1
$\begingroup$

Assume that we want to derive the standard error of $\beta_2$ in: $$y=\beta_1 + \beta_2x_2 +u$$

$u$ iid with mean $0$ variance $\sigma ^2$. The way I know how to do it, is convert it in matrix form, $$y=X\beta +u$$

then simply derive the variance of the OLS estimator of $\beta$ and take the lower right value of that matrix.

But surely there is a way to do this directly, without converting it in matrix form. What is the way to do this with a low amount of algebraic steps?

$\endgroup$
  • $\begingroup$ Are you asking about a mathematical derivation or for steps to simply calculate? $\endgroup$ – Alexis May 30 '18 at 14:34
  • $\begingroup$ @Alexis, I'm not sure what the difference is? Let's say a mathematical derivation. $\endgroup$ – user56834 May 30 '18 at 14:40
  • $\begingroup$ So I would call this a way to calculate: $s^{2}_{\beta_{2}}= \left(\frac{\sqrt{\frac{\sum_{i=1}^{n}{\varepsilon_{i}^{2}}}{n-2}}}{s_{x}\sqrt{n-1}}\right)^{2}$, where $\varepsilon = \left[y_{i} - \left(\beta_{1} + \beta_{2}x_{i}\right)\right]$, but I would not call it a mathematical derivation (although it is mathematically derived). :) $\endgroup$ – Alexis May 30 '18 at 14:52
  • $\begingroup$ @Alexis, ah ok. No that's the result I'm looking to find out how to derive quickly :P so yes I mean a derivation. $\endgroup$ – user56834 May 30 '18 at 15:12
  • $\begingroup$ So "calculate" always refers to the part where you plug in the numbers and get an actual number as a result? I tend to use "calculate" and "derive" interchangeably, but maybe that's wrong. $\endgroup$ – user56834 May 30 '18 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.