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

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?

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.