Markowitz Modern Portfolio Theory / Mean Variance Framework

We derive the efficient frontier, with and without risk free asset, formulae below. We also give characterization of the minimum variance portfolio.

Efficient Frontier of Risky Assets

Let W represent the vector of the portfolio weights, R the vector of the returns, and $$\Sigma$$ the Variance-Covariance matrix. Then the mean and variance of the portfolio can be expressed in matrix notation as follows:

$$r_{p}= W^{'}R = R^{'}W$$ $${\sigma_{p}}^2=W^{'}\Sigma W$$

The portfolio optimisation problem is then to minimise the variance of the portfolio for a given level of return and subject to budget constraint (sum of weights=1):

$$L \left( W, \lambda, \gamma \right)=\frac{1}{2} W^{'}\Sigma W + \lambda \left( r_{p}-R^{'}W\right)+\gamma \left(1- \mathbf{1}^{'} W \right)$$

Differentiating and setting the derivatives equal to zero, we get

$$\frac{\partial L \left( W, \lambda, \gamma \right)}{\partial W}=\Sigma W - \lambda R -\gamma \mathbf{1}=0$$ $$\frac{\partial L \left( W, \lambda, \gamma \right)}{\partial \lambda}=r_{p}-R^{'}W=0$$ $$\frac{\partial L \left( W, \lambda, \gamma \right)}{\partial \gamma}=1- \mathbf{1} ^{'} W =0$$

We need to solve the three equations for the three unknowns, with W being the ultimate prize. Solving the first equation for W, we get

$$\Sigma W - \lambda R -\gamma \mathbf{1}=0$$ $$W=\lambda \Sigma^{-1}R + \gamma \Sigma^{-1} \mathbf{1}$$

And substituting into the second and the third equations,

$$r_{p}-R^{'}W=0$$ $$r_{p}-R^{'} \left( \lambda \Sigma^{-1}R + \gamma \Sigma^{-1} \mathbf{1} \right)=0$$ $$r_{p}=\lambda R^{'} \Sigma^{-1}R + \gamma R^{'} \Sigma^{-1} \mathbf{1}$$

$$1- \mathbf{1}^{'}W =0$$ $$1- \mathbf{1}^{'} \left( \lambda \Sigma^{-1}R + \gamma \Sigma^{-1} \mathbf{1} \right) =0$$ $$1= \lambda \mathbf{1}^{'}\Sigma^{-1}R + \gamma \mathbf{1}^{'} \Sigma^{-1} \mathbf{1}$$

The above equations look complicated, but fear not because the intimidating vector/matrix multiplications in the above equations -e.g., $$R^{'} \Sigma^{-1}R$$ - produce scalars, and treating them that way by substituting: $$a=\mathbf{1}^{'}\Sigma^{-1}R$$, $$b= R^{'} \Sigma^{-1}R$$, and $$c=\mathbf{1}^{'} \Sigma^{-1} \mathbf{1}$$, we get 2 simple equations for 2 unknowns:

$$r_{p}= b \lambda + a \gamma$$ $$1=a \lambda + c \gamma$$

Subtracting b times the second equation from a times the first equation, and solving for $$\gamma$$ , we get

$$a r_{p}-b = a b \lambda -a b \lambda + a^{2} \gamma - b c \gamma$$ $$\gamma =\frac{a r_{p}-b}{a^{2} - b c }$$

And subtracting a times the second equation from c times the first equation, and solving for $$\gamma$$ , we get

$$c r_{p}-a = b c \lambda -a^{2} \lambda + a c \gamma - a c \gamma$$ $$\lambda=\frac{c r_{p}-a}{b c - a^{2} }$$

Thus:

$$W=\lambda \Sigma^{-1}R + \gamma \Sigma^{-1} \mathbf{1}$$ $$W=\frac{c r_{p}-a}{b c - a^{2} } \Sigma^{-1}R + \frac{a r_{p}-b}{a^{2} - b c } \Sigma^{-1} \mathbf{1}$$

Where $$a=\mathbf{1}^{'}\Sigma^{-1}R$$, $$b= R^{'} \Sigma^{-1}R$$, and $$c=\mathbf{1}^{'} \Sigma^{-1} \mathbf{1}$$