# Why $\frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}\sim N(0,I)?$

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 $$\sqrt{(X^TX)^{-1}}$$?

• $$\frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}\sim N(0,I)$$ This makes no sense to me. In the numerator and right hand side you have a matrix as variance. Do you refer to a multivariate normal distribution? Jan 9 at 17:10
• You have mixed up vectors and scalars. Jan 9 at 18:49
• @SextusEmpiricus But if we want to show that is t-distribution, we need to the numerator is standard normal. Jan 10 at 0:49
• We do not "divide" by a matrix, we multiply by its inverse. Jan 10 at 2:10

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.

• My question is that how to show that $\frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}\sim N(0,I)?$ Jan 10 at 0:47
• @quasAliki I responded to your sentence that ended in a question mark. You can not divide by the product $X^TX$ because it is a matrix. Jan 10 at 0:58
• Regarding your question why we have $$\frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}\sim N(0,I)?$$ The answer is that we do not have this. The expression makes no sense. Or at least for me it is unclear what you mean by the expression $N(0, (X^TX)^{-1})$ which is not a typical expression. Jan 10 at 1:00
• If not, how to prove that $\frac{\hat{\beta}_j-\beta_j}{\sqrt{(X^TX)^{-1}_{jj}\hat{\sigma}^2}}\sim t_{n-p}$? Jan 10 at 1:07
• @quasAliki in your question you use $$\hat{\beta}-\beta\sim N(0, \sigma^2(X^TX)^{-1})$$ that is a multivariate distribution. For the derivation/proof you can use the distribution of only the $j$-th component of that distribution $$\hat{\beta}_j-\beta_j \sim N(0, \sigma^2(X^TX)_{jj}^{-1})$$ Jan 10 at 1:22