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For standard Brownian motion started at zero in ${\mathbb{R}}^n$, every fixed orthogonal transformation preserves the process law. Its transformed increments remain independent centred Gaussian random vectors with covariance $(t-s)I$, and its paths remain continuous. Consequently its exit law from any centred sphere is normalized surface measure. Translations give the same assertion for spheres centred at its starting point.

Construct a synthetic local calibration to show the calculation independently of the historical data. Choose reference pairs $(\Delta {\lambda }_i,p_i)=(1.00,2.69),(2.00,5.42),(3.00,8.09)$, with shifts in $\mathrm{nm}$ and pressures in $\mathrm{GPa}$. The reference zero is already subtracted. Treat the shifts as exact for this fit and give the three reference pressures equal weights. Minimizing the ordinary least squares objectiveand setting its derivative to zero givesThe regression residuals are $(-0.01,0.02,-0.01)\mathrm{GPa}$. Their weighted moment ${\sum }_i\Delta {\lambda }_i(p_i-a\Delta {\lambda }_i)$ vanishes, as required by the least-squares normal equations. An unknown reference offset requires fitting an intercept instead; forcing it to zero can bias the slope. If shift errors matter, this elementary fit is insufficient.

Use the local measurement modelHere $p$ denotes pressure relative to the reference, $a$ is calibrated sensitivity, and $k_T$ is a local wavelength sensitivity to temperature at fixed pressure. Choose $a=2.70\mathrm{GPa}\mathrm{n}{\mathrm{m}}^{-1}$, $k_T=0.010\mathrm{nm}{\mathrm{K}}^{-1}$ and a reading $\lambda -{\lambda }_0=2.00\mathrm{nm}$ at $T=T_0$. The inferred pressure is $5.40\mathrm{GPa}$. These coefficients and readings are stipulated, not measured ruby constants. With the same pressure, a $10\mathrm{K}$ increase in temperature would add $0.100\mathrm{nm}$ to the shift in this model; neglecting it would overestimate pressure by $0.270\mathrm{GPa}$.

Assign independent standard uncertainties $u_a=0.020\mathrm{GPa}\mathrm{n}{\mathrm{m}}^{-1}$, $u_{\lambda }=u_0=0.010\mathrm{nm}$ and $u_T=0.50\mathrm{K}$ for the temperature difference. Here $u_a$ is an independently assigned calibration uncertainty, not an estimate from the regression residuals above; treat $k_T$ as exact in this example. First-order propagation of uncertainty follows by differentiating the measurement model:The four standard contributions are $0.0400,0.0270,0.0270,0.0135\mathrm{GPa}$, giving $u_p=0.0569\mathrm{GPa}$. This is a standard uncertainty, not automatically a $95\%$ interval. Correlated reference and loaded wavelength readings require the corresponding covariance term; common wavelength-axis drift may cancel in their difference. Uncertain $k_T$, nonzero $T-T_0$, model inadequacy and unresolved deviatoric stress require a broader budget. Taking more optical readings reduces some random errors while leaving a common reference-scale error.

For a known zero mean, ${\hat{\gamma }}_n(h)=n^{-1}{\sum }_{t=1}^{n-h}{Y}_t{Y}_{t+h}$ for $0\le h<n$. Weakly stationary process assumptions give expectation $(1-h/n)\gamma (h)$. Dividing by $n-h$ instead gives an unbiased estimate at each individual lag, although the resulting collection need not form a positive semidefinite covariance sequence.

Fix $N>1$ real population values $y_1,\dots ,y_N$. Define the finite population mean and two normalizations of the finite population variance byThe values are fixed; only the selected subset $A$ is random. For its sample mean ${\bar{y}}_A=n^{-1}{\sum }_{i\in A}{y}_i$, the exact design expectation and variance areThe sample mean is an unbiased estimator of the finite population mean. At $n=N$, every unit is observed and ${\bar{y}}_A=\bar{Y}$ exactly, so the variance is zero. The $N=1$ case is deterministic and is handled directly, since $S^2$ is undefined then.^[ref]

The finite population correction depends on the baseline. With the $N-1$ normalization it is $1-n/N$ multiplying $S^2/n$. With the $N$ normalization it is $(N-n)/(N-1)$ multiplying ${\sigma }_N^2/n$, the exact variance for independent uniform sampling with replacement. The standard error uses the square root of the relevant variance factor. Mixing the two normalizations gives a wrong answer even at $n=1$, when replacement cannot matter. For the six-value example, $\bar{Y}=7$, ${\sum }_i(y_i-7{)}^2=70$, $S^2=14$ and ${\sigma }_N^2=35/3$, reproducing $14/3$ without replacement and $35/6$ with replacement.

A short proof exposes the statistical dependence. Let $X_r$ be the value at position $r$ in uniform sequential sampling without replacement, so each $X_r$ has expected value $\bar{Y}$ and variance ${\sigma }_N^2$. Write $d_i=y_i-\bar{Y}$. Every ordered pair of distinct labels is equally likely, and ${\sum }_i{d}_i=0$, hence for $r\ne s$,Linearity of expectation proves unbiasedness without requiring independence. The variance of a sum includes all $n(n-1)$ ordered covariance terms, givingFor the original population, each pair of distinct draw positions has covariance $-7/3$. Drawing a high value leaves a lower mean among the remaining values, balancing subsequent draws. Equal marginal probability distributions therefore coexist with negative covariance.

When $2\le n\le N$, the sample variance $s_A^2=(n-1{)}^{-1}{\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2$ is an unbiased estimator of $S^2$. To check this independently, use ${\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2={\sum }_{i\in A}(y_i-\bar{Y}{)}^2-n({\bar{y}}_A-\bar{Y}{)}^2$: its expected value is $n{\sigma }_N^2-n\mathrm{Var}({\bar{y}}_A)=(n-1)S^2$. Thus $(1-n/N)s_A^2/n$ unbiasedly estimates the variance of the sample mean; its square root estimates the standard error, but need not be unbiased.^[ref]

The diagram compares variance with the expected fraction of distinct population labels observed. Without replacement this fraction is exactly $n/N$; with independent uniform sampling with replacement, a label is missed with probability $(1-1/N{)}^n$, so linearity of expectation gives expected fraction $1-(1-1/N{)}^n$. At six draws, replacement covers only about $66.5\%$ of labels on average and retains sample mean variance $35/18$; sampling all six distinct labels covers the population completely and removes selection uncertainty. This population coverage is distinct from confidence interval coverage probability.

The correction matters when the sampling fraction is appreciable. It calibrates uncertainty about this fixed population under the specified selection mechanism. It does not correct unequal selection weights, missing observations or measurement errors; observing every label also does not eliminate uncertainty about a larger unobserved population. Moreover, an exact variance alone does not guarantee the coverage probability of a proposed confidence interval: a small discrete sampling distribution can make a normal approximation inaccurate.

Sampling without replacement selects distinct labelled units: after a unit is selected it is unavailable for the remaining draws. Distinct labels can carry equal numerical values, so repeated values do not necessarily imply a repeated unit. The rule alone does not specify selection probabilities; a procedure favoring particular units is still sampling without replacement but need not be simple random sampling.

In uniform sequential sampling without replacement, the next label is chosen uniformly among those remaining. For a population of $N$ labels, any ordered sequence of $n$ distinct labels has probability $1/[N(N-1)\cdots (N-n+1)]$. Each size-$n$ subset has $n!$ possible orders, so its probability is ${\left(\genfrac{}{}{0ex}{}{N}{n}\right)}^{-1}$. This gives the fixed-size simple random sampling design. Equivalently, take the first $n$ labels of a uniformly random permutation. Restricting a uniformly random permutation to a specified label set preserves uniformity; biased modulo mappings should not replace uniform selection.

The draw positions have identical marginal probability distributions, but exhibit statistical dependence. If the first label is $i$, the conditional probability that the second label is $i$ is zero, whereas its marginal probability is $1/N$. Thus these draws are not independent and identically distributed when $N>1$ and $n>1$. For a marked subset of labels, counting size-$n$ subsets with binomial coefficients gives the hypergeometric distribution. The resulting negative covariance explains the finite population correction for both counts and sample means.

For an estimating equation with population root $\theta =T(F)$, let $A={\mathbb{E}}_F{\partial }_{\theta }\psi (X,\theta )$ be nonsingular and $B={\mathbb{E}}_F[\psi {\psi }^T]$. Differentiating the contaminated population equation gives influence function $-A^{-1}\psi (x,\theta )$. A consistent sample root with differentiable local expansion satisfies $\sqrt{n}(\hat{\theta }-\theta )=-A^{-1}{n}^{-1/2}{\sum }_i\psi (X_i,\theta )+o_p(1)$, so the central limit theorem gives the displayed covariance. In the scalar case this is $\mathbb{E}{\psi }^2/A^2$.

For the state-space model $S_t=G{S}_{t-1}+w_t$, $X_t=F{S}_t+v_t$, assume independent Gaussian process noise and measurement noise of variances $W,V$, independent of the prior state and previous observations. A normal distribution for the filtered state, with mean $m$ and variance $P$, predicts mean $Gm$ and variance $R=G^{2}P+W$. Here $R$ denotes predicted state variance, rather than the observation-noise covariance notation used in the vector equations above. Gaussian conditional expectation then gives posterior mean $Gm+K(X-FGm)$ and posterior variance $RV/(F^{2}R+V)$, where the Kalman gain is $K=FR/(F^{2}R+V)$.

Mere weak white noise assumptions do not imply normal posterior distributions: independent process noise with a Rademacher distribution makes the next state a mixture of two shifted normal distributions. The Kalman filter can still describe a best affine state estimate under appropriate second-order assumptions, but that is distinct from an exact Gaussian conditional law.

The Schrödinger equation governs the evolution of a quantum state between instances of measurement in quantum mechanics. For a particle of mass $m>0$ moving in a prescribed real potential energy $V(\mathbf{r},t)$, its position wavefunction satisfiesHere $h$ is the Planck constant, ${\nabla }^2$ is the Laplace operator, and the Hamiltonian operator combines kinetic energy with potential energy. Specify an initial wavefunction, an operator domain and any boundary conditions: the differential expression alone does not completely specify a physical model. The Born rule then turns the evolving probability amplitude into position probabilities through $\rho =|\psi {|}^2$ and integration over a region.

This scalar Schrödinger operator describes nonrelativistic quantum mechanics, with relevant momenta much smaller than $mc$, where $c$ is the speed of light. It omits the constant rest energy $m{c}^2$, assumes fixed particle number, and suppresses active spin dynamics and coupling to a magnetic vector potential. A free electron with a fixed, uncoupled spin state can still use this scalar model; the Pauli equation supplies the additional spin components and magnetic coupling when needed. A prescribed real $V$ describes a closed evolution model or an approximation in which the source of that potential energy is treated as external. Dissipation, particle creation and a general measurement in quantum mechanics need further modelling.

For a time-independent self-adjoint Hamiltonian operator, the spectral theorem defines unitary time evolution byInitial states in the operator domain satisfy the differential equation in the Hilbert space sense; the unitary operator also evolves all other states in that Hilbert space. Time-dependent Hamiltonian operators need suitable regularity and domain hypotheses; their evolution is not generally the exponential of $-itH(t)/\hslash$.

The local reason for conservation is especially tangible. Multiply the Schrödinger equation by the complex conjugate ${\psi }^{\ast }$, subtract its conjugate multiplied by $\psi$, and use the fact that the real potential energy terms cancel. For smooth wavefunctions this gives the probability continuity equationThe divergence theorem identifies the change in probability inside a region with minus the outward probability current. Decay at infinity, or appropriate probability-conserving boundary conditions, preserves wavefunction normalization. Conservation of total probability does not require a stationary probability density.

Its usefulness comes from being a linear partial differential equation: quantum superpositions of solutions are solutions, while different quantum phases can produce interference. For a fixed Hamiltonian operator, the time-independent Schrödinger equation finds the energy eigenstates; each then acquires just its own time-dependent quantum phase. One eigenvalue problem therefore organizes both allowed energies and the evolution of arbitrary quantum superpositions, with a continuous spectrum included when required. This is the attractive bridge between a spatial wave problem and a clock: energy differences determine the rate at which relative quantum phases change. It makes atomic spectral lines, quantum tunnelling and quantum scattering calculable from a specified model.

For an explicit example without confinement, take a free particle on the whole line, $V=0$, initially in a Gaussian wave packet with position standard deviation ${\sigma }_0$ and mean momentum $\hslash k_0$:Unlike a single plane wave, this wavefunction satisfies wavefunction normalization. A Fourier transform writes it as a quantum superposition of plane waves. Their dispersion relation is $\omega (k)=\hslash k^2/(2m)$, so each Fourier component gains the factor $e^{-i\hslash k^{2}t/(2m)}$. Completing the square in the resulting Gaussian integral giveswith the square root chosen continuously from $t=0$. Its position probability density and standard deviation areThe centre moves with the group velocity ${\omega }^{\prime }(k_0)=v_0$ and the wave packet spreads because different wavenumbers have different group velocities. This is unitary wave dispersion, without a random force or loss of probability. Independently, the initial momentum standard deviation is $\Delta p=\hslash /(2{\sigma }_0)$ and the initial position-momentum covariance vanishes. Free evolution gives $x(t)=x(0)+p(0)t/m$ in the Heisenberg picture, henceThis independently checks the spreading law. The Gaussian integral of $\rho$ is one at every time; the decreasing peak is exactly offset by increasing width.

As hypothetical preparation parameters, choose a free electron with ${\sigma }_0=1\mathrm{nm}$ and $k_0=2{\mathrm{nm}}^{-1}$. Using $\hslash \simeq 1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$ and $m\simeq 9.1093837\times 10^{-31}\mathrm{kg}$ givesThe rounded electron mass uses the 2022 adjustment.^[ref] At $t=t_s$, the centre is $4\mathrm{nm}$ and ${\sigma }_t=\sqrt{2}\mathrm{nm}$; at $t=2t_s$, they are $8\mathrm{nm}$ and $\sqrt{5}\mathrm{nm}$. The mean kinetic energy staysIt includes both mean motion and the momentum spread. The mean speed is about $7.72\times 10^{-4}c$ and the momentum spread corresponds to $1.93\times 10^{-4}c$, making the nonrelativistic model accurate for the appreciably occupied part of this ideal Gaussian distribution. Its mathematically unbounded tails are an idealization.

The complex width also explains where the probability goes. Write $\psi =\sqrt{\rho }{e}^{i\theta }$, where $\theta$ is the dimensionless quantum phase. Differentiating the exact solution givesRelative to the moving centre, the probability current $j_{\mathrm{rel}}=j-v_0\rho$ points outward on both sides for $t>0$. This is a flux statement, not an additional assumption about individual particle trajectories. At $t=t_s$ and $x-v_{0}t=\pm {\sigma }_0$, the relative flow speeds $j_{\mathrm{rel}}/\rho$ are $\pm {\sigma }_0/(2t_s)$, about $\pm 2.8942\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$. The probability within a moving interval of half-width ${\sigma }_0$ falls from about $0.6827$ initially to $0.5205$ at $t_s$, exactly because of this outward probability current.

Historically, the 1926 Quantisierung als Eigenwertproblem (Erste Mitteilung) by Erwin Schrödinger began with the stationary hydrogen atom problem. It distinguishes discrete negative energies from a positive continuous spectrum; it is not yet the general evolution equation displayed above.^[ref] The opening of quantum mechanics discusses its original vibrating-string analogy.

In the 12 December 1933 lecture The fundamental idea of wave mechanics by Erwin Schrödinger, Erwin Schrödinger explains the move from independent rays to waves through diffraction, then applies that analogy to atomic physics. His closing pages explicitly acknowledge the difficulty of reconciling wave language with observed particle tracks. This retrospective physical argument motivates the model; it does not derive the modern Born rule from classical mechanics.^[ref]

The dates separate proposal from measurement: the de Broglie wavelength belongs to the 1924 thesis, the Schrödinger equation to 1926, and the Davisson–Germer experiment's electron diffraction beams to 1927.^[ref] That linked experiment describes the prepared nickel crystal, evacuated electron gun, angular collector scans, retarding-potential selection and heating controls. Its measured collector electric current tests wave-dependent scattering probabilities; it neither measures a wavefunction directly nor establishes this equation uniquely.

The second moment method uses a random variable's expected value and second moment to establish that a counted or measured quantity is positive with substantial probability. It is useful when a positive average alone could be caused by extremely large values on very rare outcomes. Controlling the variance rules out too much of that concentration in rare outcomes.

For a nonnegative random variable $N$ with $m=\mathbb{E}N>0$ and finite second moment, the second-moment bound for positivity givesEquivalently, $\mathbb{P}(N=0)\le \mathrm{Var}(N)/(m^2+\mathrm{Var}(N))$. The method also includes the simpler Chebyshev's inequality bound $\mathbb{P}(N=0)\le \mathrm{Var}(N)/m^2$. Thus a sequence of such random variables with $\mathrm{Var}(N)/(\mathbb{E}N{)}^2\to 0$ has $N>0$ with probability tending to one. These statements require no independence assumption. The variance argument and its use for random counts are developed in Zhao's lecture notes. ^[ref].

For an original toy count, let $N$ count successes in ten independent Bernoulli trials with success probability $p=1/10$. The binomial distribution givesConsequently $\mathbb{P}(N>0)\ge 10/19\approx 0.526316$. The exact probability is $1-(9/10{)}^{10}\approx 0.651322$. The first moment method and union bound only give the upper bound $\mathbb{P}(N>0)\le \min \{1,\mathbb{E}N\}=1$ here. First and second moments therefore answer complementary existence questions.

For a sum of indicator random variables, the variance of a sum isIndependence makes the covariances vanish, but the formula itself also covers dependent counts. For the same ten success probabilities, let every trial share a single Bernoulli random variable $B$: then $N=10B$, with mean one, variance nine, and $\mathbb{P}(N>0)=1/10$. The second-moment lower bound is exactly $1/10$. Ignoring the positive covariances would produce the false independent-count conclusion. Zhao's discussion of sums of indicators explains why overlapping counted structures require these pairwise terms. ^[ref].

Draw one random variable $R$, taking values zero and two with equal probabilities. Conditional on $R$, generate a Poisson point process on the real line with constant rate $R$. This is a Cox process directed by the random measure $Rdx$: the same rate is shared by every window. It is a locally finite simple point process and a stationary point process. The law of total expectation gives $\mu (A)=\mathbb{E}[R]|A|=|A|$; the law of total variance and law of total covariance give, for bounded Borel sets $A,B$ with $A\cap B=\varnothing$,The conditional covariance vanishes by independence, but the shared conditional means fluctuate together. This distinguishes conditional from unconditional independence.^[ref]

For a window of length $L$, the exact probability mass function isIn particular, its void probability functional is $(1+e^{-2L})/2$, compared with $e^{-L}$ for the unit-rate Poisson point process. Whole realizations can be empty or comparatively crowded, while their mean density remains one. This example demonstrates shared-rate statistical dependence, rather than a model of short-range offspring clusters.

For one selected polarization of waves, write the detected complex fields as $E_1(t),E_2(t+\tau )$ and their light intensities as $I_j=|E_j{|}^2$, absorbing a common conversion constant. With stationary ensemble averages and nonzero ${\bar{I}}_j=⟨I_j⟩$, defineHere second-order optical coherence concerns distinct detector channels, with the corresponding normal ordering in the quantum formulation. Independent detection has $g^{(2)}=1$. Subtracting the means gives the normalized covarianceThus multiplying raw photocurrents contains a large mean-product contribution; multiplying their fluctuations isolates the small excess. Independent detector shot noise does not supply a mean cross-channel excess, although it increases measurement uncertainty.

The displayed Siegert relation requires a zero-mean, jointly circular complex Gaussian random field, the usual chaotic thermal light model. Circular means invariance under a common arbitrary complex phase rotation; in particular $⟨E_i{E}_j⟩=0$. Apply the Gaussian moment pairing theorem to the real and imaginary parts:The remaining pairing vanishes by circularity. Division proves the Siegert relation. Gaussianity is an assumption about fourth moments, not a consequence of merely measuring first-order optical coherence; a coherent state or photon antibunching cannot be interpreted by this thermal formula.

For equal-intensity beams combined in a common optical mode with matched wave polarizations, interferometric visibility is $V=|g_{12}^{(1)}|$. Consequently ideal thermal stellar intensity interferometry measures $g^{(2)}-1=V^2$, while the full $g^{(2)}$ includes the independent-count reference one. Unpolarized thermal light with two independent equally populated wave polarizations has half this excess: adding their covariances doubles the numerator while multiplying total means quadruples the denominator. Finite detector response further suppresses the peak and must be calibrated.^[ref]

The sill is the large-distance level of a semivariogram when that limit exists. For covariance tending to zero, the sill is the total marginal variance; a nugget plus a structured component has total sill ${\tau }^2+{\sigma }^2$. A nonzero constant covariance component is invisible to the semivariogram.

Simple kriging assumes a known mean function. If $m=(m(s_1),\dots ,m(s_n){)}^T$ and $m_0=m(s_0)$ are known, center the observations, predict the residual, and add back the target mean:For known zero mean, these reduce to $c^T{\Sigma }^{-1}z$ and $v_0-c^T{\Sigma }^{-1}c$. Completing the covariance quadratic form giveswhich proves minimum error variance among predictors of the form $m_0+w^T(Z-m)$. No sum-to-one constraint is needed: the known mean already supplies the offset. In the exponential example from ordinary kriging, imposing known zero mean gives weights $(0,2/5,2/5{)}^T$, prediction $2$, and error variance $3/5$. The smaller error variance uses additional knowledge of the mean; it is not a free improvement when the mean is unknown.

If $(Z_0,Z)$ has a joint multivariate normal distribution, the displayed prediction is exactly $\mathbb{E}[Z_0\mid Z=z]$, and $s_{\mathrm{SK}}^2$ is its conditional variance. A conditional expectation minimizes mean squared error among all square-integrable functions of the observations. Without a joint multivariate normal distribution, simple kriging remains the best predictor in the specified affine class, but the conditional expectation may be nonlinear and its conditional variance may depend on $z$. Thus Gaussian process regression and the second-moment kriging construction agree under these assumptions, while having different starting points.^[ref]

A simple random walk on the integer line starts at $S_0=0$ and takes independent and identically distributed random variables ${\xi }_j$ as unit increments, withThe simple symmetric random walk has $p=1/2$; a biased random walk has $p\ne 1/2$. At $p=0$ or $p=1$ every step is deterministic, giving $S_n=-n$ or $S_n=n$ respectively.

Count the rightward steps using indicator functions:Thus the endpoint is a shifted and rescaled random variable with a binomial distribution. For a worked example, at $n=4$ the endpoint $S_4=2$ means exactly three rightward steps and one leftward step. There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ possible orders, so its probability is $4p^3(1-p)$, equal to $1/4$ in the fair case. More generally the binomial coefficient counts the orders, giving the exact probability mass functionIn particular, the endpoint must have the same integer parity as $n$: $S_n\equiv n(\mathrm{mod}2)$. For $0<p<1$ the support of a probability distribution is exactly $\{-n,-n+2,\dots ,n\}$. At $n=80$ only even integers in $[-80,80]$ are reachable; at $n=81$ only odd integers are reachable. This restriction is easy to lose if one draws a smooth curve instead of discrete probability masses. The fair-walk counting argument is also presented in Grinstead and Snell, Introduction to Probability, section 12.1.

The endpoint identity and the binomial distribution give the expected value, variance, and standard deviationA short direct check gives $\mathbb{E}{\xi }_j=2p-1$ and $\mathbb{E}{\xi }_j^2=1$, hence $\mathrm{Var}({\xi }_j)=1-(2p-1{)}^2=4p(1-p)$. Then apply linearity of expectation and the variance of a sum of independent random variables.

The absolute width and the width relative to the number of steps answer different questions. In the fair case the standard deviation grows from $\sqrt{20}\approx 4.472$ to $\sqrt{80}\approx 8.944$, while ${\sigma }_n/n$ shrinks from $0.2236$ to $0.1118$. At $n=80,p=0.65$, the standard deviation is $\sqrt{72.8}\approx 8.532$, or $0.9539$ times the fair width, and ${\sigma }_n/n\approx 0.1067$. Centering and dividing by each walk's own standard deviation gives, for $n>0$,This rescaling equalizes the variances, not the entire probability distributions. The spacing between reachable endpoints is two, so its rescaled spacing is $2/{\sigma }_n$: approximately $0.4472$ for the fair $20$-step walk, $0.2236$ for the fair $80$-step walk, and $0.2344$ for the biased $80$-step walk.

The positions themselves are dependent. For $1\le m\le n$, $S_n=S_m+(S_n-S_m)$, where the second term is independent of $S_m$. Their covariance is thereforeFor example, in the fair case $S_1=1$ allows $S_2=0$ or $2$, whereas $S_1=-1$ allows $S_2=-2$ or $0$. Nevertheless the position process is a Markov chain: given its current position, the next fresh increment has no dependence on its past. Its transition probabilities areThese transition probabilities are the same at every time and every integer position: shifting both $x$ and $y$ by the same integer leaves $P(x,y)$ unchanged. Thus the Markov property and independent increments coexist with dependent positions.

For independent $p$-variate multivariate normal observations with unknown mean and unrestricted positive-definite covariance matrix, the centered scatter has rank at most $n-1$. If it is singular, choose the mean to be the sample mean and shrink the covariance on a null direction of the scatter. The quadratic likelihood term stays fixed while the determinant tends to zero, so the density likelihood tends to infinity. Thus a nonsingular covariance maximum likelihood estimate exists only when the centered scatter is positive definite; for a nonsingular Gaussian population this occurs almost surely exactly when $n>p$.

For a Gaussian free field with boundary heights $-\lambda$ and $+\lambda$ on the two sides of the starting point, its distinguished zero-height interface has chordal ${\mathrm{SLE}}_4$ law. Conditional on an initial curve segment, the field is a zero-boundary field in the slit domain plus $m_t(z)=\lambda -(2\lambda /\pi )\arg (g_t(z)-W_t)$. Its mean bracket equals minus the variation of its conditional Green covariance. Subtracting the initial harmonic function gives the corresponding coupling to a zero-boundary field.

Filtering multiplies the time-series spectral density by the squared modulus of the frequency response. Insert the spectral integral into the covariance double sum and interchange sums using absolute summability of the filter. This holds even when the input covariances are not absolutely summable, provided the spectral measure has a density.

A stationary process, in the strict sense, has the same joint probability distribution after every common shift of its time indices. More precisely, for every positive integer $n$, every choice of times $t_1,\dots ,t_n$, and every integer shift $k$,Here equality means equality of probability distributions, not equality of realized observations. A sample path can fluctuate, stay positive for a while, or show accidental runs while its generating law remains stationary. Strong white noise is a basic example: shifting time preserves every joint law of its independent and identically distributed random variables.

The distinction between strict stationarity and weak stationarity is useful. The former preserves all finite joint laws; the latter preserves only the finite mean and covariance structure. Strict stationarity with finite second moments implies weak stationarity, but strict stationarity itself does not require finite variance. A Gaussian process with weak stationarity is also strictly stationary, because each of its finite-dimensional distributions, a joint normal distribution, is determined by its mean vector and covariance matrix.

Two random variables are statistically dependent when their joint probability distribution differs from the product of their marginal probability distributions. Equivalently, some events determined by the two variables violate the product rule that defines independence. Nonzero covariance implies statistical dependence when the moments exist, but zero covariance does not in general imply independence.

This tensor integrates the local pressure tensor, where ${\sigma }_{ij}=⟨(v_i-u_i)(v_j-u_j)⟩$ is the velocity covariance. Half its trace is the random kinetic energy. It enters the tensor virial theorem with coefficient one.