What's the asymptotic variance for OLS estimates of intercept and slope of homoskedastic simple linear regression?

Question

Suppose data is generated by $Y_i=\beta_0+\beta_1X_i+U_i$ satisfying $E(U_i|X_i)=0$ and $E(U_i^2|X_i)=\sigma^2$. Suppose I have a random sample $\{Y_i,X_i\}_{i=1}^{n}$, and obtained OLS estimates $\widehat{\beta}_0$ and $\widehat{\beta}_1$. Suppose conditions for CLT of the OLS estimates are satisfied. What's the close-form asymptotic variance for each one of them? I calculated the following results: asymptotic variance for $\widehat{\beta}_0$ is $\frac{\sigma^2[E(X)]^2}{Var(X)}+\sigma^2$, and asymptotic variance for $\widehat{\beta}_1$ is $\frac{\sigma^2}{Var(X)}$. Does this look correct?

Answer 1

I think you have to divide $\sigma^2$ by the sample size $n$ in each of your equations. From the formula for the conditional variance $$var(\hat\beta_1\vert x)=\frac{\sigma^2}{\sum (x_i-\bar x)^2}$$ one has to multiply with 1/n in the nominator and denominator to get $$var(\hat\beta_1\vert x)=\frac{\frac{\sigma^2}{n}}{\frac{1}{n}\sum (x_i-\bar x)^2} = \frac{\frac{\sigma^2}{n}}{\hat \sigma_x^2}$$ Since $\hat \sigma_x^2=\frac{1}{n}\sum (x_i-\bar x)^2$ is a consistent estimator for the variance of $x$, the above expression converges to $$ \frac{\frac{\sigma^2}{n}}{\sigma_x^2}$$ which is the same as yours just divided by $n$. The same should be true for the formula for the intercept.