Portfolio management with benchmark related incentives under mean reverting processes

Abstract

We study the problem of a fund manager whose compensation depends on the relative performance with respect to a benchmark index. In particular, the fund manager’s risk-taking incentives are induced by an increasing and convex relationship of fund flows to relative performance. We consider a dynamically complete market with N risky assets and the money market account, where the dynamics of the risky assets exhibit mean reversions, either in the drift or in the volatility. The manager optimizes the expected utility of the final wealth, with an objective function that is non-concave. The optimal solution is found by using the martingale approach and a concavification method. The optimal wealth and the optimal strategy are determined by solving a system of Riccati equations. We provide a semi-closed solution based on the Fourier transform.

Appendix

1.1 Solutions of the Riccati equations for the one-dimensional models

We show the solutions $A(\tau ;z)$, $B(\tau ;z)$ and $C(\tau ;z)$, with $\tau =T-t$, of the Riccati equations arising from our models, when the number of assets is $N=1$, both in the case of stochastic market price of risk of Sect. 3.1 and in the case of stochastic volatility of Sect. 3.2.

The basic result is that the solution of the equation

$$\begin{aligned}&\frac{\partial C}{\partial \tau } = a C^2+b C+c \nonumber \\&C(0) = 0 \end{aligned}$$

(41)

where $a, b, c \in {\mathbb {C}}$, with $a \ne 0$, is given by

$$\begin{aligned} C(\tau ) = \frac{\alpha _+\alpha _-(e^{\alpha \tau }-1)}{\alpha _+e^{\alpha \tau }-\alpha _-} \end{aligned}$$

where

$$\begin{aligned} \alpha _{\pm } = \frac{-b\pm \alpha }{2a} \end{aligned}$$

and

$$\begin{aligned} \alpha = \sqrt{b^2-4ac} \end{aligned}$$

where $b^2-4ac \in {\mathbb {C}} - {\mathbb {R}}_-$ and $\sqrt{\cdot }$ denotes the analytic extension of the real square root to ${\mathbb {C}} - {\mathbb {R}}_-$. Moreover

$$\begin{aligned} \int _0^s C(u)du= \alpha _+ s + \frac{\alpha _- - \alpha _+}{\alpha } \ln \left( \frac{\alpha _+ e^{\alpha s}-\alpha _-}{\alpha _+ -\alpha _-} \right) . \end{aligned}$$

Such a result is a particular case of Lemma 10.12 in Filipović (2009) and it can be obtained by standard methods for differential equations. For convenience of the reader, we show how to integrate C(u) through the change of variable $x=e^{\alpha u}$:

$$\begin{aligned} \int _0^s C(u)du= & {} \frac{\alpha _+ \alpha _-}{\alpha } \int _1^{e^{\alpha s}} \frac{x-1}{x(\alpha _+ x-\alpha _-)} dx \\= & {} \frac{\alpha _+ \alpha _-}{\alpha } \left( \int _1^{e^{\alpha s}} \frac{1}{\alpha _- x} dx + \int _1^{e^{\alpha s}} \frac{\alpha _- - \alpha _+}{\alpha _-(\alpha _+ x-\alpha _-)} dx \right) \\= & {} \alpha _+ s + \frac{\alpha _- - \alpha _+}{\alpha } \ln \left( \frac{\alpha _+ e^{\alpha s}-\alpha _-}{\alpha _+ -\alpha _-} \right) . \end{aligned}$$

Equation (41) is an equation for $C(\tau )$ obtained when setting to zero the quadratic terms of the partial differential equation for $H_t (z)$ for both models. In the model with stochastic market price of risk this is Eq. (23), hence in this case the coefficients of Eq. (41) are

$$\begin{aligned} a= & {} \sigma _X^2, \\ b= & {} -2( \lambda _X+(1+z) \sigma _X),\\ c= & {} z (z+1). \end{aligned}$$

In the model with stochastic volatility this is Eq. (34), hence the coefficients are

$$\begin{aligned} a= & {} \sigma _v^2,\\ b= & {} -2( \lambda _v - ( \gamma z +1) \sigma _v \beta ),\\ c= & {} \gamma z ( \gamma z + 1) \beta ^2. \end{aligned}$$

The linear terms give a linear differential equation of the first order for $B(\tau )$ in both models:

$$\begin{aligned} \frac{\partial B}{\partial \tau }= & {} B g(\tau ) + f(\tau ) \nonumber \\ B(0)= & {} 0, \end{aligned}$$

(42)

where

$$\begin{aligned} g(\tau )= \frac{b}{2} + a C (\tau ). \end{aligned}$$

Also, we can write

$$\begin{aligned} f(\tau )= a_f + b_f C(\tau ) \end{aligned}$$

with coefficients $a_f, b_f \in {\mathbb {C}}$ that depend on the model. Namely, from Eq. (24), in the case of mean reverting market price of risk, we have

$$\begin{aligned} a_f= & {} -z( 1+ z \gamma ) \sigma \beta \\ b_f= & {} \lambda _X {{\bar{X}}}+ (1+ z \gamma ) \sigma _X \sigma \beta \end{aligned}$$

and from Eq. (35) for the model with mean reverting volatility

$$\begin{aligned} a_f= & {} -z (1+ z \gamma ) \beta X \\ b_f= & {} \lambda _v {{\bar{v}}}-(1+z) \sigma _v X. \end{aligned}$$

We can now proceed to the computation of $B(\tau )$ for both models at once. The solution of (42) is

$$\begin{aligned} B(\tau ) = e^{\int _0^{\tau }g(u)du}\int _0^{\tau }f(s)e^{-\int _0^s g(u)du}ds. \end{aligned}$$

(43)

We have

$$\begin{aligned} \int _0^s g(u)du= & {} \frac{b}{2} s + a \int _0^s C(u)du \\= & {} \left( \frac{b}{2}+ a \alpha _+\right) s + a \frac{\alpha _- - \alpha _+}{\alpha } \ln \left( \frac{\alpha _+ e^{\alpha s}-\alpha _-}{\alpha _+ -\alpha _-} \right) \\= & {} \frac{\alpha }{2}s - \ln \left( \frac{\alpha _+ e^{\alpha s}-\alpha _-}{\alpha _+ -\alpha _-} \right) . \end{aligned}$$

Therefore

$$\begin{aligned} e^{-\int _0^s g(u)du} = \frac{\alpha _+ e^{\frac{\alpha }{2} s}-\alpha _-e^{-\frac{\alpha }{2} s}}{\alpha _+ -\alpha _-}. \end{aligned}$$

Next we compute

$$\begin{aligned} \int _0^{\tau }f(s)e^{-\int _0^s g(u)du}ds= & {} \int _0^{\tau }\left( a_f + b_f \frac{\alpha _+\alpha _-(e^{\alpha s}-1)}{\alpha _+e^{\alpha s}-\alpha _-}\right) \frac{\alpha _+ e^{\frac{\alpha }{2} s}-\alpha _-e^{-\frac{\alpha }{2} s}}{\alpha _+ -\alpha _-}ds \\= & {} \frac{2a_f}{\alpha (\alpha _+-\alpha _-)}\left( \alpha _+\left( e^{\frac{\alpha \tau }{2}}-1\right) +\alpha _-\left( e^{\frac{-\alpha \tau }{2}}-1\right) \right) \\&+\, \frac{2b_f \alpha _+ \alpha _-}{\alpha (\alpha _+-\alpha _-)}\left( e^{\frac{\alpha \tau }{2}}+e^{\frac{-\alpha \tau }{2}}-2\right) . \end{aligned}$$

Substituting in Eq. (43), we obtain:

$$\begin{aligned} B(\tau )= & {} \frac{2a_f}{\alpha }\left( \frac{\alpha _+\left( e^{\frac{\alpha \tau }{2}}-1\right) +\alpha _-\left( e^{\frac{-\alpha \tau }{2}}-1\right) }{\alpha _+ e^{\frac{\alpha }{2} \tau }-\alpha _-e^{-\frac{\alpha }{2} \tau }}\right) \\&+\, \frac{2b_f \alpha _+ \alpha _-}{\alpha }\left( \frac{e^{\frac{\alpha \tau }{2}}+e^{\frac{-\alpha \tau }{2}}-2}{\alpha _+ e^{\frac{\alpha }{2} \tau }-\alpha _-e^{-\frac{\alpha }{2} \tau }}\right) . \end{aligned}$$

Finally, we get $A(\tau )$ in both models by direct integration.