Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Multivariate continuous distributions ▷ Multivariate normal distribution ▷ Expectation of a quadratic form

Theorem: Let $X$ follow a multivariate normal distribution:

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

Then, the expectation of the quadratic form $X^\mathrm{T} A X$ is

\[\label{eq:mvn-meanqf} \mathrm{E}\left[ X^\mathrm{T} A X \right] = \mu^\mathrm{T} A \mu + \mathrm{tr}(A \Sigma)\]

where $A$ is an $n \times n$ matrix.

Proof: For any random vector $X \in \mathbb{R}^n$, this expectation is a function of mean and covariance:

\[\label{eq:mean-qf} \mathrm{E}\left[ X^\mathrm{T} A X \right] = \mathrm{E}(X)^\mathrm{T} A \mathrm{X} + \mathrm{tr}(A \, \mathrm{Cov}(X)) \; .\]

For a multivariate normal random vector $X \sim \mathcal{N}(\mu, \Sigma)$, mean and covariance are

\[\label{eq:mvn-mean-cov} \mathrm{E}(X) = \mu \quad \text{and} \quad \mathrm{Cov}(X) = \Sigma \; .\]

Combing \eqref{eq:mean-qf} with \eqref{eq:mvn-mean-cov}, we have:

\[\label{eq:mvn-meanqf-qed} \mathrm{E}\left[ X^\mathrm{T} A X \right] = \mu^\mathrm{T} A \mu + \mathrm{tr}(A \Sigma)\]
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Metadata: ID: P499 | shortcut: mvn-meanqf | author: JoramSoch | date: 2025-05-21, 13:24.