B1 hat= sigma (Xi - Xbar)(Yi-Y bar) / sigma (Xi-Xbar)^2 —— (1)
Y bar = B0 + B1 hat X bar + e bar——(2)
Yi = B0 + B1Xi + ei
Substituting (2) and (3) into (1) we get (i skipped the intermediate steps. i believe you should be able to manage basic algebraic manipulations)
B1 hat = B1 + sigma (Xi - X bar)ei/ sigma (Xi - X bar)^2 - sigma (Xi-X bar)e bar / sigma (Xi-X bar)^2
The last term = 0 because you can write it as e bar * sigma (Xi-X bar) / sigma (Xi-X bar)^2 and sigma (Xi-X bar ) = 0
Therefore we get B1 hat = B1 + sigma (Xi - X bar)ei/ sigma (Xi - X bar)^2
Taking variance conditional on Xi on both sides
Var (B1 Hat | Xi) = var(B1 + sigma (Xi - X bar)ei/ sigma (Xi - X bar)^2 | Xi)
Since it is conditional on Xi, this means that sigma (Xi - X bar)^2 can be treated as a constant and so we get
Var (B1 Hat | Xi) = var(B1 + sigma (Xi - X bar)ei/ sigma (Xi - X bar)^2 | Xi)
=[sigma (Xi - X bar)^2]^-2 var (sigma (Xi - X bar)ei) —–(1)
var (sigma (Xi - X bar)ei) = ??
To solve for the question marks we rely on the basic definition of Var (x) = E(X - E(X))^2.
E(sigma (Xi - X bar)ei) = sigma (Xi - X bar ) E (ei) = 0 since E(ei)=0 [ take note i am actually making use of conditional expectations here, i.e. i am taking expectations conditional on Xi so sigma (Xi- Xbar) is actually a constant).
Therefore, var (sigma (Xi - X bar)ei) = E(sigma (Xi - X bar)ei)^2 = E[(sigma (Xi - X bar))^2*(ei)^2)] + cross product terms. = sigma (Xi - X bar )^2 E(ei^2)+E (cross pdt terms) and tjhe last term = 0 because of the assumption that ei~iid [ei and ej are independent and so cov (ei,ej)=0]
Therefore var (sigma (Xi - X bar)ei) =sigma (Xi - X bar )^2 *E(ei^2)=sigma (Xi - X bar )^2 *sigma ^2 [from homoskedasticity assumption]
hence, var (b1 hat) = sigma (Xi - X bar )^2 *sigma ^2*[sigma (Xi - X bar)^2]^-2 = sigma^2/sigma (Xi - X bar )^2 (PLUgging the above into (1))