Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Multivariate continuous distributions ▷ Multivariate normal distribution ▷ Drawing samples

Theorem: Let $X \in \mathbb{R}^n$ be a random vector with all entries independently following a standard normal distribution. Moreover, let $A$ be an $n \times n$ matrix, such that $A A^\mathrm{T} = \Sigma$.

Then, $Y = \mu + A X$ follows a multivariate normal distribution with mean $\mu$ and covariance $\Sigma$:

\[\label{eq:mvn-samp} Y = \mu + A X \sim \mathcal{N}(\mu, \Sigma) \; .\]

Proof: If all entries of $X$ are independent and standard normally distributed

\[\label{eq:Xi-dist} X_i \overset{\mathrm{i.i.d.}}{\sim} \mathcal{N}(0, 1) \quad \text{for all} \quad i = 1,\ldots,n \; ,\]

this implies a multivariate normal distribution with diagonal covariance matrix:

\[\label{eq:X-dist} X \sim \mathcal{N}\left( 0_n, I_n \right) \; .\]

Thus, with the linear transformation theorem for the multivariate normal distribution, it follows that

\[\label{eq:mvn-samp-qed} \begin{split} Y = \mu + A X &\sim \mathcal{N}\left(A 0_n + \mu, A I_n A^\mathrm{T} \right) \\ &\sim \mathcal{N}\left(\mu, A A^\mathrm{T} \right) \\ &\sim \mathcal{N}\left(\mu, \Sigma \right) \; . \end{split}\]

Thus, given $X$ defined by \eqref{eq:Xi-dist}, $Y$ defined by \eqref{eq:mvn-samp} is a sample from $\mathcal{N}\left( \mu, \Sigma \right)$.

∎
Sources:

Metadata: ID: P501 | shortcut: mvn-samp | author: JoramSoch | date: 2025-05-21, 16:14.