1
$\begingroup$

Let's have a look at this post: Standard error for the sum of regression coefficients when the covariance is negative

We have: $SE_{b_{2+3}} = \sqrt{SE_2^2 + SE_3^2+2Cov(\beta_2,\beta_3)}$

But why do we mix SE, which is the SD/sqrt(N) with covariance (not divided by the sqrt(N))? I know this formula for variances, but SE is not the sqrt(variance).

Could I kindly ask someone to show me, using algebra, how is this valid?

I mean - why can I take the SE from the model coefficients and use the variance-covariance matrix without any additional steps?

Improve this question
$\endgroup$
3
  • $\begingroup$ Re "but SE is not the sqrt(variance)": in this setting, yes it is. You are confusing two formulas applicable in two different situations. No algebra is needed. $\endgroup$
    – whuber ♦
    Jun 10, 2021 at 12:47
  • $\begingroup$ Thank you. So, the variance-covariance matrix contains squared standard errors (diagonal) and covariances (off-diagonal)? Not just variances? Are these covariances, then, the standard covariances, or covariances divided by the sqrt(n)? I'm sorry for dumb questions, but I try to figure it out and learn. $\endgroup$
    – Krapatik321
    Jun 10, 2021 at 14:07
  • $\begingroup$ The division by $\sqrt{n}$ has, in effect, already been performed when the covariance matrix of the estimates is constructed. $\endgroup$
    – whuber ♦
    Jun 10, 2021 at 14:40

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.