# 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 $y. \qquad$ Nov 15, 2018 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. Nov 15, 2018 at 17:23
• Let $z = x + u$. If you rewrite in terms of observable $z$, your problem is $\operatorname{E}[u \mid z] \neq 0$ Nov 15, 2018 at 18:16
• Is this a homework question? Nov 15, 2018 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. Nov 15, 2018 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 $$u$$.