4
$\begingroup$

Consider the model $$y_i = x_{i}'\beta + e_i.$$

In the discussion of instrumental variables estimator it can be shown that OLS estimator $b$ is biased and inconsistent estimator of $\beta$. Nonetheless, $b$ does estimate something: $plim_{n\to\infty}b = \beta + Q^{-1}\gamma = \theta$ with $Q = \mathbb{E}(x_i'x_i)$ and $\mathbb{E}(x_ie_i) = \gamma \neq 0.$
Assume data is i.i.d. and that $\mathbb{E}(x_i'x_i)$ has full rank.

  1. Show that $b$ is asymptotically normally distributed.
  2. Derive the asymptotic covariance matrix of $b$.

My attempt. Write $b - \beta = \left(\frac{1}{n}\sum_ix_i'x_i\right)^{-1}\left(\frac{1}{n}\sum_ix_ie_i\right)$. Since $\mathbb{E}(x_ie_i) \neq 0$, we can't use CLT directly on the RHS. But observe that $$\sqrt{n}\left(\frac{1}{n}\sum_ix_ie_i - \mathbb{E}(x_ie_i)\right) \xrightarrow{d} N(0, V),$$ where $V = Var(x_ie_i)$. Now transform expression for $b - \beta$ :

\begin{equation} \begin{aligned} b - \beta - \left(\frac{1}{n}\sum_ix_i'x_i\right)^{-1}\gamma &= \left(\frac{1}{n}\sum_ix_i'x_i\right)^{-1}\left(\frac{1}{n}\sum_ix_ie_i\right) - \left(\frac{1}{n}\sum_ix_i'x_i\right)^{-1}\gamma \\ & = \left(\frac{1}{n}\sum_ix_i'x_i\right)^{-1}\left(\frac{1}{n}\sum_ix_ie_i - \gamma\right) \end{aligned} \end{equation}

Multiply both sides by $\sqrt{n}$ to get $$\sqrt{n}\left[(b - \beta - \left(\frac{1}{n}\sum_ix_i'x_i\right)^{-1}\gamma)\right] = \left(\frac{1}{n}\sum_ix_i'x_i\right)^{-1}\sqrt{n}\left(\frac{1}{n}\sum_ix_ie_i - \gamma\right)$$

$\left(\frac{1}{n}\sum_ix_i'x_i\right)^{-1} \xrightarrow{p} Q^{-1} = \mathbb{E}(x_i'x_i)$ and $\sqrt{n}\left(\frac{1}{n}\sum_ix_ie_i - \gamma\right) \xrightarrow{d} N(0, V)$. Then, by Slutsky theorem $$\sqrt{n}\left[(b - \beta - \left(\frac{1}{n}\sum_ix_i'x_i\right)^{-1}\gamma)\right] \xrightarrow{d} N(0, Q^{-1}VQ^{-1})$$


I am not quite sure that the result is correct and makes sense. Any hints or suggestions are welcome.

$\endgroup$
2
$\begingroup$

The result is ok. What you have done is to find a specific function of $b$ (the LHS of your last expression) that does converge asymptotically to a zero-mean normal with the stated variance.

Note also that both the LHS and the RHS contain non-estimable parameters - the mean $\gamma$ and the variance $V$ of $x_ie_i$ (although for the variance there usually are some assumptions that permit you to work a bit further). This is the reason why the consistency property is so crucial -it allows us to estimate/use asymptotic results.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.