Mean-Expectile Portfolio Selection

Abstract

We consider a mean-expectile portfolio selection problem in a continuous-time diffusion model. We exploit the close relationship between expectiles and the Omega performance measure to reformulate the problem as the maximization of the Omega measure, and show the equivalence between the two problems. After showing that the solution for the mean-expectile problem is not attainable but that the value function is finite, we modify the problem by introducing a bound on terminal wealth and obtain the solution by Lagrangian duality. The global expectile minimizing portfolio and efficient frontier with a terminal wealth bound are also discussed.

Proof of Part (b) and Part (c) of Proposition 2

In this appendix, we provide the proof of parts (b) and (c) of Proposition 2. The proof essentially consists of a series of lemmas adapted from Section 5 of Jin et al. [15], in which a similar result is shown.

Recall that $0<x_{0}e^{rT} \le K < d$ is assumed for both parts (b) and (c). To proceed, we let $Y := Z - d$ and $y_{0} := x_{0} - de^{-rT} < 0$, so that the problem (16) can be equivalently cast as

(46)

where $Y\in {\mathcal {F}}_{T}$ means that Y is ${\mathcal {F}}_{T}$ measurable. Consider the optimization problem that relaxes the constraint on ${\mathbb {E}}[Y]$.

(47)

The following lemma is due to Cvitanić and Karatzas [11].

Lemma 6

Assuming $0<x_{0} \le Ke^{-rT}$ or equivalently $y_{0}\in (-de^{-rT},Ke^{-rT} - de^{-rT}]$, an optimal solution to problem (47) is given by

$$\begin{aligned} Y^* = Z^*-d = K{\mathbf {1}}_{\{ \beta ^*\xi _{T}\le 1 \}} - d, \end{aligned}$$

(48)

where $\beta ^* = \exp \left\{ \Vert \zeta \Vert \sqrt{T}\varPhi ^{-1}\left( 1 - \frac{y_{0}e^{rT}+d}{K} \right) + \left( r - \frac{1}{2}\Vert \zeta \Vert ^{2}\right) T \right\}$. The corresponding value function, denoted by $h(y_{0})$, is

$$\begin{aligned} h(y_{0}) = K\varPhi \left( \varPhi ^{-1}\left( 1- \frac{y_{0}e^{rT}+d}{K}\right) - \Vert \zeta \Vert \sqrt{T} \right) . \end{aligned}$$

(49)

In Lemma 6, when $x_{0} = Ke^{-rT}$, i.e. $y_{0}e^{rT} + d = K$, we have $\beta ^* = 0$, $Y^* = K - d$ and $h(y_0) = 0$, which means that the optimal solution to problem (47) is to invest only in the risk-free asset, and the optimal value is zero.

It is obvious that $h(y_{0})$ is strictly decreasing with respect to $y_{0}\in (-de^{-rT},Ke^{-rT} - de^{-rT}]$.

Lemma 7

For any sufficiently small $\varepsilon >0$ and $y_{0}\in (-de^{-rT},Ke^{-rT} - de^{-rT}]$, there exists a feasible solution Y to problem (47) such that $h(y_{0})\le {\mathbb {E}}\left[ \left( K - d - Y\right) _{+}\right] = h(y_{0}) + \frac{\varepsilon }{2}$ and ${\mathbb {E}}\left[ \xi _{T}Y\right] = y_{0}$.

Proof

For any feasible solution Y of problem (47), $h(y_{0})\le {\mathbb {E}}\left[ \left( K - d - Y\right) _{+}\right]$ by definition. Consider $Y_{\varepsilon }$ defined as follows:

$$\begin{aligned} Y_{\varepsilon }&= \left( K - d - \frac{\varepsilon }{2b{\mathbb {E}}\left[ \xi _{T}{\mathbf {1}}_{\{ \beta ^*\xi _{T}\le 1 \}}\right] }\right) {\mathbf {1}}_{\{ \beta ^*\xi _{T}\le 1 \}} \nonumber \\&\quad + \left( \frac{\varepsilon }{2b{\mathbb {E}}\left[ \xi _{T}{\mathbf {1}}_{\{ \beta ^*\xi _{T}>1 \}}\right] } - d\right) {\mathbf {1}}_{\{ \beta ^*\xi _{T}> 1 \}}, \end{aligned}$$

(50)

where $b = \frac{1}{{\mathbb {E}}\left[ \xi _{T} | \beta ^*\xi _{T} \le 1 \right] } - \frac{1}{{\mathbb {E}}\left[ \xi _{T} | \beta ^*\xi _{T} > 1 \right] }\ge 0$ and $\beta ^*$ is given in Lemma 6. For $\varepsilon >0$ small enough, $Y_{\varepsilon }\ge -d$ a.s. It can be verified that ${\mathbb {E}}[\xi _{T} Y_{\varepsilon }] = y_{0}$ and

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left[ \left( K-d-Y_{\varepsilon }\right) _{+}\right]&= {\mathbb {E}}\left[ \left( K - \frac{\varepsilon }{2b{\mathbb {E}}\left[ \xi _{T}{\mathbf {1}}_{\{ \beta ^*\xi _{T}>1 \}}\right] }\right) {\mathbf {1}}_{\{ \beta ^*\xi _{T}> 1 \}}\right] \\&\quad + {\mathbb {E}}\left[ \frac{\varepsilon }{2b{\mathbb {E}}\left[ \xi _{T}{\mathbf {1}}_{\{ \beta ^*\xi _{T} \le 1 \}}\right] }{\mathbf {1}}_{\{ \beta ^*\xi _{T}\le 1 \}}\right] \\&=h(y_{0}) + \frac{\varepsilon }{2b}\left( \frac{{\mathbb {P}}(\beta ^*\xi _{T}\le 1)}{{\mathbb {E}}\left[ \xi _{T}{\mathbf {1}}_{\{\beta ^*\xi _{T} \le 1\}} \right] } - \frac{{\mathbb {P}}(\beta ^*\xi _{T}>1)}{{\mathbb {E}}\left[ \xi _{T}{\mathbf {1}}_{\{\beta ^*\xi _{T} > 1\}}\right] } \right) \\&= h(y_{0}) + \frac{\varepsilon }{2b}b = h(y_{0})+\frac{\varepsilon }{2}. \end{aligned} \end{aligned}$$

Therefore, $Y_{\varepsilon }$ constructed in (50) meets the requirement. $\square$

The following is Lemma 5.2 in Jin et al. [15].

Lemma 8

For any $\alpha >0$, $\delta > 0$, and $0< \beta <\alpha \delta$, there exists a bounded random variable $\tilde{Y}\ge 0$ such that ${\mathbb {E}}[\tilde{Y}] = \alpha$, ${\mathbb {E}}[\xi _{T}\tilde{Y}] = \beta$ and $\tilde{Y} = 0$ on the set $\{ \xi _{T}\ge \delta \}$.

Lemma 9

For any sufficiently small $\varepsilon >0$ and $y_{0}\in (-de^{-rT},Ke^{-rT} - de^{-rT}]$, given the feasible solution $Y_{\varepsilon }$ in (50) to problem (47) such that $h(y_{0})\le {\mathbb {E}}\left[ \left( K - d - Y_{\varepsilon }\right) _{+}\right] = h(y_{0}) + \frac{\varepsilon }{2}$ and ${\mathbb {E}}\left[ \xi _{T}Y_{\varepsilon }\right] = y_{0}$, we have the following.

• (a) There exists a unique $\delta _{0}(a)$ for any $a\in (-de^{-rT}, y_{0}]$ such that

  $$\begin{aligned} {\mathbb {E}}\left[ \frac{a}{y_{0}}\xi _{T}Y_{\varepsilon }{\mathbf {1}}_{\{ \xi _{T}\ge \delta _{0}(a) \}}\right] = y_{0}. \end{aligned}$$

• (b) $\lim \limits _{a\nearrow y_{0}}\delta _{0}(a) = 0$.

• (c) There exists a $\delta _{1}(a)$ such that $0<\delta _{1}(a)<\delta _{0}(a)$ and

  $$\begin{aligned} \frac{{\mathbb {E}}\left[ \frac{a}{y_{0}}Y_{\varepsilon }{\mathbf {1}}_{\{ \xi _{T}\ge \delta _{1}(a)\}}\right] }{{\mathbb {E}}\left[ \xi _{T}\frac{a}{y_{0}}Y_{\varepsilon } {\mathbf {1}}_{\{\xi _{T}\ge \delta _{1}(a)\}}\right] -y_{0}} > \frac{1}{\delta _{1}(a)}, \end{aligned}$$

• (d) $\lim \limits _{a\nearrow y_{0}}\delta _{1}(a) = 0$.

Proof

• (a) From (50), notice that ${\mathbb {E}}\left[ \frac{a}{y_{0}}\xi _{T}Y_{\varepsilon }\right] =a$ and $Y_{\varepsilon }\le 0$ a.s. for any sufficiently small $\varepsilon >0$. Define $X_{\beta } := \frac{a}{y_{0}}\xi _{T}Y_{\varepsilon }{\mathbf {1}}_{\{ \xi _{T}\ge \beta \}}$ and $H(\beta ) := {\mathbb {E}}(X_{\beta })= {\mathbb {E}}\left[ \frac{a}{y_{0}}\xi _{T}Y_{\varepsilon }{\mathbf {1}}_{\{ \xi _{T}\ge \beta \}}\right]$ for $\beta >0$. We observe that $X_{\beta }$ increases in $\beta$ and tends to 0 and $\xi _{T}\frac{a}{y_{0}}Y_{\varepsilon }$ a.s. as $\beta$ tends to $\infty$ and 0 respectively. Furthermore, $\left| X_{\beta '}\right| \le \left| \xi _{T}\frac{a}{y_{0}}Y_{\varepsilon }\right|$ for all $\beta$, so the Dominated Convergence Theorem implies that $H(\beta )$ is continuous on $(0,\infty )$ with $\lim \limits _{\beta \rightarrow \infty }H(\beta ) = 0$ and $\lim \limits _{\beta \rightarrow 0}H(\beta ) = a<0$. The existence of $\delta _{0}(a)$ follows. Uniqueness follows from the strict monotonicity of H, since for $\beta _{1}>\beta _{2}>0$, we have

  $$\begin{aligned} H(\beta _{1}) -H(\beta _{2})= & {} {\mathbb {E}}\left[ \frac{a}{y_{0}}\xi _{T}Y_{\varepsilon }{\mathbf {1}}_{\{ \xi _{T}\ge \beta _{1} \}}\right] - {\mathbb {E}}\left[ \frac{a}{y_{0}}\xi _{T}Y_{\varepsilon }{\mathbf {1}}_{\{ \xi _{T}\ge \beta _{2} \}}\right] \\= & {} {\mathbb {E}}\left[ \frac{a}{y_{0}}\xi _{T}(-Y_{\varepsilon }){\mathbf {1}}_{\{ \beta _{2}\le \xi _{T} < \beta _{1} \}}\right] >0. \end{aligned}$$

• (b) It is clear that $\delta _{0}(y_{0}) = 0$. Continuity of $\delta _{0}(a)$ follows from the continuity and strict monotonicity of H.

• (c) Define $G(\lambda ) = {\mathbb {E}}\left[ \frac{a}{y_{0}}(-\lambda Y_{\varepsilon }){\mathbf {1}}_{\{ \xi _{T}\ge \lambda \}}\right] - \left( y_{0} - {\mathbb {E}}\left[ \xi _{T}\frac{a}{y_{0}}Y_{\varepsilon } {\mathbf {1}}_{\{\xi _{T}\ge \lambda \}}\right] \right)$ for $\lambda \in (0, \delta _{0}(a))$. The continuity of $G(\lambda )$ with respect to $\lambda$ can be proved in the same way as in part (a). Both random variables inside the corresponding expectations are integrable, with magnitudes bounded by $\left| \xi _{T}\frac{a}{y_{0}}Y_{\varepsilon }\right|$. Dominated Convergence then yields

  $$\begin{aligned} \begin{aligned} \lim \limits _{\lambda \nearrow \delta _{0}(a)}G(\lambda )&= {\mathbb {E}}\left[ \frac{a}{y_{0}} (-\delta _{0}(a) Y_{\varepsilon }){\mathbf {1}}_{\{ \xi _{T}\ge \delta _{0}(a)\}}\right] - \left( y_{0} - {\mathbb {E}}\left[ \xi _{T}\frac{a}{y_{0}}Y_{\varepsilon } {\mathbf {1}}_{\{\xi _{T}\ge \delta _{0}(a)\}}\right] \right) \\&= \delta _{0}(a) {\mathbb {E}}\left[ \frac{a}{y_{0}}(-Y_{\varepsilon }){\mathbf {1}}_{\{ \xi _{T}\ge \delta _{0}(a)\}}\right] > 0 \end{aligned} \end{aligned}$$

  where the second equality follows from part (a). The continuity of G implies that there exists a $0<\delta _{1}(a)<\delta _{0}(a)$ such that $G(\delta _{1}(a)) > 0$. Notice that for such a $\delta _{1}(a)$, we can obtain that $\delta _{1}(a) {\mathbb {E}}\left[ \frac{a}{y_{0}}(-Y_{\varepsilon }){\mathbf {1}}_{\{ \xi _{T}\ge \delta _{1}(a)\}}\right] > 0$ and $y_{0} - {\mathbb {E}}\left[ \xi _{T}\frac{a}{y_{0}}Y_{\varepsilon } {\mathbf {1}}_{\{\xi _{T}\ge \delta _{1}(a)\}}\right] > 0$, where the latter inequality follows from strict monotonicity of H from part (a). The result follows by rearranging $G(\delta _{1}(a))>0$.

• (d) With $0<\delta _{1}(a)<\delta _{0}(a)$ and $\lim \limits _{a\nearrow y_{0}}\delta _{0}(a) = 0$, the claim follows by Squeeze Theorem.

$\square$

Lemma 10

For any sufficiently small $\varepsilon >0$ and $y_{0}\in (-de^{-rT},Ke^{-rT} - de^{-rT}]$, there exists a feasible solution $Y_{\varepsilon }^{*}$ to problem (46) such that ${\mathbb {E}}\left[ \left( K-d-Y_{\varepsilon }^{*}\right) _{+}\right] <h(y_{0}) + \varepsilon$.

Proof

Using Lemma 8, we define

$$\begin{aligned} {\bar{Y}}_{a} = \frac{a}{y_{0}} Y_{\varepsilon } {\mathbf {1}}_{\{\xi _{T}\ge \delta _{1}(a)\}} + \tilde{Y}_{a}{\mathbf {1}}_{\{\xi _{T}<\delta _{1}(a)\}} \end{aligned}$$

(51)

where $Y_{\varepsilon }$ is defined in (50) and $\tilde{Y}_{a}\ge 0$ a.s. is such that $\tilde{Y}_{a}=0$ on the set $\{\xi _{T}\ge \delta _{1}(a)\}$ and

$$\begin{aligned} \left\{ \begin{aligned} {\mathbb {E}}\left[ \tilde{Y}_{a}\right]&= {\mathbb {E}}\left[ \tilde{Y}_{a}{\mathbf {1}}_{\{\xi _{T}<\delta _{1}(a)\}}\right] = -{\mathbb {E}}\left[ \frac{a}{y_{0}}Y_{\varepsilon }{\mathbf {1}}_{\{ \xi _{T}\ge \delta _{1}(a)\}}\right]>0\\ {\mathbb {E}}\left[ \xi _{T}\tilde{Y}_{a}\right]&= {\mathbb {E}}\left[ \xi _{T}\tilde{Y}_{a}{\mathbf {1}}_{\{\xi _{T}<\delta _{1}(a)\}}\right] = y_{0} - {\mathbb {E}}\left[ \xi _{T}\frac{a}{y_{0}}Y_{\varepsilon } {\mathbf {1}}_{\{\xi _{T}\ge \delta _{1}(a)\}}\right] >0 \end{aligned} \right. \end{aligned}$$

where $\delta _{1}(a)>0$ and the two inequalities follow from proof of part (c) in Lemma 9. Consequently, ${\mathbb {E}}\left[ Y_{a}\right] = 0$ and ${\mathbb {E}}\left[ \xi _{T}Y_{a}\right] = y_{0}$. For ${\bar{Y}}_{a}$, we have:

$$\begin{aligned} \begin{aligned} {\mathbb {E}}\left[ \left( K-d-{\bar{Y}}_{a}\right) _{+}\right]&= {\mathbb {E}}\left[ \left( K-d-\frac{a}{y_{0}} Y_{\varepsilon } \right) _{+} {\mathbf {1}}_{\{\xi _{T}\ge \delta _{1}(a)\}}\right] \\&\quad + {\mathbb {E}}\left[ \left( K-d-\tilde{Y}_{a}\right) _{+}{\mathbf {1}}_{\{\xi _{T}<\delta _{1}(a)\}}\right] \\&= {\mathbb {E}}\left[ \left( K-d-\frac{a}{y_{0}} Y_{\varepsilon } \right) _{+} {\mathbf {1}}_{\{\xi _{T}\ge \delta _{1}(a)\}}\right] . \end{aligned} \end{aligned}$$

since $\tilde{Y}_{a}\ge 0$ a.s. Since $\left| \left( K-d-\frac{a}{y_{0}} Y_{\varepsilon } \right) _{+} {\mathbf {1}}_{\{\xi _{T}\ge \delta _{1}(a)\}} \right| \le \left| K-d-\frac{a}{y_{0}}Y_{\varepsilon } \right| \le K + d + \frac{a}{y_{0}}|Y_{\varepsilon }|$ and $|Y_{\varepsilon }|$ is integrable from (50), the Dominated Convergence Theorem implies:

$$\begin{aligned} \lim \limits _{a\nearrow y_{0}}{\mathbb {E}}\left[ \left( K-d-{\bar{Y}}_{a}\right) _{+}\right] = {\mathbb {E}}\left[ \left( K-d-Y_{\varepsilon }\right) _{+}\right] = h(y_{0})+\frac{\varepsilon }{2} \end{aligned}$$

where the second equality is due to the definition of $Y_{\varepsilon }$ (50) in Proposition 7. Thus, we can find an $a<y_{0}$ such that ${\mathbb {E}}\left[ \left( K-d-{\bar{Y}}_{a}\right) _{+}\right] <h(y_{0}) + \varepsilon$.

$\square$

Lemma 11

Given $y_{0}<0$, for any feasible solution Y of (46), ${\mathbb {E}}\left[ \left( K-d-Y\right) _{+}\right] >h(y_{0})$.

Proof

Note that for any Y feasible for problem (46), ${\mathbb {E}}\left[ \xi _{T}Y_{+}\right] >0$, since otherwise $Y_{+} = 0$ a.s., and then ${\mathbb {E}}[Y] = 0$ implies $Y=0$ a.s., and thus ${\mathbb {E}}[\xi _{T}Y]=0 > y_{0}$, contradicting feasibility.

Let $Y_{-}=\max (-Y,0)$. Then $b:={\mathbb {E}}[\xi _{T}(-(Y_{-}))] \le y_{0} - {\mathbb {E}}[\xi _{T}Y_{+}] < y_{0}$, and $-(Y_{-}) \ge -d$. Thus ${\tilde{Y}} := -(Y_{-})$ is a feasible solution to (47), and we have:

$$\begin{aligned} {\mathbb {E}}\left[ \left( K - d - Y\right) _{+}\right] \ge {\mathbb {E}}\left[ \left( K-d - \tilde{Y}\right) _{+}\right] \ge h(b)>h(y_{0}). \end{aligned}$$

using the fact that h is strictly decreasing by Lemma 6. $\square$

Lemmas 10 and 11 yield the claims in both part (b) and part (c) of Proposition 2.