This question has a very detailed and helpful answer, but is pretty old. I would like to understand why \begin{equation} \frac{\widehat{\beta}_{i}-\beta_{i}}{\sigma\sqrt{(X^{T}X)^{-1}_{ii}}} \sim N(0,1) \end{equation} holds in the approved answer and am thankful for your help.
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why the formula on the left holds, or why it's standard Normally distributed? – develarist Nov 29 '19 at 15:58
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This is because the vector $\hat\beta$ with components $\hat\beta_i$ is distributed as $\hat\beta\sim N(\beta,\sigma^2(X'X)^{-1})$. As $\hat\beta$ is multivariate normal, each of its components has a normal distribution as written in your post. – StubbornAtom Nov 29 '19 at 16:09
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You need to be more specific concerning what kind of answer you are looking for, given that you acknowledge the duplicate already has an answer. – whuber Nov 29 '19 at 17:23