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Regression diagnostics check assumptions and influential observations in a fitted regression function. Residual-versus-fitted plots can reveal omitted mean structure or nonconstant variance; a quantile-quantile plot checks a specified error distribution. Regression leverage measures unusual predictor configurations, and Cook's distance combines leverage and residual size to measure coefficient sensitivity. Independence may require checking collection order and the study design in addition to residual plots.

A regression outlier has a response unusually far from its fitted conditional mean. A standardized regression residual adjusts for regression leverage; an externally studentized residual also estimates noise variance with the observation deleted. Selecting the largest residual among many observations changes its reference distribution. An outlier need not be an influential observation, and unusual valid data should not be discarded solely for being unusual.

The lowest even quantum harmonic oscillator quantum state has density $\rho (y)=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$, and the lowest odd quantum state has density $2y^2\rho (y)/{\tau }^2$. Since ${\rho }^{\prime }=-2y\rho /{\tau }^2$, integration by parts gives ${\mathbb{E}}_{\mathrm{F}}[f]={\mathbb{E}}_{\mathrm{B}}[f+y{f}^{\prime }]$ whenever the boundary terms vanish. For $f(y)=y^n$, this proves the displayed identity. Odd moments vanish in both quantum states by parity; even moments quantify the larger spatial spread in the odd quantum state. In particular the relative-coordinate variance is three times as large, and the odd density vanishes at particle coincidence.

A spatial image after a chosen dark interval is a different readout from continuously recording arrivals at a fixed probe. Here is an original idealised fluorescence imaging model: release a dilute cloud, let it move freely, then apply a short uniform probe pulse. Neglect collisions, residual forces, attenuation and motion during the pulse. Use each image's cloud centre to remove mean drift. Uniform gravity shifts that centre without changing the centred variance.

For each atom let $x(t)=x_0+v_{0}t$ along a horizontal imaging axis. Expanding the square about the expected values gives exactlyNo Gaussian distribution is needed for this variance identity. A jointly Gaussian distribution of initial positions and velocities does make the later density Gaussian. For a separable three-dimensional cloud viewed along $z$, integration along the viewing direction gives the column densityHere $N$ is atom number. Integrating this image along $y$ leaves a one-dimensional Gaussian distribution with standard deviation ${\sigma }_x(t)$. This is a projected density measurement, not a three-dimensional radius or a direct velocity histogram. Under the uniform, optically thin probe assumptions, expected fluorescence is proportional to this column density. A known, independent point spread function with finite variance adds its own variance to the measured width; probe duration and intensity gradients need separate treatment.

For a Maxwell-Boltzmann distribution, $T_x=m{\sigma }_v^2/k_B$, where $m$ is atomic mass and $k_B$ the Boltzmann constant. For a nonthermal distribution this defines a directional second-moment temperature, which need not describe every feature of the distribution. A covariance between position and velocity can initially narrow the cloud even though ${\sigma }_v^2$ is positive. The physical covariance matrix requires $C_{xv}^2\le {\sigma }_{x0}^2{\sigma }_v^2$.

Choose ${\sigma }_{x0}=0.60\mathrm{mm}$, ${\sigma }_v=0.060\mathrm{m}{\mathrm{s}}^{-1}$ and $C_{xv}=-0.018\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-1}$, giving correlation coefficient $-1/2$. With $t$ in milliseconds, the numerical variance in square millimetres is $0.36-0.036t+0.0036t^2$. Its minimum occurs at $5\mathrm{ms}$, with standard deviation $\sqrt{0.27}\mathrm{mm}$. These chosen values satisfy the covariance matrix constraint; early contraction is compatible with free motion.

Generate thirteen synthetic variance observations at equally spaced times from $0$ to $30\mathrm{ms}$, adding independent Gaussian noise of standard deviation $0.012\mathrm{m}{\mathrm{m}}^2$. Linear least squares fits all three coefficients in $a+bt+c{t}^2$. The example yields $a\simeq 0.37026\mathrm{m}{\mathrm{m}}^2$, $b\simeq -0.036413\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-1}$ and $c\simeq 0.0035820\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-2}$. With the chosen approximate sodium mass $m=3.82\times 10^{-26}\mathrm{kg}$, the true $T_x$ is $9.96\mu \mathrm{K}$ and the fitted value is $9.91\mu \mathrm{K}$. This is a fit to generated width data, not historical observations.

If the same data are fitted to $a+c{t}^2$, forcing $C_{xv}=0$, the inferred $T_x$ falls to $6.79\mu \mathrm{K}$. The omitted term is partly absorbed into both fitted coefficients. In noiseless linear least squares with a free intercept, its precise effect on the quadratic coefficient iswhere the subscript denotes equal-weight moments over the chosen observation times, using one consistent unit system. Since the ratio is positive for distinct nonnegative times, negative $C_{xv}$ biases this estimate downwards. An incorrect release-time origin can also generate a linear term, so fitting it does not by itself identify a physical initial covariance. Timing calibration, spatial resolution, residual magnetic fields and probe-induced recoil energy remain separate controls. Neither a cold-looking image nor a single long expansion time identifies all three coefficients.

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.

Replication in experimental design repeats a treatment on multiple independent experimental units, so that variation between those units can inform a treatment contrast and its standard error. Repeated measurements on one experimental unit can improve measurement precision without providing additional independent treatment replication. The experimental unit is identified by where the treatment is applied, rather than by how many responses are recorded.^[ref]

For a constructed example, grow one plant in each of eight separate pots. Apply treatment A to four pots and treatment B to four pots, without sharing drainage water between pots. Measure the length of three leaves per pot. The pot is the experimental unit; the leaf is the observational unit. There are four treatment replicates per group, eight experimental units overall, and $8\cdot 3=24$ measurements. The three leaves in one pot are nested measurements, because they share the same plant and treatment application. The setup assumes that pots do not affect one another; simply putting them in separate containers does not establish statistical independence.

Randomization chooses which experimental units receive which treatments; replication supplies multiple experimental units receiving the same treatment. These solve different problems. In the balanced allocation above, choose the four A pots uniformly among the $\left(\genfrac{}{}{0ex}{}{8}{4}\right)=70$ possible subsets. A particular pot has probability $1/2$ of receiving A. Given that one pot receives A, a second receives A with conditional probability $3/7$, rather than $1/2$. Thus the assignments are not independent random variables, even though the experimental units can supply independent responses conditional on those assignments in a suitable statistical model.

An actual ecology experiment illustrates why the distinction matters. In Stuart H. Hurlbert's account, twelve experimental ponds initially received zero, fifty or 450 fish, with five, four and three ponds respectively. Preparation included assessing variation between ponds before introducing fish; allocation considered both location and existing differences between ponds. The zero-fish ponds supplied the scientific control. These were repeated treatment applications to ponds, not twelve samples from one pond.^[ref]

The original article followed the southern California ponds for ten months. Its readouts included zooplankton populations and total phytoplankton; it reported decreases in some zooplankton populations and increased total phytoplankton.^[ref] Repeated sampling within a pond helps describe its response through time, while ponds supply treatment replication. Unequal group sizes and initial pond variation matter for the statistical model; the treatment counts alone do not establish equal variances or independent responses.

For a correctly specified full-rank fixed-design Gaussian linear regression with $p<n$, dividing the residual sum of squares by the residual degrees of freedom $n-p$ gives an unbiased estimator of the noise variance. The projection removes the whole deterministic mean, so the numerator's expectation is $(n-p){\sigma }^2$. A misspecified subset model need not have this property.

For independent errors of common unknown variance in $Y_i=\beta x_i+{\epsilon }_i$, the ordinary least squares estimate is $\hat{\beta }={\sum }_i{x}_i{Y}_i/{\sum }_i{x}_i^2$. Fitting one parameter leaves $n-1$ residual degrees of freedom. Therefore $s^2={\sum }_i(Y_i-\hat{\beta }{x}_i{)}^2/(n-1)$ estimates the common error variance, and $s^2/{\sum }_i{x}_i^2$ estimates the slope variance. For errors with a Gaussian distribution, the slope's standardized deviation follows a Student's t-distribution with $n-1$ degrees of freedom. A common calibration bias does not appear as independent residual scatter.

A residual plot displays regression residuals against fitted responses, predictor values or observation order. Curvature can suggest an incorrect regression function; changing spread can suggest unequal error variances; ordered runs can suggest dependence. A plot is a diagnostic, not a proof of independence. Dependence confined to higher moments, as for adjacent products of independent Gaussian variables, may call for lagged squared regression residuals rather than a basic scatter plot. Fitting also induces dependence among regression residuals even under independent original errors.

Original reanalysis of longitudinal trajectories; within-class variance and slope restrictions affect classification and estimated class sizes.

Connect one physical resistor $R$ across an ideal lossless capacitor $C$, with no applied source and an ideal readout that draws no electric current. Represent the resistor by its noisy Thévenin equivalent. The resulting RC low-pass frequency response gives the stationary capacitor voltage power spectral densityIts mean square follows by the effective noise bandwidth calculation:This is also the equipartition theorem: the mean capacitor energy is $\frac{1}{2}C⟨v_C^2⟩=\frac{1}{2}{k}_{\mathrm{B}}T$. Raising $R$ makes the low-frequency spectrum taller and its width narrower in exactly compensating proportions. It changes the equilibration time $RC$, not the final variance.

The integral to infinity belongs to the ideal classical model. For a physical electrical circuit, assume thermal equilibrium, linear components with negligible extra dissipation, and a corner far below the frequencies where parasitic effects, frequency-dependent electrical resistance, or $hf\sim k_{\mathrm{B}}T$ invalidate the model; the omitted high-frequency tail must be negligible. A finite-bandwidth instrument measures only part of this variance. Its input capacitance contributes to the total $C$, and its finite input electrical resistance introduces both loading and its own thermal electrical noise.

For $T=300\mathrm{K}$ and $C=1\mathrm{nF}$, the predicted equilibrium variance is $4.141947\times 10^{-12}{\mathrm{V}}^2$ and the root mean square voltage is $2.0352\mu \mathrm{V}$. At $R=10\mathrm{k}\Omega$, $RC=10\mu \mathrm{s}$ and $B_{\mathrm{n}}=25\mathrm{kHz}$. At $R=100\mathrm{k}\Omega$, $RC=100\mu \mathrm{s}$ and $B_{\mathrm{n}}=2.5\mathrm{kHz}$; the equilibrium root mean square voltage is identical.

To obtain the growth curve, prepare $v_C(0)=0$ exactly and connect the resistor at $t=0$ while maintaining its temperature. The RC low-pass filter's causal impulse response is $h(t)=e^{-t/(RC)}/(RC)$ for $t\ge 0$. Classical ideal white noise with two-sided power spectral density $2k_{\mathrm{B}}TR$ givesThe factor $2k_{\mathrm{B}}TR$ here is the two-sided time-domain noise intensity; the equilibrium spectral integral above used the one-sided $4k_{\mathrm{B}}TR$. They give the same long-time limit. The variance approaches equilibrium with time constant $RC/2$, although a deterministic initial voltage decays with time constant $RC$. If the capacitor already starts in thermal equilibrium, its initial variance supplies the missing exponentially decaying term and the total variance stays constant.

Applying a measurable per-claim retention $g$ to independent claim sizes preserves the compound Poisson distribution form. The transformed severity is the pushforward measure of the original severity law. With count parameter $\lambda$, the retained aggregate has expected value $\lambda \mathbb{E}g(X)$ and variance $\lambda \mathbb{E}[g(X{)}^2]$. Zero retained marks may be removed by Poisson thinning.

If $S$ has exponential distribution with expected value $\mu$, the tail integral formula for moments gives $\mathbb{E}\min (S,M)=\mu (1-e^{-m})$ and $\mathrm{Var}(\min (S,M))={\mu }^2(1-2m{e}^{-m}-e^{-2m})$, where $m=M/\mu$. At matching retained expected value, the excess variance under quota share reinsurance is $2{\mu }^2{e}^{-m}(m-1+e^{-m})\ge 0$.

Free bending has two zero-energy filament modes, so it does not define an unrestricted normalizable thermal height distribution. Fix the mean displacement and linear tilt, or project out that affine kernel. On the positive complement, the equipartition theorem gives independent Gaussian mode variances $k_{B}T/(A{k}_n^4)$ for normalized modes. The resulting covariance is the bending operator pseudoinverse multiplied by $k_{B}T$. It depends on how the rigid motion is fixed.

The Robertson uncertainty principle bounds the product of the standard deviations of two observables in the same quantum state. For a normalized pure state $|\psi ⟩$ and self-adjoint operators $A,B$, write their expectation values and variances asand similarly for $B$, where $I$ is the identity operator. The quantum uncertainties $\Delta A,\Delta B$ then obeyHere $[A,B]$ is the commutator, and every expectation value is taken in the same $|\psi ⟩$. These standard deviations describe the outcome spreads when $A$ and $B$ are measured on separate ensembles prepared in that same quantum state. They are not instrument errors or the change caused by a preceding measurement in quantum mechanics. The definition and bound are developed in Barton Zwiebach, Uncertainty Principle and Compatible Observables, 2013, §§1–2.

For unbounded self-adjoint operators, the operator domains matter. Assume $\psi \in D(A)\cap D(B)$, so that both second moments are finite; $⟨A^2⟩$ here means the second moment of the measured probability distribution, equal to $\Vert A\psi {\Vert }^2$, even if $A^2\psi$ is undefined. To use the displayed operator commutator, also assume $\psi \in D(AB)\cap D(BA)$. Without these product-domain assumptions, the always meaningful version on $D(A)\cap D(B)$ uses $|\mathrm{Im}⟨A\psi |B\psi ⟩|$ as its right-hand side. These qualifications prevent a formal commutator identity from being applied to an inadmissible quantum state.

The brief proof is Cauchy-Schwarz inequality applied to $u=(A-⟨A⟩I)\psi$ and $v=(B-⟨B⟩I)\psi$:The final equality uses self-adjointness and the product-domain assumptions. Thus a geometric statement about inner products becomes a constraint on experimental spreads. For nonzero $\Delta A,\Delta B$, equality requires $u,v$ to be proportional with a purely imaginary proportionality factor; the equality case of the Heisenberg uncertainty relation turns this condition into a Gaussian wave packet.

The bound is useful because computing a single commutator and its expectation value can constrain two entire outcome distributions. Its right-hand side depends on the quantum state: the spin uncertainty state dependence example gives both a positive saturated bound and a zero bound for the same noncommuting observables. Robertson's paper, titled "The Uncertainty Principle", appeared on 1 July 1929 in Physical Review, volume 34, starting at page 163. ^[ref].

Rotational diffusion is the random change of a particle's rigid body orientation through Brownian motion. A suspended ellipsoid can turn while its centre hardly moves; translational diffusion instead spreads its position. Random molecular collisions supply fluctuating torques as well as forces. The rotational diffusion coefficient $D_r$ measures how quickly angular uncertainty grows, rather than a steady spin speed. Its units are ${\mathrm{s}}^{-1}$, often written $\mathrm{ra}{\mathrm{d}}^2{\mathrm{s}}^{-1}$ to distinguish angular spreading from translational spreading, whose diffusion coefficient $D_t$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. The radian is dimensionless, but writing angular variance in $\mathrm{ra}{\mathrm{d}}^2$ remains useful.

First identify what the image can reveal. A marked end of an elongated particle defines an arrow: turning it through $\pi$ reverses that arrow. An unmarked ellipsoid with identical ends reveals only its long axis, a director: angles $\varphi$ and $\varphi +\pi$ describe the same visible state. An unmarked perfect sphere has no visible rigid body orientation at all. Tracking its centre cannot measure its rotational diffusion; a surface mark or another identifiable internal direction is needed.

This distinction makes rotational diffusion useful: it measures how long a particle remembers its pointing direction. For an elongated particle, turning also changes which laboratory direction is easiest for translational diffusion. Randomization of rigid body orientation can therefore erase directional preferences in the observed displacement, even though the particle's shape has not changed.

For $M>1$ completed-data estimates $Q_m$ with variances $U_m$, pool using $\overline{Q}=M^{-1}{\sum }_m{Q}_m$, $\overline{U}=M^{-1}{\sum }_m{U}_m$, $B=(M-1{)}^{-1}{\sum }_m(Q_m-\overline{Q}{)}^2$, and $T=\overline{U}+(1+M^{-1})B$. The total variance includes within-imputation uncertainty, between-imputation uncertainty and finite-$M$ simulation uncertainty.

For $n\ge 1$ independent and identically distributed Bernoulli random variables with success probability $p$, the sum $K={\sum }_i{X}_i$ has a binomial distribution. Every length-$n$ pattern with $k$ successes has probability $p^k(1-p{)}^{n-k}$, and the binomial coefficient counts its $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ possible success positions. ThereforeThe usual endpoint convention includes $p=0$ and $p=1$ as deterministic cases. For $p=1/2$ and $n=4$, the five probability masses are $1,4,6,4,1$ divided by $16$. The sample mean takes values $0,1/4,1/2,3/4,1$, with variance $1/16$ and standard deviation $1/4$. At $n=64$, its variance is $1/256$ and its standard deviation is $1/16$.

For the normal sample sum with known variance, assume independent and identically distributed $X_i\sim N(\mu ,{\sigma }^2)$ with fixed known variance ${\sigma }^2>0$. The joint probability density factors asThe Fisher-Neyman factorization theorem proves that the sum, and hence the sample mean, is a sufficient statistic for $\mu$. Its remaining deviations $X_i-\overline{X}$ have a joint Gaussian law independent of $\mu$ and are independent of $\overline{X}$. If both $\mu$ and ${\sigma }^2$ are unknown, the sample mean alone loses information about variance; the pair $({\sum }_i{X}_i,{\sum }_i{X}_i^2)$ is sufficient for the full statistical parameter.

A discrete counterexample shows why a sample mean is not automatically sufficient for arbitrary statistical models. Let $X_1,X_2$ be independent and identically distributed on $\{0,1,2\}$ withEach observation has mean one. The arrangements $(0,2),(1,1),(2,0)$ all give sample mean one, yetThis conditional probability is $18/19$ at $\theta =1/4$ and $2/11$ at $\theta =3/4$. The conditional distribution therefore still depends on the unknown statistical parameter: this sample mean is not a sufficient statistic. The useful reduction depends on the family of full-data laws, even when its mean stays fixed.

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.