For a general model $$y_{i} = \alpha + \beta_{1}X_{1} + \beta_{2}X_{2} + \epsilon_{i}$$ regressing $y_{i}$ on $X_{1}$ alone will result in $\beta_{1}$ being biased given by: $$plim \: \widehat{\beta}_{1} = \beta_{1} + \beta_{2}\frac{Cov(X_{1},X_{2})}{Var(X_{1})}$$

Now I stumbled across a paper which has a model of the form $$S_{i} = \alpha D_{i} + \beta P_{i} + u_{i}$$ without any other variables and no intercept. The true value of $\beta$ is supposed to be zero because the event $S$ precedes event $P$. The authors then state the omitted variable bias of the two coefficients as: $$\widehat{\alpha}_{OLS} = \alpha + \frac{Cov(D,u)Var(P) - Cov(D,P)Cov(P,u)}{Var(D)Var(P) - Cov(D,P)^{2}}$$ and $$\widehat{\beta}_{OLS} = \frac{Cov(P,u)Var(D) - Cov(D,P)Cov(D,u)}{Var(D)Var(P) - Cov(D,P)^{2}}$$

I have tried for a long time now to reproduce this result but I cannot figure out how they arrived at these two bias expressions. The problematic point is that I apparently cannot understand from which OVB formula they start because in their setting it cannot be the usual one for a single variable written above. The paper is by Adda et al. (2011) for which the posed problem can be found on PDF page 52.

If someone could point towards the right direction I would be most grateful.

  • $\begingroup$ The high "viewed" figure for this question suggests this is high in search engine results (it's certainly first page for google in "omitted variable bias multiple variables" or "omitted variable bias more than one variables"); I wonder if those viewers are getting quite what they expected with this question, given that the bias in the paper wasn't actually OVB? $\endgroup$
    – Silverfish
    Commented Mar 27, 2015 at 14:26
  • $\begingroup$ One possibility might be retitling the question so that future readers (or people who see the link on their search engine) just see "bias in Adda et al (2011)" or something similar, instead of a direct reference to OVB. Or making a direct reference to it being bias due to endogeneity instead? $\endgroup$
    – Silverfish
    Commented Mar 27, 2015 at 14:27
  • $\begingroup$ @Silverfish I see the point. If you want you can make the edit as it seems a meaningful change of the title. $\endgroup$
    – Andy
    Commented Mar 27, 2015 at 14:39
  • $\begingroup$ @Silverfish considering the fact that the question is on the site for almost two years the number of views is also not that high $\endgroup$
    – Andy
    Commented Mar 27, 2015 at 15:01
  • 1
    $\begingroup$ Feel free to revert/re-edit if you don't like what I've done. $\endgroup$
    – Silverfish
    Commented Mar 27, 2015 at 15:28

1 Answer 1


The "bias" they are mentioning in appendix A of the article is not the omitted variables bias, but a bias due to possible endogeneity. (They mention it several times in the article, and specifically in page 11 of the PDF, in the bottom part).
The estimator formulas that are mentioned in the Appendix are the regular OLS estimator formulas for 2 variables, as can be found in page 8 here:

For $S_{i} = \alpha D_{i} + \beta P_{i} + u_{i}$:

$\widehat{\alpha}_{OLS} = \frac{Cov(D,S)Var(P) - Cov(D,P)Cov(P,S)}{Var(D)Var(P) - Cov(D,P)^{2}} = \alpha + \frac{Cov(D,u)Var(P) - Cov(D,P)Cov(P,u)}{Var(D)Var(P) - Cov(D,P)^{2}}$

$\widehat{\beta}_{OLS} = \frac{Cov(P,S)Var(D) - Cov(D,P)Cov(D,S)}{Var(D)Var(P) - Cov(D,P)^{2}} = \beta+ \frac{Cov(P,u)Var(D) - Cov(D,P)Cov(D,u)}{Var(D)Var(P) - Cov(D,P)^{2}}$

They stated the $\beta$ equals 0, and that's how they got the formulas in the article. Again - the possible bias that they are mentioning is because of a possible endogeneity, not because any ommision.


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