Possibilistic mean–variance portfolios versus probabilistic ones: the winner is...

Abstract

In this paper, we compare the mean–variance portfolio modeling based on the possibilistic representation of the future stock returns to the one based on the classical probabilistic modelization of the same returns. There exist several different definitions of possibilistic mean, possibilistic variance and possibilistic covariance. In this paper, we consider definitions recently proposed in the literature for modeling portfolio selection problems: the possibilistic mean and variance à la Carlsson–Fullér–Majlender, the lower possibilistic mean and variance, and the upper possibilistic mean and variance. In particular, we mean to answer to the following research questions: first, to check whether, from a methodological and theoretical standpoint, it is possible to detect elements of superiority of one of the two approaches with respect to the other one; then, to check whether, from an operational point of view, one of the two approaches is more effective than the other one in terms of virtual-future performances. We disclosed that, on the basis of the results we obtained, the winner is the probabilistic approach.

Appendices

Appendix A: A recall on the trapezoidal fuzzy numbers

A fuzzy number $A = (a, b, \alpha , \beta )$ is called trapezoidal with tolerance interval [a, b], in which $a, b \in \mathbb {R}$ with $a \le b$, with left width $\alpha \ge 0$ and with right width $\beta \ge 0$, if its membership function $A({\gamma })$ has the form

$$\begin{aligned} A({\gamma }) = \left\{ \begin{array}{ll} \displaystyle {1 - \frac{a - {\gamma }}{\alpha }} &{}\quad \mathrm{if } \; a - \alpha< {\gamma }\le a \\ 1 &{}\quad \mathrm{if } \; a< {\gamma }\le b \\ \displaystyle {1 - \frac{{\gamma }- b}{\beta }} &{}\quad \mathrm{if } \; b < {\gamma }\le b + \beta \\ 0 &{}\quad \mathrm{otherwise} \end{array} \right. . \end{aligned}$$

In Fig. 2, the membership function of a generic trapezoidal fuzzy number is graphically represented.

Fig. 2

Membership function of a generic trapezoidal number

Given these definitions, it can be simply proved that the ${\gamma }$-level set $[A]^{\gamma }$ of the possibilistic number A is

$$\begin{aligned}{}[A]^{\gamma }= \left[ a - \left( 1 - {\gamma }\right) \alpha , b + \left( 1 - {\gamma }\right) \beta \right] , \end{aligned}$$

with $0 \le {\gamma }\le 1$. This ${\gamma }$-level set is also indicated as

$$\begin{aligned}{}[A]^{\gamma }= \left[ a_1({\gamma }), a_2({\gamma }) \right] , \end{aligned}$$

where, of course, $a_1({\gamma }) = a - (1 - {\gamma }) \alpha$ and $a_2({\gamma }) = b + (1 - {\gamma }) \beta$.

Notice that if $a = b$, $\alpha > 0$ and $\beta > 0$, as special case of a trapezoidal fuzzy number one obtains a triangular fuzzy number, whereas if $a < b$ and $\alpha = \beta = 0$ as special case of a trapezoidal fuzzy number, one obtains an interval fuzzy number.

Appendix B: Proofs

Proof of Proposition 3.1

From formula 1, it is easy to verify that

$$\begin{aligned} {\text {Var}}_\mathrm{CFM}(A) = 0 \Longleftrightarrow a=b \wedge \alpha =0 \wedge \beta =0. \end{aligned}$$

Given this parametrization, one has

$$\begin{aligned}{}[A]^{\gamma }= [a, a] = [b, b], \text { with } {\gamma }\in [0,1], \end{aligned}$$

which is the representation of A as a crisp number. (Of course, for such a number there is no variability for any value of ${\gamma }$.) $\square$

Proof of Proposition 3.2

From formula 3, it is easy to verify that

$$\begin{aligned} {\text {Var}}_* = 0 \Longleftrightarrow \alpha =0. \end{aligned}$$

Given this parametrization, one has

$$\begin{aligned}{}[A]^{\gamma }= \left[ a, b + (1 - {\gamma }) \beta \right] , \text { with } {\gamma }\in [0,1] , \end{aligned}$$

which represents a not-degenerate finite interval for any value of ${\gamma }$. (Of course, for such a ${\gamma }$-level set there is variability of A for any value of ${\gamma }$).⁶$\square$

Proof of Proposition 3.3

From formula 5, it is easy to verify that

$$\begin{aligned} {\text {Var}}^* = 0 \Longleftrightarrow \beta =0. \end{aligned}$$

Given this parametrization, one has

$$\begin{aligned}{}[A]^{\gamma }= \left[ a - (1 - {\gamma }) \alpha , b \right] , \text { with } {\gamma }\in [0,1] , \end{aligned}$$

which represents a not-degenerate finite interval for any value of ${\gamma }$. (Of course, for such a ${\gamma }$-level set there is variability of A for any value of ${\gamma }$).⁷$\square$

Proof of Proposition 3.4

Recalling that

$$\begin{aligned} \rho _\mathrm{CFM}(A_1, A_2) = \frac{{\text {Cov}}_\mathrm{CFM}(A_1, A_2)}{\sqrt{{\text {Var}}_\mathrm{CFM}(A_1)} \sqrt{{\text {Var}}_\mathrm{CFM}(A_2)}} , \end{aligned}$$

from formula 2 it is easy to verify that ${\text {Cov}}_\mathrm{CFM}(A_1, A_2) > 0$ by construction (it is given by the sum of four positive addends), and from formula 1 it is as much easy to verify that ${\text {Var}}_\mathrm{CFM}(A_1) > 0$ and ${\text {Var}}_\mathrm{CFM}(A_2) > 0$ again by construction (they are given each by sum of three positive addends.) Therefore, $\rho _\mathrm{CFM}(A_1, A_2) > 0$. $\square$

Proof of Proposition 3.5

Substituting in ${\text {Cov}}_\mathrm{CFM}(A_1, A_2)$, in ${\text {Var}}_\mathrm{CFM}(A_1)$ and in ${\text {Var}}_\mathrm{CFM}(A_2)$ the equalities $a_1 = b_1$ and $a_2 = b_2$, one has

$$\begin{aligned} \rho _\mathrm{CFM}(A_1, A_2) = \frac{\displaystyle {\frac{1}{24} (\alpha _1 + \beta _1) (\alpha _2 + \beta _2)}}{\sqrt{\displaystyle {\frac{1}{24} (\alpha _1 + \beta _1)^2}} \sqrt{\displaystyle {\frac{1}{24} (\alpha _2 + \beta _2)^2}}} = +1. \end{aligned}$$

$\square$

Proof of Proposition 3.6

Substituting in ${\text {Cov}}_\mathrm{CFM}(A_1, A_2)$, in ${\text {Var}}_\mathrm{CFM}(A_1)$ and in ${\text {Var}}_\mathrm{CFM}(A_2)$ the equalities $\alpha _1 = \beta _1 = 0$ and $\alpha _2 = \beta _2 = 0$, one has

$$\begin{aligned} \rho _\mathrm{CFM}(A_1, A_2) = \frac{\displaystyle {\frac{1}{4} (b_1 - a_1) (b_2 - a_2)}}{\sqrt{\displaystyle {\frac{1}{4} (b_1 - a_1)^2}} \sqrt{\displaystyle {\frac{1}{4} (b_2 - a_2)^2}}} = +1. \end{aligned}$$

$\square$

Proof of Proposition 3.7

Recalling that

$$\begin{aligned} \rho _*(A_1, A_2) = \frac{{\text {Cov}}_*(A_1, A_2)}{\sqrt{{\text {Var}}_*(A_1)} \sqrt{{\text {Var}}_*(A_2)}} , \end{aligned}$$

substituting in $\rho _*(A_1, A_2)$ formulas 4 and 3, one has

$$\begin{aligned} \rho _*(A_1, A_2) = \frac{\displaystyle {\frac{1}{18}} \alpha _1 \alpha _2}{\sqrt{\displaystyle {\frac{1}{18} \alpha _1^2}} \sqrt{\displaystyle {\frac{1}{18} \alpha _2^2}}} = +1. \end{aligned}$$

$\square$

Proof of Proposition 3.8

Recalling that

$$\begin{aligned} \rho ^*(A_1, A_2) = \frac{{\text {Cov}}^*(A_1, A_2)}{\sqrt{{\text {Var}}^*(A_1)} \sqrt{{\text {Var}}^*(A_2)}} , \end{aligned}$$

substituting in $\rho ^*(A_1, A_2)$ formulas 6 and 5, one has

$$\begin{aligned} \rho ^*(A_1, A_2) = \frac{\displaystyle {\frac{1}{18} \beta _1 \beta _2}}{\sqrt{\displaystyle {\frac{1}{18} \beta _1^2}} \sqrt{\displaystyle {\frac{1}{18}} \beta _2^2}} = +1. \end{aligned}$$

$\square$

Proof of Proposition 3.9

This proof is articulated in two parts. In the first one, we prove that ${{\varvec{V}}}_\mathrm{CFM}$ has rank at least equal to two or to one. In the second part, we prove that ${{\varvec{V}}}_\mathrm{CFM}$ has rank at most equal to two or to one, respectively.

First part—In this part, we consider a $2 \times 2$-dimensional variance–covariance matrix ${\varvec{V}}_\mathrm{CFM}$ and show under what conditions its rank is at least equal to 2 or to 1. In order to make simpler the notation, we put $A_h = b_h - a_h$ and $B_h = \alpha _h + \beta _h$, with $h \in \{ 1, 2 \}$. Therefore, the considered variance–covariance matrix can be written as

$$\begin{aligned} {{\varvec{V}}}_\mathrm{CFM} = \left( \begin{array}{l@{\qquad }l} \displaystyle {\frac{1}{4} A_1^2 + \frac{1}{6} A_1 B_1 + \frac{1}{24} B_1^2} &{} \displaystyle {\frac{1}{4} A_1 A_2 + \frac{1}{12} A_1 B_2 +} \\ &{}\quad \displaystyle {+ \frac{1}{12} A_2 B_1 + \frac{1}{24} B_1 B_2} \\ \displaystyle {\frac{1}{4} A_2 A_1 + \frac{1}{12} A_2 B_1 +} &{}\quad \displaystyle {\frac{1}{4} A_2^2 + \frac{1}{6} A_2 B_2 + \frac{1}{24} B_2^2} \\ \quad \displaystyle {+ \frac{1}{12} A_1 B_2 + \frac{1}{24} B_2 B_1} &{} \\ \end{array} \right) . \end{aligned}$$

Then, in order to make simpler the next arrangements, we put $A_2 = a_{1,2} A_1$, with $a_{1,2} = A_2/A_1$, and $B_2 = b_{1,2} B_1$, with $b_{1,2} = B_2/B_1$. So, ${{\varvec{V}}}_\mathrm{CFM}$ can be rewritten as

$$\begin{aligned} {{\varvec{V}}}_\mathrm{CFM} = \left( \begin{array}{cc} \displaystyle {\frac{1}{4} A_1^2 + \frac{1}{6} A_1 B_1 +} &{} \displaystyle {\frac{a_{1,2}}{4} A_1^2 + \frac{a_{1,2} + b_{1,2}}{12} A_1 B_1 +} \\ \displaystyle {+ \frac{1}{24} B_1^2} &{} \displaystyle {+ \frac{b_{1,2}}{24} B_1^2} \\ &{} \\ \displaystyle {\frac{a_{1,2}}{4} A_1^2 + \frac{a_{1,2} + b_{1,2}}{12} A_1 B_1 +} &{} \displaystyle {\frac{a_{1,2}^2}{4} A_1^2 + \frac{a_{1,2} b_{1,2}}{6} A_1 B_1 +} \\ \displaystyle {+ \frac{b_{1,2}}{24} B_1^2} &{} \displaystyle {+ \frac{b_{1,2}^2}{24} B_1^2} \\ \end{array} \right) . \end{aligned}$$

After a few arrangements, one has

$$\begin{aligned} \det \left( {{\varvec{V}}}_\mathrm{CFM}\right) = \displaystyle {\frac{A_1^2 B_1^2 \left( a_{1,2} - b_{1,2} \right) ^2}{288}}. \end{aligned}$$

Concluding, if $a_{1,2} \ne b_{1,2}$, then $\det \left( {\varvec{V}}_\mathrm{CFM}\right) \ne 0$ and then the rank of ${{\varvec{V}}}_\mathrm{CFM}$ is (at least) equal to 2, while if $a_{1,2} = b_{1,2}$, as ${{\varvec{V}}}_\mathrm{CFM}(1,1) \ne 0$ (it is given by the sum of three positive addends), then the rank of ${{\varvec{V}}}_\mathrm{CFM}$ is (at least) equal to 1.

Second part—In this part, we consider a $N \times N$-dimensional variance–covariance matrix ${\varvec{V}}_\mathrm{CFM}$, with $N \ge 3$, and show under what conditions any column (or row) of this matrix can be expressed as a linear combination of any other two its columns (or rows)—in which case the rank of ${{\varvec{V}}}_\mathrm{CFM}$ would be at most equal to two—or of any other its single column (or row)—in which case the rank of ${{\varvec{V}}}_\mathrm{CFM}$ would be at most equal to one–

We start by considering a generic $3 \times 3$-dimensional submatrix of ${{\varvec{V}}}_\mathrm{CFM}$ specified by the intersection of the i-th, j-th and k-th rows of ${{\varvec{V}}}_\mathrm{CFM}$, with $i \in \{ 1, \ldots , N\}$, $j \in \{ 1, \ldots , N \} {\setminus } \{ i \}$ and $k \in \{ 1, \ldots , N \} {\setminus } \{ i, j \}$, with the l-th, m-th and n-th columns of ${{\varvec{V}}}_\mathrm{CFM}$, with $l \in \{ 1, \ldots , N\}$, $m \in \{ 1, \ldots , N \} {\setminus } \{ l \}$ and $n \in \{ 1, \ldots , N \} {\setminus } \{ l, m \}$. We indicate such a submatrix by $\widetilde{{{\varvec{V}}}}_\mathrm{CFM}$. In order to make simpler the notation, we put again $A_h = b_h - a_h$ and $B_h = \alpha _h + \beta _h$, with $h \in \{ i, j, k, l, m, n \}$. Therefore, the generic element of the submatrix $\widetilde{{{\varvec{V}}}}_\mathrm{CFM}$ has the form

$$\begin{aligned} \widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( h_1, h_2 \right) = \displaystyle {\frac{1}{4} A_{h_1} A_{h_2} + \frac{1}{12} A_{h_1} B_{h_2} + \frac{1}{12} A_{h_2} B_{h_1} + \frac{1}{24} B_{h_1} B_{h_2}}, \end{aligned}$$

with $h_1 \in \{ i, j, k \}$ and $h_2 \in \{ l, m, n \}$. Notice that, if $h_1 = h_2$, then $\widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( h_1, h_2 \right)$ is the variance of the return of the $h_1$-th, or $h_2$-th, stock, while if $h_1 \ne h_2$ then $\widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( h_1, h_2 \right)$ is the covariance between the returns of the $h_1$-th stock and of the $h_2$-th one.

We continue by considering the first row of $\widetilde{{\varvec{V}}}_\mathrm{CFM}$, that is

$$\begin{aligned} \widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( i, \cdot \right) ^\top = \left( \begin{array}{c} \displaystyle {\frac{1}{4} A_i A_l + \frac{1}{12} A_i B_l + \frac{1}{12} A_l B_i + \frac{1}{24} B_i B_l} \\ \\ \displaystyle {\frac{1}{4} A_i A_m + \frac{1}{12} A_i B_m + \frac{1}{12} A_m B_i + \frac{1}{24} B_i B_m} \\ \\ \displaystyle {\frac{1}{4} A_i A_n + \frac{1}{12} A_i B_n + \frac{1}{12} A_n B_i + \frac{1}{24} B_i B_n} \\ \end{array} \right) . \end{aligned}$$

Notice that the result we prove starting from this first row of $\widetilde{{{\varvec{V}}}}_\mathrm{CFM}$ holds also for the second and the third rows of $\widetilde{{{\varvec{V}}}}_\mathrm{CFM}$, $\widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( j, \cdot \right)$ and $\widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( k, \cdot \right)$, respectively. In fact, the three rows of $\widetilde{{\varvec{V}}}_\mathrm{CFM}$ have the same formal structure, with the obvious difference of the indexes. Because of that, we perform the proof only with reference to the first row $\widetilde{{\varvec{V}}}_\mathrm{CFM} \left( i, \cdot \right)$.

Coming back to the proof, as first step, in order to make simpler the next arrangements, we put $A_l = a_{i,l} A_i$, with $a_{i,l} = A_l / A_i$, $A_m = a_{i,m} A_i$, with $a_{i,m} = A_m / A_i$, $A_n = a_{i,n} A_i$, with $a_{i,n} = A_n / A_i$, $B_l = b_{i,l} B_i$, with $b_{i,l} = B_l / B_i$, $B_m = b_{i,m} B_i$, with $b_{i,m} = B_m / B_i$ and $B_n = b_{i,n} B_i$, with $b_{i,n} = B_n / B_i$. So, $\widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( i, \cdot \right)$ can be rewritten as

$$\begin{aligned} \widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( i, \cdot \right) ^\top = \left( \begin{array}{c} \displaystyle {\frac{a_{i,l}}{4} A_i^2 + \frac{a_{i,l} + b_{i,l}}{12} A_i B_i + \frac{b_{i,l}}{24} B_i^2} \\ \\ \displaystyle {\frac{a_{i,m}}{4} A_i^2 + \frac{a_{i,m} + b_{i,m}}{12} A_i B_i + \frac{b_{i,m}}{24} B_i^2} \\ \\ \displaystyle {\frac{a_{i,n}}{4} A_i^2 + \frac{a_{i,n} + b_{i,n}}{12} A_i B_i + \frac{b_{i,n}}{24} B_i^2} \\ \end{array} \right) . \end{aligned}$$

As the second step, in order to detect the possible linear dependence among, or between, the columns of the variance–covariance matrix ${{\varvec{V}}}_\mathrm{CFM}$, we investigate the solution(s) of the following equation:

$$\begin{aligned} c_1 \widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( i, l \right) + c_2 \widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( i, m \right) = \widetilde{{{\varvec{V}}}}_\mathrm{CFM} \left( i, n \right) , \end{aligned}$$

(9)

where $c_1$ and $c_2$ are the unknowns. Of course, on the basis of the possible solution(s) of this equation, we can give an answer to the question of the possible linear dependence in ${{\varvec{V}}}_\mathrm{CFM}$.

After a few arrangements, Eq. 9 can be rewritten as

$$\begin{aligned} \begin{array}{c} \displaystyle { \left( \frac{a_{i,l}}{4} c_1 + \frac{a_{i,m}}{4} c_2 \right) A_i^2 + \left( \frac{a_{i,l} + b_{i,l}}{12} c_1 + \frac{a_{i,m} + b_{i,m}}{12} c_2 \right) A_i B_i +} \\ \\ \displaystyle { + \left( \frac{b_{i,l}}{24} c_1 + \frac{b_{i,m}}{24} c_2 \right) B_i^2 = \frac{a_{i,n}}{4} A_i^2 + \frac{a_{i,n} + b_{i,n}}{12} A_i B_i + \frac{b_{i,n}}{24} B_i^2} \end{array} . \end{aligned}$$

Equating the terms of the two members of the equation having the same degree in $A_i^{2-h} B_i^h$, with $h \in \{0, 1, 2 \}$, after some arrangements, one has the following linear equation system:

$$\begin{aligned} \left( \begin{array}{cc} a_{i,l} &{} a_{i,m} \\ \\ a_{i,l} + b_{i,l} &{} a_{i,m} + b_{i,m} \\ \\ b_{i,l} &{} b_{i,m} \\ \end{array} \right) \left( \begin{array}{c} c_1 \\ \\ c_2 \\ \end{array} \right) = \left( \begin{array}{c} a_{i,n} \\ \\ a_{i,n} + b_{i,n} \\ \\ b_{i,n} \end{array} \right) . \end{aligned}$$

(10)

To solve this linear equation system, we exploit the Theorem of Rouché–Capelli. So, first we determine the rank of the incomplete matrix, then we determine the rank of the complete matrix, and finally, we compare them.

As regards the determination of the rank of the incomplete matrix of the system 10, we calculate the determinants of all the possible $2 \times 2$-dimensional submatrices that can be extracted by the incomplete matrix, that is

$$\begin{aligned}&A_{1,2} = \left( \begin{array}{cc} a_{i,l} &{} a_{i,m} \\ \\ a_{i,l} + b_{i,l} &{} a_{i,m} + b_{i,m} \\ \end{array} \right) , A_{1,3} = \left( \begin{array}{cc} a_{i,l} &{} a_{i,m} \\ \\ b_{i,l} &{} b_{i,m} \\ \end{array} \right) \\&\text {and } A_{2,3} = \left( \begin{array}{cc} a_{i,l} + b_{i,l} &{} a_{i,m} + b_{i,m} \\ \\ b_{i,l} &{} b_{i,m} \\ \end{array} \right) . \end{aligned}$$

All these submatrices have the same determinant, that is

$$\begin{aligned} \det \left( A_{1,2} \right) = \det \left( A_{1,3} \right) = \det \left( A_{2,3} \right) = a_{i,l} b_{i,m} - a_{i,m} b_{i,l} . \end{aligned}$$

Therefore, if $a_{i,l} b_{i,m} \ne a_{i,m} b_{i,l}$, then $\det \left( A_{1,2} \right) \ne 0$, $\det \left( A_{1,3} \right) \ne 0$ and $\det \left( A_{2,3} \right) \ne 0$, and the rank of the incomplete matrix is equal to 2, while if $a_{i,l} b_{i,m} = a_{i,m} b_{i,l}$, as $a_{i,l} \ne 0$ (it is given by a ratio between two positive numbers), then the rank of the incomplete matrix is equal to 1.

As concerns the determination of the rank of the $3 \times 3$-dimensional complete matrix of the system 10, we calculate its determinant which turns out to be equal to 0. Therefore, the rank of the complete matrix is equal to the rank of the incomplete one.

Concluding, the linear equation system 9 admits always a unique solution, which implies the presence of linear dependence in the variance–covariance matrix ${{\varvec{V}}}_\mathrm{CFM}$. In particular, if $a_{i,l} b_{i,m} \ne a_{i,m} b_{i,l}$, then the rank of ${{\varvec{V}}}_\mathrm{CFM}$ is (at most) equal to 2, while if $a_{i,l} b_{i,m} = a_{i,m} b_{i,l}$, then the rank of ${\varvec{V}}_\mathrm{CFM}$ is (at most) equal to 1. $\square$

Proof of Proposition 3.10

Recalling from Proof of Proposition 3.5 that the covariance between two triangular fuzzy numbers is ${\text {Cov}}_\mathrm{CFM} \left( A_1, A_2 \right) = (1/24) (\alpha _1 + \beta _1) (\alpha _2 + \beta _2)$, we consider the two following generic j-th and k-th columns of the variance–covariance matrix of triangular fuzzy numbers ${\varvec{V}}_\mathrm{CFM}$, with $j \in \{1, \ldots , N \}$ and $k \in \{1, \ldots , N \} {\setminus } \{ j \}$:⁸

$$\begin{aligned} \begin{aligned} {{\varvec{V}}}_\mathrm{CFM} (\cdot , j) = \left( \begin{array}{c} \displaystyle { \frac{1}{24} \left( \alpha _1 + \beta _1 \right) \left( \alpha _j + \beta _j \right) } \\ \vdots \\ \displaystyle { \frac{1}{24} \left( \alpha _i + \beta _i \right) \left( \alpha _j + \beta _j \right) } \\ \vdots \\ \displaystyle { \frac{1}{24} \left( \alpha _N + \beta _N \right) \left( \alpha _j + \beta _j \right) } \\ \end{array} \right) \end{aligned}\\ \begin{aligned} \text {and } {{\varvec{V}}}_\mathrm{CFM} (\cdot , k) = \left( \begin{array}{c} \displaystyle { \frac{1}{24} \left( \alpha _1 + \beta _1 \right) \left( \alpha _k + \beta _k \right) } \\ \vdots \\ \displaystyle { \frac{1}{24} \left( \alpha _i + \beta _i \right) \left( \alpha _k + \beta _k \right) } \\ \vdots \\ \displaystyle { \frac{1}{24} \left( \alpha _N + \beta _N \right) \left( \alpha _k + \beta _k \right) } \\ \end{array} \right) . \end{aligned} \end{aligned}$$

${{\varvec{V}}}_\mathrm{CFM} (\cdot , k)$ can be written as a linear combination of ${{\varvec{V}}}_\mathrm{CFM} (\cdot , j)$ in the following way:

$$\begin{aligned} {{\varvec{V}}}_\mathrm{CFM} (\cdot , k) = \frac{\alpha _k + \beta _k}{\alpha _j + \beta _j} {{\varvec{V}}}_\mathrm{CFM} (\cdot , j) . \end{aligned}$$

Therefore, any column of the variance–covariance matrix of triangular fuzzy numbers ${{\varvec{V}}}_\mathrm{CFM}$ can be expressed as a linear combination of any other its single column. $\square$

Proof of Proposition 3.11

Recalling from Proof of Proposition 3.6 that the covariance between two interval fuzzy numbers is ${\text {Cov}}_\mathrm{CFM} \left( A_1, A_2 \right) = (1/4) (b_1 - a_1) (b_2 - a_2)$, we consider the two following generic j-th and k-th columns of the variance–covariance matrix of triangular fuzzy numbers ${{\varvec{V}}}_\mathrm{CFM}$, with $j \in \{1, \ldots , N \}$ and $k \in \{1, \ldots , N \} {\setminus } \{ j \}$:⁹

$$\begin{aligned} \begin{aligned} {{\varvec{V}}}_\mathrm{CFM} (\cdot , j) = \left( \begin{array}{c} \displaystyle { \frac{1}{4} \left( b_1 - a_1 \right) \left( b_j - a_j \right) } \\ \vdots \\ \displaystyle { \frac{1}{4} \left( b_i - a_i \right) \left( b_j - a_j \right) } \\ \vdots \\ \displaystyle { \frac{1}{4} \left( b_N - a_N \right) \left( b_j - a_j \right) } \\ \end{array} \right) \end{aligned}\\ \begin{aligned} \text {and } {{\varvec{V}}}_\mathrm{CFM} (\cdot , k) = \left( \begin{array}{c} \displaystyle { \frac{1}{4} \left( b_1 - a_1 \right) \left( b_k - a_k \right) } \\ \vdots \\ \displaystyle { \frac{1}{4} \left( b_i - a_i \right) \left( b_k - a_k \right) } \\ \vdots \\ \displaystyle { \frac{1}{4} \left( b_N - a_N \right) \left( b_k - a_k \right) } \\ \end{array} \right) . \end{aligned} \end{aligned}$$

${{\varvec{V}}}_\mathrm{CFM} (\cdot , k)$ can be written as a linear combination of ${{\varvec{V}}}_\mathrm{CFM} (\cdot , j)$ in the following way:

$$\begin{aligned} {{\varvec{V}}}_\mathrm{CFM} (\cdot , k) = \frac{b_k - a_k}{b_j - a_j} {{\varvec{V}}}_\mathrm{CFM} (\cdot , j). \end{aligned}$$

Therefore, any column of the variance–covariance matrix of interval fuzzy numbers ${{\varvec{V}}}_\mathrm{CFM}$ can be expressed as a linear combination of any other its single column. $\square$

Proof of Proposition 3.12

We start by considering the two following generic j-th and k-th columns of the variance–covariance matrix ${{\varvec{V}}}_*$, with $j \in \{1, \ldots , N \}$ and $k \in \{1, \ldots , N \} {\setminus } \{ j \}$:¹⁰

$$\begin{aligned} {{\varvec{V}}}_* (\cdot , j) = \left( \begin{array}{c} \displaystyle { \frac{1}{18} \alpha _1 \alpha _j } \\ \vdots \\ \displaystyle { \frac{1}{18} \alpha _i \alpha _j } \\ \vdots \\ \displaystyle { \frac{1}{18} \alpha _N \alpha _j } \\ \end{array} \right) \text { and } {{\varvec{V}}}_* (\cdot , k) = \left( \begin{array}{c} \displaystyle { \frac{1}{18} \alpha _1 \alpha _k } \\ \vdots \\ \displaystyle { \frac{1}{18} \alpha _i \alpha _k } \\ \vdots \\ \displaystyle { \frac{1}{18} \alpha _N \alpha _k } \\ \end{array} \right) . \end{aligned}$$

${{\varvec{V}}}_* (\cdot , k)$ can be written as a linear combination of ${{\varvec{V}}}_* (\cdot , j)$ in the following way:

$$\begin{aligned} {{\varvec{V}}}_* (\cdot , k) = \frac{\alpha _k}{\alpha _j} {{\varvec{V}}}_* (\cdot , j). \end{aligned}$$

Therefore, any column of ${{\varvec{V}}}_*$ can be expressed as a linear combination of any other its single column. $\square$

Proof of Proposition 3.13

We start by considering the two following generic j-th and k-th columns of the variance–covariance matrix ${{\varvec{V}}}^*$, with $j \in \{1, \ldots , N \}$ and $k \in \{1, \ldots , N \} {\setminus } \{ j \}$:¹¹

$$\begin{aligned} {{\varvec{V}}}^* (\cdot , j) = \left( \begin{array}{c} \displaystyle { \frac{1}{18} \beta _1 \beta _j } \\ \vdots \\ \displaystyle { \frac{1}{18} \beta _i \beta _j } \\ \vdots \\ \displaystyle { \frac{1}{18} \beta _N \beta _j } \\ \end{array} \right) \text { and } {{\varvec{V}}}^* (\cdot , k) = \left( \begin{array}{c} \displaystyle { \frac{1}{18} \beta _1 \beta _k } \\ \vdots \\ \displaystyle { \frac{1}{18} \beta _i \beta _k } \\ \vdots \\ \displaystyle { \frac{1}{18} \beta _N \beta _k } \\ \end{array} \right) . \end{aligned}$$

${{\varvec{V}}}^* (\cdot , k)$ can be written as a linear combination of ${{\varvec{V}}}^* (\cdot , j)$ in the following way:

$$\begin{aligned} {{\varvec{V}}}^* (\cdot , k) = \frac{\beta _k}{\beta _j} {\varvec{V}}^* (\cdot , j). \end{aligned}$$

Therefore, any column of ${{\varvec{V}}}^*$ can be expressed as a linear combination of any other its single column. $\square$