Multivariate normal distribution The FIM for a $N$-variate multivariate normal distribution, $X \sim N(\mu(\theta), \Sigma(\theta))$ has a special form. Let the $K$-dimensional vector of parameters be $\theta=\left[\begin{array}{lll}\theta_{1} & \ldots & \theta_{K}\end{array}\right]^{\top}$ and the vector of random normal variables be $X=\left[\begin{array}{lll}X_{1} & \ldots & X_{N}\end{array}\right]^{\top} .$ Assume that the mean values of these random variables are $\mu(\theta)=\left[\mu_{1}(\theta) \quad \ldots \quad \mu_{N}(\theta)\right]^{\top}$, and let $\Sigma(\theta)$ be the covariance matrix. Then, for $1 \leq m, n \leq K$, the $(m, n)$ entry of the FIM is: $^{[16]}$ $$ \mathcal{I}_{m, n}=\frac{\partial \mu^{\top}}{\partial \theta_{m}} \Sigma^{-1} \frac{\partial \mu}{\partial \theta_{n}}+\frac{1}{2} \operatorname{tr}\left(\Sigma^{-1} \frac{\partial \Sigma}{\partial \theta_{m}} \Sigma^{-1} \frac{\partial \Sigma}{\partial \theta_{n}}\right) $$ where $(\cdot)^{\top}$ denotes the transpose of a vector, $\operatorname{tr}(\cdot)$ denotes the trace of a square matrix, and: $$ \begin{aligned} \frac{\partial \mu}{\partial \theta_{m}} &=\left[\begin{array}{lll} \frac{\partial \mu_{1}}{\partial \theta_{m}} & \frac{\partial \mu_{2}}{\partial \theta_{m}} & \cdots & \frac{\partial \mu_{N}}{\partial \theta_{m}} \end{array}\right]^{\top} \\ \frac{\partial \Sigma}{\partial \theta_{m}} &=\left[\begin{array}{cccc} \frac{\partial \Sigma_{1,1}}{\partial \theta_{m}} & \frac{\partial \Sigma_{1,2}}{\partial \theta_{m}} & \cdots & \frac{\partial \Sigma_{1, N}}{\partial \theta_{m}} \\ \frac{\partial \Sigma_{2,1}}{\partial \theta_{m}} & \frac{\partial \Sigma_{2,2}}{\partial \theta_{m}} & \cdots & \frac{\partial \Sigma_{2, N}}{\partial \theta_{m}} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial \Sigma_{N, 1}}{\partial \theta_{m}} & \frac{\partial \Sigma_{N, 2}}{\partial \theta_{m}} & \cdots & \frac{\partial \Sigma_{N, N}}{\partial \theta_{m}} \end{array}\right] \end{aligned} $$
$$\mu(\theta)=\left[\mu_{1}(\theta) \quad \ldots \quad \mu_{N}(\theta)\right]^{\top} \implies \mu: \mathbb{R}^K \rightarrow \mathbb{R}^N.$$
I think it is false that $\mu: \mathbb{R}^K \rightarrow \mathbb{R}^N.$
So, $$\nexists \left[\mu_{1}(\theta) \quad \ldots \quad \mu_{N}(\theta)\right]^{\top} $$
The $\mu$ is not supposed to be a function but rather a vector independent of $\Sigma$ $$ \implies \mu \in \mathbb{R}^N \implies \mu \neq \left[\mu_{1}(\theta) \quad \ldots \quad \mu_{N}(\theta)\right]^{\top} \implies \mu = \left[\mu_{1}\quad \ldots \quad \mu_{N}\right]^{\top}. $$