Stock and Watson express the variance of $\hat{\beta _0}$ like $\hat{\sigma }^2_\hat{\beta _0}=\frac{E({X_{i}}^{2})}{n\sigma _{X}^{2}}\sigma ^{2}$, but starting from variance of $\hat{\beta _1}=\frac{\sigma^{2}}{n\sigma_{X}^{2}}$ i proved only that $\hat{\sigma }^2_\hat{\beta _0}=\frac{1}{n}\sigma^{2}(1+\frac{\bar{X}^2}{\sigma_{X}^{2}})$, that is the same that is showed here.

How can i prove that are similar forms?

| improve this question | | | | |

They are certainly not the same, since the first contains an expected value while the second does not contain an expected value but a sample mean.

Of course, they are very closely connected, and one can see this if one writes down the definition of the variance of a random variable.

| improve this answer | | | | |
  • $\begingroup$ In every case, i've been told that first formula ($\hat{\sigma }^2_\hat{\beta _0}=\frac{E({X_{i}}^{2})}{n\sigma _{X}^{2}}\sigma ^{2}$) is derived from this formula: $\sigma_{\hat{\beta_0}}^2=\frac{1}{n}\frac{var(H_iu_i)}{[E(H_{i}^{2})]^2}$, where $H_i=1-[\frac{\mu_X}{E(X_i^{2})}]X_i$, that is for heteroskedastic case. Could you help me to understand how derivate it? $\endgroup$ – Francesco Totti Jan 5 '18 at 10:28

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.