The Gauss-Markov theorem states that for a linear model

$$y = X \beta + \epsilon $$

if both of the conditions are true

$$\operatorname E[\epsilon \mid X] = 0$$ $$\operatorname{Var}(\epsilon) = \sigma^2 I < \infty $$

then the standard OLS estimator $(X'X)^{-1}X'y$ is the best linear unbiased estimator.

Now suppose we measure $X$ with errors. Then we have

$$y = (X + \mu)\beta + \epsilon = X\beta + \mu\beta+\epsilon$$

If $\mu$ is of mean $0$ with constant variance, both assumptions still hold. Why then is the OLS estimator biased?

  • 3
    $\begingroup$ I would say the OLS estimator is $(X'X)^{-1}X'y,$ not just $(X'X)^{-1}X'.$ In particular, the form $(X'X)^{-1}X'y$ shows you why the word "linear" is used: It's linear as a function of $y. \qquad$ $\endgroup$ Nov 15, 2018 at 16:54
  • 3
    $\begingroup$ $(X'X)^{-1} X' y$ is still the best linear unbiased estimator. $((X+\mu)'(X+\mu))^{-1}(X+\mu)' y$ is not. $\endgroup$
    – Frank
    Nov 15, 2018 at 17:23
  • 2
    $\begingroup$ Let $z = x + u$. If you rewrite in terms of observable $z$, your problem is $\operatorname{E}[u \mid z] \neq 0$ $\endgroup$ Nov 15, 2018 at 18:16
  • $\begingroup$ Is this a homework question? $\endgroup$ Nov 15, 2018 at 20:16
  • $\begingroup$ No, I thought that noise in the independent variables shouldn't bias the coefficients if it's independent but everything I've read online told me otherwise. Couldn't figure out why. $\endgroup$
    – badmax
    Nov 15, 2018 at 20:52

1 Answer 1


Let's say the true data generating process is: $$ y_i = x_i \beta + \epsilon_i $$ But we don't observe $x_i$, instead we observe $ z_i = x_i + u_i$. We can write the above using observables ($z_i, y_i$): $$ y_i = z_i \beta + v_i $$ Where the error term is $ v_i = \epsilon_i - \beta u_i$. Is the strict exogeneity requirement $\operatorname{E}[v \mid z] = 0$ satisfied? No.

  • If $\operatorname{E}[v \mid z] = 0$ then $\operatorname{E}[vz]=0$, but $\operatorname{E}[vz]=\operatorname{E}[(\epsilon - \beta u)(x + u)] = - \beta \operatorname{E}[u^2] $. $\bot$

The underlying cause is that $\operatorname{E}[u \mid x + u] \neq 0$. The precise story depends on the distribution of $x$ and $u$, but loosely speaking, above average measurements $z$ are going to be associated with positive measurement error $u$.


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