Ordinary kriging can have larger error than simple kriging

Each $U_i$ has mean zero and second moment one by [F5] and [F6]. For $i\mathrm{\ne }j$, each sign pair occurs equally often, so [F7] gives $\mathrm{Cov}(U_i,U_j)=0$. Consequently, $\mathrm{\Sigma }=I_2$, $c_0=(0,0{)}^{\mathrm{\top }}$, and $v_0=1$.

When $\theta$ is known, [F3] and step 1.1 give simple weight ${\mathrm{\Sigma }}^{-1}{c}_0=0$, so the simple minimizing predictor is $\theta$. Its error is $U_0$ and its mean-square error is $\mathbb{E}[U_0^2]=1$. For the ordinary model, the unbiasedness constraint is ${\lambda }_1+{\lambda }_2=1$ by [F2].

The ordinary system in [F4] applies because step 1.1 gives $\mathrm{\Sigma }=I_2$ and the constant-mean design has full rank. Directly, the ordinary prediction error is $U_0-{\lambda }_1{U}_1-{\lambda }_2{U}_2$. Expanding its square, all cross terms have expectation zero by step 1.1, so its mean-square error is $1+{\lambda }_1^2+{\lambda }_2^2$. Substituting ${\lambda }_2=1-{\lambda }_1$ gives $\frac{3}{2}+2({\lambda }_1-\frac{1}{2}{)}^2$, whose unique minimum is $3\mathrm{/}2$ at ${\lambda }_1={\lambda }_2=1\mathrm{/}2$. Since $3\mathrm{/}2>1$, ordinary kriging has the larger error in this example.