Portfolio Selection with Contrarian Strategy

Abstract

Compared with the extensive empirical literature on contrarian strategy, we develop a dynamic mean-variance model with geometric value-reversion asset prices, which implies a contrarian strategy. The model is solved (semi) explicitly under three asset price evaluations: constant valuation, exponential-varying valuation, and geometric average valuation. From a mathematical perspective, it is nontrivial to solve the extended HJB equations under stochastic opportunities. We demonstrate that our strategy exhibits the same monotonicity as that of the traditional constant relative risk-averse utility, and the welfare loss of using the dynamic mean-variance criterion is rather small, supporting that our model is a good approximation to the constant relative risk-averse utility. Empirical tests show that our strategies can help an investor achieve a less volatile wealth trajectory.

Appendices

Appendix 1 Proof of Theorem 2.1

(i) We assume that V and g are seperable² in R, i.e., they have the following structures:

$$\begin{aligned} V(t,P,R)&=\tilde{V}(t,P)+R+r\left( T-t\right) ;\end{aligned}$$

(A.1)

$$\begin{aligned} g\left( t,P,R\right)&=f\left( t,P\right) +R+r\left( T-t\right) . \end{aligned}$$

(A.2)

Substituting (A.1) and (A.2) into PDE (2.6) and PDE (2.7), we obtain:

$$\begin{aligned} \underset{\pi }{\max }\left\{ \mathcal {D}\tilde{V}\left( t,P\right) +\left( \mu -r+\lambda \left( \alpha -\ln P\right) \right) \pi -\frac{1}{2}\sigma ^{2}\pi ^{2}\right.&\\ -\left. \frac{\gamma }{2}\left[ \sigma ^{2}P^{2}(\frac{\partial f}{\partial P})^{2}+2\sigma ^{2}\pi P\frac{\partial f}{\partial P}+\sigma ^{2}\pi ^{2}\right] \right\}&=0,\\ \mathcal {D}^{*}f\left( t,P\right) +\left( \mu -r+\lambda \left( \alpha -\ln P\right) \right) \pi -\frac{1}{2}\sigma ^{2}\pi ^{2}&=0, \end{aligned}$$

subject to $\tilde{V}(T,P)=0$, $f(T,P)=0$. By the first-order condition, we derive a representation of the equilibrium $\hat{\pi }$ as (2.8). It follows putting $\hat{\pi }$ into f that, the function f satisfies PDE (2.9), subject to terminal condition $f\left( T,P\right) =0$. Then, following the line of Björk et al. (2017), it is easy to verify that the strategy (2.10) is really an equilibrium of the original problem.

(ii) First we make a transform on the variable $x=\ln P$. Then PDE (2.9) becomes

$$\begin{aligned} \left\{ \begin{array}{r} \frac{\partial f}{\partial t} +\left( \mu +\lambda \left( \alpha -x\right) -\frac{\gamma ^{2}(\mu -r+\lambda \left( \alpha -x\right) )}{\left( 1+\gamma \right) ^{2}}\right) \frac{\partial f}{\partial x}\\ +\frac{1}{2} \sigma ^{2}\left( -\frac{\partial f}{\partial x}+\frac{\partial ^{2}f}{\partial x^{2}}-\frac{\gamma ^{2}}{\left( 1+\gamma \right) ^{2}}(\frac{\partial f}{\partial x})^{2}\right) \\ +\frac{\left( 2\gamma +1\right) (\mu -r+\lambda \left( \alpha -x\right) )^{2} }{2\left( \gamma +1\right) ^{2}\sigma ^{2}}=0,\\ f(T,x)=0. \end{array} \right. \end{aligned}$$

(A.3)

Let $f\left( t,x\right) =-\frac{\left( 1+\gamma \right) ^{2}}{\gamma ^{2} }\ln u\left( t,x\right)$. Then PDE (A.3) is rewritten as

$$\begin{aligned} \left\{ \begin{array}{r} \frac{\partial u}{\partial t}+\left( \mu +\lambda \left( \alpha -x\right) -\frac{\gamma ^{2}(\mu -r+\lambda \left( \alpha -x\right) )}{\left( 1+\gamma \right) ^{2}}-\frac{1}{2}\sigma ^{2}\right) \frac{\partial u}{\partial x}\\ +\frac{1}{2}\sigma ^{2}\frac{\partial ^{2}u}{\partial x^{2}} -\frac{\left( 2\gamma +1\right) \gamma ^{2}(\mu -r+\lambda \left( \alpha -x\right) )^{2}}{2\left( \gamma +1\right) ^{4}\sigma ^{2}}u=0,\\ u\left( T,x\right) =1. \end{array} \right. \end{aligned}$$

(A.4)

Let $\omega =\alpha +\frac{\mu -r}{\lambda }-x$, $\tilde{u}(t,\omega )=u(t,x).$ Putting $\frac{\partial u}{\partial x}=-\frac{\partial \tilde{u}}{\partial \omega }$, $\frac{\partial ^{2}u}{\partial x^{2}}=\frac{\partial ^{2}\tilde{u}}{\partial \omega ^{2}}$ into PDE (A.4), we get

$$\begin{aligned} \left\{ \begin{array}{c} \frac{\partial \tilde{u}}{\partial t}-\left( r-\frac{1}{2}\sigma ^{2} +\frac{(1+2\gamma )\lambda }{\left( 1+\gamma \right) ^{2}}\omega \right) \frac{\partial \tilde{u}}{\partial \omega }+\frac{1}{2}\sigma ^{2}\frac{\partial ^{2}\tilde{u}}{\partial \omega ^{2}}-\frac{\left( 2\gamma +1\right) \lambda ^{2}\gamma ^{2}}{2\left( \gamma +1\right) ^{4}\sigma ^{2}}\omega ^{2}\tilde{u}=0,\\ \tilde{u}\left( T,\omega \right) =1. \end{array} \right. \end{aligned}$$

Set $\tilde{u}=\exp \left\{ k(T-t)+\eta \omega +\theta \omega ^{2}\right\} V(t,\omega )$. Substituting

$$\begin{aligned} \tilde{u}_{t} =&\ (-kV+V_{t})\exp \left\{ k(T-t)+\eta \omega +\theta \omega ^{2}\right\} ,\\ \tilde{u}_{\omega } =&\ \left[ (\eta +2\theta \omega )V+V_{\omega }\right] \exp \left\{ k(T-t)+\eta \omega +\theta \omega ^{2}\right\} ,\\ \tilde{u}_{\omega \omega } =&\ [ (\eta ^{2}+2\theta +4\theta ^{2}\omega ^{2}+4\theta \eta \omega )V\\ {}&+2(\eta +2\theta \omega )V_{\omega }+V_{\omega \omega }] \exp \left\{ k(T-t)+\eta \omega +\theta \omega ^{2}\right\} , \end{aligned}$$

into the above PDE, we obtain

$$\begin{aligned} \begin{aligned}&\left( -k+(\frac{\sigma ^{2}}{2}-r)\eta +\frac{1}{2}\sigma ^{2}\eta ^{2}+\sigma ^{2}\theta \right) V+\left( 2(\frac{\sigma ^{2}}{2}-r)\theta -\frac{(1+2\gamma )\lambda }{(1+\gamma )^{2}}\eta +2\sigma ^{2}\eta \theta \right) \omega V\\&+\left( 2\sigma ^{2}\theta ^{2}-2\frac{\lambda (1+2\gamma )}{(1+\gamma )^{2} }\theta -\frac{\lambda ^{2}\gamma ^{2}(1+2\gamma )}{2(1+\gamma )^{4}\sigma ^{2} }\right) \omega ^{2}V+V_{t}\\&+\left( (-r+\frac{1}{2}\sigma ^{2}+\sigma ^{2}\eta )+(2\sigma ^{2}\theta -\frac{(1+2\gamma )\lambda }{(1+\gamma )^{2}})\omega \right) V_{\omega }+\frac{1}{2}\sigma ^{2}V_{\omega \omega }=0. \end{aligned} \end{aligned}$$

(A.5)

Let the coefficient of V be 0 by separating the variable $\omega$, we get the following algebraic equations:

$$\begin{aligned} \left\{ \begin{array}{l} -k+a\eta +\frac{1}{2}\sigma ^{2}\eta ^{2}+\sigma ^{2}\theta =0,\\ 2a\theta -b\eta +2\sigma ^{2}\eta \theta =0,\\ 2\sigma ^{2}\theta ^{2}-2b\theta -c=0, \end{array} \right. \end{aligned}$$

(A.6)

where

$$\begin{aligned} a=\frac{\sigma ^{2}}{2}-r\text {, }b=\frac{\lambda (1+2\gamma )}{(1+\gamma )^{2} }\text {, }c=\frac{\lambda ^{2}\gamma ^{2}(1+2\gamma )}{2(1+\gamma )^{4}\sigma ^{2} }\text {.} \end{aligned}$$

(A.7)

From the last and the second equations, we deduce that

$$\begin{aligned} \left\{ \begin{array}{l} \theta =\frac{b\pm q}{2\sigma ^{2}},\\ \eta =\mp \frac{a(b\pm q)}{\sigma ^{2}q}, \end{array} \right. \end{aligned}$$

(A.8)

where $q=\sqrt{b^{2}+2\sigma ^{2}c}=\frac{\lambda \sqrt{1+2\gamma }}{(1+\gamma )}.$ Thus (A.5) satisfies

$$\begin{aligned} \left\{ \begin{array}{c} \frac{\partial V}{\partial t}+\left( (-r+\frac{1}{2}\sigma ^{2}+\sigma ^{2} \eta )+(2\sigma ^{2}\theta -\frac{(1+2\gamma )\lambda }{(1+\gamma )^{2}} )\omega \right) \frac{\partial V}{\partial \omega }+\frac{1}{2}\sigma ^{2} \frac{\partial ^{2}V}{\partial \omega ^{2}}=0,\\ V\left( T,\omega \right) =e^{-\eta \omega -\theta \omega ^{2}}. \end{array} \right. \end{aligned}$$

(A.9)

Then we denote $\hat{a}:=a+\sigma ^{2}\eta =\mp \frac{ab}{q}$, $\hat{b}:=2\sigma ^{2}\theta -b=\pm q$. So (A.9) is rewritten as

$$\begin{aligned} \left\{ \begin{array}{l} \frac{\partial V}{\partial t}+\left( \hat{a}+\hat{b}\omega \right) \frac{\partial V}{\partial \omega }+\frac{1}{2}\sigma ^{2}\frac{\partial ^{2} V}{\partial \omega ^{2}}=0,\\ V\left( T,\omega \right) =e^{-\eta \omega -\theta \omega ^{2}}. \end{array} \right. \end{aligned}$$

(A.10)

From the terminal structure of $V\left( T,\omega \right) ,$ it is natural to suppose $V\left( t,\omega \right) =\exp \{A\left( t\right) +B\left( t\right) \omega +C\left( t\right) \omega ^{2}\}$. Substituting $V_{t}=\left( A_{t}+B_{t}\omega +C_{t}\omega ^{2}\right) \cdot V$, $V_{\omega }=\left( B+2C\omega \right) \cdot V$, $V_{\omega \omega }=\left( B^{2}+2C+4BC\omega +4C^{2}\omega ^{2}\right) \cdot V$ into (A.10) and by separating variables, we obtain the following system of ordinary differential equations (ODEs):

$$\begin{aligned} \left\{ \begin{array}{ll} A_{t}+\hat{a}B+\frac{1}{2}\sigma ^{2}\left( B^{2}+2C\right) =0\text {,} &{} A\left( T\right) =0,\\ B_{t}+\hat{b}B+2\hat{a}C+2\sigma ^{2}BC=0\text {,} &{} B\left( T\right) =-\eta ,\\ C_{t}+2\hat{b}C+2\sigma ^{2}C^{2}=0\text {,} &{} C\left( T\right) =-\theta . \end{array} \right. \end{aligned}$$

(A.11)

Solving the last equation, we have

$$\begin{aligned} C\left( t\right) =-\frac{\hat{b}\theta e^{2\hat{b}\left( T-t\right) } }{\theta \sigma ^{2}\left( e^{2\hat{b}\left( T-t\right) }-1\right) +\hat{b} }. \end{aligned}$$

Then

$$\begin{aligned} \bar{\Theta }&:=\theta +C\\&=\frac{\left( e^{2\hat{b}\left( T-t\right) }-1\right) \theta \left( \theta \sigma ^{2}-\hat{b}\right) }{\frac{1}{2}\left( b\pm q\right) \left( e^{2\hat{b}\left( T-t\right) }-1\right) \pm q}=\frac{-c\left( e^{\pm 2q\left( T-t\right) }-1\right) }{\left( b\pm q\right) \left( e^{\pm 2q\left( T-t\right) }-1\right) \pm 2q}\\&=\frac{-c\left( e^{2q\left( T-t\right) }-1\right) }{\left( b+q\right) \left( e^{2q\left( T-t\right) }-1\right) +2q}\le 0. \end{aligned}$$

Denote $\bar{B}=\eta +B$. Then it follows (A.11) that,

$$\begin{aligned} \begin{array}{ll} \bar{B}_{t}+\left( 2\sigma ^{2}\bar{\Theta }-b\right) \bar{B}+2a\bar{\Theta }=0\text {,}&\bar{B}\left( T\right) =0. \end{array} \end{aligned}$$

(A.12)

Note that the coefficient of $\bar{B}$ and the nonhomogeneous term are functions of time t, it is not easy to get the solution of ODE (A.12). Since (A.12) is a linear equation, we can get the solution first in integral form, then by a quite daunting and tedious calculation, the explicit solution is obtained as follows:

$$\begin{aligned} \bar{B}\left( t\right) =\frac{-2ac}{q}\left( e^{q\left( T-t\right) }-1\right) ^{2}\left[ \left( b+q\right) \left( e^{2q\left( T-t\right) }-1\right) +2q\right] ^{-1}. \end{aligned}$$

From the deformations of the above series of formulas, it yields

$$\begin{aligned} \frac{\partial f}{\partial P}=\frac{(1+\gamma )^{2}}{\gamma ^{2}P}[\bar{B}(t)+2\bar{\Theta }(t)(\alpha +\frac{\mu -r}{\lambda }-\ln P)]. \end{aligned}$$

(A.13)

Then the equilibrium strategy (2.8) can be expressed explicitly by

$$\begin{aligned} \hat{\pi }(t,P)=\frac{\mu -r+\lambda \left( \alpha -\ln P\right) }{(1+\gamma )\sigma ^{2}}-\frac{2(1+\gamma )(\alpha +\frac{\mu -r}{\lambda }-\ln P)}{\gamma }\bar{\Theta }(t)-\frac{1+\gamma }{\gamma }\bar{B}\left( t\right) . \end{aligned}$$

We then rewrite the above $\hat{\pi }(t,P)$ as (2.10).

The proof is complete. $\square$

Appendix 2 Proof of Corollary 2.1

(i) From the expression of $\hat{\pi }$, it’s not difficult to find that $\hat{\pi }$ decreases as $\ln P$ increases.

(ii) When $\ln P<\alpha +\frac{\mu -r}{\lambda }$ and $r<\frac{\sigma ^{2}}{2}$, we find that $\frac{\lambda \gamma (1+2\gamma )\left( \mu -r+\lambda (\alpha -\ln P)\right) }{(1+\gamma )^{3}\sigma ^{2}}>0$ and $\frac{\lambda ^{2} \gamma (1+2\gamma )(\sigma ^{2}-2r)}{2(1+\gamma )^{3}\sigma ^{2}}>0$. On the contrary, when $\ln P>\alpha +\frac{\mu -r}{\lambda }$ and $r>\frac{\sigma ^{2} }{2}$, it is easy to get that $\frac{\lambda \gamma (1+2\gamma )\left( \mu -r+\lambda (\alpha -\ln P)\right) }{(1+\gamma )^{3}\sigma ^{2}}<0$ and $\frac{\lambda ^{2}\gamma (1+2\gamma )(\sigma ^{2}-2r)}{2(1+\gamma )^{3}\sigma ^{2} }<0$. At the same time, $\hat{\Theta }(\tau )$ is increasing in the time horizon $\tau .$ Moreover,

$$\begin{aligned} \hat{B}\left( \tau \right)&=\frac{\left( e^{q\tau }-1\right) ^{2} }{q\left[ (b+q)(e^{2q\tau }-1)+2q\right] }\\&=\frac{1}{q\left[ (b+q)(1+\frac{2}{e^{q\tau }-1})+\frac{2q}{(e^{q\tau }-1)^{2}}\right] } \end{aligned}$$

is also increasing in the time horizon $\tau$. Consequently, we obtain the desired results. $\square$

Appendix 3 Proof of Corollary 2.2

(i) In order to calculate the limit of $\hat{\pi }(t,P)$ when $\tau$ tends to infinity, we first calculate the limits of $\hat{\Theta } (\tau )$ and $\hat{B}\left( \tau \right)$.

Note that

$$\begin{aligned} \underset{\tau \rightarrow \infty }{\lim }\hat{\Theta }(\tau )&=\underset{\tau \rightarrow \infty }{\lim }\frac{e^{2q\tau }-1}{(b+q)(e^{2q\tau }-1)+2q} =\frac{1}{b+q},\\ \underset{\tau \rightarrow \infty }{\lim }\hat{B}\left( \tau \right)&=\underset{\tau \rightarrow \infty }{\lim }\frac{\left( e^{q\tau }-1\right) ^{2} }{q\left[ (b+q)(e^{2q\tau }-1)+2q\right] }\\ {}&=\underset{\tau \rightarrow \infty }{\lim }\frac{1}{q[(b+q)(1+\frac{2}{e^{q\tau }-1})+\frac{2q}{(e^{q\tau }-1)^{2} }]}=\frac{1}{q(b+q)}. \end{aligned}$$

Therefore,

$$\begin{aligned} \underset{\tau \rightarrow \infty }{\lim }\hat{\pi }(t,P) =&\ \frac{\mu -r+\lambda \left( \alpha -\ln P\right) }{(1+\gamma )\sigma ^{2}}+\frac{\lambda \gamma (1+2\gamma )\left( \mu -r+\lambda (\alpha -\ln P)\right) }{(1+\gamma )^{3}\sigma ^{2}}\frac{1}{b+q}\\&+\frac{\lambda ^{2}\gamma (1+2\gamma )(\sigma ^{2}-2r)}{2(1+\gamma )^{3}\sigma ^{2}}\frac{1}{q(b+q)}\\ =&\ \frac{\mu -r+\lambda \left( \alpha -\ln P\right) }{(1+\gamma )\sigma ^{2} }\\ {}&+\frac{\lambda ^{2}\gamma (1+2\gamma )}{[(1+\gamma )^{3}\sigma ^{2}](b+q)}\left( \left( \alpha +\frac{\mu -r}{\lambda }-\ln P\right) +\frac{\sigma ^{2}-2r}{2q}\right) . \end{aligned}$$

(ii) Denote

$$\begin{aligned} I=&\ \frac{\lambda \gamma (1+2\gamma )}{(1+\gamma )^{2}}\frac{e^{2q\tau } -1}{(b+q)(e^{2q\tau }-1)+2q}\\ {}&+\frac{\lambda \gamma (1+2\gamma )(\sigma ^{2} -2r)}{2(1+\gamma )^{2}\left( \alpha +\frac{\mu -r}{\lambda }-\ln P\right) } \frac{\left( e^{q\tau }-1\right) ^{2}}{q\left[ (b+q)(e^{2q\tau }-1)+2q\right] }. \end{aligned}$$

Hence,

$$\begin{aligned} \hat{\pi }(\tau ,P)=\frac{\mu -r+\lambda \left( \alpha -\ln P\right) }{(1+\gamma )\sigma ^{2}}\left[ 1+I\right] , \end{aligned}$$

and

$$\begin{aligned} \hat{R}^{\infty }(\tau ):=\underset{\gamma \rightarrow \infty }{\lim }\frac{\hat{H}(\tau )}{\hat{\pi }(\tau )}=\underset{\gamma \rightarrow \infty }{\lim }\frac{I}{I+1}. \end{aligned}$$

To compute the hedging ratio $\frac{I}{I+1}$, we first calculate the limit of I

$$\begin{aligned} \underset{\gamma \rightarrow \infty }{\lim }I&=\underset{\gamma \rightarrow \infty }{\lim }(\frac{\lambda \gamma (1+2\gamma )}{(1+\gamma )^{2}}\frac{e^{2q\tau }-1}{(b+q)(e^{2q\tau }-1)+2q}\\&\ \ \ +\frac{\lambda \gamma (1+2\gamma )(\sigma ^{2}-2r)}{2(1+\gamma )^{2}\left( \alpha +\frac{\mu -r}{\lambda }-\ln P\right) }\frac{\left( e^{q\tau }-1\right) ^{2}}{q\left[ (b+q)(e^{2q\tau }-1)+2q\right] })\\&=\underset{\gamma \rightarrow \infty }{\lim }\left( \frac{2\lambda \tau }{(b+q)\tau +1}+\frac{\lambda (\sigma ^{2}-2r)\tau ^{2}}{2\left[ (b+q)\tau +1\right] \left( \alpha +\frac{\mu -r}{\lambda }-\ln P\right) }\right) \\&=2\lambda \tau +\frac{\lambda ^{2}(\sigma ^{2}-2r)\tau ^{2}}{2(\mu -r+\lambda \left( \alpha -\ln P\right) )}\\&=\frac{\lambda \tau \left[ \lambda \tau (\sigma ^{2}-2r)+4(\mu -r+\lambda \left( \alpha -\ln P\right) )\right] }{2(\mu -r+\lambda \left( \alpha -\ln P\right) )}. \end{aligned}$$

Thus,

$$\begin{aligned} \underset{\gamma \rightarrow \infty }{\lim }\frac{I}{I+1}=\frac{\lambda \tau \left[ \lambda \tau (\sigma ^{2}-2r)+4(\mu -r+\lambda \left( \alpha -\ln P\right) )\right] }{\lambda \tau \left[ \lambda \tau (\sigma ^{2}-2r)+4(\mu -r+\lambda \left( \alpha -\ln P\right) )\right] +2(\mu -r+\lambda \left( \alpha -\ln P\right) )}. \end{aligned}$$

The proof is complete. $\square$

Appendix 4 Proof of Proposition 4.1

Benth and Karlsen (2005) provide a semi-explicit solution for the CRRA optimization with geometric mean reversion and a risk aversion coefficient $\tilde{\gamma }\in \left( 0,1\right)$. In our paper, $\tilde{\gamma }=\gamma +1$ should be greater than 1 because $\gamma \ge 0$ in the MV model. Here, we outline the proof and provide an explicit solution. Utilizing the classical dynamic programming principle, we derive the HJB equation for the value function:

$$\begin{aligned} \begin{aligned} \frac{\partial v}{\partial t}&+\underset{\pi }{\max }\left\{ \left( \left( \mu -r+\lambda \left( \alpha -\ln P\right) \right) \pi +r\right) W\frac{\partial v}{\partial W}\right. \\ {}&+\left. \frac{1}{2}\sigma ^{2}\pi ^{2}W^{2}\frac{\partial ^{2}v}{\partial W^{2}}+\sigma ^{2}\pi WP\frac{\partial ^{2}v}{\partial WP}\right\} \\&+\left( \mu +\lambda \left( \alpha -\ln P\right) \right) P\frac{\partial v}{\partial P}+\frac{1}{2}\sigma ^{2}P^{2}\frac{\partial ^{2}v}{\partial P^{2} }=0, \end{aligned} \end{aligned}$$

(D.1)

subject to $v(T,W,P)=\frac{W^{1-\tilde{\gamma }}-1}{1-\tilde{\gamma }}.$

Assuming there exists a sufficiently smooth solution v to the HJB equation, the first-order condition for an optimal strategy gives

$$\begin{aligned} \pi ^{*}=-\frac{\left( \mu -r+\lambda \left( \alpha -\ln P\right) \right) \frac{\partial v}{\partial W}+\sigma ^{2}P\frac{\partial ^{2}v}{\partial W\partial P}}{\sigma ^{2}W\frac{\partial ^{2}v}{\partial W^{2}}}. \end{aligned}$$

Then substituting $\pi ^{*}$ into (D.1), we get the following PDE:

$$\begin{aligned} \begin{aligned} \frac{\partial v}{\partial t}&+rw\frac{\partial v}{\partial W}+\left( \lambda \left( \alpha -\ln P\right) +\mu \right) P\frac{\partial v}{\partial P}\\ {}&+\frac{1}{2}\sigma ^{2}P^{2}\frac{\partial ^{2}v}{\partial P^{2}}-\frac{1}{2}\sigma ^{2}P^{2}\left( \frac{\partial ^{2}v}{\partial W\partial P}\right) ^{2}/\frac{\partial ^{2}v}{\partial W^{2}}\\&-\frac{\left( \lambda \left( \alpha -\ln P\right) +\mu -r\right) ^{2} }{2\sigma ^{2}}\left( \frac{\partial v}{\partial W}\right) ^{2} /\frac{\partial ^{2}v}{\partial W^{2}}\\ {}&-\left( \lambda \left( \alpha -\ln P\right) +\mu -r\right) P\frac{\partial v}{\partial W}\frac{\partial ^{2} v}{\partial W\partial P}/\frac{\partial ^{2}v}{\partial W^{2}}=0. \end{aligned} \end{aligned}$$

(D.2)

As in Benth and Karlsen (2005), we make the assumption that the solution of (D.2) takes the form

$$\begin{aligned} v(t,W,P)=\frac{W^{1-\tilde{\gamma }}}{1-\tilde{\gamma }}g(t,P)^{\tilde{\gamma } }. \end{aligned}$$

Then from (D.2) we get

$$\begin{aligned} \left\{ \begin{array}{r} \frac{\partial g}{\partial t}+\frac{1}{\tilde{\gamma }}\left( \lambda \left( \alpha -\ln P\right) +\mu -\left( 1-\tilde{\gamma }\right) r\right) P\frac{\partial g}{\partial P}+\frac{1}{2}\sigma ^{2}P^{2}\frac{\partial ^{2} g}{\partial P^{2}}\\ +\left( \frac{\left( ^{1-\tilde{\gamma }}\right) r}{\tilde{\gamma }}+\frac{\left( 1-\tilde{\gamma }\right) \left( \lambda \left( \alpha -\ln P\right) +\mu -r\right) ^{2}}{2\sigma ^{2}\tilde{\gamma }^{2} }\right) g=0,\\ g(T,P)=\left( \frac{W^{1-\tilde{\gamma }}-1}{W^{1-\tilde{\gamma }}}\right) ^{\frac{1}{\tilde{\gamma }}}, \end{array} \right. \end{aligned}$$

(D.3)

which indeed is a one-dimensional linear parabolic equation. Similar to the baseline model in Section 2, we assume $g(t,P)=\exp \left( f_{0}(t)+f_{1}(t)\ln P+f_{2}(t)(\ln P)^{2}\right) .$ Then the optimal control can be rewritten as:

$$\begin{aligned} \pi ^{*}(t,P)=\frac{\mu -r+\lambda (\alpha -\ln P)}{\sigma ^{2}\tilde{\gamma } }+f_{1}(t)+2f_{2}(t)\ln P, \end{aligned}$$

where $f_{1},$ $f_{2}$ satisfy the following system of ODEs:

$$\begin{aligned} \left\{ \begin{array}{ll} f_{1}^{\prime }-\frac{\lambda }{\tilde{\gamma }}f_{1}-\left( \sigma ^{2} -2r-\frac{2\left( \alpha \lambda +\mu -r\right) }{\tilde{\gamma }}\right) f_{2}+2\sigma ^{2}f_{1}f_{2}-\frac{\lambda \left( 1-\tilde{\gamma }\right) \left( \alpha \lambda +\mu -r\right) }{\sigma ^{2}\tilde{\gamma }^{2}}=0, &{} f_{1}\left( T\right) =0,\\ f_{2}^{\prime }+2\sigma ^{2}f_{2}^{2}-\frac{2\lambda }{\tilde{\gamma }}f_{2} +\frac{\lambda ^{2}\left( 1-\tilde{\gamma }\right) }{2\sigma ^{2}\tilde{\gamma }^{2}}=0, &{} f_{2}\left( T\right) =0. \end{array} \right. \end{aligned}$$

Through a series of calculations, we get

$$\begin{aligned} f_{1}(t)&=\frac{\lambda (\tilde{\gamma }-1)\left( \alpha \lambda +\mu -r\right) }{\sigma ^{2}\tilde{\gamma }^{2}}\hat{\Theta }\left( t\right) +\frac{\lambda ^{2}\left( \sigma ^{2}-2r\right) (\tilde{\gamma }-1)}{2\sigma ^{2}\tilde{\gamma }^{2}}\hat{B}\left( t\right) ,\\ f_{2}(t)&=-\frac{\lambda ^{2}(\tilde{\gamma }-1)}{2\sigma ^{2}\tilde{\gamma }^{2}}\hat{\Theta }\left( t\right) , \end{aligned}$$

where $\hat{\Theta }\left( t\right) ,$ $\hat{B}\left( t\right)$ are defined by (2.12), (2.13) with $b=\frac{\lambda }{\tilde{\gamma }}$, $q=\frac{\lambda }{\sqrt{\tilde{\gamma }}}$. $\square$

Appendix 5 Proof of Theorem 3.1

Let $\alpha _{t}:=\ln \bar{P}_{t}=\ln \bar{P}_{0}+\beta t$. Similar to the baseline model in Section 2, we can get an equilibrium strategy by

$$\begin{aligned} \hat{\pi }(t,P)=\frac{\mu -r+\lambda \left( \alpha _{t}-\ln P\right) }{(1+\gamma )\sigma ^{2}}-\frac{\gamma P}{1+\gamma }\frac{\partial f}{\partial P}, \end{aligned}$$

(E.1)

where the function f satisfies the following PDE:

$$\begin{aligned} \left\{ \begin{array}{r} \frac{\partial f}{\partial t}+\left( \mu +\lambda \left( \alpha _{t}-\ln P\right) -\frac{\gamma ^{2}\left( \mu -r+\lambda \left( \alpha _{t}-\ln P\right) \right) }{\left( 1+\gamma \right) ^{2}}\right) P\frac{\partial f}{\partial P}\\ +\frac{1}{2}\sigma ^{2}P^{2}\left( \frac{\partial ^{2}f}{\partial P^{2}}-\frac{\gamma ^{2}}{\left( 1+\gamma \right) ^{2}}(\frac{\partial f}{\partial P})^{2}\right) \\ +\frac{\left( 2\gamma +1\right) \left( \mu -r+\lambda \left( \alpha _{t}-\ln P\right) \right) ^{2}}{2\left( \gamma +1\right) ^{2}\sigma ^{2}}=0,\\ f\left( T,P\right) =0. \end{array} \right. \end{aligned}$$

(E.2)

First we make a transform on the variable $x=\ln P-\ln \bar{P}_{0}$. Then PDE (E.2) adapts to

$$\begin{aligned} \left\{ \begin{array}{r} \frac{\partial f}{\partial t}+\left( \mu +\lambda \left( \beta t-x\right) -\frac{\gamma ^{2}(\mu -r+\lambda \left( \beta t-x\right) )}{\left( 1+\gamma \right) ^{2}}\right) \frac{\partial f}{\partial x}\\ +\frac{1}{2} \sigma ^{2}\left( -\frac{\partial f}{\partial x}+\frac{\partial ^{2}f}{\partial x^{2}}-\frac{\gamma ^{2}}{\left( 1+\gamma \right) ^{2}}(\frac{\partial f}{\partial x})^{2}\right) \\ +\frac{\left( 2\gamma +1\right) (\mu -r+\lambda \left( \beta t-x\right) )^{2}}{2\left( \gamma +1\right) ^{2}\sigma ^{2}}=0,\\ f(T,x)=0. \end{array} \right. \end{aligned}$$

(E.3)

Let f(x) $=-\frac{\left( 1+\gamma \right) ^{2}}{\gamma ^{2}}\ln u\left( t,x\right)$. Then PDE (E.3) satisfies

$$\begin{aligned} \left\{ \begin{array}{r} \frac{\partial u}{\partial t}+\left( \mu +\lambda \left( \beta t-x\right) -\frac{\gamma ^{2}(\mu -r+\lambda \left( \beta t-x\right) )}{\left( 1+\gamma \right) ^{2}}-\frac{1}{2}\sigma ^{2}\right) \frac{\partial u}{\partial x}\\ +\frac{1}{2}\sigma ^{2}\frac{\partial ^{2}u}{\partial x^{2}} -\frac{\left( 2\gamma +1\right) \gamma ^{2}(\mu -r+\lambda \left( \beta t-x\right) )^{2}}{2\left( \gamma +1\right) ^{4}\sigma ^{2}}u=0,\\ u\left( T,x\right) =1. \end{array} \right. \end{aligned}$$

(E.4)

Let $\omega =\frac{\mu -r}{\lambda }-x$, $\tilde{u}(t,\omega )=u(t,x).$ Then it yields

$$\begin{aligned} \left\{ \begin{array}{r} \frac{\partial \tilde{u}}{\partial t}-\left( r-\frac{1}{2}\sigma ^{2} +\frac{(1+2\gamma )\lambda }{\left( 1+\gamma \right) ^{2}}(\omega +\beta t)\right) \frac{\partial \tilde{u}}{\partial \omega }\\ +\frac{1}{2}\sigma ^{2} \frac{\partial ^{2}\tilde{u}}{\partial \omega ^{2}}-\frac{\left( 2\gamma +1\right) \lambda ^{2}\gamma ^{2}}{2\left( \gamma +1\right) ^{4}\sigma ^{2} }(\omega +\beta t)^{2}\tilde{u}=0,\\ \tilde{u}\left( T,\omega \right) =1. \end{array} \right. \end{aligned}$$

Set $\tilde{u}=\exp \left\{ k(T-t)+\eta (\omega +\beta t)+\theta (\omega +\beta t)^{2}\right\} V(t,\omega )$. We obtain

$$\begin{aligned} \begin{aligned}&(-k+(\frac{\sigma ^{2}}{2}-r+\beta )\eta +\frac{1}{2}\sigma ^{2}\eta ^{2} +\sigma ^{2}\theta )V+(2(\frac{\sigma ^{2}}{2}-r+\beta )\theta \\&\quad -\frac{(1+2\gamma )\lambda }{(1+\gamma )^{2}}\eta +2\sigma ^{2}\eta \theta )(\omega +\beta t)V +(2\sigma ^{2}\theta ^{2}-2\frac{\lambda (1+2\gamma )}{(1+\gamma )^{2}} \theta \\ {}&\quad -\frac{\lambda ^{2}\gamma ^{2}(1+2\gamma )}{2(1+\gamma )^{4}\sigma ^{2} })(\omega +\beta t)^{2}V+V_{t}+\frac{1}{2}\sigma ^{2}V_{\omega \omega }\\&\left( (-r+\frac{1}{2}\sigma ^{2}+\sigma ^{2}\eta )+(2\sigma ^{2}\theta -\frac{(1+2\gamma )\lambda }{(1+\gamma )^{2}})(\omega +\beta t)\right) V_{\omega }=0. \end{aligned} \end{aligned}$$

(E.5)

Let the coefficient of V be 0 by separating the variable $(\omega +\beta t)$. Then we get the following ODE systems:

$$\begin{aligned} \left\{ \begin{array}{c} -k+(\frac{\sigma ^{2}}{2}-r+\beta )\eta +\frac{1}{2}\sigma ^{2}\eta ^{2}+\sigma ^{2}\theta =0,\\ 2(\frac{\sigma ^{2}}{2}-r+\beta )\theta -\frac{(1+2\gamma )\lambda }{(1+\gamma )^{2}}\eta +2\sigma ^{2}\eta \theta =0,\\ 2\sigma ^{2}\theta ^{2}-2\frac{\lambda (1+2\gamma )}{(1+\gamma )^{2}}\theta -\frac{\lambda ^{2}\gamma ^{2}(1+2\gamma )}{2(1+\gamma )^{4}\sigma ^{2}}=0. \end{array} \right. \end{aligned}$$

Set

$$\begin{aligned} a=\frac{\sigma ^{2}}{2}-r+\beta \text {, }b\text { and }c\text { are the same as in (A.7)}. \end{aligned}$$

Then we get the ODE system the same as (A.6). Hence $\theta$, $\eta$ have the same forms with (A.8). Consequently, (E.5) is reduced to

$$\begin{aligned} \left\{ \begin{array}{c} \frac{\partial V}{\partial t}+\left( (-r+\frac{1}{2}\sigma ^{2}+\sigma ^{2} \eta )+(2\sigma ^{2}\theta -\frac{(1+2\gamma )\lambda }{(1+\gamma )^{2}} )(\omega +\beta t)\right) \frac{\partial V}{\partial \omega }+\frac{1}{2} \sigma ^{2}\frac{\partial ^{2}V}{\partial \omega ^{2}}=0,\\ V\left( T,\omega \right) =e^{-\eta (\omega +\beta T)-\theta (\omega +\beta T)^{2}}. \end{array} \right. \end{aligned}$$

Then we denote $\hat{a}:=a+\sigma ^{2}\eta =\mp \frac{ab}{q}$, $\hat{b}:=2\sigma ^{2}\theta -b=\pm q$. Hence the above PDE becomes

$$\begin{aligned} \left\{ \begin{array}{l} \frac{\partial V}{\partial t}+\left( \hat{a}-\beta +\hat{b}(\omega +\beta t)\right) \frac{\partial V}{\partial \omega }+\frac{1}{2}\sigma ^{2} \frac{\partial ^{2}V}{\partial \omega ^{2}}=0,\\ V\left( T,\omega \right) =e^{-\eta (\omega +\beta T)-\theta (\omega +\beta T)^{2}}. \end{array} \right. \end{aligned}$$

(E.6)

We assume V has the following structure:

$$\begin{aligned} V\left( t,\omega \right) =\exp \left\{ A\left( t\right) +B\left( t\right) (\omega +\beta t)+C\left( t\right) (\omega +\beta t)^{2}\right\} . \end{aligned}$$

Substituting it into (E.6), and by separating variables, we get the same ODE system as (A.11). By solving the ODE system, we get an equilibrium strategy. The proof is complete. $\square$

Appendix 6 Proof of Theorem 3.2

By the general extended HJB in Björk et al. (2017), we have,

$$\begin{aligned} \underset{\pi }{\max }\left\{ \mathcal {D}V\left( t,P,\bar{P},R\right) -\frac{\gamma }{2}\left[ \sigma ^{2}P^{2}(\frac{\partial g}{\partial P} )^{2}+\sigma ^{2}\pi ^{2}(\frac{\partial g}{\partial R})^{2}+2\sigma ^{2}\pi P\frac{\partial g}{\partial P}\frac{\partial g}{\partial R}\right] \right\}&=0,\end{aligned}$$

(F.1)

$$\begin{aligned} \mathcal {D}^{*}g\left( t,P,\bar{P},R\right)&=0, \end{aligned}$$

(F.2)

subject to $V(T,P,\bar{P},R)=R$, $g(T,P,\bar{P},R)=R$, where $\mathcal {D},\mathcal {D}^{*}$ are the infinitesimal operators³. Now we assume that V and g are separable in R, i.e., they have the following structures:

$$\begin{aligned} V(t,P,\bar{P},R))&=\tilde{V}(t,P,\bar{P})+R+r\left( T-t\right) ;\end{aligned}$$

(F.3)

$$\begin{aligned} g\left( t,P,\bar{P},R)\right)&=f\left( t,P,\bar{P}\right) +R+r\left( T-t\right) . \end{aligned}$$

(F.4)

Substituting (F.3) and (F.4) into PDE (F.1) and PDE (F.2), it yields that

$$\begin{aligned}&\underset{\pi }{\max }\left\{ \mathcal {D}\tilde{V}\left( t,P,\bar{P}\right) +\left( \mu -r+\lambda \left( \ln \bar{P}-\ln P\right) \right) \pi -\frac{1}{2}\sigma ^{2}\pi ^{2}\right. \\ {}&\quad \qquad -\left. \frac{\gamma }{2}\left[ \sigma ^{2}P^{2}(\frac{\partial f}{\partial P})^{2}+2\sigma ^{2}\pi P\frac{\partial f}{\partial P}+\sigma ^{2}\pi ^{2}\right] \right\} =0,\\&\mathcal {D}^{*}f\left( t,P,\bar{P}\right) +\left( \mu -r+\lambda \left( \ln \bar{P}-\ln P\right) \right) \hat{\pi }-\frac{1}{2}\sigma ^{2}\hat{\pi }^{2} =0, \end{aligned}$$

subject to $\tilde{V}(T,P,\bar{P})=0$, $f(T,P,\bar{P})=0$.

Similar to the baseline model in Section 2, by the first-order condition, we get an equilibrium strategy by

$$\begin{aligned} \hat{\pi }(t,P,\bar{P})=\frac{\mu -r+\lambda \left( \ln \bar{P}-\ln P\right) }{(1+\gamma )\sigma ^{2}}-\frac{\gamma P}{1+\gamma }\frac{\partial f}{\partial P}, \end{aligned}$$

where the function f satisfies the following PDE

$$\begin{aligned} \begin{aligned} \frac{\partial f}{\partial t}&+\left( \mu +\lambda \left( \ln \bar{P}-\ln P\right) -\frac{\gamma ^{2}\left( \mu -r+\lambda \left( \ln \bar{P}-\ln P\right) \right) }{\left( 1+\gamma \right) ^{2}}\right) P\frac{\partial f}{\partial P}\\ {}&+\frac{1}{2}\sigma ^{2}P^{2}\left( \frac{\partial ^{2}f}{\partial P^{2}}-\frac{\gamma ^{2}}{\left( 1+\gamma \right) ^{2}}(\frac{\partial f}{\partial P})^{2}\right) \\&+\bar{P}\left[ \frac{\ln P-\ln \bar{P}}{t}\right] \frac{\partial f}{\partial \bar{P}}+\frac{\left( 2\gamma +1\right) \left( \mu -r+\lambda \left( \ln \bar{P}-\ln P\right) \right) ^{2}}{2\left( \gamma +1\right) ^{2}\sigma ^{2}}=0, \end{aligned} \end{aligned}$$

(F.5)

subject to terminal condition $f\left( T,\bar{P},P\right) =0$.

Let $f\left( t,\bar{P},P\right) =-\frac{\left( 1+\gamma \right) ^{2}}{\gamma ^{2}}\ln u\left( t,\bar{P},P\right)$. Then PDE (F.5) becomes

$$\begin{aligned} \frac{\partial u}{\partial t}&+\left( \mu +\lambda \left( \ln \bar{P}-\ln P\right) -\frac{\gamma ^{2}\left( \mu -r+\lambda \left( \ln \bar{P}-\ln P\right) \right) }{\left( 1+\gamma \right) ^{2}}\right) P\frac{\partial f}{\partial P}+\frac{1}{2}\sigma ^{2}P^{2}\frac{\partial ^{2}u}{\partial P^{2} }\\&+\bar{P}\left[ \frac{\ln P-\ln \bar{P}}{t}\right] \frac{\partial f}{\partial \bar{P}}-\frac{\left( 2\gamma +1\right) \gamma ^{2}\left( \mu -r+\lambda \left( \ln \bar{P}-\ln P\right) \right) ^{2} }{2\left( \gamma +1\right) ^{4}\sigma ^{2}}u=0, \end{aligned}$$

with terminal condition $u\left( T,\bar{P},P\right) =1$. Let $x=\ln \bar{P}-\ln P$, $\tilde{u}(t,x):=u(t,\bar{P},P)$. Then we deduce

$$\begin{aligned} \left\{ \begin{aligned} \frac{\partial \tilde{u}}{\partial t}&-\left( \mu +\lambda x-\frac{\gamma ^{2}(\mu -r+\lambda x)}{\left( 1+\gamma \right) ^{2}}-\frac{1}{2}\sigma ^{2}+\frac{x}{t}\right) \frac{\partial \tilde{u}}{\partial x}\\&+\frac{1}{2}\sigma ^{2}\frac{\partial ^{2}\tilde{u}}{\partial x^{2}}-\frac{\left( 2\gamma +1\right) \gamma ^{2}(\mu -r+\lambda x)^{2}}{2\left( \gamma +1\right) ^{4}\sigma ^{2}}\tilde{u}=0, \quad \tilde{u}(T,x)=1. \end{aligned} \right. \end{aligned}$$

(F.6)

As in the baseline model, we set $\tilde{u}(t,x)=\exp (A(t)+B(t)x+C(t)x^{2}).$ Substituting it into PDE (F.6) and by separating variables, we obtain

$$\begin{aligned} \left\{ \begin{array}{ll} A_{t}+\hat{a}B+\frac{1}{2}\sigma ^{2}\left( B^{2}+2C\right) -\frac{\gamma ^{2}(2\gamma +1)(\mu -r)^{2}}{2(\gamma +1)^{4}\sigma ^{2}}=0\text {,} &{} A\left( T\right) =0\\ B_{t}+(\hat{b}-\frac{1}{t})B+2\hat{a}C+2\sigma ^{2}BC-\frac{\lambda \gamma ^{2}(2\gamma +1)(\mu -r)}{(\gamma +1)^{4}\sigma ^{2}}=0\text {,} &{} B\left( T\right) =0\\ C_{t}+2(\hat{b}-\frac{1}{t})C+2\sigma ^{2}C^{2}-\frac{\lambda ^{2}\gamma ^{2}(2\gamma +1)}{2(\gamma +1)^{4}\sigma ^{2}}=0\text {,} &{} C\left( T\right) =0 \end{array} \right. \end{aligned}$$

(F.7)

where we denote $\hat{a}=-(\mu -\frac{\gamma ^{2}(\mu -r)}{(1+\gamma )^{2}} -\frac{1}{2}\sigma ^{2})$, $\hat{b}=-\frac{\lambda (1+2\gamma )}{(1+\gamma )^{2}}$.

From the above transformations, it’s easy to get that

$$\begin{aligned} f(t,\bar{P},P)=-\frac{\left( 1+\gamma \right) ^{2}}{\gamma ^{2}}\left[ A(t)+B(t)(\ln \bar{P}-\ln P)+C(t)(\ln \bar{P}-\ln P)^{2}\right] , \end{aligned}$$

(F.8)

where B(t) and C(t) satisfy (F.7). Therefore,

$$\begin{aligned} \hat{\pi }(t,P,\bar{P})=\frac{\mu -r+\lambda \left( \ln \bar{P}-\ln P\right) }{(1+\gamma )\sigma ^{2}}-\frac{1+\gamma }{\gamma }[B(t)+2C(t)(\ln \bar{P}-\ln P)]. \end{aligned}$$

The proof is complete. $\square$