= Solution
Write $z=(Z_{s_1},\ldots,Z_{s_n})^T$, let $\Sigma_{ij}=C(\|s_i-s_j\|)$, let $c_i=C(\|s_i-s_0\|)$, and put $v_0=C(0)$. Assume the <covariance matrix> $\Sigma$ is invertible. Because the mean is known to be zero, every linear predictor $w^Tz$ is unbiased. Its <mean squared prediction error> is
$$
\mathbb E(Z_{s_0}-w^Tz)^2=v_0-2w^Tc+w^T\Sigma w.
$$
Completing the square yields
$$
v_0-c^T\Sigma^{-1}c+(w-\Sigma^{-1}c)^T\Sigma(w-\Sigma^{-1}c).
$$
The <positive-definite matrix> property makes the last term nonnegative, proving the <simple kriging> formulas
$$
\boxed{\widehat Z_{s_0}=c^T\Sigma^{-1}z,\qquad v_K=v_0-c^T\Sigma^{-1}c.}
$$
The residual is uncorrelated with $z$. Since the joint law is <multivariate normal>, it is independent of $z$, so the same predictor and <variance> describe the <conditional distribution>. Consequently a 95-percent <prediction interval> is
$$
\boxed{c^T\Sigma^{-1}z\ \pm\ q_{0.025}\sqrt{v_0-c^T\Sigma^{-1}c}.}
$$
There is no constraint that the weights sum to one: that is required for an unknown constant mean in <ordinary kriging>, not for the stated known-zero-mean problem. If $\Sigma$ is singular, use a <Moore-Penrose inverse> with the corresponding <covariance> compatibility condition. At an already observed location prediction of that very same field value has zero error; predicting an independent noisy replicate is a different target. Estimated <covariance> parameters make this a plug-in interval and generally add uncertainty that the formula does not include.
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