Similar question to this one: An Analytic Derivation of the Efficient Portfolio Frontier question (from Merton 1972)
Here's the paper itself: http://www.stat.ucla.edu/~nchristo/statistics_c183_c283/analytic_derivation_frontier.pdf
My question is with regard to step (9), which is the result of plugging the results of (8) into (4), yielding:
$(9) \qquad x_k = \frac{E\, \sum_{j = 1}^m v_{kj}\, (CE_j - A)+ \sum_{j = 1}^m v_{kj}\, (B - AE_j)}{D}, \qquad k = 1, ..., m$
I can't figure out why the $E$s that appear in parentheses have an index. In step (8) we solve for $\gamma_1$ and $\gamma_2$ both of which include an $E$ but the aggregation of all the $E_j$s, not one in particular. Why do the indices appear in (9)?
Can anyone see what's going on here? Thank you.
Equation (4) obtained from applying first-order conditions is
$$\tag{4} x_k = \gamma_1 \sum_j v_{kj} E_j + \gamma_2\sum_j v_{kj}$$
After substituting for the Lagrange multipliers in (4) with $\gamma_1 = (CE- A)/D$ and $\gamma_2 = (B- AE)/D$ we obtain
$$\begin{align}x_k &= \frac{CE-A}{D}\sum_j v_{kj} E_j + \frac{B-AE}{D}\sum_j v_{kj}\\ &= \frac{CE}{D}\sum_j v_{kj} E_j - \frac{A}{D}\sum_j v_{kj} E_j + \frac{B}{D}\sum_j v_{kj}- \frac{AE}{D}\sum_j v_{kj}\\ &=\underbrace{\frac{CE}{D}\sum_j v_{kj} E_j - \frac{AE}{D}\sum_j v_{kj} }_{= \frac{E}{D}\sum_j v_{kj}(CE_j-A)}+ \underbrace{\frac{B}{D}\sum_j v_{kj} - \frac{A}{D}\sum_j v_{kj}E_j}_{=\frac{1}{D}\sum_j v_{kj}(B- AE_j)}\\ &= \frac{E\sum_j v_{kj}(CE_j-A)+\sum_j v_{kj}(B- AE_j) }{D} \end{align}$$
This result matches Equation (9) in the paper.