Omitted variable bias that pulls the slope down

Question

The true model is y = 5 x1 - 4 x2 + \varepsilon with \varepsilon exogenous. You omit x2 and regress y on x1. The omitted x2 regresses on x1 with slope \delta = 0.25. Find the value your x1 coefficient converges to and the size and sign of the bias.

Accepted Answer

Apply the formula. \hat\beta1 \xrightarrow{p} \beta1 + \beta2 \delta = 5 + (-4)(0.25) = 5 - 1 = 4. Size and sign. 	ext{bias} = \beta2 \delta = (-4)(0.25) = -1. The bias is negative because \beta2 is negative while \delta is positive, so their product is negative. Your slope reads 4 when the true direct effect is 5, a downward bias of 1. Reading the sign rule. Bias is positive when \beta2 and \delta share a sign, and negative when they differ. Here they differ, hence a downward pull. Takeaway. With \beta2 = -4 and \delta = 0.25, the bias is -1, so the slope converges to 5 - 1 = 4: opposite si…

Omitted variable bias that pulls the slope down

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