5
$\begingroup$

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.

$\endgroup$
3
  • 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 $\endgroup$ Commented Nov 7, 2020 at 7:54
  • $\begingroup$ If that doesn’t help you I can elaborate in an answer $\endgroup$ Commented Nov 7, 2020 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
    Commented Nov 8, 2020 at 1:36

1 Answer 1

5
$\begingroup$

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

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

$\endgroup$
4
  • $\begingroup$ 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
    Commented Nov 8, 2020 at 7:57
  • 1
    $\begingroup$ Yep, thanks. Corrected it. $\endgroup$ Commented Nov 8, 2020 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
    Commented Nov 10, 2020 at 11:33
  • $\begingroup$ It is the inverse of the FOC. $\endgroup$ Commented Nov 10, 2020 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.