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The stationary exponential covariance function isThe corresponding Gaussian process is Markov and is a stationary Ornstein-Uhlenbeck process.

Ordinary kriging assumes $\mathbb{E}[Z(s)]=\mu$ for an unknown constant $\mu$. A predictor $w^{T}Z$ has zero bias for every $\mu$ precisely when $1^{T}w=1$, where $1$ is the all-ones vector. Ordinary kriging minimizes the error variance $Q(w)$ subject to this constraint, giving the best linear unbiased predictor. Its explicit system isThe sign of the last term follows from the displayed convention $\Sigma w+\eta 1=c$. To derive the system, differentiate $Q(w)+2\eta (1^{T}w-1)$. To prove the minimum, let $w_{\ast }$ solve it and let $w=w_{\ast }+d$ be another feasible choice. Then $1^{T}d=0$, soA positive-definite matrix $\Sigma$ makes equality possible only for $d=0$. This argument uses second moments, not a normal distribution assumption. Ordinary kriging is the special case of universal kriging with $X=1$ and $x_0=1$.

For an original one-dimensional teaching example, observe $z=(1,3,2{)}^T$ at $s=(0,1,3)$, and predict at $s_0=2$. Choose unit marginal variance, exact observations without measurement noise, an unknown constant mean, and the exponential covariance functionThe exponential covariance function corresponds to a stationary Ornstein-Uhlenbeck process, so it gives a valid covariance model. Its semivariogram is $1-2^{-|h|}$. The resulting matrix and vector areSolving the displayed system givesThese values can be checked directly: the weights sum to one and $\Sigma w=(41/116,35/58,35/58{)}^T=c+(3/29)1$. Substitution into $Q(w)$ gives $18/29$. Nearest-neighbor interpolation must choose between the two equally close samples at one and three, returning either three or two; their error variances under this model are both $1+1-2(1/2)=1$. Thus the benefit is quantified under the chosen model, without knowing the unobserved target value.

The upper panel applies the same ordinary kriging calculation at each target location. The lower panel shows the square root of its error variance, in the units of the observations. It is zero at the exact observation locations and rises between and beyond them. No confidence interval is inferred from a variance alone; a probability distribution assumption is needed to attach a coverage level.

The weights are not required to be positive. Keep the same observations and target, but replace the covariogram by the smoother Gaussian kernel $C(h)=2^{-h^2}$. ThenThe negative coefficient compensates for dependence between samples; the constraint eliminating bias still holds. Kriging is not generally a convex combination, and its prediction need not stay between the observed values. A smaller model-based error variance does not establish that one covariance function is better supported by these three observations.

Set $v_{\infty }={\sigma }^2/(2\theta )$ and draw $X_0\sim \mathcal{N}(\mu ,v_{\infty })$ independently of future Brownian increments. The Ornstein-Uhlenbeck process is then a strictly stationary process, withEvery time has the same normal distribution, and the covariance function depends only on the time lag. For a Gaussian process these properties also make all its finite-dimensional distributions invariant under a common time shift. The stationary distribution balances the continual random forcing against the contracting drift.^[ref]

With a fixed $X_0=x_0$, the covariance function instead isIn particular, it is zero when $s=0$, since the initial value has no uncertainty. The second term records the fixed start; it is the general scaling of the zero-start Ornstein-Uhlenbeck covariance. The pure exponential covariance function displayed above assumes stationary initialization. For a fixed start, that function emerges as the late-time limit at fixed lag, while the one-time probability distributions approach $\mathcal{N}(\mu ,v_{\infty })$ by convergence in distribution.

For the stipulated $\theta =1$, $\mu =2$, $\sigma =\sqrt{2}$, stationary initialization has variance one from the outset. Its autocorrelation is $e^{-|h|}$, so the correlation time is one: a lag of one reduces the autocorrelation to $e^{-1}\approx 0.3679$. Its raw second moment $\mathbb{E}[X_s{X}_t]=4+e^{-|t-s|}$ includes the squared expected value; it must not be confused with the covariance.

The upper panel shows three simulated sequences with independent stationary initial values and exact transitions at spacing $0.05$. Joining their points only guides the eye; it does not simulate a continuous sample path between those points. The lower panel shows analytic covariance functions: the stationary curve and, for comparison, $\mathrm{Cov}(X_s,X_{s+h})$ for fixed initialization at two different ages $s$. These are simulated samples and model curves, not measured trajectories or empirical covariance estimates.

$\boxed{(a)⟷P2}$. The exponential covariance function gives a stationary Ornstein-Uhlenbeck process. At a small lag $h$,The variance of a difference quotient therefore diverges as $h\to 0$, so the process has no mean-square derivative. Its continuous version has rough, nondifferentiable paths, matching the jagged P2 trace in the original PDF.

For $t_1<t_2<t_3$, the condition from part a becomesIt holds for arbitrary positive time gaps exactly when $\eta =1$. The resulting exponential covariance function is the covariance of a stationary Ornstein-Uhlenbeck process, hence has the Markov property.

Writing $a_{32}=e^{-(t_3-t_2)/\tau }$, the predictive law isAs $t_3\to \infty$, $a_{32}\to 0$, so the predictive mean tends to the stationary mean $\mu$ and the predictive variance tends to the stationary variance $1$.

Put $\rho =e^{-\Delta t/\tau }=e^{-x}$ and retain $R=k(0,0)$, which equals one for the stated exponential covariance function. The covariance matrix of $(y_1,y_2,y_3)$ isConditioning a multivariate normal distribution and simplifying givesThe absence of $y_1$ from both conditional moments shows that $y_3$ and $y_1$ are conditionally independent given $y_2$. Equivalently, the exponential-kernel Gaussian process is the stationary Ornstein-Uhlenbeck process, which is Markov.