A Stein-type shrinkage estimator of the covariance matrix for portfolio selections

Abstract

The covariance matrix plays a crucial role in portfolio optimization problems as the risk and correlation measure of asset returns. An improved estimation of the covariance matrix can enhance the performance of the portfolio. In this paper, based on the Cholesky decomposition of the covariance matrix, a Stein-type shrinkage strategy for portfolio weights is constructed under the mean-variance framework. Furthermore, according to the agent’s maximum expected utility value, a portfolio selection strategy is proposed. Finally, simulation experiments and an empirical study are used to test the feasibility of the proposed strategy. The numerical results show our portfolio strategy performs satisfactorily.

Appendix

1.1 Some preliminaries

In order to construct our Stein-type shrinkage strategy for portfolio selection, Lemmas 4 and 5, needed to establish the main results in this paper, are presented and followed by the proofs of Lemmas 1–3, Theorem 1 and Corollary 1.

Lemma 4

(Haff 1979) Let A follow a Wishart distribution with n degrees of freedom and covariance matrix $\varSigma$, denoted as $A \sim W_{p}(n,\varSigma )$, where $n\ge p+4$. Then

$$\begin{aligned} E(A^{-1}VA^{-1})= & {} \frac{tr(\varSigma ^{-1}V)}{(n-p)(n-p-1)(n-p-3)}\varSigma ^{-1}\\&+\,\frac{1}{(n-p)(n-p-3)}\varSigma ^{-1}V\varSigma ^{-1}, \end{aligned}$$

where $V\ge 0$ is a $p\times p$ non-random matrix.

Lemma 5

(Muirhead 1982, Theorem 3.2.14) Let $A \sim W_{p}(n,I_{p})$, where $n\ge p$ and $I_{p}$ is a $p\times p$ identity matrix, and $A=C'C$, where C is an upper triangular $p\times p$ matrix with positive diagonal elements. Then the elements $c_{ij}(1\le i\le j\le p)$ of C are all independent, $c_{ii}^{2}\sim \chi _{n-i+1}^{2}(i=1,\ldots ,p)$, and $c_{ij}\sim N(0,1) (1\le i<j\le p)$.

1.2 Proof of Lemma 1

We recall that $\hat{\mu }=\frac{1}{n}\sum _{t=1}^{n}R_{t}$ and $\widehat{\mu }\sim N(\mu ,\varSigma /n)$. $\varGamma$ is the upper triangular matrix from the Cholesky decomposition of $\varSigma$, denoted as $\varSigma =\varGamma '\varGamma$. $\widetilde{\mu }=(\varGamma ')^{-1}\widehat{\mu }$. F is the upper triangular matrix from the Cholesky decomposition of $n\hat{\varSigma }$, denoted as $n\hat{\varSigma }=F'F$. $\tilde{F}=(\tilde{f}_{ij})=F\varGamma ^{-1}$. So, $E(\widehat{\mu }\widehat{\mu }')=\varSigma /n+\mu \mu '$, and $\widehat{\mu }$ and $\hat{\varSigma }$ are independent of each other.

Substituting $\hat{w}^{s}=\frac{1}{\gamma }F^{-1}H(F')^{-1}\hat{\mu }$ into (1), we have

$$\begin{aligned} U(\hat{w}^{s})=\frac{1}{\gamma }\widehat{\mu }'F^{-1}H(F')^{-1}\mu -\frac{1}{2\gamma }\widehat{\mu }'F^{-1}H(F')^{-1}\varSigma F^{-1}H(F')^{-1}\widehat{\mu }. \end{aligned}$$

Taking the expectation of $U(\hat{w}^{s})$, we obtain

$$\begin{aligned} E(U(\hat{w}^{s}))=\frac{1}{\gamma }E(\widehat{\mu }'F^{-1}H(F')^{-1}\mu )-\frac{1}{2\gamma }E(\widehat{\mu }'F^{-1}H(F')^{-1}\varSigma F^{-1}H(F')^{-1}\widehat{\mu }). \end{aligned}$$

We calculate the two parts of $E(U(\hat{w}^{s}))$ as follows:

$$\begin{aligned}&E(\widehat{\mu }'F^{-1}H(F')^{-1}\mu )\\&\quad =E(\widehat{\mu }')E(F^{-1}H(F')^{-1})\mu =\mu 'E(F^{-1}H(F')^{-1})\mu \\&\quad =E(\mu '\varGamma ^{-1}\widetilde{F}^{-1}H(\widetilde{F}')^{-1}(\varGamma ')^{-1}\mu )=\mu '\varGamma ^{-1}A_{1}(\varGamma ')^{-1}\mu ,\\&\qquad E(\widehat{\mu }'F^{-1}H(F')^{-1}\varSigma F^{-1}H(F')^{-1}\widehat{\mu })\\&\quad =E(tr(\widehat{\mu }'F^{-1}H(F')^{-1}\varSigma F^{-1}H(F')^{-1}\widehat{\mu }))\\&\quad =E(tr(F^{-1}H(F')^{-1}\varSigma F^{-1}H(F')^{-1}\widehat{\mu }\widehat{\mu }'))\\&\quad =tr(E(F^{-1}H(F')^{-1}\varSigma F^{-1}H(F')^{-1})E(\widehat{\mu }\widehat{\mu }'))\\&\quad =\frac{1}{n}tr(E(F^{-1}H(F')^{-1}\varSigma F^{-1}H(F')^{-1}\varSigma ))\\&\qquad +E(\mu 'F^{-1}H(F')^{-1}\varSigma F^{-1}H(F')^{-1}\mu )\\&\quad =\frac{1}{n}tr(A_{2})+\mu '\varGamma ^{-1}A_{2}(\varGamma ')^{-1}\mu . \end{aligned}$$

The proof is complete. $\square$

1.3 Proof of Lemma 2

1.3.1 Step 1

In order to compute $A_{1}$, we need $\tilde{F}$ and H to be decomposed as

$$\begin{aligned} \tilde{F}=\left( \begin{array}{cc} \tilde{f}_{11} &{}\quad \tilde{f}_{21} \\ \mathbf 0 &{}\quad \tilde{F}_{22} \end{array} \right) , \ \ H=\left( \begin{array}{cc} h_{1} &{}\quad \mathbf 0 ' \\ \mathbf 0 &{} \quad H_{2} \end{array} \right) , \end{aligned}$$

where $\tilde{f}_{11}$ and $h_{1}$ are scalars, $\tilde{F}_{22}$ and $H_{2}$ are the $(p-1)\times (p-1)$ submatrices of $\tilde{F}$ and H, respectively.

Then $A_{1}$ can be expressed as

$$\begin{aligned} \begin{array}{lll} A_{1} &{} =&{}E[\tilde{F}^{-1}H(\tilde{F}')^{-1}] \\ &{} = &{}E\left( \begin{array}{cc} \tilde{f}_{11}^{-2}h_{1}+\tilde{f}_{11}^{-2}\tilde{f}_{21}\tilde{F}_{22}^{-1}H_{2}(\tilde{F}_{22}')^{-1}\tilde{f}_{21}'\quad &{}\quad -\tilde{f}_{11}^{-1}\tilde{f}_{21}\tilde{F}_{22}^{-1}H_{2}(\tilde{F}_{22}')^{-1} \\ -\tilde{f}_{11}^{-1}\tilde{F}_{22}^{-1}H_{2}(\tilde{F}_{22}')^{-1}\tilde{f}_{21}'&{} \tilde{F}_{22}^{-1}H_{2}(\tilde{F}_{22}')^{-1} \end{array} \right) .\\ \end{array} \end{aligned}$$

Note that $\tilde{f}_{11}$, $\tilde{f}_{21}$ and $\tilde{F}_{22}$ are mutually independently distributed as $\tilde{f}_{11}\sim \chi _{n-1}^{2}$, $\tilde{f}_{21}\sim N(0,I_{p-1})$ and $\tilde{F}_{22}'\tilde{F}_{22}\sim W_{p-1}(n-2)$ according to Lemma 5. Calculating the means of the inverse Chi-square and normal distributions, we obtain

$$\begin{aligned} \begin{array}{lll} &{} A_{1}=&{}\left( \begin{array}{cc} \frac{h_{1}}{n-3}+\frac{1}{n-3}tr[E(\tilde{F}_{22}^{-1}H_{2}(\tilde{F}_{22}')^{-1})]&{} \mathbf 0 ' \\ \mathbf 0 &{} E(\tilde{F}_{22}^{-1}H_{2}(\tilde{F}_{22}')^{-1}) \end{array} \right) . \end{array} \end{aligned}$$

1.3.2 Step 2

$E(\tilde{F}_{22}^{-1}H_{2}(\tilde{F}_{22}')^{-1})$ needs to be calculated.

Decomposing $\tilde{F}_{22}$ and $H_{2}$ as follows

$$\begin{aligned} \tilde{F}_{22}=\left( \begin{array}{cc} \tilde{f}_{22}&{}\quad \tilde{f}_{32} \\ \mathbf 0 &{}\quad \tilde{F}_{33} \end{array} \right) , \ \ H_{2}=\left( \begin{array}{cc} h_{2}&{}\quad \mathbf 0 ' \\ \mathbf 0 &{}\quad H_{3} \end{array} \right) , \end{aligned}$$

where $\tilde{f}_{22}$ and $h_{2}$ are scalars, $\tilde{F}_{33}$ and $H_{3}$ are the $(p-2)\times (p-2)$ lower right submatrices of $\tilde{F}$ and H, respectively, we get $E(\tilde{F}_{22}^{-1}H_{2}(\tilde{F}_{22}')^{-1})$ expressed as

$$\begin{aligned} \begin{array}{lll} &{}&{}E(\tilde{F}_{22}^{-1}H_{2}(\tilde{F}_{22}')^{-1}) \\ &{}&{}\quad = E\left( \begin{array}{cc} \tilde{f}_{22}^{-2}h_{2}+\tilde{f}_{22}^{-2}\tilde{f}_{32}\tilde{F}_{33}^{-1}H_{3}(\tilde{F}_{33}')^{-1}\tilde{f}_{32}'\quad &{}\quad -\tilde{f}_{22}^{-1}\tilde{f}_{32}\tilde{F}_{33}^{-1}H_{3}(\tilde{F}_{33}')^{-1} \\ -\tilde{f}_{22}^{-1}\tilde{F}_{33}^{-1}H_{3}(\tilde{F}_{33}')^{-1}\tilde{f}_{32}'&{} \tilde{F}_{33}^{-1}H_{3}(\tilde{F}_{33}')^{-1} \end{array} \right) .\\ \end{array} \end{aligned}$$

Note that $\tilde{f}_{22}$, $\tilde{f}_{32}$ and $\tilde{F}_{33}$ are mutually independently distributed as $\tilde{f}_{22}\sim \chi _{n-2}^{2}$, $\tilde{f}_{32}\sim N(0,I_{p-2})$ and $\tilde{F}_{33}'\tilde{F}_{33}\sim W_{p-2}(n-3)$ according to Lemma 5. Calculating the means of inverse Chi-square and normal distributions, we get

$$\begin{aligned} E(\tilde{F}_{22}^{-1}H_{2}(\tilde{F}_{22}')^{-1})=\left( \begin{array}{cc} \frac{h_{2}}{n-4}+\frac{1}{n-4}tr[E(\tilde{F}_{33}^{-1}H_{3}(\tilde{F}_{33}')^{-1})]&{} \mathbf 0 ' \\ \mathbf 0 &{} E(\tilde{F}_{33}^{-1}H_{3}(\tilde{F}_{33}')^{-1}) \end{array} \right) . \end{aligned}$$

1.3.3 Step 3

Decomposing $\tilde{F}_{33}$ and $H_{3}$ similarly and repeating the same procedure, we get

$$\begin{aligned} E(\tilde{F}_{33}^{-1}H_{3}(\tilde{F}_{33}')^{-1})=\left( \begin{array}{cc} \frac{h_{3}}{n-5}+\frac{1}{n-5}tr[E(\tilde{F}_{44}^{-1}H_{4}(\tilde{F}_{44}')^{-1})]&{} \mathbf 0 ' \\ \mathbf 0 &{} E(\tilde{F}_{44}^{-1}H_{4}(\tilde{F}_{44}')^{-1}) \end{array} \right) . \end{aligned}$$

1.3.4 Step p-1

$$\begin{aligned} \begin{array}{rcl} &{}&{}E(\tilde{F}_{(p-1)\times (p-1)}^{-1}H_{p-1}(\tilde{F}_{(p-1)\times (p-1)}')^{-1})\\ &{}&{}\quad = \left( \begin{array}{cc} \frac{h_{p-1}}{n-p-1}+\frac{1}{n-p-1}E(\tilde{f}_{p\times p}^{-1}h_{p}(\tilde{f}_{_{p\times p}}')^{-1})&{} \mathbf 0 ' \\ \mathbf 0 &{} E(\tilde{f}_{_{p\times p}}^{-1}h_{p}(\tilde{f}_{_{p\times p}}')^{-1}) \end{array} \right) \\ &{}&{}\quad = \left( \begin{array}{cc} \frac{h_{p-1}}{n-p-1}+\frac{h_{p}}{(n-p-1)(n-p-2)}&{} 0 \\ 0&{} \frac{h_{p}}{n-p-2} \end{array} \right) . \end{array} \end{aligned}$$

1.3.5 Step p

Based on the computations, we have

$$\begin{aligned} A_{1}=\left( \begin{array}{cccc} \frac{h_{1}}{n-3}+\sum _{j=2}^{p}\frac{h_{j}}{(n-j-1)(n-j-2)} &{} 0 &{} \cdots &{} 0 \\ 0 &{} \frac{h_{2}}{n-4}+\sum _{j=3}^{p}\frac{h_{j}}{(n-j-1)(n-j-2)} &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} \vdots \\ 0 &{} 0 &{} 0 &{} \frac{h_{p}}{(n-p-2)}\\ \end{array} \right) . \end{aligned}$$

The proof is completed. $\square$

1.4 Proof of Lemma 3

Proof

Using $\tilde{F}'\tilde{F}\sim W_{p}(n-1)$ and Lemma 4, we can easily get the above result after some calculations. $\square$

1.5 Proof of Theorem 1

Proof

$E[U(\hat{w}^{s})]$ can be rewritten as follows:

$$\begin{aligned} E[U(\hat{w}^{s})]=\frac{1}{\gamma }\mu '\varGamma ^{-1}\left( A_{1}-\frac{1}{2}A_{2}\right) (\varGamma ')^{-1}\mu -\frac{1}{2n\gamma }tr\{A_{2}\}. \end{aligned}$$

Similarly,

$$\begin{aligned} E(U(\bar{w}_{3}))=\frac{1}{\gamma }\mu '\varGamma ^{-1}\left( \frac{k_{2}}{2}I_{p}\right) (\varGamma ')^{-1}\mu -\frac{1}{2\gamma }k_{3}, \end{aligned}$$

where $k_{2}=(\frac{n}{n+1})(2-\frac{n(n-2)(n-p-2)}{(n+1)(n-p-1)(n-p-4)})$ and $k_{3}=\frac{np(n-2)(n-p-2)}{(n+1)^{2}(n-p-1)(n-p-4)}$.

If $E[U(\hat{w}^{s})]\ge E(U(\bar{w}_{3}))$, we need the following conditions to be satisfied:

$$\begin{aligned} \begin{array}{rcl} A_{1}-\frac{1}{2}A_{2} &{}\ge &{} \frac{k_{2}}{2}I_{p}\\ \frac{k_{3} }{2\gamma }&{}\ge &{} \frac{tr(A_{2})}{2n\gamma } \end{array}. \end{aligned}$$

Using Lemma 2, the above conditions can be rewritten as

$$\begin{aligned} \begin{array}{rcl} A_{1}-\frac{(h_{m})^{2}(n-2)}{2(n-p-1)(n-p-2)(n-p-4)}I_{p}&{}\ge &{} \frac{n}{n+1}I_{p}-\frac{n^{2}(n-p-2)(n-2)}{2(n+1)^{2}(n-p-1)(n-p-4)}I_{p}\\ \frac{n^{2}(n-p-2)(n-2)p}{(n+1)^{2}(n-p-1)(n-p-4)}&{} \ge &{}\frac{(h_{m})^{2}(n-2)p}{(n-p-1)(n-p-2)(n-p-4)} \end{array}.(*) \end{aligned}$$

Using the condition in $(*)$ and the restriction $h_{i}>0$, we get $0<h_{m}\le \frac{n(n-p-2)}{n+1}$.

1.5.1 Step 1

From Lemma 2 and the first condition in $(*)$, we have

$$\begin{aligned}&\frac{h_{p}}{n-p-2}-\frac{(h_{m})^{2}(n-2)}{2(n-p-1)(n-p-2)(n-p-4)}\\&\quad \ge \frac{n}{n+1}-\frac{n^{2}(n-2)(n-p-2)}{2(n+1)^{2}(n-p-1)(n-p-4)}. \end{aligned}$$

Simplifying the above inequality, we have

$$\begin{aligned} h_{p}\ge \frac{n(n-p-2)}{n+1}-\frac{n^{2}(n-2)(n-p-2)^{2}}{2(n+1)^{2}(n-p-1)(n-p-4)}. \end{aligned}$$

Then, $max(0,\frac{n(n-p-2)}{n+1}-\frac{n^{2}(n-2)(n-p-2)^{2}}{2(n+1)^{2}(n-p-1)(n-p-4)})\le h_{p} \le \frac{n(n-p-2)}{n+1}$.

1.5.2 Step 2

From the first condition in $(*)$, we know

$$\begin{aligned}&\frac{h_{p-1}}{n-p-1}+\frac{h_{p}}{(n-p-1)(n-p-2)}-\frac{(h_{m})^{2}(n-2)}{2(n-p-1)(n-p-2)(n-p-4)}\\&\quad \ge \frac{n}{n+1}-\frac{n^{2}(n-2)(n-p-2)}{2(n+1)^{2}(n-p-1)(n-p-4)}. \end{aligned}$$

Because of $h_{p} \le \frac{n(n-p-2)}{n+1}$, we obtain

$$\begin{aligned} max(0,\frac{n(n-p-2)}{n+1}-\frac{n^{2}(n-2)(n-p-2)}{2(n+1)^{2}(n-p-4)})\le h_{p-1}\le \frac{n(n-p-2)}{n+1}. \end{aligned}$$

1.5.3 Step p

From the first condition in $(*)$, we know

$$\begin{aligned}&\frac{h_{1}}{n-3}+\cdots +\frac{h_{p}}{(n-p-1)(n-p-2)}-\frac{(h_{m})^{2}(n-2)}{2(n-p-1)(n-p-2)(n-p-4)}\\&\quad \ge \frac{n}{n+1}-\frac{n^{2}(n-2)(n-p-2)}{2(n+1)^{2}(n-p-1)(n-p-4)}. \end{aligned}$$

Using the upper limit of $h_{i}$ $(i=2,\ldots ,p)$, we get

$$\begin{aligned} max(0,\frac{n(n-p-2)}{n+1}-\frac{n^{2}(n-2)(n-p-2)(n-3)}{2(n+1)^{2}(n-p-1)(n-p-4)})\le h_{1}\le \frac{n(n-p-2)}{n+1}. \end{aligned}$$

The proof can be completed by repeating the above manipulation several times. $\square$

1.6 Proof of Corollary 1

Proof

Using $h_{1}=\cdots =h_{k}>h_{k+1}=\cdots =h_{p}$ where $1<k<p$, the sufficient condition in Theorem 1 can be easily simplified and the above result is obtained after some computations. $\square$