Measurement errors

Solution

\[b_2= β_2+ \frac{∑\left( x_i-\bar{x} \right)ε_i}{∑{\left( x_i-\bar{x} \right)}^2}\]

so

\[plim b_2=β_2+plim \frac{∑\left( x_i-\bar{x} \right)ε_i}{∑{\left( x_i-\bar{x} \right)}^2} \]

We cannot do

\[plim \frac{∑\left( x_i-\bar{x} \right)ε_i}{∑{\left( x_i-\bar{x} \right)}^2}= \frac{plim ∑\left( x_i-\bar{x} \right)ε_i}{plim ∑{\left( x_i-\bar{x} \right)}^2}???\]

since both plims are infinite. We can do

\[plim \frac{∑\left( x_i-\bar{x} \right)ε_i}{∑{\left( x_i-\bar{x} \right)}^2}=plim \frac{n^{-1}∑\left( x_i-\bar{x} \right)ε_i}{n^{-1}∑{\left( x_i-\bar{x} \right)}^2}= \frac{plim n^{-1}∑\left( x_i-\bar{x} \right)ε_i}{plim n^{-1}∑{\left( x_i-\bar{x} \right)}^2}\]

Now, \(n^{-1}∑\left( x_i-\bar{x} \right)ε_i\) is the sample covariance between \(x\) and \(ε\) . As \(n→∞\) , this will converge to the true covariance, \(cov\left( x_i,ε_i \right)\) (which is the same for all \(i\) due to random sampling). Thus,

\[plim n^{-1}∑\left( x_i-\bar{x} \right)ε_i=-β_2σ_u^2\]

\(n^{-1}∑{\left( x_i-\bar{x} \right)}^2\) is the sample variance in \(x\) which converges to the true variance \(Var\left( x_i \right)\) and

\[Var\left( x_i \right)=Var\left( w_i+u_i \right)=Var\left( w_i \right)+Var\left( u_i \right)=σ_w^2+σ_u^2\]

The second equality follows since \(w_i\) and \(u_i\) are independent (no covariance term). We have

\[plim b_2=β_2+ \frac{-β_2σ_u^2}{σ_w^2+σ_u^2}=β_2\left( 1- \frac{σ_u^2}{σ_w^2+σ_u^2} \right)\]