A rescaling technique to improve numerical stability of portfolio optimization problems

Abstract

This paper analyzes the numerical stability of Markowitz portfolio optimization model, by identifying and studying a source of instability, that strictly depends on the mathematical structure of the optimization problem and its constraints. As a consequence, it is shown how standard portfolio optimization models can result in an unstable model also when the covariance matrix is well conditioned and the objective function is numerically stable. This depends on the fact that the linear equality constraints of the model very often suffer of almost collinearity and/or bad scaling. A theoretical approach is proposed that exploiting an equivalent formulation of the original optimization problem considerably reduces such structural component of instability. The effectiveness of the proposal is empirically certified through applications on real financial data when numerical optimization approaches are needed to compute the optimal portfolio. Gurobi and MATLAB’s solvers quadprog and fmincon are compared in terms of convergence performances.

Appendix

In this section, the proofs of the main theoretical results stated in the paper are presented.

Let’s start recalling some useful results of linear algebra, whereas classical definitions and well-known concepts will be widely used without giving explicit references; for proofs and more details on the topic, the reader is referred to the textbooks Bini and Pan (1994) and Golub and Van Loan (1989). The first result, Theorem 1, is a technicality on the singular values of a matrix product (see Wang and Xi (1997), Theorem 2). To state it, the following notation is added. Let $A\in {\mathrm{Mat}}_{n\times m}({\mathbb {R}})$ and $\sigma _i(A)$ be its ith singular value. In particular, $\sigma _p(A)=0$ for each positive integer p, with $p > \mathrm{rank}(A)$.

Theorem 1

(Wang and Xi (1997), Theorem 2) Let $A\in {\mathrm{Mat}}_n({\mathbb {R}})$ and $B\in {\mathrm{Mat}}_{n\times m}({\mathbb {R}})$. Let r be a positive real number, k be a positive integer and $i_1, \ldots , i_k$ be k positive integers such that $1 \le i_1< \ldots < i_k \le n$. Then,

$$\begin{aligned} \sum _{t=1}^k \sigma _{i_t}^r(AB) \ge \sum _{t=1}^k \sigma _{i_t}^r(A)\sigma _{n-t+1}^r(B) \end{aligned}$$

(10)

The notion of square root of an Hermitian positive definite matrix and some of its properties are also recalled as follows.

Definition 2

Let $A\in {\mathrm{Mat}}_n({\mathbb {R}})$ be an Hermitian positive definite matrix and $\lambda _1(A), \ldots , \lambda _n(A)$ be the positive eigenvalues of A. The square root of A is the matrix $\sqrt{A} \in {\mathrm{Mat}}_n({\mathbb {R}})$ defined by:

$$\begin{aligned} \sqrt{A} := P^t \text {diag}(\sqrt{\lambda _1(A)}, \ldots , \sqrt{\lambda _n(A)}) P \end{aligned}$$

where $P \in {\mathrm{Mat}}_n({\mathbb {R}})$ is orthogonal and satisfies

$$\begin{aligned} A = P^t \text {diag}(\lambda _1(A), \ldots , \lambda _n(A)) P. \end{aligned}$$

Proposition 5

Let $A\in {\mathrm{Mat}}_n({\mathbb {R}})$ be an Hermitian positive definite matrix with singular values $\sigma _1(A), \ldots , \sigma _n(A)$. Let $\sqrt{A} \in {\mathrm{Mat}}_n({\mathbb {R}})$ be the square root of A. Then, the following properties hold true:

• 1. $\sqrt{A} \sqrt{A} = A$;

• 2. $\sqrt{A}$ is Hermitian positive definite matrix;

• 3. $\sqrt{A^{-1}} = \left( \sqrt{A}\right) ^{-1}$;

• 4. $\sqrt{\sigma _1(A)}, \ldots , \sqrt{\sigma _n(A)}$ are the singular values of $\sqrt{A}$.

In the following proposition, a minimization problem for the condition number of a $2 \times n$ real matrix, with $n \ge 2$, is solved.

Proposition 6

Let $n \ge 2$ and $v_1, v_2 \in \text {Mat}_{2 \times n}(\mathbb R)$ be two linearly independent row vectors. Let c be a positive real number and $A_c = \left( \begin{matrix} c v_1 \\ v_2 \end{matrix}\right) \in \text {Mat}_{2 \times n}({\mathbb {R}})$. The minimum value of the condition number of $A_c$ is attained at ${\overline{c}} = \frac{\Vert v_2\Vert _2}{\Vert v_1\Vert _2}$, that is

$$\begin{aligned} \min _{c >0} k (A_c)&= k\left( \left( \begin{matrix} {\overline{c}} v_1 \\ v_2 \end{matrix}\right) \right) \\&= \sqrt{1 + 2\frac{|\langle v_1,v_2\rangle |}{\Vert v_1\Vert _2 \Vert v_2\Vert _2 - |\langle v_1,v_2\rangle |}} \end{aligned}$$

Proof

By the hypotheses on $v_1$, $v_2$ and the real number $c>0$, it follows that $A_c$ has full row rank 2 and that $A_c A_c^t \in {\mathrm{Mat}}_2({\mathbb {R}})$ is an Hermitian positive definite matrix. Let $\sigma _1(A_c) \ge \sigma _2(A_c) >0$ be the singular values of $A_c$ and $\lambda _1(c) \ge \lambda _2(c) > 0$ be the eigenvalues of $A_c A_c^t$. By Definition 1 and well-known properties of the singular values, the condition number of $A_c$ can be expressed as $k(A_c) = \frac{\sigma _1(A_c)}{\sigma _2(A_c)} = \frac{\sqrt{\lambda _1(c)}}{\sqrt{\lambda _2(c)}}$, and so the minimization problem can be stated as follows:

$$\begin{aligned} \min _{c>0} k (A_c) = \min _{c>0} \frac{\sigma _1(A_c)}{\sigma _2(A_c)} = \min _{c>0} \sqrt{\frac{\lambda _1(c)}{\lambda _2(c)}} = \sqrt{\min _{c >0} \frac{\lambda _1(c)}{\lambda _2(c)}} \end{aligned}$$

The expressions of $\lambda _1(c)$ and $\lambda _2(c)$ are explicitly computed through the characteristic polynomial $p_{A_c}(\lambda )$ of the matrix $A_c A_c^t$:

$$\begin{aligned} p_{A_c}(\lambda )= & {} \det (A_c A_c^t - \lambda I_2) \\= & {} (c^2 \Vert v_1\Vert _2^2-\lambda )(\Vert v_2\Vert _2^2-\lambda ) - c^2 \langle v_1,v_2\rangle ^2\\= & {} \lambda ^2 -(c^2 \Vert v_1\Vert _2^2 + \Vert v_2\Vert _2^2)\lambda + c^2(\Vert v_1\Vert _2^2 \Vert v_2\Vert _2^2 \\&\quad - \langle v_1,v_2\rangle ^2) \end{aligned}$$

The equation $p_{A_c}(\lambda )=0$ has the real solutions $\lambda _1(c) \ge \lambda _2(c)>0$:

$$\begin{aligned}&\lambda _1(c)\nonumber \\&= \frac{c^2 \Vert v_1\Vert _2^2 + \Vert v_2\Vert _2^2 + \sqrt{(c^2 \Vert v_1\Vert _2^2 - \Vert v_2\Vert _2^2)^2 +4c^2\langle v_1,v_2\rangle ^2}}{2} \end{aligned}$$

(11)

$$\begin{aligned}&\lambda _2(c)\nonumber \\&= \frac{c^2 \Vert v_1\Vert _2^2 + \Vert v_2\Vert _2^2 - \sqrt{(c^2 \Vert v_1\Vert _2^2 - \Vert v_2\Vert _2^2)^2 +4c^2\langle v_1,v_2\rangle ^2}}{2}. \end{aligned}$$

(12)

Note that, in the particular case $\langle v_1,v_2 \rangle = 0$, the two eigenvalues are $\lambda _1(c) = c^2 \Vert v_1\Vert _2^2$ and $\lambda _2(c) = \Vert v_2\Vert _2^2$.

Let $g: {\mathbb {R}}_{>0} \rightarrow {\mathbb {R}}$ be the real-valued function defined by $g(c) = \frac{\lambda _1(c)}{\lambda _2(c)}$. Let’s start with the case $\langle v_1,v_2 \rangle = 0$: It is easy to verify that $\min _{c>0} g(c)$ is equal to 1 and it is attained at ${{\bar{c}}} = \frac{\Vert v_2\Vert _2}{\Vert v_1\Vert _2}$. In the general case, that is when $\langle v_1,v_2 \rangle \ne 0$, using (11) and (12), g(c) can be rewritten as follows:

$$\begin{aligned}&g(c) \nonumber \\&= \frac{c^2 \Vert v_1\Vert _2^2 + \Vert v_2\Vert _2^2 + \sqrt{(c^2 \Vert v_1\Vert _2^2 - \Vert v_2\Vert _2^2)^2 +4c^2\langle v_1,v_2\rangle ^2}}{c^2 \Vert v_1\Vert _2^2 + \Vert v_2\Vert _2^2 - \sqrt{(c^2 \Vert v_1\Vert _2^2 - \Vert v_2\Vert _2^2)^2 +4c^2\langle v_1,v_2\rangle ^2}}\nonumber \\&= 1+ 2 \frac{1}{\sqrt{\frac{(c^2 \Vert v_1\Vert _2^2 + \Vert v_2\Vert _2^2)^2}{(c^2 \Vert v_1\Vert _2^2 - \Vert v_2\Vert _2^2)^2 +4 \frac{\langle v_1,v_2\rangle ^2}{\Vert v_1\Vert _2^2} c^2 \Vert v_1\Vert _2^2}} - 1}\nonumber \\&= 1+ 2 \frac{1}{\sqrt{h(c)}-1} \end{aligned}$$

(13)

where, obviously, $h: {\mathbb {R}}_{>0} \rightarrow {\mathbb {R}}$ is the real-valued function defined by

$$\begin{aligned} h(c)=\frac{(c^2 \Vert v_1\Vert _2^2 + \Vert v_2\Vert _2^2)^2}{(c^2 \Vert v_1\Vert _2^2 - \Vert v_2\Vert _2^2)^2 +4 \frac{\langle v_1,v_2\rangle ^2}{\Vert v_1\Vert _2^2} c^2 \Vert v_1\Vert _2^2} \end{aligned}$$

Let’s observe that h(c) can be rewritten as $h(c)=h_1(h_2(c))$ where $h_i: {\mathbb {R}}_{>0} \rightarrow {\mathbb {R}}$, $i=1,2$, are real-valued functions defined as follows:

$$\begin{aligned} h_1(c)= & {} \frac{(c + \Vert v_2\Vert _2^2)^2}{(c - \Vert v_2\Vert _2^2)^2 +4 \frac{\langle v_1,v_2\rangle ^2}{\Vert v_1\Vert _2^2} c}\\ h_2(c)= & {} \Vert v_1\Vert _2^2 c^2 \end{aligned}$$

Obviously, $h_2$ is a strictly increasing function; further, by some easy computations, the expression of $h_1'(c)$ is given by:

$$\begin{aligned} h_1'(c)= & {} \frac{4(c + \Vert v_2\Vert _2^2) \left( \Vert v_2\Vert _2^2 -\frac{\langle v_1,v_2\rangle ^2}{\Vert v_1\Vert _2^2} \right) \left( \Vert v_2\Vert _2^2-c\right) }{\left( (c - \Vert v_2\Vert _2^2)^2 +4 \frac{\langle v_1,v_2\rangle ^2}{\Vert v_1\Vert _2^2} c \right) ^2} \end{aligned}$$

Since $|\langle v_1, v_2\rangle | - \Vert v_1\Vert _2\Vert v_2\Vert _2 <0$ (recall that $v_1, v_2$ are linearly independent by hypothesis), it follows that $h_1'(c) \ge 0$ for $c \le \Vert v_2\Vert _2^2$. In particular, $h_1$ is a strictly increasing function in $(0, \Vert v_2\Vert _2^2)$ with a relative maximum at ${\bar{c}}=\Vert v_2\Vert _2^2$. Now, by composition, it follows that the function h is strictly increasing in $(0, \frac{\Vert v_2\Vert _2}{\Vert v_1\Vert _2})$ with a relative maximum at ${\bar{c}}=\frac{\Vert v_2\Vert _2}{\Vert v_1\Vert _2}$. Finally, using expression (13), it follows that the function g is strictly decreasing in $\big (0, \frac{\Vert v_2\Vert _2}{\Vert v_1\Vert _2}\big )$ with a relative minimum at ${\bar{c}}=\frac{\Vert v_2\Vert _2}{\Vert v_1\Vert _2}$. Therefore,

$$\begin{aligned} \min _{c \in {\mathbb {R}}} g(c) = g({\bar{c}}) =1 +2 \frac{|\langle v_1, v_2 \rangle |}{\Vert v_1\Vert _2\Vert v_2\Vert _2-|\langle v_1, v_2 \rangle |}, \end{aligned}$$

so the result follows. $\square$

In the following, the proofs of Propositions 1, 2 and 3 are provided.

Proof of Proposition 1

Let $\sqrt{{\mathbf{V}}^{-1}}$ be the square root of the Hermitian positive definite matrix ${\mathbf{V}}^{-1}$ (see Definition 2). From the expression of $\mathbf{A}$ and using properties 1., 2. and 3. of Proposition 5, $\mathbf{A}$ can be rewritten as follows:

$$\begin{aligned} \mathbf{A}= & {} \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ^t \sqrt{{\mathbf{V}}^{-1}} \sqrt{{\mathbf{V}}^{-1}} \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] \\= & {} \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ^t (\sqrt{{\mathbf{V}}^{-1}})^t \sqrt{{\mathbf{V}}^{-1}} \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] \\= & {} \left( \sqrt{{\mathbf{V}}^{-1}} \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] \right) ^t \sqrt{{\mathbf{V}}^{-1}} \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] = \mathbf{B}^t \mathbf{B} \end{aligned}$$

where, obviously, $\mathbf{B} \in \text {Mat}_{n \times 2}({\mathbb {R}})$ is defined by the formula $\mathbf{B}:=\sqrt{{\mathbf{V}}^{-1}} \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right]$. Note that the matrix $\mathbf{B}$ has full rank equal to 2 (it follows from the equality $\text {rank}(\mathbf{B}^t \mathbf{B})=\text {rank}(\mathbf{B})$ and the fact that $\mathbf{A}$ has full rank). Let $\sigma _1(\mathbf{A}) \ge \sigma _2(\mathbf{A})>0$ be the singular values of $\mathbf{A}$ and $\sigma _1(\mathbf{B}) \ge \sigma _2(\mathbf{B})>0$ be the singular values of $\mathbf{B}$. Then, it is easy to prove that $\sigma _i(\mathbf{A}) = \sigma _i^2(\mathbf{B})$, for each $i=1,2$. An estimate of the singular values of $\mathbf{B}$ is provided. Firstly, let’s consider $\sigma _1(\mathbf{B})$; the properties of the singular values yield:

$$\begin{aligned} \sigma _1(\mathbf{B})= & {} \sigma _1(\sqrt{{\mathbf{V}}^{-1}} \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ) = \Vert \sqrt{{\mathbf{V}}^{-1}} \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] \Vert _2 \\\le & {} \Vert \sqrt{{\mathbf{V}}^{-1}}\Vert _2 \Vert \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] \Vert _2 = \sigma _1(\sqrt{{\mathbf{V}}^{-1}}) \sigma _1(\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ). \end{aligned}$$

In addition, combining with property 4. of Proposition 5, it follows

$$\begin{aligned} \sigma _1(\mathbf{B}) \le \sqrt{\sigma _1({\mathbf{V}}^{-1})} \sigma _1(\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ) = \frac{ \sigma _1(\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] )}{\sqrt{\sigma _n({\mathbf{V}})}}. \end{aligned}$$

(14)

In order to give an estimate for $\sigma _2(\mathbf{B})$, Theorem 1 is applied to the matrix $\mathbf{B} = \sqrt{{\mathbf{V}}^{-1}} \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right]$, with indices $i_t=t+1$, for $t=1,\ldots , n-1$, and $r=1$:

$$\begin{aligned} \sigma _2(\mathbf{B}) \ge \sigma _n(\sqrt{{\mathbf{V}}^{-1}}) \sigma _2(\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ) \end{aligned}$$

Combining again with property 4. of Proposition 5, the previous inequality becomes:

$$\begin{aligned} \sigma _2(\mathbf{B}) \ge \sqrt{\sigma _n({\mathbf{V}}^{-1})} \sigma _2(\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ) = \frac{\sigma _2(\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] )}{\sqrt{\sigma _1({\mathbf{V}})}} \end{aligned}$$

(15)

Recalling Definition 1, and inequalities (14) and (15), it holds

$$\begin{aligned} k(\mathbf{A})&= \frac{\sigma _1^2(\mathbf{B})}{\sigma _2^2(\mathbf{B})} \le \frac{ \sigma _1^2(\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] )}{\sigma _n({\mathbf{V}})} \frac{\sigma _1({\mathbf{V}})}{\sigma _2^2(\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] )} \\&=k({\mathbf{V}}) \left( k(\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] )\right) ^2, \end{aligned}$$

so the result follows. $\square$

Proof of Proposition 2

It is sufficient to show a case in which equality holds in formula (4). To this end, let’s consider the case ${\mathbf{V}}= \delta \mathbf{I}_n \in \text {Mat}_n({\mathbb {R}})$, with $\delta >0$. Then, the matrix $\mathbf{A} = \frac{1}{\delta }\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ^t \left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right]$; using properties of the maximum and minimum singular values, it follows that the condition number of $\mathbf{A}$ is $k(\mathbf{A}) = \left( k(\left[ \begin{matrix} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] )\right) ^2$. Then, the equality in bound (4) of Proposition 1 follows from the fact that $k({\mathbf{V}})= k(\delta \mathbf{I}_n) = 1$. $\square$

Proof of Proposition 3

Since ${\mathbf{V}}=\delta \mathbf{I}_n$, then the matrix ${{\mathbf{A}}_{\mathbf{c}}} = \frac{1}{\delta c^2}\left[ \begin{matrix} c{\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ^t \left[ \begin{matrix} c{\mathbf{R}}&{\mathbf{1}}\end{matrix}\right]$ and its condition number is:

$$\begin{aligned} k({{\mathbf{A}}_{\mathbf{c}}}) = k\left( \left[ \begin{matrix} c{\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ^t \left[ \begin{matrix} c{\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] \right) = \left( k\left( \left[ \begin{matrix} c{\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ^t\right) \right) ^2 \end{aligned}$$

Note that, by hypothesis, not all elements of ${\mathbf{R}}$ are equal, so $c{\mathbf{R}}$ and ${\mathbf{1}}$ are linearly independent vectors for any value of $c>0$. Applying Proposition 6 to the matrix $\left[ \begin{matrix} c{\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ^t$, it follows that $\min _{c >0} k \left( \left[ \begin{matrix} c{\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ^t \right)$ is attained at ${{\bar{c}}} = \frac{\Vert {\mathbf{1}}\Vert _2}{\Vert {\mathbf{R}}\Vert _2}$, and so

$$\begin{aligned} \min _{c>0} k({{\mathbf{A}}_{\mathbf{c}}})= & {} k({{\mathbf{A}}_{\bar{\mathbf{c}}}}) = \left( k \left( \left[ \begin{matrix} {{\bar{c}}} {\mathbf{R}}&{\mathbf{1}}\end{matrix}\right] ^t \right) \right) ^2\\= & {} 1 + 2\frac{|\langle {\mathbf{R}}, {\mathbf{1}} \rangle |}{\Vert {\mathbf{R}}\Vert _2 \Vert {\mathbf{1}}\Vert _2 - |\langle {\mathbf{R}}, {\mathbf{1}} \rangle |}\\= & {} 1 + 2\frac{| \sum _{i=1}^n R_i |}{ \sqrt{n \sum _{i=1}^n R_i^2} - |\sum _{i=1}^n R_i|} \end{aligned}$$

; therefore, the result follows. $\square$