Acceptability indexes via $g$-expectations: an application to liquidity risk

Abstract

Recently, several authors focused their attention on acceptability indexes (AI) and their applications in Finance. The AI notion turns out to be quite flexible and several applications in different directions have been proposed. In particular, in Cherny and Madan (Int J Theor Appl Finance 13(8):1149–1177, 2010) illiquid markets are modeled via AI and bid and ask prices are described through a “Conic Finance” approach. A different approach of dynamic type to bid and ask prices has been suggested by Bion-Nadal (J Math Econ 45(11):738–750, 2009), taking into account both transaction costs and liquidity risk, and based on time consistent pricing procedures. The purpose of the present paper is to suggest a further link between AI and risk measures based on the notion of $g$-expectation, and by this powerful tool to fill the gap between the static description of liquidity introduced by Corcuera et al. (Int J Portfolio Anal Manag 1(1):80–91, 2012) and the dynamic description provided by Bion-Nadal (J Math Econ 45(11):738–750, 2009).

Appendix: Proof of Proposition 3

Proof of Proposition 3

(i) $\Rightarrow$ (ii): For any $x>0$ and $m\in \mathbb R$, set

$$\begin{aligned} \mathcal A _{x}^{m}\triangleq \{Y\in L^{\infty }:\alpha \left( m+Y\right) \ge x\}. \end{aligned}$$

Hence, $\mathcal A _{x}\equiv \mathcal A _{x}^{0}$. Obviously, $\alpha \left( X\right) =\sup \{x>0:X\in \mathcal A _{x}\}$.

Fix now an arbitrary $x>0$. For any $m\in \mathbb R$, the set $\mathcal A _{x}^{m}$ satisfies the following properties.

Monotonicity of $\mathcal A _{x}^{m}$. By monotonicity of $\alpha$ , it is easy to check that: (a) if $Y\in \mathcal A _{x}^{m}$ and $m^{\prime }\ge m$, then also $Y\in \mathcal A _{x}^{m^{\prime }}$; (b) if $Y\in \mathcal A _{x}^{m}$ and $Y^{\prime }\ge Y$, then also $Y^{\prime }\in \mathcal A _{x}^{m}$.

Convexity of $\mathcal A _{x}^{m}$. By quasi-concavity of $\alpha$ : if $Y_{1}$ and $Y_{2}$ belong to $\mathcal A _{x}^{m}$ (i.e. $\alpha \left( m+Y_{1}\right) \ge x$ and $\alpha \left( m+Y_{2}\right) \ge x$), then $\alpha \left( m+aY_{1}+\left( 1-a\right) Y_{2}\right) =\alpha \left( a\left( m+Y_{1}\right) +\left( 1-a\right) \left( m+Y_{2}\right) \right) \ge x$ for any $a\in \left( 0,1\right)$. Hence, also $aY_{1}+\left( 1-a\right) Y_{2}\in \mathcal A _{x}^{m}$ and convexity of any $\mathcal A _{x}^{m}$ follows.

$\lambda \mathcal A _{x}^{m}+\left( 1-\lambda \right) \mathcal A _{x}^{m^{\prime }}\subseteq \mathcal A _{x}^{\lambda m+\left( 1-\lambda \right) m^{\prime }}$ for any $m,m^{\prime }\in \mathbb R$ and $\lambda \in \left( 0,1\right)$. Take indeed any $Y\in \mathcal A _{x}^{m}$ and $Y^{\prime }\in \mathcal A _{x}^{m^{\prime }}$. By quasi-concavity of $\alpha$ it follows that $\alpha \left( \lambda m+\left( 1-\lambda \right) m^{\prime }+\lambda Y \right.\left.+\left( 1-\lambda \right) Y^{\prime }\right) =\alpha \left( \lambda \left( m+Y\right) +\left( 1-\lambda \right) \left( m^{\prime }+Y^{\prime }\right) \right) \ge x$, hence the thesis.

Right-continuity, i.e. $\mathcal A _{x}^{m}=\bigcap _{n>m}\mathcal A _{x}^{n}$. On one hand, $\mathcal A _{x}^{n}\supseteq \mathcal A _{x}^{m}$ for $n>m$ (by monotonicity of $\mathcal A _{x}^{m}$), so $\bigcap _{n>m} \mathcal A _{x}^{n}\supseteq \mathcal A _{x}^{m}$. On the other hand, for any $Y\in \bigcap _{n>m}\mathcal A _{x}^{n}$ it holds $\alpha \left( n+Y\right) \ge x$ for any $n>m$. By (CFA) of $\alpha$, it follows $\alpha \left( m+Y\right) =\inf _{n>m}\alpha \left( n+Y\right) \ge x$. Hence, $Y\in \mathcal A _{x}^{m}$ and so $\bigcap _{n>m}\mathcal A _{x}^{n}\subseteq \mathcal A _{x}^{m}$.

By Theorem 1.8 and Proposition 1.14 of Drapeau and Kupper [17] (or Proposition 4 of Föllmer and Schied [20]),

$$\begin{aligned} \rho _{x}^\mathcal{A }\left( X\right) \triangleq \inf \{m\in \mathbb R :X\in \mathcal A _{x}^{m}\}=\inf \{m\in \mathbb R :\alpha \left( m+X\right) \ge x\} \end{aligned}$$

(16)

is a convex and monotone risk measure with $\mathcal A _{\rho _\mathcal{A }}\equiv \mathcal A$. Hence

$$\begin{aligned} \alpha \left( X\right) =\inf \{x>0:X\in \mathcal A _{x}\}=\inf \{x>0:\rho _{x}\left( X\right) \le 0\}. \end{aligned}$$

It is straightforward to check that $\rho _{x}:L^{\infty }\rightarrow \mathbb R$ and that $\rho _{x}$ is increasing in $x>0$.

Continuity from above of $\rho _{x}$ . Take an arbitrary sequence $\left( X_{n}\right) _{n\in \mathbb N }\subseteq L^{\infty }$ such that $X_{n}\searrow X$. Then $\rho _{x}\left( X_{n}\right)$ is increasing in $n$ and

$$\begin{aligned} \sup _{n\in \mathbb N }\rho _{x}\left( X_{n}\right)&= \sup _{n\in \mathbb N }\inf \{m\in \mathbb R :\alpha \left( m+X_{n}\right) \ge x\} \\&= \inf \{m\in \mathbb R :\alpha \left( m+X\right) \ge x\}=\rho _{x}\left( X\right), \end{aligned}$$

since, by continuity from above of $\alpha ,\, \inf \{m\in \mathbb R :\alpha (m+X_{n})\ge x\}\nearrow _{n}\inf \{m\in \mathbb R :\alpha \left( m+X\right) \ge x\}$. Continuity from above of $\rho _{x}$ is therefore proved.

By (WEC) of $\alpha$, one deduces that for any $c\in \mathbb R$

$$\begin{aligned} \rho _{x}\left( c\right) =\inf \{m\in \mathbb R :\alpha \left( m+c\right) \ge x\}=-c, \end{aligned}$$

because $\alpha \left( m+c\right) =0$ for any $m<-c$, while $\alpha \left( m+c\right) =+\infty$ otherwise.

(ii) $\Longleftrightarrow$ (iii): Since $\rho _{x}$ is a convex, monotone and continuous from above risk measure with $\rho _{x}\left( 0\right) =0$, by Föllmer and Schied [20] and Frittelli and Rosazza Gianin [24] $\rho _{x}$ can be represented as

$$\begin{aligned} \rho _{x}\left( X\right) =\sup _{Q\in \mathcal P }\left\{ E_{Q}\left[ -X \right] -F_{x}\left( Q\right) \right\} , \end{aligned}$$

where

$$\begin{aligned} F_{x}\left( Q\right) \triangleq \sup _{Z\in L^{\infty }}\left\{ E_{Q}\left[ Z \right] -\rho _{x}\left( -Z\right) \right\} . \end{aligned}$$

Furthermore, $F_{x}\left( Q\right)$ is decreasing in $x$ (since $\rho _{x}$ is increasing in $x$).

By (16) and monotonicity of $\alpha$,

$$\begin{aligned} \alpha \left( X\right) \ge x \quad \Longleftrightarrow \quad \rho _{x}\left( X\right) \le 0\quad \Longleftrightarrow \quad E_{Q}\left[ -X \right] -F_{x}\left( Q\right) \le 0 \text{ for} \text{ any} Q\in \mathcal P . \end{aligned}$$

As a consequence, $\alpha \left( X\right) =\sup \left\{ x>0:\alpha \left( X\right) \ge x\right\} =\sup \left\{ x>0:\rho _{x}\left( X\right) \le 0\right\}$.

(ii) $\Rightarrow$ (i): By (2), it is clear that $\alpha \left( X\right) \in \left[ 0,+\infty \right]$ for any $X\in L^{\infty }$. Moreover, by definition of $\alpha$ and by monotonicity of $\rho _{x}$ in $x$:

$$\begin{aligned} \rho _{x}\left( X\right) \le 0 \quad \Longleftrightarrow \quad \alpha \left( X\right) \ge x. \end{aligned}$$

(17)

Weak expectation consistency of $\alpha$ . Since $\rho _{x}(c)=-c$ for any $c\in \mathbb R$ and $\sup \emptyset =0$ by convention, for any $c<0$ it follows $\alpha (c)=\sup \{x>0:\rho _{x}(c)\le 0\}=\sup \emptyset =0$ while $\alpha \left( 0\right) =\sup \left\{ x>0:\rho _{x}\left( 0\right) \le 0\right\} =+\infty$.

Monotonicity and Quasi-convexity of $\alpha$ follow easily from monotonicity and convexity of $\rho _{x}$.

For completeness, we will show quasi-convexity of $\alpha$. Suppose that $\alpha \left( X\right) \ge x$ and $\alpha \left( Y\right) \ge x$ (hence $X,Y\in \mathcal A _{x}$) for $x>0$. This is equivalent to $\rho _{x}\left( X\right) \le 0$ and $\rho _{x}\left( Y\right) \le 0$. By convexity of any $\rho _{x}$, it follows that $\rho _{x}\left( aX+\left( 1-a\right) Y\right) \le 0$ for any $a\in \left[ 0,1\right]$, hence $\alpha \left( aX+\left( 1-a\right) Y\right) \ge x$.

Continuity from above of $\alpha$ . Take an arbitrary sequence $\left( X_{n}\right) _{n\in \mathbb N }\subseteq L^{\infty }$ such that $X_{n}\searrow ~X$. Since $\alpha \left( X_{n}\right) \ge x$ (with $x>0$ ) if and only if $\rho _{x}\left( X_{n}\right) \le 0$ and $\rho _{x}\left( X_{n}\right) \nearrow _{n}\rho _{x}\left( X\right)$ (by (CFA) of $\rho _{x}$ ), it follows that $\alpha \left( X\right) \ge x$. For $x=0$, it is trivial to check that $\alpha \left( X_{n}\right) \ge 0$ implies $\alpha \left( X\right) \ge 0$. So, $\alpha (X_n)\ge y$ implies $\alpha (X)\ge y$ for any $y\ge 0$.

Take now $\tilde{x}=\inf _{n}\alpha \left( X_{n}\right) \ge 0$. On one hand, $\alpha (X_{n})\ge \tilde{x}$, so also $\alpha (X)\ge \tilde{x}$. On the other hand, $\alpha \left( X_{n}\right) \ge \alpha \left( X\right)$ (by (MON) of $\alpha$), hence $\alpha \left( X\right) \le \inf _{n}\alpha \left( X_{n}\right) =\tilde{x}$. So, $\alpha \left( X\right) =\inf _{n}\alpha \left( X_{n}\right)$.

Finally, the former equality in (4) follows immediately by

$$\begin{aligned} F_x(Q)=\sup _{Z \in L^{\infty }} \{E_Q [-Z] -\rho _x(Z) \}= \sup _{Z \in L^{\infty }: \rho _x(Z) \le 0} E_Q [-Z] \end{aligned}$$

(see Föllmer and Schied [20]), while the latter is due to (17).$\square$