i.e. it is common to assume that $$\begin{aligned}
    d^2(\bm{a} - \bm{a}') = (\bm{a} - \bm{a}')^\mathrm{t} \bm{M} (\bm{a} - \bm{a}'),
    \label{eq:metric}\end{aligned}$$ for positive-semidefinite matrix $\bm{M}$.

The most common covariance function is the *squared-exponential* (SE), $$k_{\mathrm{SE}}(\bm{a}, \bm{a}')     = \sigma_{\mathrm{SE}}^2 \exp \left(-\dfrac{1}{2} d^2(\bm{a} - \bm{a}') \right),
    \label{eq:cov_SE}$$ where $\sigma_{\mathrm{SE}}^{2}$ is the *signal variance* (the value of $k_\mathrm{SE}$ when $\bm{a}' = \bm{a}$). Furthermore, it is normally assumed that the pseudo-metric matrix is diagonal, i.e. that $\bm{M} = {\operatorname{diag}}(m_{1}, m_{2}, \ldots , m_{D})$. Using equations \[eq:metric\] and  \[eq:cov\_SE\] we may show [see, for example, @loeppky_choosing_2009] that the mean squared gradient is $${\operatorname{E}}\left( \frac{\upartial Y(\bm{a})}{\upartial a_i} \right)^2 = 2 \sigma_\mathrm{SE}^2 m_i,$$ and therefore call $m_i$ the *sensitivity* of the model to the $i$-th parameter. Because $\bm{M}$ is positive-semidefinite it has a unique positive-semidefinite inverse, which in turn has a unique square root, i.e. there exists a unique matrix $\bm{L}$ (not to be confused with the likelihood, $L$) such that $\bm{M} = \bm{L}^{-2}$. If $\bm{M}$ is diagonal then so is $\bm{L}$ and $\bm{L} = {\operatorname{diag}}(l_1, l_2, \ldots , l_D)$ where $m_i = l_i^{-2}$ for all $i$. We can see that equation \[eq:cov\_SE\] is formally identical to a Gaussian with covariance $\bm{M}^{-1}$. We therefore identify $\bm{L}^{2}$ as a 