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A vacuum fluctuation is a nonzero quantum variance in a vacuum state. For a single electromagnetic normal mode, dimensionless optical quadratures $a+a^{†}$ and $-i(a-a^{†})$ have zero means and unit variances in the vacuum state, despite zero photon number. The mode's zero-point energy is $\hslash \omega /2$.

The variance measures the mean squared distance of a random variable from its expected value. For a real square-integrable random variable $X$,The square prevents positive and negative deviations from cancelling. It also gives the variance squared units: if $X$ measures a length in metres, its variance has units of square metres. Its nonnegative square root is the standard deviation, which has the same units as $X$.

For example, let $X$ take $-1$, $0$ and $1$ with equal probabilities. Its expected value is zero, but its variance is $[(-1{)}^2+0^2+1^2]/3=2/3$, and its standard deviation is $\sqrt{2/3}$. The constant random variable zero has the same expected value and zero variance: a mean alone does not describe the spread. The variance is zero exactly when $X=\mathbb{E}X$ almost surely.

The variance is useful because squared deviations interact neatly with linear combinations. The covariance supplies the cross terms when combining different random variables, and a covariance matrix organizes those terms for many variables. The characterization of variance as the minimum mean squared error of a constant also explains why the expected value is the best constant prediction under squared error.

Put $Q={\sum }_i{q}_i$, $M={\sum }_i{q}_i{\mu }_i$, and $A={\sum }_i{q}_i\mathbb{E}{Y}_i^2$. A compound Poisson distribution with count mean $Q$ and severity law ${\sum }_i(q_i/Q)F_i$ has variance $V_{\mathrm{P}}=A$. A compound binomial distribution with $n$ trials, probability $Q/n$, and that same severity law has variance $V_{\mathrm{B}}=A-M^2/n$. The individual-policy variance is $V=A-{\sum }_i(q_i{\mu }_i{)}^2$. The Cauchy-Schwarz inequality gives ${\sum }_i(q_i{\mu }_i{)}^2\ge M^2/n$, proving the ordering. Equality of the first two variances requires constant $q_i{\mu }_i$; it need not imply equality of the distributions. If all probabilities and severity laws are identical, the binomial compound approximation is exactly the individual model.

For independent identically distributed random vectors $Q,Q^{\prime }$ with finite second absolute moments, expand $|Q-Q^{\prime }{|}^2=|Q{|}^2+|Q^{\prime }{|}^2-2Q\cdot Q^{\prime }$. Their independence makes $\mathbb{E}(Q\cdot Q^{\prime })=|\mathbb{E}Q{|}^2$, proving the identity. The right side is the sum of the coordinate variances. Applied to the displacement in an optimal transport plan, it separates motion of the barycenter of a measure from relative particle motion.

For exponential distribution claims of mean $\mu$ and compound Poisson distribution count parameter $\lambda$, the total party variance under excess of loss reinsurance is $g(M)=2\lambda {\mu }^2(1-(M/\mu )e^{-M/\mu })$. Its derivative is $2\lambda e^{-M/\mu }(M-\mu )$, so $M=\mu$ is the unique global minimum, with value $2\lambda {\mu }^2(1-e^{-1})$.

For any aggregate with positive finite variance $v$, quota share reinsurance with retained fraction $\alpha$ has total party variance $v({\alpha }^2+(1-\alpha {)}^2)=v/2+2v(\alpha -1/2{)}^2$. Equal sharing uniquely minimizes this objective. The two party totals remain dependent; adding their individual variances is a different objective from the fixed variance of their sum.

For a Pareto distribution of shape three and minimum $d>0$ with count rate $\lambda >0$, the sum of insurer and reinsurer aggregate variances is $V(M)=\lambda (3d^2-3dM+2M^2)$ for $0\le M\le d$, and $\lambda (3d^2-d^3/M)$ for $M\ge d$. The first branch has its unique minimum at $3d/4$; the second is increasing. The branches agree in value and slope at $d$, giving the displayed global minimum. This criterion minimizes the sum of party variances; the total portfolio variance includes their covariance and remains $3\lambda d^2$.

For a full-column-rank design matrix in a normal linear model, a fitted mean at covariate vector $v$ is $v^T\hat{\beta }$. The covariance matrix of the least-squares estimator immediately gives the displayed variance. Replacing ${\sigma }^2$ by the independent unbiased residual variance gives its squared standard error and a Student t confidence interval on $n-p$ statistical degrees of freedom. A future response instead contributes an additional error variance ${\sigma }^2$ when its error is independent of the fitted data.

// Target: probability-and-statistics.bigb

For stationary covariance $C(t)=(k_{B}T/k)e^{-|t|/{\tau }_k}$, the mean over duration $\mathcal{T}$ has variance $2(k_{B}T/k)[{\tau }_k/\mathcal{T}-({\tau }_k/\mathcal{T}{)}^2(1-e^{-\mathcal{T}/{\tau }_k})]$. At $\mathcal{T}\gg {\tau }_k$ this becomes $2(k_{B}T/k){\tau }_k/\mathcal{T}$. Closely sampled positions remain correlated and cannot be counted as independent.

The variance of an estimator is the variance of its sampling distribution under the parameter value $\theta$.

Let $X_1,\dots ,X_n$ be an IID sample of real-valued random variables with finite expected value $\mu$ and finite variance ${\sigma }^2$, where $n\ge 1$. Its sample mean $\overline{X}=n^{-1}{\sum }_i{X}_i$ satisfiesLinearity of expectation gives the first identity. The variance of a sum and vanishing cross covariances give $\mathrm{Var}({\sum }_i{X}_i)=n{\sigma }^2$; scaling by $1/n$ scales the variance by $1/n^2$. Pairwise independent random variables with the same expected value and variance suffice for these identities, even without full independence. Grinstead and Snell state the corresponding finite-variance independent-trials result.^[ref]

The standard deviation of the sample mean is $\sigma /\sqrt{n}$. Four times as many observations halve this scale; they do not divide it by four. In the fair Bernoulli distribution example, $\sigma =1/2$, so sample sizes $4$ and $64$ give standard deviations $1/4$ and $1/16$, respectively. The common-coin averaging example shows why merely repeating observations with a common marginal law does not justify this calculation.

Chebyshev's inequality turns the variance identity into a probability bound: for $\epsilon >0$,This proves convergence in probability of the sample mean to $\mu$ in this finite-variance setting, a form of the weak law of large numbers. It is an upper bound, so values above one convey no improvement over the universal probability bound one. Cauchy sample mean stability lies outside these hypotheses.

On a cyclic sequence of $n$ independent Bernoulli random variables with success probability $p$, count all runs of $k$ consecutive successes, allowing overlaps and taking $1\le k\le n/2$. The count has expected value $n{p}^k$. Its variance is the displayed expression: each starting indicator has variance $p^k(1-p^k)$; two starts at cyclic distance $j<k$ have joint probability $p^{k+j}$; nonoverlapping starts are independent. There are $2n$ ordered overlapping pairs at each such distance, because the two ranges do not meet when $k\le n/2$.

For $p\le 1/2$, summing a geometric series gives $\mathrm{Var}X\le n{p}^k(1+2p/(1-p))\le 3n{p}^k$. Markov's inequality therefore proves absence with high probability when $n{p}^k\to 0$, and Chebyshev's inequality proves existence and relative concentration when $n{p}^k\to \infty$. This remains valid for varying probabilities eventually at most one half.

The variance of hypothetical means in the Bühlmann model is the between-risk variance of the conditional claim expected value. It creates the shared covariance $a$ between different years of the same risk and determines how much individual experience should influence the premium.

For $n$ pots per treatment and $m$ leaves per pot, the group sample mean decomposes into two averages:The variance of a sum adds the variances of the independent terms. Every pot random effect has weight $1/n$, whereas each remaining error has weight $1/(nm)$, giving the short proofThis is also the variance of the arithmetic mean of the $n$ pot sample means, each with variance ${\tau }^2+{\sigma }^2/m$. Equal numbers of leaves per pot make the two averaging procedures identical.

If all $nm$ leaf measurements were falsely treated as independent, the predicted variance would instead be $({\tau }^2+{\sigma }^2)/(nm)$. The ratio of correct to false variance isThus pseudoreplication can underestimate a standard error by the square root of this ratio. Equality holds when $m=1$ or ${\tau }^2=0$; sharing a pot creates an error here precisely when the shared influence has positive variance.

Choose ${\tau }^2=4$, ${\sigma }^2=1$, $n=4$ and $m=3$ as constructed model parameters in squared length units. These are assumptions rather than fitted experimental values. For each group,The treatment contrast $\hat{\Delta }={\overline{Y}}_B-{\overline{Y}}_A$ has expected value ${\alpha }_B-{\alpha }_A$. Since the group errors are independent conditional on the allocation, its variance is twice the group variance, namely $13/6$, whereas the false calculation gives $5/6$. Its correct standard error is $\sqrt{13/6}\approx 1.472$ length units; the false value is $\sqrt{5/6}\approx 0.913$. The variance ratio is $13/5=2.6$.

With four pots fixed, increasing $m$ drives the error contribution ${\sigma }^2/(4m)$ towards zero, but the limit of the group variance remains ${\tau }^2/4=1$. Measuring twelve leaves on each of four pots gives $49/48$. Using the same forty-eight measurements on sixteen pots, with three leaves per pot, gives $13/48$. More independent experimental units reduce both contributions. Extra nested measurements are useful for characterizing each experimental unit, but cannot average away its shared random effect. This is why sample collection effort and the number of independent treatment replicates answer different questions.^[ref]

The moments of the mutation count on a neutral coalescent tree give this variance for ${\hat{\theta }}_W=S/H_{n-1}$. Since $H_{n-1}\sim \log n$ and the sum of inverse squares is bounded, it tends to zero. Together with unbiasedness, this gives mean-square convergence to the mutation parameter, so it is a consistent estimator.

If a response has mean $\mu$ and variance $V(\mu )$, the delta method gives $\mathrm{Var}(g(Y))\approx g^{\prime }(\mu {)}^{2}V(\mu )$. Choosing $g^{\prime }(\mu )$ proportional to $V(\mu {)}^{-1/2}$ makes this first-order variance independent of $\mu$. For $V(\mu )=k{\mu }^2$ on positive means, the resulting transformation is the natural logarithm. This is a local approximation and requires the transformation to be defined on the observations; it does not make arbitrary normal responses positive or preserve an exact normal distribution.

The unnormalized autocorrelation function of a stationary velocity. For zero mean, $\chi (0)$ is the velocity variance. It is different from the normalized autocorrelation function $\chi (u)/\chi (0)$ often used in time series.

If $X(t)-X(0)={\int }_0^{t}V(s)ds$, integrating the two-time covariance over the square $[0,t{]}^2$ givesThis identity connects a mean squared displacement to the Green-Kubo formula for diffusion.

The planar Mestel disk has $F(E)\propto e^{-E/v_0^2}$, giving independent centered Gaussian distributions for $v_R,v_{\varphi }$ with velocity dispersions ${\sigma }_R={\sigma }_{\varphi }=v_0$. Its streaming velocity is zero. A maximally prograde stellar distribution instead has mean $\sqrt{2/\pi }{v}_0$ and azimuthal variance $v_0^2(1-2/\pi )$, while all second raw moments remain unchanged.

For data $d=aT+n$, known complex source template $T$, real amplitude $a$, and known noise covariance $C$, generalized least squares gives the displayed amplitude. This convention takes $C$ in the quadratic objective $(d-aT{)}^{†}{C}^{-1}(d-aT)$; factors of two in the amplitude variance depend on whether complex-component or real-component covariance is used. Source position, shape and contaminants must be fitted or marginalized when uncertain. An extended-source complex visibility template accounts for resolved-out flux rather than assuming the source is unresolved.

For fixed particle number and temperature, a bubble with surface tension $\Gamma$ and area $\alpha V^{2/3}$ has effective availability $A(V)=F_{\mathrm{gas}}(T,V,N)+\Gamma \alpha V^{2/3}+p_{R}V$. Its stationary condition is $p_{\mathrm{gas}}-p_R=(2/3)\Gamma \alpha V^{-1/3}$. Its curvature is $A^{\prime \prime }=V^{-1}[{\kappa }_T^{-1}-(2/9)\Gamma \alpha V^{-1/3}]$. Positive curvature is required for stable Gaussian fluctuation approximation. The variance follows from availability curvature and equilibrium fluctuations. For an ideal gas, the restoring denominator becomes $p_R+(4/9)\Gamma \alpha V^{-1/3}>0$ for positive $p_R$. Treating the internal pressure as externally fixed instead of fixing the gas particle number changes the stability problem.