# Market Portfolio Optimization

Consider the minimization problem

$$\min\left\{\frac{1}{2}x^T\Sigma x - \lambda(\mu-r_f)^Tx\right\}$$

and assume the CAPM model, i.e.

$$r_i-r_f = \beta_i(r_m-r_f) + \varepsilon_i$$

Assuming $$\Sigma$$ is invertible, prove

$$x_i \propto \frac{\beta_i}{\textrm{Var}(\varepsilon_i)}$$

It seems like lambda must stay in the minimization problem after solving for $$x$$, which is probably why we're only solving for proportionality, but I still cannot find a way to go about tackling this. Solving the Lagrangian yields

$$x=\lambda\Sigma^{-1}(\mu-r_f)$$

and we know

$$(\mu-r_f)^Tx=0$$

but this doesn't seem to help me. Where does the quadratic term yielding variance in the solution come from?

• What is the relation between $\mu_i$ and $r_i$? Jan 13 '19 at 20:22
• The notation is not well set up. $\lambda$ is a parameter which expresses a preference between risk and return, it is not a "lagrange multiplier" related to a constraint (so the statement after "and we know..." is false). It might help to write the CAPM as $r_{it}-r_f=\beta_i(r_{mt}-r_f)+\epsilon_{it}$ with $\mu_i=E_t[r_{it}]$ Jan 14 '19 at 13:20

Assuming the $$\epsilon_i$$ are zero mean, you should find that $$\mu - r_f = \beta \left(E[r_m] - r_f\right).$$ Further assuming the $$\epsilon_i$$ are independent of each other, though possibly with different variances, let $$\Gamma$$ be the diagonal matrix with the variances of $$\epsilon_i$$ on the diagonal. Then you are to find (under the more usual MVO formulation) $$\max_x \,\, x^{\top}\beta \left(E[r_m] - r_f\right) - \frac{1}{2\lambda} x^{\top}\left(\beta \beta^{\top}\sigma^2 + \Gamma\right)x.$$ (I am keeping your $$\lambda$$ associated with the mean, though usually it is risk aversion and so you would see $$\lambda/2$$.)
Now use the Lagrange Multiplier technique to find the solution, which should be something like $$x \propto \left(\beta\beta^{\top}\sigma^2 + \Gamma\right)^{-1}\beta,$$ and then use the Sherman-Morrison-Woodbury formula to simplify the matrix inverse.