Let $\mathbf{Z}$ be the design matrix of a linear regression.

Where $\mathbf{Z}^{T}\mathbf{Z}$ is symmetric.

Set $\mathbf{V} = (\mathbf{Z'Z})^{\frac{1}{2}}(E(\hat{\beta}) - \beta)$

Where $\hat{\beta} , \beta$ is the true and estimated parameter vectors of a linear regression model.

Thus we can say that:

$Cov(\mathbf{V}) = (\mathbf{Z^{T}\:Z})^{\frac{1}{2}} Cov(\hat{{\bf{\beta}}}) ((\mathbf{Z^{T}\:Z})^{\frac{1}{2}})^{T} =$

$(\mathbf{Z^{T}\:Z})^{\frac{1}{2}}\sigma^{2}(\mathbf{Z^{T}\:Z})^{-1}(\mathbf{Z^{T}\:Z})^{\frac{1}{2}} = \sigma^{2}\mathbf{I}$

Why is this? and from what matrix algebra rule does it follow?


1 Answer 1


Think about adding the powers of matrix.

For example the following identity holds:

$\mathbf{A}^{a}\mathbf{A}^{b} = \mathbf{A}^{a + b}$

Applied to your problem:

$\mathbf{A}^{0.5}\mathbf{A}^{0.5}\mathbf{A}^{-1} = \mathbf{A}^{0} = \mathbf{I}$


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