What really happens if the positive definiteness requirement on the covariance matrix of returns is relaxed in efficient portfolio selection?

Abstract

The Markowitz critical line method for mean–variance portfolio construction has remained highly influential today, since its introduction to the finance world six decades ago. The Markowitz algorithm is so versatile and computationally efficient that it can accommodate any number of linear constraints in addition to full allocations of investment funds and disallowance of short sales. For the Markowitz algorithm to work, the covariance matrix of returns, which is positive semi-definite, need not be positive definite. As a positive semi-definite matrix may not be invertible, it is intriguing that the Markowitz algorithm always works, although matrix inversion is required in each step of the iterative procedure involved. By examining some relevant algebraic features in the Markowitz algorithm, this paper is able to identify and explain intuitively the consequences of relaxing the positive definiteness requirement, as well as drawing some implications from the perspective of portfolio diversification. For the examination, the sample covariance matrix is based on insufficient return observations and is thus positive semi-definite but not positive definite. The results of the examination can facilitate a better understanding of the inner workings of the highly sophisticated Markowitz approach by the many investors who use it as a tool to assist portfolio decisions and by the many students who are introduced pedagogically to its special cases.

Appendices

Appendix A: Analytical details of two special cases

1.1 A.1 The first special case

For the first special case, where there are no upper investment limits on any of the n individual securities considered, the Lagrangian is

$$\begin{aligned} L=\sigma _{p}^{2}-\lambda \mu _{p}-\theta \left( \sum \limits _{i=1}^{n}x_{i}-1\right) , \end{aligned}$$

(A1)

where $\theta$ is an unknown Lagrange multiplier. The optimality conditions, in addition to Eq. (1), are

$$\begin{aligned} \frac{\partial L}{\partial x_{i}}-\delta _{i}=0,\text { }x_{i}\ge 0,\text { } \delta _{i}\ge 0,\text { and }x_{i}\delta _{i}=0,\text { for }i=1,2,\ldots ,n, \end{aligned}$$

(A2)

where each $\delta _{i}$ is a slack variable. For computational purposes, such conditions can be captured collectively by the matrix equation

$$\begin{aligned} \varvec{WZ}=\varvec{H}+\lambda \varvec{K} \end{aligned}$$

(A3)

or, equivalently,

$$\begin{aligned} \varvec{Z}=\varvec{W}^{-1}\varvec{H}+\lambda \varvec{W}^{-1} \varvec{K}, \end{aligned}$$

(A4)

provided that the $(n+1)\times (n+1)$ matrix $\varvec{W}$ is invertible. Here,

$$\begin{aligned} \varvec{H}=\left[ \begin{array}{ccccc} 0&0&\cdots&0&1 \end{array} \right] ^{\prime } \end{aligned}$$

(A5)

and

$$\begin{aligned} \varvec{K}=\left[ \begin{array}{ccccc} \mu _{1}&\mu _{2}&\cdots&\mu _{n}&0 \end{array} \right] ^{\prime }, \end{aligned}$$

(A6)

where the prime denotes matrix transposition, are $(n+1)$-element column vectors; so is $\varvec{Z}.$

For explicit expressions of $\varvec{Z}$ and $\varvec{W},$ we need to know which of the n securities considered are in for a given value of $\lambda .$ For ease of exposition, suppose for now that there is an iterative step where all n securities are in. In this hypothetical situation, we have

$$\begin{aligned} \varvec{Z}=\left[ \begin{array}{ccccc} x_{1}&\quad x_{2}&\quad \cdots&\quad x_{n}&\quad \theta \end{array} \right] ^{\prime } \end{aligned}$$

(A7)

and

$$\begin{aligned} \varvec{W}=\left[ \begin{array}{ccccc} 2\sigma _{11} &{}\quad 2\sigma _{12} &{}\quad \cdots &{}\quad 2\sigma _{1n} &{}\quad -1 \\ 2\sigma _{21} &{}\quad 2\sigma _{22} &{}\quad \cdots &{}\quad 2\sigma _{2n} &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\sigma _{n1} &{}\quad 2\sigma _{n2} &{}\quad \cdots &{}\quad 2\sigma _{nn} &{}\quad -1 \\ 1 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 \end{array} \right] . \end{aligned}$$

(A8)

The $n+1$ unknown elements of $\varvec{Z}$ can be solved directly in terms of $\lambda$ by using Eq. (A4).

However, there is no need for the above situation to exist at all. If any security i is out, the ith element of $\varvec{Z}$ is substituted by a slack variable $\delta _{i}$ that satisfies the conditions of $x_{i}\ge 0, \delta _{i}\ge 0,$ and $x_{i}\delta _{i}=0.$ Further, column i of $\varvec{W}$ is substituted by the negative of the corresponding column of an $(n+1)\times (n+1)$ identity matrix. To illustrate, suppose that, in an iterative step, securities 1 and 3 are out and all of the remaining securities are in. In such a step, we have

$$\begin{aligned} \varvec{Z}=\left[ \begin{array}{ccccccc} \delta _{1}&\quad x_{2}&\quad \delta _{3}&\quad x_{4}&\quad \cdots&\quad x_{n}&\quad \theta \end{array} \right] ^{\prime } \end{aligned}$$

(A9)

and

$$\begin{aligned} \varvec{W}=\left[ \begin{array}{ccccccc} -1 &{}\quad 2\sigma _{12} &{}\quad 0 &{}\quad 2\sigma _{14} &{}\quad \cdots &{}\quad 2\sigma _{1n} &{}\quad -1 \\ 0 &{}\quad 2\sigma _{22} &{}\quad 0 &{}\quad 2\sigma _{24} &{}\quad \cdots &{}\quad 2\sigma _{2n} &{}\quad -1 \\ 0 &{}\quad 2\sigma _{32} &{}\quad -1 &{}\quad 2\sigma _{34} &{}\quad \cdots &{}\quad 2\sigma _{3n} &{}\quad -1 \\ 0 &{}\quad 2\sigma _{42} &{}\quad 0 &{}\quad 2\sigma _{44} &{}\quad \cdots &{}\quad 2\sigma _{4n} &{} -1 \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 2\sigma _{n2} &{}\quad 0 &{}\quad 2\sigma _{n4} &{}\quad \cdots &{}\quad 2\sigma _{nn} &{}\quad -1 \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 \end{array} \right] \end{aligned}$$

(A10)

for use in Eq. (A4). It is implicit that $x_{1}=\delta _{2}=x_{3}=\delta _{4}=\delta _{5}=\cdots =\delta _{n}=0.$

As the above example illustrates, once the in/out status of each of the n securities considered is known, there are only n unknowns among $x_{1},x_{2},\ldots ,x_{n}$ and $\delta _{1},\delta _{2},\ldots ,\delta _{n}.$ The remaining unknown is $\theta ,$ for a total of $n+1$ unknowns to be solved in terms of $\lambda$ by using Eq. (A4). For notational convenience, let $z_{i}$ be the unknown between $x_{i}$ and $\delta _{i},$ for $i=1,2,\ldots ,n.$ Specifically, if security i is in, as $\delta _{i}=0,$ we have $z_{i}=x_{i};$ if security i is out, as $x_{i}=0,$ we have $z_{i}=\delta _{i}$ instead.

Equation (A4) can be written equivalently as

$$\begin{aligned} z_{i}=a_{i}+b_{i}\lambda ,\text { for }i=1,2,\ldots ,n, \end{aligned}$$

(A11)

and

$$\begin{aligned} \theta =a_{n+1}+b_{n+1}\lambda , \end{aligned}$$

(A12)

where $a_{i}$ and $b_{i}$ are the ith elements of the $(n+1)$-element column vectors $\varvec{W}^{-1}\varvec{H}$ and $\varvec{W}^{-1} \varvec{K},$ respectively. Supposing that the selected security for initiating the iterative procedure is security j,  we must have $x_{j}=z_{j}=a_{j}=1$ and $b_{j}=0.$ Now, we allow $\lambda$ to decrease from its initial value, which is infinitely large. As the feasibility of the portfolio decision requires that $\delta _{i}=z_{i}=a_{i}+b_{i}\lambda \ge 0$ for each security $i\ne j,$ violations will occur as soon as $\lambda$ is low enough to make any computed $z_{i}$ from Eq. (A11) negative.

The critical value of $\lambda$ from this iterative step is the highest of $-a_{i}/b_{i}$ among all cases of $b_{i}>0,$ for $i=1,2,\ldots ,n.$ Cases where $b_{i}\le 0$ need not be considered, as decreasing a positive $\lambda$ will not cause the corresponding $z_{i}=a_{i}+b_{i}\lambda$ to decrease and to become negative. The security responsible for the onset of a violation will have its in/out status revised for the iterative step that follows. The same procedure involving updated versions of Eq. (A4) is repeated in the search of more critical values of $\lambda ,$ until $\lambda =0$ is reached.

1.2 A.2 The second special case

The imposition of upper investment limits on individual securities leads to one more slack variable $\phi _{i}$ for each security i. The optimality conditions, in addition to Eq. (1), are captured collectively by

$$\begin{aligned} \frac{\partial L}{\partial x_{i}}-\delta _{i}+\phi _{i}= & {} 0, \nonumber \\ 0\le & {} x_{i}\le c_{i},\text { }\delta _{i}\ge 0,\text { }\phi _{i}\ge 0,\nonumber \\ x_{i}\delta _{i}= & {} 0,\text { and\ }(c_{i}-x_{i})\phi _{i}=0,\text { for } i=1,2,\ldots ,n. \end{aligned}$$

(A13)

Once the status of security i is known in an iterative step, only one of the variables $x_{i}, \delta _{i},$ and $\phi _{i}$ is unknown; default values are assigned to the remaining two. Specifically, if security i is out, we let $x_{i}=\phi _{i}=0.$ If security i is in, we let $\delta _{i}=\phi _{i}=0.$ If security i is up, we let $x_{i}=c_{i}$ and $\delta _{i}=0.$

Equation (A4) can still be used to search for critical values of $\lambda$ and to compute efficient portfolio weights in term of $\lambda .$ As before, if security i is out in an iterative step, the ith element of $\varvec{Z}$ is substituted by the slack variable $\delta _{i},$ and column i of $\varvec{W},$ by the negative of the corresponding column of an $(n+1)\times (n+1)$ identity matrix. If security i is up instead, the ith element of $\varvec{Z}$ is substituted by the slack variable $\phi _{i},$ and column i of $\varvec{W},$ by the corresponding column of an $(n+1)\times (n+1)$ identity matrix. With $c_{i}<1$ being the upper investment limit on security i,  the two variables among $x_{i}, \delta _{i},$ and $\phi _{i}$ that are not the ith element of $\varvec{Z}$ are assigned their corresponding default values.

To illustrate, suppose that, in an iterative step, securities 1 and 3 are out, security 4 and 5 are up, and all of the remaining securities are in. In this step, we use

$$\begin{aligned} \varvec{Z}=\left[ \begin{array}{ccccccccc} \delta _{1}&\quad x_{2}&\quad \delta _{3}&\quad \phi _{4}&\quad \phi _{5}&\quad x_{6}&\quad \cdots&\quad x_{n}&\quad \theta \end{array} \right] ^{\prime }, \end{aligned}$$

(A14)

$$\begin{aligned} \varvec{W}=\left[ \begin{array}{ccccccccc} -1 &{}\quad 2\sigma _{12} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 2\sigma _{16} &{}\quad \cdots &{}\quad 2\sigma _{1n} &{}\quad -1 \\ 0 &{}\quad 2\sigma _{22} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 2\sigma _{26} &{}\quad \cdots &{}\quad 2\sigma _{2n} &{}\quad -1 \\ 0 &{}\quad 2\sigma _{32} &{}\quad -1 &{}\quad 0 &{}\quad 0 &{}\quad 2\sigma _{36} &{}\quad \cdots &{}\quad 2\sigma _{3n} &{}\quad -1 \\ 0 &{}\quad 2\sigma _{42} &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 2\sigma _{46} &{}\quad \cdots &{}\quad 2\sigma _{4n} &{}\quad -1 \\ 0 &{}\quad 2\sigma _{52} &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad 2\sigma _{56} &{}\quad \cdots &{}\quad 2\sigma _{5n} &{}\quad -1 \\ 0 &{}\quad 2\sigma _{62} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 2\sigma _{66} &{}\quad \cdots &{}\quad 2\sigma _{6n} &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 0 &{}\quad 2\sigma _{n2} &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 2\sigma _{n6} &{}\quad \cdots &{}\quad 2\sigma _{nn} &{}\quad -1 \\ 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 \end{array} \right] , \end{aligned}$$

(A15)

and

$$\begin{aligned} \varvec{H}=\left[ \begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{array} \right] -c_{4}\left[ \begin{array}{c} 2\sigma _{14} \\ 2\sigma _{24} \\ 2\sigma _{34} \\ 2\sigma _{44} \\ 2\sigma _{54} \\ 2\sigma _{64} \\ \vdots \\ 2\sigma _{n4} \\ 1 \end{array} \right] -c_{5}\left[ \begin{array}{c} 2\sigma _{15} \\ 2\sigma _{25} \\ 2\sigma _{35} \\ 2\sigma _{45} \\ 2\sigma _{55} \\ 2\sigma _{65} \\ \vdots \\ 2\sigma _{n5} \\ 1 \end{array} \right] . \end{aligned}$$

(A16)

for Eq. (A4) to determine the next critical value of $\lambda$ and the status change of the security involved. It is implicit that $x_{1}=x_{3}=0, x_{4}=c_{4},$ and $x_{5}=c_{5},$ as well as $\delta _{2}=\delta _{4}=\delta _{5}=\cdots =\delta _{n}=0$ and $\phi _{1}=\phi _{2}=\phi _{3}=\phi _{6}=\phi _{7}=\cdots =\phi _{n}.$

To use Eq. (A11) in each iterative step to establish the next critical value of $\lambda ,$ we look for not only the highest of $-a_{i}/b_{i}$ among all cases of $b_{i}>0,$ but also the highest of $(c_{i}-a_{i})/b_{i}$ among all cases of $b_{i}<0$ if security i is in, for $i=1,2,\ldots ,n.$ The greater value between the two, if positive, will give us the next critical value of $\lambda ,$ as well as whether the status change for the corresponding security is out to in, in to out, in to up, or up to in. The iterative procedure continues until $\lambda =0$ is reached.

1.3 A.3 Matrix invertibility

As illustrated in Eqs. (A10) and (A15) for the two special cases, $\varvec{W}$ has the following feature: The variances and covariances of returns are present only in the columns of $\varvec{W}$ corresponding to those securities that are in. By using a sample covariance matrix of returns based on insufficient return observations, we examine below how the invertibility of $\widehat{\varvec{W}},$ the sample estimate of $\varvec{W},$ is affected by the number of securities that are in, among the n securities considered.

Suppose that the $n\times n$ covariance matrix $\varvec{V}$ is estimated with a set of m return observations. Suppose also that, in an iterative step, q of the n securities are in. Let us first explore the consequences if the situation of $q>m$ ever arises. This is the situation where the number of securities that are in exceeds the number of return observations for estimating the covariance matrix. As the condition of $q\le n$ always holds, the situation of $q>m$ can never arise if $m\ge n.$ Accordingly, the remainder of this subsection is confined to cases where $m<n.$

For ease of exposition, let us relabel the n securities for this iterative step in such a way that the q securities that are in are securities $1,2,\ldots ,q.$ For the purpose of verifying the invertibility of $\widehat{\varvec{W}}$ for the case where there are upper investment limits on individual securities, whether the remaining $n{-}q$ securities are also grouped into respective columns according to the out/up status is unimportant. After such relabeling, we can write

$$\begin{aligned} \widehat{\varvec{W}}=\left[ \begin{array}{ccccccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{12} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{1q} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{22} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{2q} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{q1} &{}\quad 2\widehat{\sigma }_{q2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{qq} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 2\widehat{\sigma }_{q+1,1} &{}\quad 2\widehat{\sigma }_{a+1,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{q+1,q} &{}\quad \pm 1 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 2\widehat{\sigma }_{q+2,1} &{}\quad 2\widehat{\sigma }_{q+2,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{q+2,q} &{}\quad 0 &{}\quad \pm 1 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{nq} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad \pm 1 &{}\quad -1 \\ 1 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \end{array} \right] .\nonumber \\ \end{aligned}$$

(A17)

Here, the diagonal elements in columns $q+1,q+2,\ldots ,n$ for securities that are up or out are displayed as $\pm 1$ for notational convenience; however, it is $+1$ if the corresponding security is up, and is $-1$ otherwise.

Given that the $n\times n$ sample covariance matrix of returns is based on m observations, where $m<q,$ each row there can be replicated by a linear combination of any $m{-}1$ rows of the remaining $n{-}1$ rows. Thus, by confining our attention to the first q columns of the $n\times n$ sample covariance matrix, we can write

$$\begin{aligned} \widehat{\sigma }_{kj}=\sum \limits _{i=1}^{m{-}1}\beta _{ki}\widehat{\sigma } _{ij},\text { for }j=1,2,\ldots ,q\text { and }k=m,m+1,\ldots ,n. \end{aligned}$$

(A18)

Here, $\beta _{k1},\beta _{k2},\ldots ,\beta _{k,m{-}1}$ are $m{-}1$ specific parameters for establishing the linear combination for each row k. These parameters can be deduced from

$$\begin{aligned} \left[ \begin{array}{c} \widehat{\sigma }_{k1} \\ \widehat{\sigma }_{k2} \\ \vdots \\ \widehat{\sigma }_{k,m{-}1} \end{array} \right] =\left[ \begin{array}{cccc} \widehat{\sigma }_{11} &{}\quad \widehat{\sigma }_{12} &{}\quad \cdots &{}\quad \widehat{\sigma } _{1,m{-}1} \\ \widehat{\sigma }_{21} &{}\quad \widehat{\sigma }_{22} &{}\quad \cdots &{}\quad \widehat{\sigma } _{2,m{-}1} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ \widehat{\sigma }_{m-1,1} &{}\quad \widehat{\sigma }_{m-1,2} &{}\quad \cdots &{}\quad \widehat{ \sigma }_{m-1,m-1} \end{array} \right] \left[ \begin{array}{c} \beta _{k1} \\ \beta _{k2} \\ \vdots \\ \beta _{k,m-1} \end{array} \right] \end{aligned}$$

(A19)

or, more directly, from

$$\begin{aligned} \left[ \begin{array}{c} \beta _{k1} \\ \beta _{k2} \\ \vdots \\ \beta _{k,m-1} \end{array} \right] =\left[ \begin{array}{cccc} \widehat{\sigma }_{11} &{}\quad \widehat{\sigma }_{12} &{}\quad \cdots &{}\quad \widehat{\sigma } _{1,m-1} \\ \widehat{\sigma }_{21} &{}\quad \widehat{\sigma }_{22} &{}\quad \cdots &{}\quad \widehat{\sigma } _{2,m-1} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ \widehat{\sigma }_{m-1,1} &{}\quad \widehat{\sigma }_{m-1,2} &{}\quad \cdots &{}\quad \widehat{ \sigma }_{m-1,m-1} \end{array} \right] ^{-1}\left[ \begin{array}{c} \widehat{\sigma }_{k1} \\ \widehat{\sigma }_{k2} \\ \vdots \\ \widehat{\sigma }_{k,m-1} \end{array} \right] . \end{aligned}$$

(A20)

For a given set of risky securities, as long as the only cause for the non-invertibility of a sample covariance matrix is the use of insufficient return observations for its estimation, the above $(m{-}1)\times (m{-}1)$ matrix, which is a submatrix of the $n\times n$ matrix $\widehat{\varvec{ V}},$ is invertible. As such a submatrix can be viewed as an $(m{-}1)\times (m{-}1)$ sample covariance matrix based on m return observations, its invertibility is assured. Thus, the deduced values of $\beta _{k1},\beta _{k2},\ldots ,\beta _{k,m{-}1}$ are unique for each $k=m,m+1,\ldots ,n.$

The determinant of $\widehat{\varvec{W}},$ labeled as $\left| \widehat{\varvec{W}}\right| ,$ remains unchanged if we subtract from any row a linear combination of some or all of the remaining rows there. In view of such an algebraic property, we can subtract $\beta _{m1}$ times row $1, \beta _{m2}$ times row $2, \ldots , \beta _{m,m{-}1}$ times row $m{-}1$ from row m,  without affecting the value of $\left| \widehat{\varvec{W}}\right| .$ Likewise, we can subtract $\beta _{m+1,1}$ times row $1, \beta _{m+1,2}$ times row $2,\ldots ,\beta _{m+1,m{-}1}$ times row $m{-}1$ from row $m+1,$ also without affecting the value of $\left| \widehat{\varvec{W}}\right| .$

These two row operations lead to

$$\begin{aligned} \left| \widehat{\varvec{W}}\right| =\left| \begin{array}{ccccccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{12} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{1q} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{22} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{2q} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{m{-}1,1} &{}\quad 2\widehat{\sigma }_{m{-}1,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m{-}1,q} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad \gamma _{m} \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad \gamma _{m+1} \\ 2\widehat{\sigma }_{m+2,1} &{}\quad 2\widehat{\sigma }_{m+2,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+2,q} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{q+1,1} &{}\quad 2\widehat{\sigma }_{q+1,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{q+1,q} &{}\quad \pm 1 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 2\widehat{\sigma }_{q+2,1} &{}\quad 2\widehat{\sigma }_{q+2,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{q+2,q} &{}\quad 0 &{}\quad \pm 1 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{nq} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad \pm 1 &{}\quad -1 \\ 1 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \end{array}\right| , \nonumber \\ \end{aligned}$$

(A21)

where

$$\begin{aligned} \gamma _{m}=-1+\sum \limits _{i=1}^{m{-}1}\beta _{mi} \end{aligned}$$

(A22)

and

$$\begin{aligned} \gamma _{m+1}=-1+\sum \limits _{i=1}^{m{-}1}\beta _{m+1,i}. \end{aligned}$$

(A23)

Notice that both rows m and $m+1$ have almost all zero elements, with the only exceptions being the $(m,n+1)$ element and the $(m+1,n+1)$ element, which are labeled as $\gamma _{m}$ and $\gamma _{m+1},$ respectively. Notice also that the expression of $\left| \widehat{\varvec{W}} \right|$ in Eq. (A21) pertains to situations where $m+1<q.$ If $m+1=q$ instead, then any displayed row below row $m+1,$ but above row $q+1,$ ought to be omitted. Such omissions, if necessary, are for displaying properly the matrix elements affected by these two row operations. However, regardless of how $\left| \widehat{\varvec{W}} \right|$ is displayed, $\widehat{\varvec{W}}$ is still an $(n+1)\times (n+1)$ matrix.

The next step is to subtract $\gamma _{m}/\gamma _{m+1}$ times row $m+1$ from row m. As the resulting row m is a row of all zeros, $\left| \widehat{\varvec{W}}\right|$ must be zero, implying that $\widehat{ \varvec{W}}$ is not invertible. Thus, if a set of m return observations is used to estimate the $n\times n$ covariance matrix of returns, where $m<n,$ but $m+1$ or more securities are in during an iterative step, Eq. (A4)—which requires the existence of $\widehat{\varvec{W}}^{-1}$ when $\widehat{\varvec{W}}$ is used for $\varvec{W}$ in its implementation—will be unable to perform its intended task. As the existence of $\widehat{\varvec{W}}^{-1}$ is required for the completion of each iterative step, the situation of $q>m$ can never arise throughout the entire iterative procedure.

However, if $q=m$ (that is, if only m securities are in) instead, Eq. (A17) becomes

$$\begin{aligned} \widehat{\varvec{W}}=\left[ \begin{array}{ccccccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{12} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{1m} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{22} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{2m} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{m1} &{}\quad 2\widehat{\sigma }_{m2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{mm} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 2\widehat{\sigma }_{m+1,1} &{}\quad 2\widehat{\sigma }_{m+1,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+1,m} &{}\quad \pm 1 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 2\widehat{\sigma }_{m+2,1} &{}\quad 2\widehat{\sigma }_{m+2,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+2,m} &{}\quad 0 &{}\quad \pm 1 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{nm} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad \pm 1 &{}\quad -1 \\ 1 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \end{array} \right] .\nonumber \\ \end{aligned}$$

(A24)

Noting that row m of the $n\times n$ sample covariance matrix can be expressed as a linear combination of the first $m{-}1$ rows there, we can subtract from row m of the $(n+1)\times (n+1)$ matrix $\widehat{ \varvec{W}}$ the same linear combination of its first $m{-}1$ rows without changing the value of $\left| \widehat{\varvec{W}}\right| .$ Specifically, we can reach

$$\begin{aligned} \left| \widehat{\varvec{W}}\right| =\left| \begin{array}{ccccccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{12} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{1m} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{22} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{2m} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{m{-}1,1} &{}\quad 2\widehat{\sigma }_{m{-}1,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m{-}1,m} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad \gamma _{m} \\ 2\widehat{\sigma }_{m+1,1} &{}\quad 2\widehat{\sigma }_{m+1,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+1,m} &{}\quad \pm 1 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ 2\widehat{\sigma }_{m+2,1} &{}\quad 2\widehat{\sigma }_{m+2,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+2,m} &{}\quad 0 &{}\quad \pm 1 &{}\quad \cdots &{}\quad 0 &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{nm} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad \pm 1 &{}\quad -1 \\ 1 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 \end{array} \right| .\nonumber \\ \end{aligned}$$

(A25)

Row m in the above expression has almost all zero elements, with the only exception being the $(m,n+1)$ element, which is $\gamma _{m}$ as defined in Eq. (A22). Given that the $(m{-}1)\times (m{-}1)$ covariance matrix for securities $1,2,\ldots ,m{-}1$ is positive definite, no row operations for $\left| \widehat{\varvec{W}}\right|$ can reduce row m (or any other row) to a row of all zeros. Thus, if $q=m,$ as $\left| \widehat{ \varvec{W}}\right|$ is nonzero, $\widehat{\varvec{W}}$ is invertible for the iterative step involved.

1.4 A.4 The case of nearly sufficient observations

The special case where $m=n$ is an interesting case. It is the case where the $n\times n$ sample covariance matrix itself is not invertible. If frictionless short sales are allowed, the determination of efficient portfolio weights for a given expected portfolio return requires the inversion of $\widehat{\varvec{V}}.$ As $\widehat{\varvec{V}}$ here is based on n return observations, any of its rows can always be replicated by a linear combination of the remaining $n{-}1$ rows. Accordingly, the determinant of $\widehat{\varvec{V}}$ is zero. Then, as $\widehat{\varvec{V}}^{-1}$ does not exist, the corresponding task of portfolio construction will fail.

In contrast, for efficient portfolio selection without short sales, with or without investment limits on individual securities, the corresponding augmented matrix $\widehat{\varvec{W}}$ is always invertible. This specific $m(=n)$ is the lowest number of return observations for estimating the $n\times n$ covariance matrix, which still ensures that $\widehat{\varvec{W}}$ be invertible throughout the entire iterative procedure, regardless of whether all or only some of the n securities are in during an iterative step. Then, what makes the $(n+1)\times (n+1)$ matrix $\widehat{\varvec{W}}$ invertible, while the underlying $n\times n$ sample covariance matrix $\widehat{\varvec{V}}$ is not invertible?

For an answer, let us consider the situation where all n securities are in. If there are investment limits on individual securities, this situation is where none of the n securities are up or out. With all n securities in, the corresponding $\widehat{ \varvec{W}}$ is simply $2\widehat{\varvec{V}}$ augmented by an extra row and an extra column. Specifically, we have

$$\begin{aligned} \widehat{\varvec{W}}=\left[ \begin{array}{ccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{12} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{1n} &{}\quad -1 \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{22} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{2n} &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{nn} &{}\quad -1 \\ 1 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 \end{array} \right] , \end{aligned}$$

(A26)

which is Eq. (A8) where all matrix elements are substituted by their sample estimates.

To see that the extra row and the extra column are responsible for making $\widehat{\varvec{W}}$ invertible, let us first perform Laplace expansion for $\left| \widehat{\varvec{W}}\right|$ along row $n+1,$ which leads to

$$\begin{aligned} \left| \widehat{\varvec{W}}\right|= & {} (-1)^{n+2}\left| \begin{array}{cccc} 2\widehat{\sigma }_{12} &{}\quad \cdots &{}\quad 2\widehat{\sigma }_{1n} &{}\quad -1 \\ 2\widehat{\sigma }_{22} &{}\quad \cdots &{}\quad 2\widehat{\sigma }_{2n} &{}\quad -1 \\ \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n2} &{}\quad \cdots &{}\quad 2\widehat{\sigma }_{nn} &{}\quad -1 \end{array} \right| \nonumber \\&+\,(-1)^{n+3}\left| \begin{array}{ccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{13} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{1n} &{}\quad -1 \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{23} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{2n} &{}\quad -1 \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n3} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{nn} &{}\quad -1 \end{array} \right| \nonumber \\&+\,(-1)^{n+4}\left| \begin{array}{cccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{12} &{}\quad 2\widehat{\sigma }_{14} &{}\quad \cdots &{}\quad 2\widehat{\sigma }_{1n} &{}\quad -1 \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{22} &{}\quad 2\widehat{\sigma }_{24} &{}\quad \cdots &{}\quad 2\widehat{\sigma }_{2n} &{}\quad -1 \\ \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n2} &{}\quad 2\widehat{\sigma }_{n4} &{}\quad \cdots &{}\quad 2\widehat{\sigma }_{nn} &{}\quad -1 \end{array} \right| \nonumber \\&+\,\cdots +(-1)^{2n+1}\left| \begin{array}{cccc} 2\widehat{\sigma }_{11} &{}\quad \cdots &{}\quad 2\widehat{\sigma }_{1,n-1} &{}\quad -1 \\ 2\widehat{\sigma }_{21} &{}\quad \cdots &{}\quad 2\widehat{\sigma }_{2,n-1} &{}\quad -1 \\ \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad \cdots &{}\quad 2\widehat{\sigma }_{n,n-1} &{}\quad -1 \end{array} \right| . \end{aligned}$$

(A27)

We can perform Laplace expansion again, now for each of the n determinants in Eq. (A27) along its column n (which is the column where each element is ${-}1).$ Each of the $n^{2}$ resulting determinants is $2^{n{-}1}$ times the determinant of an $(n{-}1)\times (n{-}1)$ submatrix of $\widehat{\varvec{V}},$ obtained by deleting a row and a column from $\widehat{\varvec{V}}.$ With the signs attached to the individual determinants accounted for, $\left| \widehat{\varvec{W}}\right|$ is $2^{n{-}1}$ times the sum of the $n^{2}$ cofactors (signed minors) of $\widehat{\varvec{V}}.$

In this special case where $m=n,$ as none of the rows in each of the $(n{-}1)\times (n{-}1)$ submatrices can be replicated by a linear combination of the remaining rows, the corresponding determinants are nonzero. Further, as the $n^{2}$ submatrices are different, so are the corresponding cofactors. With $\left| \widehat{\varvec{W}}\right|$ being nonzero, the inverse of $\widehat{\varvec{W}}$ exists, even if all n securities considered are in. Thus, in this special case, Eq. (A4) can always perform its intended task.

Appendix B: Analytical details of more general cases

This appendix extends the two special cases in Appendix A by imposing additional linear equality constraints for allocations of investment funds. For the purpose of the analytical examination here, a crucial algebraic feature is that each additional constraint will result in an extra row and an extra column for the augmented matrix involved. It is the presence of such extra row and column that allows an additional security to be in without making the augmented matrix involved singular.

For extensions to more general cases, let us start with an extension to the first special case—where disallowance of short sales and full allocations of investment funds are the only constraints—by adding a linear equality constraint that Eq. (2) provides. The Lagrangian L in Eq. (A1) will have an extra additive term; specifically, it is $w\left( \sum \nolimits _{i=1}^{n}h_{i}x_{i}-s\right) ,$ where w is the corresponding Lagrange multiplier. As w is an extra unknown to be determined for each iterative step, the four matrices in Eq. (A4) will have to be augmented. Specifically, the original $(n+1)$-element column vectors $\varvec{H}, \varvec{K},$ and $\varvec{Z}$ now become $(n+2)$-element column vectors with the last (augmented) elements being s, 0,  and w,  respectively, and the original $(n+1)\times (n+1)$ matrix $\varvec{W}$ now becomes an $(n+2)\times (n+2)$ matrix with the last row being $h_{1},h_{2},\ldots ,h_{n},0,0$ and with the last column being $-h_{1},-h_{2},\ldots ,-h_{n},0,0.$

This extension is also applicable to the second special case, where there are upper investment limits on individual securities in addition to all the constraints in the first special case. However, if any of the securities considered are up during an iterative step, the $(n+2)$-element column vector $\varvec{H}$ will have to be revised. The revised vector is as follows:

$$\begin{aligned} \varvec{H}=\left[ \begin{array}{c} 0 \\ \vdots \\ 0 \\ 1 \\ s \end{array} \right] -\sum \limits _{i\in U}c_{i}\left[ \begin{array}{c} 2\sigma _{1i} \\ \vdots \\ 2\sigma _{ni} \\ 1 \\ h_{i} \end{array} \right] . \end{aligned}$$

(B1)

Here, $\sum \nolimits _{i\in U}$ stands for the summation over all securities that are up.

Given the linear equality constraints in Eqs. (1) and (2), the determination of the initial portfolio to start the iterative procedure in either case is by exhausting the securities with the highest expected returns among the n securities considered. We can still use the same iterative procedure to search for critical values of $\lambda$ successively, as described in Appendix A. Suppose that, as before, the $n\times n$ covariance matrix of returns is estimated with m securities. We are interested in knowing whether $\widehat{\varvec{W}}$ is invertible if q of the n securities considered are in during an iterative step, where $q>m.$

To seek an answer, we start with $q=m+2.$ Implicitly, as the condition of $q\le n$ always holds in any iterative step, we also impose that $m+2\le n$ or, equivalently, $m<n{-}1.$ Once again, for ease of exposition, let us relabel the n securities for an iterative step in such a way that the q securities that are in are securities $1,2,\ldots ,q.$ After relabeling, we can write

$$\begin{aligned} \widehat{\varvec{W}}=\left[ \begin{array}{cccccccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{12} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{1,m+2} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{1} \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{22} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{2,m+2} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{2} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \cdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{m+2,1} &{}\quad 2\widehat{\sigma }_{m+2,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+2,m+2} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m+2} \\ 2\widehat{\sigma }_{m+3,1} &{}\quad 2\widehat{\sigma }_{m+3,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+3,m+2} &{}\quad \pm 1 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m+3} \\ 2\widehat{\sigma }_{m+4,1} &{}\quad 2\widehat{\sigma }_{m+4,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+4,m+2} &{}\quad 0 &{}\quad \pm 1 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m+4} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{n,m+2} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad \pm 1 &{}\quad -1 &{}\quad -h_{n} \\ 1 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ h_{1} &{}\quad h_{2} &{}\quad \cdots &{}\quad h_{m+2} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{array} \right] . \nonumber \\ \end{aligned}$$

(B2)

Here, each diagonal element of $\widehat{\varvec{W}}$ in columns $m+3,m+4,\ldots ,n,$ if equal to $+1,$ indicates that the corresponding security is up; if equal to $-1$ instead, the corresponding security is out.

As each row of the $n\times n$ sample covariance matrix $\widehat{ \varvec{V}}$ can be replicated by a linear combination of $m{-}1$ of the remaining rows, we can write the determinant of $\widehat{\varvec{W}}$ as

$$\begin{aligned} \left| \widehat{\varvec{W}}\right| =\left| \begin{array}{cccccccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{12} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{1,m+2} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{1} \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{22} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{2,m+2} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{2} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \cdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{m{-}1,1} &{}\quad 2\widehat{\sigma }_{m{-}1,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m{-}1,m+2} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m{-}1} \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad \gamma _{m} &{}\quad \kappa _{m} \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad \gamma _{m+1} &{}\quad \kappa _{m+1} \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad \gamma _{m+2} &{}\quad \kappa _{m+2} \\ 2\widehat{\sigma }_{m+3,1} &{}\quad 2\widehat{\sigma }_{m+3,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+3,m+2} &{}\quad \pm 1 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m+3} \\ 2\widehat{\sigma }_{m+4,1} &{}\quad 2\widehat{\sigma }_{m+4,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+4,m+2} &{}\quad 0 &{}\quad \pm 1 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m+4} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{n,m+2} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad \pm 1 &{}\quad -1 &{}\quad -h_{n} \\ 1 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ h_{1} &{}\quad h_{2} &{}\quad \cdots &{}\quad h_{m+2} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{array} \right| , \nonumber \\ \end{aligned}$$

(B3)

by subtracting from each of rows $m, m+1,$ and $m+2$ the corresponding linear combination of the first $m{-}1$ rows of $\widehat{\varvec{W}}.$ Here, $\gamma _{m}$ and $\gamma _{m+1}$ are as shown in Eqs. (A22) and (A23), and $\gamma _{m+2}, \kappa _{m}, \kappa _{m+1},$ and $\kappa _{m+2}$ are given by

$$\begin{aligned} \gamma _{m+2}=-1+\sum \limits _{i=1}^{m{-}1}\beta _{m+2,i} \end{aligned}$$

(B4)

and

$$\begin{aligned} \kappa _{m+j}=-h_{m+j}+\sum \limits _{i=1}^{m{-}1}h_{i}\beta _{m+j,i},\text { for }j=0,1,\text { and }2. \end{aligned}$$

(B5)

Subtracting $\gamma _{m}/\gamma _{m+1}$ times row $\gamma _{m+1}$ from row m will change row m into a row of almost all zeros, with the only exception being the last element there. So will subtracting $\gamma _{m+1}/\gamma _{m+2}$ times row $\gamma _{m+2}$ from row $m+1.$ Letting $\eta _{m}$ and $\eta _{m+1}$ be the last elements of the new rows m and $m+1,$ respectively, we can subtract $\eta _{m}/\eta _{m+1}$ times the new row $m+1$ from the new row m to reach a row of all zeros for row m. With $\left| \widehat{\varvec{W}}\right|$ being zero for the situation where $q=m+2, \widehat{\varvec{W}}$ is not invertible. As the status change is for one security at a time, situations where $q>m+2$—which requires the invertibility of $\widehat{\varvec{W}}$ in a preceding iterative step where $q=m+2$—can never occur during the iterative procedure.

Now, suppose that $q=m+1$ instead in an iterative step. In this situation, Eqs. (B2) and (B3) become

$$\begin{aligned} \widehat{\varvec{W}}=\left[ \begin{array}{cccccccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{12} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{1,m+1} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{1} \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{22} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{2,m+1} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{2} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \cdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{m+1,1} &{}\quad 2\widehat{\sigma }_{m+1,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+1,m+1} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m+1} \\ 2\widehat{\sigma }_{m+2,1} &{}\quad 2\widehat{\sigma }_{m+2,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+2,m+1} &{}\quad \pm 1 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m+2} \\ 2\widehat{\sigma }_{m+3,1} &{}\quad 2\widehat{\sigma }_{m+3,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+3,m+1} &{}\quad 0 &{}\quad \pm 1 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m+3} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{n,m+1} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad \pm 1 &{}\quad -1 &{}\quad -h_{n} \\ 1 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ h_{1} &{}\quad h_{2} &{}\quad \cdots &{}\quad h_{m+1} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{array} \right] \nonumber \\ \end{aligned}$$

(B6)

and

$$\begin{aligned} \left| \widehat{\varvec{W}}\right| =\left| \begin{array}{cccccccccc} 2\widehat{\sigma }_{11} &{}\quad 2\widehat{\sigma }_{12} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{1,m+1} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{1} \\ 2\widehat{\sigma }_{21} &{}\quad 2\widehat{\sigma }_{22} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{2,m+1} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{2} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \cdots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{m{-}1,1} &{}\quad 2\widehat{\sigma }_{m{-}1,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m{-}1,m+1} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m{-}1} \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad \gamma _{m} &{}\quad \kappa _{m} \\ 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad \gamma _{m+1} &{}\quad \kappa _{m+1} \\ 2\widehat{\sigma }_{m+2,1} &{}\quad 2\widehat{\sigma }_{m+2,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+2,m+1} &{}\quad \pm 1 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m+2} \\ 2\widehat{\sigma }_{m+3,1} &{}\quad 2\widehat{\sigma }_{m+3,2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{m+3,m+1} &{}\quad 0 &{}\quad \pm 1 &{}\quad \cdots &{}\quad 0 &{}\quad -1 &{}\quad -h_{m+3} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots &{}\quad \vdots &{}\quad \vdots \\ 2\widehat{\sigma }_{n1} &{}\quad 2\widehat{\sigma }_{n2} &{}\quad \cdots &{}\quad 2\widehat{ \sigma }_{n,m+1} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad \pm 1 &{}\quad -1 &{}\quad -h_{n} \\ 1 &{}\quad 1 &{}\quad \cdots &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ h_{1} &{}\quad h_{2} &{}\quad \cdots &{}\quad h_{m+1} &{}\quad 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 0 &{}\quad 0 \end{array} \right| ,\nonumber \\ \end{aligned}$$

(B7)

respectively. As no row operations can result in a row of all zeros, we can conclude that, if $m+1$ securities are in, the corresponding $\widehat{\varvec{W}}$ is still invertible. If less than $m+1$ securities are in, the same conclusion can also be reached.

In the two special cases considered in Appendix A, the $(n+1)\times (n+1)$ matrix $\widehat{\varvec{W}}$ is invertible if $q\le m$ or, equivalently, if $q<m+1.$ That is, if the number of securities that is in does not exceed the number of return observations for estimating the $n\times n$ covariance matrix. In contrast, in the above extension, where an additional linear equality constraint is also imposed on the portfolio choice, the corresponding $(n+2)\times (n+2)$ matrix $\widehat{ \varvec{W}}$ is invertible if $q\le m+1$ or, equivalently, if $q<m+2.$

This result reveals the following pattern: The presence of $\upsilon$ linear equality constraints, including Eq. (1), will lead to an $(n+\upsilon )\times (n+\upsilon )$ matrix $\widehat{\varvec{W}},$ where the last $\upsilon$ columns can be partitioned into an $n\times \upsilon$ block with nonzero elements and a $\upsilon \times \upsilon$ block of zeros. The $(n+\upsilon )\times (n+\upsilon )$ matrix $\widehat{ \varvec{W}}$ is invertible if $q\le m+\upsilon -1$ or, equivalently, if $q<m+\upsilon .$

As long as $m<n,$ row operations to $\widehat{\varvec{W}}$ can lead to a matrix containing one or more rows of nearly all zeros, with the corresponding row elements in the $n\times \upsilon$ block being the exceptions. Further row operations to reach a row with all zeros, without affecting the determinant of $\widehat{\varvec{W}},$ will require $q=m+\upsilon .$ Given that, for a sample covariance matrix of returns (which is always positive semi-definite), the Markowitz algorithm always works as intended, the situation where the number of securities that are in exceeds $m+\upsilon -1$ in an iterative step will never occur. As the condition of $q\le n$ always holds, we must have $m+\upsilon -1\le n$ or, equivalently, $m\le n{-}\upsilon +1,$ in order to make it possible for the maximum number of securities to be in during an iterative step, while maintaining the invertibility of the corresponding $(n+\upsilon )\times (n+\upsilon )$ matrix $\widehat{\varvec{W}}$.