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


and we know


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

  • 1
    $\begingroup$ What is the relation between $\mu_i$ and $r_i$? $\endgroup$
    – Bob Jansen
    Commented Jan 13, 2019 at 20:22
  • $\begingroup$ 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 $\mu_i=E_t[r_{it}]$ $\endgroup$
    – Alex C
    Commented Jan 14, 2019 at 13:20

1 Answer 1


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.


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