## 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 $$c=\mathbf{1}^{'} \Sigma^{-1} \mathbf{1}$$