Quadratic form (statistics)


In multivariate statistics, if is a vector of random variables, and is an -dimensional symmetric matrix, then the scalar quantity is known as a quadratic form in.

Expectation

It can be shown that
where and are the expected value and variance-covariance matrix of, respectively, and tr denotes the trace of a matrix. This result only depends on the existence of and ; in particular, normality of is not required.
A book treatment of the topic of quadratic forms in random variables is that of Mathai and Provost.

Proof

Since the quadratic form is a scalar quantity,.
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

Variance in the Gaussian case

In general, the variance of a quadratic form depends greatly on the distribution of. However, if does follow a multivariate normal distribution, the variance of the quadratic form becomes particularly tractable. Assume for the moment that is a symmetric matrix. Then,
In fact, this can be generalized to find the covariance between two quadratic forms on the same :

Computing the variance in the non-symmetric case

Some texts incorrectly state that the above variance or covariance results hold without requiring to be symmetric. The case for general can be derived by noting that
so
is a quadratic form in the symmetric matrix, so the mean and variance expressions are the same, provided is replaced by therein.

Examples of quadratic forms

In the setting where one has a set of observations and an operator matrix, then the residual sum of squares can be written as a quadratic form in :
For procedures where the matrix is symmetric and idempotent, and the errors are Gaussian with covariance matrix, has a chi-squared distribution with degrees of freedom and noncentrality parameter, 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 estimates with no bias, then the noncentrality is zero and follows a central chi-squared distribution.