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In a Linear Regression model like $Y=\beta_0 +\beta_1X+u$ ,How we can prove that: $\mathrm{Var}(\hat\beta_0)={\bar X^2 \sigma^2\over\Sigma x^2}$ which $x=X-\bar X$

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up vote 1 down vote accepted

It's easiest to work in matrices.

$\hat{\beta} =\bigl(\begin{smallmatrix} \hat\beta_0\\ \hat\beta_1 \end{smallmatrix} \bigr)=(X'X)^{-1}X'Y=\beta+(X'X)X'\mu$

where $X = \bigl(\begin{smallmatrix} \bf{1}'\\ \bf{x}' \end{smallmatrix} \bigr)$ , $\bf{1}$ is a nx1 vector of 1's and $\bf{x}$ is an nx1 vector of the x's.

Assuming homoskedasticity, we have that

$Var(\hat{\beta})=(X'X)^{-1}\sigma^2$

where $(X'X)^{-1} = (\begin{smallmatrix} n&\sum x_i\\ \sum x_i&\sum x^2_i \end{smallmatrix} \bigr)^{-1}$

The variance you want is the (1,1) element in this matrix which is $\frac{\sum x^2_i}{n\sum x_i^2 -(\sum x_i)^2}$ Divide both sides by $n^2$ and you get your answer

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