Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I do not get this connection.

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

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

share|improve this question
    
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
add comment

1 Answer 1

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.