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Aumann–Serrano index of risk in portfolio optimization

Abstract

The paper is devoted to study the portfolio optimization problem for an investor who aims to minimize the exposure to equity markets measured by the Aumann–Serrano index of riskiness. The ARMA–GARCH model with normal variance–mean mixture innovations is employed to capture the stylized facts of stock returns. Using a two-step scheme, we convert the high-dimensional optimization problem into a two-dimensional one. We further prove that the dimension reduction technique preserves the convexity of the problem as long as the risk measure is convex and monotonic. In the empirical study, we observe that the optimal portfolio outperforms benchmarks based on a 10-year backtesting window covering the financial crisis.

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Fig. 1
Fig. 2

Notes

  1. 1.

    It is more precise to say GH and NTS distributions are semi-heavy-tailed because both have exponentially decaying tails.

  2. 2.

    The lower bound of \(\mu \) is derived by setting the expectation, or the derivative of MGF (13) at \(t=0\), to be zero. The upper bound is derived by setting \(M_X(l)\ge 1\) in (15).

  3. 3.

    Typically, the first step of QMLE estimation of ARMA–GARCH parameters is performed assuming normal innovations. The QMLE with Student’s t innovations is proved to be practically useful in Kim et al. (2010), Anand et al. (2016) and Goode et al. (2015).

  4. 4.

    Based on the definition formula (1), we have \(cov({\varvec{X}}) = \mathbb {E}(T){\varvec{\varSigma }} + var(T){\varvec{\beta }}{\varvec{\beta }}'\).

  5. 5.

    The results in this empirical analysis are mainly for illustration purpose. To address practitioners’ concerns, one should incorporate transaction cost in the optimization. As we rebalance the portfolio on a daily basis, the impact of transaction cost is not negligible.

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Acknowledgements

We thank Abhinav Anand, Tetsuo Kurosaki and Xiang Shi for valuable discussions over the course of writing this paper. We thank two anonymous reviewers whose comments and suggestions helped improve and clarify the manuscript. The work was supported by National Natural Science Foundation of China (Grant No. 72071132, 72173089, 71773079) and Shenzhen University (Grant No. SZJR003).

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Correspondence to Fumin Zhu.

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Appendices

Appendices

A Proof of Proposition 1

Proof

To solve problem (28), we need to prove the following first

$$\begin{aligned} \underset{{\varvec{w}}}{\text {argmin }} R({\tilde{\mu }}+{\tilde{\beta }}T+\sqrt{T}\sigma ^{(p)}Z) = \underset{{\varvec{w}}}{\text {argmin }}\sigma ^{(p)}. \end{aligned}$$
(42)

Let \(R(\sigma ):=R({\tilde{\mu }} + {\tilde{\beta }} T + \sigma \sqrt{T}Z)\). Notice that \(\sigma Z\) and \(-\sigma Z\) have exactly the same distribution, where Z is a univariate standard Gaussian random variable. The domain of \(R(\sigma )\) can be extended from \(\mathbb {R}^+\) to \(\mathbb {R}\), and it is trivially true that \(R(\sigma )\) is symmetric to \(\sigma =0\).

Since \(R(\cdot )\) is a convex risk measure, for any \(0< \lambda <1\) and \(\sigma _1, \sigma _2\in \mathbb {R}\) we have

$$\begin{aligned}&R\left( \lambda \sigma _1 + (1-\lambda )\sigma _2\right) \\&\quad = R\left( {\tilde{\mu }} + {\tilde{\beta }} T + [\lambda \sigma _1 + (1-\lambda )\sigma _2]\sqrt{T}Z\right) \\&\quad = R\left( \lambda [{\tilde{\mu }} + {\tilde{\beta }} T + \sigma _1\sqrt{T}Z] + (1-\lambda )[{\tilde{\mu }} + {\tilde{\beta }} T + \sigma _2\sqrt{T}Z] \right) \\&\quad \le \lambda R({\tilde{\mu }} + {\tilde{\beta }} T + \sigma _1\sqrt{T}Z) + (1-\lambda )R({\tilde{\mu }} + {\tilde{\beta }} T + \sigma _2\sqrt{T}Z) \\&\quad = \lambda R(\sigma _1) + (1-\lambda )R(\sigma _2). \end{aligned}$$

Therefore, \(R(\sigma )\) is a convex function of \(\sigma \). If there exist \(0<\sigma _1<\sigma _2\), and \(R(\sigma _1)>R(\sigma _2)\), then find \(\lambda \) such that \(\lambda (-\sigma _2) + (1-\lambda )\sigma _2 = \sigma _1\). However,

$$\begin{aligned} \lambda R(-\sigma _2) + (1-\lambda ) R(\sigma _2) = R(\sigma _2) < R(\sigma _1), \end{aligned}$$

which contradicts with the convexity of \(R(\sigma )\). In other words, \(R(\sigma )\) is nondecreasing on \(\mathbb {R}^+\), so (42) holds.

Recall that \(\sigma ^{(p)}:=\sqrt{{\varvec{w}}'{\varvec{\varSigma }} {\varvec{w}}}\), and also by (42), problem (28) is therefore equivalent to the linear-quadratic programming below

$$\begin{aligned} \begin{aligned}&\underset{{\varvec{w}}}{\text {min}}&\frac{1}{2}{\varvec{w}}'{\varvec{\varSigma }} {\varvec{w}} \\&\text {s.t.}&{\varvec{w}}'{\varvec{e}} = 1, \\&{\varvec{w}}'{\varvec{\mu }} = {\tilde{\mu }}, \\&{\varvec{w}}'{\varvec{\beta }} = {\tilde{\beta }}. \end{aligned} \end{aligned}$$
(43)

A simple application of Lagrange multiplier will prove the proposition. \(\square \)

B Proof of Proposition 2

To prove Proposition 2, we need to prove the following lemma first.

Lemma 1

If \({\varvec{A}}\) is positive definite, then \(f({\varvec{x}}) = \sqrt{{\varvec{x}}'{\varvec{A}}{\varvec{x}}}\) is convex on \(\mathbb {R}^d\).

Proof

Define an inner product space \(\langle \cdot ,\cdot \rangle : \mathbb {R}^{d} \times \mathbb {R}^d \rightarrow \mathbb {R}\) such that \(\langle {\varvec{x}},{\varvec{y}} \rangle = {\varvec{x}}'{\varvec{A}}{\varvec{y}}\), for any \( {\varvec{x}}, {\varvec{y}} \in \mathbb {R}^d\). This inner product space is well-defined since for any \({\varvec{x}},{\varvec{y}},{\varvec{z}} \in \mathbb {R}^d\),

  1. 1.

    \(\langle {\varvec{x}},{\varvec{y}} \rangle = {\varvec{x}}'{\varvec{A}}{\varvec{y}}=({\varvec{x}}'{\varvec{A}}{\varvec{y}})'={\varvec{y}}'{\varvec{A}}{\varvec{x}}=\langle {\varvec{y}},{\varvec{x}} \rangle \)

  2. 2.

    \(\langle a{\varvec{x}},{\varvec{y}} \rangle =a{\varvec{x}}'{\varvec{A}}{\varvec{y}}=a\langle {\varvec{x}},{\varvec{y}} \rangle \), \(\langle {\varvec{x}}+{\varvec{y}},{\varvec{z}} \rangle =({\varvec{x}}+{\varvec{y}})'{\varvec{A}}{\varvec{z}}={\varvec{x}}'{\varvec{A}}{\varvec{z}}+{\varvec{y}}'{\varvec{A}}{\varvec{z}}=\langle {\varvec{x}},{\varvec{z}} \rangle +\langle {\varvec{y}},{\varvec{z}} \rangle \)

  3. 3.

    \(\langle {\varvec{x}},{\varvec{x}} \rangle = {\varvec{x}}'{\varvec{A}}{\varvec{x}}>0, \ \forall \ {\varvec{x}} \in \mathbb {R}^d \setminus \{{\varvec{0}}\}\)

A simple inequality follows from the Cauchy–Schwarz inequality:

$$\begin{aligned} \sqrt{\langle \lambda {\varvec{x}} + (1-\lambda ){\varvec{y}}, \lambda {\varvec{x}} + (1-\lambda ){\varvec{y}} \rangle } \le \lambda \sqrt{\langle {\varvec{x}}, {\varvec{x}} \rangle } + (1-\lambda )\sqrt{\langle {\varvec{y}}, {\varvec{y}} \rangle }, \ \forall \lambda \in [0,1], \end{aligned}$$

which proves the convexity. \(\square \)

Now we can prove Proposition 2.

Proof

Let \(h({\tilde{\mu }},{\tilde{\beta }}, {\tilde{\gamma }}) = ({\tilde{\mu }}\ {\tilde{\beta }}\ {\tilde{\gamma }}){\varvec{\varOmega }}^{-1}({\tilde{\mu }}\ {\tilde{\beta }}\ {\tilde{\gamma }})'\) where \({\varvec{\varOmega }} = ({\varvec{\mu }}\ {\varvec{\beta }}\ {\varvec{e}})'{\varvec{\varSigma }}^{-1}({\varvec{\mu }}\ {\varvec{\beta }}\ {\varvec{e}})\), and \({\varvec{\varSigma }}\) and \({\varvec{\varSigma }}^{-1}\) are positive definite. By Lemma 1, for any \(0\le \lambda \le 1\),

$$\begin{aligned}&\sqrt{h\left( \lambda {\tilde{\mu }}_1 + (1-\lambda ) {\tilde{\mu }}_2,\lambda {\tilde{\beta }}_1 + (1-\lambda ) {\tilde{\beta }}_2,\lambda {\tilde{\gamma }}_1 + (1-\lambda ) {\tilde{\gamma }}_2 \right) } \\&\quad \le \ \lambda \sqrt{h({\tilde{\mu }}_1,{\tilde{\beta }}_1, {\tilde{\gamma }}_1 )} + (1-\lambda )\sqrt{h({\tilde{\mu }}_2,{\tilde{\beta }}_2, {\tilde{\gamma }}_2 )}. \end{aligned}$$

Since \(h({\tilde{\mu }},{\tilde{\beta }}, 1) = g({\tilde{\mu }},{\tilde{\beta }})\), we obtain that

$$\begin{aligned}&\sqrt{g\left( \lambda {\tilde{\mu }}_1 + (1-\lambda ) {\tilde{\mu }}_2,\lambda {\tilde{\beta }}_1 + (1-\lambda ) {\tilde{\beta }}_2 \right) } \\&\quad \le \ \lambda \sqrt{g({\tilde{\mu }}_1,{\tilde{\beta }}_1 )} + (1-\lambda )\sqrt{g({\tilde{\mu }}_2,{\tilde{\beta }}_2 )}. \end{aligned}$$

Therefore, \(\sqrt{g({\tilde{\mu }},{\tilde{\beta }})}\) is convex on \(\mathbb {R}^2\).

We denote \(R({\tilde{\mu }},{\tilde{\beta }}) = R\left( {\tilde{\mu }}+{\tilde{\beta }}T+\sqrt{g({\tilde{\mu }},{\tilde{\beta }})}\sqrt{T}Z\right) \), for any \(0\le \lambda \le 1\),

$$\begin{aligned}&\ \lambda R({\tilde{\mu }}_1,{\tilde{\beta }}_1)+(1-\lambda )R({\tilde{\mu }}_2,{\tilde{\beta }}_2) \nonumber \\&\quad = \ \lambda R\left( {\tilde{\mu }}_1+{\tilde{\beta }}_1T+\sqrt{g({\tilde{\mu }}_1,{\tilde{\beta }}_1)}\sqrt{T}Z\right) +(1-\lambda ) R\left( {\tilde{\mu }}_2+{\tilde{\beta }}_2T+\sqrt{g({\tilde{\mu }}_2,{\tilde{\beta }}_2)}\sqrt{T}Z\right) \nonumber \\&\quad \ge \ R\left( {\bar{\mu }}+{\bar{\beta }}T+\left[ \lambda \sqrt{g({\tilde{\mu }}_1,{\tilde{\beta }}_1)}+(1-\lambda )\sqrt{g({\tilde{\mu }}_2,{\tilde{\beta }}_2)}\right] \sqrt{T}Z\right) , \end{aligned}$$
(44)

where \({\bar{\mu }} = \lambda {\tilde{\mu }}_1+(1-\lambda ) {\tilde{\mu }}_2\) and \({\bar{\beta }} = \lambda {\tilde{\beta }}_1+(1-\lambda ) {\tilde{\beta }}_2\), and the last inequality holds by the convexity of \(R(\cdot )\).

If we further denote \(\alpha =\lambda \sqrt{g({\tilde{\mu }}_1,{\tilde{\beta }}_1)}+(1-\lambda )\sqrt{g({\tilde{\mu }}_2,{\tilde{\beta }}_2)}\) and \(\gamma = \sqrt{g\left( {\bar{\mu }},{\bar{\beta }}\right) }\), then \(\alpha \ge \gamma \ge 0\) by the convexity of \(\sqrt{g(\cdot , \cdot )}\). Consider two normal variance–mean mixtures, \(X_{\alpha } = {\bar{\mu }}+{\bar{\beta }}T+\alpha \sqrt{T}Z\) and \(X_{\gamma } = {\bar{\mu }}+{\bar{\beta }}T+\gamma \sqrt{T}Z\). \(F_{X_{\alpha }}\) and \(F_{X_{\gamma }}\) are CDFs of \(X_{\alpha }\) and \(X_{\gamma }\) respectively. For any \(l \in \mathbb {R}\), we have

$$\begin{aligned}&\int _{-\infty }^l \left( F_{X_{\alpha }}(k)-F_{X_{\gamma }}(k)\right) \mathrm {d}k \\&\quad = \int _{-\infty }^l \left( \mathbb {P}({\bar{\mu }}+{\bar{\beta }}T+\alpha \sqrt{T}Z \le k)-\mathbb {P}({\bar{\mu }}+{\bar{\beta }}T+\gamma \sqrt{T}Z \le k) \right) \mathrm {d}k \\&\quad = \int _{-\infty }^l \mathbb {E}\left[ \mathbb {P}({\bar{\mu }}+{\bar{\beta }}T+\alpha \sqrt{T}Z \le k)-\mathbb {P}({\bar{\mu }}+{\bar{\beta }}T+\gamma \sqrt{T}Z \le k) | T\right] \mathrm {d}k\\&\quad = \int _{-\infty }^l \left( \int _0^{\infty } \left[ \mathbb {P}({\bar{\mu }}+{\bar{\beta }}T+\alpha \sqrt{t}Z \le k)-\mathbb {P}({\bar{\mu }}+{\bar{\beta }}T+\gamma \sqrt{t}Z \le k)\right] f_T(t)\mathrm {d}t \right) \mathrm {d}k\\&\quad = \int _0^{\infty } \left( \int _{-\infty }^l \left[ \mathbb {P}({\bar{\mu }}+{\bar{\beta }}t+\alpha \sqrt{t}Z \le k)-\mathbb {P}({\bar{\mu }}+{\bar{\beta }}t+\gamma \sqrt{t}Z \le k)\right] \mathrm {d}k \right) f_T(t)\mathrm {d}t\\&\quad = \int _0^{\infty } \left( \int _{-\infty }^l \left[ \mathbb {P} \left( \alpha Z \le \frac{k-{\bar{\mu }}-{\bar{\beta }}t}{\sqrt{t}}\right) -\mathbb {P} \left( \gamma Z \le \frac{k-{\bar{\mu }}-{\bar{\beta }}t}{\sqrt{t}} \right) \right] \mathrm {d}k \right) f_T(t)\mathrm {d}t\\&\quad = \int _0^{\infty } \left( \int _{-\infty }^{\frac{l-{\bar{\mu }}-{\bar{\beta }}t}{\sqrt{t}}} \left[ P \left( \alpha Z \le m \right) -P \left( \gamma Z \le m \right) \right] \mathrm {d}m \right) \sqrt{t} f_T(t)\mathrm {d}t\\&\quad \ge 0, \end{aligned}$$

where the interchangeability of integrals is guaranteed by Fubini’s Theorem. By definition, this means \(X_{\gamma }\) second-order stochastically dominates \(X_{\alpha }\). Therefore \(R(X_{\gamma }) \le R(X_{\alpha })\) by the monotonicity of \(R(\cdot )\), which in combination with (44) proves the convexity of \(R({\tilde{\mu }},{\tilde{\beta }})\) on \(\mathbb {R}^2\). \(\square \)

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Li, T., Kim, Y.S., Fan, Q. et al. Aumann–Serrano index of risk in portfolio optimization. Math Meth Oper Res 94, 197–217 (2021). https://doi.org/10.1007/s00186-021-00753-x

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Keywords

  • Aumann–Serrano index of riskiness
  • Portfolio optimization
  • Normal variance–mean mixture
  • Convex risk measure
  • Average value-at-risk

Mathematics Subject Classification

  • 46N10
  • 91B28
  • 91B84