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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?

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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.

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  • $\begingroup$ 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

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