I am trying to find a closed form solution for the constrained MVO problem below.

$\max_w w'\mu - \frac{\lambda}{2}w'\Sigma w $
s.t. $w'$1 = 1

The Lagrange for the objective is $L(w, \gamma) = w'\mu - \frac{\lambda}{2}w'\Sigma w -\gamma(w'$1 - 1).

The first order conditions are:

$\frac{\partial L}{\partial w} = \mu - \lambda \Sigma w - \gamma$ 1 = 0
Hence, $w = \frac{1}{\lambda}\Sigma^{-1}(\mu-\gamma$ 1) $\hspace{1cm} [1]$

$\frac{\partial L}{\partial \gamma} = w'$1 - 1 = 0
Hence, $w$'1 = 1 $\hspace{1cm} [2]$

Just to clarify, $\gamma$ is a real number; 1, $\mu$ have dimensions of Nx1; $\Sigma$ has dimension of NxN.

Substitute [1] into [2]
$[\frac{1}{\lambda}\Sigma^{-1}(\mu-\gamma$ 1$)]'$1 = 1
$(\mu-\gamma 1)'\Sigma^{-1} $ 1 = $\lambda$

I am unable to express $\gamma \text{ in terms of } \Sigma, \mu, \lambda$, to substitute it into [1] to write a solution for $w$.

Any help is very much appreciated.

  • 1
    $\begingroup$ Rewrite your FOC as a linear equation system in matrix notation. Solve for $x$ using the blockwise matrix inversion theorem en.m.wikipedia.org/wiki/Block_matrix $\endgroup$ Nov 7 '20 at 7:54
  • $\begingroup$ If that doesn’t help you I can elaborate in an answer $\endgroup$ Nov 7 '20 at 8:04
  • $\begingroup$ @Kermittfrog could you please elaborate? I tried writing down the FOCs but couldn't see where I should apply the inversion theorem. Thank you so much for your help. $\endgroup$
    – vpy
    Nov 8 '20 at 1:36

Let's stick with the nomenclature in the literature and let $\gamma$ denote the decision maker's risk aversion coefficient. The optimization problem is

$$ \max_{\mathrm{w}} \mathrm{w}^T\mathrm{\mu}-\frac{1}{2}\gamma \mathrm{w}^T\mathrm{\Sigma}\mathrm{w} \quad s.t. \mathrm{w}^T\mathrm{e}=1 $$ where $e$ denotes a vector of ones. The corresponding Lagrangean reads: $$ L(\mathrm{w},\lambda)= \mathrm{w}^T\mathrm{\mu}-\frac{1}{2}\gamma \mathrm{w}^T\mathrm{\Sigma}\mathrm{w} -\lambda\left( \mathrm{w}^T\mathrm{e}-1\right) $$

The first order conditions are linear: \begin{align} L_w&=\mathrm{\mu}-\gamma\mathrm{\Sigma}\mathrm{w}-\lambda\mathrm{e}=!=0\\ L_{\lambda}&=\mathrm{e}^T\mathrm{w}=!=1 \end{align}

We can formulate this as a linear system as: \begin{equation} \begin{pmatrix} \gamma\mathrm{\Sigma} & \mathrm{e}\\ \mathrm{e}^T & 0\end{pmatrix} \begin{pmatrix} \mathrm{w}\\ \lambda\end{pmatrix}=\begin{pmatrix}\mathrm{\mu} \\ 1\end{pmatrix} \end{equation} and hence

\begin{equation} \begin{pmatrix} \mathrm{w^*}\\ \lambda^*\end{pmatrix}= \begin{pmatrix} \gamma\mathrm{\Sigma} & \mathrm{e}\\ \mathrm{e}^T & 0\end{pmatrix}^{-1} \begin{pmatrix}\mathrm{\mu} \\ 1\end{pmatrix}=\begin{pmatrix}c_{11} & c_{12} \\ c_{21} & c_{22}\end{pmatrix}\begin{pmatrix}\mathrm{\mu} \\ 1\end{pmatrix} \end{equation}

This is where we use the block matrix inversion theorem . We know that $w^*$ is given by the first row of the inverted matrix times the constraint vector,

$$ \mathrm{w}^*=c_{11}\mathrm{\mu}+c_{12} $$ Looking up the two inversions from wiki, we find

$$ c_{11}=\frac{1}{\gamma}\mathrm{\Sigma}^{-1}-\frac{1}{\gamma}\mathrm{\Sigma}^{-1}\mathrm{e}\left(\mathrm{e}^T\frac{1}{\gamma}\mathrm{\Sigma}^{-1}\mathrm{e}\right)^{-1}\mathrm{e}^T\frac{1}{\gamma}\mathrm{\Sigma}^{-1} $$ and $$ c_{12}=\frac{1}{\gamma}\mathrm{\Sigma^{-1}}\mathrm{e}\left(\mathrm{e}^T\frac{1}{\gamma}\mathrm{\Sigma}^{-1}\mathrm{e}\right)^{-1}=\frac{\mathrm{\Sigma^{-1} e}}{\mathrm{e}^T\mathrm{\Sigma^{-1} e}} $$

Let us introduce the following

\begin{align} a&=\mathrm{e}^T\mathrm{\Sigma^{-1}}\mathrm{e}\\ b&=\mathrm{e}^T\mathrm{\Sigma^{-1}}\mathrm{\mu}\\ w_{MVP}&=\frac{\mathrm{\Sigma^{-1} e}}{\mathrm{e}^T\mathrm{\Sigma^{-1} e}}\\ w_{Tangency}&=\frac{\mathrm{\Sigma^{-1} \mu}}{\mathrm{e}^T\mathrm{\Sigma^{-1} \mu}} \end{align} Note that, conveniently, the two portfolios are the minimum-variance portfolio and the tangency portfolio. Then, $c_{12}\mu$ simplifies to $$ c_{12}\mathrm{\mu}=\frac{1}{\gamma}bw_{Tangency}-\frac{1}{\gamma}w_{MVP}b $$

and thus

$$ w^* = w_{MVP} + \frac{1}{\gamma}b\left(w_{Tangency}-w_{MVP}\right) $$

Finally, we note that $E(MVP)=\frac{b}{a}$, and $V(MVP)=\frac{1}{a}$. Thus, we may ultimately replace $b$ in equation above and arrive at

$$ w^* = w_{MVP} + \frac{1}{\gamma}\frac{\mu_{MVP}}{\sigma_{MVP}^2}\left(w_{Tangency}-w_{MVP}\right) $$

  • $\begingroup$ FOC $L_w=\mathrm{\mu}-\mathrm{\Sigma}\mathrm{w}-\lambda\mathrm{e}=!=0$ doesn't match the first row of the linear system that follows because it missed $\gamma$ $\endgroup$
    – develarist
    Nov 8 '20 at 7:57
  • 1
    $\begingroup$ Yep, thanks. Corrected it. $\endgroup$ Nov 8 '20 at 8:27
  • $\begingroup$ i see the "inverted matrix" and "constraint vector", but is there a name for the matrix consisting of $c$'s, and what its elements are for?. it looks like it is the inverted matrix $\endgroup$
    – develarist
    Nov 10 '20 at 11:33
  • $\begingroup$ It is the inverse of the FOC. $\endgroup$ Nov 10 '20 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.