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For $t<1$ this stochastic integral solves $d{X}_t=d{B}_t-X_t(1-t{)}^{-1}dt$. The Itô product rule also gives $X_t=B_t-(1-t){\int }_0^t{B}_s(1-s{)}^{-2}ds$. The process is centered Gaussian, with covariance $\min (s,t)-st$, and has a continuous extension $X_1=0$. Time reversal of a Brownian bridge proves endpoint continuity without assuming it at the start.

If $Z$ is nondegenerate, $g$ is strictly increasing and the relevant products are integrable, an independent copy $Z^{\prime }$ gives $2\mathrm{Cov}(g(Z),Z)=\mathbb{E}[(g(Z)-g(Z^{\prime }))(Z-Z^{\prime })]$. The integrand is positive exactly when $Z\ne Z^{\prime }$, so the covariance is strictly positive. The same argument applies within each conditional law and yields a positive terminal stock hedge for an attainable increasing payoff.

A noise-driven barrier crossing between adjacent phase-locked wells. The Kramers escape rate is proportional to $e^{-\Delta V/T_{\mathrm{eff}}}$ when a mobility-one Langevin equation has white-noise covariance $2T_{\mathrm{eff}}\delta (t-t^{\prime })$. Both the barrier and local relaxation time determine whether an intrawell phase autocorrelation approximation is appropriate.

A time series is a sequence of observations indexed by time: hourly temperatures, daily sales, or successive measurements of a fluctuating signal. A probabilistic model treats these observations as one sample path of a stochastic process $(X_t)$. For equally spaced observations, take $t\in \mathbb{Z}$ and measure a lag of a time series in observation intervals. The order matters: shuffling a record preserves its observed values but destroys information about how one value follows another.

Two basic ways to model temporal dependence are an autoregressive model, which feeds past observations into the next value, and a moving-average model, which combines present and past white noise. An AR(1) carries a disturbance forward with diminishing weights; an MA(1) carries it into just two observations. Their autocorrelation functions make this difference visible. A weakly stationary process supplies a common mean and a lag-dependent autocovariance with which to describe that dependence.

These models help forecast future observations and quantify the uncertainty of averages. Successive measurements need not provide as much information as the same number of independent random variables: the long-run variance of a stationary process accounts for their accumulated covariances. Trend and seasonality require attention before fitting a stationary model; differencing and a seasonal difference operator can help, but do not automatically make every process stationary. See Peter Bartlett, Introduction to Time Series Analysis, lecture 1: objectives and models.

The three illustrated sample paths have zero mean and unit variance, so the visual comparison concerns dependence rather than scale. The AR(1) paths start from their stationary normal distribution, independently of future discrete Gaussian white noise; the MA(1) path includes the noise preceding its first observation. These are finite random examples, not exact pictures of an ACF. Compare their theoretical autocorrelation functions in autocorrelation.

Historically, Yule's 1927 study investigated disturbed periodicities in sunspot observations; it should not be read as the discovery of the particular AR(1) examples here. Slutsky studied how summing random inputs could produce wave-like records, a warning against inferring a periodic cause from appearance alone. The editorial note to his 1937 English paper identifies the 1927 Russian original. See G. Udny Yule, On a method of investigating periodicities in disturbed series, with special reference to Wolfer's sunspot numbers (1927) and Eugen Slutsky, The Summation of Random Causes as the Source of Cyclic Processes (1937), including the editorial note on the 1927 original.

The total differential is the linear map that combines the first-order responses to simultaneous changes in all coordinates. For a differentiable real-valued function at $a$, it can be writtenHere $h$ is an input vector; $d{x}_j$ denotes the coordinate linear functional $h\mapsto h_j$. The coordinate formula and the Fréchet derivative describe the same linear map. The defining approximation isThe little o notation says the remaining approximation error becomes small relative to the input norm, across every direction of approach.^[ref]

For $f=x^2+xy+2y^2$ at $a=(1,1)$, $d{f}_a(h,k)=3h+5k$. A simultaneous change $(h,k)=(0.02,-0.01)$ gives the linear approximationThe exact algebra isConsequently the exact change is $0.0104$, and the approximation error is $0.0004$. If both input changes are halved, the linear prediction becomes $0.005$ and the error becomes $0.0001$: the first-order response halves, while the quadratic error falls by a factor of four. The right panel under directional derivative displays this comparison as the displacement is scaled.

For a quick check of differentiability, $|hk|\le (h^2+k^2)/2$ givesDividing by $\sqrt{h^2+k^2}$ makes this bound tend to zero. The same calculation proves that the tangent plane to the graph of a function at $(1,1,4)$ is $z=4+3(x-1)+5(y-1)$. It approximates the entire local surface, rather than only two coordinate slices.

The total differential is useful for estimating the combined effect of small input changes without reevaluating a complicated function. For example, if $|h|\le 0.02$ and $|k|\le 0.01$, then $|d{f}_a(h,k)|\le 0.11$. The exact remainder bound from the displayed polynomial is at most $0.0008$, so this example has $|\Delta f|\le 0.1108$ throughout that input rectangle. These are deterministic bounds; a variance calculation additionally needs a joint probability distribution and its covariances. A first-order estimate is also local: large changes require control of the remainder, not merely knowledge of the partial derivatives at one point.

The bulk tangent covariance is twice the inverse Fourier transform of $k_{B}T/(A{q}^2+f)$, one term for each transverse direction. Equivalently the inverse of $-A{\partial }_r^2+f$ is $e^{-|r|/\xi }/(2\sqrt{Af})$. This gives the displayed correlation and identifies a force-dependent correlation length, different from the zero-force persistence length. Finite chains require endpoint conditions or a periodic convention.

For the triangle count in a binomial random graph, distinct triangle indicators are independent unless they share an edge. A shared-edge pair has covariance $p^5(1-p)$, and there are $12\left(\genfrac{}{}{0ex}{}{n}{4}\right)$ ordered pairs of that type. The individual indicator variances give the first term. Equivalently,by counting ordered pairs according to the number of shared vertices.

For $S=A\cos (q\theta )$ with independent standard uncertainties, linear propagation gives ${\sigma }_S^2={\cos }^2(q\theta ){\sigma }_A^2+[qA\sin (q\theta ){]}^2{\sigma }_{\theta }^2$. Angular uncertainty must be in radians when using this derivative. Covariance adds cross terms for correlated inputs. Strict error bounds instead add absolute first-order contributions; they do not justify Gaussian significance tests.

Here is an explicit joint distribution, for $0<p<0.8$: assign mass $(1+p)e^{-p}-p$ at $(0,0)$, mass $p(1-e^{-p})$ at $(1,0)$, mass $p{e}^{-p}$ at $(1,1)$, and mass $e^{-p}{p}^k/k!$ at $(0,k)$ for $k\ge 2$; all other masses are zero. These are nonnegative: the first is decreasing in $p$, and at $p=0.8$ it is still positive. Summing over rows and columns gives a Bernoulli distribution with mean $p$ and a Poisson distribution with mean $p$, so the joint mass sums to one.

Only $(1,1)$ contributes to the mixed first moment, giving $\mathbb{E}[XY]=p{e}^{-p}$ and the covariance above. The strictly decreasing function $e^{-p}-p$ has a unique zero in $(0,0.8)$, approximately $0.567143$. At that parameter the variables are uncorrelated. They are nevertheless dependent, because $\mathbb{P}(X=1\mid Y=1)=1$ whereas $\mathbb{P}(X=1)=p<1$. The Bernoulli independence from zero covariance result requires both variables to be Bernoulli and does not apply here.

Two square-integrable random variables $X,Y$ are uncorrelated random variables when $\mathrm{Cov}(X,Y)=0$. Independent random variables with finite second moments are uncorrelated random variables, but zero covariance does not generally imply independence.

For an explicit counterexample, let $X$ take $-1$, $0$, $1$ with equal probabilities, and let $Y=X^2$. Then $\mathbb{E}X=\mathbb{E}{X}^3=0$, so $\mathrm{Cov}(X,Y)=\mathbb{E}{X}^3-\mathbb{E}X\mathbb{E}{X}^2=0$. Nevertheless, $Y$ is completely determined by $X$: $\mathbb{P}(Y=0\mid X=0)=1$, while $\mathbb{P}(Y=0)=1/3$. Thus these are uncorrelated random variables with statistical dependence. In this example, opposite contributions from the odd function $x^3$ cancel; the even function $x^2$ still retains information about $|X|$. The last panel of the covariance diagram shows the relationship.

There are useful special cases. Uncorrelated jointly normal variables are independent, and the Bernoulli independence from zero covariance result supplies another setting where the converse holds. Their distribution assumptions do the extra work; zero covariance alone does not.

Place $n$ labelled balls independently into $m$ labelled bins, each with probability $1/m$. The vector of bin counts has a multinomial distribution, and each count separately has the binomial distribution $\mathrm{Bin}(n,1/m)$. Distinct bin counts have covariance $-n/m^2$, so they are dependent for $n>0$ and $m>1$. Counts of bins of a specified size are sums of indicator variables; their expectations follow from the linearity of expectation, without any independence between bins.

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.

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 $3\lambda d^2$.

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.

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.

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 velocity ellipsoid represents the positive velocity covariance tensor by its orthogonal principal axes and eigenvalue square roots. The uncentered second-moment tensor differs from covariance when a population streams. Vanishing mixed second moments align the second-moment ellipsoid with the chosen frame; for pure azimuthal streaming in a phase-mixed axisymmetric model, the off-diagonal covariance also vanishes.

For a particle of mass $m$ with linear drag $-\zeta v$, the velocity relaxation time is the exponential relaxation time ${\tau }_v=m/\zeta$. Under a constant applied force $F$ and zero-mean random forcing, the averaged Langevin equation is $md⟨v⟩/dt=-\zeta ⟨v⟩+F$. Solving givesThus the initial mean-velocity deviation decays by a factor $e^{-1}$ after time ${\tau }_v$. Since the force-per-velocity coefficient $\zeta$ has units $\mathrm{kg}{\mathrm{s}}^{-1}$, $m/\zeta$ has units of time. If the damping parameter is instead defined as a rate $\gamma =\zeta /m$, the same time is $1/\gamma$. For a sphere obeying Stokes's law, ${\tau }_v=m/(6\pi \eta R)$. The fluctuation-dissipation relation for a Langevin particle gives the same decay time for its equilibrium velocity covariance.

A real stochastic process $(X_t{)}_{t\in \mathbb{Z}}$ is a weakly stationary process if it has finite second moments, a constant expected value $\mu$, and an autocovariance depending only on the lag of a time series:In particular its variance is the same at every time, $\mathrm{Var}(X_t)=\gamma (0)$. This is often called second-order stationarity. It makes observations from different parts of a record comparable through the same autocorrelation function. Bartlett uses this second-order meaning of stationarity in Introduction to Time Series Analysis, lecture 2: mean, covariance, and stationarity.

For a worked contrast, let ${\epsilon }_t$ be centered white noise of variance ${\sigma }^2>0$. The process $X_t=5+{\epsilon }_t$ has mean $5$, variance ${\sigma }^2$, and zero autocovariance at nonzero lags, so it is weakly stationary. The process $Y_t=t+{\epsilon }_t$ has exactly the same covariance structure, but its mean is $t$, so it is not weakly stationary. Constant variance alone is insufficient.

Weak stationarity need not imply strict stationarity. For example, choose independent random variables at all integer times, using a standard normal distribution at even times and a Rademacher distribution at odd times. Every coordinate has zero mean and unit variance, and distinct coordinates have zero covariance. Nevertheless a shift by one changes the marginal probability distribution from continuous to two-point. This also illustrates why weak white noise need not be identically distributed.

There exists a probability measure on $C_0([0,\infty ),\mathbb{R})$ under which the coordinate process is standard Brownian motion, and its distribution there is unique. Construct consistent centered Gaussian finite-dimensional laws with covariance $\min (s,t)$ and use the Kolmogorov extension theorem. The fourth increment moment $3|t-s{|}^2$ supplies a continuous modification by the Kolmogorov continuity theorem. Its independent increments and normal increment distributions are unchanged. Uniqueness follows because coordinate evaluations at rational times generate the Borel sigma-algebra of continuous-path space.