# Which of the Gauss-Markov assumptions does error-in-variables violate?

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?

• 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 $y. \qquad$ – Michael Hardy Nov 15 '18 at 16:54
• $(X'X)^{-1} X' y$ is still the best linear unbiased estimator. $((X+\mu)'(X+\mu))^{-1}(X+\mu)' y$ is not. – Frank Nov 15 '18 at 17:23
• Let $z = x + u$. If you rewrite in terms of observable $z$, your problem is $\operatorname{E}[u \mid z] \neq 0$ – Matthew Gunn Nov 15 '18 at 18:16
• Is this a homework question? – Matthew Gunn Nov 15 '18 at 20:16
• 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. – badmax Nov 15 '18 at 20:52

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 $$u$$.