Regression – Understanding $\frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}\sim N(0,I)$

Question

I try to prove one result from linear regression model as follows.

Let $\beta_j$ denote the $j$th element of $\beta$ and $A_{ii}^{-1}$ the $i$th diagonal element of $A^{-1}$. Then we have
$$\frac{\hat{\beta}_j-\beta_j}{\sqrt{(X^TX)^{-1}_{jj}\hat{\sigma}^2}}\sim t_{n-p}
$$
where
$$
\hat{{\sigma}^2}=\frac{1}{n-p}(Y-\hat{Y})^T(Y-\hat{Y})$$

Indeed, $\hat{\beta}-\beta\sim N(0, \sigma^2(X^TX)^{-1})$ and is independent of $\hat{\sigma}^2$. Then
$$\frac{\hat{\beta}-\beta}{\hat{\sigma}}\sim\frac{\hat{\beta}-\beta}{\sqrt{\frac{\sigma^2\chi^2_{n-p}}{n-p}}}\sim \frac{N(0, (X^TX)^{-1})}{\sqrt{\frac{\chi^2_{n-p}}{n-p}}}
$$

But I am confused that why
$$
\frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}\sim N(0,I)?
$$
Isn't it divided by $\sqrt{(X^TX)^{-1}}$?

Answer 1

Isn't it divided by $\sqrt{(X^TX)^{-1}}$?

$\sqrt{(X^TX)^{-1}}$ is not a single number.

This is because $X$ is a $n$ by $p$ matrix and $X^T X$ will be a $p$ by $p$ matrix of which you need the $j$-th diagonal element.

But I am confused that why $$ \frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}\sim N(0,I)?$$

This is not the case.

• Yes, you do have that each individual $\hat\beta_j - \beta_j$ is normal distributed

• But no, the multivariate case does not hold. The scaled $\hat\beta_j - \beta_j$ will be correlated.

In addition the expression $$\frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}$$ does not seem quite right. The arithmetics/operation, how you divide the multivariate normal distribution, is not very clear and defined.