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I do not get this connection.

Is is reliable to divide this equation by $\sum x^2 _t$ to get just $\sigma^2$ ?

$$\sum x^2 _t \times E(\hat \beta - \beta)^2=\sum x^2 _t \times \text{Var}(\beta)=\frac{\sum x^2 _t \times \sigma^2}{ \sum x^2 _t} = \sigma^2$$

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What is the context? What is $\beta$? – Bravo Mar 27 '13 at 17:36
$\beta$ is an the slope of the linear regression model. However I is shown that $E(\sum x^2 _t \times (\hat \beta - \beta)^2)=\sigma^2$ – Le Chifre Mar 27 '13 at 18:15

We know that $\hat{\beta} \sim \mathcal{N}(\beta,\dfrac{\sigma^2}{S_{xx}})$ where $S_{xx}=\sum(x_t-\bar x)^2$.

So, we have $\sum x^2 _t \times Var(\hat\beta)=\sum x^2 _t \cdot \dfrac{ \sigma^2}{S_{xx}}$. Now if $\bar x=0$ then $$\sum x^2 _t \times Var(\hat\beta)=\sum x^2 _t \cdot \dfrac{ \color{red}{\sigma^2}}{\sum x^2 _t}=\color{red}{\sigma^2}\cdot\dfrac{\sum x^2 _t}{\sum x^2 _t} = \sigma^2.$$ Since $\sigma^2$ is constant.

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