# An Algorithm for the Fisher Information Matrix of a VARMAX Process
## 5 Numerical Experiments  
The approach developed above can be used for any VARMAX model, for instance, the one produced by [42] obtained using a least-squares-based iterative estimation method. To save space, we will use examples based on the simplest case, i.e., \(n=m=2\), \(C_{0}=I_{2}\), and \(p=r=q=1\). We denote simply \(A=A_{1}\), \(B=B_{1}\), and \(C=C_{1}\). For our first three examples, we will have:  
\[R_{x}=\left( \begin{array}{cc}2.0&0.0\\ 0.0&3.0\end{array} \right),\quad\Sigma=I_{2}.\]  
**Example 1**. : _Let \(A\), \(B\), and \(C\) be defined by_  
\[A=\left( \begin{array}{cc}-0.8&0.0\\ -0.5&-a\end{array} \right),\quad B=\left( \begin{array}{cc}-b&0.0\\ -0.5&-0.6\end{array} \right),\quad C=\left( \begin{array}{cc}-a&0.0\\ -0.5&-0.7\end{array} \right),\]  
_where \(a\) and \(b\) are constants. The eigenvalues of \(A(z)\), \(B(z)\), and \(C(z)\) are, respectively, the pairs \((0.8,a)\), \((0.6,b)\), and \((0.7,a)\) so that, whatever \(a\) and \(b\), there is a common eigenvalue for \(A(z)\) and \(C(z)\). Clearly, the model is identifiable except if \(a=b=0.8\) because then the factor \(1-0.8z\) can be simplified on the first row of the system equation. In this case, because of the presence of the symbolic constants \(a\) and \(b\), the algorithm needs to be changed by adding a command Assuming, as shown in Figure 5, with the mention that \(a\) and \(b\) are restricted to the interval \([-1,1]\). It is of course not possible to show here more of the matrix, see Supplementary Materials._  
_Take \(b=0.8\) to simplify the discussion. Then, using Mathematica, it can be seen that the Fisher information matrix has a determinant proportional to \((4-5a)\) with a strictly positive factor, and indeed it is \(0\) if, and only if, \(a=0.8\). If \(a=0.8\), the 2 nd, 6 th, and 10 th rows of the Fisher information matrix contain the fractions_  
\[Row2=\left(\frac{1125}{416},\frac{75}{16},0,0,-\frac{375}{208},-\frac{25}{8},0, 0,-\frac{375}{416},-\frac{25}{16},0,0\right),\]  
\[Row6=\left(-\frac{375}{208},-\frac{25}{8},0,0,\frac{375}{208},\frac{25}{8},0, 0,0,0,0\right),\]  
\[Row10=\left(-\frac{375}{416},-\frac{25}{16},0,0,0,0,0,\frac{375}{416},\frac{25 }{16},0,0\right),\]  
_and it is easy to check that \(-Row2=-Row6-Row10\). Since the asymptotic information matrix is singular, there is no need to show it._  
_As we said, the computations described in Section 3 were implemented as matrix computations in Mathematica with coefficients given in the fractional form (e.g., 8/10 instead of 0.8) to have exact results. The complete Mathematica notebook (Example51.nb) is provided as Supplementary Materials, with its PDF output (Example51.pdf)._  
**Example 2**. : _Let \(A\), \(B\), and \(C\) be defined by:_  
\[A=\left( \begin{array}{cc}0.6&0.2\\ 0.0&0.0\end{array} \right),\quad B=\left( \begin{array}{cc}0.5&0.76\\ 0.0&0.0\end{array} \right),\quad C=\left( \begin{array}{cc}0.8&0.0\\ 0.0&0.0\end{array} \right).\]  
_This example is a generalization of a VARMA model considered by [24] based on an example in [43]. The determinants of \(A(z)\), \(B(z)\), and \(C(z)\) have degree \(1\), so we need to consider the reciprocal matrix polynomials \(\tilde{A}(z)\), \(\tilde{B}(z)\), and \(\tilde{C}(z)\) which have respective roots \((-0.6,0)\), \((-0.5,0)\), and \((-0.8,0)\). Hence, we can consider the singularity of the Fisher information matrix. This is confirmed by  
Figure 5: Partial snapshot of the Mathematica notebook to treat Example 5.1. Note that the integration command is prefixed by a constraining assumption. computation with Mathematica which shows that the matrix has rank 10. Moreover, rows 3, 4, 7, 8, 11, and 12 of the Fisher information matrix contain the fractions_  
\[