For omitted variable bias to occur when a variable is left out of a regression, there is one axiom and one condition that must be fulfilled:

  1. (Axiom) By definition, the coefficient of the variable has to not be equal to zero (or it can't even be considered an omitted variable).
  2. (Condition) The omitted variable must be correlated with some regressor, which means the regressor will be correlated with the error term, violating gauss markov assumptions and generating bias.

Problem: Consider a model of the most basic form:

$$y_i = \alpha + \beta x_i + e_i$$

Now, we all know that leaving out alpha (corrsponding variable which is 1 for all of the observation), will cause bias in the beta parameter. However, a constant can't be correlated with $x_i$, so something is off here. Where is the error?

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    $\begingroup$ Omitted variables cause bias but not all bias is caused by omitted variables. The intercept is the expected mean of $y$ when $x = 0$. If you omit the intercept, you assume the expected mean is 0, thus introducing bias in your estimation if this expected mean is not 0. $\endgroup$ Apr 8, 2016 at 9:22
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    $\begingroup$ To say the same as @Antoine Vernet, one of the conditions for the BLUE estimators is that the $\epsilon$ has mean zero, now if you drop the constant $\alpha$ while there should be one (i.e. $\alpha \ne 0$), then $E(\epsilon)=\alpha\ne0$, so the condition that mean of $\epsilon$ is zero is violated. $\endgroup$
    – user83346
    Apr 8, 2016 at 9:31
  • $\begingroup$ @AntoineVernet So are you saying that what we have here is not omitted variable bias, but something else? A bit unusual definition given that there is an omitted variable causing the bias. $\endgroup$
    – Dole
    Apr 8, 2016 at 14:19
  • $\begingroup$ The intercept is not a variable, it is a constant. So yes, what I mean is the bias is not coming from an omitted variable but from the fact that if you omit the constant, you implicitly make the assumption that the mean of $y$ is 0 when $x = 0$, which introduces bias in the case where the mean of $y$ is different from 0. @fcop gives a more complete explanation of why this is the case. $\endgroup$ Apr 8, 2016 at 14:35
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    $\begingroup$ @AntoineVernet Sure it is a variable, a vector of ones. $\endgroup$
    – Dole
    Apr 8, 2016 at 14:42

2 Answers 2


One of the assumptions in gauss markov theory is strong exogeneity: $E(u|x)=0$. So that, conditional on x, error term u should have zero mean.

When $x=0$, the above assumption means $E(u|x=0)=0$. Therefore, when the true constant term is not zero but been omitted, the assumption that $E(u|x=0)=0$ is not satisfied.


All right, time to attempt answering my own question. My preferred explanation here is that the conditions for the omitted variable bias are in fact, not complete. If it is assumed that the constant is axiomatically included in the regression, then the result is correct. Of course you may also not consider the constant a variable, which is a strange interpretation as it behaves exactly like a variable. This was a prevailing opinion in the comments none-the-less.

Let's go through the math, using the simple regression model (one variable + constant being the independent variables in the population function). Let's estimate the model using only the one variable (and leaving the constant out):

$$\hat \beta =(x'x)^{-1}x'y$$

Our x is simply a vector of the values of the single variable (not a matrix with the constant included). This is why it is denoted with small x and not with the large x as you will commonly find

Now let's plug the population regression function into y:

$$\hat \beta=(x'x)^{-1}x'(x\beta +1\alpha+u)$$

Just as a note the 1 is the vector of ones, while alpha is the parameter value for the constant. Simplifying we have:

$$\hat \beta = \beta +(x'x)^{-1}x'1a+(x'x)^{-1}xu$$

Finally, taking the expected value:

$$\text E(\hat \beta) = \beta +\text E ((x'x)^{-1}x'1a)+\text E ((x'x)^{-1}xu)$$

The final term on the RH side is zero since the expected value of u is still 0 (contrary to comments). However, the 2nd term on the RH side is not necessarily zero even if cov(x,1) is zero and a is not zero. Such result is only true for demeaned variables. Adding the constant of course demeans every variable, and as such should be a other axiom for the result ("there is a constant / only demeaned variables in the model"). Interestingly, when there is no constant even the beta parameter omitted variable bias can exist without the covariance being zero, so claiming that constant is not a variable doesn't address the issue.


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