On the impact of semidefinite positive correlation measures in portfolio theory

Abstract

In this paper potential usage of different correlation measures in portfolio problems is studied. We characterize especially semidefinite positive correlation measures consistent with the choices of risk-averse investors. Moreover, we propose a new approach to portfolio selection problem, which optimizes the correlation between the portfolio and one or two market benchmarks. We also discuss why should correlation measures be used to reduce the dimensionality of large scale portfolio problems. Finally, through an empirical analysis, we show the impact of different correlation measures on portfolio selection problems and on dimensionality reduction problems. In particular, we compare the ex post sample paths of several portfolio strategies based on different risk measures and correlation measures.

Appendix

Proof of Theorem 1 Suppose that properties 1 and 2 of the theorem are satisfied. Cauchy–Schwarz inequality applied to V guarantees property 1 and 2 of correlation measures, i.e. $\rho :H\times H\rightarrow [-1,1]$ and $\rho (X,X)=1;$ $\rho (X,-X)=-\rho (X,X)=-1.$ Consider $g(X,Y)=(v_{1,g},v_{2,g}).$ Properties 3 and 4 of correlation measures are logical consequences of $\langle g(X,Y)\rangle =\langle g(Y,X)\rangle$ and $\langle g(-X,Y)\rangle =\langle g(X,-Y)\rangle =-\langle g(X,Y)\rangle ,$ while Property 5 of the correlation measure follows by property 2 of the theorem. Moreover, since $\rho (X,Y)$ can be seen as an inner product of the vectors $\frac{v_{1,g}}{\sqrt{\langle g(X,X)\rangle }}$ and $\frac{v_{2,g}}{\sqrt{\langle g(Y,Y)\rangle }},$ we deduce from the Gram representation theorem that the correlation matrix that satisfies properties 1 and 2 of the theorem is semidefinite positive.

Viceversa, let $\rho$ be a semidefinite positive correlation measure. Then any correlation matrix $Q=[\rho _{i,j}]$ should be semidefinite positive. Consider the correlation matrix defined on the random variables belonging to $H_{1}=\left\{ X_{1},\ldots ,X_{n}\right\}$. From the Gram representation theorem we know that an n-dimensional correlation matrix $Q=[\rho _{i,j}]$ is semidefinite positive if, and only if, there exists n vectors $v_{1}$, $v_{2},\ldots ,v_{n}$, in a vectorial space V such that $\rho (X_{i},X_{j})=\rho _{i,j}=\langle v_{i},v_{j}\rangle$. Since $\rho :H_{1}\times H_{1}\rightarrow [-1,1]$ and $\rho (X_{i},X_{i})=1,$ then $\langle v_{i},v_{i}\rangle =1.$ Consider the function $g(\pm X_{i},\pm X_{j})=\left( \pm k_{i}v_{i},\pm k_{j}v_{j}\right)$ for some positive real $k_{i},k_{j}$ $\left( i,j=1,\ldots ,n\right) .$ Therefore we obtain $\rho _{i,j}=\frac{\langle g(X_{i},X_{j})\rangle }{\sqrt{\langle g(X_{i},X_{i})}\rangle \langle g(X_{j},X_{j})\rangle }.$ Since $\rho (-X_{i},X_{j})=\rho (X_{i},-X_{j})=-\rho (X_{i},X_{j})$ it follows that $\langle g(-X_{i},X_{j})\rangle =\langle g(X_{i},-X_{j})\rangle =-\langle g(X_{i},X_{j})\rangle$. If X and Y are independent random variables $\rho (X,Y)=0$ and thus $\langle g(X,Y)\rangle =0$.

Proof of Proposition 1 We first prove Case M1. Observe that $V_{p/2}(aX+b)=aV_{p/2}(X)+b$ holds for any constant a and b and the functional

$$\begin{aligned} \mathrm {cvp}(X,Y)=\mathbb {E}\left( \left( X-V_{p/2}(X)\right) ^{\langle p/2\rangle }\left( Y-V_{p/2}(Y)\right) ^{\langle p/2\rangle }\right) ^{\langle \min (2/p,2)\rangle } \end{aligned}$$

is an inner product in the $L^{p}$ space. Thus using the same notation of Theorem 1 we can assume that $\langle g(X,Y)\rangle =\mathrm {cvp}(X,Y)$ and then we easily prove that $\mathrm {cvp}$ satisfies the properties of Theorem 1. In particular, $\mathrm {cvp}(X,Y)=\mathrm {cvp}(Y,X)$ and $\mathrm {cvp}(aX+b,Y)= \mathrm {cvp}(X,Y)a^{\langle \min (p,1)\rangle }.$ Moreover, if X is independent of Y, then

$$\begin{aligned} \mathrm {cvp}(X,Y)^{\frac{1}{\langle \min (2/p,2)\rangle }}=\mathbb {E}\left( \left( X-V_{p/2}(X)\right) ^{\langle p/2\rangle }\right) \mathbb {E}\left( \left( Y-V_{p/2}(Y)\right) ^{\langle p/2\rangle }\right) =0. \end{aligned}$$

If $\left| \rho _{p}(X,Y)\right| =1$, then $\left| \mathrm {cvp} (X,Y)\right| =\left\| X-V_{p/2}(X)\right\| _{p}\left\| Y-V_{p/2}(Y)\right\| _{p}$, so that there exists a real c such that $\left( X-V_{p/2}(X)\right) ^{\langle p/2\rangle }=c\left( Y-V_{p/2}(Y)\right) ^{\langle p/2\rangle }$ and the thesis follows. Regarding Case M2, we can easily prove that $Y=X$ a.s. implies $X_{1}=Y_{1}$ a.s. and consequently $\tau _{p}(X,X)=1$. Since $(X_{1},Y_{1})\overset{d}{=}(X,Y),$ the assumption $Y=X$ a.s. implies that for any $x_{n}\in \mathbb {Q}$; $\Pr (X_{1}\le x_{n},Y_{1}>x_{n})+\Pr (X_{1}>x_{n},Y_{1}\le x_{n})=0.$ Consider $\widetilde{\varOmega }=\left( \cup _{n}\varOmega _{n}\right) ^{C}$ (where $\varOmega _{n}=\left\{ w|X_{1}(w)\le x_{n},Y_{1}(w)>x_{n}\text { or }X_{1}(w)>x_{n},Y_{1}(w)\le x_{n}\right\} )$. Then $\Pr (\widetilde{\varOmega })=1$ and $X_{1}(w)=Y_{1}(w)$ for any $w\in \widetilde{\varOmega }$ (otherwise if $X_{1}(w)>Y_{1}(w)$—or $X_{1}(w)\langle Y_{1}(w)$—there exists $x_{n}\in \mathbb {Q}$ such that $X_{1}(w)>x_{n}\ge Y_{1}(w)$—or $X_{1}(w)\le x_{n}\langle Y_{1}(w)$ contrary to the assumption that $w\in \widetilde{\varOmega }$ ). Moreover since $\left( X-X_{1}\right)$ and $\left( Y-Y_{1}\right)$ are symmetric random variables $\mathbb {E}\left( \left( X-X_{1}\right) ^{\langle p/2\rangle }\right) =\mathbb {E}\left( \left( Y-Y_{1}\right) ^{\langle p/2\rangle }\right) =0,$ when X and Y are independent random variables $\tau _{p}(X,Y)=0.$ All the other properties of Theorem 1 are satisfied by the measures $\tau _{p}(X,Y)$. Regarding Case M3, we observe that the random variable $Z=(X^{\langle p/2\rangle }-E(X^{\langle p/2\rangle }|\mathfrak {I}_{1})),$ where $X\in L^{p}(\varOmega ,\mathfrak {I},\Pr )$ is orthogonal to $L^{2}(\varOmega ,\mathfrak {I}_{1},\Pr )$ because for any $V\in L^{2}(\varOmega ,\mathfrak {I}_{1},\Pr )$ then $E(VZ)=E(VE(Z|\mathfrak {I}_{1}))=0.$ Since the measure $O_{p,\mathfrak {I}_{1}}$ is the Pearson measure applied to the space of the random variables of the type $Z=(X^{\langle p/2\rangle }-E(X^{\langle p/2\rangle }|\mathfrak {I}_{1}))$ orthogonal to $L^{2}(\varOmega ,\mathfrak {I}_{1},\Pr )$ the measure M3 is a correlation measure on this space. On the one hand, the measure M3 is not law invariant correlation measure on the whole space of random variables $L^{p}(\varOmega ,\mathfrak {I},\Pr ).$ As a matter of fact, if $X_{(1)},Y_{(1)}\in L^{p},$ $(X_{(1)},Y_{(1)})\overset{d}{=}\left( X_{(2)},Y_{(2)}\right)$ not necessarily$\ \left( Z_{1,(1)},Z_{2,(1)}\right) \overset{d}{=}\left( Z_{1,(2)},Z_{2,(2)}\right)$ where $Z_{1,(i)}=(X_{(i)}^{\langle p/2\rangle }-E(X_{(i)}^{\langle p/2\rangle }|\mathfrak {I}_{1}))$ and $Z_{2,(i)}=(Y_{(i)}^{\langle p/2\rangle }-E(Y_{(i)}^{\langle p/2\rangle }|\mathfrak {I}_{1}))$ ($i=1,2$) and it may be $O_{p,\mathfrak {I}_{1}}(Z_{1,(1)},Z_{2,(1)})\ne O_{p,\mathfrak {I}_{1}}\left( Z_{1,(2)},Z_{2,(2)}\right)$. On the other hand, measure M3 satisfies concordance property (v) on the whole space $L^{p},$ because if X is independent from Y then $O_{p,\mathfrak {I}_{1}}(Z_{1},Z_{2})=0.$

Proof of Proposition 2 Under the assumptions of the proposition every portfolio of returns is uniquely determined by the mean and the variance or the mean and by another deviation measure. Moreover, as a consequence of Theorem 5 in Ortobelli (2001), if we use another deviation measure $\sigma _{X}$ to characterize these distribution families then $\sigma _{X}=a\sqrt{var(X)}$ where a is a positive constant for all the random variables X of the two parametric family. Suppose that the vector of returns $z=(z_{1},z_{2},z_{3})^{\prime }$ is distributed with mean vector $\mu =[\mu _{1},\mu _{2},\mu _{1} \overline{x}+(1-\overline{x})\mu _{2}]$ for a given $\overline{x}\in (0,1)$ and a given variance covariance matrix

$$\begin{aligned} Q=\left[ \begin{array}{c@{\quad }c@{\quad }c} \text {var}(z_{1}) &{} \text {cov}(z_{1},z_{2}) &{} 0 \\ \text {cov}(z_{1},z_{2}) &{} \text {var}(z_{2}) &{} 0 \\ 0 &{} 0 &{} \text {var}(z_{3}) \end{array} \right] , \end{aligned}$$

where $\hbox {var}(z_{3})=\overline{x}^{2}\hbox {var}(z_{1})+(1-\overline{x})^{2}\hbox {var} (z_{2})+2\overline{x}(1-\overline{x})\hbox {cov}(z_{1},z_{2}).$ Then the distribution of portfolio $[\overline{x},(1-\overline{x}),0]z$ is equal to the distribution of $z_{3}.$ Therefore, if $\rho$ is a semidefinite positive correlation measure and $\sigma$ is a deviation measure, we obtain

$$\begin{aligned} d_{\rho ,\sigma }([0,0,1]z)^{2}=\sigma _{z_{3}}^{2}=a^{2}var(z_{3})=a^{2}var( \overline{x},(1-\overline{x}),0]z) \end{aligned}$$

and

$$\begin{aligned} d_{\rho ,\sigma }([\overline{x},(1-\overline{x}),0]z)^{2}= & {} \overline{x} ^{2}a^{2}var(z_{1})+a^{2}(1-\overline{x})^{2}var(z_{2}) \\&+\,2\overline{x}(1-\overline{x})a^{2}\sqrt{var(z_{1})var(z_{2})}\rho (z_{1},z_{2}), \end{aligned}$$

where a is the positive constant defined as a consequence of Theorem 5 in Ortobelli (2001) applied to the family of these distributions. Therefore $d_{\rho ,\sigma }([0,0,1]z)^{2}=a^{2}var(\overline{x},(1-\overline{x} ),0]z)=d_{\rho ,\sigma }([\overline{x},(1-\overline{x}),0]z)^{2}$ if and only if

$$\begin{aligned} \sqrt{var(z_{1})var(z_{2})}\rho (z_{1},z_{2})=\text {cov}(z_{1},z_{2}). \end{aligned}$$

Moreover, let $\rho$ be the Pearson correlation measure and let $z=\left[ z_{1},z_{2},\ldots ,z_{n}\right] ^{\prime }$ be the vector of the returns. If two portfolios $x^{\prime }z$ and $y^{\prime }z$ have the same distributions and $\sigma$ is a deviation measure, then $\sigma _{z_{i}}=a \sqrt{var(z_{i})}$ (where a is a positive constant) for any $i=1,\ldots ,n$ and

$$\begin{aligned} d_{\rho ,\sigma }(x^{\prime }z)^{2}= & {} \sum \limits _{i=1}^{n}x_{i}^{2}\sigma _{z_{i}}^{2}+2\sum \limits _{i=1}^{n}\sum \limits _{j=i+1}x_{i}x_{j}\sigma _{z_{i}}\sigma _{z_{i}}\rho _{i,j}=a^{2}var(x^{\prime }z) \\= & {} a^{2}var(y^{\prime }z)=\sum \limits _{i=1}^{n}y_{i}^{2}\sigma _{z_{i}}^{2}+2\sum \limits _{i=1}^{n}\sum \limits _{j=i+1}y_{i}y_{j}\sigma _{z_{i}}\sigma _{z_{i}}\rho _{i,j}=d_{\rho ,\sigma }(y^{\prime }z)^{2}. \end{aligned}$$

Proof of Proposition 3 Under these assumptions, the matrix $Q_{\rho ,\sigma }=[\sigma _{z_{j}}\sigma _{z_{i}}\rho _{i,j}]=[\langle g(z_{i},z_{j})\rangle ].$ Moreover, from bilinearity we deduce $x^{\prime }Q_{\rho ,\sigma }x=\langle g(x^{\prime }z,x^{\prime }z)\rangle .$ If we have two portfolios $x^{\prime }z$ and $y^{\prime }z$ with the same distribution

$$\begin{aligned} y^{\prime }Q_{\rho ,\sigma }y=\langle g(y^{\prime }z,y^{\prime }z)\rangle =\sigma _{y^{\prime }z}^{2}=\sigma _{x^{\prime }z}^{2}=\langle g(x^{\prime }z,x^{\prime }z)\rangle =x^{\prime }Q_{\rho ,\sigma }x, \end{aligned}$$

where we used the invariance in law of the variability measure $\sigma _{X}= \sqrt{\langle g(X,X) \rangle }$. Thus $d_{\rho ,\sigma }(x^{\prime }z)$ is invariant in law.

Proof of Proposition 4 As a consequence of (7) the measure (8) is defined on the marginals by:

$$\begin{aligned} \rho (z_{i},z_{j})=\frac{v_{ij}}{\sqrt{v_{ii}v_{jj}}}=\frac{ A(\alpha ,p)E\left( \left( z_{i}-E(z_{i})\right) \left( z_{j}-E(z_{j})\right) ^{\left\langle p-1\right\rangle }\right) }{A(\alpha ,1)A(\alpha ,p-1)E\left( \left| z_{i}-E(z_{i})\right| \right) E\left( \left| z_{j}-E(z_{j})\right| ^{p-1}\right) } \end{aligned}$$

for any $p\in [1,\alpha ).$ Moreover, considering that the matrix Q is semidefinite positive, we can easily verify that $\rho$ is a linear correlation measure (for further details on stable sub-Gaussian distributions see Samorodnitsky and Taqqu 1994). Under the assumptions of the proposition, every portfolio of gross returns $x^{\prime }z$ is uniquely determined by the mean $x^{\prime }\mu$ (since $\alpha >1$) and the dispersion $x^{\prime }Qx$ of this elliptical distribution or the mean and the deviation measure $\sigma _{x^{\prime }z}$. Therefore, as a consequence of Theorem 5 in Ortobelli (2001), we can essentially repeat the proof of Proposition 2 and the measure $d_{\rho ,\sigma }(x^{\prime }z)$ is invariant in law if and only if $\rho$ is defined by (8).

Proof of Corollary 1 Let us consider $\rho (X,Y)=\sum _{i=1}^{m}a_{i}\rho _{i}(X,Y)$ such that $a_{i}\ge 0; \sum _{i=1}^{m}a_{i}=1.$ Clearly, $\rho (X,Y)$ is a concordance (correlation) measure if all $\rho _{i}(X,Y)$ satisfy the seven (five) properties of concordance (correlation) measures. Similarly, if $\rho _{i}$ for $i=1,\ldots ,m$ are semidefinite positive correlation measures, then also $\rho =\sum _{i=1}^{m}a_{i}\rho _{i}$ is semidefinite positive because any correlation matrix $Q=\sum _{i=1}^{m}a_{i}Q_{i}$ derived from $\rho$ is the convex combination of the correlation matrices $Q_{i}$ derived from the measures $\rho _{i}.$ Observe that if $\rho _{i}(X,Y)$ $i=1,\ldots ,m$ are linear correlation measures $\langle g_{i}(X,X)\rangle _{i}\ge 0$ and $\langle g_{i}(X,X)\rangle _{i}=0$ if and only if X is a constant. Thus $\langle X-\mathbb {E}(X),X-\mathbb {E}(X)\rangle \ge 0$ and it is equal to zero if and only if $X-\mathbb {E}(X)=0$. Since $\langle g_{i}(X,Y)\rangle _{i}=\langle g_{i}(Y,X)\rangle _{i}$ then $\left\langle X-\mathbb {E}(X),Y- \mathbb {E}(Y)\right\rangle =\left\langle Y-\mathbb {E}(Y),X-\mathbb {E} (X)\right\rangle .$ In addition, for any $a,b\in \mathbb {R}$ and $X,Y,Z\in H$, $\langle g_{i}(aX+bZ,Y)\rangle _{i}=a\langle g_{i}(X,Y)\rangle _{i}+b\langle g_{i}(Z,Y)\rangle _{i},$ thus $\langle aX+bZ,Y\rangle =a\langle X,Y\rangle +b\langle Z,Y\rangle$ and $\langle X,Y\rangle$ is an inner product in the class of centered random variables belonging to H.

Proof of Proposition 5 The matrix $Q_{\rho ,\sigma }$ is still semidefinite positive since $x^{\prime }Q_{\rho ,\sigma }x=y^{\prime }Q_{\rho }y\ge 0$ where $y=(x_{1} \sigma _{z_{1}},x_{2}\sigma _{z_{2}},\ldots ,x_{n}\sigma _{z_{n}})^{\prime }.$ Moreover, Bauerle and Müller (2006) have proved that in a finite probability space where the probability $\Pr$ is uniform, any invariant in law, convex measure D (i.e. $D( aX+( 1-a) Y) \le aD(X)+(1- a )D(Y)$ for any $a\in [0,1])$ is consistent with the choices of risk-averse investors. The measure $x^{\prime }Q_{\rho ,\sigma }x$ (or its estimator $x^{\prime }\widetilde{Q}_{\rho ,\sigma }x$ with semidefinite positive matrix $\widetilde{Q}_{\rho ,\sigma }$) is convex in the class of portfolio returns $x^{\prime }z$ since the function $f(x)=x^{\prime }Q_{\rho , \sigma }x$ ($f(x)=x^{\prime }\widetilde{Q}_{\rho ,\sigma }x$) is a convex function because the Hessian of function f(x) is $Q_{\rho ,\sigma }$ ($\widetilde{Q}_{\rho ,\sigma }$). Then, for any $a\in [0,1]$:

$$\begin{aligned} \left( ax+(1- a )y\right) ^{\prime }Q_{\rho ,\sigma }\left( ax+(1- a )y\right) \le ax^{\prime }Q_{\rho ,\sigma }x+(1- a )y^{\prime }Q_{\rho ,\sigma }y. \end{aligned}$$

The measure $x^{\prime }Q_{\rho ,\sigma }x$ (or its estimator $x^{\prime } \widetilde{Q}_{\rho ,\sigma }x)$ is strictly convex when matrix $Q_{\rho ,\sigma }$ ($\widetilde{Q}_{\rho ,\sigma }$) is definite positive. According to Bauerle and Müller (2006), if $w^{\prime }z$ is dominated in the sense of the convex order by $y^{\prime }z,$ then $w^{\prime }Q_{\rho ,\sigma }w\le y^{\prime }Q_{\rho ,\sigma }y.$