Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Normal distribution ▷ Special case of multivariate normal distribution

Theorem: The normal distribution is a special case of the multivariate normal distribution with number of variables $n = 1$, i.e. random vector $x \in \mathbb{R}$, mean $\mu \in \mathbb{R}$ and covariance matrix $\Sigma = \sigma^2$.

Proof: The probability density function of the multivariate normal distribution is

\[\label{eq:mvn-pdf} \mathcal{N}(x; \mu, \Sigma) = \frac{1}{\sqrt{(2 \pi)^n |\Sigma|}} \cdot \exp \left[ -\frac{1}{2} (x-\mu)^\mathrm{T} \Sigma^{-1} (x-\mu) \right] \; .\]

Setting $n = 1$, such that $x, \mu \in \mathbb{R}$, and $\Sigma = \sigma^2$, we obtain

\[\label{eq:norm-pdf} \begin{split} \mathcal{N}(x; \mu, \sigma^2) &= \frac{1}{\sqrt{(2 \pi)^1 |\sigma^2|}} \cdot \exp \left[ -\frac{1}{2} (x-\mu)^\mathrm{T} (\sigma^2)^{-1} (x-\mu) \right] \\ &= \frac{1}{\sqrt{(2\pi) \sigma^2}} \cdot \exp\left[-\frac{1}{2 \sigma^2} (x-\mu)^2 \right] \\ &= \frac{1}{\sqrt{2\pi} \sigma} \cdot \exp \left[ -\frac{1}{2} \left( \frac{x-\mu}{\sigma} \right)^2 \right] \end{split}\]

which is equivalent to the probability density function of the normal distribution.

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Metadata: ID: P331 | shortcut: norm-mvn | author: JoramSoch | date: 2022-08-19, 19:41.