Normal distribution notation I am wondering... is saying
$\mathcal{N}\left(0,\begin{bmatrix} 0.1 & 0.02 \\ 0.02 & 0.3 \end{bmatrix}\right)$ equivalent to 
$\mathcal{N}\left(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 0.1 & 0.02 \\ 0.02 & 0.3 \end{bmatrix}\right)$ 
 A: The multivariate normal distribution in $\mathbb{R}^{d}$ is denoted by $\mathcal{N}(\mu,\Sigma)$ where $\mu \in \mathbb{R}^{d}$ and $\Sigma \in \mathcal{M}_{d}(\mathbb{R})$ is as symmetric positive matrix. The associated probability density function is :
$$ x \in \mathbb{R}^{d} \, \longmapsto \, \frac{1}{(\sqrt{2\pi})^d} \frac{1}{\sqrt{\det(\Sigma)}} \exp \Big( -\frac{1}{2} (x-\mu)^{\top} \Sigma^{-1} (x-\mu) \Big) $$
As a consequence, only $\mathcal{N}\Big( \begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0.1 & 0.02 \\ 0.02 & 0.3 \end{bmatrix} \Big)$ makes sense, since the mean must be a vector in $\mathbb{R}^{2}$.
A: If this is supposed to represent a bivariate normal distribution then the mean should be a $1\times 2$ vector and the covariance matrix should be symmetric, such as   $\mathcal{N}\left(\begin{bmatrix} 0 \\ 0  \end{bmatrix},\begin{bmatrix} 0.1 & 0.02 \\ 0.02 & 0.3 \end{bmatrix}\right)$  
A: In mathematics, $0$ is often used to mean the zero vector in any vector space. In this case, the covariance matrix is a $2\times 2$ matrix, then the mean is a vector in $\mathbb{R}^2$, so the zero vector is $\begin{bmatrix}0\\0\end{bmatrix}$.
