Quadratic form (statistics)

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If ε is a vector of n random variables, and Λ is an n-dimensional symmetric square matrix, then the scalar quantity ε'Λε is known as a quadratic form in ε.

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[edit] Expectation

It can be shown that

\operatorname{E}\left[\epsilon'\Lambda\epsilon\right]=\operatorname{tr}\left[\Lambda \Sigma\right] + \mu'\Lambda\mu

where μ and Σ are the expected value and variance-covariance matrix of ε, respectively, and tr denotes the trace of a matrix. This result only depends on the existence of μ and Σ; in particular, normality of ε is not required.

[edit] Variance

In general, the variance of a quadratic form depends greatly on the distribution of ε. However, if ε does follow a multivariate normal distribution, the variance of the quadratic form becomes particularly tractable. Assume for the moment that Λ is a symmetric matrix. Then,

\operatorname{var}\left[\epsilon'\Lambda\epsilon\right]=2\operatorname{tr}\left[\Lambda \Sigma\Lambda \Sigma\right] + 4\mu'\Lambda\Sigma\Lambda\mu

In fact, this can be generalized to find the covariance between two quadratic forms on the same ε (once again, Λ1 and Λ2 must both be symmetric):

\operatorname{cov}\left[\epsilon'\Lambda_1\epsilon,\epsilon'\Lambda_2\epsilon\right]=2\operatorname{tr}\left[\Lambda _1\Sigma\Lambda_2 \Sigma\right] + 4\mu'\Lambda_1\Sigma\Lambda_2\mu

[edit] Computing the variance in the non-symmetric case

Some texts incorrectly state the above variance or covariance results without enforcing Λ to be symmetric. The case for general Λ can be derived by noting that

ε'Λ'ε = ε'Λε

so

\epsilon'\Lambda\epsilon=\epsilon'\left(\Lambda+\Lambda'\right)\epsilon/2

But this is a quadratic form in the symmetric matrix \tilde{\Lambda}=\left(\Lambda+\Lambda'\right)/2, so the mean and variance expressions are the same, provided Λ is replaced by \tilde{\Lambda} therein.

[edit] Examples of quadratic forms

In the setting where one has a set of observations y and an operator matrix H, then the residual sum of squares can be written as a quadratic form in y:

\textrm{RSS}=y'\left(I-H\right)'\left(I-H\right)y

For procedures where the matrix H is symmetric and idempotent, and the errors are Gaussian with covariance matrix σ2I, RSS / σ2 has a chi-square distribution with k degrees of freedom and noncentrality parameter λ, where

k=\operatorname{tr}\left[\left(I-H\right)'\left(I-H\right)\right]
\lambda=\mu'\left(I-H\right)'\left(I-H\right)\mu/2

may be found by matching the first two central moments of a noncentral chi-square random variable to the expressions given in the first two sections. If Hy estimates μ with no bias, then the noncentrality λ is zero and RSS / σ2 follows a central chi-square distribution.

[edit] See also