j^\prime}$ is an indicator function that equals 1 if $j=j^\prime$ and 0 otherwise. This parameterization ensures that the implied error variances, $\psi_j^2$, are positive, even though all parameters are reals.

The Jacobian of this model is $$\Delta_{\text{Shapiro}} = \kbordermatrix{
        &\lambda_1 & \lambda_2 &\lambda_3 & \psi_1 & \psi_2& \psi_3\\
\sigma_{11} & 2 \lambda_1 & 0& 0& 2 \psi_1 & 0& 0\\
\sigma_{21} & \lambda_2  & \lambda_1 & 0& 0& 0& 0\\
\sigma_{31} & \lambda_3  & 0 & \lambda_1 & 0& 0& 0\\
\sigma_{22} & 0 & 2 \lambda_2 & 0& 0 & 2 \psi_2& 0\\
\sigma_{32} & 0  & \lambda_3 & \lambda_2 & 0& 0& 0\\
\sigma_{33} & 0 & 0 & 2 \lambda_3 & 0 & 0& 2 \psi_3\\
},$$ so that simply setting any $\psi_j = 0$ will lead to a rank deficiency with null space equal to a “one-hot” indicator vector, e.g. setting $\psi_3=0$ gives $\text{Null}(\Delta_{\text{Shapiro}}) = [\begin{array}{cccccc}
0&0&0&0&0 &1
\end{array}]^T$. However, @shapiro_investigation_1983 pointed out that even with $\psi_3 = 0$, the system of equations $\boldsymbol{\sigma}(\boldsymbol{\theta}) = \boldsymbol{\sigma}$ can still be solved. Thus, similarly to the unidimensional function $f(x) = \text{sign}(x)\cdot x^3$ shown in Figure \[fig:fx\] having a unique maximum, all parameters of the model, including $\psi_3$, are 