**Appendix A. Fisher Information for Multivariate Gaussian**

Here, we consider the structure of the Fisher information matrix for *n*-dimensional multivariate Gaussian with density given by

$$f(\mathbf{x}\_{\text{n}} : \boldsymbol{\mu}\_{\text{n}}, \boldsymbol{\Sigma}) = 2\pi^{-\frac{\mu}{2}} \det(\boldsymbol{\Sigma})^{-\frac{1}{2}} \exp - \frac{(\boldsymbol{X} - \boldsymbol{\mu})^{T} \boldsymbol{\Sigma}^{-1} (\boldsymbol{X} - \boldsymbol{\mu})}{2} \tag{A1}$$

where *X* is a data vector, *μ* = [*μ*1, *μ*2, ..., *μn*] is the *n*-dimensional mean vector of the distribution and Σ is the *n* × *n* covariance matrix.

These define an *n*-dimensional multivariate Gaussian. This distribution has (*n*+3)*<sup>n</sup>* 2 unique parameters, which we will capture as a single vector *θ* such that

$$\theta = \{ \mu\_1, \mu\_2, \dots \mu\_n, \sigma\_{1,1'}^2, \sigma\_{1,2'}^2, \dots, \sigma\_{n,n}^2 \}. \tag{A2}$$

2

The log-likelihood associated with Equation (A1) is

$$\log f = L(\theta) = \frac{n}{2} \ln 2\pi - \frac{1}{2} \ln \det \Sigma - \frac{1}{2} (\mathbf{x} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} (\mathbf{x} - \boldsymbol{\mu}). \tag{A3}$$

As stated, we will find an equation that will yield each element of the Fisher information matrix using the following definition

$$\log g\_{i\rangle}(\theta) = E\left[\frac{\partial}{\partial \theta\_i} \log f(\mathbf{x}; \theta) \frac{\partial}{\partial \theta\_j} \log f(\mathbf{x}; \theta)\right]. \tag{A4}$$

Equation (A4) requires the partial derivative of each parameter in the log-likelihood defined in Equation (A3).

$$\begin{split} \frac{\partial L}{\partial \theta\_i} &= -\frac{1}{2} \text{tr} \left[ \boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_i} \right] + \frac{1}{2} (\boldsymbol{x} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_i} \boldsymbol{\Sigma}^{-1} (\boldsymbol{x} - \boldsymbol{\mu}) \\ &+ \frac{\partial \boldsymbol{\mu}}{\partial \theta\_i}^T \boldsymbol{\Sigma}^{-1} (\boldsymbol{x} - \boldsymbol{\mu}) \\ &\mathbb{C} \end{split} \tag{A5}$$

Similarly, we can take the partial derivative with respect to a different parameter, *θ<sup>j</sup>* and obtain the same result, indexed with *j* instead of *i*. The below equation labels *A*.*B*, *C* will provide clarity in the proof.

To find each *gij* in the Fisher information matrix, we need to find the expectation of the product of every combination two partial derivatives which will result in nine terms. However, upon taking the expectation, some of these terms will vanish to 0, because the expectation of data vector *x* approaches the mean vector *μ*. Specifically, let us denote *Ai*, *Bi*, *Ci* to be the terms of *<sup>∂</sup><sup>L</sup> ∂θ<sup>i</sup>* and *Aj*, *Bj*, *Cj* to be the terms of *<sup>∂</sup><sup>L</sup> ∂θ<sup>j</sup>* . Upon taking the expectation of the product, *AiCj* = *CiAj* = *BiCj* = *BjCi* = 0. Ignoring these, we will look individually at each of the remaining terms. Starting with *AiAj*

$$\begin{split} A\_i A\_j &= \left( -\frac{1}{2} tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_i} \right] \right) \left( -\frac{1}{2} tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_j} \right] \right) \\ &= \frac{1}{4} tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_i} \right] tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_j} \right] \end{split} \tag{A6}$$

Next, calculating *BiBj*,

$$\begin{split} B\_{i}B\_{j} &= \left[\frac{1}{2}(\boldsymbol{\chi}-\boldsymbol{\mu})^{T}\boldsymbol{\Sigma}^{-1}\frac{\partial\boldsymbol{\Sigma}}{\partial\theta\_{i}}\boldsymbol{\Sigma}^{-1}(\boldsymbol{\chi}-\boldsymbol{\mu})\right] \left[\frac{1}{2}(\boldsymbol{\chi}-\boldsymbol{\mu})^{T}\boldsymbol{\Sigma}^{-1}\frac{\partial\boldsymbol{\Sigma}}{\partial\theta\_{j}}\boldsymbol{\Sigma}^{-1}(\boldsymbol{\chi}-\boldsymbol{\mu})\right] \\ &= \frac{1}{4}\left[(\boldsymbol{\chi}-\boldsymbol{\mu})^{T}\boldsymbol{\Sigma}^{-1}\frac{\partial\boldsymbol{\Sigma}}{\partial\theta\_{i}}\boldsymbol{\Sigma}^{-1}(\boldsymbol{\chi}-\boldsymbol{\mu})(\boldsymbol{\chi}-\boldsymbol{\mu})^{T}\boldsymbol{\Sigma}^{-1}\frac{\partial\boldsymbol{\Sigma}}{\partial\theta\_{j}}\boldsymbol{\Sigma}^{-1}(\boldsymbol{\chi}-\boldsymbol{\mu})\right] \end{split} \tag{A7}$$

We are required to take the expectation of this, which is a fourth moment of the multivariable normal distribution. The result of this is

$${}^{B}B\_{j} = \frac{1}{4} \left[ tr \left( \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{i}} \right) tr \left( \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{j}} \right) \right] + 2tr \left[ \left( \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{i}} \right) \left( \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{j}} \right) \right]. \tag{A8}$$

Turning attention to *CiCj*,

$$\begin{split} \mathbb{C}\_{i}\mathbb{C}\_{j} &= \frac{\partial\mu}{\partial\theta\_{i}}^{T}\Sigma^{-1}(\boldsymbol{x}-\boldsymbol{\mu})(\boldsymbol{x}-\boldsymbol{\mu})^{T}\Sigma^{-1}\frac{\partial\mu}{\partial\theta\_{j}} \\ &= \frac{\partial\mu}{\partial\theta\_{i}}^{T}\Sigma^{-1}\Sigma\Sigma^{-1}\frac{\partial\mu}{\partial\theta\_{j}} \\ &= \frac{\partial\mu}{\partial\theta\_{i}}^{T}\Sigma^{-1}\frac{\partial\mu}{\partial\theta\_{j}}. \end{split} \tag{A9}$$

The final set of non-vanishing terms, *AiBj* + *BiAj* are considered simultaneously.

$$\begin{split} A\_i B\_j + B\_i A\_j &= \left( -\frac{1}{2} tr \left[ \boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_i} \right] \right) \left( \frac{1}{2} (\boldsymbol{x} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_j} \boldsymbol{\Sigma}^{-1} (\boldsymbol{x} - \boldsymbol{\mu}) \right) \\ &+ \left( \frac{1}{2} (\boldsymbol{x} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_i} \boldsymbol{\Sigma}^{-1} (\boldsymbol{x} - \boldsymbol{\mu}) \right) \left( -\frac{1}{2} tr \left[ \boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_j} \right] \right) \\ &= -\frac{1}{4} \left\{ \left( tr \left[ \boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_i} \right] \right) \left( (\boldsymbol{x} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_j} \boldsymbol{\Sigma}^{-1} (\boldsymbol{x} - \boldsymbol{\mu}) \right) \\ &+ \left( (\boldsymbol{x} - \boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_i} \boldsymbol{\Sigma}^{-1} (\boldsymbol{x} - \boldsymbol{\mu}) \right) \left( tr \left[ \boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_j} \right] \right) \right\} \end{split} \tag{A10}$$

Now, we use the identity

$$b^T A b = \text{tr}(b b^T A) \tag{A11}$$

on all terms of Equation (A10) that do not yet involve the trace of a matrix. Doing so, Equation (A10) becomes

$$\begin{split} A\_{i}B\_{j} + B\_{i}A\_{j} &= -\frac{1}{4} \left\{ \left( tr\left[\Sigma^{-1}\frac{\partial\Sigma}{\partial\theta\_{i}}\right] \right) \left( tr\left[(x-\mu)(x-\mu)^{T}\Sigma^{-1}\frac{\partial\Sigma}{\partial\theta\_{j}}\Sigma^{-1}\right] \right) \\ &+ \left( tr\left[(x-\mu)(x-\mu)^{T}\Sigma^{-1}\frac{\partial\Sigma}{\partial\theta\_{i}}\Sigma^{-1}\right] \right) \left( tr\left[\Sigma^{-1}\frac{\partial\Sigma}{\partial\theta\_{j}}\right] \right) \right\} \\ &= -\frac{1}{4} \left\{ \left( tr\left[\Sigma^{-1}\frac{\partial\Sigma}{\partial\theta\_{i}}\right] \right) \left( tr\left[\Sigma\Sigma^{-1}\frac{\partial\Sigma}{\partial\theta\_{j}}\Sigma^{-1}\right] \right) \right. \\ &\left. + \left( tr\left[\Sigma\Sigma^{-1}\frac{\partial\Sigma}{\partial\theta\_{i}}\Sigma^{-1}\right] \right) \left( tr\left[\Sigma^{-1}\frac{\partial\Sigma}{\partial\theta\_{j}}\right] \right) \right\} \\ &= -\frac{1}{4} \left\{ \left( tr\left[\Sigma^{-1}\frac{\partial\Sigma}{\partial\theta\_{i}}\right] \right) \left( tr\left[\frac{\partial\Sigma}{\partial\theta\_{j}}\Sigma^{-1}\right] \right) \right. \\ &\left. + \left( tr\left[\frac{\partial\Sigma}{\partial\theta\_{i}}\Sigma^{-1}\right] \right) \right) \left( tr\left[\Sigma^{-1}\frac{\partial\Sigma}{\partial\theta\_{j}}\right] \right) \right\}. \end{split} \tag{A12}$$

Once again, we took the expectation as required to find the Fisher information. Finally, we use the commutative property of trace to clean up the expression in Equation (A12)

$$\begin{split} A\_{i}B\_{j} + B\_{i}A\_{j} &= -\frac{1}{4} \Biggl\{ \left( tr\left[\boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_{l}}\right] \right) \left( tr\left[\boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_{j}}\right] \right) + \left( tr\left[\boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_{l}}\right] \right) \left( tr\left[\boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_{l}}\right] \right) \right\} \\ &= -\frac{1}{2} \left( tr\left[\boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_{l}}\right] \right) \left( tr\left[\boldsymbol{\Sigma}^{-1} \frac{\partial \boldsymbol{\Sigma}}{\partial \theta\_{l}}\right] \right) . \end{split} \tag{A13}$$

Combining Equations (A6), (A7, (A9 and (A13) into Equation (A5), we obtain

$$\begin{split} g\_{i,j}(\theta) &= \frac{1}{4} tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{i}} \right] tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{j}} \right] \\ &+ \frac{1}{4} \left[ tr \left( \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{i}} \right) tr \left( \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{j}} \right) \right] + 2tr \left[ \left( \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{i}} \right) \left( \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{j}} \right) \right] \\ &+ \frac{\partial \mu^{T}}{\partial \theta\_{i}} \Sigma^{-1} \frac{\partial \mu}{\partial \theta\_{j}} \\ &+ -\frac{1}{2} \left( tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{i}} \right] \right) \left( tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{j}} \right] \right) \\ &= \frac{1}{2} tr \left[ \left( \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{i}} \right) \left( \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta\_{j}} \right) \right] + \frac{\partial \mu^{T}}{\partial \theta\_{i}} \Sigma^{-1} \frac{\partial \mu}{\partial \theta\_{j}}. \end{split} \tag{A14}$$

#### **Appendix B. Fisher Information of a 2-Dimensional Gaussian**

The usefulness of Equation (A14) lies in the ability of calculating the individual terms in the equation. Here, the inverse of a covariance matrix is extremely illusive for a highdimensional Gaussian distribution, even after leveraging its symmetric properties.

Considering just a 2 × 2 covariance matrix, Equation (A14) is tractable, since its inverse is known exactly and is reasonably manageable. Furthermore, we will collect all the parameters of a general bivariate Gaussian into a single vector, to facilitate the calculation of each element of the Fisher information matrix.

$$\begin{aligned} \theta &= \left[ \theta\_1, \theta\_1, \theta\_1, \theta\_1, \theta\_1 \right] \\ &= \left[ \mu\_{1\prime} \mu\_{2\prime} \sigma\_{1\prime}^2 \sigma\_{2\prime}^2 \sigma\_{12} \right] \end{aligned} \tag{A15}$$

Starting with the diagonal elements, *g*<sup>11</sup> and *g*<sup>22</sup> share similar structures. Focusing just on *g*11, and consider a 2-dimensional Gaussian with mean vector *μ<sup>T</sup>* = [*μ*1, *μ*2] and covariance matrix

$$
\Sigma = \begin{pmatrix} \sigma\_1^2 & \sigma\_{12} \\ \sigma\_{12} & \sigma\_2^2 \end{pmatrix},
$$

which will be indexed according to Equation (A15). We now have, using the standard definition of the inverse

$$
\Sigma^{-1} = \frac{1}{\sigma\_1^2 \sigma\_2^2 - \sigma\_{12}^2} \begin{pmatrix} \sigma\_2^2 & -\sigma\_{12} \\ -\sigma\_{12} & \sigma\_1^2 \end{pmatrix}.
$$

For succinctness, we will let *k* = <sup>1</sup> *σ*2 <sup>1</sup> *<sup>σ</sup>*<sup>2</sup> 2−*σ*<sup>2</sup> 12 .

Conveniently, the means are not involved in the covariance matrix, so the first term of Equation (A14) vanishes. To find *g*<sup>11</sup> we need

$$\begin{split} g\_{11} &= \frac{1}{k} \left[ \frac{\partial \mu}{\partial \mu\_1}^T \begin{pmatrix} \sigma\_2^2 & -\sigma\_{12} \\ -\sigma\_{12} & \sigma\_1^2 \end{pmatrix} \frac{\partial \mu}{\partial \mu\_1} \right] \\ &= \frac{1}{k} \left[ \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} \sigma\_2^2 & -\sigma\_{12} \\ -\sigma\_{12} & \sigma\_1^2 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right] \\ &= \frac{\sigma\_2^2}{k} \\ &= \frac{\sigma\_2^2}{\sigma\_1^2 \sigma\_2^2 - \sigma\_{12}^2} .\end{split} \tag{A16}$$

Finding *g*<sup>22</sup> easily follows the above, resulting in

$$g\_{22} = \frac{\sigma\_1^2}{\sigma\_1^2 \sigma\_2^2 - \sigma\_{12}^2}. \tag{A17}$$

Finding the remaining diagonal elements involve the just the first term of Equation (A14). For the variance of the first variable, we will need to find

$$\begin{split} g\_{33} &= \frac{1}{2} \left( tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \sigma\_1^2} \right]^2 \right) \\ &= \frac{1}{2} \left( tr \left[ \frac{1}{k} \begin{pmatrix} \sigma\_2^2 & -\sigma\_{12} \\ -\sigma\_{12} & \sigma\_1^2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right]^2 \right) \\ &= \frac{1}{2k^2} \left( tr \left[ \begin{pmatrix} \sigma\_2^2 & 0 \\ -\sigma\_{12} & 0 \end{pmatrix} \right]^2 \right) \\ &= \frac{1}{2k^2} \left( tr \left[ \begin{pmatrix} (\sigma\_2^2)^2 & 0 \\ (\sigma\_{12} \sigma\_2^2)^2 & 0 \end{pmatrix} \right] \right) \\ &= \frac{1}{2k^2} (\sigma\_2^2)^2 \\ &= \frac{1}{2} \left( \frac{\sigma\_2^2}{\sigma\_1^2 \sigma\_2^2 - \sigma\_{12}^2} \right)^2 . \end{split} \tag{A18}$$

Once again, the element of the Fisher information matrix for the variance of the second variable mirrors the above exactly.

$$\mathbf{g}\_{44} = \frac{1}{2} \left( \frac{\sigma\_1^2}{\sigma\_1^2 \sigma\_2^2 - \sigma\_{12}^2} \right)^2. \tag{A19}$$

The covariance component will complete the diagonal elements of the Fisher information matrix.

$$\begin{split} g\_{55} &= \frac{1}{2} \left( tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \sigma\_{12}} \right]^2 \right) \\ &= \frac{1}{2} \left( tr \left[ \frac{1}{k} \begin{pmatrix} \sigma\_2^2 & -\sigma\_{12} \\ -\sigma\_{12} & \sigma\_1^2 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right]^2 \right) \\ &= \frac{1}{2k^2} \left( tr \left[ \begin{pmatrix} -\sigma\_{12} & \sigma\_2^2 \\ \sigma\_1^2 & -\sigma\_{12} \end{pmatrix} \right]^2 \right) \\ &= \frac{1}{2k^2} \left( tr \left[ \begin{pmatrix} \sigma\_2^2 \sigma\_2^2 + \sigma\_{12}^2 & 2\sigma\_2^2 \sigma\_{12} \\ 2\sigma\_1^2 \sigma\_{12} & \sigma\_1^2 \sigma\_2^2 + \sigma\_{12}^2 \end{pmatrix} \right] \right) \\ &= \frac{1}{2k^2} 2(\sigma\_1^2 \sigma\_2^2 + \sigma\_{12}^2) \\ &= \frac{\sigma\_1^2 \sigma\_2^2 + \sigma\_{12}^2}{(\sigma\_1^2 \sigma\_2^2 - \sigma\_{12}^2)^2} . \end{split} \tag{A20}$$

The off-diagonal elements are only slightly more involved. However, because the terms in Equation (A14) involve the partial derivatives, and because the mean vector and the covariance matrix have no overlapping terms, many of the off-diagonal elements vanish, specifically the ones that involve both a mean component and a variance component, i.e., *gij* = 0 for *i* ∈ (1, 2) and *j* ∈ (3, 4, 5). For the other off-diagonal components, we will employ all the conveniences of symmetry to complete the Fisher information matrix.

Turning our attention to the *g*12, the element concerning the two means:

$$\begin{split} \mathcal{g}\_{12} &= \frac{\partial \mu}{\partial \theta\_1}^T \Sigma^{-1} \frac{\partial \mu}{\partial \theta\_2} \\ &= \frac{1}{k} \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} \sigma\_2^2 & -\sigma\_{12} \\ -\sigma\_{12} & \sigma\_1^2 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ &= -\frac{1}{k} \sigma\_{12} \\ &= -\frac{\sigma\_{12}}{\sigma\_1^2 \sigma\_2^2 - \sigma\_{12}^2} = \mathcal{g}\_{21}. \end{split} \tag{A21}$$

Next, we consider the elements of the Fisher information matrix involving both variances, *g*<sup>34</sup>

$$\begin{split} \mathcal{R}34 &= \frac{1}{2} \left( tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \sigma\_1^2} \Sigma^{-1} \frac{\partial \Sigma}{\partial \sigma\_2^2} \right] \right) \\ &= \frac{1}{2k^2} \left( tr \left[ \begin{pmatrix} \sigma\_2^2 & -\sigma\_{12} \\ -\sigma\_{12} & \sigma\_1^2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} \sigma\_2^2 & -\sigma\_{12} \\ -\sigma\_{12} & \sigma\_1^2 \end{pmatrix} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right] \right) \\ &= \frac{1}{2k^2} \left( tr \left[ \begin{pmatrix} \sigma\_2^2 & 0 \\ -\sigma\_{12} & 0 \end{pmatrix} \begin{pmatrix} 0 & -\sigma\_{12} \\ 0 & \sigma\_1^2 \end{pmatrix} \right] \right) \\ &= \frac{1}{2k^2} \left( tr \left[ \begin{pmatrix} 0 & -\sigma\_2^2 \sigma\_{12} \\ 0 & \sigma\_{12}^2 \end{pmatrix} \right] \right) \\ &= \frac{1}{2} \left( \frac{\sigma\_{12}}{\sigma\_1^2 \sigma\_2^2 - \sigma\_{12}^2} \right)^2 = \mathfrak{g} \mathfrak{a} \cdot \text{d} \end{split} \tag{A22}$$

The variance/covariance elements of the Fisher information matrix will all have similar structures. We calculate one of them below

$$\begin{split} g\_{35} &= \frac{1}{2} \left( tr \left[ \Sigma^{-1} \frac{\partial \Sigma}{\partial \sigma\_1^2} \Sigma^{-1} \frac{\partial \Sigma}{\partial \sigma\_{12}} \right] \right) \\ &= \frac{1}{2k^2} \left( tr \left[ \begin{pmatrix} \sigma\_2^2 & -\sigma\_{12} \\ -\sigma\_{12} & \sigma\_1^2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} \sigma\_2^2 & -\sigma\_{12} \\ -\sigma\_{12} & \sigma\_1^2 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \right] \right) \\ &= \frac{1}{2k^2} \left( tr \left[ \begin{pmatrix} \sigma\_2^2 & 0 \\ -\sigma\_{12} & 0 \end{pmatrix} \begin{pmatrix} -\sigma\_{12} & \sigma\_2^2 \\ \sigma\_1^2 & -\sigma\_{12} \end{pmatrix} \right] \right) \\ &= \frac{1}{2k^2} \left( tr \left[ \begin{pmatrix} -\sigma\_{12}\sigma\_2^2 & \sigma\_{12}^2 \\ \sigma\_{12}^2 & -\sigma\_{12}\sigma\_2^2 \end{pmatrix} \right] \right) \\ &= -\frac{\sigma\_{12}\sigma\_2^2}{\left( \sigma\_1^2 \sigma\_2^2 - \sigma\_{12}^2 \right)^2} = g\_{53} \text{s} \end{split} \tag{A23}$$

Similarly, the element involving the second variance with the covariance is

$$\mathfrak{g}\_{45} = \mathfrak{g}\_{54} = -\frac{\sigma\_{12}\sigma\_{1}^{2}}{\left(\sigma\_{1}^{2}\sigma\_{2}^{2} - \sigma\_{12}^{2}\right)^{2}}.\tag{A24}$$
