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2

I just want to add that there is also a connection to the Rayleigh quotient, which for the real symmetric $k \times k$ Matrix $\hat{\Sigma}$ is defined as $\frac{x^\prime \hat{\Sigma} x }{x^\prime x}, \,x \in \mathbb{R}^{k},\, x \neq 0$ . By the Courant–Fischer Theorem, we obtain $$\lambda_{min} = \min_{x \neq 0} \frac{x^\prime \hat{\Sigma} x }{x^\prime ... 3 I arrive at a slightly different result and am curious about your thoughts (no square of the minimal eigenvalue). Some experimentation suggests that the inequality (if correct) would be stronger, as the minimal eigenvalue of X'X/n generally is less than one. First, since the quadratic form is a scalar, it equals its trace,$$(\hat{\beta}-\beta)'\hat{\Sigma}...

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It's a textbook rather than notes, but I found a great resource that covers this matieral in exactly the way you describe is Econometric Theory and Methods by Davidson and MacKinnon. See here: http://qed.econ.queensu.ca/ETM/. There are many PDF versions floating around the internet if you have a quick look for them. This would have been a comment but I don't ...

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Write $\hat{\mathbf\beta} = (\hat\beta_0,\hat\beta_1,\hat\beta_2,\hat\beta_3,)'$. You have $$\hat{\mathbf\beta} = (X'X)^{-1}X'y$$ You're interested in the variance-covariance matrix of $\hat{\mathbf\beta}$, which I will denote by $\text{cov}(\hat{\mathbf\beta} )$:  \text{cov}(\hat{\mathbf\beta} ) = (X'X)^{-1}X'\text{var(y)}X(X'X)^{-1} = \text{var(y)}(X'X)...

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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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