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What I got was

$\text{Var}(\beta) = \sigma^2(X'X)^{-1}$
$\text{Var}(\beta_1) = \sigma ^ 2(X_1'X_1)^{-1}$
$\text{Var}(\beta_2) = \sigma ^ 2(X_2'X_2)^{-1}$
$\text{Var}(\tilde{\beta}) = \frac{1}{4}\left(\text{Var}(\beta_1) + \text{Var}(\beta_2)\right)$

and then I don't know what to do next.

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    $\begingroup$ Please add the self-study tag. And tell us what you have tried. $\endgroup$ – Roland Sep 23 '18 at 21:56
  • $\begingroup$ For (a), start by expressing the variance of $\hat \beta$, $\hat \beta_1$, $\hat \beta_2$ and then $\tilde \beta$. Try to edit the question with what you find then, and where you're stuck. $\endgroup$ – Roland Sep 23 '18 at 23:17
  • $\begingroup$ Remember that you are dealing with a simple regression model (not multiple), so formulas are simpler. For instance, $Var(\hat \beta) = \frac{\sigma^2}{\sum (X_i-\bar X)^2}$. You can also write it: $Var(\hat \beta) = \frac{\sigma^2}{vn}$, where $v = \frac{1}{n}\sum (X_i-\bar X)^2$. You can write the same expression for $Var(\hat \beta_1)$ and $Var(\hat \beta_2)$ and you should see what you need to assume to get the equality. $\endgroup$ – Roland Sep 24 '18 at 8:13
  • $\begingroup$ $\frac{\sigma^2}{\sum{(X_i - \bar{X})^2}} = \frac{\frac{\sigma^2}{\sum{(X_{1i} - \bar{X})^2}} + \frac{\sigma^2}{\sum{(X_{2i} - \bar{X})^2}}}{4}$ so it means vn1 = vn2 = 0.5 * vn ? If it is the case, why do we need the $\bar{X_1} = \bar{X_2}$ Thank you! $\endgroup$ – wcLin Sep 24 '18 at 16:11
  • $\begingroup$ Solved part (a). But still no idea about part (b) $\endgroup$ – wcLin Sep 24 '18 at 22:43

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