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Stock and Watson say nothing about that, so I ask you if you know how to derive the variance of the estimator of the intercept in a simple regression, $\beta_0$, i.e. $$\sigma^2_{\hat{\beta}_0}=\frac{1}{n}\frac{var[H_iu_i]}{[E(X_{i}^{2})]^2}.$$

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First, there is a mistake in your expression; it should be $$\sigma^2_{\hat{\beta}_0}=\frac{1}{n}\frac{var[H_iu_i]}{[E(H_{i}^{2})]^2}.$$ I will depart from the general expression for the asymptotic variance covariance matrix of the OLS estimator under conditional heteroskedasticity: \begin{equation*} \sqrt{n}(b-\beta)\to_d N(0,\Sigma_{xx}^{-1}S\Sigma_{xx}^{-1}). \end{equation*} A proof of this result may be found in, e.g., Hayashi, Econometrics, Proposition 2.1. In case of a simple linear regression with intercept $$ \Sigma_{xx}:=E(\mathbf{x}_i\mathbf{x}_i') $$ reduces to $$\Sigma_{xx}=\begin{pmatrix}1&\mu_X\\ \mu_X & E(X_i^2)\end{pmatrix}$$ Next, $$S=\begin{pmatrix}E(u_i^2)&E(X_iu_i^2)\\ E(X_iu_i^2)& E(X_i^2u_i^2)\end{pmatrix}$$ in this case.

So the task is to find the $1,1$-element of $\Sigma_{xx}^{-1}S\Sigma_{xx}^{-1}$. A tedious task, which you may speed up like this, to get \begin{eqnarray*} Avar(\hat\beta_0)&=&\frac{1}{\sigma^4_X}\{E(X_i^2)[E(u_i^2)E(X_i^2)-\\&&\qquad\mu_XE(X_iu_i^2)]-\mu_X[E(X_iu_i^2)E(X_i^2)-\mu_XE(X_i^2u_i^2)]\} \end{eqnarray*} We then need to check that this is (up to $1/n$, which results from using the asymptotic variance to approximate the finite-sample one) the same as your result.

Here, $$ H_i=1-\left(\frac{\mu_X}{E(X_i^2)}\right)X_i. $$ As in your related question, we may always write $$ var(H_iu_i)=E(H_i^2u_i^2)-[E(H_iu_i)]^2 $$ Under the maintained assumption of correct specification, $E(u_i|X_i)=0$ and hence, by the law of iterated expecations, as $H_i$ only depends on $X_i$, $$ E(H_iu_i)=E(H_iE(u_i|X_i))=0 $$ Hence, $$ var(H_iu_i)=E(H_i^2u_i^2) $$ Here, $$ H_i^2=\frac{E(X_i^2)^2-2\mu_XX_iE(X_i^2)+\mu_X^2X_i^2}{[E(X_i^2)]^2}=\frac{[E(X_i^2)-\mu_XX_i]^2}{[E(X_i^2)]^2} $$ so that $$ E(H_i^2)=\frac{\sigma^2_X}{E(X_i^2)} $$ and $$ [E(H_i^2)]^2=\frac{\sigma^4_X}{[E(X_i^2)]^2} $$ as well as $$ E(H_i^2u_i^2)=\frac{E\{[E(X_i^2)-\mu_XX_i]^2u_i^2\}}{[E(X_i^2)]^2} $$ giving $$ Avar(\hat\beta_0)=\frac{E\{[E(X_i^2)-\mu_XX_i]^2u_i^2\}}{\sigma^4_X}.$$ Solving the numerator verifies that this is the same expression as the one in the above numerator of $Avar(\hat\beta_0)$.

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