So I was trying to prove that $Var[\hat{\beta_o}]=\dfrac{\sigma^2n^{-1}\sum{(x_i)^2}}{\sum{(x_i-\bar{x}})^2}$

And I got stuck with the part $\dfrac{ -2\bar{x}}{\sum{(x_i-\bar{x})^2}}\sum[{(x_i - \bar{x})}E(u_i\bar{u})]$

Now, I need to show that this whole expression equals 0 but I don't know how to go about it. I'm mainly confused as to how I should be treating the $E(u_i\bar{u})$ term. I have read that $u_i$ and $u_j$ are uncorrelated so $E(u_iu_j)=0$ so long as $i \ne j$. But I don't know how I can use that fact to reduce the $E(u_i\bar{u})$ part to something. The $u_i$ I'm assuming would go on forever i.e. $i=1,2,3........$ but the $\bar{u}$ would only have the values of $u$ that go from $1$ to $n$ like $u_1+u_2+....+u_n$, so how should I go about this?

Note: $u_i's$ are NOT the residual terms. These are the actual $u's$ from the line/function that models the relationship between $y$ and $x$ And $\bar{u}$ would be defined to be actual $u's$ from $1$ to $n$ whole divided by $n$

Can anyone help simplify this? Thank you.


I'm assuming that this for SLR.

So at some point, you should get something like $V(\hat{\beta_0}) = V(\bar{Y}) + \bar(x)^2V(\hat{\beta}_1) - 2\bar{x}\text{Cov}(\bar{Y}, \beta_1)$.

I'm assuming that you are asking about the last term, but what you should get is $\frac{\sigma^2}{n} \sum(\frac{x_i - \bar{x}}{S_{xx}})$, which is slightly different than what you have, and it is the $\sum({x_i - \bar{x}}) = 0$ that causes the whole term to go to 0.

For more details, for example, look at the answer here.

how to calculate variance


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