# Does omitted variable bias 'contaminate' coefficients for regressors not correlated with the omitted variable?

Suppose we have a model that is:

$$y = X_1\beta_1 + X_2\beta_2 + X_3\beta_3 + \varepsilon$$

where $$X_1$$ is independent of $$X_2$$ and $$X_3$$ but $$X_2$$ and $$X_3$$ are correlated with each other.

Suppose we don't have data on $$X_3$$ so we omit it and run the regression:

$$y = X_1\gamma_1 + X_2\gamma_2 + \nu$$

Econometric theory on OVB states that $$\gamma_2$$ will be biased.

My questions:

1. Will $$\gamma_1$$ be biased in this case as well? I ran a simulation with some data I created and I do get a biased result.

2. If it is unbiased then how do I prove it? I use FWL theorem to find $$\gamma_1$$ and show that its expectation equals $$\beta_1$$ but get stuck here:

$$\hat{\gamma_1} = \beta_1 + (X_1'M_{X_2}X_1)^{-1}(X_1'M_{X_2}X_3\beta_3) + (X_1'M_{X_2}X_1)^{-1}(X_1'M_{X_2}\varepsilon)$$

The third term will become 0 when we take the expectation but not sure how to make the second term 0.

• Are you sure you've done your simulation correctly? If X1 and X3 are independent then gamma1 shouldn't be biased since Cov(X1,X3)=0. This also implies that Cov(X1Mx2X3)=0. So the the second term in part two of your question goes to zero when X1 and X3 are independent of one another. – Andrew M Nov 30 '18 at 11:05
• $\gamma_1$ won't be biased, given $\mathbb{C}(X_1, X_2) = 0 = \mathbb{C}(X_1, X_3)$ – Pedro Cavalcante Dec 7 '18 at 17:37

I would suspect that if $$X_1$$ is independent of $$X_3$$, then it must imply $$E[X_1*X_3]$$ is equal to $$E[X1] * E[X_3]$$, given this you can conclude that the Covariance is 0. $$Cov(X_1,X_3) = E[X_1*X_3] - E[X_1]* E[X_3]$$. If the Covariance is 0 between these two variables then so is the correlation. Thus I would believe that removing $$X_3$$ from the regression and having it's effect enter the error term would not result in having $$γ_1$$ be biased as the $$X_3$$ in the residual term does not have correlation with $$γ_1$$.

Are you looking for a rigorous mathematical proof?

Your approach leads to a biased estimator. Nevertheless, there exists a way to obtain an unbiased estimator for $$\beta_1$$. We asssume all the regularity conditions and exogeneity of regressors with respect to the error $$\varepsilon$$.

1) Center the variables on their sample means, $$y_c = y- \bar y, X_{c1} = X_1 - \bar X_1$$, etc.

2) Specify the regression

$$y_c = X_{c1}\delta + u$$

Then

$$\hat \delta_{OLS} = \left(X'_{c1} X_{c1}\right)^{-1}X'_{c1}y_c$$

$$=\left(X'_{c1} X_{c1}\right)^{-1}X'_{c1}X_{c1}\beta_1 + \left(X'_{c1} X_{c1}\right)^{-1}X'_{c1}X_{c2}\beta_2 \\+ \left(X'_{c1} X_{c1}\right)^{-1}X'_{c1} X_{c3}\beta_3 + \left(X'_{c1} X_{c1}\right)^{-1}X'_{c1}\varepsilon_c$$

Simplifying and taking expectations conditional on $$X_{c1}$$ we get

$$\mathbb E(\hat \delta_{OLS} \mid X_{c1})= \beta_1 + \left(X'_{c1} X_{c1}\right)^{-1}X'_{c1}\mathbb E(X_{c2} \mid X_{c1})\beta_2 \\+\left(X'_{c1} X_{c1}\right)^{-1}X'_{c1} \mathbb E(X_{c3}\mid X_{c1})\beta_3 + \left(X'_{c1} X_{c1}\right)^{-1}X'_{c1}\mathbb E(\varepsilon_c\mid X_{c1})$$

By the regressor independence and exogeneity assumptions, as well as due to the centering we have

$$\mathbb E(X_{c2} \mid X_{c1}) = \mathbb E(X_{c2} ) = 0$$ $$\mathbb E(X_{c3} \mid X_{c1}) = \mathbb E(X_{c3} ) = 0$$ $$\mathbb E(\varepsilon_c \mid X_{c1}) = 0$$

So we are left with

$$\mathbb E(\hat \delta_{OLS} \mid X_{c1})= \beta_1 \implies \mathbb E(\hat \delta_{OLS}) = \beta_1$$