Edit: Background information: I have two variables of interest, $y$ and $x$ that are linearly related via the following: $y = a + bx + u$, where "$a$" and "$b$" are fixed parameters to solve for, and "$u$" is the error term and captures the fact that there are other factors that also affect $y$, which aren't captured by "$a+bx$" alone. In this dataset, there would be 9 errors terms $u$ for each of the 9 $(x,y)$ pairs below.

We are told in textbooks that if we assume $$E(u|x) = 0$$ then the population regression function can be interpreted as $E(y|x) = a + bx$, where "$a$" and "$b$" are the population parameters. However I can construct a counterexample where this doesn't hold: $$\begin{array}{c|c|c|} & \text{x} & \text{y} \\ \hline \text{} & 1 & 12 \\ \hline \text{} & 2 & 14 \\ \hline \text{} & 3 & 16\\ \hline \text{} & 4 & 20\\ \hline \text{} & 5 & 25\\ \hline \text{} & 6 & 29\\ \hline \text{} & 7 & 31\\ \hline \text{} & 8 & 40\\ \hline \text{} & 9 & 20\\ \hline \end{array}$$

Suppose the above is a population dataset (I'm assuming I know exactly the population, there is no sample). $E(y|x) = y$, for instance $E(y|x=1) = 12$ since $x=1$ has only 1 $y$ value which is equal to 12. If this is true whats the population regression line that relates "$y$" to "$x$" in a linear form $y=a+bx+u$ where $E(u|x)=0$ actually holds?

If I solve for "$a$" and "$b$" using a regression calculator gives us $Y = 10.87 + 2.41x$. However, this does not satisfy the property of $E(u|x)=0$, clearly for $x=1$, the predicted value is not equal to 12, there is an error. So how is it that linear regression satisfies $E(u|x)$?

Does this mean $E(u|x) = 0$ doesn't hold in the population and is just an assumption?

Even under the assumption that $E(u|x)$ is true, with the population dataset above I cannot find a linear equation that satisfies this. What are the implications of this in real world examples when it doesn't seem to hold?

  • $\begingroup$ Welcome to the site! A few aspects of your question are currently not clear to me. What is $u$? What is "the population regression function"? Keep in mind that people come from different fields and use different notation, and make sure everything in your question is understandable even if someone is not looking at the exact same textbook you're reading. :-) $\endgroup$
    – Eike P.
    Jul 22, 2020 at 12:53
  • $\begingroup$ for the model you fit, the residuals sum to zero, $\sum_i u_i=0$ and are uncorrelated with the covariate $\sum_i u_ix_i=0$. So if you were to fit a regression of $u_i=c+dx_i$ then the coefficients $c=d=0$ $\endgroup$ Jul 22, 2020 at 12:59
  • 1
    $\begingroup$ "Suppose the above is a population dataset (I'm assuming I know exactly the population, there is no sample)." Statistical models are for drawing inferences and making predictions based on samples. If you know the population, you don't need statistics, so why would you expect linear regression to make sense? $\endgroup$
    – Eoin
    Jul 22, 2020 at 13:22
  • $\begingroup$ Thank you "Jhin". I have made edits, hopefully it is clear. "probabilityislogic" that makes sense, if d=0 then indeed the error u is uncorrelated with x. How can we show that $E(ui|xi) = 0$, in my example above it doesn't seem to hold? I guess my issue is how we can reconcile your answer and the result I got above, i.e $E(u|x=1) = 1.28$, not zero. $\endgroup$
    – A.L
    Jul 22, 2020 at 13:45
  • $\begingroup$ "Eoin", I understand, this is a hypothetical experiment. I suppose one may still use regression on a population to describe the relationship between the variables, even as a line of best fit. $\endgroup$
    – A.L
    Jul 22, 2020 at 13:48

1 Answer 1


This is a great example that illustrates why, in regression models, (i) the assumption $E(u | x) = 0$ should not be used, and (ii) the "population" framework should not be used.

Rather than the assumption $E(u | x) =0$, it would make much more sense to state the assumption in the equivalent form $E(y | X=x) = \beta_0 +\beta_1 x$. This assumption states that the means of the conditional distributions fall exactly on a line of the form $\beta_0 + \beta_1 x$, for some $\beta_0$, $\beta_1$.

As the OP notes, the conditional distributions in the population framework are all degenerate, so that the mean is just equal to the single $y$ value. For example, the distribution of $Y | X = 9$ is given by $Pr(Y = 20 | X=9) = 1$, with $Pr(Y = y | X=9) = 0$, for all $y \neq 20$. The mean of this distribution is clearly 20.

Since these conditional mean values do not all fall precisely on a straight line, the assumption $E(y | X=x) = \beta_0 +\beta_1 x$ is violated. This explains the OP's finding that $E(u | x) \neq 0$.

Here is an example "population" where it works.

$$\begin{array}{c|c|c|} & \text{x} & \text{y} \\ \hline \text{} & 1 & 12 \\ \hline \text{} & 1 & 14 \\ \hline \text{} & 1 & 16\\ \hline \text{} & 2 & 20\\ \hline \text{} & 2 & 24\\ \hline \text{} & 2 & 28\\ \hline \text{} & 3 & 31\\ \hline \text{} & 3 & 33\\ \hline \text{} & 3 & 38\\ \hline \end{array}$$

Here the conditional means are 14, 24, 34, falling on a linear function of $x = 1,2,3$. Consider the distribution of $y | X=3$:

$$\begin{array}{c|c|c|} & \text{p(y|x)} & \text{y} \\ \hline \text{} & 1/3 & 31\\ \hline \text{} & 1/3 & 33\\ \hline \text{} & 1/3 & 38\\ \hline \end{array}$$

The distribution of $u$ is obtained by replacing the $y$ values with $y-34$, so $E(u | X = 3) = (1/3)(31 - 34) + (1/3)(33-34) + (1/3)(38-34) = 0.$

If the means of the distributions are configured so that they do not fall exactly on a line, then $E(u | X = x) \neq 0$ for some $x$.

This example also illustrates the point that the "population model" should not be used to define the regression model. As the examples illustrate, the conditional means are not really true means in the scientific, generalizable sense, they are instead quite noisy due to small sample sizes in the subpopulation defined by the "$| x$." In some cases, there may be no observations whatsoever in such subsets of the population, even when the population is large. This problem is magnified multiplicatively in the case of multiple regression.

  • $\begingroup$ Yet it’s super common in econometrics to write $E(u\vert x)=0$… $\endgroup$
    – Dave
    Feb 2 at 13:05

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