From Handwiki In multivariate statistics, if [math]\displaystyle{ \varepsilon }[/math] is a vector of [math]\displaystyle{ n }[/math] random variables, and [math]\displaystyle{ \Lambda }[/math] is an [math]\displaystyle{ n }[/math]-dimensional symmetric matrix, then the scalar quantity [math]\displaystyle{ \varepsilon^T\Lambda\varepsilon }[/math] is known as a quadratic form in [math]\displaystyle{ \varepsilon }[/math].
It can be shown that[1]
where [math]\displaystyle{ \mu }[/math] and [math]\displaystyle{ \Sigma }[/math] are the expected value and variance-covariance matrix of [math]\displaystyle{ \varepsilon }[/math], respectively, and tr denotes the trace of a matrix. This result only depends on the existence of [math]\displaystyle{ \mu }[/math] and [math]\displaystyle{ \Sigma }[/math]; in particular, normality of [math]\displaystyle{ \varepsilon }[/math] is not required.
A book treatment of the topic of quadratic forms in random variables is that of Mathai and Provost.[2]
Since the quadratic form is a scalar quantity, [math]\displaystyle{ \varepsilon^T\Lambda\varepsilon = \operatorname{tr}(\varepsilon^T\Lambda\varepsilon) }[/math].
Next, by the cyclic property of the trace operator,
Since the trace operator is a linear combination of the components of the matrix, it therefore follows from the linearity of the expectation operator that
A standard property of variances then tells us that this is
Applying the cyclic property of the trace operator again, we get
In general, the variance of a quadratic form depends greatly on the distribution of [math]\displaystyle{ \varepsilon }[/math]. However, if [math]\displaystyle{ \varepsilon }[/math] does follow a multivariate normal distribution, the variance of the quadratic form becomes particularly tractable. Assume for the moment that [math]\displaystyle{ \Lambda }[/math] is a symmetric matrix. Then,
In fact, this can be generalized to find the covariance between two quadratic forms on the same [math]\displaystyle{ \varepsilon }[/math] (once again, [math]\displaystyle{ \Lambda_1 }[/math] and [math]\displaystyle{ \Lambda_2 }[/math] must both be symmetric):
In addition, a quadratic form such as this follows a generalized chi-squared distribution.
The case for general [math]\displaystyle{ \Lambda }[/math] can be derived by noting that
so
is a quadratic form in the symmetric matrix [math]\displaystyle{ \tilde{\Lambda}=\left(\Lambda+\Lambda^T\right)/2 }[/math], so the mean and variance expressions are the same, provided [math]\displaystyle{ \Lambda }[/math] is replaced by [math]\displaystyle{ \tilde{\Lambda} }[/math] therein.
In the setting where one has a set of observations [math]\displaystyle{ y }[/math] and an operator matrix [math]\displaystyle{ H }[/math], then the residual sum of squares can be written as a quadratic form in [math]\displaystyle{ y }[/math]:
For procedures where the matrix [math]\displaystyle{ H }[/math] is symmetric and idempotent, and the errors are Gaussian with covariance matrix [math]\displaystyle{ \sigma^2I }[/math], [math]\displaystyle{ \textrm{RSS}/\sigma^2 }[/math] has a chi-squared distribution with [math]\displaystyle{ k }[/math] degrees of freedom and noncentrality parameter [math]\displaystyle{ \lambda }[/math], where
may be found by matching the first two central moments of a noncentral chi-squared random variable to the expressions given in the first two sections. If [math]\displaystyle{ Hy }[/math] estimates [math]\displaystyle{ \mu }[/math] with no bias, then the noncentrality [math]\displaystyle{ \lambda }[/math] is zero and [math]\displaystyle{ \textrm{RSS}/\sigma^2 }[/math] follows a central chi-squared distribution.
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Categories: [Statistical theory] [Quadratic forms]