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I would like to find a Derivation of the efficient frontier set for the markowitz problem: enter image description here

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To solve this constraint minimization problem, first form the Lagrangian Function \begin{align} L(w,\lambda_1,\lambda_2)=w'\Sigma w + \lambda_1(w'\boldsymbol{\mu}-m) + \lambda_2 (w'\boldsymbol{1}-1). \end{align}

The first order conditions for a minimum are then given by \begin{align} \frac{\delta L(w,\lambda_1,\lambda_2)}{\delta w}&=2 \Sigma w + \lambda_1 \boldsymbol{\mu} + \lambda_2 \boldsymbol{1}=\boldsymbol{0} \\ \frac{\delta L(w,\lambda_1,\lambda_2)}{\lambda_1}&=w'\boldsymbol{\mu}-m=0 \\ \frac{\delta L(w,\lambda_1,\lambda_2)}{\lambda_2}&=w'\boldsymbol{1}-1=0. \end{align}

This system of linear equations using matrix algebra can be represented as \begin{align} \begin{bmatrix} 2\Sigma & \boldsymbol{\mu} & \boldsymbol{1} \\ \boldsymbol{\mu}' & 0 & 0 \\ \boldsymbol{1}' & 0 & 0 \end{bmatrix} \begin{bmatrix} w \\ \lambda_1 \\ \lambda_2 \end{bmatrix}= \begin{bmatrix} \boldsymbol{0} \\ m \\ 1 \end{bmatrix}, \end{align} or \begin{align} \boldsymbol{A}\boldsymbol{z}=\boldsymbol{b}, \end{align} where

\begin{align} \boldsymbol{A}:=\begin{bmatrix} 2\Sigma & \boldsymbol{\mu} & \boldsymbol{1} \\ \boldsymbol{\mu}' & 0 & 0 \\ \boldsymbol{1}' & 0 & 0 \end{bmatrix}, \boldsymbol{z}:= \begin{bmatrix} w \\ \lambda_1 \\ \lambda_2 \end{bmatrix} \boldsymbol{b}:= \begin{bmatrix} \boldsymbol{0} \\ m \\ 1 \end{bmatrix}. \end{align} The solution for $\boldsymbol{z}$ is then given by (A has full rank and is thus invertible)

\begin{align} \boldsymbol{z}=\boldsymbol{A}^{-1} \boldsymbol{b} \end{align}

The first element of $\boldsymbol{z}$ gives you the set of efficient portfolios varying m.

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  • $\begingroup$ You might add that one can analytically block invert $A$. This means that if all you really want is the first element of $z$, then you can save yourself some calculations. $\endgroup$ – John Mar 22 '17 at 20:29

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