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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?

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    $\begingroup$ What is the relation between $\mu_i$ and $r_i$? $\endgroup$ – Bob Jansen Jan 13 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 Jan 14 at 13:20
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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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