Consistent estimator - consistent with what exactly?

Question

Lets assume, that the real DGP (real world data) is generated from the model:

$$y_i = \beta_0 + \beta_1x_{1i} + \beta_2x_{2i} + \varepsilon_i$$

Lets further assume, that $x_1$ and $x_2$ are correlated. Precisely, $x_1$ is a confounder variable, that causes $x_2$:

$$x_{2i} = \alpha_0 + \alpha_1 x_{1i} + u_i$$

Researcher does not know above information, he is sure, that the true model has only one variable, and assumes following functional form:

$$ y_i = \gamma_0 + \gamma_2x_{2i} + v_i $$

What can we, who know everything, tell about the consistency of the estimator $\hat \gamma_2$?

• It is inconsistent, because consistent estimator has limit in the 'real world parameter', which in this case is $\beta_2$.
• It is consistent, because consistent estimator has limit in the parameters of 'assumed model'. In this case $\gamma_2$. It is the model, which does not fit real world, not the estimator.

I see these two possibilities. Which one is (more) true, and what is most important - why?

Answer 1

Neither. An estimator is consistent for some parameter, so in this case the answer is

• Yes, $\hat\gamma_2$ is consistent for $\beta_2$
• No, $\hat\gamma_2$ is not consistent for $\gamma_2$ (or for $\beta_0$ or lots of other things).

In this case, the causal assumptions suggest you'd be more interested in whether it was consistent for $\gamma_2$, but you still need to say "consistent for $\gamma_2$", not just "consistent". The same is true for 'biased' and 'unbiased': an estimator is biased or unbiased for a parameter

Sometimes there is genuinely only one interesting limit, and it's a reasonable abuse of notation to leave it implied, but a claim of consistency does require specifying the limit.