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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]

Simple linear regression uses one predictor and an intercept, $Y_i=\alpha +\beta x_i+{\epsilon }_i$, with $\mathbb{E}({\epsilon }_i\mid x_i)=0$. The regression coefficients describe a straight conditional-mean line: $\alpha$ is its value at zero and $\beta$ its rate of change per unit of the predictor. When zero is outside the observed range, predictor centering can make the intercept refer to a more representative predictor value.

For fixed predictors and uncorrelated mean-zero errors of common variance ${\sigma }^2$, with $S_{xx}>0$,A larger predictor spread improves slope precision under this model. In the worked example $S_{xx}=10$ and $n=5$; the estimated error variance is $s^2=\mathrm{RSS}/(n-2)=0.8/3$. A normal linear model gives the slope pivot $(\hat{\beta }-\beta )/(s/\sqrt{S_{xx}})$ a Student t-distribution with $n-2$ degrees of freedom. The distinction between a fitted line and a reliable uncertainty statement is precisely the role of the error assumptions.

Here simple random sampling means a random unordered subset $A$ of exactly $n$ distinct labels from $\{1,\dots ,N\}$, with every size-$n$ subset equally likely. The sample size is fixed in advance, $1\le n\le N$, and sampling without replacement prevents duplicate units. This is SRSWOR; some texts also use "simple random sampling" for independent uniform draws with replacement, which is a different design.^[ref]

Every label has inclusion probability $n/N$, but equal inclusion probabilities alone are insufficient. For example, from six labels choose either $\{1,2,3\}$ or $\{4,5,6\}$ with probability $1/2$. Each label is included with probability $1/2$, yet eighteen of the twenty size-three subsets are impossible: this is a two-group cluster sampling design. Unequal probability sampling deliberately changes inclusion probabilities, often to improve precision for a particular estimand, and generally requires an appropriately weighted estimator rather than the plain sample mean.

Consider the original mathematical population with labelled values $(2,4,6,8,10,12)$, and let $n=2$. The fifteen equiprobable subsets, followed by their sample means, areTheir sample means sum to $105$, giving expected value $105/15=7$. Their squared deviations from $7$ sum to $70$, giving variance $70/15=14/3$. This is an enumeration over hypothetical selections from six stipulated numbers, not measured survey data.

With independent uniform sampling with replacement, the $6^2=36$ ordered outcomes each have probability $1/36$. The two orders $(2,4)$ and $(4,2)$ are distinct outcomes, and $(2,2)$ is now possible. Forgetting order does not make the resulting unordered multisets equiprobable: $\{2,4\}$ has probability $2/36$ but $\{2,2\}$ has probability $1/36$. The thirty-six sample means still have expected value $7$, but their squared deviations sum to $210$, giving variance $210/36=35/6$. Avoiding duplicate labels improves precision for this fixed population.

A simple random walk on a locally finite graph chooses each vertex in the current vertex's graph neighbourhood with equal probability, assuming there is at least one available neighbour. On the integer line there are two choices, one unit left or right, so each has probability $1/2$. This is the one-dimensional simple symmetric random walk. Terminology varies: a simple random walk on the integer line is also often allowed unequal step probabilities, with “symmetric” specifying the fair case.

For a simple symmetric random walk starting at $S_0=0$, each increment has the Rademacher distribution, with expected value zero and variance one. Thus linearity of expectation and the variance of a sum of independent random variables giveFor example, after $80$ steps the expected value is zero, the variance is $80$, and the standard deviation is $\sqrt{80}\approx 8.944$. Doubling the typical distance scale takes four times as many steps. This square-root growth explains why a random walk can wander far even when there is no preferred direction.

The colored curves are illustrative sample paths, not averages over all possible walks. The dashed curves show the expected values; the shaded bands show the expected value plus or minus one standard deviation at each time. These bands are a scale of spread, not confidence intervals, and give no guarantee that an entire sample path stays inside them.

Zero expected value does not make this a stationary process: its one-time probability distribution changes with $n$, as the increasing variance already proves. It does have stationary increments and independent increments, because blocks of fresh steps have the same probability distribution at any starting time and disjoint blocks use independent random variables. This separation of position from increment is one reason the random walk is such a useful first example of a stochastic process.

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 a known mean-zero noise law with finite variance ${\sigma }^2$, observing only $Y=a+ϵ$ with unrestricted $a\in \mathbb{R}$ has minimax risk ${\sigma }^2$ under squared-error loss. The observation itself attains this risk. For the lower bound, use a flat prior distribution on $[-A,A]$ and condition an easier experiment on $|ϵ|\le M$. Away from the prior endpoints, the Bayesian posterior noise has its original truncated law. Let $A\to \infty$ and then $M\to \infty$ in the resulting Bayes risk lower bound. No normal distribution or density for the noise is required.

The Skorokhod embedding theorem represents a centered probability distribution $\mu$ on the real numbers with finite variance as $B_T\sim \mu$ for a stopping time $T$ of Brownian motion. One may choose $T$ so that the stopped process $(B_{t\wedge T})$ is uniformly integrable and $\mathbb{E}T=\int x^2\mu (dx)$.

For centered multivariate normal distributions represented by vectors $U,V$ with equal corresponding variances and $\mathbb{E}[U_i{U}_j]\ge \mathbb{E}[V_i{V}_j]$ for all distinct indices, $\mathbb{P}(U_i\le u_i\text{ for all }i)\ge \mathbb{P}(V_i\le u_i\text{ for all }i)$ for every threshold vector. In particular the supremum of $U$ is stochastically smaller than the supremum of $V$. Equal corresponding variances are essential to this formulation. The related Sudakov-Fernique inequality compares expected values of maxima using variances of increments and does not require equal pointwise variances.

A smoothing bandwidth $h>0$ determines the scale over which a regression kernel averages data, or a kernel for density estimation spreads each observation. It has the same units as the observations. In kernel density estimation, changing $h$ changes the visible detail while preserving the total integral one. For a Gaussian density kernel, each bump has standard deviation $h$ and its individual contribution has peak height $1/(nh\sqrt{2\pi })$.

If the smoothing bandwidth is too small relative to the spacings in a statistical sample, the kernel density estimator shows narrow spikes tied to individual observations: apparent peaks may be sampling accidents. If it is too large relative to features of the unknown probability density function, it blurs distinct groups together and spreads fitted probability far from the observations. An intermediate smoothing bandwidth can reveal broad structure, but visual smoothness alone cannot establish accuracy.

The first three panels use exactly $(-1.8,-1.4,-1.1,-0.7,0.7,1.1,1.4,1.8)$ with smoothing bandwidths $0.08$, $0.35$ and $1.2$, respectively. Every curve has area one; the two groups in the observed statistical sample become less distinct as $h$ increases. No true probability density function is specified in this illustration, so the middle panel is illustrative, not a certified optimal fit. The fourth panel uses $(0.1,0.4,0.9)$ and $h=0.5$ to illustrate boundary bias of a kernel density estimator.

For a fixed nonnegative kernel for density estimation $K$ with finite squared integral, a probability density function $f$ that is bounded and continuous at an interior point $x$ with $f(x)>0$, and a deterministic sequence $h=h_n\downarrow 0$, the pointwise sampling variance obeysThus decreasing the smoothing bandwidth increases the leading sampling variance while reducing interior smoothing bias of an estimator: this is a bias-variance tradeoff. The conditions $h_n\to 0$ and $n{h}_n\to \infty$ express two goals, increasingly fine resolution and enough observations contributing locally. Practical methods such as plug-in bandwidth selection use the data to choose $h$; pointwise optimal kernel bandwidth depends on local features of the unknown probability density function. The smoothing bandwidth remains a tuning choice in nonparametric statistics, and knowing the support of a probability distribution can require more than a width adjustment near a boundary.

Use a complex envelope $A(t)$ normalized so that $|A(t){|}^2=P(t)$; its units are ${\mathrm{W}}^{1/2}$. The optical carrier convention is $e^{-i{\omega }_{0}t}$: the physical electric field is proportional to the real part of $A(t)e^{-i{\omega }_{0}t}$ for a fixed transverse mode. With angular frequency offset $\Omega =\omega -{\omega }_0$, fix the Fourier transform conventionWrite $\tilde{A}=B{e}^{i\varphi }$ with real $B\ge 0$. A dispersive element adds spectral phaseConstant phase rotates the envelope; the linear term delays it by ${\varphi }_1$. The group delay is $d\varphi /d\Omega$. The quadratic coefficient ${\varphi }_2$ is group delay dispersion, abbreviated GDD, with units of time squared. Positive GDD makes higher frequencies arrive later with this convention. It is accumulated dispersion of a component; material group velocity dispersion is its corresponding coefficient per unit length.

To undo a quadratic stretcher spectral phase, the compressor must supply opposite GDD. Include the amplifier and all intervening optics in the sum:Setting this sum to zero does not cancel third-order dispersion, ${\varphi }_3$, or still higher orders. Residual spectral phase can produce asymmetry and satellite pulses even when the quadratic term vanishes. A diffraction grating system can make path length depend on frequency; matching the complete spectral phase over the occupied spectrum is more demanding than merely choosing opposite quadratic coefficients.

There is a useful proof of what phase correction can accomplish. By Parseval's identity,For a smooth decaying spectrum, the Fourier transform of $tA$ is $-id\tilde{A}/d\Omega$. Consequently the power-weighted mean arrival time and variance obeyAt fixed spectral amplitude, constant group delay minimizes this root mean square duration. Such a pulse is a transform-limited pulse, up to a time shift and constant phase. The nonnegative second term measures temporal broadening from unequal arrival times; the first term remains even after perfect phase correction.

A centered weakly stationary process has a representation $X_t={\int }_{-1/2}^{1/2}{e}^{2\pi it\omega }dZ(\omega )$ with orthogonal random increments. Their variance measure is the spectral measure. When it has density $f$, the autocovariance is $\gamma (h)=\int e^{2\pi ih\omega }f(\omega )d\omega$.

For spin one-half, choose the normalized spin eigenstates $|+z⟩,|-z⟩$ with $S_z|\pm z⟩=\pm (\hslash /2)|\pm z⟩$. The spin one-half along an axis pure state in the $xz$ plane isIts norm is one because the two eigenstates are orthogonal and ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$. Applying the Born rule to a $z$ projective measurement givesConsequently the expectation value, second moment, and variance areAt $\theta =\pi /3$, the outcome probabilities are $3/4$ and $1/4$, the mean is $\hslash /4$, and the variance is $3{\hslash }^2/16$. Every individual outcome remains $\pm \hslash /2$. At $\theta =\pi /2$, the mean is zero and the variance is maximal, ${\hslash }^2/4$. At either pole, one outcome has probability one and the variance vanishes.

The diagonal mixed state ${\rho }_{\mathrm{diag}}=p_{+}|+z⟩⟨+z|+p_{-}|-z⟩⟨-z|$ has the same $S_z$ probabilities, expectation value, and variance as $|{\psi }_{\theta }⟩$. It is nevertheless a different quantum state for $0<\theta <\pi$: the displayed pure state has $⟨S_x⟩=\hslash \sin \theta /2$, while ${\rho }_{\mathrm{diag}}$ has $⟨S_x⟩=0$. A single component's mean does not reveal all relative quantum phases or establish purity of a density operator.

Under the conditional-variance and overshoot assumptions of the absorption-time bound from conditional variance and overshoot, $\mathbb{P}(T>n)\le 2\tau /(\sigma \sqrt{n})$. Bound the probability by $\mathbb{P}(T_R<\infty )+\mathbb{E}(T\wedge T_R)/n$ and optimize $1/R+{\tau }^{2}R/({\sigma }^{2}n)$. If the resulting bound is below one, its minimizing threshold is greater than one; otherwise the probability bound is trivial.

Applying a single-mode squeeze operator to a thermal state gives the displayed principal optical-quadrature variances. Their uncertainty product exceeds the pure-state minimum when $n_{\mathrm{th}}>0$. Sufficient squeezing is needed to put the quiet optical quadrature below the vacuum reference.

The standard deviation is the nonnegative square root ${\sigma }_X=\sqrt{\mathrm{Var}(X)}$ of the variance.

A Gaussian random vector with independent normal coordinates of expectation zero and variance one. Its distribution is invariant under orthogonal transformations. In particular, its direction in a fixed two-dimensional plane is uniform, which underlies the Gaussian sign-correlation identity.

With fixed positive scale estimates $s_j$, this is Euclidean distance in standardized coordinates. It removes marginal units but ignores correlations. Small estimated variances can greatly amplify measurement noise; a zero scale must be handled separately.

In a normal linear model, the $i$th regression residual has variance ${\sigma }^2(1-h_{ii})$, where $h_{ii}$ is a diagonal entry of the hat matrix. Replacing $\sigma$ by the residual standard deviation $s$ gives the standardized residual used in the normal Q-Q plot of R's linear-model diagnostic plots.

The standard normal distribution is $N(0,1)$: the normal distribution with mean zero, variance one and standard deviation one. Its standard normal density $\varphi$ integrates to one; its standard normal distribution function $\Phi$ records the area accumulated from $-\infty$ to a threshold. The distinction matters: $\varphi (0)=1/\sqrt{2\pi }\approx 0.398942$, $\Phi (0)=1/2$, and $\mathbb{P}(Z=0)=0$ for a standard normal random variable $Z$.

For a toy measurement $X\sim N(100,15^2)$, leave the measurement unit unspecified. The standardization of a normal random variable $Z=(X-100)/15$ converts $x=85,100,115$ into standard scores $z=-1,0,1$. Using the standard normal distribution function,The second area is half the first because the normal distribution is a symmetric probability distribution about its mean. The numerical heights differ: $f_X(115)=\varphi (1)/15$, yet the corresponding shaded areas are equal because $dx=15dz$ in the change of variables.