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

  • 1
    $\begingroup$ $$\frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}\sim N(0,I)$$ 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? $\endgroup$ Jan 9 at 17:10
  • 2
    $\begingroup$ You have mixed up vectors and scalars. $\endgroup$ Jan 9 at 18:49
  • $\begingroup$ @SextusEmpiricus But if we want to show that is t-distribution, we need to the numerator is standard normal. $\endgroup$
    – Hermi
    Jan 10 at 0:49
  • $\begingroup$ We do not "divide" by a matrix, we multiply by its inverse. $\endgroup$
    – usεr11852
    Jan 10 at 2:10

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.

  • $\begingroup$ My question is that how to show that $\frac{N(0, (X^TX)^{-1})}{\sqrt{(X^TX)^{-1}_{jj}}}\sim N(0,I)? $ $\endgroup$
    – Hermi
    Jan 10 at 0:47
  • $\begingroup$ @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. $\endgroup$ Jan 10 at 0:58
  • $\begingroup$ 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. $\endgroup$ Jan 10 at 1:00
  • $\begingroup$ If not, how to prove that $\frac{\hat{\beta}_j-\beta_j}{\sqrt{(X^TX)^{-1}_{jj}\hat{\sigma}^2}}\sim t_{n-p} $? $\endgroup$
    – Hermi
    Jan 10 at 1:07
  • $\begingroup$ @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 $$\hat{\beta}_j-\beta_j \sim N(0, \sigma^2(X^TX)_{jj}^{-1})$$ $\endgroup$ Jan 10 at 1:22

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