# Consistency of estimator

Regression: Wage=b0+b1collegegrad, where collegegrad is a dummy variable. Suppose you want to estimate the wage ratio between college graduates and non-college graduates. Is the estimator theta=b0/b1 consistent?

My thinking is that if we could increase our sample size to infinity, we would cover the entire population and thus get the true ratio, so the estimator is consistent. Am I correct, or am I missing something?

• Doesn't it just follow from Slutsky? – Glen_b -Reinstate Monica Oct 1 '12 at 4:52
• Your intuition is correct in some sense (you have to be a bit careful when talking about what exactly that makes up the population), but probably won't do as an answer to the homework. As @Glen_b suggested, Slutsky's theorem might be of help. – MånsT Oct 1 '12 at 5:54
• Hint: what does $b_0/b_1$ estimate? (Strong hint: it does not estimate the wage ratio.) – whuber Oct 1 '12 at 14:14
• Oh, I see my mistake now. Would the estimator b0/(b0+b1) be an appropriate consistent estimator instead? Since b0 is the wage for non college graduates and b0+b1 is the wage for college graduates? – user14386 Oct 2 '12 at 16:27

To close this one: We have the regression $$w = b_0+b_1d + u$$ where $d$ is a binary $1/0$ random variable, and $u$ is an independent error term, with $E(u\mid d) = 0$ and so also $E(u)=0$. The size of the i.i.d. sample is $n$ and we denote by $n_1$ the number of observations for which $d=1$ and by $n_0$ the number of the rest.

One can find that the OLS estimator gives

$$\hat b_0 = \frac 1{n_0}\sum_{d_i=0}w_i = \frac 1{n_0}\left(n_0b_0+\sum_{d_i=0}u_i\right) = b_0 + \frac 1{(n_0/n)}\left(\frac 1n\sum_{d_i=0}u_i\right) \tag{1}$$

and

$$\hat b_1 = \frac{n}{n_1n_0}\sum_{d_i=0}w_i - \frac 1{n_0}\sum_{i=1}^nw_i$$

$$=\frac{n}{n_1n_0}\left(n_1b_0+n_1b_1+\sum_{d_i=1}u_i\right) - \frac 1{n_0}\left(nb_0+n_1b_1+\sum_{i=1}^nu_i\right)$$

$$=b_1 +\frac{n}{n_1n_0}\sum_{d_i=1}u_i - \frac 1{n_0}\sum_{i=1}^nu_i$$

$$\Rightarrow \hat b_1 = b_1 +\frac 1{(n_1/n)}\frac 1n\sum_{d_i=1}u_i - \frac 1{(n_0/n)}\frac 1n\sum_{d_i=0}u_i \tag{2}$$

So

$$\text{plim}\hat b_0 = b_0 + \text{plim}\left[\frac 1{(n_0/n)}\right]\cdot \text{plim}\left(\frac 1n\sum_{d_i=0}u_i\right)$$

and by the Law of Large Numbers

$$= b_0 + [1/P(d=0)]\cdot \left(\frac 1n\sum_{d_i=0}E(u_i)\right) = b_0$$

since $E(u_i)=0$, and analogously for $\hat b_1$. So both coefficient estimators are consistent therefore the expression

$$\frac {\hat b_0}{ \hat b_0 + \hat b_1} \xrightarrow{p} \frac {b_0}{b_0 + b_1}$$

since the probability limit distributes, when it is a constant, as a consequence of the Continuous Mapping (Mann-Wald) Theorem.