If $epsilon$ is a vector of $n$ random variables, and $Lambda$ is an $n$-dimensional symmetric square matrix, then the scalar quantity $epsilon\text{'}Lambdaepsilon$ is known as a quadratic form in $epsilon$.

Expectation

It can be shown that

:$operatorname\left\{E\right\}left \left[epsilon\text{'}Lambdaepsilon ight\right] =operatorname\left\{tr\right\}left \left[Lambda Sigma ight\right] + mu\text{'}Lambdamu$

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

Variance

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

:$operatorname\left\{var\right\}left \left[epsilon\text{'}Lambdaepsilon ight\right] =2operatorname\left\{tr\right\}left \left[Lambda SigmaLambda Sigma ight\right] + 4mu\text{'}LambdaSigmaLambdamu$

In fact, this can be generalized to find the covariance between two quadratic forms on the same $epsilon$ (once again, $Lambda_1$ and $Lambda_2$ must both be symmetric):

:$operatorname\left\{cov\right\}left \left[epsilon\text{'}Lambda_1epsilon,epsilon\text{'}Lambda_2epsilon ight\right] =2operatorname\left\{tr\right\}left \left[Lambda _1SigmaLambda_2 Sigma ight\right] + 4mu\text{'}Lambda_1SigmaLambda_2mu$

Computing the variance in the non-symmetric case

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

:$epsilon\text{'}Lambda\text{'}epsilon=epsilon\text{'}Lambdaepsilon$

so

:$epsilon\text{'}Lambdaepsilon=epsilon\text{'}left\left(Lambda+Lambda\text{'} ight\right)epsilon/2$

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

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 $y$:

:$extrm\left\{RSS\right\}=y\text{'}left\left(I-H ight\right)\text{'}left\left(I-H ight\right)y$

For procedures where the matrix $H$ is symmetric and idempotent, and the errors are Gaussian with covariance matrix $sigma^2I$, $extrm\left\{RSS\right\}/sigma^2$ has a chi-square distribution with $k$ degrees of freedom and noncentrality parameter $lambda$, where

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

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 $mu$ with no bias, then the noncentrality $lambda$ is zero and $extrm\left\{RSS\right\}/sigma^2$ follows a central chi-square distribution.

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