usually when I see derivations of ommited variable bias, I see something of the sort: from y=xb + $\eta$, and looking at the for formula for the slope estimate:

  1. $cov(x,y)$$/var(x)$
  2. $cov(x,xb+\eta )$$/var(x)$=
  3. b + $cov(x,\eta)$$/var(x)$

my question is, how does intuitively step 2 make sense? plugging in y in to the covariance formula? Is it that we are taking y=xb + $\eta$ literally of how Y is being generated, and so any y we see in the population is theoretically equivalent to xb + $\eta$?

  • $\begingroup$ I don't get your question, because the gist of it reads, "assuming $y=xb+\eta,$ why is it valid to substitute $xb+\eta$ for $y$ in a covariance calculation?" There seems to be nothing to say. What am I missing? $\endgroup$
    – whuber
    Jul 30, 2020 at 15:34
  • $\begingroup$ I think I should have worded 'assuming' as part of the question- just meant its usually stated in that way, so is that saying that y literally equal to xb +$\eta$? $\endgroup$
    – Steve
    Jul 30, 2020 at 17:19
  • $\begingroup$ Equations are meant pretty literally, so yes. $\endgroup$
    – whuber
    Jul 30, 2020 at 17:23


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