covariance matrix and call the element $l_i$ the *correlation length* (also *scale length*) for the $i$-th parameter. The elements of $\bm{M}$ together with $\sigma_{\mathrm{SE}}^2$ and the coefficients of our basis functions form the hyperparameter vector, $\bm{\theta} = \{\sigma_{\mathrm{SE}}^2, \mu_{1}, \ldots, \mu_{N}, m_{1}, \ldots , m_{D}\}$, of the Gaussian-process estimator. One advantage of assuming that $r$ is the zero function is that this hyperparameter vector becomes shorter: $\bm{\theta} = \{\sigma_{\mathrm{SE}}^2, m_{1}, \ldots , m_{D}\}$. This means that the dimension of the domain of the hyperparameter likelihood is smaller and hence the hyperparameter vector is easier to optimize.

If $r$ is expanded according to equation \[eq:gpe\_basis\] and we use the squared-exponential covariance function then the mean-square error of the estimator, i.e. the mean-square difference of $\hat{r}$ and $r$ (Sacks et al. 1989), is $$\begin{aligned}
    \begin{split}
    &\operatorname{MSE}(Y(\bm{a}) \,|\, (\bm{Y} = \bm{y})) = \\
    &\sigma_\mathrm{SE}^2
    \bigg(1 -
    \begin{bmatrix}
        \bm{\xi}^{\mathrm{t}}(\bm{a}) & \sigma_\mathrm{SE}^{-2} \bm{k}^{\mathrm{t}}(\bm{a})
    \end{bmatrix}
    \begin{bmatrix}
        \bm{0}      & \bm{\Xi}^{\mathrm{t}}\\
        \bm{\Xi}   & \sigma_\mathrm{SE}^{-2} \bm{K}
    \end{bmatrix}^{\mathrm{-1}}
    \begin{bmatrix}
        \bm{\xi}(\bm{a})\\
        \sigma_\mathrm{SE}^{-2} \bm{k}(\bm{a})
    \end{bmatrix}
    \bigg)
    \end{split}
    \label{eq:mse}\end{aligned}$$ where $[\bm{\Xi}]_{i} = \bm{\xi}^{\mathrm{t}}(\bm{a})$. In the case that $\bm{\xi}(\bm{a}) = \bm{0}$ (as we assume here) this expression reduces to the estimator variance (equation \[eq:gpe\_var\]). We may therefore construct a confidence interval for the estimator, 