# Normal distribution for a high dimensional case

Let $X \sim N(\mu,\Sigma)$ be a Gaussian vector of dimension $d$, where $\Sigma$ is known but $\mu$ is unknown. Hence $\theta =\mu \in \mathbb R^d$.

For $X\in \mathbb R$ we get:

$$p(x_i;\theta)=\frac{1}{\sqrt{2\pi \Sigma^2}}e^{-\frac{(x_i-\mu)^2}{2\Sigma^2}}$$

From my lecture notes for $X\in \mathbb R^d$ we get:

$$p(x_i;\theta)=\frac{1}{\sqrt{2\pi^d |\Sigma|}}\exp\left\{-\frac{1}{2}(x_i-\mu)^T \Sigma^{-1}(x_i-\mu)\right\}$$

How do I expand $\frac{1}{\sqrt{2\pi \Sigma^2}}e^{-\frac{(x_i-\mu)^2}{2\Sigma^2}}$ to $\frac{1}{\sqrt{2\pi^d |\Sigma|}}\exp\left\{-\frac{1}{2}(x_i-\mu)^T \Sigma^{-1}(x_i-\mu)\right\}$ ?

If $x \in \mathbb{R}^d$, thus $(x-\mu)'\Sigma^{-1}(x - \mu)$ is a quadratic form, i.e., denote $W = \Sigma^{-1}$ hence \begin{align} (x-\mu)'W(x - \mu) &= \sum_{i=1}^d \sum_{j=1}^d w_{ij} (x_i - \mu)(x_j - \mu)\\ &= 2\sum_{i > j}^d w_{ij} (x_i - \mu)(x_j - \mu) + \sum_{i=1}^d w_{ii}(x_i - \mu)^2. \end{align}