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After fitting the regression model, $$y=\beta_0 + \beta_1 x_1 + \beta_2 x_2 + \beta_3 x_3 + \epsilon$$ on 15 cases, it is found that the mean square errors $$s^2 = 3$$ and

$$(X^T X)^{-1}= \begin{bmatrix} 0.5 & 0.3 & 0.2 & 0.6 \\ 0.3 & 6.0 & 0.5 & 0.4 \\ 0.2 & 0.5 & 0.2 & 0.7 \\ 0.6 & 0.4 & 0.7 & 3.0 \\ \end{bmatrix}$$

This is a practice problem for my midterm... I cannot figure out how I would figure out the estimated correlation between $$\hat\beta_1 \mbox{ and } \hat\beta_3.$$

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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 $\text{cov}(\hat{\mathbf\beta} )$:

$$ \text{cov}(\hat{\mathbf\beta} ) = (X'X)^{-1}X'\text{var(y)}X(X'X)^{-1} = \text{var(y)}(X'X)^{-1}X'X(X'X)^{-1} = \text{var(y)}(X'X)^{-1} $$

You'll get an estimate for $\text{var(y)}$ by using $s^2$ appropriately, and you'll get the covariance between $\hat\beta_1$ and $\hat\beta_3$ by looking at the element $(2,4)$ of the covariance matrix $\text{cov}(\hat{\mathbf\beta} )$ (this is not element $(1,3)$ because the index $i$ in $\hat\beta_i$ starts at 0 for the intercept...).

From there, you can compute the correlation between $\hat\beta_1$ and $\hat\beta_3$ by standardizing appropriately, using the variances of $\hat\beta_1$ and $\hat\beta_3$, which you get from the elements $(2,2)$ and $(4,4)$ of $\text{cov}(\hat{\mathbf\beta} )$.

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  • $\begingroup$ +1 Note, though, that the references to $s^2$ are only distracting from the ideas because the scale factor $s$ does not affect any correlations and can be ignored. $\endgroup$
    – whuber
    Nov 21, 2021 at 18:02
  • $\begingroup$ Can you explain how you know which index in the matrix to look at? $\endgroup$ Nov 21, 2021 at 23:29
  • $\begingroup$ Look at the dimension of the matrix $X$: it is $4\times 4$. This is because $X$ is a $15\times 4$ matrix. The confusion comes from the fact that while you have only 3 variables $x_1, x_2, x_3$, you also have a constant in the first column, associated with the coefficient $\beta_0$. So all the indices are shifted by 1.... for instance the variance of $\beta_1$ appears in $\text{cov}(\hat\beta)_{2,2}$. $\endgroup$
    – wiwh
    Nov 22, 2021 at 8:46

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