How is the formula for the variance and covariance of residuals derived?

Question

The first formula I have is as follows : $\operatorname{var}(\widehat\varepsilon )=\sigma^2(−)$ where $=(^)^{−1}$ and $\widehat\varepsilon$ is the residual. Furthermore, I am having trouble understanding how $\operatorname{cov}((−))=\sigma^2(−)$ as well. Here, $(−)=−̂$ Any help would be greatly appreciated.

Answer 1

It's not clear what you do and don't know about all this, so I will answer only a bit of this:

The matrix $I-H$ is symmetric and idempotent. "Symmetric" means it's its own transpose: $(I-H)'=I-H.$ "Idempotent" means it's its own square: $(I-H)^2=I-H.$

Generally if $U\in\mathbb R^{n\times1}$ is random vector and $\operatorname{var}(U) = \Sigma\in\mathbb R^{n\times n}$ and $A\in\mathbb R^{k\times n}$ is a constant (i.e. non-random) matrix, then $\operatorname{var}(AU) = A\Sigma A',$ where $A'$ is the transpose of $A.$

So you have \begin{align} \operatorname{var}(\widehat{\varepsilon\,}) & = \operatorname{var}((I-H)Y) \\[6pt] & = (I-H)(\sigma^2 I) (I-H)' \\[6pt] & = \sigma^2(I-H)(I-H)' \\[6pt] & = \sigma^2 (I-H)(I-H) \\[6pt] & = \sigma^2(I-H). \end{align}