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Benchmark Beating with the Increasing Convex Orderthanks: Supported by the National Key R&D Program of China (2020YFA0712700) and NSFC (12071146).

Jianming Xia RCSDS, NCMIS, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, China; Email: [email protected].
Abstract

In this paper we model benchmark beating with the increasing convex order (ICX order). The mean constraint in the mean-variance theory of portfolio selection can be regarded as beating a constant. We then investigate the problem of minimizing the variance of a portfolio with ICX order constraints, based on which we also study the problem of beating-performance-variance efficient portfolios. The optimal and efficient portfolios are all worked out in closed form for complete markets.

Keywords: portfolio selection, increasing convex order, benchmark beating, beating-performance-variance efficiency, mean-variance efficiency

1 Introduction

Portfolio selection involves a trade-off between return and risk. In contrast to the expected utility theory which models return seeking and risk aversion of an investor with the monotonicity and the concavity of the utility function, the mean-variance theory of Markowitz (1952) models the return and the risk of a portfolio with its mean and variance. Thereby the trade-off between the return and the risk is explicitly formulated in Markowitz’s mean-variance theory.

In spite of the popularity of the mean-variance theory in both academia and industry, variance has been frequently criticized as a risk measure for its main shortcoming: it treats the volatile positive returns as a part of risk. While the return is measured by mean, some alternative risk measures have been proposed to replace variance in portfolio selection, e.g., value-at-risk (VaR) (Campbell et al. (2001), Alexander and Baptista (2002)), conditional VaR (Rockafellar and Uryasev (2000)), weighted VaR (He et al. (2015)), entropic VaR (Ahmadi-Javid and Fallah-Tafti (2019)), and expectile (Wagner and Uryasev (2019), Lin et al. (2021)); see He et al. (2015) and Lin et al. (2021) for reviews on the development along this line.

Instead of the risk aversion, we focus on the other side—the return seeking—of portfolio selection in this paper. In the mean-variance theory, the return of a portfolio is measured by its mean, which, unlike variance as a risk measure, has been seldom criticized. Moreover, a mean-variance efficient portfolio can be found out by minimizing the variance under the constraint that the return is no less than a benchmark level, which is modeled by a constant. We consider an extension of the mean constraint such that the benchmark is modeled by a random variable, which is much more flexible in modeling return seeking than a constant. To this end, we use a stochastic order.

Given a random variable X0X_{0}, which is called a benchmark, we say a random variable XX beats the benchmark X0X_{0} and write XicxX0X\succeq_{\mathrm{icx}}X_{0} if XX dominates X0X_{0} in the increasing convex order (ICX order), i.e.,

E[f(X)]E[f(X0)]\mathrm{E}[f(X)]\geq\mathrm{E}[f(X_{0})] (1.1)

for all increasing convex functions ff provided that the expectations are well defined. It is well known that XicxX0X\succeq_{\mathrm{icx}}X_{0} if and only if

t1QX(s)𝑑st1QX0(s)𝑑s,t(0,1),\int_{t}^{1}Q_{X}(s)ds\geq\int_{t}^{1}Q_{X_{0}}(s)ds,\quad t\in(0,1), (1.2)

where QXQ_{X} and QX0Q_{X_{0}} are quantile functions of XX and X0X_{0}.

The financial meaning of (1.2) is clearer when X0X_{0} is discretely distributed:

(X0=ai)=pi,1in,\mathbb{P}(X_{0}=a_{i})=p_{i},\quad 1\leq i\leq n, (1.3)

where n1n\geq 1, a1<a2<<ana_{1}<a_{2}<\dots<a_{n}, pi(0,1]p_{i}\in(0,1] for all i=1,,ni=1,\dots,n, and i=1npi=1\sum_{i=1}^{n}p_{i}=1. Let

q0=0,qi=p1++pi,i=1,,n1.q_{0}=0,\quad q_{i}=p_{1}+\dots+p_{i},\quad i=1,\dots,n-1.

It is easy to see that both sizes of (1.2) are concave w.r.t. tt. Moreover, the right hand size of (1.2) is piecewise linear in the special case (1.3). Therefore, (1.2) is equivalent to

qi1QX(s)𝑑sqi1QX0𝑑s,i=0,,n1.\int_{q_{i}}^{1}Q_{X}(s)ds\geq\int_{q_{i}}^{1}Q_{X_{0}}ds,\quad i=0,\dots,n-1. (1.4)

In (1.4), the constraint with i=0i=0 amounts to E[X]E[X0]\mathrm{E}[X]\geq\mathrm{E}[X_{0}]. For each i=1,,n1i=1,\dots,n-1, qi1QX(s)𝑑s\int_{q_{i}}^{1}Q_{X}(s)ds is the average gain of XX in the best (100×qi)%(100\times q_{i})\% of cases and ignores the values of XX in the worst (100×(1qi))%(100\times(1-q_{i}))\% of cases. It is called expected upside, expected gain or tail gain; see Bacon (2013) for example.111In contrast, the expected shortfall or conditional value at risk with confidence level α\alpha is the average loss of XX in the worst 100×α%100\times\alpha\% of cases. The ratio of expected upside and expected shortfall gives the Rachev ratio in finance, see Biglova et al. (2004). Therefore (1.4) consists of a constraint on the mean of XX and some other constraints on the expected upsides of XX. The ICX order imposes not only the usual mean constraint as in the mean-variance theory but also constraints on the right tail. The constraints on the right tail makes the ICX order appropriate for modeling benchmark beating (right-tail reward control), since by beating a benchmark we expect that XX is more profitable than the benchmark X0X_{0}.

For a constant zz, XicxzX\succeq_{\mathrm{icx}}z if and only if E[X]z\mathrm{E}[X]\geq z. Therefore, the mean constraint E[X]z\mathrm{E}[X]\geq z can be regarded as beating constant zz. The aim of the present paper is to extend the mean constraint in the mean-variance theory to the ICX order constraint. More precisely, we consider the following problem:

minimizeX𝒳Var[X]subject to XicxX0,\operatorname*{minimize\,}_{X\in\mathscr{X}}\mathrm{Var}[X]\quad\text{subject to }X\succeq_{\mathrm{icx}}X_{0},

where 𝒳\mathscr{X} is a feasible set.

As initiating work, we study the problem of portfolio selection with ICX order constraint within a complete market. The original problem is generally non-convex since the set {XXicxX0}\{X\mid X\succeq_{\mathrm{icx}}X_{0}\} is generally non-convex. But the set {QXXicxX0}\{Q_{X}\mid X\succeq_{\mathrm{icx}}X_{0}\} is always convex. Then by the recently developed quantile formulation (e.g., Schied (2004), Carlier and Dana (2006), Jin and Zhou (2008), and He and Zhou (2011)), we transform the original non-convex problem into a convex one and finally provide the optimal solution in closed form. Moreover, for a payoff XX, let its performance of beating X0X_{0} be defined as

ψ(X)sup{mXmicxX0}with sup=.\psi(X)\triangleq\sup\{m\in\mathbb{R}\mid X-m\succeq_{\mathrm{icx}}X_{0}\}\quad\text{with }\sup\emptyset=-\infty.

A payoff X𝒳X\in\mathscr{X} is called beating-performance-variance (BPV) efficient in 𝒳\mathscr{X} if there is no Y𝒳Y\in\mathscr{X} such that

ψ(Y)ψ(X)andVar[Y]Var[X]\psi(Y)\geq\psi(X)\quad\text{and}\quad\mathrm{Var}[Y]\leq\mathrm{Var}[X]

with at least one inequality holding strictly. We then also investigate the problem of BPV efficient portfolio and the BPV efficient portfolios are worked out in closed form. In the special case X0=0X_{0}=0, Ψ(X)=E[X]\Psi(X)=\mathrm{E}[X] and hence BPV efficiency reduces to mean-variance efficiency.

The ICX order is closely related to another stochastic order, the increasing concave order (ICV order), which is also called the second-order stochastic dominance (SSD). A random random variable XX dominates X0X_{0} in the ICV order and write XicvX0X\succeq_{\mathrm{icv}}X_{0} if (1.1) holds for all increasing concave functions ff provided that the expectations are well defined. It is well known that XicvX0X\succeq_{\mathrm{icv}}X_{0} if and only if

0tQX(s)𝑑s0tQX0(s)𝑑s,t(0,1).\int_{0}^{t}Q_{X}(s)ds\geq\int_{0}^{t}Q_{X_{0}}(s)ds,\quad t\in(0,1).

The financial meaning of ICX order is significantly different from that of ICV order. Actually, for a constant zz, XicxzX\succeq_{\mathrm{icx}}z amounts to E[X]z\mathrm{E}[X]\geq z, whereas XicvzX\succeq_{\mathrm{icv}}z amounts to XzX\geq z a.s. Moreover, by Dentcheva and Ruszczyński (2003, Proposition 3.2), for a discretely distributed benchmark X0X_{0} satisfying (1.3), XicvX0X\succeq_{\mathrm{icv}}X_{0} if and only if

Xa1 a.s. and E[(aiX)+]E[(aiX0)+],i=2,,n.X\geq a_{1}\text{ a.s. and }\mathrm{E}[(a_{i}-X)^{+}]\leq\mathrm{E}[(a_{i}-X_{0})^{+}],\ i=2,\dots,n. (1.5)

The constraints in (1.5) pay attention to the left-tail risk of XX measured by essinfX\operatorname*{ess\,inf}X and E[(aiX)+\mathrm{E}[(a_{i}-X)^{+} (i=2,,ni=2,\dots,n). Therefore, the ICV order is appropriate for modeling risk control (left-tail risk control). In a series of papers, Dentcheva and Ruszczyński (2003, 2004, 2006a, 2006b) introduced a stochastic optimization model with ICV order constraints. As an application, they also investigated the problem of portfolio selection with risk control by ICV order instead of variance. For further developments along this line, see Rudolf and Ruszczyński (2008), Fábián et al. (2011) and Dentcheva et al. (2016). In particular, Wang and Xia (2021) investigate the problem of expected utility maximization with the ICV order constraint.

The rest of this paper is organized as follows. Sections 2 and 3 introduce the basic model and the quantile formulation, respectively. Section 4 establishes the duality between the primal and the dual problems. Section 5 investigates the dual problem and gives the dual optimizer in closed form. Section 6 studies the primal optimizer. Section 7 presents the optimal solution for the special case of two-point distributed benchmark. Section 8 discusses some problems including BPV efficient portfolios, multi-benchmark beating, and mean-variance portfolio selection with ICX order constraint, The Appendix collects some proofs.

2 Problem Formulation

Consider a complete nonatomic probability space (Ω,,)\left(\Omega,\mathcal{F},\mathbb{P}\right). Let L2L^{2} (resp., LL^{\infty}) denote all squarely integrable (resp., essentially bounded) \mathcal{F}-measurable random variables. The (upper) quantile function QXQ_{X} of a random variable XX is defined by

QX(t)inf{x(Xx)>t},t[0,1).Q_{X}(t)\triangleq\inf\{x\in\mathbb{R}\mid\mathbb{P}(X\leq x)>t\},\quad t\in[0,1).

By convention, we extend QXQ_{X} by letting QX(1)QX(1)=limt1QX(t)Q_{X}(1)\triangleq Q_{X}(1-)=\lim_{t\uparrow 1}Q_{X}(t) and QX(0)0Q_{X}(0-)\triangleq 0 if needed. For more details about quantile functions, see, e.g., Föllmer and Schied (2016, Appendix A.3).

Let XX and YY be two random variables. We say that XX dominates YY in the increasing convex order (ICX order) and write XicxYX\succeq_{\mathrm{icx}}Y if E[f(X)]E[f(Y)]\mathrm{E}[f(X)]\geq\mathrm{E}[f(Y)] for all increasing convex functions ff provided that the expectations exist. It is well known that222Throughout the paper, for t1<t2t_{1}<t_{2}, we use t1t2\int_{t_{1}}^{t_{2}} to denote the integration on the open interval (t1,t2)(t_{1},t_{2}), that is, t1t2=(t1,t2)\int_{t_{1}}^{t_{2}}=\int_{(t_{1},t_{2})}.

XicxY\displaystyle X\succeq_{\mathrm{icx}}Y\Longleftrightarrow t1QX(s)𝑑st1QY(s)𝑑s for all t[0,1]\displaystyle\int_{t}^{1}Q_{X}(s)ds\geq\int_{t}^{1}Q_{Y}(s)ds\text{ for all }t\in[0,1] (2.1)
\displaystyle\Longleftrightarrow [0,1]QX(s)𝑑w(s)[0,1]QY(s)𝑑w(s) for all w𝒲icx,\displaystyle\int_{[0,1]}Q_{X}(s)dw(s)\geq\int_{[0,1]}Q_{Y}(s)dw(s)\text{ for all }w\in\mathscr{W}^{\mathrm{icx}},

where

𝒲icx{w:[0,1][0,)w is increasing and convex with w(0)=0}.\mathscr{W}^{\mathrm{icx}}\triangleq\{w:[0,1]\to[0,\infty)\mid w\text{ is increasing and convex with }w(0)=0\}.

By convention, we set w(0)0w(0-)\triangleq 0 for w𝒲icxw\in\mathscr{W}^{\mathrm{icx}}. We can identify each w𝒲icxw\in\mathscr{W}^{\mathrm{icx}} as a finite measure on the measurable space ([0,1],[0,1])\left([0,1],\mathcal{B}_{[0,1]}\right), where [0,1]\mathcal{B}_{[0,1]} denotes all Borel subsets of [0,1][0,1]. For each w𝒲icxw\in\mathscr{W}^{\mathrm{icx}}, the measure of {0}\{0\} is 0 and the measure of {1}\{1\} is w(1)w(1)w(1)-w(1-). Furthermore, for each w𝒲icxw\in\mathscr{W}^{\mathrm{icx}} and XL2X\in L^{2}, by the convexity of ww and QXL2([0,1))Q_{X}\in L^{2}([0,1)), we know that [0,1]QX(s)𝑑w(s)=(0,1]QX(s)𝑑w(s)\int_{[0,1]}Q_{X}(s)dw(s)=\int_{(0,1]}Q_{X}(s)dw(s) is well defined and [0,1]QX(s)𝑑w(s)>\int_{[0,1]}Q_{X}(s)dw(s)>-\infty. For more details about the increasing convex order, see, e.g., Shaked and Shanthikumar (2007, Chapter 4).

Consider an arbitrage-free market. Assume that the market is complete and has a unique stochastic discount factor (SDF) ρL2\rho\in L^{2} with (ρ>0)=1\mathbb{P}(\rho>0)=1 and variance Var[ρ]>0\mathrm{Var}[\rho]>0.

Let us consider an investor with initial capital x\mathit{x}, who wants to minimize the risk measured by the payoff’s variance Var[X]\mathrm{Var}[X] while beating a benchmark payoff X0LX_{0}\in L^{\infty} in the sense of the ICX order. The problem of the investor is thus

minimizeXL2Var[X]subject to E[ρX]x,XicxX0.\begin{split}&\operatorname*{minimize\,}_{X\in L^{2}}\mathrm{Var}[X]\\ &\text{subject to }\mathrm{E}[\rho X]\leq x,\;X\succeq_{\mathrm{icx}}X_{0}.\end{split} (2.2)
Remark 2.1.

Obviously, for every zz\in\mathbb{R}, XicxzX\succeq_{\mathrm{icx}}z if and only if E[X]z\mathrm{E}[X]\geq z. Then for X0=zX_{0}=z a.s., problem (2.2) reduces to

minimizeXL2Var[X]subject to E[ρX]x,E[X]z,\begin{split}&\operatorname*{minimize\,}_{X\in L^{2}}\mathrm{Var}[X]\\ &\text{\rm subject to }\mathrm{E}[\rho X]\leq x,\;\mathrm{E}[X]\geq z,\end{split}

which arises from the classical problem of mean-variance portfolio selection. Therefore, the classical mean-variance portfolio selection is to minimize variance while beating a constant zz.

Let

𝒳icx(x,X0){XL2|E[ρX]x and XicxX0}.\mathscr{X}_{\mathrm{icx}}(x,X_{0})\triangleq\{X\in L^{2}\>|\>\mathrm{E}[\rho X]\leq x\text{ and }X\succeq_{\mathrm{icx}}X_{0}\}.

A payoff XX is called variance-minimal in 𝒳icx(x,X0)\mathscr{X}_{\mathrm{icx}}(x,X_{0}) if it solves problem (2.2).

From (2.1), the advantage of ICX order is obvious: it pays more attention to the right tail (the gain part) than to the left tail (the loss part) of a random variable. This advantage makes the ICX order appropriate for modeling benchmark beating, since by beating a benchmark we expect that the payoff XX has more profit than the benchmark X0X_{0}. The mean, however, pays equal attention to the right and the left tails.

3 Quantile Formulation

In general, the set 𝒳icx(x,X0)\mathscr{X}_{\mathrm{icx}}(x,X_{0}) is not convex, which leads to a difficulty of the problem. The appropriate technique for overcoming this difficulty is the well-developed “quantile formulation”: changing the decision variable of the problem from the random variable XX to its quantile function QXQ_{X}; see Schied (2004), Carlier and Dana (2006), Jin and Zhou (2008), and He and Zhou (2011). This formulation recovers the implicit convexity (in terms of quantile functions) of 𝒳icx(x,X0)\mathscr{X}_{\mathrm{icx}}(x,X_{0}).

Let

𝒬{Q:[0,1)|Qis increasing, right-continuous and01Q2(s)𝑑s<}.\mathscr{Q}\triangleq\left\{Q:[0,1)\rightarrow\mathbb{R}\>\bigg{|}\>Q\>\text{is increasing, right-continuous and}\>\int_{0}^{1}Q^{2}(s)ds<\infty\right\}.

Obviously, 𝒬\mathscr{Q} is the set of quantile functions of random variables XL2\mathit{X}\in L^{2}, that is,

𝒬={QX|XL2}.\mathscr{Q}=\{Q_{X}\>|\>X\in L^{2}\}.

For any Q1,Q2𝒬Q_{1},Q_{2}\in\mathscr{Q}, we write Q1icxQ2Q_{1}\succeq_{\mathrm{icx}}Q_{2} if

t1Q1(s)𝑑st1Q2(s)𝑑s for all t[0,1].\int_{t}^{1}Q_{1}(s)ds\geq\int_{t}^{1}Q_{2}(s)ds\quad\text{ for all }t\in[0,1].

Obviously,

Q1icxQ2[0,1]Q1(s)𝑑w(s)[0,1]Q2(s)𝑑w(s) for all w𝒲icx.Q_{1}\succeq_{\mathrm{icx}}Q_{2}\Longleftrightarrow\int_{[0,1]}Q_{1}(s)dw(s)\geq\int_{[0,1]}Q_{2}(s)dw(s)\text{ for all }w\in\mathscr{W}^{\mathrm{icx}}.

For any Q𝒬Q\in\mathscr{Q}, we have

QicxQ0infw𝒲icx([0,1](Q(s)Q0(s))𝑑w(s))=0,QicxQ0infw𝒲icx([0,1](Q(s)Q0(s))𝑑w(s))=.\displaystyle\begin{aligned} &Q\succeq_{\mathrm{icx}}Q_{0}\Longleftrightarrow\inf_{w\in\mathscr{W}^{\mathrm{icx}}}\left(\int_{[0,1]}(Q(s)-Q_{0}(s))dw(s)\right)=0,\\ &Q\not\succeq_{\mathrm{icx}}Q_{0}\Longleftrightarrow\inf_{w\in\mathscr{W}^{\mathrm{icx}}}\left(\int_{[0,1]}(Q(s)-Q_{0}(s))dw(s)\right)=-\infty.\end{aligned} (3.1)

For notational simplicity, we write Q0\mathit{Q_{\mathrm{0}}} instead of QX0\mathit{Q_{X_{\mathrm{0}}}}. Let

𝒬icx(x,Q0){Q𝒬|01Q(s)Qρ(1s)𝑑sxandQicxQ0}.\mathscr{Q}_{\mathrm{icx}}(x,Q_{0})\triangleq\left\{Q\in\mathscr{Q}\>\Big{|}\>\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\leq x\>\mathrm{and}\>Q\succeq_{\mathrm{icx}}Q_{0}\right\}.

Obviously, the set 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}) is convex and closed in L2([0,1))L^{2}([0,1)).

With abuse of notation, we let

E[Q]01Q(s)𝑑s and Var[Q]01Q2(s)𝑑s(01Q(s)𝑑s)2,Q𝒬.\mathrm{E}[Q]\triangleq\int_{0}^{1}Q(s)ds\ \mbox{ and }\mathrm{Var}[Q]\triangleq\int_{0}^{1}Q^{2}(s)ds-\left(\int_{0}^{1}Q(s)ds\right)^{2},\quad Q\in\mathscr{Q}.

A quantile function Q𝒬icx(x,Q0)Q\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}) is called variance-minimal in 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}) if it solves the following problem:

minimizeQ𝒬Var[Q]subject to 01Q(s)Qρ(1s)𝑑sx,QicxQ0.\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}}\mathrm{Var}[Q]\\ &\text{subject to }\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\leq x,\;Q\succeq_{\mathrm{icx}}Q_{0}.\end{split} (3.2)

Let v(x)v^{\circ}(x) denote the minimal value of problem (3.2). Obviously, vv^{\circ} is convex.

The next proposition shows that finding variance-minimal payoffs is equivalent to finding variance-minimal quantile functions.

Proposition 3.1.

If XX is variance-minimal in 𝒳icx(x,X0)\mathscr{X}_{\mathrm{icx}}(x,X_{0}), then so is QX\mathit{Q}_{X} in 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}). Conversely, if Q\mathit{Q} is variance-minimal in 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}), then so is X=Q(1ξ)\mathit{X}=Q(1-\xi) in 𝒳icx(x,X0)\mathscr{X}_{\mathrm{icx}}(x,X_{0}), where333For the existence of such a ξ\xi, see, e.g., Föllmer and Schied (2016, Lemma A.28). Moreover, let ξ\xi be a random variable uniformly distributed on (0,1)(0,1). Then by Xu (2014, Theorem 5), ξΞ\xi\in\Xi if and only if (ξ,ρ)(\xi,\rho) is comonotonic.

ξΞ{ξ|ξ is uniformly distributed on (0,1) and ρ=Qρ(ξ) a.s.}.\xi\in\Xi\triangleq\{\xi\>|\>\xi\text{ is uniformly distributed on }(0,1)\text{ and }\rho=Q_{\rho}(\xi)\text{ a.s.}\}.

Proof. It can be proved by the standard arguments in the aforementioned quantile formulation literature. \Box

Hereafter, we focus on studying variance-minimal quantile functions.

4 Duality

The following theorem implies that 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0})\neq\emptyset for any xx\in\mathbb{R}.

Theorem 4.1.

Assume that X0LX_{0}\in L^{\infty}. Then infQicxQ001Q(s)Qρ(1s)𝑑s=\inf_{Q\succeq_{\mathrm{icx}}Q_{0}}\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds=-\infty.

Proof. See Appendix B.1. \Box

The following proposition shows that the problem is trivial when Q0(1)E[ρ]xQ_{0}(1)\mathrm{E}[\rho]\leq x.

Proposition 4.2.

Assume that X0LX_{0}\in L^{\infty}. If Q0(1)E[ρ]xQ_{0}(1)\mathrm{E}[\rho]\leq x, then v(x)=0v^{\circ}(x)=0 and the variance-minimal quantile functions in 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}) are constants in the interval [Q0(1),xE[ρ]]\left[Q_{0}(1),{x\over\mathrm{E}[\rho]}\right].

Proof. Assume Q0(1)E[ρ]xQ_{0}(1)\mathrm{E}[\rho]\leq x. In this case, for every c[Q0(1),xE[ρ]]c\in\left[Q_{0}(1),{x\over\mathrm{E}[\rho]}\right], we have that c𝒬icx(x,Q0)c\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}), which implies that v(x)=0v^{\circ}(x)=0 and cc is variance-minimal in 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}). On the other hand, if QQ^{\circ} is variance-minimal in 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}), then Var[Q]=0\mathrm{Var}[Q^{\circ}]=0 and hence Qc0Q^{\circ}\equiv c_{0} for some c0c_{0}\in\mathbb{R}. Then by c0𝒬icx(x,Q0)c_{0}\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}), we have c0[Q0(1),xE[ρ]]c_{0}\in\left[Q_{0}(1),{x\over\mathrm{E}[\rho]}\right]. \Box

Hereafter, we always make the following assumption unless otherwise stated.

Assumption 4.3.

X0LX_{0}\in L^{\infty} and Q0(1)E[ρ]>xQ_{0}(1)\mathrm{E}[\rho]>x.

The next theorem establishes the existence and uniqueness of the variance-minimal solution.

Theorem 4.4.

Under Assumption 4.3, v(x)>0v^{\circ}(x)>0 and there exists a unique variance-minimal quantile function in 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}).

Proof. See Appendix B.2. \Box

It is well known that Var[Q]\mathrm{Var}[Q] is convex in QQ. By Theorem 4.1, there exists some Q𝒬Q\in\mathscr{Q} such that QicxQ0Q\succeq_{\mathrm{icx}}Q_{0} and 01Q(s)Qρ(1s)𝑑s<x\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds<x. Therefore, the Slater condition holds for the budget constraint in (3.2). Then by the standard results in convex optimization theory (see, e.g., Luenberger (1969, Sections 8.3–8.5)), we have the following proposition.

Proposition 4.5.

Assume that X0LX_{0}\in L^{\infty}. Then for every xx\in\mathbb{R}, QQ^{\circ} solves problem (3.2) if and only if there exists some λ0\lambda^{\circ}\geq 0 such that QQ^{\circ} solves the following problem

minimizeQ𝒬Var[Q]+λ01Q(s)Qρ(1s)𝑑ssubject to QicxQ0\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}}\quad\mathrm{Var}[Q]+\lambda^{\circ}\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\\ &\text{\rm subject to }\quad Q\succeq_{\mathrm{icx}}Q_{0}\end{split}

and satisfies

λ(01Q(s)Qρ(1s)𝑑sx)=0.\lambda^{\circ}\left(\int_{0}^{1}Q^{\circ}(s)Q_{\rho}(1-s)ds-x\right)=0.

If that is the case, then λv(x)-\lambda^{\circ}\in\partial v^{\circ}(x), where v\partial v^{\circ} denotes the subdifferential of vv^{\circ}.

Remark 4.6.

If Q0(1)E[ρ]>xQ_{0}(1)\mathrm{E}[\rho]>x, then λ>0\lambda^{\circ}>0. Actually, suppose on the contrary that λ=0\lambda^{\circ}=0. Then QQ^{\circ} solves the problem

minimizeQ𝒬Var[Q]subject to QicxQ0.\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}}\quad\mathrm{Var}[Q]\\ &\text{\rm subject to }\quad Q\succeq_{\mathrm{icx}}Q_{0}.\end{split}

Obviously, every constant cQ0(1)c\geq Q_{0}(1) also solves the above problem. Therefore, v(x)=Var[Q]=0v^{\circ}(x)=\mathrm{Var}[Q^{\circ}]=0, which is impossible by Theorem 4.4.

We now consider, for any fixed λ>0\lambda>0, the following problem:

minimizeQ𝒬Var[Q]+λ01Q(s)Qρ(1s)𝑑ssubject to QicxQ0,\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}}\quad\mathrm{Var}[Q]+\lambda\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\\ &\text{subject to }\quad Q\succeq_{\mathrm{icx}}Q_{0},\end{split} (4.1)

which is equivalent to

minimizeQ𝒬minβ01(Q(s)β)2𝑑s+λ01Q(s)Qρ(1s)𝑑ssubject to QicxQ0.\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}}\quad\min_{\beta\in\mathbb{R}}\int_{0}^{1}(Q(s)-\beta)^{2}ds+\lambda\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\\ &\text{subject to }\quad Q\succeq_{\mathrm{icx}}Q_{0}.\end{split} (4.2)

To solve problem (4.1) or (4.2), we first consider, for any fixed λ>0\lambda>0 and β\beta\in\mathbb{R}, the following problem:

minimizeQ𝒬01(Q(s)β)2𝑑s+λ01Q(s)Qρ(1s)𝑑ssubject to QicxQ0.\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}}\quad\int_{0}^{1}(Q(s)-\beta)^{2}ds+\lambda\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\\ &\text{subject to }\quad Q\succeq_{\mathrm{icx}}Q_{0}.\end{split} (4.3)

Let v(β,λ)v(\beta,\lambda) denote the optimal value of problem (4.3). Let

L(Q,w;β,λ)\displaystyle L(Q,w;\beta,\lambda)\triangleq 01(Q(s)β)2𝑑s+λ01Q(s)Qρ(1s)𝑑s\displaystyle\int_{0}^{1}(Q(s)-\beta)^{2}ds+\lambda\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds
([0,1]Q(s)𝑑w(s)[0,1]Q0(s)𝑑w(s)),Q𝒬,w𝒲icx.\displaystyle-\left(\int_{[0,1]}Q(s)dw(s)-\int_{[0,1]}Q_{0}(s)dw(s)\right),\quad Q\in\mathscr{Q},\;w\in\mathscr{W}^{\mathrm{icx}}.

In view of (3.1), we know that

v(β,λ)=infQ𝒬supw𝒲icxL(Q,w;β,λ).\displaystyle v(\beta,\lambda)=\inf_{Q\in\mathscr{Q}}\sup_{w\in\mathscr{W}^{\mathrm{icx}}}L(Q,w;\beta,\lambda).

Moreover, we have the following proposition.

Proposition 4.7.

Assume that X0LX_{0}\in L^{\infty}. Let Q𝒬Q^{*}\in\mathscr{Q}. Then QQ^{*} solves problem (4.3) if and only if there exists some w𝒲icxw^{*}\in\mathscr{W}^{\mathrm{icx}} such that (Q,w)(Q^{*},w^{*}) is a saddle point of L(,;β,λ)L(\cdot\;,\cdot\;;\beta,\lambda) (with respect to minimizing in QQ and maximizing in ww).

Proof. See Appendix B.3. \Box

Similarly to Lemma B.1(a), problem (4.3) has a unique optimal solution Qβ,λQ^{*}_{\beta,\lambda}. Then by Proposition 4.7, there exists some wβ,λ𝒲icxw^{*}_{\beta,\lambda}\in\mathscr{W}^{\mathrm{icx}} such that (Qβ,λ,wβ,λ)(Q^{*}_{\beta,\lambda},w^{*}_{\beta,\lambda}) is a saddle point of L(,;β,λ)L(\cdot\;,\cdot\;;\beta,\lambda). In this case, wβ,λw^{*}_{\beta,\lambda} is an optimal solution to the following dual optimization problem:

maximizew𝒲icxinfQ𝒬L(Q,w;β,λ),\operatorname*{maximize\,}_{w\in\mathscr{W}^{\mathrm{icx}}}\inf_{Q\in\mathscr{Q}}L(Q,w;\beta,\lambda), (4.4)

which will be discussed in the next section.

5 Dual Optimization

Note that each w𝒲icxw\in\mathscr{W}^{\mathrm{icx}} is continuous on [0,1)[0,1) and can be discontinuous at the right end-point 11. The next lemma shows that the dual optimizer is continuous on [0,1][0,1].

Lemma 5.1.

Assume that X0LX_{0}\in L^{\infty}. Let w𝒲icxw\in\mathscr{W}^{\mathrm{icx}}. If w(1)>w(1)w(1)>w(1-), then infQ𝒬L(Q,w;β,λ)=\inf_{Q\in\mathscr{Q}}L(Q,w;\beta,\lambda)=-\infty.

Proof. See Appendix B.4. \Box.

The next lemma shows that ww solves the dual optimization problem only if wL2([0,1))w^{\prime}\in L^{2}([0,1)).

Lemma 5.2.

Assume that X0LX_{0}\in L^{\infty}. Let w𝒲icxw\in\mathscr{W}^{\mathrm{icx}}. If w(1)=w(1)w(1)=w(1-) and wL2([0,1))w^{\prime}\notin L^{2}([0,1)), then infQ𝒬L(Q,w;β,λ)=\inf_{Q\in\mathscr{Q}}L(Q,w;\beta,\lambda)=-\infty.

Proof. See Appendix B.5. \Box

Let

𝒲c,2icx\displaystyle\mathscr{W}^{\mathrm{icx}}_{c,2}\triangleq {w𝒲icxw is continuous and wL2([0,1))}\displaystyle\{w\in\mathscr{W}^{\mathrm{icx}}\mid w\text{ is continuous and }w^{\prime}\in L^{2}([0,1))\}
=\displaystyle= {w:[0,1][0,)|w is increasing, continuous, and convex,wL2([0,1)),w(0)=0}.\displaystyle\left\{w:[0,1]\to[0,\infty)\;\left|\;\begin{aligned} &w\text{ is increasing, continuous, and convex,}\\ &w^{\prime}\in L^{2}([0,1)),w(0)=0\end{aligned}\right.\right\}.

Then by Lemmas 5.1 and 5.2, the dual optimization problem (4.4) is equivalent to

maximizew𝒲c,2icxinfQ𝒬L(Q,w;β,λ).\operatorname*{maximize\,}_{w\in\mathscr{W}^{\mathrm{icx}}_{c,2}}\inf_{Q\in\mathscr{Q}}L(Q,w;\beta,\lambda). (5.1)

To solve problem (5.1), we first consider, for any given w𝒲c,2icxw\in\mathscr{W}^{\mathrm{icx}}_{c,2}, the inner optimization problem

minimizeQ𝒬L(Q,w;β,λ).\operatorname*{minimize\,}_{Q\in\mathscr{Q}}L(Q,w;\beta,\lambda). (5.2)

The next lemma completely solves the inner optimization problem for w𝒲c,2icxw\in\mathscr{W}^{\mathrm{icx}}_{c,2}.

Lemma 5.3.

Assume that X0LX_{0}\in L^{\infty}. Let w𝒲c,2icxw\in\mathscr{W}^{\mathrm{icx}}_{c,2}. Then the optimal solution to problem (5.2) is given by

Q(s)=β+w(s)λQρ((1s))2,s[0,1).Q(s)=\beta+{w^{\prime}(s)-\lambda Q_{\rho}((1-s)-)\over 2},\quad s\in[0,1). (5.3)

Proof. For every w𝒲c,2icxw\in\mathscr{W}^{\mathrm{icx}}_{c,2} and Q𝒬Q\in\mathscr{Q}, we have

L(Q,w;β,λ)=\displaystyle L(Q,w;\beta,\lambda)= 01(Q(s)β)2𝑑s+λ01Q(s)Qρ(1s)𝑑s\displaystyle\int_{0}^{1}(Q(s)-\beta)^{2}ds+\lambda\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds
01Q(s)w(s)𝑑s+01Q0(s)w(s)𝑑s.\displaystyle-\int_{0}^{1}Q(s)w^{\prime}(s)ds+\int_{0}^{1}Q_{0}(s)w^{\prime}(s)ds.

The pointwise optimizer satisfies the following first-order condition:

2(Q(s)β)+λQρ((1s))w(s)=0,s[0,1),2(Q(s)-\beta)+\lambda Q_{\rho}((1-s)-)-w^{\prime}(s)=0,\quad s\in[0,1),

which is equivalent to (5.3). Obviously, the pointwise optimizer is a quantile function in 𝒬\mathscr{Q} and hence is a true optimal solution to problem (5.2). \Box

For every w𝒲c,2icxw\in\mathscr{W}^{\mathrm{icx}}_{c,2}, by Lemma 5.3 and by plugging (5.3) into L(Q,w;β,λ)L(Q,w;\beta,\lambda), we know that

minQ𝒬L(Q,w;β,λ)\displaystyle\min_{Q\in\mathscr{Q}}L(Q,w;\beta,\lambda)
=\displaystyle= 1401(w(s)λQρ(1s))2𝑑s+01(Q0(s)β)w(s)𝑑s+λβE[ρ]\displaystyle-{1\over 4}\int_{0}^{1}(w^{\prime}(s)-\lambda Q_{\rho}(1-s))^{2}ds+\int_{0}^{1}(Q_{0}(s)-\beta)w^{\prime}(s)ds+\lambda\beta\mathrm{E}[\rho]
=\displaystyle= 01(14(w(s))2+(12λQρ(1s)+Q0(s)β)w(s))𝑑s14λ2E[ρ2]+λβE[ρ].\displaystyle\int_{0}^{1}\left(-{1\over 4}(w^{\prime}(s))^{2}+\left({1\over 2}\lambda Q_{\rho}(1-s)+Q_{0}(s)-\beta\right)w^{\prime}(s)\right)ds-{1\over 4}\lambda^{2}\mathrm{E}[\rho^{2}]+\lambda\beta\mathrm{E}[\rho].

Therefore, the dual optimization problem (4.4) reduces to

maximizew𝒲c,2icx01(12(w(s))2(2βλQρ(1s)2Q0(s))w(s))𝑑s,\operatorname*{maximize\,}_{w\in\mathscr{W}^{\mathrm{icx}}_{c,2}}\int_{0}^{1}\left(-{1\over 2}(w^{\prime}(s))^{2}-\left(2\beta-\lambda Q_{\rho}(1-s)-2Q_{0}(s)\right)w^{\prime}(s)\right)ds,

or, equivalently,

maximizew𝒲c,2icx0112(w(s))2ds01w(s)𝑑Hβ,λ(s),\operatorname*{maximize\,}_{w\in\mathscr{W}^{\mathrm{icx}}_{c,2}}\int_{0}^{1}-{1\over 2}(w^{\prime}(s))^{2}ds-\int_{0}^{1}w^{\prime}(s)dH_{\beta,\lambda}(s), (5.4)

where

Hβ,λ(s)0s(2βλQρ(1t)2Q0(t))𝑑t,s[0,1].H_{\beta,\lambda}(s)\triangleq\int_{0}^{s}(2\beta-\lambda Q_{\rho}(1-t)-2Q_{0}(t))dt,\quad s\in[0,1]. (5.5)

Problem (5.4) can be reformulated as

maximizeG𝒬0112(G(s))2ds01G(s)𝑑Hβ,λ(s)subject to G(s)0,s[0,1).\begin{split}&\operatorname*{maximize\,}_{G\in\mathscr{Q}}\int_{0}^{1}-{1\over 2}(G(s))^{2}ds-\int_{0}^{1}G(s)dH_{\beta,\lambda}(s)\\ &\text{subject to }\quad G(s)\geq 0,s\in[0,1).\end{split} (5.6)

Problem (5.6) is similar to a problem arising in rank-dependent utility maximization. If Hβ,λH_{\beta,\lambda} is concave, then its right derivative Hβ,λH_{\beta,\lambda}^{\prime} is decreasing. In this case, the solution to problem (5.6) is given by the pointwise optimizer G(s)=(Hβ,λ(s))+G^{*}(s)=(-H_{\beta,\lambda}^{\prime}(s))^{+}. In general, Hβ,λH_{\beta,\lambda} is potentially non-concave, and therefore (Hβ,λ(s))+(-H_{\beta,\lambda}^{\prime}(s))^{+} is potentially not increasing, which violates the constraint G𝒬G\in\mathscr{Q}. Thus, the problem cannot be solved straightforwardly by pointwise optimization in general. To overcome such an obstacle, in the context of rank-dependent utility maximization and by calculus of variations, Xia and Zhou (2016) showed that the solution is given explicitly by the concave envelope. Then Xu (2016) provided the concave envelope relaxation approach to the solution, which is more straightforward. For more applications of concave envelope relaxation, see Rogers (2009), Wei (2018), and Wang and Xia (2021).

Let Hinvbreveβ,λ\invbreve{H}_{\beta,\lambda} denotes the concave envelope of Hβ,λH_{\beta,\lambda}, that is, Hinvbreveβ,λ\invbreve{H}_{\beta,\lambda} is the smallest concave function on [0,1][0,1] that is no less than Hβ,λH_{\beta,\lambda}:

Hinvbreveβ,λ(s)inf{H(s)|H is concave and HHβ,λ on [0,1]},s[0,1].\invbreve{H}_{\beta,\lambda}(s)\triangleq\inf\{H(s)\,|\,H\mbox{ is concave and }H\geq H_{\beta,\lambda}\mbox{ on }[0,1]\},\quad s\in[0,1].

By Lemma A.2,

Hinvbreveβ,λ(0)=Hβ,λ(0)=0,Hinvbreveβ,λ(1)=Hβ,λ(1)=2βλE[ρ]2E[X0],Hinvbreveβ,λH,\invbreve{H}_{\beta,\lambda}(0)=H_{\beta,\lambda}(0)=0,\ \invbreve{H}_{\beta,\lambda}(1)=H_{\beta,\lambda}(1)=2\beta-\lambda\mathrm{E}[\rho]-2\mathrm{E}[X_{0}],\ \invbreve{H}_{\beta,\lambda}\geq H,

Hinvbreveβ,λ\invbreve{H}_{\beta,\lambda} is continuous on [0,1][0,1], and the right derivative Hinvbreveβ,λ\invbreve{H}^{\prime}_{\beta,\lambda} is flat on [Hinvbreveβ,λ>Hβ,λ][\invbreve{H}_{\beta,\lambda}>H_{\beta,\lambda}].

We replace Hβ,λH_{\beta,\lambda} in (5.6) with its concave envelope Hinvbreveβ,λ\invbreve{H}_{\beta,\lambda} and consider the problem

maximizeG𝒬0112(G(s))2ds01G(s)𝑑Hinvbreveβ,λ(s)subject to G(s)0,s[0,1).\begin{split}&\operatorname*{maximize\,}_{G\in\mathscr{Q}}\int_{0}^{1}-{1\over 2}(G(s))^{2}ds-\int_{0}^{1}G(s)d\invbreve{H}_{\beta,\lambda}(s)\\ &\text{subject to }\quad G(s)\geq 0,s\in[0,1).\end{split} (5.7)

We have the following lemma for problems (5.6) and (5.7).

Lemma 5.4.

Assume that X0LX_{0}\in L^{\infty}. Then for every β\beta\in\mathbb{R} and λ>0\lambda>0, we have that H𝑖𝑛𝑣𝑏𝑟𝑒𝑣𝑒β,λ(1)>\invbreve{H}^{\prime}_{\beta,\lambda}(1-)>-\infty. Let

Gβ,λ(s)=(Hinvbreveβ,λ(s))+,s[0,1).G^{*}_{\beta,\lambda}(s)=\left(-\invbreve{H}_{\beta,\lambda}^{\prime}(s)\right)^{+},\quad s\in[0,1). (5.8)

Then Gβ,λG^{*}_{\beta,\lambda} is bounded and uniquely solves both problems (5.6) and (5.7). Moreover, problems (5.6) and (5.7) have the same optimal value.

Proof. Obviously, Hβ,λ(1)=2βλQρ(0)2Q0(1)>H^{\prime}_{\beta,\lambda}(1-)=2\beta-\lambda Q_{\rho}(0)-2Q_{0}(1-)>-\infty. Then by Lemma A.3, we have Hinvbreveβ,λ(1)>\invbreve{H}^{\prime}_{\beta,\lambda}(1-)>-\infty, which implies that Gβ,λG^{*}_{\beta,\lambda} is bounded. The proof of the rest part is standard in the concave envelope relaxation literature; see, e.g., Xu (2016), Wei (2018), and Wang and Xia (2021, Lemma 5.5). \Box

From Lemma 5.4, we have the following theorem, which provides the solution to the dual optimization problem in closed form.

Theorem 5.5.

Assume that X0LX_{0}\in L^{\infty}. Then for every β\beta\in\mathbb{R} and λ>0\lambda>0, the dual optimization problem (4.4) has a unique solution wβ,λw^{*}_{\beta,\lambda}, which satisfies

(wβ,λ)(s)=(Hinvbreveβ,λ(s))+,s[0,1),(w^{*}_{\beta,\lambda})^{\prime}(s)=\left(-\invbreve{H}_{\beta,\lambda}^{\prime}(s)\right)^{+},\quad s\in[0,1),

where Hβ,λH_{\beta,\lambda} is defined in (5.5) and H𝑖𝑛𝑣𝑏𝑟𝑒𝑣𝑒β,λ\invbreve{H}_{\beta,\lambda} is the concave envelope of Hβ,λH_{\beta,\lambda}. Moreover, (wβ,λ)(w^{*}_{\beta,\lambda})^{\prime} is bounded and wβ,λ𝒲c,2icxw^{*}_{\beta,\lambda}\in\mathscr{W}^{\mathrm{icx}}_{c,2}.

6 Primal Optimizer

The next theorem provides, for every (β,λ)(\beta,\lambda), the optimal solution to problem (4.3) in closed form, which is an immediate consequence of combining Theorem 5.5, Lemma 5.3, and Proposition 4.7.

Theorem 6.1.

Assume that X0LX_{0}\in L^{\infty}. For every β\beta\in\mathbb{R} and λ>0\lambda>0, let

Qβ,λ(s)β+(Hinvbreveβ,λ(s))+λQρ((1s))2,s[0,1),Q^{*}_{\beta,\lambda}(s)\triangleq\beta+{\left(-\invbreve{H}_{\beta,\lambda}^{\prime}(s)\right)^{+}-\lambda Q_{\rho}((1-s)-)\over 2},\quad s\in[0,1),

where Hβ,λH_{\beta,\lambda} is defined in (5.5) and H𝑖𝑛𝑣𝑏𝑟𝑒𝑣𝑒β,λ\invbreve{H}_{\beta,\lambda} is the concave envelope of Hβ,λH_{\beta,\lambda}. Then Qβ,λQ^{*}_{\beta,\lambda} is the unique solution to problem (4.3).

Recall that, for any β\beta and λ\lambda,

Hβ,λ(s)=2βsNλ(s),s[0,1],H_{\beta,\lambda}(s)=2\beta s-N_{\lambda}(s),\quad s\in[0,1],

where

Nλ(s)0s(λQρ(1t)+2Q0(t))𝑑t,s[0,1].N_{\lambda}(s)\triangleq\int_{0}^{s}(\lambda Q_{\rho}(1-t)+2Q_{0}(t))dt,\quad s\in[0,1].

Let N˘λ\breve{N}_{\lambda} denote the convex envelope of NλN_{\lambda}. Then

Hinvbreveβ,λ(s)=2βsN˘λ(s),s[0,1]\invbreve{H}_{\beta,\lambda}(s)=2\beta s-\breve{N}_{\lambda}(s),\quad s\in[0,1]

and

Hinvbreveβ,λ=2βN˘λ.\invbreve{H}^{\prime}_{\beta,\lambda}=2\beta-\breve{N}^{\prime}_{\lambda}.

Therefore,

Qβ,λ(s)=β+(N˘λ(s)2β)+λQρ((1s))2,s[0,1),Q^{*}_{\beta,\lambda}(s)=\beta+{\left(\breve{N}_{\lambda}^{\prime}(s)-2\beta\right)^{+}-\lambda Q_{\rho}((1-s)-)\over 2},\quad s\in[0,1), (6.1)

By Theorem 6.1,

v(β,λ)=\displaystyle v(\beta,\lambda)= 01(Qβ,λ(s)β)2𝑑s+λ01Qβ,λ(s)Qρ(1s)𝑑s\displaystyle\int_{0}^{1}(Q^{*}_{\beta,\lambda}(s)-\beta)^{2}ds+\lambda\int_{0}^{1}Q^{*}_{\beta,\lambda}(s)Q_{\rho}(1-s)ds
=\displaystyle= 1401((N˘λ(s)2β)+)2𝑑s+λβE[ρ]λ24E[ρ2].\displaystyle{1\over 4}\int_{0}^{1}\left(\left(\breve{N}_{\lambda}^{\prime}(s)-2\beta\right)^{+}\right)^{2}ds+\lambda\beta\mathrm{E}[\rho]-{\lambda^{2}\over 4}\mathrm{E}[\rho^{2}].

The function x(x+)2x\mapsto(x^{+})^{2} is convex and continuously differentiable. So is vv with respect to β\beta and

v(β,λ)β=λE[ρ]01(N˘λ(s)2β)+𝑑sh(β,λ).{\partial v(\beta,\lambda)\over\partial\beta}=\lambda\mathrm{E}[\rho]-\int_{0}^{1}\left(\breve{N}_{\lambda}^{\prime}(s)-2\beta\right)^{+}ds\triangleq h(\beta,\lambda). (6.2)

Finally, we can find the variance-minimal quantile function according to the following two steps.

(i)

For any fixed λ>0\lambda>0, consider the problem

minimizeβv(β,λ).\operatorname*{minimize\,}_{\beta\in\mathbb{R}}v(\beta,\lambda).

The minimizer βλ\beta_{\lambda} is determined by h(βλ,λ)=0h(\beta_{\lambda},\lambda)=0, where hh is given by (6.2). Let QλQβλ,λQ^{*}_{\lambda}\triangleq Q^{*}_{\beta_{\lambda},\lambda}. Then QλQ^{*}_{\lambda} solves problem (4.1)/(4.2).

(ii)

Let

𝒳(λ)01Qλ(s)Qρ(1s)𝑑s,λ(0,).\displaystyle\mathcal{X}(\lambda)\triangleq\int_{0}^{1}Q^{*}_{\lambda}(s)Q_{\rho}(1-s)ds,\quad\lambda\in(0,\infty). (6.3)

By Proposition 4.5, 𝒳\mathcal{X} is decreasing on (0,)(0,\infty). Moreover, a combination of Theorem 4.4, Proposition 4.5 and Remark 4.6 implies that, for any x<Q0(1)E[ρ]x<Q_{0}(1)\mathrm{E}[\rho], 𝒳(λ)=x\mathcal{X}(\lambda^{\circ})=x for some λ(0,)\lambda^{\circ}\in(0,\infty). Therefore, 𝒳\mathcal{X} is continuous on (0,)(0,\infty), limλ0𝒳(λ)=Q0(1)E[ρ]\lim_{\lambda\downarrow 0}\mathcal{X}(\lambda)=Q_{0}(1)\mathrm{E}[\rho], and limλ𝒳(λ)=\lim_{\lambda\uparrow\infty}\mathcal{X}(\lambda)=-\infty. The monotonicity and continuity of 𝒳\mathcal{X} makes it easy to search for the desired Lagrange multiplier λ\lambda^{\circ} for any given budget level x<Q0(1)E[ρ]x<Q_{0}(1)\mathrm{E}[\rho] by solving the equation 𝒳(λ)=x\mathcal{X}(\lambda)=x. Once λ\lambda^{\circ} has been determined, then Q=QλQ^{\circ}=Q^{*}_{\lambda^{\circ}} is the desired variance-minimal quantile function in 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}).

7 Special Case

7.1 The Classical Case: Constant Benchmark

Now we consider the classical case when X0=zX_{0}=z a.s. for a constant z>xE[ρ]z>{x\over\mathrm{E}[\rho]}. In this case,

Nλ(s)=λ0sQρ(1s)𝑑s+2zs,s[0,1].N_{\lambda}(s)=\lambda\int_{0}^{s}Q_{\rho}(1-s)ds+2zs,\quad s\in[0,1].

Obviously, NλN_{\lambda} is concave and hence

N˘Nλ(1)=λE[ρ]+2z.\breve{N}^{\prime}\equiv N_{\lambda}(1)=\lambda\mathrm{E}[\rho]+2z.

Then

h(β,λ)={2β2zif 2β<λE[ρ]+2z,λE[ρ]if 2βλE[ρ]+2z.h(\beta,\lambda)=\begin{cases}2\beta-2z&\text{if }2\beta<\lambda\mathrm{E}[\rho]+2z,\\ \lambda\mathrm{E}[\rho]&\text{if }2\beta\geq\lambda\mathrm{E}[\rho]+2z.\end{cases}

For any λ>0\lambda>0, we have βλ=z\beta_{\lambda}=z and hence

Qλ(s)=z+λ2E[ρ]λ2Qρ((1s)),s[0,1).Q^{*}_{\lambda}(s)=z+{\lambda\over 2}\mathrm{E}[\rho]-{\lambda\over 2}Q_{\rho}((1-s)-),\quad s\in[0,1).

It is easy to see that

𝒳(λ)=zE[ρ]λ2Var[ρ].\mathcal{X}(\lambda)=z\mathrm{E}[\rho]-{\lambda\over 2}\mathrm{Var}[\rho].

Therefore, λ2=zE[ρ]xVar[ρ]{\lambda^{\circ}\over 2}={z\mathrm{E}[\rho]-x\over\mathrm{Var}[\rho]} and hence

Q(s)=z+zE[ρ]xVar[ρ]E[ρ]zE[ρ]xVar[ρ]Qρ((1s)),s[0,1).Q^{\circ}(s)=z+{z\mathrm{E}[\rho]-x\over\mathrm{Var}[\rho]}\mathrm{E}[\rho]-{z\mathrm{E}[\rho]-x\over\mathrm{Var}[\rho]}Q_{\rho}((1-s)-),\quad s\in[0,1).

7.2 Two-Point Distributed Benchmark

We now consider the special case when X0X_{0} is discretely distributed:

(X0=a)=p=1(X0=b),\mathbb{P}(X_{0}=a)=p=1-\mathbb{P}(X_{0}=b), (7.1)

where aba\leq b and p(0,1)p\in(0,1). In this case,

Q0(s)=a𝟏s<p+b𝟏sp,s[0,1).Q_{0}(s)=a\mathbf{1}_{s<p}+b\mathbf{1}_{s\geq p},\quad s\in[0,1).

In Assumption 4.3, condition Q0(1)E[ρ]>xQ_{0}(1)\mathrm{E}[\rho]>x amounts to bE[ρ]>xb\mathrm{E}[\rho]>x.

We will use the following notation:

A11p0pQρ(1s)𝑑s,A211pp1Qρ(1s)𝑑s.A_{1}\triangleq{1\over p}\int_{0}^{p}Q_{\rho}(1-s)ds,\quad A_{2}\triangleq{1\over 1-p}\int_{p}^{1}Q_{\rho}(1-s)ds.

Obviously, A1>A2A_{1}>A_{2}.

The next proposition explicitly provides the variance-minimal quantile function.

Proposition 7.1.

For a benchmark X0X_{0} that satisfies (7.1), the variance-minimal quantile function QQ^{\circ} is given as follows.

(a)

If xE[X0]E[ρ]baA1A2Var[ρ]x\leq\mathrm{E}[X_{0}]\mathrm{E}[\rho]-{b-a\over A_{1}-A_{2}}\mathrm{Var}[\rho], then

Q(s)=E[X0]+E[X0]E[ρ]xVar[ρ]E[ρ]E[X0]E[ρ]xVar[ρ]Qρ((1s)),s[0,1).Q^{\circ}(s)=\mathrm{E}[X_{0}]+{\mathrm{E}[X_{0}]\mathrm{E}[\rho]-x\over\mathrm{Var}[\rho]}\mathrm{E}[\rho]-{\mathrm{E}[X_{0}]\mathrm{E}[\rho]-x\over\mathrm{Var}[\rho]}Q_{\rho}((1-s)-),\quad s\in[0,1).
(b)

If E[X0]E[ρ]baA1A2Var[ρ]<xbE[ρ]baA1((1p)E[ρ2]+p2A12(1p)2A22)\mathrm{E}[X_{0}]\mathrm{E}[\rho]-{b-a\over A_{1}-A_{2}}\mathrm{Var}[\rho]<x\leq b\mathrm{E}[\rho]-{b-a\over A_{1}}\left((1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}\right), then

Q(s)={a+apA1+b(1p)A2xE[ρ2]pA12(1p)A22A1apA1+b(1p)A2xE[ρ2]pA12(1p)A22Qρ((1s))if s[0,p),b+apA1+b(1p)A2xE[ρ2]pA12(1p)A22A2apA1+b(1p)A2xE[ρ2]pA12(1p)A22Qρ((1s))if s[p,1).Q^{\circ}(s)=\begin{cases}a+{apA_{1}+b(1-p)A_{2}-x\over\mathrm{E}[\rho^{2}]-pA_{1}^{2}-(1-p)A_{2}^{2}}A_{1}-{apA_{1}+b(1-p)A_{2}-x\over\mathrm{E}[\rho^{2}]-pA_{1}^{2}-(1-p)A_{2}^{2}}Q_{\rho}((1-s)-)&\text{if }s\in[0,p),\\ b+{apA_{1}+b(1-p)A_{2}-x\over\mathrm{E}[\rho^{2}]-pA_{1}^{2}-(1-p)A_{2}^{2}}A_{2}-{apA_{1}+b(1-p)A_{2}-x\over\mathrm{E}[\rho^{2}]-pA_{1}^{2}-(1-p)A_{2}^{2}}Q_{\rho}((1-s)-)&\text{if }s\in[p,1).\\ \end{cases}
(c)

If bE[ρ]baA1((1p)E[ρ2]+p2A12(1p)2A22)<x<bEρ]b\mathrm{E}[\rho]-{b-a\over A_{1}}\left((1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}\right)<x<b\mathrm{E}\rho], then

Q(s)={bp(bE[ρ]x)(1p)E[ρ2]+p2A12(1p)2A22A1(1p)(bE[ρ]x)(1p)E[ρ2]+p2A12(1p)2A22Qρ((1s))if s[0,p),b+(1p)(bE[ρ]x)(1p)E[ρ2]+p2A12(1p)2A22A2(1p)(bE[ρ]x)(1p)E[ρ2]+p2A12(1p)2A22Qρ((1s))if s[p,1).Q^{\circ}(s)=\begin{cases}b-{p(b\mathrm{E}[\rho]-x)\over(1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}}A_{1}-{(1-p)(b\mathrm{E}[\rho]-x)\over(1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}}Q_{\rho}((1-s)-)\\ \hskip 142.26378pt\text{if }s\in[0,p),\\ b+{(1-p)(b\mathrm{E}[\rho]-x)\over(1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}}A_{2}-{(1-p)(b\mathrm{E}[\rho]-x)\over(1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}}Q_{\rho}((1-s)-)\\ \hskip 142.26378pt\text{if }s\in[p,1).\\ \end{cases}

Proof. See Appendix B.6. \Box

We can reformulate Proposition 7.1 as follows.

Proposition 7.2.

For a benchmark X0X_{0} that satisfies (7.1), the variance-minimal payoff XX^{\circ} is given as follows.

(a)

If xE[X0]E[ρ]baA1A2Var[ρ]x\leq\mathrm{E}[X_{0}]\mathrm{E}[\rho]-{b-a\over A_{1}-A_{2}}\mathrm{Var}[\rho], then

X=E[X0]+E[X0]E[ρ]xVar[ρ]E[ρ]E[X0]E[ρ]xVar[ρ]ρ a.s.X^{\circ}=\mathrm{E}[X_{0}]+{\mathrm{E}[X_{0}]\mathrm{E}[\rho]-x\over\mathrm{Var}[\rho]}\mathrm{E}[\rho]-{\mathrm{E}[X_{0}]\mathrm{E}[\rho]-x\over\mathrm{Var}[\rho]}\rho\text{ a.s.}
(b)

If E[X0]E[ρ]baA1A2Var[ρ]<xbE[ρ]baA1((1p)E[ρ2]+p2A12(1p)2A22)\mathrm{E}[X_{0}]\mathrm{E}[\rho]-{b-a\over A_{1}-A_{2}}\mathrm{Var}[\rho]<x\leq b\mathrm{E}[\rho]-{b-a\over A_{1}}\left((1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}\right), then

X={a+apA1+b(1p)A2xE[ρ2]pA12(1p)A22A1apA1+b(1p)A2xE[ρ2]pA12(1p)A22ρif ρ>Qρ((1p)),b+apA1+b(1p)A2xE[ρ2]pA12(1p)A22A2apA1+b(1p)A2xE[ρ2]pA12(1p)A22ρif ρQρ((1p)).X^{\circ}=\begin{cases}a+{apA_{1}+b(1-p)A_{2}-x\over\mathrm{E}[\rho^{2}]-pA_{1}^{2}-(1-p)A_{2}^{2}}A_{1}-{apA_{1}+b(1-p)A_{2}-x\over\mathrm{E}[\rho^{2}]-pA_{1}^{2}-(1-p)A_{2}^{2}}\rho&\text{if }\rho>Q_{\rho}((1-p)-),\\ b+{apA_{1}+b(1-p)A_{2}-x\over\mathrm{E}[\rho^{2}]-pA_{1}^{2}-(1-p)A_{2}^{2}}A_{2}-{apA_{1}+b(1-p)A_{2}-x\over\mathrm{E}[\rho^{2}]-pA_{1}^{2}-(1-p)A_{2}^{2}}\rho&\text{if }\rho\leq Q_{\rho}((1-p)-).\\ \end{cases}
(c)

If bE[ρ]baA1((1p)E[ρ2]+p2A12(1p)2A22)<x<bEρ]b\mathrm{E}[\rho]-{b-a\over A_{1}}\left((1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}\right)<x<b\mathrm{E}\rho], then

X={bp(bE[ρ]x)(1p)E[ρ2]+p2A12(1p)2A22A1(1p)(bE[ρ]x)(1p)E[ρ2]+p2A12(1p)2A22ρif ρ>Qρ((1p)),b+(1p)(bE[ρ]x)(1p)E[ρ2]+p2A12(1p)2A22A2(1p)(bE[ρ]x)(1p)E[ρ2]+p2A12(1p)2A22ρif ρQρ((1p)).X^{\circ}=\begin{cases}b-{p(b\mathrm{E}[\rho]-x)\over(1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}}A_{1}-{(1-p)(b\mathrm{E}[\rho]-x)\over(1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}}\rho\\ \hskip 142.26378pt\text{if }\rho>Q_{\rho}((1-p)-),\\ b+{(1-p)(b\mathrm{E}[\rho]-x)\over(1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}}A_{2}-{(1-p)(b\mathrm{E}[\rho]-x)\over(1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}}\rho\\ \hskip 142.26378pt\text{if }\rho\leq Q_{\rho}((1-p)-).\\ \end{cases}
Example 7.3.

The SDF ρ\rho is log-normal: logρN(μ,σ2)\log\rho\sim N(\mu,\sigma^{2}) with μ=0.1\mu=-0.1 and σ=0.34\sigma=0.34. The initial capital x=1.0x=1.0.

(a)

The benchmark payoff X0X_{0} satisfies (X0=1.1δ)=(X0=1.1+δ)=0.5\mathbb{P}(X_{0}=1.1-\delta)=\mathbb{P}(X_{0}=1.1+\delta)=0.5 with δ=0.10,0.15,0.20,0.40\delta=0.10,0.15,0.20,0.40. The variance-minimal payoffs XX^{\circ} v.s. SDF ρ\rho are plotted in Figure 1.

(b)

The benchmark payoff X0X_{0} satisfies (X0=α0.30)=(X0=α+0.30)=0.5\mathbb{P}(X_{0}=\alpha-0.30)=\mathbb{P}(X_{0}=\alpha+0.30)=0.5 with α=0.7432,1.0,1.15,1.20\alpha=0.7432,1.0,1.15,1.20. The variance-minimal payoffs XX^{\circ} v.s. SDF ρ\rho are plotted in Figure 2. Here, α=0.7432\alpha=0.7432 is the solution of Q0(1)E[ρ]=(α+0.30)E[ρ]=xQ_{0}(1)\mathrm{E}[\rho]=(\alpha+0.30)\mathrm{E}[\rho]=x; in this case, XX^{\circ} is a constant.

Refer to caption
Figure 1: Variance-minimal payoffs v.s. SDF
Refer to caption
Figure 2: Variance-minimal payoffs v.s. SDF

8 Discussions

8.1 Beating-Performance-Variance Efficient Payoffs

Consider a given benchmark payoff X0LX_{0}\in L^{\infty}. For any payoff XL2X\in L^{2}, its performance of beating X0X_{0} is444 The benchmark-beating performance is similar to but different from the SSD-based risk measure of Fábián et al. (2011), which is also called benchmark-based expected shortfall as a special case of adjusted expected shortfall in Burzoni et al. (2022). Given a benchmark payoff X0X_{0}, the SSD-based risk measure or the benchmark-adjusted expected shortfall of a payoff XX is (X)inf{mXmicvX0}\mathcal{R}(X)\triangleq\inf\{m\in\mathbb{R}\mid X-m\succeq_{\mathrm{icv}}X_{0}\}.

ψ(X)sup{mXmicxX0}with sup=.\psi(X)\triangleq\sup\{m\in\mathbb{R}\mid X-m\succeq_{\mathrm{icx}}X_{0}\}\quad\text{with }\sup\emptyset=-\infty.

Obviously, the beating performance ψ\psi satisfies the following properties.

  • Monotonicity: ψ(X)ψ(Y)\psi(X)\geq\psi(Y) if XYX\geq Y a.s.

  • Translation Invariance: ψ(X+c)=ψ(X)+c\psi(X+c)=\psi(X)+c for all XL2X\in L^{2} and cc\in\mathbb{R}.

  • Law Invariance: ψ(X)=ψ(Y)\psi(X)=\psi(Y) if XX and YY are identically distributed.

Moreover, ψ\psi has the following representation:

ψ(X)=inft(0,1)(1t1t1(QX(s)Q0(s))𝑑s),XL2.\displaystyle\psi(X)=\inf_{t\in(0,1)}\left({1\over t}\int_{1-t}^{1}(Q_{X}(s)-Q_{0}(s))ds\right),\quad X\in L^{2}.

Clearly, ψ(X)[,)\psi(X)\in[-\infty,\infty) for all XL2X\in L^{2} and

ψ(c)=cQ0(1)for all c.\psi(c)=c-Q_{0}(1)\quad\text{for all }c\in\mathbb{R}.

It is easy to see that, for all XL2X\in L^{2} and zz\in\mathbb{R},

ψ(X)zXicxX0+z.\psi(X)\geq z\Longleftrightarrow X\succeq_{\mathrm{icx}}X_{0}+z.

In particular, ψ(X)0\psi(X)\geq 0 if and only if XicxX0X\succeq_{\mathrm{icx}}X_{0}.

For any budget level xx\in\mathbb{R}, let

𝒳(x){XL2E[ρX]x}.\mathscr{X}(x)\triangleq\{X\in L^{2}\mid\mathrm{E}[\rho X]\leq x\}.
Definition 8.1.

A payoff X𝒳(x)X\in\mathscr{X}(x) is called beating-performance-variance (BPV) efficient in 𝒳(x)\mathscr{X}(x) if there is no Y𝒳(x)Y\in\mathscr{X}(x) such that

ψ(Y)ψ(X)andVar[Y]Var[X]\psi(Y)\geq\psi(X)\quad\text{and}\quad\mathrm{Var}[Y]\leq\mathrm{Var}[X]

with at least one inequality holding strictly.

Remark 8.2.

If X0=0X_{0}=0, then ψ(X)=E[X]\psi(X)=\mathrm{E}[X] for all XL2X\in L^{2}. In this case, the BPV efficiency reduces to the mean-variance efficiency.

Proposition 8.3.

Assume X0LX_{0}\in L^{\infty}. Then XX^{*} is BPV efficient in 𝒳(x)\mathscr{X}(x) if and only if XX^{*} is variance-minimal in 𝒳icx(x,X0+z)\mathscr{X}_{\mathrm{icx}}(x,X_{0}+z) for some zxE[ρ]Q0(1)z\geq{x\over\mathrm{E}[\rho]}-Q_{0}(1).

Proof. Assume that XX^{*} is BPV efficient in 𝒳(x)\mathscr{X}(x). Let z=ψ(X)z=\psi(X^{*}). By xE[ρ]𝒳(x){x\over\mathrm{E}[\rho]}\in\mathscr{X}(x) and Var[xE[ρ]]=0Var(X)\mathrm{Var}\left[{x\over\mathrm{E}[\rho]}\right]=0\leq\mathrm{Var}(X^{*}) and by the BPV efficiency of XX^{*}, we have

z=ψ(X)ψ(xE[ρ])=xE[ρ]Q0(1).z=\psi(X^{*})\geq\psi\left({x\over\mathrm{E}[\rho]}\right)={x\over\mathrm{E}[\rho]}-Q_{0}(1).

Now we show that XX^{*} is variance-minimal in 𝒳icx(x,X0+z)\mathscr{X}_{\mathrm{icx}}(x,X_{0}+z). Suppose, on the contrary, that there exists some Y𝒳icx(x,X0+z)Y\in\mathscr{X}_{\mathrm{icx}}(x,X_{0}+z) such that Var(Y)<Var(X)\mathrm{Var}(Y)<\mathrm{Var}(X^{*}). Then ψ(Y)z=ψ(X)\psi(Y)\geq z=\psi(X^{*}) and Y𝒳(x)Y\in\mathscr{X}(x), which contradicts the BPV efficiency of XX^{*}.

Conversely, assume that XX^{*} is variance-minimal in 𝒳icx(x,X0+z)\mathscr{X}_{\mathrm{icx}}(x,X_{0}+z) for some zxE[ρ]Q0(1)z\geq{x\over\mathrm{E}[\rho]}-Q_{0}(1). We are going to show XX^{*} is BPV efficient in 𝒳(x)\mathscr{X}(x). The discussion is divided into two cases.

(a)

Assume z=xE[ρ]Q0(1)z={x\over\mathrm{E}[\rho]}-Q_{0}(1), i.e., (Q0(1)+z)E[ρ]=x(Q_{0}(1)+z)\mathrm{E}[\rho]=x. In this case, Proposition 4.2 implies that Var[X]=0\mathrm{Var}[X^{*}]=0 and hence X=cX^{*}=c a.s. for some c𝒳icx(x,X0+z)c\in\mathscr{X}_{\mathrm{icx}}(x,X_{0}+z). Then, cE[ρ]xc\mathrm{E}[\rho]\leq x and cicxX0+zc\succeq_{\mathrm{icx}}X_{0}+z. Therefore, cQ0(1)+zc\geq Q_{0}(1)+z and hence cE[ρ](Q0(1)+z)E[ρ]=xc\mathrm{E}[\rho]\geq(Q_{0}(1)+z)\mathrm{E}[\rho]=x, implying cE[ρ]=xc\mathrm{E}[\rho]=x. Suppose on the contrary that XX^{*} is not BPV efficient in 𝒳(x)\mathscr{X}(x), i.e., there exists some Y𝒳(x)Y\in\mathscr{X}(x) such that

Var(Y)Var(X)=0andψ(Y)ψ(X)\mathrm{Var}(Y)\leq\mathrm{Var}(X^{*})=0\quad\text{and}\quad\psi(Y)\geq\psi(X^{*})

with at least one inequality holds strictly. Then Y=c0Y=c_{0} a.s. for some c0c_{0}\in\mathbb{R} and ψ(c0)>ψ(X)=ψ(c)\psi(c_{0})>\psi(X^{*})=\psi(c). Therefore, c0>cc_{0}>c and hence c0E[ρ]>cE[ρ]=xc_{0}\mathrm{E}[\rho]>c\mathrm{E}[\rho]=x, which is impossible since Y=c0𝒳(x)Y=c_{0}\in\mathscr{X}(x). Thus, XX^{*} is BPV efficient in 𝒳(x)\mathscr{X}(x).

(b)

Assume z>xE[ρ]Q0(1)z>{x\over\mathrm{E}[\rho]}-Q_{0}(1), i.e., (Q0(1)+z)E[ρ]>x(Q_{0}(1)+z)\mathrm{E}[\rho]>x. Let Y𝒳(x)Y\in\mathscr{X}(x) and ψ(Y)ψ(X)\psi(Y)\geq\psi(X^{*}). Then ψ(Y)ψ(X)z\psi(Y)\geq\psi(X^{*})\geq z and hence Y𝒳icx(x,X0+z)Y\in\mathscr{X}_{\mathrm{icx}}(x,X_{0}+z). By the uniqueness of the variance-minimal payoff (Theorem 4.4), Var(Y)>Var(X)\mathrm{Var}(Y)>\mathrm{Var}(X^{*}) unless Y=XY=X^{*} a.s. Thus, XX^{*} is BPV efficient in 𝒳(x)\mathscr{X}(x). \Box

Proposition 8.4.

Assume X0LX_{0}\in L^{\infty}. Let zxE[ρ]Q0(1)z\geq{x\over\mathrm{E}[\rho]}-Q_{0}(1). If XX^{*} is variance-minimal in 𝒳icx(x,X0+z)\mathscr{X}_{\mathrm{icx}}(x,X_{0}+z), then ψ(X)=z\psi(X^{*})=z.

Proof. Consider the case when z=xE[ρ]Q0(1)z={x\over\mathrm{E}[\rho]}-Q_{0}(1). By the proof of Proposition 8.3, we know X=xE[ρ]X^{*}={x\over\mathrm{E}[\rho]} a.s. and hence ψ(X)=xE[ρ]Q0(1)=z\psi(X^{*})={x\over\mathrm{E}[\rho]}-Q_{0}(1)=z.

Now we consider the case when z>xE[ρ]Q0(1)z>{x\over\mathrm{E}[\rho]}-Q_{0}(1), i.e., (Q0(1)+z)E[ρ]>x(Q_{0}(1)+z)\mathrm{E}[\rho]>x. In this case, Var[X]>0\mathrm{Var}[X^{*}]>0 by Theorem 4.4. Obviously, ψ(X)z\psi(X^{*})\geq z by XicxX0+zX^{*}\succeq_{\mathrm{icx}}X_{0}+z. It is left to show ψ(X)z\psi(X^{*})\leq z. Suppose on the contrary that ψ(X)>z\psi(X^{*})>z. Then there exists some α>0\alpha>0 such that XαicxX0+zX^{*}-\alpha\succeq_{\mathrm{icx}}X_{0}+z. Therefore,

Xε(1ε)(Xα)+ε(Q0(1)+z)𝒳icx(x,X0+z)X^{\varepsilon}\triangleq(1-\varepsilon)(X^{*}-\alpha)+\varepsilon(Q_{0}(1)+z)\in\mathscr{X}_{\mathrm{icx}}(x,X_{0}+z)

for all sufficiently small ε>0\varepsilon>0. But Var[Xε]=(1ε)2Var[X]<Var[X]\mathrm{Var}[X^{\varepsilon}]=(1-\varepsilon)^{2}\mathrm{Var}[X^{*}]<\mathrm{Var}[X^{*}], contradicting the variance-minimality of XX^{*}. Therefore, ψ(X)=z\psi(X^{*})=z. \Box

Based on Propositions 8.3 and 8.4, the beating-performance-standard-deviation (BPSD) efficient frontier is

{(Var(X(z),z)|zxE[ρ]Q0(1)},\left\{\Big{(}\sqrt{\mathrm{Var}(X^{*}(z)},z\Big{)}\,\left|\,z\geq{x\over\mathrm{E}[\rho]}-Q_{0}(1)\right.\right\},

where, for every zz, X(z)X^{*}(z) is the variance-minimal payoff in 𝒳icx(x,X0+z)\mathscr{X}_{\mathrm{icx}}(x,X_{0}+z).

Example 8.5.

The SDF ρ\rho and the initial capital xx are same to Example 7.3. The benchmark payoff X0X_{0} satisfies (X0=δ)=(X0=δ)=0.5\mathbb{P}(X_{0}=-\delta)=\mathbb{P}(X_{0}=\delta)=0.5 with δ=0.0,0.2,0.5\delta=0.0,0.2,0.5. The BPSD efficient frontiers are plotted in Figure 3. In particular, δ=0.0\delta=0.0 refers to the classical mean-standard-deviation efficient frontier, which is a straight line.

Refer to caption
Figure 3: BPSD frontiers

8.2 Multi-Benchmark Beating

In the previous discussion, we considered the problem with beating only one benchmark. Now we consider the problem with beating multiple benchmarks. It turns out that the multi-benchmark case can be reduced to the single-benchmark case. Actually, in the quantile formulation, we can write beating multiple benchmarks as

t1Q(s)𝑑st1Qj(s)𝑑sfj(t)for all t[0,1] and 1jk,\int_{t}^{1}Q(s)ds\geq\int_{t}^{1}Q_{j}(s)ds\triangleq f_{j}(t)\quad\mbox{for all }t\in[0,1]\mbox{ and }1\leq j\leq k,\\

where Qj𝒬Q_{j}\in\mathscr{Q} are bounded for all 1jk1\leq j\leq k. This system of kk constraints is equivalent to

t1Q(s)𝑑smax{t1Q1(s)𝑑s,,t1Qk(s)𝑑s}g(t)for all t[0,1].\int_{t}^{1}Q(s)ds\geq\max\left\{\int_{t}^{1}Q_{1}(s)ds,\ldots,\int_{t}^{1}Q_{k}(s)ds\right\}\triangleq g(t)\quad\mbox{for all }t\in[0,1].

Obviously, fjgf_{j}\leq g, and fj(1)=g(1)=0f_{j}(1)=g(1)=0 for all jj. There exists a sequence {tn}n1(0,1)\{t_{n}\}_{n\geq 1}\subset(0,1) such that limntn=1\lim_{n\to\infty}t_{n}=1 and

limng(tn)g(1)tn1=lim inft1g(t)g(1)t1.\lim_{n\to\infty}{g(t_{n})-g(1)\over t_{n}-1}=\liminf_{t\to 1}{g(t)-g(1)\over t-1}.

There exists some j0j_{0} such that g(tn)=fj0(tn)g(t_{n})=f_{j_{0}}(t_{n}) for infinitely many nn. In this case,

limng(tn)g(1)tn1lim infnfj0(tn)fj0(1)tn1=fj0(1)=Qj0(1).\displaystyle\lim_{n\to\infty}{g(t_{n})-g(1)\over t_{n}-1}\geq\liminf_{n\to\infty}{f_{j_{0}}(t_{n})-f_{j_{0}}(1)\over t_{n}-1}=f^{\prime}_{j_{0}}(1-)=-Q_{j_{0}}(1-).

Then

lim inft1g(t)g(1)t1Qj0(1)>.\liminf_{t\to 1}{g(t)-g(1)\over t-1}\geq-Q_{j_{0}}(1-)>-\infty.

Similarly, lim supt0g(t)g(0)t<\limsup_{t\to 0}{g(t)-g(0)\over t}<\infty.

Let ginvbreve\invbreve{g} be the concave envelope of gg. By Lemma A.3, >ginvbreve(0)ginvbreve(1)>\infty>\invbreve{g}^{\prime}(0)\geq\invbreve{g}^{\prime}(1-)>-\infty. Therefore, ginvbreve-\invbreve{g}^{\prime} is bounded and increasing. Let

Q¯(t)=ginvbreve(t),t[0,1).\bar{Q}(t)=-\invbreve{g}^{\prime}(t),\quad t\in[0,1).

Then Q¯\bar{Q} is the quantile function of some X¯L\bar{X}\in L^{\infty} and ginvbreve(t)=t1Q¯(s)𝑑s\invbreve{g}(t)=\int_{t}^{1}\bar{Q}(s)ds for all t[0,1]t\in[0,1].

For any Q𝒬Q\in\mathscr{Q}, let f(t)=t1Q(s)𝑑sf(t)=\int_{t}^{1}Q(s)ds, t[0,1]t\in[0,1]. Obviously, ff is a concave function. By the definition of concave envelope, we know that

fgfginvbreve.f\geq g\Longleftrightarrow f\geq\invbreve{g}.

On the other hand,

QicxQj,jfgQ\succeq_{\mathrm{icx}}Q_{j},\;\forall j\Longleftrightarrow f\geq g

and

QicxQ¯fginvbreve.Q\succeq_{\mathrm{icx}}\bar{Q}\Longleftrightarrow f\geq\invbreve{g}.

Therefore,

QicxQj,jQicxQ¯.Q\succeq_{\mathrm{icx}}Q_{j},\;\forall j\Longleftrightarrow Q\succeq_{\mathrm{icx}}\bar{Q}.

This shows that the multi-benchmark case can be reduced to the single-benchmark case.

8.3 Mean-Variance Efficient Payoff

Now we consider the following problem of mean-variance portfolio selection with the increasing convex order constraint:

minimizeQ𝒬icx(x,Q0)Var[Q] subject to E[Q]z,\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0})}\quad\mathrm{Var}[Q]\\ &\text{ subject to }\quad\mathrm{E}[Q]\geq z,\end{split} (8.1)

where zz\in\mathbb{R}. Such a mean-variance problem can be covered by the analysis in the previous sections.

Actually, in the case when zE[Q]z\leq\mathrm{E}[Q^{\circ}] for some variance-minimal quantile function Q𝒬icx(x,Q0)Q^{\circ}\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}), QQ^{\circ} solves problem (8.1).

Now assume that z>E[Q]z>\mathrm{E}[Q^{\circ}] for all variance-minimal quantile function Q𝒬icx(x,Q0)Q^{\circ}\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}). In this case, by the convexity of variance and by the method of Lagrangian multiplier, QQ^{*} solves problem (8.1) if and only if, for some γ0\gamma^{*}\geq 0, QQ^{*} solves the following problem

minimizeQ𝒬icx(x,Q0)Var[Q]2γE[Q]\operatorname*{minimize\,}_{Q\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0})}\quad\mathrm{Var}[Q]-2\gamma^{*}\mathrm{E}[Q] (8.2)

and

γ(E[Q]z)=0.\gamma^{*}(\mathrm{E}[Q^{*}]-z)=0.

It is easy to see that γ=0\gamma^{*}=0 is impossible. Then γ>0\gamma^{*}>0 and hence E[Q]=z\mathrm{E}[Q^{*}]=z. In this case, problem (8.1) is equivalent to

minimizeQ𝒬icx(x,Q0)Var[Q]=01Q2(s)𝑑sz2 subject to E[Q]=z,\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0})}\quad\mathrm{Var}[Q]=\int_{0}^{1}Q^{2}(s)ds-z^{2}\\ &\text{ subject to }\quad\mathrm{E}[Q]=z,\end{split} (8.3)

which is further equivalent to

minimizeQ𝒬icx(x,Q0)01Q2(s)𝑑s2γ01Q(s)𝑑s.\begin{split}\operatorname*{minimize\,}_{Q\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0})}\quad\int_{0}^{1}Q^{2}(s)ds-2\gamma^{*}\int_{0}^{1}Q(s)ds.\end{split} (8.4)

By the method of Lagrangian multiplier once again, problem (8.4) can be transformed to, for some λ0\lambda\geq 0,

minimizeQ𝒬01Q2(s)𝑑s2γ01Q(s)𝑑s+λ01Q(s)Qρ(1s)𝑑ssubject to QicxQ0.\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}}\quad\int_{0}^{1}Q^{2}(s)ds-2\gamma^{*}\int_{0}^{1}Q(s)ds+\lambda\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\\ &\text{subject to }\quad Q\succeq_{\mathrm{icx}}Q_{0}.\end{split} (8.5)

It is just problem (4.3) with β=γ\beta=\gamma^{*}.

As Remark 2.1 indicates, we can also read E[Q]z\mathrm{E}[Q]\geq z as QicxzQ\succeq_{\mathrm{icx}}z and hence the mean-variance problem (8.1) is equivalent to a variance minimizing problem with two benchmarks: X0X_{0} and zz. Then it can be reduced to a variance minimizing problem with one benchmark, as discussed in Section 8.2.

Appendix A Some Technical Results

The results in this section may be unoriginal but are provided here for convenience nevertheless.

A.1 Saddle Point

The following lemma is essentially an abstract version of the proof of Wang and Xia (2021, Proposition 5.3). It is provided here for the convenience of its applications, not only in the proof of Proposition B.2 here but also elsewhere.

Lemma A.1.

Let XX be a nonempty set and Y1Y_{1} be a nonempty subset of a linear space. Let YY be the cone generated by Y1Y_{1}, that is, Y={μyμ0,yY1}Y=\{\mu y\mid\mu\geq 0,\;y\in Y_{1}\}. Consider a function f:X×Yf:X\times Y\to\mathbb{R}. Let functions g:X×Y1×[0,)g:X\times Y_{1}\times[0,\infty)\to\mathbb{R} and h:X×[0,)h:X\times[0,\infty)\to\mathbb{R} be given by

g(x,y,μ)=f(x,μy),xX,yY1,μ0,\displaystyle g(x,y,\mu)=f(x,\mu y),\quad x\in X,y\in Y_{1},\mu\geq 0,
h(x,μ)=supyY1g(x,y,μ)=supyY1f(x,μy),xX,μ0.\displaystyle h(x,\mu)=\sup_{y\in Y_{1}}g(x,y,\mu)=\sup_{y\in Y_{1}}f(x,\mu y),\quad x\in X,\mu\geq 0.

Assume that both of the following two conditions are satisfied.

(a)

For every μ0\mu\geq 0, there exists some y1(μ)Y1y_{1}(\mu)\in Y_{1} such that

infxXg(x,y1(μ),μ)=supyY1infxXg(x,y,μ)=infxXsupyY1g(x,y,μ).\inf_{x\in X}g(x,y_{1}(\mu),\mu)=\sup_{y\in Y_{1}}\inf_{x\in X}g(x,y,\mu)=\inf_{x\in X}\sup_{y\in Y_{1}}g(x,y,\mu).
(b)

(x,μ)(x^{*},\mu^{*}) is a saddle point of hh, that is,

h(x,μ)h(x,μ)h(x,μ),xX,μ0.h(x,\mu^{*})\geq h(x^{*},\mu^{*})\geq h(x^{*},\mu),\quad\forall x\in X,\mu\geq 0.

Let y=μy1(μ)y^{*}=\mu^{*}y_{1}(\mu^{*}). Then (x,y)(x^{*},y^{*}) is a saddle point of ff, that is,

f(x,y)f(x,y)f(x,y),xX,yY.f(x,y^{*})\geq f(x^{*},y^{*})\geq f(x^{*},y),\quad\forall x\in X,y\in Y.

Proof. We will frequently use conditions (a) and (b). Firstly, we have

h(x,μ)=supμ0infxXh(x,μ)=supμ0infxXsupyY1g(x,y,μ)\displaystyle h(x^{*},\mu^{*})=\sup_{\mu\geq 0}\inf_{x\in X}h(x,\mu)=\sup_{\mu\geq 0}\inf_{x\in X}\sup_{y\in Y_{1}}g(x,y,\mu)
=\displaystyle= supμ0supyY1infxXg(x,y,μ)=supμ0supyY1infxXf(x,μy)=supyYinfxXf(x,y).\displaystyle\sup_{\mu\geq 0}\sup_{y\in Y_{1}}\inf_{x\in X}g(x,y,\mu)=\sup_{\mu\geq 0}\sup_{y\in Y_{1}}\inf_{x\in X}f(x,\mu y)=\sup_{y\in Y}\inf_{x\in X}f(x,y).

Secondly, we have

h(x,μ)=infxXh(x,μ)=infxXsupyY1g(x,y,μ)=infxXg(x,y1(μ),μ)=infxXf(x,y).\displaystyle h(x^{*},\mu^{*})=\inf_{x\in X}h(x,\mu^{*})=\inf_{x\in X}\sup_{y\in Y_{1}}g(x,y,\mu^{*})=\inf_{x\in X}g(x,y_{1}(\mu^{*}),\mu^{*})=\inf_{x\in X}f(x,y^{*}).

Thirdly, we have

h(x,μ)=infxXsupμ0h(x,μ)=infxXsupμ0supyY1f(x,μy)=infxXsupyYf(x,y).\displaystyle h(x^{*},\mu^{*})=\inf_{x\in X}\sup_{\mu\geq 0}h(x,\mu)=\inf_{x\in X}\sup_{\mu\geq 0}\sup_{y\in Y_{1}}f(x,\mu y)=\inf_{x\in X}\sup_{y\in Y}f(x,y).

Fourthly, we have

h(x,μ)=supμ0h(x,μ)=supμ0supyY1f(x,μy)=supyYf(x,y).\displaystyle h(x^{*},\mu^{*})=\sup_{\mu\geq 0}h(x^{*},\mu)=\sup_{\mu\geq 0}\sup_{y\in Y_{1}}f(x^{*},\mu y)=\sup_{y\in Y}f(x^{*},y).

From the above discussion, we have

infxXf(x,y)=supyYinfxXf(x,y)=infxXsupyYf(x,y)=supyYf(x,y).\inf_{x\in X}f(x,y^{*})=\sup_{y\in Y}\inf_{x\in X}f(x,y)=\inf_{x\in X}\sup_{y\in Y}f(x,y)=\sup_{y\in Y}f(x^{*},y).

Therefore, (x,y)(x^{*},y^{*}) is a saddle point of ff. \Box

A.2 Concave Envelope

For a proof of the following lemma, see, e.g., Wang and Xia (2021, Appendix A.3).

Lemma A.2.

Assume that H:[0,1]H:[0,1]\to\mathbb{R} is upper semi-continuous. Let H𝑖𝑛𝑣𝑏𝑟𝑒𝑣𝑒\invbreve{H} be the concave envelope of HH, i.e.,

Hinvbreve(s)inf{G(s)|G is concave and GH on [0,1]},s[0,1].\invbreve{H}(s)\triangleq\inf\{G(s)\,|\,G\mbox{ is concave and }G\geq H\mbox{ on }[0,1]\},\quad s\in[0,1].

We have the following assertions:

(a)

Hinvbreve(0)=H(0)\invbreve{H}(0)=H(0) and Hinvbreve(1)=H(1)\invbreve{H}(1)=H(1);

(b)

Hinvbreve\invbreve{H} is continuous on [0,1][0,1];

(c)

Hinvbreve\invbreve{H} is affine on [Hinvbreve>H][\invbreve{H}>H].

Lemma A.3.

Assume that H:[0,1]H:[0,1]\to\mathbb{R} is upper semi-continuous. Let H𝑖𝑛𝑣𝑏𝑟𝑒𝑣𝑒\invbreve{H} be the concave envelope of HH. Then we have the following two assertions.

(a)

If lim inft1H(t)H(1)t1>\liminf_{t\to 1}{H(t)-H(1)\over t-1}>-\infty, then Hinvbreve(1)>\invbreve{H}^{\prime}(1-)>-\infty.

(b)

If lim supt0H(t)H(0)t<\limsup_{t\to 0}{H(t)-H(0)\over t}<\infty, then Hinvbreve(0)<\invbreve{H}^{\prime}(0)<\infty.

Proof. We only prove assertion (b), since the proof of assertion (a) is similar. Assume that there exists some ε(0,1)\varepsilon\in(0,1) such that

(ε,1)[Hinvbreve>H].(\varepsilon,1)\subseteq[\invbreve{H}>H]. (A.1)

In this case, Hinvbreve\invbreve{H}^{\prime} is constant and finite on (ε,1)(\varepsilon,1). Therefore, Hinvbreve(1)>\invbreve{H}^{\prime}(1-)>-\infty. Assume that (A.1) holds for no ε(0,1)\varepsilon\in(0,1). In this case, there exits a sequence {tn}n1(0,1)\{t_{n}\}_{n\geq 1}\subset(0,1) such that limntn=1\lim_{n\to\infty}t_{n}=1 and Hinvbreve(tn)=H(tn)\invbreve{H}(t_{n})=H(t_{n}) for all n1n\geq 1. Then by Hinvbreve(1)=H(1)\invbreve{H}(1)=H(1), we have that

Hinvbreve(1)=limnHinvbreve(tn)Hinvbreve(1)tn1=limnH(tn)H(1)tn1lim inft1H(t)H(1)t1>.\displaystyle\invbreve{H}^{\prime}(1-)=\lim_{n\to\infty}{\invbreve{H}(t_{n})-\invbreve{H}(1)\over t_{n}-1}=\lim_{n\to\infty}{H(t_{n})-H(1)\over t_{n}-1}\geq\liminf_{t\to 1}{H(t)-H(1)\over t-1}>-\infty.

\Box

Appendix B Some Proofs

B.1 Proof of Theorem 4.1

For any ε(0,12)\varepsilon\in\left(0,{1\over 2}\right), let

aε=0εQρ(s)𝑑s=1ε1Qρ(1s)𝑑s,ε(0,12).a_{\varepsilon}=\int_{0}^{\varepsilon}Q_{\rho}(s)ds=\int_{1-\varepsilon}^{1}Q_{\rho}(1-s)ds,\quad\varepsilon\in\left(0,{1\over 2}\right).

For any ε(0,12)\varepsilon\in\left(0,{1\over 2}\right) and n1n\geq 1, let QnεQ^{\varepsilon}_{n} be defined by

Qnε(s)=Q0(s)+naε𝟏s>1εnaε𝟏s<ε,s(0,1),Q^{\varepsilon}_{n}(s)=Q_{0}(s)+{n\over a_{\varepsilon}}\mathbf{1}_{s>1-\varepsilon}-{n\over a_{\varepsilon}}\mathbf{1}_{s<\varepsilon},\quad s\in(0,1),

Obviously, Qnε𝒬Q^{\varepsilon}_{n}\in\mathscr{Q} and QnεicxQ0Q^{\varepsilon}_{n}\succeq_{\mathrm{icx}}Q_{0} for all ε(0,12)\varepsilon\in\left(0,{1\over 2}\right) and n1n\geq 1. Moreover, since Qρ(1)>Qρ(0)Q_{\rho}(1-)>Q_{\rho}(0), there exists some ε0(0,12)\varepsilon_{0}\in(0,{1\over 2}) and α>0\alpha>0 such that

1aε00ε0Qρ(1s)𝑑s=0ε0Qρ(1s)𝑑s0ε0Qρ(s)𝑑s1+α.{1\over a_{\varepsilon_{0}}}\int_{0}^{\varepsilon_{0}}Q_{\rho}(1-s)ds={\int_{0}^{\varepsilon_{0}}Q_{\rho}(1-s)ds\over\int_{0}^{\varepsilon_{0}}Q_{\rho}(s)ds}\geq 1+\alpha.

Therefore, for any n1n\geq 1,

01Qnε0(s)Qρ(1s)𝑑s\displaystyle\int_{0}^{1}Q^{\varepsilon_{0}}_{n}(s)Q_{\rho}(1-s)ds
=\displaystyle= 01Q0(s)Qρ(1s)𝑑s+naε01ε01Qρ(1s)𝑑snaε00ε0Qρ(1s)𝑑s\displaystyle\int_{0}^{1}Q_{0}(s)Q_{\rho}(1-s)ds+{n\over a_{\varepsilon_{0}}}\int_{1-\varepsilon_{0}}^{1}Q_{\rho}(1-s)ds-{n\over a_{\varepsilon_{0}}}\int_{0}^{\varepsilon_{0}}Q_{\rho}(1-s)ds
\displaystyle\leq 01Q0(s)Qρ(1s)𝑑s+nn(1+α)\displaystyle\int_{0}^{1}Q_{0}(s)Q_{\rho}(1-s)ds+n-n(1+\alpha)
=\displaystyle= 01Q0(s)Qρ(1s)𝑑sαn.\displaystyle\int_{0}^{1}Q_{0}(s)Q_{\rho}(1-s)ds-\alpha n.

As a consequence, infQ𝒬01Q(s)Qρ(1s)𝑑s=\inf_{Q\in\mathscr{Q}}\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds=-\infty. \Box

B.2 Proof of Theorem 4.4

It is well known that Var[X]=minβE[(Xβ)2]\mathrm{Var}[X]=\min_{\beta\in\mathbb{R}}\mathrm{E}[(X-\beta)^{2}] for every XL2X\in L^{2}. Then Var[Q]=minβ01(Q(s)β)2𝑑s\mathrm{Var}[Q]=\min_{\beta\in\mathbb{R}}\int_{0}^{1}(Q(s)-\beta)^{2}ds for every Q𝒬Q\in\mathscr{Q}. Therefore, problem (3.2) can be rewritten as

minimizeQ𝒬minβ01(Q(s)β)2𝑑ssubject to 01Q(s)Qρ(1s)𝑑sx,QicxQ0.\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}}\quad\min_{\beta\in\mathbb{R}}\int_{0}^{1}(Q(s)-\beta)^{2}ds\\ &\text{subject to }\quad\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\leq x,\;Q\succeq_{\mathrm{icx}}Q_{0}.\end{split} (B.1)

We now consider, for any fixed β\beta\in\mathbb{R}, the following problem

minimizeQ𝒬01(Q(s)β)2𝑑ssubject to 01Q(s)Qρ(1s)𝑑sx,QicxQ0.\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}}\quad\int_{0}^{1}(Q(s)-\beta)^{2}ds\\ &\text{subject to }\quad\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\leq x,\;Q\succeq_{\mathrm{icx}}Q_{0}.\end{split} (B.2)

Let v1(β)v_{1}(\beta) denote the optimal value of problem (B.2). Obviously, v(x)=infβv1(β)v^{\circ}(x)=\inf_{\beta\in\mathbb{R}}v_{1}(\beta).

Lemma B.1.

Under Assumption 4.3, we have the following assertions.

(a)

For every β\beta\in\mathbb{R}, there exists a unique optimal solution to problem (B.2).

(b)

v1v_{1} is continuous and convex on \mathbb{R} with lim|β|v1(β)=\lim_{|\beta|\to\infty}v_{1}(\beta)=\infty.

(c)

There exists some β\beta^{*}\in\mathbb{R} such that v1(β)=infβv1(β)v_{1}(\beta^{*})=\inf_{\beta\in\mathbb{R}}v_{1}(\beta).

Proof. We have known that 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}) is a nonempty, convex, and closed subset of Hilbert space L2([0,1))L^{2}([0,1)). Problem (B.2) is nothing but to find an element Q𝒬icx(x,Q0)Q\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}) which is closest to β\beta. Such an element always exists and is unique; see, e.g., Luenberger (1969, Section 3.12). Therefore, assertion (a) is proved.

Assertion (c) is an easy implication of assertion (b). We now prove assertion (b).

We firstly prove the convexity. It suffices to show v1(12β1+12β2)12v1(β1)+12v1(β2)v_{1}\left({1\over 2}\beta_{1}+{1\over 2}\beta_{2}\right)\leq{1\over 2}v_{1}(\beta_{1})+{1\over 2}v_{1}(\beta_{2}) for all β1,β2\beta_{1},\beta_{2}\in\mathbb{R}. We use \|\cdot\| to denote the norm on Hilbert space L2([0,1))L^{2}([0,1)). Let β1,β2\beta_{1},\beta_{2}\in\mathbb{R}. By assertion (a), there exist some Qi𝒬icx(x,Q0)Q_{i}\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}) such that v1(βi)=Qiβi2v_{1}(\beta_{i})=\|Q_{i}-\beta_{i}\|^{2} for i{1,2}i\in\{1,2\}. Obviously, 12(Q1+Q2)𝒬icx(x,Q0){1\over 2}(Q_{1}+Q_{2})\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}) and

v1(12β1+12β2)\displaystyle v_{1}\left({1\over 2}\beta_{1}+{1\over 2}\beta_{2}\right)\leq 12(Q1+Q2)12(β1+β2)2=12(Q1β1)+12(Q2β2)2\displaystyle\left\|{1\over 2}(Q_{1}+Q_{2})-{1\over 2}(\beta_{1}+\beta_{2})\right\|^{2}=\left\|{1\over 2}(Q_{1}-\beta_{1})+{1\over 2}(Q_{2}-\beta_{2})\right\|^{2}
\displaystyle\leq 12Q1β12+12Q2β22=12v1(β1)+12v1(β2).\displaystyle{1\over 2}\|Q_{1}-\beta_{1}\|^{2}+{1\over 2}\|Q_{2}-\beta_{2}\|^{2}={1\over 2}v_{1}(\beta_{1})+{1\over 2}v_{1}(\beta_{2}).

Therefore, v1v_{1} is convex on \mathbb{R}.

Secondly, because v1v_{1} is convex and finite on \mathbb{R}, v1v_{1} is continuous on \mathbb{R}.

Finally, we show lim|β|v1(β)=\lim_{|\beta|\to\infty}v_{1}(\beta)=\infty. To this end, for any β\beta\in\mathbb{R}, let

v0(β)=infQ𝒬01(Q(s)β)2𝑑ssubject to 01Q(s)Qρ(1s)𝑑sx.\begin{split}v_{0}(\beta)=&\inf_{Q\in\mathscr{Q}}\int_{0}^{1}(Q(s)-\beta)^{2}ds\\ &\text{subject to }\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\leq x.\end{split} (B.3)

It is a classical problem of mean-variance portfolio selection without constraint. The optimal solution to problem (B.3) is obviously given by

Q(s)=βλQρ((1s))2,s[0,1),Q(s)=\beta-\frac{\lambda Q_{\rho}((1-s)-)}{2},\quad s\in[0,1),

where λ=2(βE[ρ]x)E[ρ2]\lambda={2(\beta\mathrm{E}[\rho]-x)\over\mathrm{E}[\rho^{2}]}. Then v0(β)=(βE[ρ]x)2E[ρ2]v_{0}(\beta)=\frac{(\beta\mathrm{E}[\rho]-x)^{2}}{\mathrm{E}[\rho^{2}]} and hence

lim|β|v1(β)lim|β|v0(β)=.\lim_{|\beta|\to\infty}v_{1}(\beta)\geq\lim_{|\beta|\to\infty}v_{0}(\beta)=\infty.

Therefore, lim|β|v1(β)=\lim_{|\beta|\to\infty}v_{1}(\beta)=\infty. \Box

Proof of Theorem 4.4

Lemma B.1 guarantees the existence of a variance-minimal quantile function in 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}).

We now show v(x)>0v^{\circ}(x)>0. Otherwise, v(x)=0v^{\circ}(x)=0 and hence there exists some constant c𝒬icx(x,Q0)c\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}). Then cQ0(1)c\geq Q_{0}(1) and hence x01cQρ(1s)𝑑sQ0(1)E[ρ]x\geq\int_{0}^{1}cQ_{\rho}(1-s)ds\geq Q_{0}(1)\mathrm{E}[\rho], contradicting the assumption that Q0(1)E[ρ]>xQ_{0}(1)\mathrm{E}[\rho]>x. Therefore, v(x)>0v^{\circ}(x)>0.

It is left to show the uniqueness. Let Q1,Q2𝒬icx(x,Q0)Q_{1},Q_{2}\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}) be variance-minimal in 𝒬icx(x,Q0)\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}). We need to show Q1=Q2Q_{1}=Q_{2}.

We first show Q1E[Q1]=Q2E[Q2]Q_{1}-\mathrm{E}[Q_{1}]=Q_{2}-\mathrm{E}[Q_{2}]. Actually, Var[Q1]=Var[Q2]=v(x)\mathrm{Var}[Q_{1}]=\mathrm{Var}[Q_{2}]=v^{\circ}(x). Let Q¯=12(Q1+Q2)\bar{Q}={1\over 2}(Q_{1}+Q_{2}). Then Q¯𝒬icx(x,Q0)\bar{Q}\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}). Moreover,

v(x)\displaystyle v^{\circ}(x)\leq Var[Q¯]=E[(Q¯E[Q¯])2]\displaystyle\mathrm{Var}[\bar{Q}]=\mathrm{E}[(\bar{Q}-\mathrm{E}[\bar{Q}])^{2}]
=\displaystyle= E[(12(Q1E[Q1])+12(Q2E[Q2]))2]\displaystyle\mathrm{E}\left[\left({1\over 2}(Q_{1}-\mathrm{E}[Q_{1}])+{1\over 2}(Q_{2}-\mathrm{E}[Q_{2}])\right)^{2}\right]
\displaystyle\leq E[12(Q1E[Q1])2+12(Q2E[Q2])2]\displaystyle\mathrm{E}\left[{1\over 2}(Q_{1}-\mathrm{E}[Q_{1}])^{2}+{1\over 2}(Q_{2}-\mathrm{E}[Q_{2}])^{2}\right]
=\displaystyle= 12Var[Q1]+12Var[Q2]=v(x)\displaystyle{1\over 2}\mathrm{Var}[Q_{1}]+{1\over 2}\mathrm{Var}[Q_{2}]=v^{\circ}(x)

and hence

E[(12(Q1E[Q1])+12(Q2E[Q2]))2]=E[12(Q1E[Q1])2+12(Q2E[Q2])2].\mathrm{E}\left[\left({1\over 2}(Q_{1}-\mathrm{E}[Q_{1}])+{1\over 2}(Q_{2}-\mathrm{E}[Q_{2}])\right)^{2}\right]=\mathrm{E}\left[{1\over 2}(Q_{1}-\mathrm{E}[Q_{1}])^{2}+{1\over 2}(Q_{2}-\mathrm{E}[Q_{2}])^{2}\right].

By the strict convexity of function xx2x\mapsto x^{2}, we have Q1E[Q1]=Q2E[Q2]Q_{1}-\mathrm{E}[Q_{1}]=Q_{2}-\mathrm{E}[Q_{2}].

To show Q1=Q2Q_{1}=Q_{2}, we needs to show E[Q1]=E[Q2]\mathrm{E}[Q_{1}]=\mathrm{E}[Q_{2}]. To this end, it suffices to show 01Q1(s)Qρ(1s)𝑑s=01Q2(s)Qρ(1s)𝑑s=x\int_{0}^{1}Q_{1}(s)Q_{\rho}(1-s)ds=\int_{0}^{1}Q_{2}(s)Q_{\rho}(1-s)ds=x. Without loss of generality, suppose on the contrary that 01Q1(s)Qρ(1s)𝑑s<x\int_{0}^{1}Q_{1}(s)Q_{\rho}(1-s)ds<x. For ε(0,1)\varepsilon\in(0,1), let QεQ^{\varepsilon} be given by

Qε=(1ε)Q1+εQ0(1).Q^{\varepsilon}=(1-\varepsilon)Q_{1}+\varepsilon Q_{0}(1).

Then for all sufficiently small ε>0\varepsilon>0, Qε𝒬icx(x,Q0)Q^{\varepsilon}\in\mathscr{Q}_{\mathrm{icx}}(x,Q_{0}). By Var[Q1]=v(x)>0\mathrm{Var}[Q_{1}]=v^{\circ}(x)>0, we have Var[Qε]=(1ε)2Var[Q1]<Var[Q1]\mathrm{Var}[Q^{\varepsilon}]=(1-\varepsilon)^{2}\mathrm{Var}[Q_{1}]<\mathrm{Var}[Q_{1}]. It is impossible since Q1Q_{1} is variance-minimal. Therefore, 01Q1(s)Qρ(1s)𝑑s=01Q2(s)Qρ(1s)𝑑s=x\int_{0}^{1}Q_{1}(s)Q_{\rho}(1-s)ds=\int_{0}^{1}Q_{2}(s)Q_{\rho}(1-s)ds=x. \Box

B.3 Proof of Proposition 4.7

Let

{m:[0,1][0,)m is increasing and right-continuous}.\displaystyle\mathscr{M}\triangleq\left\{m:[0,1]\to[0,\infty)\mid m\mbox{ is increasing and right-continuous}\right\}.

We can identify \mathscr{M} as the set of all finite measures on the measurable space ([0,1],[0,1])\left([0,1],\mathcal{B}_{[0,1]}\right). For each mm\in\mathscr{M}, the measure of {0}\{0\} is m(0)m(0) and the measure of {1}\{1\} is m(1)m(1)m(1)-m(1-). Let

1{mm(1)=1}.\displaystyle\mathscr{M}_{1}\triangleq\left\{m\in\mathscr{M}\mid m(1)=1\right\}.

We can identify 1\mathscr{M}_{1} as the set of all probability measures on ([0,1],[0,1])\left([0,1],\mathcal{B}_{[0,1]}\right). Consider the weak topology (the topology of weak convergence of measures) on 1\mathscr{M}_{1}, which is induced by all real-valued continuous functions on [0,1][0,1]. By Aliprantis and Border (2006, Theorem 15.11), 1\mathscr{M}_{1} is weakly compact.555In Aliprantis and Border (2006, Chapter 15), the weak* topology refers to the weak topology here. Moreover, weak compactness (resp. closedness) here refers to the compactness (resp. closedness) under the weak topology.

For any Q𝒬Q\in\mathscr{Q}, let

AQ(0)Q(1),AQ(s)1s1s1Q(t)𝑑t,s(0,1].A_{Q}(0)\triangleq Q(1-),\;A_{Q}(s)\triangleq{1\over s}\int_{1-s}^{1}Q(t)dt,\quad s\in(0,1].

Obviously, AQA_{Q} is decreasing and lower bounded on [0,1][0,1]. For any Q𝒬Q\in\mathscr{Q},

QicxQ0\displaystyle Q\succeq_{\mathrm{icx}}Q_{0}\Longleftrightarrow [0,1]AQ(s)𝑑m(s)[0,1]AQ0(s)𝑑m(s),m\displaystyle\int_{[0,1]}A_{Q}(s)dm(s)\geq\int_{[0,1]}A_{Q_{0}}(s)dm(s),\quad\forall m\in\mathscr{M}
\displaystyle\Longleftrightarrow [0,1]AQ(s)𝑑m(s)[0,1]AQ0(s)𝑑m(s),m1.\displaystyle\int_{[0,1]}A_{Q}(s)dm(s)\geq\int_{[0,1]}A_{Q_{0}}(s)dm(s),\quad\forall m\in\mathscr{M}_{1}.

Similarly to Föllmer and Schied (2016, Lemma 4.69), the identity

{w(t)=(1t,1]s1𝑑m(s),t[0,1)w(1)w(1)=m(0)\begin{cases}w^{\prime}(t)=\displaystyle\int_{(1-t,1]}s^{-1}dm(s),\quad t\in[0,1)\\ w(1)-w(1-)=m(0)\end{cases} (B.4)

defines a bijection

𝒥:𝒲icx,\mathcal{J}:\mathscr{M}\to\mathscr{W}^{\mathrm{icx}},

where ww^{\prime} denotes the right derivative. Under identity (B.4), an application of Fubini theorem implies that

[0,1]Q(t)𝑑w(t)=[0,1]AQ(t)𝑑m(t),Q𝒬.\displaystyle\int_{[0,1]}Q(t)dw(t)=\int_{[0,1]}A_{Q}(t)dm(t),\quad\forall\,Q\in\mathscr{Q}.

Let

L1(Q,m;β,λ)\displaystyle L_{1}(Q,m;\beta,\lambda)\triangleq 01(Q(s)β)2𝑑s+λ01Q(s)Qρ(1s)𝑑s\displaystyle\int_{0}^{1}(Q(s)-\beta)^{2}ds+\lambda\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds
[0,1]AQ(s)𝑑m(s)+[0,1]AQ0(s)𝑑m(s),Q𝒬,m.\displaystyle-\int_{[0,1]}A_{Q}(s)dm(s)+\int_{[0,1]}A_{Q_{0}}(s)dm(s),\quad Q\in\mathscr{Q},\;m\in\mathscr{M}.

Then, under identity (B.4),

L(Q,w;β,λ)=L1(Q,m;β,λ).L(Q,w;\beta,\lambda)=L_{1}(Q,m;\beta,\lambda).

As a consequence, Proposition 4.7 can be reformulated as the following one.

Proposition B.2.

Assume that X0LX_{0}\in L^{\infty}. Let Q𝒬Q^{*}\in\mathscr{Q}. Then QQ^{*} solves problem (4.3) if and only if there exists some mm^{*}\in\mathscr{M} such that (Q,m)(Q^{*},m^{*}) is a saddle point of L1(,;β,λ)L_{1}(\cdot\;,\cdot\;;\beta,\lambda) (with respect to minimizing in QQ and maximizing in mm).

Before the proof of Proposition B.2, we introduce two lemmas.

For every β\beta\in\mathbb{R}, λ0\lambda\geq 0, and μ0\mu\geq 0, let

K(Q,m,μ;β,λ)\displaystyle K(Q,m,\mu;\beta,\lambda)\triangleq 01(Q(s)β)2𝑑s+λ01Q(s)Qρ(1s)𝑑s\displaystyle\int_{0}^{1}(Q(s)-\beta)^{2}ds+\lambda\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds
μ[0,1]AQ(s)𝑑m(s)+μ[0,1]AQ0(s)𝑑m(s),Q𝒬,m1.\displaystyle-\mu\int_{[0,1]}A_{Q}(s)dm(s)+\mu\int_{[0,1]}A_{Q_{0}}(s)dm(s),\quad Q\in\mathscr{Q},\;m\in\mathscr{M}_{1}.
Lemma B.3.

Assume that X0LX_{0}\in L^{\infty}. For every β\beta\in\mathbb{R}, λ0\lambda\geq 0, and μ0\mu\geq 0, there exists some m1(μ)1m_{1}(\mu)\in\mathscr{M}_{1} such that

infQ𝒬K(Q,m1(μ),μ;β,λ)=supm1infQ𝒬K(Q,m,μ;β,λ)=infQ𝒬supm1K(Q,m,μ;β,λ).\begin{split}\inf_{Q\in\mathscr{Q}}K(Q,m_{1}(\mu),\mu;\beta,\lambda)=&\sup_{m\in\mathscr{M}_{1}}\inf_{Q\in\mathscr{Q}}K(Q,m,\mu;\beta,\lambda)\\ =&\inf_{Q\in\mathscr{Q}}\sup_{m\in\mathscr{M}_{1}}K(Q,m,\mu;\beta,\lambda).\end{split} (B.5)

Proof. It is obvious that KK is convex in QQ and affine in mm. By the boundedness of X0X_{0}, we know that AQ0A_{Q_{0}} is bounded and continuous on [0,1][0,1]. Therefore, the functional

1m[0,1]AQ0(s)𝑑m(s)\mathscr{M}_{1}\ni m\mapsto\int_{[0,1]}A_{Q_{0}}(s)dm(s)

is weakly continuous.666Here the weak continuity refers to the continuity under the weak topology. Similarly, the weak upper (resp. lower) semi-continuity below refers to the upper (resp. lower) semi-continuity under the weak topology. For each n1n\geq 1 and Q𝒬Q\in\mathscr{Q}, let QnQ^{n} be given by

Qn(s)=min{Q(s),n},s[0,1].Q^{n}(s)=\min\{Q(s),n\},\quad s\in[0,1].

We know that, for each n1n\geq 1, AQnA_{Q^{n}} is bounded and continuous on [0,1][0,1] and hence the functional

1m[0,1]AQn(s)𝑑m(s)\mathscr{M}_{1}\ni m\mapsto\int_{[0,1]}A_{Q^{n}}(s)dm(s)

is weakly continuous. The monotone convergence theorem implies that

[0,1]AQ(t)𝑑m(t)=supn1[0,1]AQn(t)𝑑m(t).\int_{[0,1]}A_{Q}(t)dm(t)=\sup_{n\geq 1}\int_{[0,1]}A_{Q^{n}}(t)dm(t).

Therefore, for each Q𝒬Q\in\mathscr{Q}, the functional

1m[0,1]AQ(t)𝑑m(t)\mathscr{M}_{1}\ni m\mapsto\int_{[0,1]}A_{Q}(t)dm(t)

is weakly lower semi-continuous. As a consequence of the above discussion, KK is upper semi-continuous in mm. By the weak compactness of 1\mathscr{M}_{1} and by the minimax theorem (e.g., Mertons et al. (2015, Proposition I.1.3)), there exists some m1(μ)1m_{1}(\mu)\in\mathscr{M}_{1} that satisfies (B.5). \Box

Obviously, problem (4.3) is equivalent to the following one:

minimizeQ𝒬01(Q(s)β)2𝑑s+λ01Q(s)Qρ(1s)𝑑s subject to infm1([0,1]AQ(s)𝑑m(s)[0,1]AQ0(s)𝑑m(s))0.\begin{split}&\operatorname*{minimize\,}_{Q\in\mathscr{Q}}\quad\int_{0}^{1}(Q(s)-\beta)^{2}ds+\lambda\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds\\ &\text{ subject to }\quad\inf_{m\in\mathscr{M}_{1}}\left(\int_{[0,1]}A_{Q}(s)dm(s)-\int_{[0,1]}A_{Q_{0}}(s)dm(s)\right)\geq 0.\end{split} (B.6)

Let

J(Q,μ;β,λ)\displaystyle J(Q,\mu;\beta,\lambda) 01(Q(s)β)2𝑑s+λ01Q(s)Qρ(1s)𝑑s\displaystyle\triangleq\int_{0}^{1}(Q(s)-\beta)^{2}ds+\lambda\int_{0}^{1}Q(s)Q_{\rho}(1-s)ds
μinfm1([0,1]AQ(s)𝑑m(s)[0,1]AQ0(s)𝑑m(s)),Q𝒬,μ0.\displaystyle-\mu\inf_{m\in\mathscr{M}_{1}}\left(\int_{[0,1]}A_{Q}(s)dm(s)-\int_{[0,1]}A_{Q_{0}}(s)dm(s)\right),\quad Q\in\mathscr{Q},\mu\geq 0.
Lemma B.4.

Assume that X0LX_{0}\in L^{\infty}. Let Q𝒬Q^{*}\in\mathscr{Q}. Then QQ^{*} solves problem (B.6) if and only if there exists some μ0\mu^{*}\geq 0 such that

J(Q,μ;β,λ)J(Q,μ;β,λ)J(Q,μ;β,λ)Q𝒬,μ0.\displaystyle J(Q^{*},\mu;\beta,\lambda)\leq J(Q^{*},\mu^{*};\beta,\lambda)\leq J(Q,\mu^{*};\beta,\lambda)\quad\forall\,Q\in\mathscr{Q},\mu\geq 0. (B.7)

Proof. Let Q(s)=Q0(s)+αQ(s)=Q_{0}(s)+\alpha for all s[0,1]s\in[0,1], where α>0\alpha>0. Obviously, Q𝒬Q\in\mathscr{Q} and

infm1([0,1]AQ(s)𝑑m(s)[0,1]AQ0(s)𝑑m(s))=α>0.\inf_{m\in\mathscr{M}_{1}}\left(\int_{[0,1]}A_{Q}(s)dm(s)-\int_{[0,1]}A_{Q_{0}}(s)dm(s)\right)=\alpha>0.

Thus, the Slater condition for the constraint in (B.6) is satisfied. Then the conclusion of the lemma is an implication of Luenberger (1969, Corollary 1 on p. 219 and Theorem 2 on p. 221). \Box

Proof of Proposition B.2.

Obviously, \mathscr{M} is the cone generated by 1\mathscr{M}_{1}. For all Q𝒬Q\in\mathscr{Q}, m1m\in\mathscr{M}_{1}, and μ0\mu\geq 0, we have

K(Q,m,μ;β,λ)=L1(Q,μm;β,λ),\displaystyle K(Q,m,\mu;\beta,\lambda)=L_{1}(Q,\mu m;\beta,\lambda),
J(Q,μ;β,λ)=supm1K(Q,m,μ;β,λ)=supm1L1(Q,μm;β,λ).\displaystyle J(Q,\mu;\beta,\lambda)=\sup_{m\in\mathscr{M}_{1}}K(Q,m,\mu;\beta,\lambda)=\sup_{m\in\mathscr{M}_{1}}L_{1}(Q,\mu m;\beta,\lambda).

If QQ^{*} solves problem (4.3), or, equivalently, problem (B.6). Let μ\mu^{*} be given by Lemma B.4 and m1(μ)m_{1}(\mu^{*}) be given by Lemma B.3. Set m=μm1(μ)m^{*}=\mu^{*}m_{1}(\mu^{*}). Then by Lemma A.1, we know that (Q,m)(Q^{*},m^{*}) is a saddle point of L1(,;β,λ)L_{1}(\cdot\;,\cdot\;;\beta,\lambda). Therefore, the “only if” part is proved. The “if” part is obvious. \Box

B.4 Proof of Lemma 5.1

Let w𝒲icxw\in\mathscr{W}^{\mathrm{icx}}. The boundedness of X0X_{0} implies that [0,1]Q0(s)𝑑w(s)(,)\int_{[0,1]}Q_{0}(s)dw(s)\in(-\infty,\infty) Assume that w(1)w(1)>0w(1)-w(1-)>0. For any ε(0,1)\varepsilon\in(0,1), let Qε𝒬Q^{\varepsilon}\in\mathscr{Q} be given by

Qε(s)=1ε𝟏s1ε,s[0,1].Q^{\varepsilon}(s)={1\over\sqrt{\varepsilon}}\mathbf{1}_{s\geq 1-\varepsilon},\quad s\in[0,1].

Then we have

L(Qε,w;β,λ)[0,1]Q0(s)𝑑w(s)\displaystyle L(Q^{\varepsilon},w;\beta,\lambda)-\int_{[0,1]}Q_{0}(s)dw(s)
=\displaystyle= 12βε+β2+λε1ε1Qρ(1s)𝑑sw(1)w(1ε)ε\displaystyle 1-2\beta\sqrt{\varepsilon}+\beta^{2}+{\lambda\over\sqrt{\varepsilon}}\int_{1-\varepsilon}^{1}Q_{\rho}(1-s)ds-{w(1)-w(1-\varepsilon)\over\sqrt{\varepsilon}}
\displaystyle\leq 12βε+β2+λεQρ(ε)w(1)w(1)εε0,\displaystyle 1-2\beta\sqrt{\varepsilon}+\beta^{2}+{\lambda\sqrt{\varepsilon}}Q_{\rho}(\varepsilon)-{w(1)-w(1-)\over\sqrt{\varepsilon}}\overset{\varepsilon\downarrow 0}{\longrightarrow}-\infty,

which leads to infQ𝒬L(Q,w;β,λ)=\inf_{Q\in\mathscr{Q}}L(Q,w;\beta,\lambda)=-\infty. \Box

B.5 Proof of Lemma 5.2

Assume that w(1)=w(1)w(1)=w(1-) and wL2([0,1))w^{\prime}\notin L^{2}([0,1)). Then 01(w(s))2𝑑s=\int_{0}^{1}(w^{\prime}(s))^{2}ds=\infty. For each n1n\geq 1, let QnQ_{n} be given by

Qn(s)=β+fn(s)λQρ((1s))2,s[0,1),Q_{n}(s)=\beta+{f_{n}(s)-\lambda Q_{\rho}((1-s)-)\over 2},\quad s\in[0,1),

where

fn(s)=min{w(s),n},s[0,1).f_{n}(s)=\min\{w^{\prime}(s),n\},\quad s\in[0,1).

Obviously, Qn𝒬Q_{n}\in\mathscr{Q} for all n1n\geq 1 and

01fn2(s)𝑑s01(w(s))2𝑑s= as n.\int_{0}^{1}f_{n}^{2}(s)ds\to\int_{0}^{1}(w^{\prime}(s))^{2}ds=\infty\text{ as }n\to\infty.

The boundedness of X0X_{0} implies that a101Q0(s)w(s)𝑑s(,)a_{1}\triangleq\int_{0}^{1}Q_{0}(s)w^{\prime}(s)ds\in(-\infty,\infty). Moreover,

a2\displaystyle a_{2}\triangleq 01Qρ(1s)w(s)𝑑s\displaystyle\int_{0}^{1}Q_{\rho}(1-s)w^{\prime}(s)ds
\displaystyle\leq w(12)012Qρ(1s)𝑑s+Qρ(12)(w(1)w(12))<.\displaystyle w^{\prime}\left({1\over 2}\right)\int_{0}^{1\over 2}Q_{\rho}(1-s)ds+Q_{\rho}\left({1\over 2}\right)\left(w(1)-w\left({1\over 2}\right)\right)<\infty.

By fnwf_{n}\leq w^{\prime}, we have 01Qρ(1s)fn(s)𝑑sa2\int_{0}^{1}Q_{\rho}(1-s)f_{n}(s)ds\leq a_{2}. Then

L(Qn,w;β,λ)\displaystyle L(Q_{n},w;\beta,\lambda)
=\displaystyle= 01(fn(s)λQρ(1s))24𝑑s+λ01(β+fn(s)λQρ(1s)2)Qρ(1s)𝑑s\displaystyle\int_{0}^{1}{(f_{n}(s)-\lambda Q_{\rho}(1-s))^{2}\over 4}ds+\lambda\int_{0}^{1}\left(\beta+{f_{n}(s)-\lambda Q_{\rho}(1-s)\over 2}\right)Q_{\rho}(1-s)ds
01(β+fn(s)λQρ(1s)2)w(s)𝑑s+a1\displaystyle-\int_{0}^{1}\left(\beta+{f_{n}(s)-\lambda Q_{\rho}(1-s)\over 2}\right)w^{\prime}(s)ds+a_{1}
\displaystyle\leq 01(fn(s)λQρ(1s))24𝑑s+λβE[ρ]+λa22\displaystyle\int_{0}^{1}{(f_{n}(s)-\lambda Q_{\rho}(1-s))^{2}\over 4}ds+\lambda\beta\mathrm{E}[\rho]+{\lambda a_{2}\over 2}
βw(1)1201fn(s)w(s)𝑑s+λa22+a1\displaystyle\quad-\beta w(1)-{1\over 2}\int_{0}^{1}f_{n}(s)w^{\prime}(s)ds+{\lambda a_{2}\over 2}+a_{1}
\displaystyle\leq 1401fn2(s)𝑑s+λ24E[ρ2]+λβE[ρ]+λa2βw(1)+a11201fn(s)w(s)𝑑s\displaystyle{1\over 4}\int_{0}^{1}f_{n}^{2}(s)ds+{\lambda^{2}\over 4}\mathrm{E}[\rho^{2}]+\lambda\beta\mathrm{E}[\rho]+\lambda a_{2}-\beta w(1)+a_{1}-{1\over 2}\int_{0}^{1}f_{n}(s)w^{\prime}(s)ds
\displaystyle\leq 1401fn2(s)𝑑s+λ24E[ρ2]+λβE[ρ]+λa2βw(1)+a11201fn2(s)𝑑s\displaystyle{1\over 4}\int_{0}^{1}f_{n}^{2}(s)ds+{\lambda^{2}\over 4}\mathrm{E}[\rho^{2}]+\lambda\beta\mathrm{E}[\rho]+\lambda a_{2}-\beta w(1)+a_{1}-{1\over 2}\int_{0}^{1}f^{2}_{n}(s)ds
=\displaystyle= 1401fn2(s)𝑑s+λ24E[ρ2]+λβE[ρ]+λa2βw(1)+a1\displaystyle-{1\over 4}\int_{0}^{1}f_{n}^{2}(s)ds+{\lambda^{2}\over 4}\mathrm{E}[\rho^{2}]+\lambda\beta\mathrm{E}[\rho]+\lambda a_{2}-\beta w(1)+a_{1}
\displaystyle\to  as n.\displaystyle-\infty\quad\text{ as }n\to\infty.

Therefore, infQ𝒬L(Q,w;β,λ)=\inf_{Q\in\mathscr{Q}}L(Q,w;\beta,\lambda)=-\infty. \Box

B.6 Proof of Proposition 7.1

For every λ>0\lambda>0,

Nλ(s)={λQρ((1s))+2aif s[0,p),λQρ((1s))+2bif s[p,1).N^{\prime}_{\lambda}(s)=\begin{cases}\lambda Q_{\rho}((1-s)-)+2a&\text{if }s\in[0,p),\\ \lambda Q_{\rho}((1-s)-)+2b&\text{if }s\in[p,1).\\ \end{cases}

Obviously, NλN_{\lambda} is concave on each of the intervals [0,p)[0,p) and [p,1][p,1]. Therefore, N˘λ\breve{N}^{\prime}_{\lambda} is constant on each of the two intervals.

We divide the discussion into the following two cases.

(i)

Assume that λ2b2aA1A2\lambda\geq{2b-2a\over A_{1}-A_{2}}.

In this case, Nλ(p)pNλ(1)Nλ(p)1p{N_{\lambda}(p)\over p}\geq{N_{\lambda}(1)-N_{\lambda}(p)\over 1-p}. Then

N˘λ(s)=Nλ(1)=λE[ρ]+2E[X0],s[0,1).\displaystyle\breve{N}^{\prime}_{\lambda}(s)=N_{\lambda}(1)=\lambda\mathrm{E}[\rho]+2\mathrm{E}[X_{0}],\quad s\in[0,1).

Therefore,

h(β,λ)={2β2E[X0]if 2βλE[ρ]+2E[X0],λE[ρ]if 2β>λE[ρ]+2E[X0].\displaystyle h(\beta,\lambda)=\begin{cases}2\beta-2\mathrm{E}[X_{0}]\quad&\text{if }2\beta\leq\lambda\mathrm{E}[\rho]+2\mathrm{E}[X_{0}],\\ \lambda\mathrm{E}[\rho]\quad&\text{if }2\beta>\lambda\mathrm{E}[\rho]+2\mathrm{E}[X_{0}].\end{cases}

As a consequence, βλ=E[X0]\beta_{\lambda}=\mathrm{E}[X_{0}],

Qλ(s)=E[X0]+λ2E[ρ]λ2Qρ((1s)),s[0,1)Q^{*}_{\lambda}(s)=\mathrm{E}[X_{0}]+{\lambda\over 2}\mathrm{E}[\rho]-{\lambda\over 2}Q_{\rho}((1-s)-),\quad s\in[0,1)

and

𝒳(λ)=E[X0]E[ρ]λ2Var[ρ].\displaystyle\mathcal{X}(\lambda)=\mathrm{E}[X_{0}]\mathrm{E}[\rho]-{\lambda\over 2}\mathrm{Var}[\rho].
(ii)

Assume that λ<2b2aA1A2\lambda<{2b-2a\over A_{1}-A_{2}}.

In this case, Nλ(p)p<Nλ(1)Nλ(p)1p{N_{\lambda}(p)\over p}<{N_{\lambda}(1)-N_{\lambda}(p)\over 1-p}. Then

N˘λ(s)=\displaystyle\breve{N}^{\prime}_{\lambda}(s)= {Nλ(p)pif s[0,p)Nλ(1)Nλ(p)1pif s[p,1)\displaystyle\begin{cases}{N_{\lambda}(p)\over p}&\text{if }s\in[0,p)\\ {N_{\lambda}(1)-N_{\lambda}(p)\over 1-p}&\text{if }s\in[p,1)\end{cases}
=\displaystyle= {2a+λA1if s[0,p),2b+λA2if s[p,1).\displaystyle\begin{cases}2a+\lambda A_{1}&\text{if }s\in[0,p),\\ 2b+\lambda A_{2}&\text{if }s\in[p,1).\end{cases}

Therefore,

h(β,λ)={2β2E[X0]if 2β2a+λA1,2β(1p)+λpA12b(1p)if 2a+λA1<2β2b+λA2,λE[ρ]if 2β>2b+λA2.\displaystyle h(\beta,\lambda)=\begin{cases}2\beta-2\mathrm{E}[X_{0}]&\text{if }2\beta\leq 2a+\lambda A_{1},\\ 2\beta(1-p)+\lambda pA_{1}-2b(1-p)&\text{if }2a+\lambda A_{1}<2\beta\leq 2b+\lambda A_{2},\\ \lambda\mathrm{E}[\rho]&\text{if }2\beta>2b+\lambda A_{2}.\end{cases}

As a consequence,

βλ={E[X0]if λ(2b2a)(1p)A1,bλp2(1p)A1if λ<(2b2a)(1p)A1.\displaystyle\beta_{\lambda}=\begin{cases}\mathrm{E}[X_{0}]&\text{if }\lambda\geq{(2b-2a)(1-p)\over A_{1}},\\ b-{\lambda p\over 2(1-p)}A_{1}&\text{if }\lambda<{(2b-2a)(1-p)\over A_{1}}.\end{cases}

Moreover, if

λ(2b2a)(1p)A1,\lambda\geq{(2b-2a)(1-p)\over A_{1}},

then

Qλ(s)={a+λ2A1λ2Qρ((1s))if s[0,p)b+λ2A2λ2Qρ((1s))if s[p,1)Q^{*}_{\lambda}(s)=\begin{cases}a+{\lambda\over 2}A_{1}-{\lambda\over 2}Q_{\rho}((1-s)-)&\text{if }s\in[0,p)\\ b+{\lambda\over 2}A_{2}-{\lambda\over 2}Q_{\rho}((1-s)-)&\text{if }s\in[p,1)\end{cases}

and hence

𝒳(λ)=apA1+b(1p)A2+λp2A12+λ(1p)2A22λ2E[ρ2];\displaystyle\mathcal{X}(\lambda)=apA_{1}+b(1-p)A_{2}+{\lambda p\over 2}A_{1}^{2}+{\lambda(1-p)\over 2}A_{2}^{2}-{\lambda\over 2}\mathrm{E}[\rho^{2}];

if

λ<(2b2a)(1p)A1,\lambda<{(2b-2a)(1-p)\over A_{1}},

then

Qλ(s)={bλp2(1p)A1λ2Qρ((1s))if s[0,p)b+λ2A2λ2Qρ((1s))if s[p,1)Q^{*}_{\lambda}(s)=\begin{cases}b-{\lambda p\over 2(1-p)}A_{1}-{\lambda\over 2}Q_{\rho}((1-s)-)&\text{if }s\in[0,p)\\ b+{\lambda\over 2}A_{2}-{\lambda\over 2}Q_{\rho}((1-s)-)&\text{if }s\in[p,1)\end{cases}

and hence

𝒳(λ)=bE[ρ]+λ2(1p)((1p)2A22p2A12)λ2E[ρ2].\displaystyle\mathcal{X}(\lambda)=b\mathrm{E}[\rho]+{\lambda\over 2(1-p)}\left((1-p)^{2}A_{2}^{2}-p^{2}A_{1}^{2}\right)-{\lambda\over 2}\mathrm{E}[\rho^{2}].

Summarizing the discussions in the above two cases, we have

𝒳(λ)={E[X0]E[ρ]λ2Var[ρ]if λ2b2aA1A2,apA1+b(1p)A2+λp2A12+λ(1p)2A22λ2E[ρ2]if (2b2a)(1p)A1λ<2b2aA1A2,bE[ρ]+λ2(1p)((1p)2A22p2A12)λ2E[ρ2]if 0<λ<(2b2a)(1p)A1.\displaystyle\mathcal{X}(\lambda)=\begin{cases}\mathrm{E}[X_{0}]\mathrm{E}[\rho]-{\lambda\over 2}\mathrm{Var}[\rho]\quad\text{if }\lambda\geq{2b-2a\over A_{1}-A_{2}},\\ apA_{1}+b(1-p)A_{2}+{\lambda p\over 2}A_{1}^{2}+{\lambda(1-p)\over 2}A_{2}^{2}-{\lambda\over 2}\mathrm{E}[\rho^{2}]\\ \hskip 142.26378pt\text{if }{(2b-2a)(1-p)\over A_{1}}\leq\lambda<{2b-2a\over A_{1}-A_{2}},\\ b\mathrm{E}[\rho]+{\lambda\over 2(1-p)}\left((1-p)^{2}A_{2}^{2}-p^{2}A_{1}^{2}\right)-{\lambda\over 2}\mathrm{E}[\rho^{2}]\quad\text{if }0<\lambda<{(2b-2a)(1-p)\over A_{1}}.\end{cases}

Then solving equation 𝒳(λ)=x\mathcal{X}(\lambda)=x yields

λ={2(E[X0]E[ρ]x)Var[ρ]if xE[X0]E[ρ]baA1A2Var[ρ],2(apA1+b(1p)A2x)E[ρ2]pA12(1p)A22if E[X0]E[ρ]baA1A2Var[ρ]<xbE[ρ]baA1((1p)E[ρ2]+p2A12(1p)2A22),2(1p)(bE[ρ]x)(1p)E[ρ2]+p2A12(1p)2A22if bE[ρ]baA1((1p)E[ρ2]+p2A12(1p)2A22)<x<bEρ].\displaystyle\lambda^{\circ}=\begin{cases}{2(\mathrm{E}[X_{0}]\mathrm{E}[\rho]-x)\over\mathrm{Var}[\rho]}\quad\text{if }x\leq\mathrm{E}[X_{0}]\mathrm{E}[\rho]-{b-a\over A_{1}-A_{2}}\mathrm{Var}[\rho],\\ {2(apA_{1}+b(1-p)A_{2}-x)\over\mathrm{E}[\rho^{2}]-pA_{1}^{2}-(1-p)A_{2}^{2}}\\ \quad\text{if }\mathrm{E}[X_{0}]\mathrm{E}[\rho]-{b-a\over A_{1}-A_{2}}\mathrm{Var}[\rho]<x\\ \hskip 85.35826pt\leq b\mathrm{E}[\rho]-{b-a\over A_{1}}\left((1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}\right),\\ {2(1-p)(b\mathrm{E}[\rho]-x)\over(1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}}\\ \quad\text{if }b\mathrm{E}[\rho]-{b-a\over A_{1}}\left((1-p)\mathrm{E}[\rho^{2}]+p^{2}A_{1}^{2}-(1-p)^{2}A_{2}^{2}\right)<x<b\mathrm{E}\rho].\end{cases}

Finally, a substitution leads to Proposition 7.1. \Box

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