One-period pricing strategy of ‘money doctors’ under cumulative prospect theory

Abstract

We focus on the interaction between investors and portfolio managers, employing a cumulative prospect theory approach to the investor’s preferences. In an original way, we model trust in the manager and the relative anxiety about investing in a risky asset. Moreover, we investigate how trust and anxiety affect the manager’s fee and the portfolios of cumulative prospect theory investors. The novelty of our contribution relative to previous work is that we rely on cumulative prospect theory(CPT) rather than the classical mean-variance framework. Moreover, our research differs from traditional CPT work through an improved value function that accurately characterizes the reduction in anxiety suffered by the CPT investors from bearing risk when assisted by the portfolio managers’ help relative to when they lack such assistance. Our results differ in several respects from those obtained when using on classical preferences. First, the optimal fees are not symmetric. Specially, the dominant managers obtain higher fees than subordinate managers regardless of changes in risk of risky assets (a risky asset) and changes in the dispersion of trust in the population. Another difference is that these fees are not proportional to expected returns. In particular, the optimal fees increase nonlinearly as risk of risky assets (a risky asset) increases and the dispersion of trust in the population increases.

Appendix

1.1 The proof of Proposition 22

Proof

Equation 2.7 identifies

$$\begin{array}{@{}rcl@{}} \frac{\sigma^{2}}{a}(f_{A}-f_{B})\in [-\theta, \theta]. \end{array} $$

Otherwise, only one manager earns zero profit. This manager could reduce hisfee and also make a positive profit. This condition alone implies that when 𝜃 = 0, the unique equilibriumfeatures $f_{A}^{*}=f_{B}^{*}=0$. Thus, weonly need to address 𝜃 > 0.

Let

$$x=1-\frac{\sigma^{2}}{a}(f_{A}-f_{B}).$$

We first consider f _A ≥ f _B . From Theorem 21, we propose that

$$\begin{array}{@{}rcl@{}} U_{A}(f_{A},f_{B})=f_{A} (\frac{a}{\sigma^{2}}-f_{A})\hat{k}\frac{x-1+\theta}{2\theta} \end{array} $$

and

$$\begin{array}{@{}rcl@{}} U_{B}(f_{A},f_{B})=f_{B}\hat{k}[\frac{1}{2}(\frac{a}{\sigma^{2}}-f_{B})+\frac{1}{\theta}(1+x-\frac{2\sigma^{2}}{a}f_{B})\frac{a}{\sigma^{2}}(1-x)]. \end{array} $$

Set

$$\begin{array}{@{}rcl@{}} \frac{\partial U_{A}}{\partial f_{A}}=0 \end{array} $$

and

$$\begin{array}{@{}rcl@{}} \frac{\partial U_{B}}{\partial f_{B}}=0 \end{array} $$

We obtain

$$\begin{array}{@{}rcl@{}} f_{A}=\frac{2x-1+2\theta\pm\sqrt{(2x-1+2\theta)^{2}-4(x-1+\theta)}}{2\sigma^{2}/a}. \end{array} $$

(5.1)

and

$$\begin{array}{@{}rcl@{}} f_{B}=\frac{2\theta+2-x\pm\sqrt{(2\theta+2-x)^{2}-4(\theta+\frac{1}{2}(1-x^{2}))}}{2\sigma^{2}/a}. \end{array} $$

(5.2)

Write

$$\begin{array}{@{}rcl@{}} f_{A,1}=\frac{2x-1+2\theta-\sqrt{(2x-1+2\theta)^{2}-4(x-1+\theta)}}{2\sigma^{2}/a}, \end{array} $$

$$\begin{array}{@{}rcl@{}} f_{A,2}=\frac{2x-1+2\theta+\sqrt{(2x-1+2\theta)^{2}-4(x-1+\theta)}}{2\sigma^{2}/a}, \end{array} $$

$$\begin{array}{@{}rcl@{}} f_{B,1}=\frac{2\theta+2-x-\sqrt{(2\theta+2-x)^{2}-4(\theta+\frac{1}{2}(1-x^{2}))}}{2\sigma^{2}/a} \end{array} $$

and

$$\begin{array}{@{}rcl@{}} f_{B,2}=\frac{2\theta+2-x+\sqrt{(2\theta+2-x)^{2}-4(\theta+\frac{1}{2}(1-x^{2}))}}{2\sigma^{2}/a}. \end{array} $$

Because

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2}U_{A}}{\partial {f_{A}^{2}}}(f_{A,1})<0, \end{array} $$

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2}U_{A}}{\partial {f_{A}^{2}}}(f_{A,2})>0, \end{array} $$

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2}U_{B}}{\partial {f_{B}^{2}}}(f_{B,1})<0, \end{array} $$

and

$$\begin{array}{@{}rcl@{}} \frac{\partial^{2}U_{B}}{\partial {f_{B}^{2}}}(f_{B,2})>0, \end{array} $$

f _A,1 and f _B,1 are the values thatlocally maximize U _A and U _B , respectively.

Moreover,

$$U_{A}(f_{A,1})\geq\max \{U_{A}(0),U_{A}(\frac{a}{\sigma^{2}})\}$$

and

$$U_{B}(f_{B,1})\geq\max \{U_{B}(0),U_{B}(\frac{a}{\sigma^{2}})\}.$$

Therefore, f _A,1 and f _B,1 are the values thatlocally maximize U _A and U _B , respectively. Furthermore,

$$U_{A}(f_{A,1})\geq U_{B}(f_{B,1}),$$

; thus, if manager A is in the dominant position in the financial market, he prefers to charge a higher fee f _A than the fee f _B of manager B to obtaina larger total profit U _A than the total profit U _B of manager B. This is the conclusion of Proposition 22. Using helpful software, such as Matlaband Mathmatic, we can obtain the approximately and explicitly optimal solutions of f _A and f _B

$$\begin{array}{@{}rcl@{}} f_{A}^{*}&=&2\theta + \frac{16\theta - 8(2\theta+ 1)^{1/2}+ 20\theta^{2} - 8\theta(2\theta- (2\theta+ 1)^{1/2}+ 2) + 9^{1/2}}{2\sigma^{2}/a}\\ &&+ 2(2\theta+ 1)^{1/2}- 1 \end{array} $$

(5.3)

and

$$\begin{array}{@{}rcl@{}} f_{B}^{*}\!&=&\!\frac{6\sigma^{2}/a+12\sigma^{2}\theta/a+ (16\theta- 8(2\theta+ 1)^{1/2}+ 20\theta^{2} - 8\theta(2\theta - (2\theta + 1)^{1/2}) + 2) + 9)^{1/2}}{2\sigma^{2}/a}\\ &&+\frac{2(2\theta+ 1)^{1/2}- 4\frac{\sigma^{2}}{a}(2\theta- (2\theta+ 1)^{1/2}+ 2) - 2}{2\sigma^{2}/a}. \end{array} $$

(5.4)

We can simplify the results as

$$\begin{array}{@{}rcl@{}} f_{A}^{*}=2\theta + 2(2\theta+ 1)^{1/2}- 1+\frac{a}{\sigma^{2}}\frac{1}{2}(8(\theta-1)(2\theta+ 1)^{1/2}+ 4\theta^{2}+3) \end{array} $$

(5.5)

and

$$\begin{array}{@{}rcl@{}} f_{B}^{*}&=&\frac{1}{2}(4\theta-2+4(2\theta+1)^{1/2}) +\frac{a}{\sigma^{2}}\frac{1}{2}((8\theta-6)(2\theta+1)^{1/2}+4\theta^{2}+1). \end{array} $$

(5.6)

Let

$$\begin{array}{@{}rcl@{}} C=2\theta + 2(2\theta+ 1)^{1/2}- 1, \end{array} $$

(5.7)

$$\begin{array}{@{}rcl@{}} D=\frac{1}{2}(8(\theta-1)(2\theta+ 1)^{1/2}+ 4\theta^{2}+3), \end{array} $$

(5.8)

$$\begin{array}{@{}rcl@{}} E=\frac{1}{2}(4\theta-2+4(2\theta+1)^{1/2}) \end{array} $$

(5.9)

and

$$\begin{array}{@{}rcl@{}} F=\frac{1}{2}((8\theta-6)(2\theta+1)^{1/2}+4\theta^{2}+1). \end{array} $$

(5.10)

Then,

$$\begin{array}{@{}rcl@{}} f_{A}^{*}=C + D\frac{a}{\sigma^{2}} \end{array} $$

(5.11)

and

$$\begin{array}{@{}rcl@{}} f_{B}^{*}&=&E+F\frac{a}{\sigma^{2}}. \end{array} $$

(5.12)

Since 𝜃 ∈ [0, 1], 2(2𝜃 + 1)^1/2 − 1 > 0. So, C > 0.

We have that

$$\begin{array}{@{}rcl@{}} \frac{d E}{d \theta}=\frac{1}{2}(4+4(2\theta+1)^{-1/2})>0. \end{array} $$

(5.13)

Hence, when 𝜃 = 0,we can get the minimum of E is 1. Therefore, we have E > 0(𝜃 ∈ [0, 1]).

Now, we can get Eqs. 5.13 and 5.12 .

Symmetrically, when manager B is in a dominant position and manager A isin a subordinate position in the financial market, we obtain analogous results. □

1.2 The proof of Proposition 23

Proof

Given the definition of the rate of fees, we have

$$\begin{array}{@{}rcl@{}} F_{A}=\frac{f_{A}}{W_{i}} \end{array} $$

(5.14)

and

$$\begin{array}{@{}rcl@{}} F_{B}=\frac{f_{B}}{W_{i}}. \end{array} $$

(5.15)

Because $W_{i}=v_{i}\xi =k(\frac {a\tau _{i,j}}{\sigma ^{2}}-f_{j})\xi$,we easily obtain the following:

$$\begin{array}{@{}rcl@{}} F_{A}^{*}&=&\frac{f_{A}^{*}}{v_{i}\xi}\\ &=&\frac{f_{A}^{*}}{\hat{k}(\frac{a}{\sigma^{2}}-f_{A}^{*})\xi}\\ &=&\frac{f_{A}^{*}-\frac{a}{\sigma^{2}}+\frac{a}{\sigma^{2}}}{\hat{k}(\frac{a}{\sigma^{2}}-f_{A}^{*})\xi}\\ &=&-\frac{1}{\hat{k}\xi}+\frac{\frac{a}{\sigma^{2}}}{\hat{k}(\frac{a}{\sigma^{2}}-f_{A}^{*})\xi}\\ &=&-\frac{1}{\hat{k}\xi}+\frac{\frac{a}{\sigma^{2}}}{\hat{k}(\frac{a}{\sigma^{2}}-(C+D\frac{a}{\sigma^{2}}))\xi}\\ &=&-\frac{1}{\hat{k}\xi}+\frac{1}{\hat{k}(1-C\frac{\sigma^{2}}{a}-D)\xi}. \end{array} $$

Therefore, we have

$$\begin{array}{@{}rcl@{}} \frac{\partial F_{A}^{*}}{\partial \sigma^{2}}=\frac{1}{\hat{k}\xi}\frac{C}{a}(1-\frac{C\sigma^{2}}{a}-D)^{-2}> 0. \end{array} $$

(5.16)

$f_{A}^{*}$is thus an increasingfunction of σ ².

Similarly, we have

$$\begin{array}{@{}rcl@{}} F_{B}^{*}=-\frac{1}{\hat{k}\xi}+\frac{1}{\hat{k}(1-E\frac{\sigma^{2}}{a}-F)\xi} \end{array} $$

and

$$\begin{array}{@{}rcl@{}} \frac{\partial F_{B}^{*}}{\partial \sigma^{2}}=\frac{1}{\hat{k}\xi}\frac{E}{a}(1-\frac{E\sigma^{2}}{a}-F)^{-2}> 0. \end{array} $$

(5.17)

Therefore, $f_{B}^{*}$is also anincreasing function of σ ². □

1.3 The proof of Theorem 26

Proof

We first discuss the total profit of manager A when f _A ≥ f _B .

If $V(\bar {k}, \tau _{i,A}, f_{A})\geq V(\bar {k}, \tau _{i,B}, f_{B})$,investor i prefers manager A to manager B.

From

$$\begin{array}{@{}rcl@{}} (\frac{a\tau_{i,A}}{\hat{\sigma}^{2}}-f_{A})^{\alpha} G(\hat{k})\geq (\frac{a\tau_{i,B}}{\hat{\sigma}^{2}}-f_{B})^{\alpha} G(\hat{k}), \end{array} $$

we can demonstrate that

$$\begin{array}{@{}rcl@{}} \frac{a}{\hat{\sigma}^{2}}(\tau_{i,A}-\tau_{i,B})\geq (f_{A}-f_{B}). \end{array} $$

(5.18)

Note that the right-hand side of Eq. 5.18 is not less than 0. For B-trusting investors, theleft-hand side of (5.18) is less than 0. Thus, Eq. 5.18 does not hold. That is, no B-trusting investorswill choose manager A.

For an A-trusting investor, if

$$\begin{array}{@{}rcl@{}} \tau_{i,B}\leq 1-\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B}), \end{array} $$

an A-trusting investor prefers manager B. Namely, although manager A charges a higher fee thanmanager B, some A-trusting investors have so little trust in manager B that they prefer managerA, regardless of the latter’s higher fee.

Hence, when f _A ≥ f _B , andnoting that

$$V_{i}^{*}=l_{i}^{*}\xi_{M}=\bar{k}(\frac{a}{\hat{\sigma}^{2}}-f_{A})\xi_{M},$$

we canstate that the manager A obtains total profit

$$\begin{array}{@{}rcl@{}} &&f_{A} (\frac{a}{\hat{\sigma}^{2}}-f_{A})\bar{k}\xi_{M}\cdot I{\int}_{\max[1-\theta,\frac{\hat{\sigma}^{2}}{a}f_{B}]}^{\max[1-\theta,\frac{\hat{\sigma}^{2}}{a}f_{B}, 1-\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B})]}\frac{1}{2\theta}d\tau_{i,B}\\ &=&f_{A} (\frac{a}{\hat{\sigma}^{2}}-f_{A})\bar{k}\xi_{M}\cdot I{\int}_{\max[1-\theta,\frac{\hat{\sigma}^{2}}{a}f_{B}]}^{\max[1-\theta, 1-\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B})]}\frac{1}{2\theta}d\tau_{i,B}, \end{array} $$

where the n-dimension I = (1, 1,..., 1)^′.

We can demonstrate that $1-\theta \geq \frac {\hat {\sigma }^{2}}{a}f_{B}.$ Otherwise, we would have some paradoxical results. Specifically, when $1-\theta <\frac {\hat {\sigma }^{2}}{a}f_{B}$, if there exits any $\tau _{i,B}\in [1-\theta, \frac {\hat {\sigma }^{2}}{a}f_{B}]$, this is inconsistent with Eq. 2.6 . If no τ _i,B satisfies $\tau _{i,B}\in [1-\theta, \frac {\hat {\sigma }^{2}}{a}f_{B}]$, namely, any $\tau _{i,B}\in [\frac {\hat {\sigma }^{2}}{a}f_{B}, 1],$this contradicts theassumption that τ _i,B isuniformly distributed on [1 − 𝜃, 1]. Thus, $1-\theta \geq \frac {\hat {\sigma }^{2}}{a}f_{B}$is reasonable.

Moreover, if $V(\bar {k}, \tau _{i,B}, f_{B})\geq V(\bar {k}, \tau _{i,A}, f_{A})$, namely,

$$\begin{array}{@{}rcl@{}} \frac{a}{\hat{\sigma}^{2}}(\tau_{i,B}-\tau_{i,A})\geq (f_{B}-f_{A}). \end{array} $$

(5.19)

investor i prefers manager B rather to manager A.

Since τ _i,A − τ _i,B ∈ [−𝜃,𝜃],Eq. 5.19 indicates that

$$\begin{array}{@{}rcl@{}} \frac{\hat{\sigma}^{2}}{a} (f_{B}-f_{A})\geq-\theta. \end{array} $$

Otherwise, manager A earns zero profit. Manager A could reduce his fee and earn a positive profit.Therefore,

$$1-\theta\leq1-\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B}).$$

When f _A ≥ f _B , managerA’s total profit $U_{f_{A}}(f_{A}, f_{B})$is rewritten as

$$\begin{array}{@{}rcl@{}} U_{A}(f_{A}, f_{B})=f_{A} (\frac{a}{\hat{\sigma}^{2}}-f_{A})\hat{k}\xi_{M}\cdot I{\int}_{1-\theta}^{1-\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B})}\frac{1}{2\theta}d\tau_{i,B}. \end{array} $$

Subsequently, we consider f _A < f _B .

If $V(\bar {k}, \tau _{i,A}, f_{A})\geq V(\bar {k}, \tau _{i,B}, f_{B})$,investor i prefers manager A to manager B.

From

$$\begin{array}{@{}rcl@{}} (\frac{a\tau_{i,A}}{\hat{\sigma}^{2}}-f_{A})^{\alpha} H(\bar{k})\geq (\frac{a\tau_{i,B}}{\hat{\sigma}^{2}}-f_{B})^{\alpha} H(\bar{k}), \end{array} $$

we can demonstrate that

$$\begin{array}{@{}rcl@{}} \frac{a}{\hat{\sigma}^{2}}(\tau_{i,A}-\tau_{i,B})\geq (f_{A}-f_{B}). \end{array} $$

(5.20)

If investor i is an A-trusting investor, as mentioned above, τ _i,A = 1. Thus, τ _i,A − τ _i,B ≥ 0. Since the right-hand side of Eq. 5.20 is less than 0, Eq. 5.20 always holds. Therefore, all A-trustinginvestors prefer manager A.

For a B-trusting investor and τ _i,B = 1,if

$$\begin{array}{@{}rcl@{}} \tau_{i,A}\geq 1+\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B}), \end{array} $$

the B-trusting investor will choose manager A, as he hopes to pay a lower management fee.

Therefore, when f _B > f _A , manager A’s total profit is

$$\begin{array}{@{}rcl@{}} f_{A} \bar{k} \xi_{M}\cdot I [\frac{1}{2}(\frac{a}{\hat{\sigma}^{2}}-f_{A})+{\int}_{\max[1-\theta,1+\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B})]}^{1}(\frac{a\tau_{i,A}}{\hat{\sigma}^{2}}-f_{A})\frac{1}{2\theta}d\tau_{i,A}]. \end{array} $$

(5.21)

From Eq. 2.9, we can state that

$$\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B})\geq -\theta.$$

That is,

$$\max[1-\theta,1+\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B})]=1+\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B}).$$

Therefore, when f _B > f _A , manager A’s total profit $U_{f_{A}}(f_{A}, f_{B})$is

$$\begin{array}{@{}rcl@{}} U_{A}(f_{A}, f_{B})=f_{A} \bar{k} \xi_{M}\cdot I [\frac{1}{2}(\frac{a}{\hat{\sigma}^{2}}-f_{A})+{\int}_{1+\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B})}^{1}(\frac{a\tau_{i,A}}{\hat{\sigma}^{2}}-f_{A})\frac{1}{2\theta}d\tau_{i,A}]. \end{array} $$

(5.22)

As in the above results, we obtain manager A’s total profits U _A (f _A , f _B ) by

$$U_{A}(f_{A},f_{B})=\left\{\begin{array}{ll} f_{A} (\frac{a}{\hat{\sigma}^{2}}-f_{A})\bar{k}\xi_{M}\cdot I{\int}_{1-\theta}^{1-\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B})}\frac{1}{2\theta}d\tau_{i,B}.&\text{if}\, f_{A}\geq f_{B}, \\ f_{A} \bar{k} \xi_{M}\cdot I [\frac{1}{2}(\frac{a}{\hat{\sigma}^{2}}-f_{A})+{\int}_{1+\frac{\hat{\sigma}^{2}}{a}(f_{A}-f_{B})}^{1}(\frac{a\tau_{i,A}}{\hat{\sigma}^{2}}-f_{A})\frac{1}{2\theta}d\tau_{i,A}]. &\text{if}\, f_{A}< f_{B}. \end{array}\right.$$

Similarly, the total profits U _B (f _A , f _B ) of manager B are given by

$$U_{B}(f_{A},f_{B})=\left\{\begin{array}{ll} f_{B} (\frac{a}{\hat{\sigma}^{2}}-f_{B})\bar{k}\xi_{M}\cdot I{\int}_{1-\theta}^{1-\frac{\hat{\sigma}^{2}}{a}(f_{B}-f_{A})}\frac{1}{2\theta}d\tau_{i,A}.&\text{if}\, f_{B}> f_{A},\\ f_{B} \bar{k} \xi_{M}\cdot I [\frac{1}{2}(\frac{a}{\hat{\sigma}^{2}}-f_{B})+{\int}_{1+\frac{\hat{\sigma}^{2}}{a}(f_{B}-f_{A})}^{1}(\frac{a\tau_{i,B}}{\hat{\sigma}^{2}}-f_{B})\frac{1}{2\theta}d\tau_{i,B}]. &\text{if}\, f_{B}\leq f_{A}. \end{array}\right. $$

□