Dynamic risk-sensitive fuzzy asset management with coherent risk measures derived from decision maker’s utility


A dynamic risk-sensitive portfolio optimization problem under risk constraints is discussed by the use of coherent risk measures and fuzzy random variables. Risk-sensitive expected rewards under utility functions are approximated by weighted average value-at-risks, and risk constraints are described by coherent risk measures. The coherent risk measures are represented as weighted average value-at-risks with the best risk spectrum derived from decision maker’s risk-averse utility, and the risk spectrum can inherit the risk-averse property of the decision maker’s utility as weighting. By perception-based extension, the risk-sensitive estimation and coherent risk measures are applied to fuzzy random variables. To find feasible regions of risk constraints, a one-step risk-minimizing problem is investigated by mathematical programming. Next, dynamic risk-sensitive total reward maximization under the feasible coherent risk constraints is discussed by dynamic programming. A few numerical examples are investigated to explain the significance of the obtained results. The proposed method is extremely effective for high-speed portfolio trading.

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The proofs of Lemmas 34 and 11, Theorems 3 and 4 and Corollary 1 in the previous sections are given in this appendix.

Proof of Lemma 3

Let \(\lambda\) be a risk spectrum, and let a random variable \(Z \in {\mathcal{X}}\) satisfy \(E(Z)=0\) and \(\sigma (Z)=1\). From (2) and (9), \(\kappa (p) = {\text {VaR}}_p(Z) = F_Z^{-1}(p)\) is a strictly increasing function of \(p (\in (0,1))\). Therefore \(\kappa (p) > \kappa (q)\) for \(q \in (0,p)\), and we get

$$\begin{aligned} (\kappa ^{\lambda }(p))' \! = \! \frac{\kappa (p) \lambda (p) \int _{0}^{p} \lambda (q) \, {\text {d}}q - \lambda (p) \int _{0}^{p} \kappa (q) \lambda (q) \, {\text {d}}q}{\left( \int _{0}^{p} \lambda (q) \, {\text {d}}q \right) ^2} > 0, \end{aligned}$$

for \(p \in (0,1)\). Thus (i) holds. We can easily check (iii) from Proposition 1(ii) in Yoshida (2018). We also have \({\hat{\kappa }}(1) = \int _{0}^{1} \kappa (q) \, {\text {d}}q = \int _{0}^{1} F_Z^{-1}(q) \, {\text {d}}q = E(Z) = 0.\) Finally we obtain (ii) together with (iii). \(\square\)

Sketch proof of Lemma 4

Let \(p \in (0,1)\). From (9), for \(X \in {\mathcal{X}}\), we put \({\text {VaR}}_p(X) = \mu + \kappa (p) \cdot \sigma\) with the mean \(\mu = E(X)\) and the standard deviation \(\sigma = \sigma (X)\). To discuss the minimization (15), we define Eq. (15) by \(G(\lambda )\) for risk spectra \(\lambda\). Hence let \({\underline{\lambda }}\) be a risk spectrum minimizing (15). Then \((1-s) {\underline{\lambda }} + s \varepsilon\) is also a risk spectrum for \(s \in (0,1)\) and risk spectra \(\varepsilon\). Hence we have

$$\begin{aligned} \lim _{s \downarrow 0} \frac{G((1-s){\underline{\lambda }}+s \varepsilon )-G({\underline{\lambda }})}{s} = 0 \end{aligned}$$

for any risk spectrum \(\varepsilon\). This equation implies

$$\begin{aligned} & \sum _{X \in {\mathcal{X}}} \sigma \left( f^{-1}\left( \frac{1}{p} \int \nolimits_{0}^{p} f(\mu + \kappa (q) \sigma ) \, {\text {d}}q \right) \right. \\ & \quad \left. - \int \nolimits_{0}^{p} (\mu + \kappa (q) \sigma ) \, {\underline{\lambda }}(q) \, {\text {d}}q \big / \int \nolimits_{0}^{p} {\underline{\lambda }}(q) \, {\text {d}}q \right) = 0. \end{aligned}$$

Thus we have

$$\begin{aligned}&\sum _{X \in {\mathcal{X}}} \sigma \left( f^{-1}\left( \frac{1}{p} \int \nolimits_{0}^{p} f(\mu + \kappa (q) \sigma ) \, {\text {d}}q \right) \int \nolimits_{0}^{p} {\underline{\lambda }}(q) \, {\text {d}}q \right. \\&\quad \left. - \int \nolimits_{0}^{p} (\mu + \kappa (q) \sigma ) \, {\underline{\lambda }}(q) \, {\text {d}}q \right) =0, \end{aligned}$$

for all \(p \in (0,1)\). Differentiating this equation with respect to p, we get

$$\begin{aligned} \frac{{\underline{\lambda }}(p)}{\int _{0}^{p} {\underline{\lambda }}(q) \, {\text {d}}q} = C(p), \end{aligned}$$

for all \(p \in (0,1)\), where C is defined by (17). Thus we obtain (16) from this equation, and (16) is also a unique solution of the minimization problem. \(\square\)

Proof of Theorem 3

Let \(t=1,2,\ldots ,T\). From Lemmas 9 and 10 we note an equivalence:

$$\begin{aligned}&\exists (w_t,\gamma _t) \in {\mathcal{W}}_t(\delta) \times {\mathbf{R}}\ \text{ such } \text{ that } \ E(E^{\theta }({\tilde{R}}_{t})) = \gamma _t \\&\quad \iff \exists (w_t,\gamma _t) \in {\mathcal{W}} \times {\mathbf{R}}\ \text{ such} \text{ that} \ E(E^{\theta }({\tilde{R}}_{t})) = \gamma _t \\&\qquad \text{ and } \ \rho (E^{\theta }({\tilde{R}}_{t})) \le \delta \\&\quad \iff \exists (w_t,\gamma _t) \in \overline{{\mathcal{W}}}_t(\gamma _t) \times {\mathbf{R}}\ \text{ such } \text{ that } \\ & \qquad - \gamma _t - \kappa ^{\nu }(p) \sqrt{\frac{A_t \gamma _t^2 - 2 B_t \gamma _t +C_t}{\varDelta _t}} \le \delta \\ & \quad \iff \exists (w_t,\gamma _t) \in \overline{{\mathcal{W}}}_t(\gamma _t) \times {\mathbf{R}}\ \text{ such } \text{ that } \ \gamma _t \in [\gamma _t^-, \gamma _t^+]. \end{aligned}$$

In a similar manner to Lemma 9, when \(E(E^{\theta }({\tilde{R}}_{t})) = \gamma _t\), we also have

$$\begin{aligned}&\sup _{w_t \in \overline{{\mathcal{W}}}_t(\gamma _t)} {\text {AVaR}}_1^{\lambda }(E^{\theta }({\tilde{R}}_{t})) \\&\quad = \gamma _t + \kappa ^{\lambda }(1) \sqrt{\frac{A_{t} \gamma _t^2 - 2 B_{t} \gamma _t +C_{t}}{\varDelta _{t}}}. \end{aligned}$$

Using these equalities, we obtain (57) from (37). In the same way, we also have (58) from (38). \(\square\)

Proof of Lemma 11

We check (59) by backward induction on \(t \, (=1,2,\ldots ,T)\). Let \(t=1,2,\ldots ,T-1\), and we assume \(v_{t+1} \ge 1 - \delta \ge 0\) holds. We note \(\beta \, v_{t+1} \ge 0\) since \(\beta > 0\). Together with the assumption \(\kappa ^{\nu }(p)\le \kappa ^{\lambda }(1) \le 0\), we have

$$\begin{aligned} ( 1 + \beta \, v_{t+1}) \kappa ^{\nu }(p) \le \kappa ^{\nu }(p) \le \kappa ^{\lambda }(1) \le 0. \end{aligned}$$

We also have

$$\begin{aligned} \gamma _t + \kappa ^{\nu }(p) \sqrt{\frac{A_t \gamma _t^2 - 2 B_t \gamma _t +C_t}{\varDelta _t}} \ge - \delta \end{aligned}$$

from risk constraint (55). From (57), (81), (82), Lemma 10 and the assumption \(v_{t+1} \ge 1 - \delta \ge 0\), we obtain

$$\begin{aligned} v_t\ge & {} ( 1 + \gamma _t )( 1 + \beta \, v_{t+1}) + \kappa ^{\lambda }(1) \sqrt{\frac{A_t \gamma _t^2 - 2 B_t \gamma _t +C_t}{\varDelta _t}} \\\ge & {} ( 1 + \beta \, v_{t+1}) \\&\cdot \left( ( 1 + \gamma _t ) + \kappa ^{\nu }(p) \sqrt{\frac{A_t \gamma _t^2 - 2 B_t \gamma _t +C_t}{\varDelta _t}} \right) \\\ge & {} (1 + \beta \, (1 - \delta ) ) (1 - \delta ) \\\ge & {} 1 - \delta , \end{aligned}$$

for \(\gamma _t \in [\gamma _t^-, \gamma _t^+]\), and we get \(v_t \ge 1 - \delta \ge 0\). On the other hand, from (55), (58), Lemma 10 and the assumption \(\kappa ^{\nu }(p) \le \kappa ^{\lambda }(1) \le 0\), in the same way we also have \(v_T \ge 1 - \delta \ge 0\) for \(\gamma _T \in [\gamma _T^-, \gamma _T^+]\), and (59) holds for \(t=T\). Thus by backward induction on t we get (59) for all \(t = 1,2,\ldots ,T\). \(\square\)

Proof of Theorem 4

Fix any \(t=1,2,\ldots ,T\). From (57), we let a function

$$\begin{aligned} \varPhi (\gamma _t) = ( 1 + \gamma _t )( 1 + \beta \, v_{t+1}) + \kappa ^{\lambda }(1) \sqrt{\frac{A_t \gamma _t^2 - 2 B_t \gamma _t +C_t}{\varDelta _t}}, \end{aligned}$$

for \(\gamma _t \in {\mathbf{R}}\). In a case of \(\kappa ^{\lambda }(1) \ge - \sqrt{\varDelta _t/A_t}( 1 + \beta v_{t+1} )\), we have \(A_t \kappa {^\lambda }(1)^2 - \varDelta _{t} (1 + \beta v_{t+1})^2 \le 0\). Using this inequality, we can easily check \(\varPhi '(\gamma _t)> 0\). Thus \(\varPhi\) is strictly increasing on \([\gamma _t^-, \gamma _t^+]\) and its has a maximum at \(\gamma _t^+\). In the rest of this proof, we investigate a case of \(\kappa ^{\lambda }(1) < - \sqrt{\varDelta _t/A_t}( 1 + \beta v_{t+1} )\). Then we have \(A_t \kappa {^\lambda }(1)^2 - \varDelta _{t} (1 + \beta v_{t+1})^2 > 0\). Hence \(\varPhi\) is concave since \(\varPhi ''(\gamma _t) = \frac{\kappa ^{\lambda }(1)}{\varDelta _t} \left( \frac{A_t \gamma _t^2 - 2 B_t \gamma _t +C_t}{\varDelta _t} \right) ^{-3/2} \le 0,\) and \(\varPhi\) has a maximum at \({\hat{\gamma }}_t\) given by

$$\begin{aligned} {\hat{\gamma }}_t = \frac{B_t}{A_t} + \frac{ \varDelta _{t} (1 + \beta v_{t+1}) }{A_t \sqrt{ A_t \kappa {^\lambda }(1)^2 - \varDelta _{t} (1 + \beta v_{t+1})^2} }, \end{aligned}$$

which is a solution of \(\varPhi '({\hat{\gamma }}_t ) = 0\). Hence (57) implies the maximum value of concave function \(\varPhi (\gamma _t)\) on closed interval \([\gamma _t^-, \gamma _t^+]\), and then it follows

$$\begin{aligned} \max _{\gamma _t \in [\gamma _t^-, \gamma _t^+]} \varPhi (\gamma _t) = \left\{ \begin{array}{ll} \varPhi ({\hat{\gamma }}_t) &{} \text{ if } \ \gamma _t^- \le {\hat{\gamma }}_t \le \gamma _t^+ \\ \varPhi (\gamma _t^+) &{} \text{ if } \ {\hat{\gamma }}_t > \gamma _t^+ \\ \varPhi (\gamma _t^-) &{} \text{ if } \ {\hat{\gamma }}_t < \gamma _t^-. \end{array} \right. \end{aligned}$$

However in this case \(\gamma _t^- \le {\hat{\gamma }}_t\) holds actually. To prove this inequality, we put \(\varPsi (\delta ) = {\hat{\gamma }}_t - \gamma _t^-\) and we check \(\varPsi (\delta ) \ge 0\) for \(\delta \in [{\underline{\delta }}_t(p), 1]\). Firstly we prove \(\varPsi ({\underline{\delta }}_t(p)) \ge 0\). From (56) and (83), we have

$$\begin{aligned} \varPsi (\delta )= & {} {\hat{\gamma }}_t - \gamma _t^- \\\ge & {} \frac{B_t}{A_t} + \frac{ \varDelta _{t} (1 + \beta v_{t+1}) }{A_t \sqrt{ A_t \kappa {^\lambda }(1)^2 - \varDelta _{t} (1 + \beta v_{t+1})^2} } \\&- \dfrac{B_t \kappa ^{\nu }(p)^2 + \varDelta _t \delta }{A_t \kappa ^{\nu }(p)^2 - \varDelta _t} \\= & {} \frac{ \varDelta _{t} (1 + \beta v_{t+1}) }{A_t \sqrt{ A_t \kappa {^\lambda }(1)^2 - \varDelta _{t} (1 + \beta v_{t+1})^2} } \\&- \frac{ \varDelta _t (A_t \delta + B_t)}{A_t (A_t \kappa ^{\nu }(p)^2 - \varDelta _t)}. \end{aligned}$$

Taking \(\delta = {\underline{\delta }}_t(p)\) in this equality, from (54) it follows

$$\begin{aligned} \varPsi ({\underline{\delta }}_t(p))\ge & {} {\displaystyle \frac{ \varDelta _{t} (1 + \beta v_{t+1}) }{A_t \sqrt{ A_t \kappa {^\lambda }(1)^2 - \varDelta _{t} (1 + \beta v_{t+1})^2} } } \nonumber \\&- {\displaystyle \frac{ \varDelta _t}{A_t \sqrt{A_t \kappa ^{\nu }(p)^2 - \varDelta _t}}. } \end{aligned}$$

While we also have

$$\begin{aligned} \kappa ^{\nu }(p)^2 ( 1 + \beta \, v_{t+1})^2 \ge \kappa ^{\lambda }(1)^2 \end{aligned}$$

from \(\kappa ^{\nu }(p) ( 1 + \beta \, v_{t+1}) \le \kappa ^{\nu }(p) \le \kappa ^{\lambda }(1) \le 0\) and \(1 + \beta \, v_{t+1} > 0\), and this follows

$$\begin{aligned} ( A_t \kappa ^{\nu }(p)^2 - \varDelta _{t} ) (1 + \beta v_{t+1})^2 \ge A_t \kappa ^{\lambda }(1)^2 - \varDelta _{t} (1 + \beta v_{t+1})^2. \end{aligned}$$

From (85) and this inequality, we obtain \(\varPsi ({\underline{\delta }}_t(p)) \ge 0\). Next we prove the function \(\varPsi\) is increasing on \([{\underline{\delta }}_t(p), 1]\). Hence we have \(A_t > 0\), \(\varDelta _t = A_t C_t - B_t^2 > 0\) and \(A_t \kappa ^{\nu }(p)^2 - \varDelta _t > 0\) from \(\kappa ^{\nu }(p) < - \sqrt{\varDelta _t/A_t}\). Therefore we get

$$\begin{aligned}&A_t ( \kappa ^{\nu }(p)^2 (A_t \delta + B_t)^2 \nonumber \\&\qquad - \varDelta _t (A_t \delta ^2 + 2 B_t \delta + C_t - \kappa ^{\nu }(p)^2) )\nonumber \\&\quad = ( A_t \kappa ^{\nu }(p)^2 - \varDelta ) ( (A_t \delta + B_t)^2 + \varDelta _t ) > 0, \end{aligned}$$

and \(A_t \delta ^2 + 2 B_t \delta + C_t - \kappa ^{\nu }(p)^2 > 0\). Thus (86) implies

$$\begin{aligned} \kappa ^{\nu }(p)^2 \frac{(A_t \delta + B_t)^2}{A_t \delta ^2 + 2 B_t \delta + C_t - \kappa ^{\nu }(p)^2} > \varDelta _t. \end{aligned}$$

Hence we have \(\kappa ^{\nu }(p) < 0\) and \(A_t \delta + B_t \ge A_t {\underline{\delta }}_t(p) + B_t > 0\) from (54). From these inequalities, Eq. (87) follows

$$\begin{aligned} \sqrt{\varDelta _t} + \kappa ^{\nu }(p) \frac{A_t \delta + B_t}{\sqrt{A_t \delta ^2 + 2 B_t \delta + C_t - \kappa ^{\nu }(p)^2}} < 0. \end{aligned}$$

From (56), (83) and (88), we can easily obtain

$$\begin{aligned} \varPsi '(\delta )= & {} \frac{d}{d \delta } ({\hat{\gamma }}_t - \gamma _t^- ) \\= & {} \dfrac{- \sqrt{\varDelta _t} }{A_t \kappa ^{\nu }(p)^2 - \varDelta _t} \\&\cdot \left( \sqrt{\varDelta _t} + \kappa ^{\nu }(p) \frac{A_t \delta + B_t}{\sqrt{A_t \delta ^2 + 2 B_t \delta + C_t - \kappa ^{\nu }(p)^2}} \right) \\> & {} 0. \end{aligned}$$

Thus \(\varPsi : [{\underline{\delta }}_t(p),1] \mapsto {\mathbf{R}}\) is increasing and \(\varPsi ({\underline{\delta }}_t(p)) \ge 0\). Therefore \(\varPsi (\delta ) \ge 0\) holds for all \(\delta \in [{\underline{\delta }}_t(p),1]\), and we get \({\hat{\gamma }}_t - \gamma _t^- \ge 0\). Thus (84) is reduce to

$$\begin{aligned} \max _{\gamma _t \in [\gamma _t^-, \gamma _t^+]} \varPhi (\gamma _t) = \left\{ \begin{array}{ll} \varPhi ({\hat{\gamma }}_t) &{} \text{ if } \ {\hat{\gamma }}_t \le \gamma _t^+ \\ \varPhi (\gamma _t^+) &{} \text{ if } \ {\hat{\gamma }}_t > \gamma _t^+. \end{array} \right. \end{aligned}$$

Hence we can easily check

$$\begin{aligned} \varPhi ({\hat{\gamma }}_t)= & {} \frac{(A_t + B_t) ( 1 + \beta v_{t+1} ) }{A_t } \\&- \frac{\sqrt{ A_t \kappa ^{\lambda }(1)^2 - \varDelta _{t} (1 + \beta v_{t+1})^2} }{A_t } \end{aligned}$$


$$\begin{aligned} \varPhi (\gamma _t^+)= & {} (1 + \gamma _t^+ ) (1+ \beta v_{t+1}) \\&+ \kappa ^{\lambda }(1) \sqrt{\dfrac{A_t (\gamma _t^+)^2 - 2 B_t \gamma _t^+ +C_t}{\varDelta _t}}. \end{aligned}$$

These results complete the proof of Theorem 4. \(\square\)

Proof of Corollary 1

Theorem 4 can be represented as Corollary 1(i), which comes from an equivalence between \(\gamma _t^* \le \gamma _t^+\) in (60) and \(\delta _t^+ \le \delta\) in (62). Here we also have

$$\begin{aligned} \varPhi (\gamma _t^+)= \; & (1 + \gamma _t^+ ) (1+ \beta v_{t+1}) \\&+ \kappa ^{\lambda }(1) \sqrt{\dfrac{A_t (\gamma _t^+)^2 - 2 B_t \gamma _t^+ +C_t}{\varDelta _t}} \\= \; & (1 + \gamma _t^+ ) ( 1 + \beta v_{t+1}) - \dfrac{\kappa ^{\lambda }(1)}{\kappa ^{\nu }(p)} ( \delta + \gamma _t^+ ), \end{aligned}$$

which can be checked easily. This implies (i) holds. (ii) and (iii) can be checked easily in a similar way to Theorem 2. \(\square\)

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Yoshida, Y. Dynamic risk-sensitive fuzzy asset management with coherent risk measures derived from decision maker’s utility. Granul. Comput. 6, 19–35 (2021). https://doi.org/10.1007/s41066-019-00196-0

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  • Risk-sensitive expectation
  • Risk constraint
  • Coherent risk measure
  • Weighted average value-at-risk
  • Risk averse utility
  • Fuzzy random variable
  • Perception-based extension