I am stuck on the following question*:
derive the least squares estimate:
$$ b1 = \frac{\sum X_iY_i - \frac{\sum X_i \sum Y_i}{n}} {\sum X_i^2 - \frac{( \sum X_i)^2}{n}} $$
from the normal equations:
$$ (i) \sum Y_i = nb_0 + b_1 \sum X_i $$ $$ (ii) \sum X_iY_i = b_0 \sum X_i + b_1 \sum X_i^2$$
I get:
$$ b1 = \frac { \sum X_i Y_i - \frac {\sum X_i \sum Y_i}{n} } {(\sum X_i)^2 - \frac {(\sum X_i)^2}{n}} $$
I think I am there except that I have $(\sum X_i)^2$ instead of $\sum X_i^2$ in the demonimator. I take the latter to mean the sum (X * X) rather than the sum(X) * sum(X). However I cannot see how you would get sum (X * X) from the original equations. Any help would be much appreciated.
Unless (X * X) is interpreted the matrix muliplication (X * X') where X is a column vector in which case $( \sum X_i)^2$ and $\sum X_i^2$ are equal.
(Please see below for working)
I first solve for $b_0$ with $i - (\frac {ii} {\sum X_i})$:
$(\frac {ii} {\sum X_i})$: $ \frac{\sum X_iY_i}{\sum X_i} = b_0 + b_1 \sum X_i$
$(i - \frac {ii} {\sum X_i})$: $ \sum Y_i - \frac{\sum X_iY_i}{\sum X_i} = nb_0 - b_0$
rearranging yeilds:
$$ b_o = \frac {\sum X_i \sum Y_i - \sum X_i Y_i}{\sum X_i (n -1)} $$
then substituting into (i) (rearranged to isolate b1) and cancelling:
$$ b1 = \frac { \sum Y_i - n \frac { \sum X_i \sum Y_i - \sum X_i Y_i} {\sum X_i(n-1)}} { \sum X_i} $$
$$ b1 \sum X_i = \frac { \sum X_i \sum Y_i (n-1) - n \sum X_i \sum Y_i - n\sum X_i Y_i} {\sum X_i(n-1)} $$
$$ b1 = \frac {(n \sum X_i \sum Y_i - \sum X_i \sum Y_i) - (n \sum X_i \sum Y_i - n\sum X_i Y_i)} {\sum X_i \sum X_i(n-1)} $$
$$ b1 = \frac {n \sum X_i Y_i - \sum X_i \sum Y_i } {\sum X_i \sum X_i(n-1)} $$
$$ b1 = \frac {n \sum X_i Y_i - \sum X_i \sum Y_i } {n(\sum X_i)^2 - (\sum X_i)^2} $$
multiply both sides by $\frac {\frac{1}{n}}{\frac{1}{n}}$
$$ b1 = \frac { \sum X_i Y_i - \frac {\sum X_i \sum Y_i}{n} } {(\sum X_i)^2 - \frac {(\sum X_i)^2}{n}} $$
*q2.30 from Applied Linear Statistical Models, Neter et al 3rd edition.
Your error occurs when you divide $(ii)$ by $\sum X_i$. The result should be: $$\frac{\sum X_iY_i}{\sum X_i} = b_0 + b_1 \frac{\sum X_i^2}{\sum X_i} $$ because, as you mentioned, $\sum X_i^2$ does not mean $(\sum X_i)(\sum X_i)$.