Two negatives make an upward omitted variable bias

Question

The true model is y = 10 x1 - 3 x2 + \varepsilon with \varepsilon exogenous. You omit x2 and regress y on x1. The omitted x2 is negatively related to x1: its regression on x1 has slope \delta = -0.4. Compute the limiting value of the x1 coefficient and the bias.

Accepted Answer

Apply the formula, minding both signs. \hat\beta1 \xrightarrow{p} \beta1 + \beta2 \delta = 10 + (-3)(-0.4) = 10 + 1.2 = 11.2. Size and sign. 	ext{bias} = \beta2 \delta = (-3)(-0.4) = +1.2. The bias is positive, an upward pull, because \beta2 and \delta are both negative and their product is positive. This is the sign rule at work: same sign on the effect and the loading (here both negative) yields positive bias. Your slope reads 11.2 when the true direct effect is 10. Takeaway. Two negatives multiply to a positive bias: (-3)(-0.4) = +1.2, so the slope converges to 10 + 1.2 = 11.2; the sign o…

Two negatives make an upward omitted variable bias

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