Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Multivariate continuous distributions ▷ Multivariate normal distribution ▷ Relationship to chi-squared distribution

Theorem: Let $x$ be an $n \times 1$ random vector following a multivariate normal distribution with zero mean and arbitrary covariance matrix $\Sigma$:

\[\label{eq:mvn} x \sim \mathcal{N}(0, \Sigma) \; .\]

Then, the quadratic form of $x$, weighted by $\Sigma$, follows a chi-squared distribution with $n$ degrees of freedom:

\[\label{eq:mvn-chi2} y = x^\mathrm{T} \Sigma^{-1} x \sim \chi^2(n) \; .\]

Proof: Define a new random vector $z$ as

\[\label{eq:z} z = \Sigma^{-1/2} x \; .\]

where $\Sigma^{-1/2}$ is the matrix square root of $\Sigma$. This matrix must exist, because $\Sigma$ is a covariance matrix and thus positive semi-definite. Due to the linear transformation theorem, $z$ is distributed as

\[\label{eq:z-dist} \begin{split} z &\sim \mathcal{N}\left( \Sigma^{-1/2} 0, \, \Sigma^{-1/2} \Sigma \, {\Sigma^{-1/2}}^\mathrm{T} \right) \\ &\sim \mathcal{N}\left( \Sigma^{-1/2} 0, \, \Sigma^{-1/2} \Sigma^{1/2} \Sigma^{1/2} \Sigma^{-1/2} \right) \\ &\sim \mathcal{N}(0, I_n) \; , \end{split}\]

i.e. each entry of this vector follows a standard normal distribution:

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

We further observe that $y$ can be represented in terms of $z$

\[\label{eq:y-z} y = x^\mathrm{T} \Sigma^{-1} x = \left( x^\mathrm{T} \Sigma^{-1/2} \right) \left( \Sigma^{-1/2} x \right) = z^\mathrm{T} z \; ,\]

thus $z$ is a sum of $n$ squared standard normally distributed random variables

\[\label{eq:y-z-sum} y = \sum_{i=1}^{n} z_i^2 \quad \text{where all} \quad z_i \sim \mathcal{N}(0, 1)\]

which, by definition, is chi-squared distributed with $n$ degrees of freedom:

\[\label{eq:mvn-chi2-qed} y \sim \chi^2(n) \; .\]
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Metadata: ID: P395 | shortcut: mvn-chi2 | author: JoramSoch | date: 2022-12-20, 16:27.