Optimal portfolio choice for a behavioural investor in continuous-time markets

Abstract

The aim of this work consists in the study of the optimal investment strategy for a behavioural investor, whose preference towards risk is described by both a probability distortion and an S-shaped utility function. Within a continuous-time financial market framework and assuming that asset prices are modelled by semimartingales, we derive sufficient and necessary conditions for the well-posedness of the optimisation problem in the case of piecewise-power probability distortion and utility functions. Finally, under straightforwardly verifiable conditions, we further demonstrate the existence of an optimal strategy.

Appendix: Proofs

1.1 Proofs of Section 3

1.1.1 Proof of Lemma 3.6

Let the function $f:\left[\left.0,+\infty \right)\right. \rightarrow \left[\left.0,+\infty \right)\right.$ be given by $f\!\left(x\right) \text{:=} \mathbb E \!\left[X\wedge x\right]$. It is clear that $f$ is a non-decreasing function of $x$ and that $\lim _{x \rightarrow +\infty } f\!\left(x\right)=+\infty$. Indeed, to see this let $M>0$ be arbitrary. Since $\mathbb E \!\left[X\wedge n\right]\rightarrow \mathbb E \!\left[X\right]=+\infty$ by the Monotone Convergence Theorem, there exists some $n_{0} \in \mathbb N$ such that $\mathbb E \!\left[X\wedge n\right]>M$ for all $n \ge n_{0}$. Given that $f$ is non-decreasing, we conclude $f\!\left(x\right)\ge f\!\left(n_{0}\right)>M$ for all $x\ge n_{0}$, as intended.

Moreover, $f$ is a continuous function on $\left[\left.0,+\infty \right)\right.$. In fact, let $\left\{ x_{n}\right\} _{n \in \mathbb N }$ be a sequence in $\left[\left.0,+\infty \right)\right.$ converging to some $x\ge 0$. Being convergent, $\left\{ x_{n}\right\} _{n \in \mathbb N }$ is bounded, that is, there exists some $R>0$ such that $x_{n}\le R$ for all $n$. Hence, the sequence $\left\{ X_{n}\right\} _{n \in \mathbb N }$ of random variables $X_{n} \text{:=} X\wedge x_{n}$ is dominated by the integrable random variable $X\wedge R$ and converges pointwise to $X\wedge x$ (we recall that the minimum is a continuous function). Therefore, by Lebesgue’s Dominated Convergence Theorem, $f\!\left(x_{n}\right)=\mathbb E \!\left[X\wedge x_{n}\right] \xrightarrow [n\rightarrow +\infty ]{} \mathbb E \!\left[X\wedge x\right]=f\!\left(x\right)$ and $f$ is continuous as claimed.

So let us consider $b\ge 0$ arbitrary. It is obvious that $f\!\left(b\right)\le b$. Besides, since $f\!\left(x\right)\rightarrow +\infty$ as $x\rightarrow +\infty$, there exists some $N>0$ so that $f\!\left(x\right)>b$ for all $x\ge N$. Hence, because $f\!\left(b\right)\le b < f\!\left(N\right)$ and $f$ is continuous, by the Intermediate Value Theorem (note that $N>b$, otherwise $f\!\left(N\right)\le f\!\left(b\right)\le b$ by the monotonicity of $f$) we conclude that $b=f\!\left(a\right)$ for some $a \in \left[b,N\right]$.$\square$

1.1.2 Proof of Lemma  3.12

Let $t>0$ be arbitrary (but fixed). Then, for any nonnegative random variable $X$, we have by the monotonicity of the integral that

$$\begin{aligned} \int \limits _{0}^{+\infty } \mathbb P \!\left\{ X^{b}>y\right\} ^{a} dy&= \int \limits _{0}^{+\infty } \mathbb P \!\left\{ \left(X^{s}\right)^{\frac{b}{s}}>y\right\} ^{a} dy \ge \int \limits _{0}^{t^{\frac{b}{s}}} \mathbb P \!\left\{ \left(X^{s}\right)^{\frac{b}{s}}>y\right\} ^{a} dy\\&\ge \int \limits _{0}^{t^{\frac{b}{s}}} \mathbb P \!\left\{ \left(X^{s}\right)^{\frac{b}{s}}>t^{\frac{b}{s}}\right\} ^{a} dy=t^{\frac{b}{s}} \mathbb P \!\left\{ \left(X^{s}\right)^{\frac{b}{s}}>t^{\frac{b}{s}} \right\} ^{a}, \end{aligned}$$

where the last inequality follows from the inclusion $\left\{ \left(X^{s}\right)^{\frac{b}{s}}>t^{\frac{b}{s}}\right\} \subseteq \left\{ \left(X^{s}\right)^{\frac{b}{s}}>y\right\}$ for all $0 \le y \le t^{\frac{b}{s}}$. Furthermore, $\mathbb P \!\left\{ \left(X^{s}\right)^{\frac{b}{s}}>t^ {\frac{b}{s}}\right\} =\mathbb P \!\left\{ X^{s}>t\right\}$, so

$$\begin{aligned} \mathbb P \!\left\{ X^{s}>t\right\} \le \frac{1}{t^{\frac{b}{s a}}}\left(\int \limits _{0}^{+\infty } \mathbb P \!\left\{ X^{b}>y\right\} ^{a} dy\right)^{\frac{1}{a}}. \end{aligned}$$

Hence,

$$\begin{aligned} \mathbb E _\mathbb{P }\!\left[X^{s}\right]&= \int \limits _{0}^{\infty } \mathbb P \!\left\{ X^{s}>t\right\} dt\le 1+\int \limits _{1}^{+\infty } \mathbb P \!\left\{ X^{s}>t\right\} dt\\&\le 1+\left(\int \limits _{0}^{\infty } \mathbb P \!\left\{ X^{b}>y\right\} ^{a} dy\right)^{\frac{1}{a}}\int \limits _{1}^{+\infty } \frac{1}{t^{\frac{b}{s a}}} dt, \end{aligned}$$

and setting the constant $D=\int _{1}^{+\infty } \frac{1}{t^{\frac{b}{s a}}} dt \in \left(0,+\infty \right)$ (recall that $\frac{b}{s a}>1$ by hypothesis), which depends only on the parameters, completes the proof.$\square$

1.1.3 Proof of Lemma 3.13

We start by noticing that the hypothesis $\alpha <\gamma \le 1$ implies that $\frac{1}{\gamma }<\frac{1}{\alpha \gamma }$ and $\frac{1}{\gamma }<\frac{1}{\alpha }$. Moreover, since $\alpha <\beta$ and $\delta <\beta$, there exists $\eta$ so that $\max \left\{ \alpha ,\delta \right\} <\eta <\beta$. In particular, we deduce that $\frac{\eta }{\alpha }>1$, and thus $\frac{\eta }{\alpha \gamma }>\frac{1}{\gamma }$. Hence, we choose $\lambda$ so that $\frac{1}{\gamma }<\lambda <\min \left\{ \frac{1}{\alpha },\frac{\eta }{\alpha \gamma }\right\}$. Then, given that $\frac{1}{\lambda \alpha }>1$, there exists some $p$ satisfying $1<p<\frac{1}{\lambda \alpha }$. Finally, we note that $1<\frac{\eta }{\delta }$ and $\frac{\alpha \lambda \gamma }{\delta }<\frac{\eta }{\delta }$ (because $\lambda < \frac{\eta }{\alpha \gamma }$, that is, $\alpha \gamma \lambda < \eta$), so we can take $q$ such that $\max \left\{ 1,\frac{\alpha \lambda \gamma }{\delta }\right\} <q<\frac{\eta }{\delta }$.

Since for all $y\ge 1$ we have $\mathbb{P }\left\{ \left({X}^{+}\right)^{\alpha }>y\right\} \le \frac{\mathbb{E }_{\mathbb{P }}\left[\left({X}^{+}\right)^{\alpha \lambda }\right]}{{y}^{\lambda }}$ by Chebyshev’s inequality, it follows from the monotonicity of the integral that

$$\begin{aligned} \int \limits _{0}^{+\infty } \mathbb{P }\!\left\{ \left(X^{+}\right)^{\alpha }>y\right\} ^{\gamma } dy\le 1+C_{1}\, \mathbb{E }_\mathbb{P }\!\left[\left(X^{+}\right)^{\alpha \lambda }\right]^{\gamma }, \end{aligned}$$

with the strictly positive constant $C_{1}$ being given by $C_{1}\text{:=} \int _{1}^{+\infty } 1/y^{\lambda \gamma } dy$ (we recall that $\lambda \gamma >1$). Also, applying Hölder’s inequality yields

$$\begin{aligned} \mathbb E _\mathbb{P }\!\left[\left(X^{\!+\!}\right)^{\alpha \lambda }\right]\!=\!\mathbb E _\mathbb{P }\!\left[\frac{1}{\rho ^{1/p}} \rho ^{1/p} \left(X^{\!+\!}\right)^{\alpha \lambda }\right]\le C_{2}\,\mathbb E _\mathbb{P }\!\left[\rho \left(X^{\!+\!}\right)^{\alpha \lambda p}\right]^{\frac{1}{p}}\!=\!C_{2}\,\mathbb E _\mathbb{Q } \!\left[\left(X^{\!+\!}\right)^{\alpha \lambda p}\right]^{\frac{1}{p}}, \end{aligned}$$

where the constant $C_{2}\text{:=} \mathbb E _\mathbb{P }\!\left[\frac{1}{\rho ^{1/\left(p-1\right)}} \right]^{\frac{p-1}{p}}$ is finite and strictly positive. Thus, combining the previous equation and Jensen’s inequality for concave functions (we note that $\alpha \lambda p<1$), we obtain

$$\begin{aligned} \mathbb E _\mathbb{P }\!\left[\left(X^{+}\right)^{\alpha \lambda }\right]^{\gamma }&\le C_{3}\,\mathbb E _\mathbb{Q }\!\left[\left(X^{+}\right)^{\alpha \lambda p}\right]^{\frac{\gamma }{p}}\le C_{3}\,\mathbb E _\mathbb{Q }\!\left[X^{+}\right]^{\alpha \lambda \gamma }\nonumber \\&= C_{3}\left(x_{0}+\mathbb E _\mathbb{Q }\!\left[X^{-}\right] \right)^{\alpha \lambda \gamma } \le C_{4}+C_{3}\,\mathbb E _\mathbb{Q }\!\left[X^{-}\right]^{\alpha \lambda \gamma },\quad \ \ \end{aligned}$$

(7.1)

where the last inequality follows from the trivial inequality $\left|x+y\right|^{a}\le \left|x\right|^{a}+\left|y\right|^{a}$ for $0<a\le 1$ (notice that $\lambda <\frac{1}{\alpha }\le \frac{1}{\alpha \gamma }$ implies $\alpha \lambda \gamma <1$), and $C_{3},C_{4} \in \left(0,+\infty \right)$. Now, we again use Hölder’s inequality to see that $\mathbb E _\mathbb{Q }\!\left[X^{-}\right]^{\alpha \lambda \gamma }\le C_{5}\,\mathbb E _\mathbb{P }\!\left[\left(X^{-}\right)^{q}\right] ^{\frac{\alpha \lambda \gamma }{q}}$ (here $C_{5}\text{:=} \mathbb E _\mathbb{P }\!\left[\rho ^{q/\left(q-1\right)}\right]^{\alpha \lambda \gamma \left(q-1\right)/q}$ is also strictly positive and finite). Moreover, since $\frac{\alpha \lambda \gamma }{q}<\delta <1$, we have by the trivial inequality mentioned above that $\mathbb E _\mathbb{P }\!\left[\left(X^{-}\right)^{q}\right]^ {\frac{\alpha \lambda \gamma }{q}}\le 2+\mathbb E _\mathbb{P }\!\left[\left(X^{-}\right)^{q}\right]^ {\delta }$. Therefore, these inequalities combined with (7.1) yield

$$\begin{aligned} \mathbb E _\mathbb{P }\!\left[\left(X^{+}\right)^{\alpha \lambda }\right]^{\gamma }&\le C_{6}+C_{7}\,\mathbb E _\mathbb{P }\!\left[\left(X^{-}\right)^ {q}\right]^{\delta }\nonumber \\&\le C_{6}+C_{7}\left[1+D_{1}\left(\int \limits _{0}^{+\infty } \mathbb P \!\left\{ \left(X^{-}\right)^{\eta }>y\right\} ^{\delta } dy\right)^{\frac{1}{\delta }}\right]^{\delta }\nonumber \\&\le M_{1}+M_{2}\int \limits _{0}^{+\infty } \mathbb P \!\left\{ \left(X^{-}\right)^{\eta }>y\right\} ^{\delta } dy, \end{aligned}$$

(7.2)

where to obtain the second inequality we apply Lemma 3.12 above (note that $\frac{\eta }{\delta q}>1$), and the last inequality is again due to the trivial inequality we referred to previously in this proof (with $0<\delta <1$). Furthermore, all constants $C_{6},C_{7},M_{1},M_{2}$ belong to $\left(0,+\infty \right)$.

Hence,

$$\begin{aligned} \int \limits _{0}^{+\infty } \mathbb P \!\left\{ \left(X^{+}\right)^{\alpha }>y\right\} ^{\gamma } dy \le L_{1}+L_{2}\int \limits _{0}^{+\infty } \mathbb P \!\left\{ \left(X^{-}\right)^{\eta }>y\right\} ^{\delta } dy, \end{aligned}$$

with $L_{1}$ and $L_{2}$ positive constants that do not depend on the r.v. X (only on the parameters), as intended.$\square$

1.1.4 Proof of Lemma  3.14

We start by fixing some $\chi$ satisfying $\frac{1}{s}<\chi <\frac{b}{s a}$. Such a $\chi$ exists since $1<\frac{b}{a}$ implies that $\frac{1}{s}<\frac{b}{s a}$. We also note that, because $\chi a<\frac{b}{s}$, we can choose $\xi$ so that $\chi a<\xi <\frac{b}{s}$.

Let $X$ be an arbitrary nonnegative random variable. Given that $\frac{b}{s \xi }>1$, we know from Lemma 3.12 that $\mathbb E _\mathbb{P }\!\left[X^{\xi }\right]\le 1+D \left(\int _{0}^{+\infty } \mathbb P \!\left\{ X^{b}>y\right\} ^{s} dy\right)^{1/s}$ for some strictly positive finite constant $D$ (not depending on $X$, but only on the parameters). Therefore, recalling that $s\le 1$, it follows from the trivial inequality $\left|x+y\right|^{s}\le \left|x\right|^{s}+\left|y\right|^{s}$ that

$$\begin{aligned} \mathbb E _\mathbb{P }\!\left[X^{\xi }\right]^{s}\le 1+C_{1}\int \limits _{0}^{+\infty } \mathbb P \!\left\{ X^{b}>y\right\} ^{s} dy, \end{aligned}$$

(7.3)

with $C_{1} \in \left(0,+\infty \right)$. Now, by Jensen’s inequality for concave functions (note that $\frac{a \chi }{\xi }<1$), we obtain

$$\begin{aligned} \mathbb E _\mathbb{P }\!\left[X^{a \chi }\right]=\mathbb E _\mathbb{P }\!\left[\left(X^{\xi }\right) ^{\frac{a \chi }{\xi }}\right]\le \mathbb E _\mathbb{P }\!\left[X^{\xi }\right]^{\frac{a \chi }{\xi }}. \end{aligned}$$

(7.4)

Moreover, using Chebyshev’s inequality, we get

$$\begin{aligned} \int \limits _{0}^{+\infty } \mathbb P \!\left\{ X^{a}>y\right\} ^{s} dy\le 1+C_{2}\,\mathbb E _\mathbb{P }\!\left[X^{a \chi }\right]^{s}, \end{aligned}$$

(7.5)

where the strictly positive constant $C_{2}$ is given by $C_{2}\text{:=} \int _{1}^{+\infty } \frac{1}{y^{s \chi }} dy$ (note that $s \chi >1$).

Thus, combining the inequalities (7.3), (7.4) and (7.5) above yields

$$\begin{aligned} \int \limits _{0}^{+\infty } \mathbb P \!\left\{ X^{a}>y\right\} ^{s} dy&\le 1+C_{2}\left(\mathbb E _\mathbb{P }\!\left[X^{\xi }\right]^{s}\right) ^{\frac{a \chi }{\xi }}\\&\le 1+C_{2}\left(1+C_{1}\int \limits _{0}^{+\infty } \mathbb P \!\left\{ X^{b}>y\right\} ^{s} dy\right)^{\!\!\frac{a \chi }{\xi }}\\&\le R_{1}+R_{2}\left(\int \limits _{0}^{+\infty } \mathbb P \!\left\{ X^{b}>y\right\} ^{s} dy\right)^{\!\!\frac{a \chi }{\xi }} \end{aligned}$$

where the last inequality is due to the trivial inequality mentioned above (again we recall that $\frac{a \chi }{\xi }<1$), and the positive constants $R_{1}$, $R_{2}$ depend only on the parameters. Setting $\zeta =\frac{a \chi }{\xi }$ completes the proof.$\square$

1.2 Proofs of Section 4

1.2.1 Proof of Lemma 4.2

Let $\lambda >0$ be as in the proof of Lemma 3.13. We begin by showing that

$$\begin{aligned} \sup _{n \in \mathbb N } \mathbb E _\mathbb{P }\!\left[\left(X_{n}^{+}\right)^{\alpha \lambda }\right]<+\infty . \end{aligned}$$

(7.6)

Assume by contradiction that $\sup _{n \in \mathbb N } \mathbb E _\mathbb{P }\!\left[\left(X_{n}^{+}\right)^{\alpha \lambda }\right]=+\infty$. Then we can take a subsequence of $\left\{ \mathbb E _\mathbb{P }\!\left[\left(X_{n}^{+}\right)^{\alpha \lambda }\right]\right\} _{n \in \mathbb N }$ such that $\mathbb E _\mathbb{P }\!\left[\left(X_{n_{l}}^{+}\right)^{\alpha \lambda }\right] \rightarrow +\infty$ as $l \rightarrow +\infty$. By Eq. (7.2) in the proof of Lemma 3.13, we conclude that $\int _{0}^{+\infty } \mathbb P \!\left\{ \left(X_{n_{l}}^{-}\right)^{\eta }>y\right\} ^{\delta } dy \xrightarrow [l\rightarrow +\infty ]{} +\infty$, where $\eta$ is as defined in the proof. Therefore, using Lemma 3.14 we also obtain that $V_{-}\!\left(X_{n_{l}}^{-}\right)=\int _{0}^{+\infty } \mathbb P \!\left\{ \left(X_{n_{l}}^{-}\right)^{\beta }>y\right\} ^{\delta } dy \xrightarrow [l\rightarrow +\infty ]{} +\infty$, and hence

$$\begin{aligned} V\left(X_{n_{l}}\right)&= V_{+}\!\left(X_{n_{l}}^{+}\right)-V_{-}\!\left(X_{n_{l}}^{-}\right) \le L_{1}+L_{2}\int \limits _{0}^{+\infty } \mathbb P \!\left\{ \left(X_{n_{l}}^{-}\right)^{\eta }>y\right\} ^{\delta } dy-V_{-}\!\left(X_{n_{l}}^{-}\right)\\&\le C_{1}+C_{2} \left(V_{-}\!\left(X_{n_{l}}^{-}\right)\right)^{\zeta }-V_{-}\!\left(X_{n_{l}}^{-}\right) \xrightarrow [l\rightarrow +\infty ]{} -\infty , \end{aligned}$$

where the first and second inequalities follow, respectively, from Lemma 3.13 and Lemma 3.14 ($C_{1}$ and $C_{2}$ are strictly positive constants depending only on the parameters.), and $0<\zeta <1$. But this is absurd, because $\lim _{n \in \mathbb N } V\!\left(X_{n}\right)=V^{*}>-\infty$ and therefore any subsequence of $V\!\left(X_{n}\right)$ must also converge to $V^{*}$.

Now we show that (7.6) implies that $\sup _{n \in \mathbb N } V_{-}\!\left(X_{n}^{-}\right)<+\infty$. Indeed, by Chebyshev’s inequality, for some $C_{3}\in \left(0,+\infty \right)$, $V_{+}\!\left(X_{n}^{+}\right)\le 1+C_{3}\,\mathbb E _\mathbb{P }\!\left[\left(X_{n}^{+}\right)^{\alpha \lambda }\right]^{\gamma }$ for all $n$. Using (7.6), we thus get $\sup _{n\in \mathbb N } V_{+}\!\left(X_{n}^{+}\right)<+\infty$. Furthermore, since $V\!\left(X_{n}\right) \rightarrow V^{*}$ as $n \rightarrow +\infty$ and any convergent sequence is bounded, we have that $\inf _{n \in \mathbb N } V\!\left(X_{n}\right)>-\infty$. Thus,

$$\begin{aligned} \sup _{n \in \mathbb N } V_{-}\!\left(X_{n}^{-}\right)\le \sup _{n \in \mathbb N } V_{+}\!\left(X_{n}^{+}\right)-\inf _{n \in \mathbb N } V\!\left(X_{n}\right)<+\infty , \end{aligned}$$

as intended.

Finally, recalling that $\frac{\beta }{\delta }>1$, we can choose $\xi \in \left(1,\frac{\beta }{\delta }\right)$. Therefore $\frac{\beta }{\delta \xi }>1$, and it follows from Lemma 3.12 that there exists some $D\in \left(0,+\infty \right)$ such that

$$\begin{aligned} \mathbb E _\mathbb{P }\!\left[\left(X_{n}^{-}\right)^{\xi }\right]\le 1+D\left(V_{-}\!\left(X_{n}^{-}\right)\right)^{\frac{1}{\delta }}, \end{aligned}$$

for all $n \in \mathbb N$, which implies that $\sup _{n \in \mathbb N } \mathbb E _\mathbb{P }\!\left[\left(X_{n}^{-}\right)^{\xi }\right] <+\infty$. In particular, we note that, because $g:\left[\left.0,+\infty \right)\right. \rightarrow \left[\left.0,+\infty \right)\right.$ given by $g\!\left(t\right)\text{:=} t^{\xi }$ is a nonnegative, strictly increasing and strictly convex function on $\left[\left.0,+\infty \right)\right.$ satisfying $\lim _{t \rightarrow +\infty } \frac{g\!\left(t\right)}{t}=+\infty$, as well as $\sup _{n \in \mathbb N } \mathbb E _\mathbb{P }\!\left[g\!\left(X_{n}^{-}\right)\right]<+\infty$, by de la Vallée–Poussin theorem we conclude that the family $\left\{ X_{n}^{-}\right\} _{n \in \mathbb N }$ is uniformly integrable.

So we set $\tau =\alpha \lambda \in \left(0,1\right)$. A straightforward application of Hölder’s inequality (with $p=\frac{\xi }{\tau }>1$) and of the trivial inequality $\left|x+y\right|^{\tau }\le \left|x\right|^{\tau }+\left|y\right|^{\tau }$ gives

$$\begin{aligned} \mathbb E _\mathbb{P }\!\left[\left|X_{n}\right|^{\tau }\right]\le \mathbb E _\mathbb{P }\!\left[\left(X_{n}^{+}\right)^{\tau }\right] +\mathbb E _\mathbb{P }\!\left[\left(X_{n}^{-}\right)^{\tau }\right] \le \mathbb E _\mathbb{P }\!\left[\left(X_{n}^{+}\right)^{\tau }\right] +\mathbb E _\mathbb{P }^{\frac{\tau }{\xi }}\!\left[\left(X_{n}^{-} \right)^{\xi }\right], \end{aligned}$$

hence $\sup _{n \in \mathbb N } \mathbb E _\mathbb{P }\!\left[\left|X_{n}\right|^{\tau }\right] <+\infty$.$\square$

1.2.2 Proof of Corollary 4.3

Let $\epsilon >0$ be arbitrary. By Lemma 4.2 above, $\sup _{n \in \mathbb N } \mathbb{E }_\mathbb{P }\!\left[\left|X_{n}\right|^{\tau }\right]=S \in \left[\left.0,+\infty \right)\right.$, for some $\tau \in \left(0,1\right)$. So choosing $M=M\!\left(\epsilon \right)$ such that $M>\left(\frac{S}{\epsilon }\right)^{\frac{1}{\tau }}\ge 0$, and setting $K=\left[-M,M\right]$, by Chebyshev’s inequality we obtain

$$\begin{aligned} \mathbb{P }\left\{ {{X}_{n}}\in K^{c}\right\} =\mathbb{P }\left\{ \left|{X}_{n}\right|{>}M\right\} \le \frac{\mathbb{E }_\mathbb{P }\left[\left|{X}_{n}\right|^{\tau }\right]}{M^{\tau }}<\epsilon , \end{aligned}$$

which completes the proof.$\square$