The quoted fluctuation formula holds at quadratic order in the Gaussian fluctuation approximation; it is not an exact identity for an arbitrary nonlinear $A(x)$. Higher terms can shift the mean away from its most probable value $x_0$.

Bubble equilibrium. Keep the number of air molecules fixed and include the surface energy in the system's total internal energy. For constant surface tension $\Gamma$, this contribution is $\Gamma \Omega =\Gamma \alpha V^{2/3}$, with no surface entropy under the stated approximation. Its differential isHenceIndependent thermal and volume variations give $T=T_R$ andThe air pressure is evaluated at $V_0$; since the bubble is closed it generally depends on volume, so this equation is implicit until an air equation of state is supplied. For a sphere, $\alpha =(36\pi {)}^{1/3}$, $\Omega =4\pi R^2$, and the pressure jump becomes $2\Gamma /R$, the Young–Laplace equation for a single gas–liquid interface. There is no second surface as in a thin soap film.

Isothermal volume fluctuations. With the air maintained at $T_R$, eliminate its entropy in favor of its Helmholtz free energy:The identity $(\partial F_{\mathrm{air}}/\partial V{)}_{T,N}=-p_{\mathrm{air}}$ givesUsing the curvature result,These are the volume fluctuations of a closed gas bubble. Stability requires the denominator to be positive. The negative surface contribution reflects the concavity of $V^{2/3}$; gas compression supplies the positive restoring curvature. For ideal air, ${\kappa }_T^{-1}=p_{\mathrm{air}}$, and mechanical equilibrium reduces the denominator toThe units are consistent: $D$ is pressure, so $k_B{T}_R{V}_0/D$ has units ${\mathrm{m}}^6$, the units of a volume variance. Holding $p_{\mathrm{air}}$ externally fixed while varying $V$ would remove the gas-compression term and change this stability problem; that is not the fixed-particle-number situation here.

Numerical size of the fluctuations. The volume conversion is important:Use the specified $\alpha =4.84$, $\Gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $p_R=10^5\mathrm{Pa}$. ThenWith the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$,Thus the fractional root mean square fluctuation and fractional variance are respectivelyThe absolute root mean square fluctuation is $2.01\times 10^{-19}{\mathrm{m}}^3$. The very small fractional value supports the local Gaussian fluctuation approximation. This estimate neglects exchange of dissolved air, variations of surface tension, shape fluctuations and temperature fluctuations, consistently with the prescribed fixed-number, isothermal bubble model.

By the density formula for expected value,For the standard normal distribution, the integrand in the second moment is even, soComparing these two integrals gives

The standard normal distribution has mean zero and variance one, so $\mathbb{E}[Z^2]=\mathrm{var}(Z)+(\mathbb{E}[Z]{)}^2=1$. Part (iv) therefore givesUsing the second moment from part (iii),

Complete the square in the joint probability density function:The density therefore factors as the standard normal distribution density for $Y$ times a normal distribution density in $x$ with mean $\rho y$ and variance $1-{\rho }^2$. It follows that $\mathbb{E}[Y]=0$, $\mathbb{E}[Y^2]=1$, and $\mathbb{E}[X\mid Y]=\rho Y$. Symmetry gives $\mathbb{E}[X]=0$. By the law of total expectation,Consequently the covariance of this bivariate normal distribution is $\mathrm{Cov}(X,Y)=\rho$. This calculation derives the result from the given density.

A photon-recoil lag correlation asks how often a photon record occurs at a specified time displacement from an electron record. Let $r_e,r_{\gamma }$ be observed separate-channel rates, $S$ the rate of detected genuine pairs before a timing cut, and $f(\ell )$ their normalized timing probability density function. For an original dilute, stationary model with observation time $T$, neglecting end effects and detector dead time, the expected pair density isThis counts all pairs of records, rather than at most one detector trigger per electron. It uses independent unrelated detections for the flat term and an added same-event contribution. A normal distribution for timing spread is a chosen approximation, not a claim about the historical instrument. Intrinsic emission delay, counter response and readout uncertainty can all affect the observed probability density function.

Define the accepted lag interval by $|\ell |<w/2$: $w$ is its full width. Integrating the flat baseline gives the existing accidental coincidence expectation $T{r}_e{r}_{\gamma }w$. The additional area is $TS{\int }_{-w/2}^{w/2}f(\ell )d\ell$. Moving a window of the same full width away from the peak tests the baseline under the model's stationarity assumption. It does not remove correlated instrumental backgrounds or changing rates automatically.

For the lag model, choose $T=1000\mathrm{s}$, $r_e=2{\mathrm{s}}^{-1}$, $r_{\gamma }=0.5{\mathrm{s}}^{-1}$, $S=0.1{\mathrm{s}}^{-1}$ and timing standard deviation $\sigma =1\mathrm{ms}$. A full width $w=2\mathrm{ms}$ gives $2$ expected accidental coincidences and $100\mathrm{erf}(1/\sqrt{2})\simeq 68.27$ accepted genuine pairs, using the error function. The total is approximately $70.27$, whereas the displaced interval from $4$ to $6\mathrm{ms}$ has approximately $2.003$ expected pairs. A finite sample fluctuates around these expectations; these values are not measured counts or a significance calculation.

The right panel isolates the information lost by retaining singles totals alone. In a separate idealized model, stipulate $M=10^6$ independent $1\mathrm{ms}$ bins, with Bernoulli distribution indicators $X,Y$ for electron and photon detection in a bin. Set $p_e=\mathrm{Pr}(X=1)=0.002$ and $p_{\gamma }=\mathrm{Pr}(Y=1)=0.0005$. Independence predicts $\mathrm{Pr}(X=1,Y=1)=p_e{p}_{\gamma }=10^{-6}$, hence one expected double detection. Alternatively chooseThese nonnegative probabilities sum to one and preserve both marginal probabilities, yet predict $101$ double detections in expectation. Their covariance is $0.000101-p_e{p}_{\gamma }=0.000100$. This is an exact joint probability distribution illustrating pairing; it does not model the timing spread in the left panel or reconstruct either historical hypothesis quantitatively. Its lesson is why two singles totals cannot decide whether the observed events share a cause.

Cecil Frank Powell's 1950 lecture illustrates the successive decays of pions and muons through their tracks and explains why the nearly fixed secondary muon range, allowing for range straggling, supports a single neutral decay partner. He describes its identification as a neutrino tentatively. The neutral particle itself leaves no recorded track. He contrasts this decay with the nuclear reactions caused by stopped negative pions, connecting the latter with the nuclear force.^[ref]

Here is a modern special relativity derivation for the dominant leptonic pion decay, assuming a free pion at rest, a massless muon neutrino, and no emitted photon. Write $m_{\pi }$ and $m_{\mu }$ for the pion and muon masses. By conservation of momentum the daughter momenta are opposite with equal magnitude $p$. By conservation of energy and the energy–momentum relation,Substitute $E_{\mu }=m_{\pi }{c}^2-pc$, square, and cancel $p^2{c}^2$. This givesThe resulting $p$ is fixed, although the muon's direction is not set by the incoming stopping track. Thus the kink angle relative to that earlier track is not the angle between the two daughters: in the pion rest frame, the daughters are back to back. A neutral-partner arrow in a diagram represents inferred momentum, not an observed string of grains.

For an original calculation, adopt rounded input values $m_{\pi }{c}^2=139.57\mathrm{MeV}$ and $m_{\mu }{c}^2=105.66\mathrm{MeV}$. They give $pc=29.79\mathrm{MeV}$, $T_{\mu }=4.119\mathrm{MeV}$ and $E_{\nu }=29.79\mathrm{MeV}$. Their sum satisfies $T_{\mu }+E_{\nu }=(m_{\pi }-m_{\mu })c^2=33.91\mathrm{MeV}$. These are rounded calculation inputs and outputs, not a fit to the 1947 plates.

At fixed material and particle identity, a fixed initial kinetic energy suggests a narrow particle range distribution broadened by range straggling and measurement uncertainty. The synthetic example convolves a single chosen range with a normal distribution; a second example superposes two such chosen ranges. It explains why repeated comparable stopping distances can be informative, without pretending to extract a historical spectrum.

The decay equations do not identify the neutral partner as a neutrino on their own: a massless photon would satisfy the same energy and momentum equations. Nor does any isolated kink establish the pion's identity. The inference gains force from calibrated particle range, track grain density, repeated decay patterns and independent particle properties. Historical particle names should not obscure the distinction: a muon is a lepton, whereas a pion is a meson.

If $X_{\lambda }$ has a Poisson distribution, then $\mathbb{P}(X_{\lambda }\le \lfloor \lambda \rfloor )\to 1/2$ as $\lambda \to \infty$. The Poisson central limit theorem standardizes by mean $\lambda$ and variance $\lambda$; the cutoff $(\lfloor \lambda \rfloor -\lambda )/\sqrt{\lambda }$ tends to zero. Squeezing between fixed nearby cutoffs and using continuity of the standard normal distribution function gives the limit.

For a vector parameter with finite second moments and posterior mean $m_y$, the posterior covariance matrix records its uncertainty and linear dependence after conditioning on the observations. Its diagonal entries are posterior variances. In a regular locally flat-prior normal distribution approximation it is the inverse observed information. A quadratic form has posterior expectation equal to the form at the posterior mean plus the trace of its matrix times this covariance.

For a zero-mean stationary velocity process of variance ${\sigma }_v^2$ and normalized autocorrelation $(1+|s|{)}^{-\alpha }$, integrating velocity twice gives the displayed expression, exactly rather than just asymptotically. For $\alpha >1$ the long-time diffusion coefficient is ${\sigma }_v^2/(\alpha -1)$. At $\alpha =1$, the displacement variance is asymptotic to $2{\sigma }_v^{2}t\log t$. For $0<\alpha <1$, it is asymptotic to $2{\sigma }_v^2{t}^{2-\alpha }/[(1-\alpha )(2-\alpha )]$. Linear variance growth does not by itself imply a central limit theorem or a normal distribution for displacement.

For $J\ge 2$ conditionally independent effects ${\beta }_j\sim N(\mu ,{\tau }^2)$, each pair difference has normal distribution $N(0,2{\tau }^2)$. With $m=\left(\genfrac{}{}{0ex}{}{J}{2}\right)$, an upper bound $A=R/[\sqrt{2}{\Phi }^{-1}(1-\epsilon /(2m))]$ on $\tau$ ensures, by the union bound, that $\mathbb{P}({\max }_j{\beta }_j-{\min }_j{\beta }_j>R)\le \epsilon$. Any proper scale prior distribution supported on $(0,A)$ preserves this bound after averaging. This is conservative simultaneous calibration, rather than the weaker statement about one selected pair. For log odds ratios, a bound $R$ corresponds to a ratio-of-odds-ratios bound $e^R$.

The probable error is the half-width of an interval around a specified central value containing probability $1/2$. For a normal distribution $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, its quantile function giveswhere $\Phi$ is the standard normal cumulative distribution function and $\sigma$ is the standard deviation. The factor is specific to the normal distribution. The centered Cauchy distribution with parameter $\gamma >0$ has probable error $\gamma$, despite having no finite variance. Thus a central probability width need not require a finite second moment.

The quantile function is finite at every probability strictly between zero and one, but need not be finite at one. For any unbounded-above normal distribution, $F(x)<1$ for every real $x$, so its generalized inverse at one is $+\infty$. Inverse transform sampling still produces a real random variable almost surely because a continuous uniform distribution puts zero mass at that endpoint. A real-valued implementation may discard endpoint draws or assign an arbitrary finite value there without changing the sampled law.

The quantum variational principle estimates the lowest energy by choosing a trial wavefunction and computing its expectation value, without solving the Schrödinger equation exactly. Minimizing over a manageable family of trial wavefunctions gives an upper bound; enlarging that family can improve the estimate.

Let $H$ be a self-adjoint operator bounded below, and put $E_{\ast }=\inf \sigma (H)$. For $\Vert \psi \Vert =1$ in its operator domain $D(H)$,Trial states may also lie in the form domain $Q(H)=D((H-bI{)}^{1/2})$, where $b\le E_{\ast }$. Then use the quadratic form of a positive quantum Hamiltonian shifted by $b$:This extends the expectation value beyond $D(H)$ without requiring $H\psi$ to exist. Equality for a normalized state holds exactly in the energy eigenspace at $E_{\ast }$, if that eigenvalue exists. ^[ref].

A short proof makes the direction of the bound memorable. If $H$ has a complete orthonormal basis of energy eigenstates ${\varphi }_n$, writeThe expectation value is a weighted average of allowed energies:Equality requires every occupied energy eigenstate to have energy $E_{\ast }$. ^[ref]. With continuous spectrum, the spectral theorem replaces the sum by an integral against the scalar spectral measure; the same nonnegative-difference argument works.

For an unnormalized trial wavefunction, divide by its squared norm: $R_H[\psi ]=q_H[\psi ]/\Vert \psi {\Vert }^2$ is the form version of the Rayleigh quotient. The quantum variational principle supplies an upper bound on $E_{\ast }$; it supplies neither a lower bound on $E_{\ast }$ nor, by itself, a guarantee that a bound state exists.

To see the method at work, choose the real Gaussian wave packetThe Gaussian integral gives wavefunction normalization and the two useful position averages:Thus its position probability density is a normal distribution with standard deviation $1/\sqrt{2a}$: large $a$ means narrow confinement. In units $\hslash =m=1$, integration by parts and ${\psi }_a^{\prime }=-ax{\psi }_a$ give the kinetic energyThese smooth, rapidly decaying trial wavefunctions belong to the operator domains of both polynomial Quantum Hamiltonians below. Narrowing the Gaussian wave packet increases its kinetic energy; widening it increases the potential energy in either confining potential.

For the quantum harmonic oscillator, take $\hslash =m=\omega =1$:The trial values at $a=1/2,1,2$ are respectively $0.625,0.5,0.625$. The derivative $E_{\mathrm{ho}}^{\prime }(a)=(1-a^{-2})/4$ changes sign at $a=1$, where kinetic energy and potential energy are both $1/4$. Here the trial family contains an exact ground state:The zero-point energy remains positive even at the best width. For comparison, removing the potential leaves $E(a)=a/4\to 0$ as $a\to 0$: the free Quantum Hamiltonian on the line has spectral bottom zero but no normalizable zero-energy energy eigenstate. A sequence of ever broader trial wavefunctions need not approach an actual bound state.

For a pure quartic oscillator, keep $\hslash =m=1$ and set the quartic coefficient to one:The trial values at $a=1,2,3$ are respectively $1,0.6875,0.833333\dots$. Setting the derivative to zero givesThe sign of this derivative changes from negative to positive at the optimum, giving the smallest energy within this Gaussian wave packet family:At this width, $⟨T⟩\approx 0.454280$ and $⟨V⟩\approx 0.227140$, so $2⟨T⟩=4⟨V⟩$, as expected from the power-law quantum virial theorem for a scale-optimized trial wavefunction. The number $0.681420$ is an upper estimate of the ground-state energy, not an asserted exact energy eigenvalue. Indeed,depends on $x$ for every $a>0$, so no member of this trial family is an energy eigenstate. Nevertheless, a single width parameter already turns a difficult Schrödinger equation into a useful, explicit energy estimate.

For a unit vector $u$ and $\theta >0$, a rank-one covariance spike is the covariance matrix $I+\theta u{u}^{\top }$. Its eigenvalue is $1+\theta$ in the direction $u$ and one on the orthogonal complement. It arises by adding an independent scalar normal distribution signal $Yu$ to a standard Gaussian random vector.

For $0<q\le 1$, put probability mass $1-q$ at zero and $q/2$ at each of $-1$ and $1$. This symmetric probability distribution has mean zero. Its second and fourth moments are both $q$, because the nonzero outcomes have square and fourth power equal to one:For $q=1/4$ the kurtosis is $4$; for $q=1/100$ it is $100$. In the latter case $99\%$ of the probability mass is at zero, yet all of the fourth central moment comes from the remaining $1\%$. The standard score $Z=X/\sqrt{q}$ takes values $0$ and $\pm q^{-1/2}$, with variance one. Each nonzero outcome contributesto kurtosis. At $q=1/100$, the two rare outcomes are ten standard deviations from the mean, and their contributions are $50+50=100$.

Every member has finite support of a probability distribution, so all its moments are finite. Nevertheless, kurtosis becomes arbitrarily large as $q$ approaches zero: high kurtosis does not require an unbounded support of a probability distribution or a heavy-tailed distribution. At $q=0$ the variance vanishes and kurtosis is undefined. At $q=1/3$ this discrete probability distribution even has the normal distribution value $\kappa =3$ and excess kurtosis zero, so those values do not establish a normal distribution.

The joint probability density of $X,Y$ isIn polar coordinates, $x=r\cos \theta$, $y=r\sin \theta$, and the absolute Jacobian determinant is $r$. Thus the joint density of radius and angle factors asIntegrating over angle gives the Rayleigh distribution. The same factorization shows that the angle has uniform distribution and is independent of the radius. This is a short change of variables argument: an annulus of width $dr$ has area $2\pi rdr$ to first order.

The assumptions are substantive. Unequal component variances give elliptical rather than circular density contours; a nonzero mean displaces the center. Even zero covariance and equal normal distributions do not suffice without independence. For an original counterexample, let $X\sim N(0,{\sigma }^2)$ and let $S$ be an independent Rademacher random variable. Put $Y=SX$. Then $Y$ has the same normal distribution as $X$, and $\mathrm{Cov}(X,Y)=\mathbb{E}[S{X}^2]=0$. Nevertheless, all points lie on $y=x$ or $y=-x$, and $R=\sqrt{2}|X|$ has densitywhich is positive at zero, unlike the Rayleigh distribution. This pair is not a Gaussian random vector.

For a bounded continuous adapted process $H$ and a Brownian motion $B$, ${\int }_t^{t+h}{H}_{s}d{B}_s/(B_{t+h}-B_t)$ converges in probability to $H_t$ as $h\downarrow 0$. The Itô isometry makes the error numerator small relative to $\sqrt{h}$; the normal distribution of the denominator controls its chance of being too close to zero.

Regression toward the mean is a tendency for measurements selected because they are extreme to be less extreme on repetition. A large first reading can combine a persistent signal with unusually positive measurement error; a fresh error need not repeat that contribution. This can change the selected group's mean even when the underlying signal stays fixed and no intervention occurs.

The claim concerns a conditional expectation, not a rule that every individual moves toward the population mean. Some second readings are more extreme. Nor does an observed change establish a causal effect: selection alone can produce it. The following original normal distribution model makes both distinctions quantitative.

A reparameterization gradient differentiates an expected value by representing the simulated random variable as a differentiable transformation of parameter-independent noise. For $X_{\eta }=g_{\eta }(\epsilon )$, the identity ${\nabla }_{\eta }\mathbb{E}[F_{\eta }(X_{\eta })]=\mathbb{E}[{\nabla }_{\eta }{F}_{\eta }(g_{\eta }(\epsilon ))]$ requires justified differentiation under the integral sign. For a normal distribution, use $X=m+e^{\rho }\epsilon$ with $\epsilon$ drawn from the standard normal distribution.

Imagine an instrument measuring the same fixed quantity twice on each unit. Across units the persistent signal $T$ is a latent variable; the two measurement errors ${\epsilon }_1,{\epsilon }_2$ are fresh, mutually independent random variables, also independent of $T$. Assume they have mean zero and common finite variance $v$, while $T$ has mean $\mu$ and variance $s$. ThenThus the Pearson correlation coefficient is $\rho =s/(s+v)$ when $s+v>0$. Both measurements retain the same marginal mean and variance. Repetition does not make the entire population collapse toward $\mu$; it changes what we expect after selecting on $X$.

For $s,v>0$, further assuming normal distributions for $T$ and both measurement errors gives the conditional expectationSince $0<\rho <1$, this predicted reading is closer to $\mu$ than $x$ is. Independence and zero mean errors alone do not justify this linear conditional expectation: without the normal distribution assumption the same expression is the linear least-squares projection, and the actual conditional expectation can be nonlinear. Correlated errors, a changing signal, or different error variances require their own model; a noisy reading by itself supplies no universal shrinkage factor.