1 The model-free approach to derivative pricing: problem motivation↩︎

Suppose \(S = (S_t)_{t \geq 0}\) is the price process of a risky asset in a financial market with riskless bank account paying deterministic rate of interest \(r = (r_t)_{t \geq 0}\). According to standard no-arbitrage theory the price of a call option with strike \(k\) and maturity \(T\) (i.e., a payoff of \((S_T-k)^+\) at time \(T\)) is given by \(\mathbb{E}^{\mathbb{Q}}[e^{- \int_0^T r_t dt} (S_T - k)^+]\) where \(\mathbb{Q}\) is a risk neutral measure and \(\mathbb{E}^{\mathbb{Q}}\) denotes expectations with respect to \(\mathbb{Q}\). In the classical approach we work on a filtered probability space \((\Omega,\mathcal{F},\mathbb{F}= (\mathcal{F}_t)_{0 \leq t \leq T},\mathbb{P})\) and assume that there exists an equivalent martingale measure \(\mathbb{Q}\) such that the discounted price process \(Z = (Z_t)_{0 \leq t \leq T}\), defined by \(Z_t = e^{- \int_0^t r_s ds} S_t\), is a \((\mathbb{F},\mathbb{Q})\)-martingale (and \(\mathbb{Q}\) is equivalent to \(\mathbb{P}\) on \(\mathcal{F}=\mathcal{F}_T\)). Then \(\mathbb{E}^{\mathbb{Q}}[e^{- \int_0^T r_t dt} (S_T - k)^+] = \mathbb{E}^{\mathbb{Q}}[(Z_T - {K})^+]\), where \({K} = ke^{- \int_0^T r_t dt}\) is the discounted strike. In a complete market, the price of the call can be justified as the lowest price with which it is possible to replicate the call option. As a simple example, \((\Omega,\mathcal{F},\mathbb{F}= (\mathcal{F}_t)_{0 \leq t \leq T},\mathbb{P})\) may support a Brownian motion \(W\) and if \(S\) is given by \(S_t = S_0 e^{\sigma W_t + \mu t}\) and \(r\) is constant then call prices are given by the Black-Scholes option pricing formula.

In well-functioning markets vanilla option prices are not given by a model, but rather are fixed by supply and demand. Then we may still have \(C(K,T) = \mathbb{E}^\mathbb{Q}[e^{- \int_0^T r_t dt} (S_T - K e^{\int_0^T r_t dt})^+] = \mathbb{E}^\mathbb{Q}[(Z_T - K)^+]\), but now it is the call prices which are given (as traded prices on the financial market) and the probabilistic model as represented by \(\mathbb{Q}\) which is unknown—typically we care about the risk-neutral probabilities \(\mathbb{Q}\) rather than the physical measure \(\mathbb{P}\). Nonetheless, if the set of call prices is sufficiently rich, then we can infer quantities such as \(\mathbb{Q}(Z_T>K)\). As Breeden and Litzenberger [1] conclude, this means that we do not need a model to price an option with payoff \(a(S_T)\) at time \(T\): instead we can write it as a combination of call and put payoffs whose prices are known.

What can we say about the prices of exotic or path-dependent options? Assume that we are given the prices of a class of derivatives (these become our vanilla, liquidly traded derivatives whose prices can be observed in the financial market), and that there exists a stochastic model such that in the model the discounted price process \(Z = (Z_t)_{t \geq 0}\) is a martingale under the risk-neutral measure \(\mathbb{Q}\) and the prices of vanilla derivatives are given by expectations under \(\mathbb{Q}\). Then the expected payoff under \(\mathbb{Q}\) is a candidate price for the exotic option. But, there may be many models which are consistent with the given prices of the vanilla derivatives. Then the robust derivative pricing problem becomes to find the supremum (and infimum) of the possible prices given by expectation, where the supremum (resp., infimum) is taken over all models (for which \(Z\) is a martingale) which agree with the quoted prices of the vanilla options in the sense that the expected discounted payoff under the model agrees with the traded price for each vanilla derivative. See Hobson [2] for a survey of this approach.

Suppose that the time index set is \(\mathbb{T}= \{ 0,1,2 \}\), suppose that the initial price of the risky asset is known, and that we know the prices of call and put options of all strikes with maturities \(T=1\) and \(T=2\). This is a reasonable class to take as the class of vanilla options. Then, with \(Z_1 = X\) and \(Z_2=Y\), we know \(C(K,1) = \mathbb{E}^{\mathbb{Q}}[(X-K)^+]\) and \(C(K,2) = \mathbb{E}^{\mathbb{Q}}[(Y-K)^+]\) for all \(K>0\). It follows that we know the laws of both \(X\) and \(Y\) (but note that we have no information about the joint law beyond the marginals). We denote these laws by \(\mu\) and \(\nu\), respectively. We also know that \((Z_0=z,Z_1,Z_2)\) is a martingale. It then follows that \(Z_0 = \int x \mu(dx) = \int y \nu(dy)\) and that \(\mu\) and \(\nu\) are in convex order³, denoted by \(\mu \leq_{cx} \nu\). For a given (Borel) \(c:\mathbb{R}^2\to\mathbb{R}\) and a path-dependent random payoff \(c = c(X,Y)\), the problem is to find \(\sup \mathbb{E}[ c(X,Y) ]\), where the supremum is taken over possible joint laws of \((X,Y)\) which respect the marginals \((X \sim \mu,Y \sim \nu)\) and the martingale property \(\mathbb{E}[Y|X] = X\). The case \(c(x,y)=\pm|y-x|\) corresponding to a forward start straddle was studied by Hobson and Neuberger [3] and Hobson and Klimmek [4] (see also Beiglböck and Juillet [5] and Henry- Labordère and Touzi [6], where the authors construct a model that is optimal for a certain (but large) class of cost functions \(c\)). More generally, this is the martingale optimal transport problem, as introduced by Beiglböck et al [7] and Galichon et al [8].

One fruitful approach to the martingale optimal transport problem is via the dual. In the context of the previous paragraph, the dual approach involves searching for univarite functions \(\phi, \psi\) and \(\theta\) such that \[\label{eq:mot} c(x,y) \leq \phi(x) + \psi(y) + \theta(x) (y-x),\quad x,y \in \mathbb{R}.\tag{1}\] If 1 holds and \(\mathbb{E}[Y|X]=X\), then, since \(\mathbb{E}[\theta(X)(Y-X)|X] = \theta(X)(\mathbb{E}[Y|X]-X) = 0\), we have \(\mathbb{E}[c(X,Y)] \leq \mathbb{E}[\phi(X) + \psi(Y)]\). The primal problem of finding \(\mathcal{P}= \sup \mathbb{E}[c(X,Y)]\), where the supremum is taken over joint laws with \(X \sim \mu\) and \(Y \sim \nu\) which respect the martingale property, is thus related to the dual problem of finding \(\mathcal{D}= \inf \left( \int \phi(x) \mu(dx) + \int \psi(y) \nu(dy) \right)\), where the infimum is taken over all trios \((\phi, \psi, \theta)\) for which 1 holds. The martingale optimal transport literature is concerned with formalising the above set-up, with deriving sufficient conditions for strong duality \(\mathcal{P}= \mathcal{D}\) (rather than the weak duality \(\mathcal{P}\leq \mathcal{D}\), which follows very easily) and with (explicitly constructing or) characterising the form of the primal and dual optimisers (where they exist) for particular choices of objective function \(c\).

In this article we are concerned with Bermudan-style payoffs in a two-period model. In the setting of the previous paragraph, given laws \(\mu\) and \(\nu\) in convex order, (Borel) functions \(a,b:\mathbb{R}\to\mathbb{R}\), and setting \(c(\cdot,1)=a\), \(c(\cdot,2)=b\), the primal problem is to find \[\label{eq:primal} \mathcal{P}= \mathcal{P}(\mu,\nu;a,b) = \sup_{\mathcal{M}\in M(\mu,\nu)} \sup_{\tau \in \mathcal{T}_{1,2}} \mathbb{E}^{\mathcal{M}}[ c(Z_\tau,\tau)],\tag{2}\] where \(M=M(\mu,\nu)\) is the set of \((\mu,\nu)\)-consistent models (recall, a model is a filtered probability space \((\Omega, \mathcal{F}, \mathbb{F}, \mathbb{Q})\) supporting a stochastic process \(Z=(Z_0,Z_1,Z_2)\) such that \(Z\) is a \((\mathbb{F},\mathbb{Q})\)-martingale with given marginals \(X \equiv Z_1 \sim \mu\) and \(Y \equiv Z_2 \sim \nu\)) and \(\mathcal{T}=\mathcal{T}_{1,2}\) is the set of \(\mathbb{F}\)-stopping times taking values in \(\{1,2\}\). As introduced in Neuberger [9] and Hobson and Neuberger [10], the dual problem is to find \[\label{eq:dual} \mathcal{D}= \mathcal{D}(\mu,\nu;a,b) = \inf_{\phi,\psi,\theta_1,\theta_2} \left( \int \phi(x) \mu(dx) + \int \psi(y) \nu(dy) \right),\tag{3}\] where the infimum is taken over quadruples \((\phi,\psi,\theta_1,\theta_2): \mathbb{R}\to \mathbb{R}\) such that, for all \(x,y\in\mathbb{R}\), \[\begin{align} a(x) & \leq & \phi(x) + \psi(y) + \theta_1(x)(y-x), \tag{4}\\ b(y) & \leq & \phi(x) + \psi(y) + \theta_2(x)(y-x). \tag{5} \end{align}\] Then, if 4 and 5 hold, for any \(\sigma\in\mathcal{T}\) we have that (almost surely) \[c(Z_\sigma,\sigma) = a(Z_\sigma)I_{ \{\sigma=1\} } + b(Z_\sigma) I_{ \{ \sigma= 2 \} } \leq \phi(Z_1) + \psi(Z_2) + \theta_\sigma(Z_1) (Z_2 - Z_1) .\] In particular, whatever the stopping strategy of the Bermudan option holder, a hedging strategy of

1. holding a portfolio of call and put options with maturity \(T=1\) and payoff \(\phi\),

2. holding a portfolio of call and put options with maturity \(T=2\) and payoff \(\psi\),

3. if the Bermudan option is exercised at \(t=1\), holding \(\theta_1 =\theta_1(Z_1)\) units of the risky asset between times one and two,

4. otherwise, if the Bermudan option is not exercised at \(t=1\) holding \(\theta_2=\theta_2(Z_1)\) units of the risky asset between times one and two

is a superreplicating strategy. Furthermore, under each \((\mu,\nu)\)-consistent model \(\mathcal{M}\in M(\mu,\nu)\) we have that \(\mathbb{E}^\mathcal{M}[c(Z_\sigma,\sigma)]\leq\int \phi(x) \mu(dx) + \int \psi(y) \nu(dy)\), from which the weak duality \(\mathcal{P}\leq\mathcal{D}\) follows.

Neuberger [9] and Hobson and Neuberger [10] studied the Bermudan option pricing problem for assets taking values on a lattice and showed (using linear programming methods) that there is no duality gap \(\mathcal{P}=\mathcal{D}\). One of the key insights was that the filtration matters and it is not enough to simply consider the primal problem as one of finding the optimal martingale transport in the canonical filtration for the price process. The results in [10] were re-proved and extended (e.g., to the non-lattice case) by Aksamit et al. [11], where the authors, instead of focusing on the filtration, considered the impact of enlarging the set of traded assets. They show that in a wide set of circumstances strong duality holds. Bayraktar et al. [12] also consider the robust hedging of Bermudan-style options in a discrete-time framework. They consider both upper and lower bounds, but, since they restrict attention to a setting where the filtration is the canonical filtration, they find a duality gap—subsequently Bayraktar and Zhou [13] show that this gap can be removed if the set-up is extended to allow for randomized stopping times.

Hobson and Norgilas [14] studied the Bermudan option pricing problem for the case of put options. (In the case when the risk-free interest rate is positive, a Bermudan call is trivial since the optimal strategy is to wait until maturity to exercise the call.) The authors showed that there is no duality gap, the model which achieves the highest price for the put is associated to the left-curtain martingale coupling of Beiglbock and Juillet [5], and it is possible to write down the cheapest superhedging portfolio. For a given strike for the Bermudan put, the optimal portfolio involves vanilla puts and calls with a finite number of strikes. The results in [14] are obtained under the assumption that the initial law \(\mu\) is atom-free, and, in the setting of [14], it is enough to look for models that are equipped with the canonical filtration of the price process. Later, Hobson and Norgilas [15] extended the results of [14] to the case of a general initial law \(\mu\): in this setting, the optimal model is (still) associated to the lifted left-curtain coupling, but the information generated by the price process alone is no longer sufficient and additional randomization is required.

In this paper we extend the results of [10] and [14] to general convex payoffs. Our results are in two directions. First, we show that the set of superreplicating strategies over which we search in the dual problem can be greatly simplified. Second, under some structural assumptions on the laws \(\mu\) and \(\nu\) (which are trivially satisfied in the case where \(\mu\) and \(\nu\) are (in convex order and) both normal, log-normal or shifted exponential) we characterize the optimal model and the optimal superhedge. Based on [14], one could conjecture that, in the case the marginals \((\mu,\nu)\) are atom-free, it is enough to restrict the search to the set of models that are equipped with the canonical filtration of the price process. From our results, however, it follows that even in this regular case there is no maximiser within this class. Instead a richer class of models is required.

The structural conditions that we require are that \(\mu\) and \(\nu\) satisfy the Dispersion Assumption (see Definition 36) introduced in Hobson and Klimmek [4]. This assumption states that there exist points \(e^{L}=e^L(\mu,\nu)< e^R=e^R(\mu,\nu)\) such that if \(\rho\) (resp., \(\eta\)) is the density of \(\mu\) (resp., \(\nu\)) and \((\alpha_\mu,\beta_\mu)\) (resp., \((\alpha_\nu,\beta_\nu)\)) is the support of \(\mu\) (resp., \(\nu\)), then \(\rho < \eta\) on \((\alpha_\nu,e^L) \cup (e^R,\beta_\nu)\) whereas \(\eta < \rho\) on \((e^L,e^R)\).

Notation: for a measurable function \(h\) we write \(h^+\) for its positive part, and define \(h^c\) to be the convex hull of \(h\), so that \(h^c\) is the largest convex function \(H\) satisfying \(H \leq h\); for a convex function \(g\) we write \(g'\) for its right-derivative – we could in fact use any subdifferential. Sometimes we abbreviate \(\int a(x) \mu(dx)\) to \(\int a d\mu\).

For any \(w,z\in \mathbb{R}\) with \(w < z\), and two functions \(g,h:\mathbb{R}\to \mathbb{R}\cup\{\infty\}\) with \(g(w),h(z)\in\mathbb{R}\), the line that goes through \((w,g(w))\) and \((z,h(z))\) is denoted by \(T^{w,z}_{g,h}:\mathbb{R}\to\mathbb{R}\).

We denote the point mass at \(x\) by \(\delta_x\). If \(f,g\in\mathbb{R}\) are such that \(f<x < g\), then \(\pi_x^{f,g}\), given by \(\pi_x^{f,g} = \frac{x-f}{g-f} \delta_g + \frac{g-x}{g-f} \delta_f\), is the two-point distribution with support \(\{ f,g \}\) and mean \(x\). If \(f \leq x=g\) or \(f=x < g\) then \(\pi_x^{f,g} = \delta_x\).

For a (Borel) measure \(\xi\) on \(\mathbb{R}\), the barycenter of \(\xi\) is defined by \(\bar\xi=\int_\mathbb{R}x\xi(dx)/\xi(\mathbb{R})\).

2 Simplifying the dual problem↩︎

In this section we want to study the cheapest superhedge for a Bermudan-style option which pays \(a(Z_1)\) if exercised at time-1 and \(b(Z_2)\) if exercised at time-2, where we assume that the prices of European options imply that \(X \equiv Z_1\) has law \(\mu\) and \(Y \equiv Z_2\) has law \(\nu\). Necessarily, we must have \(\mu \leq_{cx} \nu\), so that both \(\mu\) and \(\nu\) are integrable (i.e., elements of \(L^1\)) and \(\int x \mu(dx) = \int y \nu(dy)\). We also assume that both \(a,b :\mathbb{R}\to\mathbb{R}_+\) are non-negative.

The terminology is explained by the fact that if \((\phi,\psi, \theta)\) is a superhedge then \(a(Z_1) I_{ \{\tau = 1 \} } + b(Z_2)I_{ \{\tau > 1 \} } \leq \phi(Z_1) + \psi(Z_2) + \theta(Z_1) (Z_2-Z_1)\) holds almost surely, where \(\theta = I_{ \{\tau = 1 \} } \theta_1 + I_{ \{\tau = 2 \} } \theta_2\). We write \(\mathcal{S}= \mathcal{S}(a,b)\) for the set of superhedges, in the sense of Definition 1, for the Bermudan option with payoff \((a,b)\).

Note that the component \(\theta(Z_1)(Z_2-Z_1)\) does not contribute to the cost of the hedge. This corresponds to the notion that \(Z\) is the discounted price of a traded asset, and is related to the notion that \(Z\) is a martingale under any consistent martingale model.

Since \(H(\phi,\psi,\theta)\) does not depend on \(\theta\) we write \(HC(\phi,\psi)\) instead of \(HC(\phi,\psi,\theta)\).

The problem of finding the cheapest superhedging strategy is the dual problem:

It follows from Hobson and Norgilas [14] that any function \(\psi \geq b\), with \(\psi\) convex, can be used to generate a superhedge:

Let \(\tilde{\mathcal{S}}(b) = \{ \psi \geq b with \psi convex. \}\). Let \((\phi,\psi,\theta)\) be given by \((\phi,\psi,(\theta_{i})_{i=1,2}) = ((a-\psi)^+,\psi, - \psi',0)\). If follows from Lemma [lem:phifrompsi] that each \(\psi \in \tilde{\mathcal{S}}(b)\) generates an element \((\phi,\psi,\theta) \in \mathcal{S}(a,b)\). Then, for \(\psi \in {\tilde{\mathcal{S}}}(b)\) we can define \(\widetilde{HC}(\psi) = HC((a-\psi)^+,\psi)\), which is the hedging cost associated with the superhedge \((\phi,\psi, \{\theta_i \}_{i = 1,2}) = (\phi,\psi, -\psi',0)\).

Clearly \(\mathcal{D}\leq \tilde{\mathcal{D}}\). Now suppose \(b\) is convex. The main result of this section is that the cheapest superreplicating strategy is of a form generated by \(\psi \in \tilde{\mathcal{S}}(b)\).

The idea behind the proof is to take a general superhedging strategy \((\phi,\psi,\theta) \in \mathcal{S}(a,b)\) and to show that it can be modified to give another superhedging strategy which is generated by an element of \(\hat{\psi} \in \tilde{\mathcal{S}}(b)\), and which has a lower hedging cost. We do this in three stages. First we show that given \((\phi,\psi,\theta) \in \mathcal{S}(a,b)\) we can replace \(\psi\) with \(\psi^c\), so that \((\phi,\psi^c,\theta) \in \mathcal{S}(a,b)\) is still a superreplicating strategy. Clearly, this can only reduce the hedging cost. Hence, without loss of generality, we may restrict attention to superhedges for which \(\psi\) is convex. Second, we show that, given \((\phi,\psi,\theta) \in \mathcal{S}(a,b)\) with \(\psi\) convex, we can take a particular choice of \(\phi\) (namely \(\phi=\max \{ (a-\psi), (-(\psi-b)^c) \}\)) and we still have a superhedge. Again, we will show that this can only lower the hedging cost. Hence we may restrict attention to superhedges for which \(\psi\) is convex and \(\phi\) takes this particular form. Finally, we show that, given \(\psi\) convex and \(\phi\) of the particular form, we can introduce \(\hat{\psi}\) with \(\hat{\psi} = \psi - (\psi-b)^c \geq b\), and such that the hedging cost associated with the superhedge generated by \(\hat{\psi}\) is no larger than the hedging cost associated with the superhedge \((\phi,\psi,\theta)\).

We begin with some preliminaries from Beiglböck et al [16].

Also, we have the following ‘obvious’ result.

From now on we may and do assume that \(\psi\) is convex.

From now on we may assume that \(\psi\) is convex and \(\phi = (a - \psi) \vee (-(\psi-b)^c)\).

The main result of this section is Theorem 6, which says that when looking for an optimiser in the dual problem it is sufficient to search over convex functions which dominate \(b\), and then to use these convex functions to generate hedging strategies. This greatly simplifies the analysis of the dual problem. In the next few sections we exploit this simplification to find the solution of the primal problem and to show that there is no duality gap. We make a related point in Remark 20.

3 Standing assumptions↩︎

In the second part of the paper we aim to find explicit solutions to the pricing and hedging problems under some simplifying structural assumptions on \(\mu\) and \(\nu\). The goal is to find the model \(\mathcal{M}^*\) and associated stopping time \(\tau^*\) such that the highest model-based price is attained, along with the cheapest superhedege \((\phi^*,\psi^*,\theta^*)\). We find candidates for each and proceed to show that \(\mathcal{P}^* := \mathbb{E}^{\mathcal{M}^*}[c(Z_{\tau^*},\tau^*)] = \int \phi^* d \mu + \int \psi^* d \nu =: \mathcal{D}^*\) where \(\mathbb{E}^{\mathcal{M}}\) denotes expectations in the model \(\mathcal{M}=(\Omega, \mathcal{F}, \mathbb{F}, \mathbb{P})\). Then \(\mathcal{P}^* \leq \mathcal{P}\leq \mathcal{D}\leq \mathcal{D}^*\) (the two outer inequalities are by definition, and the middle one follows by weak duality). It follows that we must have equality throughout and that we have found a model under which the Bermudan option has the highest price and a superhedge with the lowest hedging cost. Moreover, there is no duality gap. We begin by stating standing assumptions on the laws \(\mu\) and \(\nu\) and the payoff functions.

Consider a Bermudan option with convex payoff functions \(a,b:\mathbb{R}\to[0,\infty]\) which satisfy \(a\geq b\) on \(\mathbb{R}\) and \(a(x_0) <\infty\) for some \(x_0\in\mathbb{R}\). We begin by ruling out a couple of cases where the result is trivial.

First, if \(a=b\) on \((\alpha_\mu,\beta_\mu)\), then by (the conditional) Jensen’s inequality and the martingale property of \((Z_0,Z_1,Z_2)\) (under any model \(\mathcal{M}\in M(\mu,\nu)\)), we have that \((C_1=a(Z_1),C_2=b(Z_2))\) forms a \(\mathcal{M}\)-submartingale, and thus it is optimal to stop at time-2. It follows that every model \(\mathcal{M}\) is optimal and \(\mathcal{P}=\int_\mathbb{R}b(y)\nu(dy)\).

Second, dropping the assumption that \(a=b\), due to the convexity (of \(a\) and \(b\)) we have that there exist intervals \(I_a,I_b\subseteq \mathbb{R}\) such that \(a<\infty\) on \(I_a\) (resp., \(b<\infty\) on \(I_b\)) and \(a=\infty\) on \(I_a^\infty:=\mathbb{R}\setminus I_a\) (resp., \(b=\infty\) on \(I_b^\infty:=\mathbb{R}\setminus I_b\)). If \(\mu(I^\infty_a)=\mu((\alpha_\mu,\beta_\mu)\cap I^\infty_a)>0\) (resp., \(\nu(I^\infty_b)=\nu((\alpha_\nu,\beta_\nu)\cap I^\infty_b)>0\)), then for any model \(\mathcal{M}\), by taking \(\tau^*=1\) (resp., \(\tau^*=2\)) we have that \(\mathbb{E}^{\mathcal{M}}[a(Z_1)I_{ \{ \tau^*=1 \} } + b(Z_2)I_{ \{ \tau^*=2 \} } ]=\mathbb{E}^\mathcal{M}[a(Z_1)]=\int_\mathbb{R}a(x)\mu(dx)=\infty\) (resp., \(\mathbb{E}^{\mathcal{M}}[a(Z_1)I_{ \{ \tau^*=1 \} } + b(Z_2)I_{ \{ \tau^*=2 \} } ]=\mathbb{E}^\mathcal{M}[b(Z_2)]=\int_\mathbb{R}b(y)\nu(dy)=\infty\)). Again, it follows that all models \(\mathcal{M}\) are optimal and \(\mathcal{P}=\infty\).

These two observations motivate the following standing assumption (the second part is a mild simplifying assumption on the payoff functions which helps in the proof of Proposition 48).

We extend the definitions of \(a\) and \(b\) to \(\mathbb{R}\) by setting \(a = \infty\) on \((-\infty,\alpha_\mu) \cup (\beta_\mu,\infty)\), \(b = \infty\) on \((-\infty,\alpha_\nu) \cup (\beta_\nu,\infty)\), \(a(\alpha_\mu)=\lim_{x\downarrow\alpha_\mu}a(x)\), \(a(\beta_\mu)=\lim_{x\uparrow\beta_\mu}a(x)\), \(b(\alpha_\nu)=\lim_{x\downarrow\alpha_\nu}b(x)\) and \(b(\beta_\nu)=\lim_{x\uparrow\beta_\nu}b(x)\). Note that the choice to extend \(a\) and \(b\) in this way does not affect the expected payoff of the Bermudan option since \(\mu((\alpha_\mu,\beta_\mu)) = 1 = \nu((\alpha_\nu,\beta_\nu))\).

Let \(L^b\) be the smallest straight line such that \(L^b \geq b\) on \((\alpha_\nu,\beta_\nu)\). Note that if both \(\alpha_\nu\) and \(\beta_\nu\) are finite and \(b(\alpha_\nu)\vee b(\beta_\nu)<\infty\) then \(L^b=T_{b,b}^{\alpha_\nu,\beta_\nu}\) is the straight line passing through \((\alpha_\nu,b(\alpha_\nu))\) and \((\beta_\nu,b(\beta_\nu))\); if \(b(\alpha_\nu)\vee b(\beta_\nu)=\infty\) then there does not exist a finite straight line with \(L^b \geq b\) in which case we set \(L^b \equiv \infty\). If \(\alpha_\nu\) and \(\beta_\nu\) are both infinite then either \(b\) is constant or again there is no finite straight line with \(L^b \geq b\) and then we set \(L^b \equiv \infty\). If exactly one of the endpoints \(\alpha_\nu,\beta_\nu\) is finite then there may or may not be a finite straight line with \(L^b \geq b\), if not then \(L^b \equiv \infty\).

The case considered in the following lemma only occurs when \(L^b\) is finite.

From now on we will exclude this case. Note that if \(b\) is constant and \(a \geq b\) then \(a \geq L^b =b\), so that in particular we are excluding the case where \(b\) is constant. This case is covered in Lemma 15.

4 Special subsets of \({\mathbb{R}}^n\), special convex functions and the main result↩︎

In this section we introduce three classes of models defined via four, five or six points and three classes of hedges, again defined by four, five or six points. The plan is to show that if we can find a model and a hedge defined by the same \(n\) points (where \(n \in \{4,5,6\}\)) then the model and hedge are both optimal and there is no duality gap. Later we show that under some reasonable assumptions it is always possible to find such a set of points.

For \(n=4,5,6\) and \(w^{(n)} = (w_1, \ldots, w_n) \in \mathbb{R}^n\) define

\(\Sigma_n= \Sigma_n^{\mu,\nu}\) by \[\begin{align} \Sigma^{\mu,\nu}_4 & = & \{ w^{(4)} \in \mathbb{R}^4 : \alpha_\nu < w_1 \leq w_2 < w_3 \leq w_4 < \beta_\nu, \alpha_\mu \leq w_2 < w_3 \leq \beta_\mu, and \\ &&(\beta_\mu - w_3)+(w_2 - \alpha_\mu)>0 \}, \\ \Sigma^{\mu,\nu}_5 & = & {\{ w^{(5)} \in \mathbb{R}^5 : \alpha_\nu < w_1 \leq w_2 \leq w_3 \leq w_4 \leq w_5 < \beta_\nu, \alpha_\mu < w_2 < w_4 <\beta_\mu, and } \\ &&{(either w_2>w_1 or w_1=w_2=w_3) and (either w_4<w_5 or w_3=w_4=w_5) \}, } \\ \Sigma^{\mu,\nu}_6 & = & \{ w^{(6)} \in \mathbb{R}^6 : \alpha_\nu < w_1 \leq w_2 \leq w_3 < w_4 \leq w_5 \leq w_6 < \beta_\nu, \alpha_\mu < w_2 < w_5 <\beta_\mu,and \\ &&{ (either w_2>w_1 or w_1=w_2=w_3) and (either w_5<w_6 or w_4=w_5=w_6) \}. } \end{align}\]

Note the different conventions over weak and strict inequalities in each case. Essentially, for \(n=4\) we need to allow for one of the points \(w_2,w_3\) to lie at the ends of the support of \(\mu\), but this case does not occur when \(n=5\) or \(n=6\).

For \(n=4\) define \(\Gamma^{\mu,\nu}_4\) by \[\begin{align} \label{eq:Gamma4def} \Gamma^{\mu,\nu}_4 & = & \{ (y_-,x_-,x_+,y_+) \in \Sigma_4 : \mu|_{(\alpha_\mu, y_-) \cup (y_+,\beta_\mu)} \leq \nu|_{(\alpha_\mu, y_-) \cup (y_+,\beta_\mu)}, \\ && \nonumber\mu|_{(y_-,x_-) \cup (x_+,y_+)} \leq_{cx} \nu|_{(y_-,y_+)} \} . \end{align}\tag{9}\]

For \(n=5\) define \(\Gamma^{\mu,\nu}_5\) by \[\begin{align} \label{eq:Gamma5def} \Gamma^{\mu,\nu}_5 & = & \{ (y_-,x_-,z,x_+,y_+) \in \Sigma_5 : \mu|_{(\alpha_\mu, y_-) \cup (y_+,\beta_\mu)} \leq \nu|_{(\alpha_\mu, y_-) \cup (y_+,\beta_\mu)}, \\ && \nonumber\mu|_{(y_-,x_-)} \leq_{cx} \nu|_{(y_-,z)}, \mu|_{(x_+,y_+)} \leq_{cx} \nu|_{(z,y_+)} \} . \end{align}\tag{10}\]

As in the case of \(n=4\) we have an identical result with a similar proof.

For \(n=6\) define \(\Gamma^{\mu,\nu}_6\) by \[\begin{align} \label{eq:Gamma6def} \Gamma^{\mu,\nu}_6 & = & \{ (y_-,x_-,z_-,z_+,x_+,y_+) \in \Sigma_6 : \mu|_{(\alpha_\mu, y_-) \cup (y_+,\beta_\mu)} \leq \nu|_{(\alpha_\mu, y_-) \cup (y_+,\beta_\mu)}, \\ && \nonumber\mu|_{(y_-,x_-)} \leq_{cx} \nu|_{(y_-,z_-)}, \mu|_{(x_+,y_+)} \leq_{cx} \nu|_{(z_+,y_+)}, \nu|_{(z_-,z_+)} \leq \mu|_{(z_-,z_+)} \} . \end{align}\tag{11}\] This time we find that

We let \(C(\alpha_\nu,\beta_\nu)\) be the set of continuous functions on \((\alpha_\nu,\beta_\nu)\) and we let \(C_{cx}(\alpha_\nu,\beta_\nu) \subset C(\alpha_\nu,\beta_\nu)\) be the set of convex functions on \((\alpha_\nu,\beta_\nu)\).

For \((y_-,x_-,x_+,y_+) \in \Sigma_4\) define \(\Psi_4 = \Psi^{y_-,x_-,x_+,y_+}_{4,a,b}:(\alpha_\nu,\beta_\nu): \to \mathbb{R}\) by \[\label{eq:psi4points} \Psi_4 (u) = \begin{cases} b(u), & u < y_-; \\ T^{y_-,y_+}_{b,b}(u), \quad \quad & y_- \leq u \leq y_+; \\ b(u), & y_+ < u. \end{cases}\tag{12}\] Our convention that \(\alpha_\nu < y_-<y_+ < \beta_\nu\) for \((y_-,x_-,x_+,y_+) \in \Sigma_4\) ensures that \(\Psi_4\) is well defined. By construction, \(\Psi_4\) is convex. Now we define a subset \(\mathcal{L}^{a,b}_4\) of \(\Sigma_4\) with the property that \(T_{b,b}^{y_-,y_+}\) takes the same value as \(a\) at \(x_{\pm}\): \[\label{eq:L4def} \mathcal{L}^{a,b}_4 = \{ (y_-,x_-,x_+,y_+) \in \Sigma_4 : a(x_--) \geq T_{b,b}^{y_-,y_+}(x_-) \geq a(x_-+), a(x_+-) \leq T_{b,b}^{y_-,y_+}(x_+) \leq a(x_++) \} .\tag{13}\] Observe that if \(x_- > \alpha_\mu\) then \(a(x_--)= a(x_-+)\), and similarly if \(x_+<\beta_\mu\) then \(a(x_+ -)= a(x_+ +)\), so that in the case when \(\alpha_\mu<x_-<x_+<\beta_\mu\) we can more simply write the condition \((y_-,x_-,x_+,y_+) \in \mathcal{L}^{a,b}_4\) as the condition that the points \((y_-,b(y_-))\), \((x_-,a(x_-))\), \((x_+,a(x_+))\) and \((y_+,b(y_+))\) all lie on the same straight line.

For \((y_-,x_-,z,x_+,y_+) \in \Sigma_5\) define \(\Psi_5=\Psi^{y_-,x_-,z,x_+,y_+}_{5,a,b}:(\alpha_\nu,\beta_\nu): \to \mathbb{R}\) by \[\label{eq:psi5points} \Psi_5 (w) = \begin{cases} b(w), & w \leq y_-; \\ T^{y_-,x_-}_{b,a}(w), \quad \quad & y_- < w \leq z; \\ T^{x_+,y_+}_{a,b}(w), & z < w \leq y_+; \\ b(w), & y_+<w. \end{cases}\tag{14}\] Note that \(\Psi_5\) need not be continuous at \(z\). Furthermore, if \(y_-=z\) (resp., \(z=y_+\)) then the second (resp., third) line in the definition of \(\Psi_5\) is redundant; otherwise, by the definition of \(\Sigma_5\) we must have that \(y_-<x_-\) (resp., \(x_+=y_+\)) so that \(T^{y_-,x_-}_{b,a}\) (resp., \(T^{x_+,y_+}_{a,b}\)) is well defined. Let \(\mathcal{L}^{a,b}_5\) be given by \[\mathcal{L}^{a,b}_5 = \{ (y_-,x_-,z,x_+,y_+) \in \Sigma_5 : \Psi_{5,a,b}^{y_-,x_-,z,x_+,y_+} \in C_{cx}(\alpha_\nu,\beta_\nu), \Psi_{5,a,b}^{y_-,x_-,z,x_+,y_+}(z) \geq a(z) \} .\] In particular, \(\Psi_{5} \in C_{cx}(\alpha_\nu,\beta_\nu)\) guarantees that \(\Psi_{5}\) is continuous.

For \((y_-,x_-,z_-,z_+,x_+,y_+) \in {\Sigma}_6\) define \(\Psi_6 = \Psi^{y_-,x_-,z_-,z_+,x_+,y_+}_{6,a,b}:(\alpha_\nu,\beta_\nu): \to \mathbb{R}\) by \[\label{eq:psi6points} \Psi_6 (w) = \begin{cases} b(w), & w \leq y_-; \\ T^{y_-,x_-}_{b,a}(w), \quad \quad & y_- < w \leq z_-; \\ a(w), & z_- < w\leq z_+;\\ T^{x_+,y_+}_{a,b}(w), & z_+ < w \leq y_+; \\ b(w), & y_+<w. \end{cases}\tag{15}\] Note that \(\Psi^{y_-,x_-,z_-,z_+,x_+,y_+}_{6,a,b}\) need not be continuous at \(w\) for \(w \in \{ z_\pm \}\). Furthermore, if \(y_-=z_-\) (resp., \(z_+=y_+\)) then the second (resp., fourth) line in the definition of \(\Psi_6\) is redundant; otherwise the definition of \(\Sigma_6\) ensures that \(T^{y_-,x_-}_{b,a}\) (resp., \(T^{x_+,y_+}_{a,b}\)) is well defined. Let \(\mathcal{L}^{a,b}_6\) be given by \[\mathcal{L}^{a,b}_6 = \{ (y_-,x_-,z_-,z_+,x_+,y_+) \in \Sigma_6 : \Psi_{6,a,b}^{y_-,x_-,z_-,z_+,x_+,y_+} \in C_{cx}(\alpha_\nu,\beta_\nu) \} .\]

The first main result says that if we can find \(w = (w_1, \ldots w_n) \in \Gamma_n^{\mu,\nu} \cap \mathcal{L}_{n}^{a,b}\) for some \(n \in \{4,5,6\}\) then there is no duality gap and \(w\) characterises both the optimal model and the optimal hedge.

The second main result says that under an assumption on the pair of laws \((\mu,\nu)\) which will be introduced below, but which is satisfied for a wide class of examples, the non-emptyness hypothesis of Theorem 21 is satisfied. It follows that under this assumption the problem is completely solved.

5 Martingale couplings and models↩︎

5.1 Canonical models and lifted models↩︎

In order to explicitly solve the primal problem 2 (by finding an optimal model and an associated optimal stopping time), we introduce two useful (and natural) subsets of the set of \((\mu,\nu)\)-consistent models \(M(\mu,\nu)\). For \(S = (\alpha,\beta) \subseteq \mathbb{R}\) let \(\mathcal{B}(S)\) denote the set of Borel subsets of \((\alpha,\beta)\) and more generally for \(S^n=(\alpha_1,\beta_1) \times \ldots \times (\alpha_n,\beta_n) \subseteq \mathbb{R}^n\) let \(\mathcal{B}(S^n)\) denote the set of Borel subsets of \(S^n\).

Let \(\Pi_M(\mu,\nu)\) be the set of martingale couplings of \(\mu\leq_{cx}\nu\). In particular, we write \(\pi\in\Pi_M(\mu,\nu)\) if \(\pi\) is a (Borel) probability measure on \((\alpha_\mu,\beta_\mu)\times(\alpha_\nu,\beta_\nu)\), if the first and second marginals of \(\pi\) are \(\mu\) and \(\nu\), respectively (i.e., \(\pi(A\times(\alpha_\nu,\beta_\nu))=\mu(A)\) and \(\pi((\alpha_\mu,\beta_\mu)\times B)=\nu(B)\) for all \(A\subseteq \mathcal{B}((\alpha_\mu,\beta_\mu)),B\subseteq\mathcal{B}((\alpha_\nu,\beta_\nu))\)), and if \(\pi(dx,dy)=\mu(dx)\pi_x(dy)\) is such that \(\int |y|\pi_x(dy)<\infty\) and \(\int y\pi_x(dy)=x\) for \(\mu\)-a.e. \(x\in(\alpha_\mu,\beta_\mu)\). Here \((\pi_x)_{x \in (\alpha_\mu,\beta_\mu)}\) is a disintegration of \(\pi\) with respect to \(x\). The following theorem is due to Strassen [17]

It follows that \(\Pi_M(\mu,\nu)\) naturally induces the set of canonical models. In particular, let \(\Omega^{can}= \{(x,y):\alpha_\mu < x < \beta_\mu,~ \alpha_\nu < y < \beta_\nu \}\), \(\mathcal{F}^{can} = \{ A \times B: A \in \mathcal{B}((\alpha_\mu,\beta_\mu)),~ B \in \mathcal{B}((\alpha_\nu,\beta_\nu))\}\), \(\mathcal{F}^{can}_0 = \{\Omega,\emptyset\}\), \(\mathcal{F}^{can}_1 = \{ A \times (\alpha_\nu,\beta_\nu) : A \in \mathcal{B}((\alpha_\mu,\beta_\mu)) \}\),  \(\mathcal{F}^{can}_2 = \mathcal{F}^{can}\) and define \[\label{eq:canonicalModels} M^{can}(\mu,\nu):=\{(\Omega^{can},\mathcal{F}^{can},\mathbb{F}^{can}=(\mathcal{F}^{can}_0,\mathcal{F}^{can}_1,\mathcal{F}^{can}_2),\mathbb{P}=\pi):\pi\in\Pi_M(\mu,\nu)\}.\tag{16}\] Then \(M^{can}(\mu,\nu)\subset M(\mu,\nu)\), and for each \(\pi\in\Pi_M(\mu,\nu)\) there are random variables \((X,Y)=(Z_1,Z_2)\) such that \(\mathbb{P}(X \in dx, Y \in dy) = \pi(dx,dy)\). The corresponding model (associated to a coupling \(\pi\in\Pi_M(\mu,\nu)\)) is denoted by \(\mathcal{M}_\pi\in M^{can}(\mu,\nu)\). Denoting by \(\mathcal{T}^{can}_{1,2}\) the set of \(\mathbb{F}^{can}\)-stopping times (taking values in \(\{1,2\}\)) set \[\begin{align} \label{eq:primalCanonical} \mathcal{P}^{can} = \mathcal{P}^{can}(\mu,\nu;a,b) &= \sup_{\mathcal{M}\in M^{can}(\mu,\nu)} \sup_{\tau \in \mathcal{T}^{can}_{1,2}} \mathbb{E}^{\mathcal{M}}[ c(Z_\tau,\tau)], \end{align}\tag{17}\] and note that \[\mathcal{P}\geq\mathcal{P}^{can}=\sup_{\pi\in\Pi_M(\mu,\nu)}\sup_{B\in \mathcal{B}((\alpha_\mu,\beta_\mu))}\left\{\int_Ba(x)\mu(dx)+\int_{((\alpha_\mu,\beta_\mu)\setminus B)\times\mathbb{R}}b(y)\pi_x(dy)\mu(dx)\right\}.\]

The second useful subset of \(M(\mu,\nu)\) requires a uniform random variable on \([0,1]\) that is independent of \(X\sim\mu\).

Let \(\lambda=\lambda_{[0,1]}\) denote the Lebesgue measure on \([0,1]\). We write \(\hat{\pi}\in\hat{\Pi}_M(\mu,\nu)\) if \(\hat{\pi}\) is a (Borel) probability measure on \((\alpha_\mu,\beta_\mu)\times(0,1)\times(\alpha_\nu,\beta_\nu)\), if \(\hat{\pi}(A\times V\times\mathbb{R})=(\mu\otimes\lambda)(A\times V)\) for all \(A\in \mathcal{B}((\alpha_\mu,\beta_\mu)), V \in \mathcal{B}((0,1))\) (i.e., \(\int_y\hat{\pi}(dx,du,dy)=(\mu\otimes\lambda)(dx,du)=\mu(dx)du\) is the product measure of \(\mu\) and \(\lambda\), so that trivially the first and second marginals of \(\hat{\pi}\) are \(\mu\) and \(\lambda\), respectively), if the third marginal of \(\hat{\pi}\) is \(\nu\), and if \(\hat{\pi}(dx,du,dy)=\mu(dx)du\hat{\pi}_{x,u}(dy)\) is such that \(\int |y|\pi_{x,u}(dy)<\infty\) and \(\int y\pi_{x,u}(dy)=x\) for \((\mu\otimes\lambda)\)-a.e. \((x,u)\in(\alpha_\mu,\beta_\mu)\times(0,1)\). We again have that \(\hat{\Pi}_M(\mu,\nu)\) is non-empty if and only if \(\mu\leq_{cx}\nu\). Indeed, take \(\pi\in\Pi_M(\mu,\nu)\) and set \(\tilde{\pi}(dx,du,dy)=\mu(dx)du\pi_x(dy)\). Then \(\tilde{\pi}\in\hat{\Pi}_M(\mu,\nu)\). Also note that, if \(\hat{\pi}\in\hat{\Pi}_M(\mu,\nu)\), then \(\int_{u \in (0,1)} d\hat{\pi}\in\Pi_M(\mu,\nu)\).

Next, we introduce the set of models that are induced by \(\hat{\Pi}_M(\mu,\nu)\). Let \(\Omega^{rand}= \{(x,u,y): \alpha_\mu < x < \beta_\mu,~0<u<1,~ \alpha_\nu < y < \beta_\nu \}\), \(\mathcal{F}^{rand}= \sigma(\{ A\times V \times B: A \in \mathcal{B}((\alpha_\mu,\beta_\mu)),~ V \in \mathcal{B}((0,1)),~B \in \mathcal{B}((\alpha_\nu,\beta_\nu))\})\), \(\mathcal{F}^{rand}_0 = \{\Omega,\emptyset\}\), \(\mathcal{F}^{rand}_1 = \sigma(\{ A\times V \times (\alpha_\nu,\beta_\nu) : A \in \mathcal{B}((\alpha_\mu,\beta_\mu)),~V \in \mathcal{B}((0,1))\}\)), \(\mathcal{F}^{rand}_2 = \mathcal{F}^{rand}\). Define \[\label{eq:UcanonicalModels} {M}^{rand}(\mu,\nu):=\{(\Omega^{rand},\mathcal{F}^{rand},{\mathbb{F}}^{rand}=(\mathcal{F}^{rand}_0,\mathcal{F}^{rand}_1,\mathcal{F}^{rand}_2),\mathbb{P}^{rand}=\hat{\pi}):\hat{\pi}\in\hat{\Pi}_M(\mu,\nu)\}.\tag{18}\] Then \(M^{rand}(\mu,\nu)\subset M(\mu,\nu)\), and for each \(\hat{\pi}\in\hat{\Pi}_M(\mu,\nu)\) there are random variables \((X,U,Y)=(Z_1,U,Z_2)\) such that \(\mathbb{P}( X \in dx,U\in du, Y \in dy) = \hat{\pi}(dx,du,dy)\) (the corresponding model is denoted by \(\mathcal{M}_{\hat{\pi}}\in M^{rand}(\mu,\nu)\)). The set of \({\mathbb{F}}^{rand}\)-stopping times (taking values in \(\{1,2\}\)) is denoted by \(\mathcal{T}^{rand}_{1,2}\). Set \[\begin{align} \label{eq:UprimalCanonical} \mathcal{P}^{rand} = \mathcal{P}^{rand}(\mu,\nu;a,b) = \sup_{\mathcal{M}\in M^{rand}(\mu,\nu)} \sup_{\tau \in \mathcal{T}^{rand}_{1,2}} \mathbb{E}^{\mathcal{M}}[ c(Z_\tau,\tau)], \end{align}\tag{19}\] and note that \[\begin{align} \mathcal{P}\geq \mathcal{P}^{rand}=\sup_{\hat{\pi}\in\hat{\Pi}_M(\mu,\nu)}\sup_{\hat{B}\in \sigma( \mathcal{B}((\alpha_\mu,\beta_\mu))\times\mathcal{B}((0,1)))}\Bigg\{&\int_{\hat{B}}a(x)\mu(dx)du\\&+\int_{\left((\alpha_\mu,\beta_\mu)\times(0,1)\right)\setminus\hat{B}}\left(\int_\mathbb{R}b(y)\hat{\pi}_{x,u}(dy)\right)\mu(dx)du\Bigg\}. \end{align}\]

Under an assumption that there exists special \(n\)-tuples (\(n=4,5,6\)), in Section 5.2 we will show that either \(\mathcal{P}^{can}=\mathcal{P}^{rand}=\mathcal{P}\) (see 4-point and 5-point constructions) or \(\mathcal{P}^{rand}=\mathcal{P}\) (see 6-point construction); this is summarized in Theorem 28. In the first case we have that the supremum in \(\mathcal{P}^{can}\) is attained. The supremum in \(\mathcal{P}^{rand}\) is always attained. We will prove our results by explicitly constructing optimal models that belong to \(M^{can}(\mu,\nu)\) or \(M^{rand}(\mu,\nu)\), respectively. Later, in Section 7, we will further show that, under an additional (but natural) assumption regarding the marginals \((\mu,\nu)\) (see Definition 36), the assumptions of Theorem 28 are satisfied (proving the strong duality for such pairs of marginals).

5.2 Useful classes of martingale models and martingale couplings↩︎

In this section we identify elements of \(\Gamma^{\mu,\nu}_n\) with martingale couplings \(\pi \in \Pi_M(\mu,\nu)\). Since the marginal distributions \(\mu\) and \(\nu\) are assumed to be continuous, in all the constructions below it is sufficient to describe the couplings in terms of behaviour on open inetrvals.

5.2.0.1 4-point construction.

Fix \(x_-,x_+\in(\alpha_\mu,\beta_\mu)\) and \(y_-,y_+\in(\alpha_\nu,\beta_\nu)\) with \(y_- \leq x_-<x_+ \leq y_+\). Define \[\label{eq:4points} \Pi^{y_-,x_-,x_+,y_+}_{M,4}(\mu,\nu):=\{\pi\in\Pi_M(\mu,\nu):\eqref{eq:4points1},~\eqref{eq:4points2},~\eqref{eq:4points3}\textrm{ hold}\},\tag{20}\] where \[\begin{align} \tag{21} \pi(A\times\mathbb{R})=\pi(A\times A)\quad& \forall A\in \mathcal{B}((\alpha_\mu,y_-)\cup(y_+,\beta_\mu)),\\ \pi(A\times\mathbb{R})=\pi(A\times(y_-,y_+))\quad&\forall A\in \mathcal{B}((y_-,x_-)\cup(x_+,y_+))\tag{22},\\ \pi(A\times\mathbb{R})=\pi(A\times(\alpha_\nu,y_-)\cup(y_+,\beta_\nu))\quad&\forall A\in \mathcal{B}((x_-,x_+)).\tag{23} \end{align}\] Note that 21 can be restated as mass in \((\alpha_\mu,y_-)\cup(y_+,\beta_\mu)\) stays where it is under \(\pi\); 22 can be restated as mass in \((y_-,x_-)\cup(x_+,y_+)\) is mapped to the set \((y_-,y_+)\) under \(\pi\); 23 can be restated as mass in \((x_-,x_+)\) is mapped to the set \((\alpha_\nu,y_-)\cup(y_+,\beta_\nu)\) under \(\pi\).

Define a candidate stopping time \(\tau^{x_-,x_+}_4\) by \(\tau^{x_-,x_+}_4= \tau^{x_-,x_+}\) where \[\label{eq:tau4point} \tau^{x_-,x_+}=\begin{cases}1,\quad \textrm{if } X=Z_1\in(\alpha_\nu,x_-]\cup[x_+,\beta_\mu),\\ 2,\quad \textrm{otherwise.} \end{cases}\tag{24}\] Note that \(\tau^{x_-,x_+}_4\in\mathcal{T}_{1,2}^{can}\).

5.2.0.2 5-point construction.

Fix \(x_-,z,x_+\in(\alpha_\mu,\beta_\mu)\) and \(y_-,y_+\in(\alpha_\nu,\beta_\nu)\) with \(y_-<x_- \leq z \leq x_+<y_+\). Define \[\label{eq:5points} \Pi^{y_-,x_-,z,x_+,y_+}_{M,5}(\mu,\nu):=\{\pi\in\Pi_M(\mu,\nu):\eqref{eq:5points1},~\eqref{eq:5points2},~\eqref{eq:5points3},~\eqref{eq:5points4}\textrm{ hold}\},\tag{25}\] where \[\begin{align} \tag{26} \pi(A\times\mathbb{R})=\pi(A\times A)\quad& \forall A\in \mathcal{B}((\alpha_\mu,y_-)\cup(y_+,\beta_\mu)),\\ \pi(A\times\mathbb{R})=\pi(A\times(y_-,z))\quad&\forall A\in \mathcal{B}((y_-,x_-))\tag{27},\\ \pi(A\times\mathbb{R})=\pi(A\times(z,y_+))\quad&\forall A \in \mathcal{B}((x_+,y_+))\tag{28},\\ \pi(A\times\mathbb{R})=\pi(A\times(\alpha_\nu,y_-)\cup(y_+,\beta_\nu))\quad&\forall A \in \mathcal{B}((x_-,x_+)).\tag{29} \end{align}\] Note that 26 can be restated as mass in \((\alpha_\mu,y_-)\cup(y_+,\beta_\mu)\) stays where it is under \(\pi\); 27 (resp., 28 ) can be restated as mass in \((y_-,x_-)\) (resp., \((x_+,y_+)\)) is mapped to the set \((y_-,z)\) (resp., \((z,y_+)\)) under \(\pi\); 29 can be restated as mass in \((x_-,x_+)\) is mapped to the set \((\alpha_\nu,y_-)\cup(y_+,\beta_\nu)\) under \(\pi\).

The proof of the following result is similar to the proof of Lemma 25 (except that in this case we use Lemma 18 in place of Lemma 17, together with Theorem 23).

In this case we take the same candidate stopping time as in the 4-point construction. In particular, set \(\tau^{x_-,x_+}_5 = \tau^{x_-,x_+}\) where \(\tau^{x_-,x_+}\) is as defined in 24

5.2.0.3 6-point construction.

Fix \(x_-,z_-,z_+,x_+\in(\alpha_\mu,\beta_\mu)\) and \(y_-,y_+\in(\alpha_\nu,\beta_\nu)\) with \(y_-<x_-\leq z_- < z_+ \leq x_+<y_+\). Define \[\label{eq:6points} \hat{\Pi}^{y_-,x_-,z_-,z_+,x_+,y_+}_{M,6}(\mu,\nu):=\{\hat{\pi}\in\hat{\Pi}_M(\mu,\nu):\eqref{eq:6points1},~\eqref{eq:6points2},~\eqref{eq:6points3},~\eqref{eq:6points4},~\eqref{eq:6points5},~\eqref{eq:6points6}\textrm{ hold}\},\tag{30}\] where, for each \(\hat{\pi}\in\hat{\Pi}(\mu,\nu)\) we write \(\hat{\pi}(dx,du,dy)=\mu(dx)\hat{\pi}_x(du,dy)\), and \[\begin{align} \hat{\pi}(A\times V\times\mathbb{R}) & = & \hat{\pi}(A\times V\times A), \tag{31} \\ & & \forall A\times V \in \mathcal{B}(((\alpha_\mu,y_-)\cup(y_+,\beta_\mu)) \times(0,1)); \nonumber \\ \hat{\pi}(A\times V\times\mathbb{R}) &= & \hat{\pi}(A\times V \times (y_-,z_-)), \tag{32} \\ && \forall A\times V \in \mathcal{B}{((y_-,x_-))} \times(0,1)); \nonumber \\ \tag{33} \hat{\pi}(A\times V\times\mathbb{R}) &=&\hat{\pi}(A\times V\times(z_+,y_+)), \\ && \forall A\times V \in \mathcal{B}((x_+,y_+) \times(0,1)); \nonumber \\ \tag{34} \hat{\pi}(A\times V\times\mathbb{R})&=&\hat{\pi}(A\times V\times(\alpha_\nu,y_-)\cup(y_+,\beta_\nu)), \\ && \forall A\times V \in \mathcal{B}(((x_-,z_-)\cup(z_+,x_+)) \times (0,1)); \nonumber \\ \tag{35} \hat{\pi}_x(V\times\mathbb{R})& = &\hat{\pi}_x(V\times\{x\}), \\ && \forall V \in \mathcal{B}\left(\left(0,\frac{\nu(dx)}{\mu(dx)}=\frac{\eta(x)}{\rho(x)}\wedge1\right)\right) for \mu-a.e. x\in(z_-,z_+); \nonumber \\ \tag{36} \hat{\pi}_x(V\times\mathbb{R}) &=&\hat{\pi}_x(V\times(\alpha_\nu,y_-)\cup(y_+,\beta_\nu)), \\ && \forall V \in \mathcal{B}\left(\left(\frac{\nu(dx)}{\mu(dx)}=\frac{\eta(x)}{\rho(x)}\wedge1,1\right)\right) for \mu-a.e. x\in(z_-,z_+). \nonumber \end{align}\] Here 31 –34 have a similar interpretation as in the 4- and 5-point construction. For 35 and 36 mass at \(x\) in \((z_-,z_+)\) either stays where it is with probability \(\frac{\eta(x)}{\rho(x)}\) or moves to the tails of \(\nu\) (i.e., outside \((y_-,y_+)\)) with probability \(1-\frac{\eta(x)}{\rho(x)}\).

Define a candidate stopping time \(\hat{\tau}^{x_-,z_-,z_+,x_+}\) by \[\label{eq:tau6point} \hat{\tau}^{x_-,z_-,z_+,x_+}=\begin{cases}1,\quad \textrm{if } X=Z_1\in(\alpha_\nu,x_-)\cup(x_+,\beta_\mu)\textrm{ or }X=Z_1\in(z_-,z_+)\textrm{ and } U\in\left(0,\frac{\eta(X)}{\rho(X)}{ \wedge1}\right),\\ 2,\quad \textrm{otherwise.} \end{cases}\tag{37}\]

We are now in a position to prove Theorem 21 which will follow as an immediate corollory of the following result and Lemmas 25, 26, and 27. Recall that \(c(\cdot,1)=a\) and \(c(\cdot,2)=b\). Also recall Problem 2 and the definition of the restricted dual value \(\tilde{\mathcal{D}}=\inf_{ \psi \in \tilde{S}(b) }\widetilde{HC}(\psi)\).

6 Shadows and the left- and right-curtain martingale couplings↩︎

6.1 Shadow couplings↩︎

We recall the notion of the shadow measure (see Beiglböck and Juillet [5] and Beiglböck et al. [16]).

Two (Borel) measures \(\xi,\chi\) on \(\mathbb{R}\) are said to be in extended convex order, denoted by \(\xi\leq_E\chi\), if \(\int f d\xi\leq\int fd\chi\) for all non-negative and convex \(f:\mathbb{R}\to\mathbb{R}_+\). (It is easy to see that if \(\mu\leq_{cx}\nu\), then any \(\tilde{\mu}\leq\mu\) satisfies \(\tilde{\mu}\leq_E\nu\).) If \(\xi\leq_E\chi\), then \(\{\theta:\xi\leq_{cx}\theta\leq\chi\}\) is non-empty and admits the minimal element with respect to \(\leq_{cx}\).

For any (Borel) measure \(\xi\) on \(\mathbb{R}\), we define the (put) potential \(P_\xi:\mathbb{R}\to\mathbb{R}\) by \(P_\xi(k)=\int_\mathbb{R}(k-x)^+\xi(dx)\), \(k\in\mathbb{R}\). It is well-known that the potenial \(P_\xi\) uniquely identifies the underlying measure (via the second (distributional) derivative of \(P_\xi''\)). Furthermore, \(\xi\leq_{cx}\chi\) if and only if \(P_\xi\leq P_\chi\) on \(\mathbb{R}\) and \(\xi,\chi\) have the same total mass and mean. See, for example, Chacon [18], Chacon and Walsh [19] or Hirsch and Roynette [20] for the relevant properties of potential functions.

Recall that given \(f:\mathbb{R}\to\mathbb{R}\) we denote the convex hull of \(f\) by \(f^c\).

One of the main structural properties of the shadow measure (see Beiglböck and Juillet [5], [21], Beiglböck et al. [16]), is the associativity property.

The following lemmas will be useful in the proofs of our main results.

6.2 Curtain couplings↩︎

The shadow measure allows us to construct martingale couplings \(\pi \in \Pi_M(\mu,\nu)\) which later can be used to construct models \(\mathcal{M}_\pi \in M^{can}(\mu,\nu)\). Recall the definition of \(\pi^{f,g}_v\) from the introduction.

In many natural cases (for example a pair of centred Gaussian distributions, or a pair of lognormal distributions with the same mean) there exists \(e^L \in [\alpha_\mu,\beta_\mu)\) such that \(\rho < \eta\) on an interval \((\alpha_\nu, e^L)\) and \(\rho>\eta\) on a small interval \((e^L,e^L+\epsilon)\) to the right of \(e^L\). Then we can define \(f^L=g^L=id\) on \((\alpha_\mu,e^L]\) and extend the functions rightward from \(e^L\) using the relationship \[\label{eq:deffg} \int_{f^L(x)}^x z^i \rho(z) dz = \int_{f^L(x)}^{g^L(x)} z^i \eta(z) dz;i=0,1,\tag{38}\] which encapsulates the idea that mass in \((f^L(x),x)\) according to the law \(\mu\) can be mapped to \((f^L(x),g^L(x))\) according to the law \(\nu\) in a way which respects the martingale property (and then at the margins, mass at \(x\) is mapped to either \(f^L(x)\) or \(g^L(x)\) as in the definition of the left-curtain coupling). Under Standing Assumption 1 the solutions to 38 can be expressed via a coupled pair of differential equations \[\label{eq:fgode} \frac{d}{dx} g^L (x) = \frac{x - f^L(x)}{g^L(x)-f^L(x)} \frac{\rho(x)}{\eta(g(x))};\frac{d}{dx} f^L (x) = - \frac{g^{L}(x) - x}{g^L(x)-f^L(x)} \frac{\rho(x)}{\eta(f^L(x))- \rho(f^L(x))}\tag{39}\] subject to \(f^L(e^L)=g^L(e^L)=e^L\). Note that \(f^L\) is decreasing (at least locally, so that by hypothesis the denominators in 39 are positive for \(x>e^L\). In many (regular) cases the solutions \(f^L, g^L\) to 39 satisfy \(g^L > id\) on \((e^L,\beta_\mu)\). In this case the left-curtain coupling has a simple structure, see Figure 1.

Similar considerations apply to the right-curtain coupling. In regular cases the densities \(\rho,\eta\) are such that there exists \(e_R>e_L\) such that \(\rho<\eta\) on \((e^R, \beta_\mu)\) and \(\rho> \eta\) on an interval \((e^R-\epsilon,e^R)\) and then \(f^R,g^R\) can be defined to the left of \(e^R\) via \[\int_x^{g^R(x)} z^i \rho(z) dz = \int_{f^R(x)}^{g^R(x)} z^i \eta(z) dz;i=0,1,\] which again have a representation via a pair of coupled differential equations. In many cases \(f^R\) defined in this way satisfies \(f^R < id\) on \((\alpha_\mu,e^R)\) and the right-curtain coupling has a simple structure.

The following assumption is a small modification of one introduced by Hobson and Klimmek [4], see also Henry-Labordère and Touzi [6]. See Figure 2. It is often satisfied when \(\mu\) and \(\nu\) are from the same family of centred, continuous distributions with a unimodal density such as centred normals, lognormals or uniforms.

For convenience, we introduce the following notation: for a continuous measure \(\xi\) (on \(\mathbb{R}\)) and for \(u \leq v\) let \(\xi_u^v = \xi|_{(u,v)}\) and \(\bar{\xi}_u^v = \xi - \xi_u^v = \xi|_{(-\infty,u)\cup(v,\infty)}\).

Under the Dispersion Assumption, the left-curtain and right-curtain martingale couplings have a simple structure; the proof of the following lemma can be found in Appendix 10.

We finish this section with a lemma that will be useful in proving our main results.

Recall the definitions and the properties (under the Dispersion Assumption, Definition 36) of \(g^L\) and \(f^R\); see Theorem 35 and Lemma 37. Note that if \(\beta_\mu<\beta_\nu\), then \((g^L)^{-1}(\beta_\mu)<\beta_\mu\), while in the case \(\beta_\mu=\beta_\nu\) we set \((g^L)^{-1}(\beta_\mu)=\beta_\mu\). Similarly, if \(\alpha_\nu<\alpha_\mu\), then \(\alpha_\mu<(f^R)^{-1}(\alpha_\mu)\), and in the case \(\alpha_\nu=\alpha_\mu\) we set \((f^R)^{-1}(\alpha_\mu)=\alpha_\mu\).

The motivation behind the next definition is that we want to separate the cases where \(S^\nu(\mu_{\alpha_\mu}^{x}) + S^\nu(\mu_y^{\beta_\mu})= S^\nu(\mu_{\alpha_\mu}^{x} + \mu_y^{\beta_\mu})\) and \(S^\nu(\mu_{\alpha_\mu}^{x}) + S^\nu(\mu_y^{\beta_\mu}) \neq S^\nu(\mu_{\alpha_\mu}^{x} + \mu_y^{\beta_\mu})\). In the former case, we have \(S^\nu(\mu_y^{\beta_\mu})=S^{\nu- S^\nu(\mu_{\alpha_\mu}^{x})}(\mu_y^{\beta_\mu})\).

Set \(\bar x=(g^L)^{-1}(e^R)\). Define \(B:(\alpha_\mu,(g^L)^{-1}(\beta_\mu))\to(\alpha_\nu,\beta_\nu)\) by \[\label{eq:B} B(x):=(f^R)^{-1}(g^L(x)),\quad x\in(\alpha_\mu,(g^L)^{-1}(\beta_\mu)).\tag{40}\] In the case \(\beta_\mu<\beta_\nu\), we extend the definition of \(B(\cdot)\) to \((\alpha_\mu,\beta_\mu)\) by setting \(B(x)=\beta_\mu\) for \(x\in[(g^L)^{-1}(\beta_\mu),\beta_\mu)\). Then, using that \(g^L(x)=x\) for \(x\leq e^L\) and \(f^R(y)=y\) for \(y\geq e^R\), we equivalently have that \[\label{eq:BoundaryB} B(x) = \begin{cases} (f^R)^{-1}(x) & \alpha_\mu < x \leq e^L \\ (f^R)^{-1}(g^L(x)) & e^L < x \leq \bar{x} \\ g^L(x) & \bar{x} < x \leq (g^L)^{-1}(\beta_\mu) \\ \beta_\mu & (g^L)^{-1}(\beta_\mu) < x < \beta_\mu. \end{cases}\tag{41}\]

The proof of Lemma 39 is deferred to Appendix 10.

7 Existence of an optimal model↩︎

The goal of this section is to show that, under the Dispersion Assumption on the pair \((\mu,\nu)\), the hypotheses of Theorem 28 are satisfied and we have found the optimal solution. In particular, we prove that if \(\mu\) and \(\nu\) satisfy the dispersion assumption then \(\cup_{n=4,5,6} (\Gamma^{\mu,\nu}_n \cap \mathcal{L}^{a,b}_n) \neq \emptyset\) and hence, by Theorem 28, we can construct optimal solutions to both the primal and dual problems and there is no duality gap.

The focus of this section is on proving Theorem 40. Many of the proofs of intermediate results are deferred to Appendix 10.

Given the structure of the left- and right-curtain couplings under the Dispersion Assumption (recall Lemma 37), it is rather easy to see that the sets \(\Gamma^{\mu,\nu}_4\), \(\Gamma^{\mu,\nu}_5\) and \(\Gamma^{\mu,\nu}_6\) (as in 9 , 10 and 11 , respectively) are non-empty. Indeed, for \(e^L<x_-<x_+<e^R\) with \(g^L(x_-)= f^R(x_+)\), we have that \(f^L(x_-)<e^L<x_-<g^L(x_-)=f^R(x_+)<x_+<e^R<g^R(x_+)\), and then, by setting \(y_-=f^L(x_-), z=g^L(x_-)=f^R(x_+),g^R(x_+)=y_+\), we have that \(\Gamma^{\mu,\nu}_4\neq\emptyset\) and \(\Gamma^{\mu,\nu}_5\neq\emptyset\). (This is an immediate consequence of the definitions of curtain couplings, Corollary 38 and Lemma 39.) Similarly, if \(e^L<x_-<x_+<e^R\) with \(g^L(x_-)< f^R(x_+)\), by setting \(y_-=f^L(x_-), z_-=g^L(x_-), z_+=f^R(x_+),g^R(x_+)=y_+\), we have that \(\Gamma^{\mu,\nu}_6\neq\emptyset\); here we use that, under the Dispersion Assumption, \(\mu\lvert_{(z_-,z_+)}\leq\nu\lvert_{(z_-,z_+)}\), since \((z_-=g^L(x_-),z_+=f^R(x_+))\subset (e^L,e^R)\).

As above, we can express the candidate points \(x_-,x_+,y_-,y_+,z_-,z,z_+\in\mathbb{R}\) in terms of \(x_-\) and \(x_+\) only. Then the key insight is that by restricting the search to such pairs \((x_-,x_+)\) (which together with the functions \(f^L,g^L,f^R,g^R\), give the remaining four points), we can find \((x_-^*,x_+^*)\), for which (at least) one of \(\Psi_4^{y^*_-,x^*_-,x^*_+,y^*_+}\), or \(\Psi_5^{y^*_-,x^*_-,z^*,x^*_+,y^*_+}\), or \(\Psi_6^{y^*_-,x^*_-,z^*_-,z^*_+,x^*_+,y^*_+}\) is convex (recall 12 , 14 , 15 ). Then using Lemmas 25, 26, 27 (depending on which case we are in) together with Theorem 28 it must be the case that we have found the solutions to the primal and dual problems and the strong duality holds. The main difficulty in finding \((x^*_-,x_+^*)\) is that a priori it is not clear which of the three types as listed in Theorem 28 we should search for. Furthermore, in the case of the 4-point construction, we often need to consider \(x_-< e^L\) or \(x_+ > e^R\), and in this case the original supporting functions \(f^L,g^L,f^R,g^R\) from the left- and right- curtain couplings are no-longer useful and then we need alternative arguments.

The remainder of this section is dedicated to proving Theorem 40. We will do it in several steps. First, in Section 7.1, for given \(\mu\leq_{cx}\nu\), we use the supporting functions \((f^L,g^L)\) and \((f^R,g^R)\) to distinguish three cases. Then we prove Theorem 40 separately in each case (see Sections 7.2, 7.3, 7.4).

Throughout this section we make the following standing assumption:

7.1 Dividing the problem into cases↩︎

For \(x\in(e^L,\beta_\mu)\) define \(T^L_x:= T_{b,a}^{f^L(x),x}\) to be the line passing through the points \((f^L(x),b(f^L(x)))\) and \((x,a(x))\). Similarly, let \(T^R_x = T^{x,g^R(x)}_{a,b}\). Define the functions \(C^L: (e^L,\beta_\nu) \to \mathbb{R}\) and \(C^R: (\alpha_\nu,e^R) \to \mathbb{R}\) by \(C^L(z)= T^L_{(g^L)^{-1}(z)}(z)\) and \(C^R(z)= T^R_{(f^R)^{-1}(z)}(z)\). We are interested in the crossing points (if any) of the curves \(C^L\) and \(C^R\) on \((e^L,e^R)\). There will be three cases, and the optimal hedges and corresponding extremal models will be different in each case.

Define \(\hat{z}^L\) by \(\hat{z}^L = \inf \{ z\in (e^L,{\beta_\mu}): C^L(z) \leq a(z) \}\) where if this set is empty we take \(\hat{z}^L = {\beta_\mu}\). Similarly, define \(\hat{z}^R = \sup \{ z \in ({ \alpha_\mu}, e^R) : C^R(z) \leq a(z) \}\) where if this set is empty we take \(\hat{z}^R = { \alpha_\mu}\). Note that \(\hat{z}^L\geq e^L\) and \(\hat{z}^R\leq e^R\). Define \(\hat{x}^L = (g^L)^{-1}(\hat{z}^L)\) (with \(\hat{x}^L=\beta_\mu\) if \(\hat{z}^L = \beta_\mu=\beta_\nu\)) and \(\hat{x}^R = (f^R)^{-1}(\hat{z}^R)\) (with \(\hat{x}^R=\alpha_\mu\) if \(\hat{z}^R = \alpha_\mu=\alpha_\nu\)).

Note that if \(e^L < \hat{z}^R < \hat{z}^L < e^R\), then by the intermediate value theorem applied to \(C^R-C^L\) on \([\hat{z}^R, \hat{z}^L]\), there exists \(z^* \in (\hat{z}^R, \hat{z}^L)\subseteq(e^L,e^R)\) such that \(C^L(z^*)=C^R(z^*)>a(z^*)\). Further, if \(\hat{z}^R=\hat{z}^L \in (e^L,e^R)\) then taking \(z^* = \hat{z}^R\) we find \(C^L(z^*)=C^R(z^*) = a(z^*)\). In either case, if there exists \(z^* \in (e^L,e^R)\) such that \(C^L(z^*)=C^R(z^*)\) then we can define \(x^*_L := (g^L)^{-1}(z^*)\), \(x^*_R := (f^R)^{-1}(z^*)\) and \(\Psi:(\alpha_\nu,\beta_\nu) \mapsto \mathbb{R}\) by \[\label{eq:Bdef} \Psi(w) = \begin{cases} b(w), & w \leq f^L(x^*_L) \\ T^L_{x^*_L}(w), \quad \quad & f^L(x^*_L) < w \leq z^*; \\ T^{R}_{x^*_R}(w), & z^* < w < g^R(x^*_R); \\ b(w), & w \geq g^R(x^*_R). \end{cases}\tag{42}\] We will have that \(\Psi\) is continuous provided there is continuity at \(f^L(x^*_L)\), \(z^*\) and \(g^R(x^*_R)\). By the definition of \(T^L_x = T^{f^L(x),x}_{b,a}\) we have that \(T^L_{x^*_L}(f^L(x^*_L)) = T^{f^L(x^*_L),x^*_L}_{b,a}(f^L(x^*_L))=b({f^L(x^*_L)})\) and continuity of \(\Psi\) at \(x^*_L\); a similar argument applies at \(g^R(x^*_R)\). At \(z^*\) continuity follows from the fact that \(T^L_{x^*_L}(z^*) = C^L(z^*) = C^R(z^*) = T^R_{x^*_R}(z^*)\). However, although \(\Psi\) is continuous, it may or may not be convex.

The three cases will be as follows:

1. 1. \(e^L <\hat{z}^R \leq \hat{z}^L < e^R\) (so that either there exists \(z^* \in (\hat{z}^R, \hat{z}^L)\) such that \(C^L(z^*)=C^R(z^*) > a(z^*)\), or \(\hat{z}^R =\hat{z}^L\) and then taking \(z^*=\hat{z}^R = \hat{z}^L\) we have \(C^L(z^*)=C^R(z^*) =a(z^*)\)); moreover \(\Psi\) defined in 42 is convex;

   2. either \(\hat{z}^R = e^R = \hat{z}^L\) or \(\hat{z}^R = e^L = \hat{z}^L\).

2. \(\hat{z}^L < \hat{z}^R\);

3. 1. \(\hat{z}^L \geq e^R\) and/or \(\hat{z}^R \leq e^L\); moreover \(\hat{z}^R<\hat{z}^L\).

   2. \(e^L < \hat{z}^R \leq \hat{z}^L < e^R\) (so that again \(C^L(z^*)=C^R(z^*) \geq a(z^*)\) for some \(z^* \in [\hat{z}^R, \hat{z}^L]\)) but \(\Psi\) defined in 42 is not convex.

It is clear that the cases (C1), (C2) and (C3) are mutually exclusive, except that in principle it is possible that there exists \(z^*_1,z^*_2 \in (e^L,e^R)\) such that \(C^L(z^*_i)= C^R(z^*_i) \geq a(z^*_i)\) for \(i=1,2\) and such that for \(z^*_1\), \(\Psi\) defined in 42 is convex and for \(z^*_2\), \(\Psi\) defined in 42 is not convex. It can be shown that this cannot happen. However, since the three cases are merely starting points of a construction, it is not crucial to the argument that the construction must start in a unique way, and therefore we omit this argument.

The goal, in each of the cases (C1), (C2) and (C3), is to choose \(\psi\) such that the total cost of the superhedging portfolio generated by \(\psi\) coincides with the model-based price of the Bermudan claim (for some model \(\mathcal{M}^* \in M(\mu,\nu)\) and the associated optimal stopping time \(\tau^*\)).

7.2 Proof of Theorem 40 in Case (C1)↩︎

In this case we take \((y^*_-,x^*_-,z^*, x^*_+,y^*_+) = (f^L(x_L^*),x^*_L,z^*,x^*_R,g^R(x^*_R))\). Note that \(g^L(x^*_L)=z^*=f^R(x^*_R)\). Then \(\Psi^{y^*_-,x^*_-,z^*, x^*_+,y^*_+}_{5,a,b}\) (defined in 14 ) and \(\Psi\) (defined in 42 ) are such that \(\Psi^{y^*_-,x^*_-,z^*, x^*_+,y^*_+}_{5,a,b} = \Psi\) (in Case (C1)(b), if \(\hat{z}^R=e^R=\hat{z}^L\) (resp., \(\hat{z}^R=e^L=\hat{z}^L\)), then \(y^*_+=x^*_+=z^*=e^R\) (resp., \(y^*_-=x^*_-=z^*=e^L\)) and thus the third (resp., second) lines in the definitions of both \(\Psi^{y^*_-,x^*_-,z^*, x^*_+,y^*_+}_{5,a,b}\) and \(\Psi\) are redundant). By construction, \(\Psi\) is convex if we are in Case (C1)(a). On the other hand, in Case (C1)(b), by Lemma 42 we have that \(a(\hat{z}^R)=b(\hat{z}^R)\), and thus \(\Psi\) is again convex. It follows that \((y^*_-,x^*_-,z^*, x^*_+,y^*_+) \in \mathcal{L}_5^{a,b}\) in Case (C1). See Figure 3.

On the other hand, since \(e^L\leq \hat{z}^R\leq z^*\leq\hat{z}^L\leq e^R\), we have that \(e^L\leq x^*_-<x^*_+\leq e^R\) and \(x^*_+=B(x^*_-)\) (where \(B(\cdot)\) is given by 41 ). Then by Lemma 39(i) we have that \(S^\nu(\bar{\mu}_{x^*_-}^{x^*_+}) = S^\nu(\mu_{\alpha_\mu}^{x^*_-})+S^{\nu}(\mu_{x^*_+}^{\beta_\mu})\). By Corollary 38(i), \(S^\nu(\mu_{\alpha_\mu}^{x^*_-})=\mu_{\alpha_\mu}^{y^*_-} + \nu_{y^*_-}^{z^*}\leq\nu\), and therefore \(\mu_{\alpha_\mu}^{y^*_-}\leq\nu_{\alpha_\mu}^{y^*_-}\) (so that \(S^\nu(\mu_{\alpha_\mu}^{y^*_-})=\mu_{\alpha_\mu}^{y^*_-}\)). By the definition and the associativity property of the shadow measure (see Proposition 32) it then follows that \(S^\nu(\mu_{\alpha_\mu}^{y^*_-})+S^{\nu-S^\nu(\mu_{\alpha_\mu}^{y^*_-})}(\mu_{y^*_-}^{x^*_-})=S^\nu(\mu_{\alpha_\mu}^{x^*_-})=\mu_{\alpha_\mu}^{y^*_-} + \nu_{y^*_-}^{z^*}\) and thus \(S^{\nu-S^\nu(\mu_{\alpha_\mu}^{y^*_-})}(\mu_{y^*_-}^{x^*_-})=\nu_{y^*_-}^{z^*}\), from which we conclude that \(\mu_{y^*_-}^{x^*_-}\leq_{cx}\nu_{y^*_-}^{z^*}\). By applying similar arguments to \((z^*,x^*_+,y^*_+)\) (together with Corollary 38(ii)) we have that \(\mu_{y^*_+}^{\beta_\mu}\leq\nu_{y^*_+}^{\beta_\mu}\) and \(\mu_{x^*_+}^{y^*_+}\leq_{cx}\nu_{z^*}^{y^*_+}\). Combining both cases shows that \((y^*_-,x^*_-,z^*, x^*_+,y^*_+)\in \Gamma_5^{\mu,\nu}\).

We conclude that \((f^L(x_L^*),x^*_L,z^*,x^*_R,g^R(x^*_R)) \in \Gamma_5^{\mu,\nu} \cap \mathcal{L}_5^{a,b}\), which finishes the proof in Case (C1).

7.3 Proof of Theorem 40 in Case (C2)↩︎

In this case we take \((\hat{y}_-,\hat{x}_-,\hat{z}_-,\hat{z}_+,\hat{x}_+,\hat{y}_+)=(f^L(\hat{x}^L),\hat{x}^L,\hat{z}^L,\hat{z}^R,\hat{x}^R,g^R(\hat{x}^R))\). Note that \(e^L\leq \hat{z}^L<\hat{z}^R\leq e^R\).

Suppose first that \(e^L< \hat{z}^L<\hat{z}^R<e^R\), so that (by Lemma 42) \(b(e^L)<a(e^L)\) and \(b(e^R)<a(e^R)\). In that case we define a candidate convex function \(\psi^{*,2}\) such that \(\psi^{*,2}\geq b\) everywhere (so that, as before, it generates a superhedge with total cost \(\int(a- \psi^{*,2})^+d\mu+\int\psi^{*,2}d\nu\))): \[\label{eq:psi1} \psi^{*,2}(x)=\begin{cases} b(x),\quad x\in(-\infty,\hat{y}_-]\cup[\hat{y}_+,\infty)\\ a(x),\quad x\in (\hat{z}_-,\hat{z}_+)\\ T^L_{\hat{x}_L}(x),\quad x\in(\hat{y}_-,\hat{z}_-]\\ T^R_{\hat{x}_R}(x),\quad x\in[\hat{z}_+,\hat{y}_+). \end{cases}\tag{43}\] Then \(\Psi^{\hat{y}_-,\hat{x}_-,\hat{z}_-,\hat{z}_+,\hat{x}_+,\hat{y}_+}_{6,a,b}\) (defined in 15 ) is such that \(\Psi^{\hat{y}_-,\hat{x}_-,\hat{z}_-,\hat{z}_+,\hat{x}_+,\hat{y}_+}_{6,a,b} = \psi^{*,2}\) (note that, due to the definitions of \(\hat{z}_-=\hat{z}^L\) and \(\hat{z}_+=\hat{z}^R\), and the fact that we are in Case (C2), \(\psi^{*,2}\) is convex). It follows that \((\hat{y}_-,\hat{x}_-,\hat{z}_-,\hat{z}_+,\hat{x}_+,\hat{y}_+) \in \mathcal{L}_6^{a,b}\).

Since \(e^L<\hat{x}_-=\hat{x}^L<g^L(\hat{x}^L)=\hat{z}^L=\hat{z}_- < \hat{z}_+=\hat{z}^R=f^R(\hat{x}^R)<\hat{x}^R=\hat{x}_+<e^R\), and similarly as in Case (1), by Lemma 39(i) we have that \(\mu_{\alpha_\mu}^{\hat{y}_-}+\mu_{\hat{y}_+}^{\beta_\mu}\leq\nu_{\alpha_\mu}^{\hat{y}_-}+\nu_{\hat{y}_+}^{\beta_\mu}\), \(\mu_{\hat{y}_-}^{\hat{x}_-}\leq_{cx}\nu_{\hat{y}_-}^{\hat{z}_-}\) and \(\mu_{\hat{x}_+}^{\hat{y}_+}\leq_{cx}\nu_{\hat{z}_+}^{\hat{y}_+}\). On the other hand, by the Dispersion Assumption (recall Definition 36), \(\nu\leq\mu\) on \((\hat{z}_-,\hat{z}_+)\). It follows that \((\hat{y}_-,\hat{x}_-,\hat{z}_-,\hat{z}_+,\hat{x}_+,\hat{y}_+)\in \Gamma_6^{\mu,\nu}\), which completes the proof in this case. See Figure [fig:case2].

If \(b(e^L)=a(e^L)\), or \(b(e^R)=a(e^R)\), or both, then the above construction simplifies. Indeed, by Lemma 42, if \(a(e^L)=b(e^L)\), then \(\hat{y}_- =\hat{x}_- =\hat{z}_-\), and therefore \(\nu_{\hat{y}_-}^{\hat{z}_-}=\nu_{\hat{z}_-}^{\hat{z}_-}\) is the zero measure, and the third line in the definition of \(\psi^{*,2}\) (see 43 ) is redundant, so that \(\psi^{*,2}=b\) on \((-\infty,e^L=\hat{z}_-]\). Similarly, if \(a(e^R)=b(e^R)\), then \(\hat{z}_+=\hat{x}_+=\hat{y}_+\), and therefore \(\nu_{\hat{z}_+}^{\hat{y}_+}=\nu_{\hat{z}_+}^{\hat{z}_+}\) is the zero measure, and the fourth line in 43 is redundant, so that \(\psi^{*,2}=b\) on \([e^R=\hat{z}_+,\infty)\). Finally, if both \(a(e^L)=b(e^L)\) and \(a(e^L)=b(e^L)\), then \((\nu_{\hat{y}_-}^{\hat{z}_-}+\nu_{\hat{z}_+}^{\hat{y}_+})\) is the zero measure, and both the third and fourth lines in 43 are redundant, so that \(\psi^{*,2}\) simplifies to \(\psi^{*,2}=a\) on \((e^L,e^R)\) and \(\psi^{*,2}=b\) otherwise. Note that in any of these cases we still have that \(\Psi^{\hat{y}_-,\hat{x}_-,\hat{z}_-,\hat{z}_+,\hat{x}_+,\hat{y}_+}_{6,a,b} = \psi^{*,2}\) is convex, and \((\hat{y}_-,\hat{x}_-,\hat{z}_-,\hat{z}_+,\hat{x}_+,\hat{y}_+) \in \mathcal{L}_6^{a,b}\cap \Gamma_6^{\mu,\nu}\), which completes the proof.

7.4 Proof of Theorem 40 in Case (C3)↩︎

Recall the definition of \(\mathcal{L}^{a,b}_4\) in 13 and note that, at least if \(x>\alpha_\mu\) and \(y<\beta_\mu\), \((w,x,y,z)\in \mathcal{L}^{a,b}_4\) if \((w,b(w))\), \((x,a(x))\), \((y,a(y))\) and \((z,b(z))\) lie on a straight line. Recall also the definition of \(\Gamma^{\mu,\nu}_4\) in 9 . In this section we write \(\mathcal{L}_4\) as shorthand for \(\mathcal{L}^{a,b}_4\) and \(\Gamma_4\) as shorthand for \(\Gamma_4^{\mu,\nu}\).

Our goal is to find \((w,x,y,z) \in \mathcal{L}_4 \cap \Gamma_4\). Then we can build a model including a coupling \(\pi\) (of \(\mu\) and \(\nu\)) such that the mass outside \((x,y)\) is mapped to its shadow, and the associated optimal strategy is to take \(\tau^*=1\) on \(X_1 \notin (x,y)\). The corresponding optimal hedge is generated by \(\psi = b \vee L\) where \(L\) is the straight line that passes through the four collinear points \((w,b(w))\), \((x,a(x))\), \((y,a(y))\) and \((z,b(z))\). See Case 1 of Theorem 28.

We will find \((w,x,y,z) \in \mathcal{L}_4 \cap \Gamma_4\) in four steps. Step 1: first we will construct maps \(w,z:(\alpha_\mu,\beta_\mu)\to(\alpha_\nu,\beta_\nu)\) such that \((w(x,y),x,y,z(x,y)) \in \Gamma_4\); see Lemma 45. Step 2: we will use the functions \(C^L\) and \(C^R\) above to find a starting point \((x_0,y_0)\) and then a modified starting point \((x_1,y_0)\). Step 3: we show that there exists a continuous function \(y^*:[x_1,\beta_\mu)\to(\alpha_\mu,\beta_\mu)\) such that \((w(x,y^*(x)),x,y^*(x),z(x,y^*(x))) \in \Gamma_4\) and the points \((x,a(x))\), \((y^*(x),a(y^*(x)))\) and \((z(x,y^*(x)),b(z(x,y^*(x)))\) lie on the same line; see Corollary 50. Step 4: finally, by adjusting \(x\), we will show that there exists \(x^*\in[x_1,\alpha_\mu)\), for which the remaining point \((w(x^*,y^*(x^*)),b(w(x^*,y^*(x^*))))\) also lies on the same line and thus \((w(x^*,y^*(x^*)),x^*,y^*(x^*),z(x^*,y^*(x^*))) \in \mathcal{L}_4\cap \Gamma_4\); see Proposition 53.

Step 1: constructing maps \(w,z\). Recall the definition of \(B:(\alpha_\mu,\beta_\mu)\to(\alpha_\mu,\beta_\mu)\) given in 41 . The following lemma is an immediate consequence of Lemma 37.

Note that the restriction of \(B(\cdot)\) to \((\alpha_\mu,(g^L)^{-1}(\beta_\mu))\) admits an inverse \(B^{-1}:((f^R)^{-1}(\alpha_\mu),\beta_\mu)\to(\alpha_\mu,(g^L)^{-1}(\beta_\mu))\). In the case \(\alpha_\nu<\alpha_\mu\), we extend the domain of \(B^{-1}(\cdot)\) to \((\alpha_\mu,\beta_\mu)\) by setting \(B^{-1}(y)=\alpha_\mu\) for all \(y\in(\alpha_\mu,(f^R)^{-1}(\alpha_\mu)]\). Then \(B^{-1}(\cdot)\) is continuous (and non-decreasing) on \((\alpha_\mu,\beta_\mu)\) and strictly increasing on \(((f^R)^{-1}(\alpha_\mu),\beta_\mu)\).

Define \(\mathcal{A}:=\{(x,y): x\in[\alpha_\mu, \beta_\mu)\cap(\alpha_\nu,\beta_\mu),~y\in(\alpha_\mu,\beta_\mu]\cap(\alpha_\mu,\beta_\nu),~x\leq y\leq B(x)\}\). Also, for \(x\in[\alpha_\mu, \beta_\mu)\cap(\alpha_\nu,\beta_\mu)\) and \(y\in(\alpha_\mu,\beta_\mu]\cap(\alpha_\mu,\beta_\nu)\), set \(\mathcal{A}_x=\{z:(x,z)\in\mathcal{A}\}\) and \(\mathcal{A}_y:=\{z:(z,y)\in\mathcal{A}\}\), respectively. Set \(\mathcal{A}^<=\{(x,y)\in\mathcal{A}:x<y\}\), \(\mathcal{A}^<_x=\{z\in\mathcal{A}_x:x<z\}\) and \(\mathcal{A}^<_y=\{z\in\mathcal{A}_y:z<y\}\).

Step 2: finding the starting point \((x_1,y_0)\in\mathcal{A}^<\). For each \((x,y) \in { \mathcal{A}^<}\) define \(\Psi_{x,y}\) by setting \[\label{eq:PsiXY} \Psi_{x,y}(u) = \begin{cases} b(u)& x \leq w(x,y) \\ T_{b,a}^{w(x,y),x}(u) & w(x,y) < u < x \\ T_{a,a}^{x,y}(u) & x \leq u \leq y \\ T_{a,b}^{y,z(x,y)}(u) & y < u < z(x,y) \\ b(u) & z(x,y) \leq u. \end{cases}\tag{44}\] Note that, if \(w(x,y)=x\) (resp., \(z(x,y)=y\)) then the second (resp., fourth) line in the definition of \(\Psi_{x,y}\) is redundant. This happens when \(x\leq e^L\) (resp., \(y\geq e^R\)) and \((x,y)\in\mathcal{A}\) lies on the boundary, i.e., \(y=B(x)\) (resp., \(x=B^{-1}(y)\)).

For \((x,y)\in\mathcal{A}^<\), let \(\tilde{S}_{b,a}^{w(x,y),x}, \tilde{S}_{a,a}^{x,y}\) and \(\tilde{S}_{a,b}^{y,z(x,y)}\) be the slope of \(\Psi_{x,y}\) on \((w(x,y),x), (x,y)\) and \((y,z(x,y))\), respectively. If \(w(x,y)=x\) (resp., \(z(x,y)=y\)) then we set \(\tilde{S}_{b,a}^{w(x,y),x}=\infty\) (resp., \(\tilde{S}_{a,b}^{y,z(x,y)}=-\infty\)).

Recall that in Case (C3) we are in one of the following situations:

1. \(\hat{z}^L \geq e^R\) and/or \(\hat{z}^R \leq e^L\); moreover \(\hat{z}^R<\hat{z}^L\);

2. \(e^L < \hat{z}^R < \hat{z}^L < e^R\), so that there exists \(z^* \in [\hat{z}^R,\hat{z}^L]\subset(e^L,e^R)\) such that \(C^L(z^*)=C^R(z^*) \geq a(z^*)\), but \(\Psi\) defined by 42 is not convex.

where the strict inequality in Case \((b)\) follows from Remark 43.

We subdivide Case \((a)\) into the disjoint union \(D_1 \cup D_2 \cup D_3 \cup D_4 \cup D_5\) where \(D_1=(\hat{z}^L \geq e^R > e^L \geq \hat{z}^R)\), \(D_2= (\hat{z}^L \geq e^R > \hat{z}^R > e^L)\), \(D_3 = (\hat{z}^L > e^R = \hat{z}^R)\), \(D_4 = (e^R > \hat{z}^L > e^L \geq \hat{z}^R)\) and \(D_5 = (\hat{z}^L = e^L > \hat{z}^R)\).

Note that on \(D_2\) we must have that either \(C^L > C^R\) on \((\hat{z}^R,e^R)\) or there exists a point \(z^* \in (\hat{z}^R,e^R)\subset(e^L,e^R)\) such that \(C^L(z^*)=C^R(z^*)\geq a(z^*)\) (indeed, if \(C^L(z)\leq C^R(z)\) for some \(z\in(\hat{z}^R,e^R)\), then the existence of \(z^*\) follows from the continuity of \(C^L,C^R\) and the fact that \(C^L(\hat{z}^R)>a(\hat{z}^R)=C^R(\hat{z}^R)\)). In the latter case, without loss of generality, we further assume that \(\Psi\) defined in 42 is not convex⁴ (if \(\Psi\) was convex then it would be like being in Case (C1) and we already know how to find an optimal model and optimal hedge in that setting). Similarly, on \(D_4\) we must have that either \(C^L < C^R\) on \((e^L,\hat{z}^L)\) or there exists a point \(z^* \in (e^L, \hat{z}^L)\subset(e^L,e^R)\) such that \(C^L(z^*)=C^R(z^*)\geq a(z^*)\) and \(\Psi\) (as in 42 ) is not convex. Moreover, we further divide \(D_1\) according as \(C^L>C^R\) on \((e^L,e^R)\), \(C^L<C^R\) on \((e^L,e^R)\) or there exists a point \(z^* \in (e^L,e^R)\) such that \(C^L(z^*)=C^R(z^*)\geq a(z^*)\). In the last case we again assume that \(\Psi\) defined in 42 is not convex.

Recombining various cases we have that Case (C3) can be divided into

1. \(\hat{z}^L \geq e^R > \hat{z}^R\) moreover, \(C^L(z) > C^R(z)\) on \((\hat{z}^R \vee e^L,e^R)\);

2. \(\hat{z}^L > e^R = \hat{z}^R\);

3. \(\hat{z}^R \leq e^L <\hat{z}^L\); moreover, \(C^L(z) < C^R(z)\) on \((e^L,\hat{z}^L \wedge e^R)\);

4. \(\hat{z}^R < e^L = \hat{z}^L\);

5. \(\hat{z}^R < \hat{z}^L\) and there exists \(z^* \in (e^L,e^R)\) such that \(C^L(z^*)=C^R(z^*) \geq a(z^*)\), but \(\Psi\) defined by 42 is not convex.

Note that if \(e^R = \hat{z}^R\) then the set \((\hat{z}^R \vee e^L,e^R)\) is empty, so we have omitted this condition from \((ii)\); a similar consideration applies in \((iv)\).

Suppose we are in Case \((ii)\). Then \(a(e^R) = b(e^R)\). We take \(z_0=e^R\) and define \((x_0 = (g^L)^{-1}(z_0),y_0=e^R)\). Then \(g^R(y_0) = y_0 = e^R=z_0\), and the fourth line in the definition of \(\Psi_{x_0,y_0}\) is redundant. Similarly, in Case \((iv)\) we take \(z_0=e^L\) and \((x_0 = e^L, y_0 =(f^R)^{-1}(z_0))\). Then \(f^L(x_0) = x_0 = e^L=z_0\) and the second line in the definition of \(\Psi_{x_0,y_0}\) is redundant.

Suppose now that we are in Case \((v)\). In this case take \(x_0 = (g^L)^{-1}(z^*)\) and \(y_0 = (f^R)^{-1}(z^*)\).

It remains to define \((x_0,y_0)\) in Cases \((i)\) and \((iii)\). We consider Case \((i)\); the construction in Case \((iii)\) follows a symmetric argument. Since \(\hat{z}^R<e^R\) we must have \(a(e^R) > b(e^R)\). For \(z \in (\hat{z}^R,e^R)\) we can define \((\check{x}(z) = (g^L)^{-1}(z),\check{y}(z) = (f^R)^{-1}(z))\). By taking \(z\) close enough to \(e^R\) (in Case \((iii)\) we take \(z\in(e^L,\hat{z}^L)\) close enough to \(e^L\)) we can ensure that \(-\infty < \tilde{S}^{\check{y}(z),z}_{a,b} < \tilde{S}^{\check{x}(z),\check{y}(z)}_{a,a}< \tilde{S}_{b,a}^{f^L(\check{x}(z)),\check{x}(z)}<\infty\). Let \(z_0\in(e^L,e^R)\) be a value with this property. Finally, let \((x_0 = \check{x}(z_0),y_0 = \check{y}(z_0))\).

Note that (by construction) in each case we have that \((x_0,y_0)\in\mathcal{A}^<\) is such that both \(x_0,y_0\) are finite, \(y_0=B(x_0)\leq\beta_\mu\) and \(x_0\leq (g^L)^{-1}(\beta_\mu)\).

Our aim is to find \(x_1\) with \(x_0 \leq x_1 < y_0\) (so that \((x_1,y_0) \in \mathcal{A}^<\)) such that either \(\Psi_{x_1,y_0}=T_{a,a}^{x_1,y_0}=T_{a,b}^{x_1,z(x_1,y_0)}\) on \((x_1,z(x_1,y_0))\) and hence \(\tilde{S}_{b,a}^{w(x_1,y_0),x_1} \geq \tilde{S}_{a,a}^{x_1,y_0}=\tilde{S}_{a,b}^{y_0,z(x_1,y_0)}\) (at least if \(w(x_1,y_0) < x_1 < y_0 < z(x_1,y_0)\)), or \(\Psi_{x_1,y_0}=T_{b,a}^{w(x_1,y_0),x_1}=T_{b,a}^{w(x_1,y_0),y_0}\) on \((w(x_1,y_0),y_0)\) and \(\tilde{S}_{b,a}^{w(x_1,y_0),x_1}=\tilde{S}_{a,a}^{x_1,y_0} \geq \tilde{S}_{a,b}^{y_0,z(x_1,y_0)}\) (again, at least if \(w(x_1,y_0) < x_1 < y_0 < z(x_1,y_0)\)).

Define \[\begin{align} \label{eq:sLaab} {\mathcal{L}}^{a,a,b}_{3} & = & \{(x,y,z) : \alpha_\mu < x < y < \beta_\mu, y \leq z < \beta_\nu, \\ &&(x,a(x)), (y,a(y)) and (z,b(z)) are collinear \} \nonumber \\ &&\cup \{(x,\beta_\mu,z) : \alpha_\mu < x < \beta_\mu < z < \beta_\nu, \nonumber \\ &&a(\beta_\mu-) \leq \frac{z-\beta_\mu}{z-x} a(x) + \frac{\beta_\mu-x}{z-x} b(z) \nonumber \}. \end{align}\tag{45}\] Then, for \((x,y,z) \in {\mathcal{L}}^{a,a,b}_{3}\) with \(y<\beta_\mu\), we have that \(T_{a,b}^{x,z}=T_{a,a}^{x,y}=T_{a,b}^{y,z}\) is the straight line passing through \((x,a(x))\), \((y,a(y))\) and \((z,b(z))\). When \(y=\beta_\mu<\infty\) we have that \(T_{a,b}^{x,z}(\beta_\mu)\) lies in the interval \((a(\beta_\mu-),\infty)\). Similarly, define \[\begin{align} \label{eq:sLbaa} {\mathcal{L}}^{b,a,a}_{3} & = & \{(w,x,y) : \alpha_\mu < x < y < \beta_\mu, \alpha_\mu < w \leq x\\ &&(w,b(w)), (x,a(x)) and (y,a(y)) are collinear \} \nonumber \\ &&\cup \{(w,\alpha_\mu,y) : \alpha_\nu < w < \alpha_\mu < y < \beta_\mu, \nonumber \\ &&a(\alpha_\mu+) \leq \frac{\alpha_\mu-w}{y-w} a(y) + \frac{y-\alpha_\mu}{y-w} b(w) \nonumber \} \end{align}\tag{46}\]

Consider Case \((ii)\) whence \((x_0,y_0) = ((g^L)^{-1}(e^R),e^R)\) and \(a(e^R)=b(e^R)\). In this case we set \(x_1=x_0\). Then \(w(x_1,y_0) = f^L(x_1) < x_1 < y_0 = z(x_1,y_0)\). It follows from the twin facts that \(a(y_0)=b(y_0)\) and \(y_0=z(x_1,y_0)\) that (tautologically) \(T_{a,a}^{x_1,y_0} = T_{a,b}^{x_1,z(x_1,y_0)}\). Moreover, from the fact that \(\hat{z}^L > e^R\) we have that \(C^L(e^R)> a(e^R)\) and hence \(\tilde{S}_{b,a}^{w(x_1,y_0),x_1} > \tilde{S}_{a,a}^{x_1,y_0=e^R}\). We find \((x_1,y_0,z(x_1,y_0))= (x_1,e^R,e^R)\) is such that both \((x_1,a(x_1))\), \((y_0,a(y_0))\) and \((z(x_1,y_0),b(z(x_1,y_0))\) lie on the same straight line (because the last two points are the same point).

Case \((iv)\) is similar. This time we adjust \(y\) and leave \(x_0=e^L\) unchanged. We find \(y_1\in (x_0,y_0)\) such that both \((w(x_0,y_1), b(w(x_0,y_1))\) and \((x_0,a(x_0))\), \((y_1,a(y_1))\) lie on the same straight line (because the last two points are the same point).

Now we consider Cases \((v)\), \((i)\) and \((iii)\). In Case \((v)\), recall that \(x_0 = (g^L)^{-1}(z^*)\) and \(y_0 = (f^R)^{-1}(z^*)\). In Cases \((i)\) and \((iii)\) we have that \(x_0 = (g^L)^{-1}(z_0)\) and \(y_0= (f^R)^{-1}(z_0)\). In all cases \(e^L < x_0 < y_0 < e^R\). For \(x\in[x_0,y_0)\), let \(\tilde{S}^L_x:=\tilde{S}_{b,a}^{w(x,y_0),x},\tilde{S}^a_x:=\tilde{S}_{a,a}^{x,y_0},\tilde{S}^R_x:=\tilde{S}_{a,b}^{y_0,z(x,y_0)}\) and note that \(\Psi_{x_0,y_0}\) (as in 44 ) is continuous and \(\infty>\tilde{S}^{L}_{x_0}> \tilde{S}^{a}_{x_0} > \tilde{S}^{R}_{x_0}>-\infty\), since we are in Case \((i)\), \((iii)\) or \((v)\). Let \(w(x,y_0)\) and \(z(x,y_0)\) be as defined in Lemma 45 Now, keeping \(y_0\) fixed, consider increasing \(x\) from \(x_0\). By Lemma 45\((ii)\), as \(x\) increases we have that \(w(x,y_0)\) decreases and \(z(x,y_0)\) increases. Then, by the convexity of \(a\) and \(b\), the slopes \(\tilde{S}^a_x\) and \(\tilde{S}^R_x\) increase, while the slope \(\tilde{S}^L_x\) may (or may not) be monotone. On the other hand, all three slopes \(x\mapsto \tilde{S}^a_x,\tilde{S}^R_x,\tilde{S}^L_x\) are continuous on \((x_0,y_0)\). We increase \(x\) until either \(\tilde{S}^L_{x} = \tilde{S}^a_{x}\) or \(\tilde{S}^a_{x}=\tilde{S}^R_{x}\). Let \(x_1 = \inf \{ x_0<x<y_0 : \tilde{S}^L_{x} \leq \tilde{S}^a_{x} or \tilde{S}^a_{x} \leq \tilde{S}^R_{x} \}\) and set \(\inf\emptyset=y_0\). We argue that we must have \(x_0<x_1 < y_0\). Note that, as \(x \uparrow y_0\) then \(\mu ((\alpha_\mu,x) \cup(y_0,\beta_\mu)) \uparrow 1\) and so we must have \(w(x,y_0) \downarrow \alpha_\nu\) and \(z(x,y_0) \uparrow \beta_\nu\).

Since \(\infty>\tilde{S}^{L}_{x_0}> \tilde{S}^{a}_{x_0} > \tilde{S}^{R}_{x_0}>-\infty\), by the continuity we have that there exists \(\epsilon>0\) such that for all \(x\in[x_0,x_0+\epsilon)\), \(\infty>\tilde{S}^{L}_{x}> \tilde{S}^{a}_{x} > \tilde{S}^{R}_{x}>-\infty\). It follows that \(x_0<x_0+\epsilon\leq {x_1}\).

Now suppose \(x_1=y_0\). Then \(\tilde{S}^{L}_{x}> \tilde{S}^{a}_{x} > \tilde{S}^{R}_{x}\) for all \(x\in[x_0,y_0)\). By construction, we then have that \[a(u)\geq T_{a,a}^{x,y_0}(u)>T_{b,a}^{w(x,y_0),x}(u)\geq b(u),\quad u\in(w(x,y_0),x)\] and \[a(u)\geq T_{a,a}^{x,y_0}(u)>T_{a,b}^{y_0,z(x,y_0)}(u)\geq b(u),\quad u\in(y_0,z(x,y_0)).\] It follows (using the continuity w.r.t. \(x\in(x_0,y_0)\)) that \[a(u)\geq \lim_{x\uparrow y_0}T_{a,a}^{x,y_0}(u)=a(y_0)+a'_-(y_0)(u-y_0)\geq b(u),\quad u\in(\alpha_\nu,\beta_\nu),\] where \(a'_-\) denotes the left derivative of a convex function \(a\). But this contradicts the Standing Assumption 3, and thus we conclude that \(x_1<y_0\).

It is possible that \(\tilde{S}^L_{x_1} = \tilde{S}^a_{x_1} = \tilde{S}^R_{x_1}\). In that (rather special) case we have found \[(w(x_1,y_0),x_1,y_0,z(x_1,y_0)) \in \mathcal{L}_4 \cap \Gamma_4\] and the proof of Theorem 40 is complete. More generally, we are in either Case \((I)\): \(\tilde{S}^L_{x_1} > \tilde{S}^a_{x_1} = \tilde{S}^R_{x_1}\) so that \((x_1,y_0,z(x_1,y_0))\in\mathcal{L}^{a,a,b}_3\); or in Case \((II)\): \(\tilde{S}^L_{x_1} = \tilde{S}^a_{x_1} > \tilde{S}^R_{x_1}\) so that \((w(x_1,y_0),x_1,y_0)\in\mathcal{L}_3^{b,a,a}\). In each case \((w(x_1,y_0),x_1,y_0,z(x_1,y_0)) \in \Gamma_4\) but only one of the points \((w=w(x_1,y_0),b(w))\) or \((z=z(x_1,y_0),b(z))\) lies on the straight line \(T_{a,a}^{x_1,y_0}\) passing through \((x_1,a(x_1))\) and \((y_0,a(y_0))\).

In all the cases we found \((x_1,y_1) \in \mathcal{A}^<\) such that either \((w(x_1,y_1),b(w(x_1,y_1)))\) lies on \(T_{a,a}^{x_1,y_1}\) (this corresponds to \((x_1,y_1)=(x_0,y_1)\) in Case \((iv)\) or \((x_1,y_1)=(x_1,y_0)\) in Case \((II)\)) or \((z(x_1,y_1),b(z(x_1,y_1)))\) lies on \(T_{a,a}^{x_1,y_1}\) (this corresponds to \((x_1,y_1)=(x_1,y_0)\) in Case \((ii)\) or \((x_1,y_1)=(x_1,y_0)\) in Case \((I)\)); if both, then we have found a point \((w(x_1,y_1),x_1,y_1,z(x_1,y_1)) \in \mathcal{L}_4 \cap \Gamma_4\) and we are done.

Without loss of generality we assume that \((z(x_1,y_1),b(z(x_1,y_1)))\) lies on \(T_{a,a}^{x_1,y_1}\) and \(T_{a,a}^{x_1,y_1}(w(x_1,y_1)) > b(w(x_1,y_1))\). In particular, we are either in Case \((ii)\) or Case \((I)\), so that (in both cases) \((x_1,y_1)=(x_1,y_0)\). This is the starting point of Step 3.

Step 3: the map \(x\mapsto y^*(x)\). Our aim is to increase \(x\) from \(x_1\) and to adjust \(y=y_x\) (starting from \(y_0\)), so that \((x,y,z(x,y)) \in \mathcal{L}^{a,a,b}_{3}\) and \((w(x,y),x,y,z(x,y)) \in \Gamma_4\), and then to keep increasing \(x\) (and simultaneously adjusting \(y=y_x\)) until \((w(x,y), x , y, z(x,y)) \in \mathcal{L}_4 \cap \Gamma_4\).

Let \(\mathcal{A}_1=\{(x,y,z):\alpha_\mu< x< y < \beta_\mu,~ y<z < \beta_\nu \}\). Define \(\hat{\Lambda}: \mathcal{A}_1 \to \mathbb{R}\) by \[\hat{\Lambda}(x,y,z) = a(y) + \frac{z-y}{y-x} (a(y)-a(x)) - b(z).\] Extend the definition to \(y=x<z\) by \(\hat{\Lambda}(x,x+,z) = a(x) + (z-x) a'(x+) - b(z)\) and \(\hat{\Lambda}(x-,x,z) = a(x) + (z-x) a'(x-) - b(z)\), and to \(y = \beta_\mu<\infty\) by taking \(a\) to have the value \(a(\beta_\mu-)\) there: for \(x<\beta_\mu < z\) \[\label{eq:Lambdabeta} \hat{\Lambda}(x,\beta_\mu,z) = a(\beta_\mu-) + \frac{z-\beta_\mu}{\beta_\mu-x} (a(\beta_\mu-)-a(x)) - b(z).\tag{47}\]

Let \(\Lambda: \mathcal{A}^< \to \mathbb{R}\) be given by \(\Lambda(x,y) = \hat{\Lambda}(x,y,z(x,y))\). Let \(\Lambda(x,{ x+}) = \lim_{y \downarrow x} \hat{\Lambda}(x,y,z(x,y)) = a(x) + \lim_{z \uparrow \beta_\nu} \{ (z-x) a'(x+) - b(z) \}\). Note that \(\Lambda\) may not be continuous at the diagonal if \(a'\) is not continuous, but \(\Lambda\) is continuous on \(\mathcal{A}^<\).

Recall the definition of \(B\) (see 41 ) and that \(\bar{x}= (g^{L})^{-1}(e^R)\).

Moreover, recall the final part of Standing Assumption 2 that there are no intervals \(I \subseteq (\alpha_\mu,\beta_\mu)\) on which \(a=b\) and \(a\) is linear. Proposition 48 is the only place where this assumption is used.

Let \(\tilde{x}= \inf \{ x \in (x_1, \beta_\mu) : \Lambda(x,x+) \geq 0 \}\) where \(\inf \emptyset = \beta_\mu\).

We now argue that \(\tilde{x}>x_1\). It is enough to show that \(\Lambda(x,x+)<0\) on some interval immediately to the right of \(x_1\). Since \(\Lambda (x_1,y_0)=0\) and \(x_1<y_0\), by the final part of Proposition 48 we have that \(\Lambda (x_1,y)<0\) for each \(y\in (x_1,y_0)\). Fix \(y \in (x_1,y_0)\). Then, by the continuity of \(\Lambda(\cdot,y)\), for small enough \(\epsilon =\epsilon(x_1,y) \in (0,y-x_1)\) we must have that \(\Lambda (x,y)<0\) for all \(x\in[x_1,x_1+\epsilon(x_1,y))\). Then using the monotonicity of \(\Lambda(x_1,\cdot)\) on \((x_1,y)\) (see the first part of Proposition 48), we have that \(\Lambda(x,x+) < 0\). Hence \(\tilde{x} \geq x_1+\epsilon > x_1\).

For \(x \in [x_1, \tilde{x})\) let \(y^*=y^*(x)\) denote the unique point in \((x,B(x)]\) such that \(\Lambda(x,y^*(x))=0\). We will show that \(y^*\) exists on a suitable subset of \([x_1, \tilde{x})\). Then we will extend the definition of \(y^*\) to the whole of \([x_1, \tilde{x})\) and show that this extended version is continuous as a map \(x\mapsto y^*(x)\).

Write \([x_1,\tilde{x})\) as a disjoint union \(\mathcal{A}^\Lambda_{>}\cup\mathcal{A}^\Lambda_{=}\cup\mathcal{A}^\Lambda_{<}\), where \(\mathcal{A}^\Lambda_{>}:=\{x\in[x_1,\tilde{x}):\Lambda(x,B(x))>0\}\), \(\mathcal{A}^\Lambda_{=}:=\{x\in[x_1,\tilde{x}):\Lambda(x,B(x))=0\}\) and \(\mathcal{A}^\Lambda_{<}:=\{x\in[x_1,\tilde{x}):\Lambda(x,B(x))<0\}\). Note that, due to Proposition 49, \(\mathcal{A}^\Lambda_{<}\subseteq ((g^L)^{-1}(\beta_\mu)\wedge\tilde{x},\tilde{x})\). Furthermore, \(x_1\in \mathcal{A}^\Lambda_{>}\cup\mathcal{A}^\Lambda_{=}\).

Step 4: finding \((w,x,y,z)\in\mathcal{L}_4\cap\Gamma_4\). Define \(\Upsilon:[x_1,\tilde{x} ) \to \mathbb{R}\) by \[\Upsilon(x) = a(x) + (w(x,y^*(x)) -x) \frac{b(z(x,y^*(x)) - a(x)}{z(x,y^*(x))-x} - b(w(x,y^*(x)))\] Note that \(\Upsilon\) has been defined in such a way that it does not depend on the value of \(a(y^*(x))\). This is important in the case where \(y^*(x)= \beta_\mu\).

Since we are in Case (C3) we have that \(\Upsilon(x_1)>0\).

It follows from the proposition that in Case (C3) there exists \(x \in (x_1,\tilde{x})\) such that \(\Upsilon(x)=0\). Then, by Lemma 52, \((w(x,y^*(x)),x,y^*(x),z(x,y^*(x))) \in \mathcal{L}_4 \cap \Gamma_4\). The proof of Theorem 40 is then complete. Further, we have constructed an optimal model and an optimal superhedge. The model and superhedge are illustrated in Figure 6.

8 \(\mathcal{P}^{can} = \mathcal{P}^{rand}\)↩︎

9 Further remarks and extensions↩︎

9.1 Dropping the assumption that \(a \geq b\)↩︎

In the main text of this section we assumed that \(a \geq b\). Here we show that this assumption is not necessary and that the results remain true without this assumption. Instead of requiring \(a\) and \(b\) are convex and \(a \geq b\) it is sufficient that \(b\) and \(a \vee b\) are convex.

First, we consider the pricing problem. Fix a model \(\mathcal{M}\in M(\mu,\nu)\) and consider the model-based price of the Bermudan claim: \[P(\mathcal{M};a,b):=\sup_{\tau \in {\mathcal{T}}_{1,2}} \mathbb{E}^{\mathcal{M}}[ a (Z_1)I_{\{{\tau}=1\}}+b(Z_2)I_{\{{\tau}=2\}}]\]

Now we turn to the dual problem and the hedging cost.

9.2 Stopping at time-0↩︎

From a financial point of view, it is natural to allow immediate exercise for Bermudan (or American) options; of course the initial distribution of the underlying asset is trivial (i.e., \(Z_0\sim\delta_{\bar\nu}\)). On the other hand, in the earlier parts of the paper the set \(\mathcal{T}\) of admissible stopping rules did not include \(t=0\). In this subsection, by allowing the option holder to stop immediately, but restricting the stopping rules to take values in \(\{0,1\}\), we show how to recover the highest no-arbitrage price of the Bermudan option, together with the cheapest superhedging strategy. In particular, our goal is to solve \[\sup_{\mathcal{M}\in M_0(\mu=\delta_{\bar\nu},\nu)} \sup_{\tau \in \mathcal{T}_{0,1}} \mathbb{E}^{\mathcal{M}}[ a(Z_0)I_{\{\tau=0\}}+b(Z_1)I_{\{\tau=1\}}],\] where \(M_0(\mu=\delta_{\bar\nu},\nu)\) is the set of consistent models (i.e., filtered probability spaces \((\Omega, \mathcal{F}, \mathbb{F}, \mathbb{Q})\)) supporting a stochastic process \(Z=(Z_0,Z_1)\) such that \(Z\) is a \((\mathbb{F},\mathbb{Q})\)-martingale with given marginals \(Z_0 \sim \mu =\delta_{\bar\nu}\) and \(Z_1 \sim \nu\), and \(\mathcal{T}_{0,1}\) is the set of \(\mathbb{F}\)-stopping times taking values in \(\{0,1\}\). We still assume that \(b\) is convex and \(\nu\) is continuous on \((\alpha_\nu,\beta_\nu)\).

Note that, by considering candidate stopping rules \(\tau_0=0\) and \(\tau_1=1\), we have that \[\label{adnrqgkm} \sup_{\mathcal{M}\in M_0(\mu=\delta_{\bar\nu},\nu)} \sup_{\tau \in \mathcal{T}_{0,1}} \mathbb{E}^{\mathcal{M}}[ a(Z_0)I_{\{\tau=0\}}+b(Z_1)I_{\{\tau=1\}}]\geq a(\bar\nu)\vee\int b(y)\nu(dy).\tag{54}\] On the other hand, the right hand side is attained if we only consider models with (canonical) filtration generated by \(Z\). Indeed, in this case \(\mathcal{F}_0\) is trivial (since \(Z_0\sim\mu={\delta_{\bar\nu}}\)), and thus \(\mathcal{T}_{0,1}=\{0,1\}\). We now show that, by considering a richer probabilistic structure (similarly as in Subsection 7.3), the inequality in [eq:t01] is strict.

Suppose that \(b\) is convex (and not linear). Further suppose that \(b(\bar\nu)<a(\bar\nu)\) (otherwise, under any model, it is always optimal to stop at time-1) and \(a(\bar{\nu})<\sup_{\beta_l < x < \bar\nu < y < \beta_r} \left\{ \frac{y-\bar\nu}{y-x} b(x) + \frac{\bar\nu - x}{y-x} b(y) \right\}\) (otherwise, under any model, it is always optimal to stop at time-0). Let \((f,g)\) with \(f < \bar\nu < g\) solve \[\label{eq:fgdef} \int_f^g (y - \bar\nu) \nu(dy) = 0;\frac{b(g)-b(f)}{g-f} = \frac{a(\bar\nu) - b(f)}{\bar\nu-f}.\tag{55}\] Set \(\Lambda = \Lambda_{\nu,a,b} = \frac{b(g)-b(f)}{g-f}\). Then also \(\Lambda = \frac{b(g) -a(\bar\nu)}{g-\bar\nu}\).

Let \(L(x)= a(\bar\nu) + \Lambda(x- \bar\nu)\) and set \(\psi=\max\{b,L\}\). Then by construction \(\psi = L\) on \([f,g]\) and \(\psi = b\) on \([\alpha_\nu,f] \cup [g,\beta_\nu]\) (note that \(\psi(\bar\nu)=L(\bar\nu)=a(\bar\nu)\)). Further, \(\psi\) is convex and \(\psi\geq b\) on \(\mathbb{R}\), and thus \(\psi\) generates a superhedge with total cost \[\begin{align} \int(a-\psi)^+d\delta_{\bar\nu}+\int\psi d\nu = \int\psi d\nu & = & \int_{(\alpha_\nu,f]\cup[g,\beta_\nu)}b(y)\nu(dy)+ \int_f^g L(y) \nu(dy) \\ & =&\int_{(\alpha_\nu,f]\cup[g,\beta_\nu)}b(y)\nu(dy)+a(\bar\nu)\nu((f,g)) \end{align}\] where we use the first part of 55 to rewrite the final term. Note that \(\int\psi d\nu>a(\bar\nu)\vee\int bd\nu\).

Then, the optimal model is obtained by stopping a \(\nu((f,g))\) amount of mass at time-0 at location \(\bar\nu\) (this is achieved by working with additional uniform random variable \(U\), and a stopping time \(\tau\) such that \(\{\tau=0\}=\{U\leq\nu((f,g))\}\)), while the remaining \(1-\nu((f,g))\) mass at \(\bar\nu\) is mapped to \(\nu\lvert_{\{(\alpha_\nu,f]\}\cup\{[g,\beta_\nu)\}}\). It follows that \[\begin{align} \sup_{\mathcal{M}\in M_0(\mu=\delta_{\bar\nu},\nu)} \sup_{\tau \in \mathcal{T}_{0,1}} &\mathbb{E}^{\mathcal{M}}[ a(Z_0)I_{\{\tau=0\}}+b(Z_1)I_{\{\tau=1\}}]\\&=\int_{(\alpha_\nu,f]\cup[g,\beta_\nu)}b(y)\nu(dy)+a(\bar\nu)\nu((f,g))> a(\bar\nu)\vee\int b(y)\nu(dy). \end{align}\]

Again, models with canonical filtration are not rich enough.

10 Proofs↩︎

Proof of Theorem 28↩︎

Case 1: 4-point construction. Write \(\Psi_4\) as shorthand for \(\Psi^{y_-,x_-,x_+,y_+}_{4,a,b}\). Since \((y_-,x_-,x_+,y_+) \in \mathcal{L}^4_{a,b}\), on \([y_-,y_+]\), \(\Psi_4=T^{y_-,y_+}_{b,b}\) is the straight-line passing through \((y_-,b(y_-))\), \((x_-,a(x_-))\), \((x_+,a(x_+))\) and \((y_+,b(y_+))\), at least if \(a\) is continuous at \(x_-\) and \(x_+\). Then, by the convexity of \(b\), we have that \(\Psi_4\geq b\) on \([y_-,y_+]\) and thus \(\Psi_4\geq b\) on \(\mathbb{R}\). It follows that \(\Psi_4\in\tilde{\mathcal{S}}(b)\). Also note that, by the convexity of \(a\), we have that \(\Psi_4\geq a\) on \([x_-,x_+]\) and \(\Psi^{y_-,x_-,x_+,y_+}\leq a\) on \(\mathbb{R}\setminus[x_-,x_+]\). To ease the notation, set \(\psi=\Psi_4=\Psi^{y_-,x_-,x_+,y_+}_{4,a,b}\), \(\phi=(a-\psi)^+\) and \(T=T^{y_-,y_+}_{b,b}\). Then \[\begin{align}\\ &=&\int_{(\alpha_\mu,y_-)\cup(y_+,\beta_\mu)}(a(u)-b(u))\mu(du)+\int_{(y_-,x_-)\cup(x_+,y_+)}(a(u)-T(u))\mu(du)\\ && \quad +\int_{(\alpha_\nu,y_-)\cup(y_+,\beta_\nu)}b(v)\nu(dv)+\int_{(y_-,y_+)}T(v)\nu(dv)\\ &=&\int_{(\alpha_\mu,x_-)\cup(x_+,\beta_\mu)}a(u)\mu(du)+\int_{(\alpha_\nu,y_-)\cup(y_+,\beta_\nu)}b(v)(\nu-\mu)(dv)\\ && \quad +\left\{\int_{(y_-,y_+)}T(v)\nu(dv)-\int_{(y_-,x_-)\cup(x_+,y_+)}T(u)\mu(du)\right\}. \end{align}\] Now, by assumption, \(\Pi_M^{y_-,x_-,x_+,y_+}(\mu,\nu)\neq\emptyset\). Then, by Lemma 25, \((y_-,x_-,x_+,y_+)\in\Gamma^{\mu,\nu}_4\). Hence (see the definition of \(\Gamma^{\mu,\nu}_4\) given in 9 ) we must have that \(\mu\leq\nu\) on \((-\alpha_\nu,y_-)\cup(y_+,\beta_\nu)\), so that, on \((-\alpha_\nu,y_-)\cup(y_+,\beta_\nu)\), \((\nu-\mu)\) is a well-defined (non-negative) measure.

We also argue that the final bracketed term vanishes. Indeed, since \(T\) is a line, it is enough to show that \(\mu|_{(y_-,x_-) \cup(x_+,y_+)}\) and \(\nu|_{(y_-,y_+)}\) have the same total mass and mean. But this is immediate from the fact that \(\mu|_{(y_-,x_-) \cup(x_+,y_+)} \leq_{cx} \nu|_{(y_-,y_+)}\) (here we again use that \((y_-,x_-,x_+,y_+)\in\Gamma^{\mu,\nu}_4\) together with 9 ).

Finally, note that, under each \(\pi\in\Pi_M^{y_-,x_-,x_+,y_+}(\mu,\nu)\), the mass in \((x_-,x_+)\) is mapped to \((-\alpha_\nu,y_-)\cup(y_+,\beta_\nu)\) and \(\mu\lvert_{(x_-,x_+)}\leq_{cx} (\nu-\mu)\lvert_{(-\alpha_\nu,y_-)\cup(y_+,\beta_\nu)}\); see Lemma 17 and 23 . In particular, \((\nu - \mu)(dv)\left[I_{ \{ v \geq y_+ \} } +I_{ \{ v \leq y_- \}} \right] = \int_{w\in(x_-,x_+)} \pi_w(dv) \mu(dw)\). It follows that \[\begin{align}\\ &=&\int_{(\alpha_\mu,x_-)\cup(x_+,\beta_\mu)}a(u)\mu(du)+\int_{(\alpha_\nu,y_-)\cup(y_+,\beta_\nu)}b(v)(\nu-\mu)(dv)\\ & = & \mathbb{E}^{\mathcal{M}_\pi}[a(Z_1) I_{ \{ \tau^{x_-,x_+}=1 \} } + b(Z_2) I_{ \{ \tau^{x_-,x_+} = 2 \} }], \end{align}\] which finishes the proof in this case.

Case 2: 5-point construction. Since, by assumption, \(\Psi_5 := \Psi_{5,a,b}^{y_-,x_-,z,x_+,y_+}\) is convex, it is continuous at each \(w\in\{y_-,x_-,z,x_+,y_+\}\), and therefore \(\Psi_5(y_-) =b(y_-)=T^{y_-,x_-}_{b,a}(y_-)\), \(\Psi_5(z)=T^{y_-,x_-}_{b,a}(z)=T^{x_+,y_+}_{a,b}(z)\) and \(\Psi_5(y_+)=b(y_+)=T^{x_+,y_+}_{a,b}(y_+)\). Then, since \(a\) is convex and (again by assumption) \(\Psi_5(z) \geq a(z)\), we have that \(\Psi_5\geq a\geq b\) on \((x_-,x_+)\) and \(\Psi_5\leq a\) on \(\mathbb{R}\setminus(x_-,x_+)\). By applying a similar argument (together with the convexity of \(b\)) we also have that \(\Psi_5\geq b\) on \((y_-,y_+)\) (recall that \(\Psi_5=b\) on \(\mathbb{R}\setminus(y_-,y_+)\)). It follows that \(\Psi_5\in\tilde{\mathcal{S}}(b)\). To ease the notation, set \(\psi=\Psi_5 = \Psi_{5,a,b}^{y_-,x_-,z,x_+,y_+}\), \(\phi=(a-\psi)^+\), \(T_-=T^{y_-,x_-}_{b,a}\) and \(T_+=T^{x_+,y_+}_{a,b}\). Then \[\begin{align} \int_\mathbb{R}\phi(w)\mu(dw)&= &\int_{\alpha_\mu}^{y_-}(a(w)-b(w))\mu(dw)+\int_{y_-}^{x_-}(a(w)-T_-(w))\mu(dw)\\ &&+\int_{x_+}^{ y_+}(a(w)-T_{+}(w))\mu(dw)+\int_{y_+}^{\beta_\mu}(a(w)-b(w))\mu(dw) \end{align}\] and \[\int_\mathbb{R}\psi(v)\nu(dv)=\int_{\alpha_\nu}^{y_-}b(v)\nu(dv)+\int_{y_-}^{z} T_{-}(v)\nu(dv) +\int_{z}^{ y_+}T_{+}(v)\nu(dv)+\int_{y_+}^{\beta_\nu} b(v)\nu(dv).\] It follows that \[\begin{align}\\ &=&\int_{(\alpha_\mu,x_-)\cup(x_+,\beta_\mu)}a(w)\mu(dw)+\int_{(\alpha_\nu,y_-)\cup( y_+,\beta_\nu)}b(v)(\nu-\mu)(dv)\\ && +\left\{\int_{y_-}^{z} T_{-}(v)\nu(dv)-\int_{y_-}^{x_-}T_{-}(w)\mu(dw)\right\} \\ &&+ \left\{\int_{z}^{ y_+} T_+(v)\nu(dv)-\int_{x_+}^{y_+} T_{+}(w)\mu(dw)\right\}. \end{align}\] However, similarly as in Case 1, the bracketed terms in the last two lines vanish, whilst \((\nu-\mu)\) is a well-defined (non-negative) measure on \(\mathbb{R}\setminus(y_-,y_+)\). For the latter, note that, since \(\Pi_M^{y_-,x_-,z,x_+,y_+}(\mu,\nu)\neq\emptyset\), by Lemma 26 we have that \((y_-,x_-,z,x_+,y_+)\in\Gamma^{\mu,\nu}_5\). Then by 10 it follows that \(\mu\leq\nu\) on \((-\alpha_\nu,y_-)\cup(y_+,\beta_\nu)\) (see also 26 ). For the former, again using that \((y_-,x_-,z,x_+,y_+)\in\Gamma^{\mu,\nu}_5\) (together with 10 ), we have both \(\mu|_{(y_-,x_-)}\leq_{cx} \nu|_{(y_-,z)}\) and \(\mu|_{(x_+,y_+)}\leq_{cx}\nu|_{(z,y_+)}\), and thus the bracketed terms vanish due to the linearity of \(T_-\) and \(T_+\). Also note that, under each \(\pi\in\Pi_M^{y_-,x_-,z,x_+,y_+}(\mu,\nu)\), the mass in \((x_-,x_+)\) is mapped to \((-\alpha_\nu,y_-)\cup(y_+,\beta_\nu)\) and \(\mu\lvert_{(x_-,x_+)}\leq_{cx} (\nu-\mu)\lvert_{(-\alpha_\nu,y_-)\cup(y_+,\beta_\nu)}\); see 29 and Lemma 18. In particular, \((\nu - \mu)(dv)\left[I_{ \{ v \geq y_+ \} } +I_{ \{ v \leq y_- \}} \right] = \int_{w\in(x_-,x_+)} \pi_w(dv) \mu(dw)\).

Finally, fix \(\pi\in\Pi^{y_-,x_-,z,x_+,y_+}_M(\mu,\nu)\) and let \(\mathcal{M}_\pi\in M^{can}(\mu,\nu)\) be the corresponding canonical model. Also, recall the definition of the candidate stopping time \(\tau^{x_-,x_+}\in\mathcal{T}^{can}_{1,2}\), see 24 . Then \[\begin{align} \\ & = & \int_{w \in (\alpha_\mu,x_-) \cup (x_+,\beta_\mu)} a(w) \pi_w(dv) \mu(dw) + \int_{w \in (x_-,x_+)} \int_{v\in\mathbb{R}} b(v) \pi_w(dv) \mu(dw) \\ & = & \mathbb{E}^{\mathcal{M}_\pi}[a(Z_1) I_{ \{ \tau^{x_-,x_+}=1 \} } + b(Z_2) I_{ \{ \tau^{x_-,x_+} = 2 \} }], \end{align}\] which finishes the proof in this case.

Proof of Lemma 37↩︎

To prove the claim, one needs to construct the functions \((f^L,g^L)\) and \((f^R,g^R)\) under the Dispersion Assumption, and then show that they satisfy the required properties. See, for example, Henry-Labordère and Touzi [6] where the authors construct the supporting functions by solving a system of coupled ODEs. In particular, under the Dispersion Assumption, define \(f^L(x)=x=g^L(x)\) on \((\alpha_\mu,e^L]\), while on \((e^L,\beta_\mu)\) let \((f^L,g^L)\) be the solutions to 39 .

Alternatively, a more general construction via potential functions is provided by Hobson and Norgilas [25]. Fix \(x\in(\alpha_\mu,\beta_\mu)\). By Theorem 35, the left-curtain coupling embeds \(\mu_{\alpha_\mu}^x\) into \(\nu\) via the shadow measure \(S^\nu(\mu_{\alpha_\mu}^x)\). Then the main insight of Hobson and Norgilas [25] is that we can recover the locations \((f^L(x),g^L(x))\) from the potential function \(P_{S^\nu(\mu_{\alpha_\mu}^x)}\). In particular, by Proposition 31 we have that \(P_{\nu-S^\nu(\mu_{\alpha_\mu}^x)}=(P_\nu-P_{\mu_{\alpha_\mu}^x})^c\). In the case \((P_\nu-P_{\mu_{\alpha_\mu}^x})^c(x)=(P_\nu-P_{\mu_{\alpha_\mu}^x})(x)\) one sets \(f^L(x)=x=g^L(x)\). On the other hand, if \((P_\nu-P_{\mu_{\alpha_\mu}^x})^c(x)<(P_\nu-P_{\mu_{\alpha_\mu}^x})(x)\), then \((P_\nu-P_{\mu_{\alpha_\mu}^x})^c\) is linear and satisfies \((P_\nu-P_{\mu_{\alpha_\mu}^x})^c<(P_\nu-P_{\mu_{\alpha_\mu}^x})\) on some open interval \(I_x\ni x\). In this case \((f^L(x),g^L(x))=I_x\), i.e., the locations of the supporting functions coincide with the end-points of \(I_x\).

Under the Dispersion Assumption, we have that \(\mu\leq\nu\) (and thus also \(\mu_{\alpha_\mu}^x\leq\mu\leq\nu\) for \(x\leq e^L\)) on \((\alpha_\mu,e^L]\). Then, for \(x\in(\alpha_\mu,e^L]\), \((P_\nu-P_{\mu_{\alpha_\mu}^x})\) is convex and therefore \((P_\nu-P_{\mu_{\alpha_\mu}^x})^c=(P_\nu-P_{\mu_{\alpha_\mu}^x})\). It follows that \(f^L(x)=x=g^L(x)\). On the other hand, for \(x\in(e^L,\beta_\mu)\), we have that \((P_\nu-P_{\mu_{\alpha_\mu}^x})^c(x)<(P_\nu-P_{\mu_{\alpha_\mu}^x})(x)\) (this follows from the fact that \((P_\nu-P_{\mu_{\alpha_\mu}^x})\) is strictly convex and increasing on \((\alpha_\mu,e^L]\), strictly concave on \((e^L,e^R\wedge x)\) and again strictly convex on \((e^R\wedge x,\beta_\mu)\); note that \((P_\nu-P_{\mu_x})\) is strictly positive and continuous on \((\alpha_\nu,\beta_\nu)\)). In this case \((f^L(x),g^L(x))=I_x\). One can show that \(f^L,g^L\) are indeed continuous and strictly monotone (since the densities \(\rho,\eta\) are strictly positive on the respective intervals). The case of the right-curtain martingale coupling can be proved using symmetric arguments.

Proof of Lemma 39↩︎

Since (by assumption) \(y<\beta_\mu\), we have that either \(x<(g^L)^{-1}(\beta_\mu)\) or \((g^L)^{-1}(\beta_\mu)\leq x<y<\beta_\mu =B(x)\). Further note that \(y>B(x)\) (resp., \(y=B(x)\)) is equivalent to \(g^L(x)< f^R(y)\) (resp., \(g^L(x)=f^R(y)\)).

We first prove Case (i). Note that by Corollary 38 we have that \(S^\nu(\mu_{\alpha_\mu}^x)=\mu_{\alpha_\mu}^{f^L(x)} + \nu_{f^L(x)}^{g^L(x)}\) and \(S^{\nu}(\mu_y^{\beta_\mu})= \nu_{f^R(y)}^{g^R(y)} + \mu_{g^R(y)}^{\beta_\nu}\). On the other hand, by the associativity of the shadow measure (recall Proposition 32), we have that \(S^\nu(\bar\mu^y_x)=S^\nu(\mu_{\alpha_\mu}^x+\mu_y^{\beta_\mu})=S^\nu(\mu_{\alpha_\mu}^x)+S^{\nu-S^\nu(\mu_{\alpha_\mu}^x)}(\mu_y^{\beta_\mu})\), and thus we are left to prove that \(S^{\nu-S^\nu(\mu_{\alpha_\mu}^x)}(\mu_y^{\beta_\mu})=S^{\nu}(\mu_y^{\beta_\mu})\). For this, observe that, since \(g^L(x)\leq f^R(y)\), the supports of the shadow measures \(S^\nu(\mu_{\alpha_\mu}^x)\) and \(S^\nu(\mu_y^{\beta_\mu})\) are disjoint, and therefore \(S^{\nu}(\mu_y^{\beta_\mu})\leq (\nu-S^\nu(\mu_{\alpha_\mu}^x))\). Then, by applying Lemma 33 with \(\xi=\mu_y^{\beta_\mu}\), \(\eta=(\nu-S^\nu(\mu_{\alpha_\mu}^x))\) and \(\chi=\nu\), we obtain \(S^{\nu-S^\nu(\mu_{\alpha_\mu}^x)}(\mu_y^{\beta_\mu})=S^{\nu}(\mu_y^{\beta_\mu})\), as claimed. Finally, \(\mu^y_x\leq_{cx}(\nu-\mu)^{f^L(x)}_{\alpha_\nu}+\nu_{g^L(x)}^{f^R(y)}+(\nu-\mu)_{g^R(y)}^{\beta_\nu}\) is a direct consequence of the associativity of the shadow measure (Proposition 32), since \(\mu^y_x\leq_{cx}(\nu-S^\nu(\bar\mu^y_x)\). This completes the proof in Case (i).

Note that, in Case (i) and with \(y=B(x)\) (i.e., \(g^L(x)=f^R(y)\)), we have that \(S^\nu(\bar{\mu}_x^y) = \mu_{\alpha_\mu}^{f^L(x)} + \nu_{f^L(x)}^{g^R(y)} + \mu_{g^R(y)}^{\beta_\nu}\). Then by taking \(w=f^L(x)\) and \(z=g^R(y)\), we immediately obtain the result in Case (ii) with \(y=B(x)\).

We now prove Case (ii) for \(y<B(x)\); note that in this case \(x\) can take values in \((\alpha_\mu,\beta_\mu)\). First, by the associativity of the shadow measure we have that \(S^\nu(\bar\mu^y_x)=S^\nu\left(\bar\mu_x^{B(x)}+\mu_y^{B(x)}\right)=S^\nu(\bar\mu_x^{B(x)})+S^{\nu-S^\nu(\bar\mu_x^{B(x)})}(\mu_y^{B(x)})\). Now, by the first part of the proof, \(S^\nu(\bar\mu_x^{B(x)})=\mu_{\alpha_\mu}^{f^L(x)}+\nu_{f^L(x)}^{g^R(B(x))}+\mu_{g^R(B(x))}^{\beta_\mu}\) and \(S^{\nu-S^\nu(\bar\mu_x^{B(x)})}(\mu_y^{B(x)})=S^{(\nu-\mu)_{\alpha_\nu}^{f^L(x)}+(\nu-\mu)_{g^R(B(x))}^{\beta_\nu}}(\mu_y^{B(x)})\). Note that \(f^L(x)\leq x<y<B(x)\leq g^R(B(x))\), and therefore \((\nu-\mu)_{\alpha_\nu}^{f^L(x)}+(\nu-\mu)_{g^R(B(x))}^{\beta_\nu}\) and \(\mu_y^{B(x)}\) have disjoint supports. Then by Lemma 34, \(S^{\nu-S^\nu(\bar\mu_x^{B(x)})}(\mu_y^{B(x)})=(\nu-\mu)_{w}^{f^L(x)}+(\nu-\mu)_{g^R(B(x))}^{z}\)for some \(w<f^L(x)\leq e^L<e^R\leq g^R(B(x))<z\) (uniqueness follows from the fact that, due to the Dispersion Assumption, \((\nu-\mu)_{\alpha_\nu}^{f^L(x)}+(\nu-\mu)_{g^R(B(x))}^{\beta_\nu}\) admits density that is strictly positive on \((\alpha_\nu,f^L(x))\cup(g^R(y),\beta_\nu)\)). It follows that \(S^\nu(\bar\mu^y_x)=\mu_{\alpha_\mu}^w + \nu_w^z + \mu_{z}^{\beta_\nu}\) and (again by the associativity of the shadow measure) \(\mu^y_x\leq_{cx}(\nu-\mu)^{w}_{\alpha_\nu}+(\nu-\mu)_{z}^{\beta_\nu}\), which finishes the proof.

Proof of Lemma 41↩︎

We only prove the claim regarding \(C^L\) (the case for \(C^R\) uses symmetric arguments). By definition, for each \(x\in(e^L,\beta_\mu)\), \(C^L( g^L(x)) = a(x) + \frac{g^L(x) - x}{x-f^L(x)} (a(x) - b(f^L(x)))\), and the continuity of \(C^L\) on \((e^L,\beta_\nu)\) follows from the continuity of \(f^L,g^L,a,b\) on \((e^L,\beta_\mu)\).

Since \(a\geq b\) on \((\alpha_\mu,\beta_\mu)\) and \(b\) is convex (and finite on \((\alpha_\nu,\beta_\nu)\supseteq(\alpha_\mu,\beta_\mu)\)), we have that for all \(x\in(e^L,\beta_\mu)\), \[C^L(g^L(x))\geq a(x)+(g^L(x)-x)\frac{b(x)-b(f^L(x))}{x-f^L(x)}\geq a(x)+(g^L(x)-x)b'(f^L(x));\] recall that \(b'(x)\) denotes the right derivative of \(b\) at \(x\). Choose \(x_1 \in (e^L,\beta_\mu)\); then for \(x<x_1\), \(f^L(x)>f^L(x_1)\) and \(b'(f^L(x)) \geq b'(f^L(x_1))\). Furthermore, \(b\) is finite on \((\alpha_\nu,\beta_\nu)\), and therefore \(b'(x)\in\mathbb{R}\) for all \(x\in(\alpha_\nu,\beta_\nu)\). Then, since \(e^L\in (\alpha_\nu,\beta_\nu)\), \[\liminf_{z\downarrow e^L}C^L(z)=\liminf_{x\downarrow e^L}C^L(g^L(x))\geq [\liminf_{x\downarrow e^L} a(x)]+b'(f^L(x_1))[\liminf_{x\downarrow e^L}(g^L(x)-x)]=a(e^L),\] which finishes the proof.

Proof of Lemma 42↩︎

We only prove the first statement. Suppose that \(a(e^L) > b(e^L)\). Then, since \(C^L( g^L(x)) = a(x) + \frac{g^L(x) - x}{x-f^L(x)} (a(x) - b(f^L(x)))\), by the continuity of \(f^L\) at \(e^L\), there exists \(\epsilon>0\) such that for \(e^L<x<e^L+\epsilon\), \(a(x) > b(f^L(x))\). Then \(C^L( g^L(x)) > a(x)\) on \((e^L,e^L+\epsilon)\) and \(\hat{z}^L>e^L+\epsilon\). It follows that if \(\hat{z}^L=e^L\) then \(a(e^L)=b(e^L)\).

Now suppose that \(a(e^L) = b(e^L)\). If \(c\) is a convex function and if \(w>v>u\) then \(c(w) \geq c(v) + (w - v) \frac{c(v)-c(u)}{v-u}\). Thus, defining \(c\) by \(c(w) = b(w)I_{ \{ w<e^L \}} + a(w) I_{ \{ w \geq e^L \} }\) and fixing \(x\in(e^L,(g^L)^{-1}(e^R)]\) (so that \(g^L(x)\in(e^L,e^R]\)), we have that \(a(g^L(x)) {\geq} a(x) + (g^L(x)-x) \frac{a(x)-b(f^L(x))}{x - f^L(x)} = C^L(g^L(x))\). Hence \(\hat{z}^L \leq g^L(x)\). Since \(x\in(e^L,(g^L)^{-1}(e^R)]\) is arbitrary and \(\lim_{x \downarrow e^L}g^L(x) = e^L\) we have that \(\hat{z}^L=e^L\).

The final statement then follows from the fact that \(f^L\) and \(g^L\) are continuous and \(g^L(e^L)=e^L\).

Proof of Lemma 45↩︎

It follows from Lemma 39(ii) that, for each \((x,y) \in \mathcal{A}^<\), \(S^\nu(\bar{\mu}_x^y) = \mu_{\alpha_\mu}^{w} + \nu_w^z + \mu_z^{\beta_\mu}\) for some (unique) \(w \leq e^L\wedge x < e^R\vee y \leq z\). We set \(w=w(x,y)\) and \(z=z(x,y)\).

Next we prove the continuity, monotonicity and limit properties of \(w\) and \(z\) in \(x\) and \(y\). We fix \(x\) and consider varying \(y \in \mathcal{A}^<_x\). The argument for fixed \(y\) and varying \(x\in\mathcal{A}^<_y\) is similar.

Suppose \(x <y< y' \leq B(x)\wedge\beta_\mu\). Then by the associativity property of the shadow measure (Proposition 32) \[S^\nu( \bar{\mu}_x^y) = S^\nu( \bar{\mu}_x^{y'}) + S^{\nu - S^\nu(\bar{\mu}_x^{y'})}(\mu_y^{y'}).\] Moreover, by Lemma 39(ii), \[S^\nu( \bar{\mu}_x^y) = \mu_{\alpha_\mu}^{w(x,y)} + \nu_{w(x,y)}^{z(x,y)} + \mu_{z(x,y)}^{\beta_\mu} \quad\textrm{and} \quad S^\nu( \bar{\mu}_x^{y'}) = \mu_{\alpha_\mu}^{w(x,y')} + \nu_{w(x,y')}^{z(x,y')} + \mu_{z(x,y')}^{\beta_\mu};\] where both \(w(x,y) \leq e^L\wedge x<e^R\vee y \leq z(x,y)\) and \(w(x,y') \leq e^L\wedge x<e^R\vee y' \leq z(x,y')\). It follows that \(\nu - S^\nu(\bar{\mu}_x^{y'})=(\nu-\mu)_{\alpha_\nu}^{w(x,y')}+(\nu-\mu)_{z(x,y')}^{\beta_\nu}\) and \(\mu^{y'}_y\) have disjoint supports, and thus, by Lemma 34, \(S^{\nu - S^\nu(\bar{\mu}_x^{y'})}(\mu_y^{y'})=(\nu-\mu)^{w(x,y')}_{\hat{w}}+(\nu-\mu)^{\hat{z}}_{z(x,y')}\) for some \(\hat{w} < w(x,y')\) and \(\hat{z}>z(x,y')\). Hence \[\mu_{\alpha_\mu}^{w(x,y)} + \nu_{w(x,y)}^{z(x,y)} + \mu_{z(x,y)}^{\beta_\mu}=S^\nu( \bar{\mu}_x^y)=S^\nu( \bar{\mu}_x^{y'}) + S^{\nu - S^\nu(\bar{\mu}_x^{y'})}(\mu_y^{y'})=\mu_{\alpha_\mu}^{\hat{w}} + \nu_{\hat{w}}^{\hat{z}} + \mu_{\hat{z}}^{\beta_\mu},\] and therefore \(w(x,y)=\hat{w} < w(x,y')\) and \(z(x,y)=\hat{z}>z(x,y')\), as required.

For the right-continuity, fix \(y\) and consider the limit \(y' \downarrow y\). We know that \[S^{\nu - S^\nu(\bar{\mu}_x^y)}(\mu_y^{y'}) = (\nu-\mu)_{w(x,y)}^{w(x,y')} + (\nu-\mu)_{z(x,y')}^{z(x,y)}.\] As \(y' \downarrow y\) we see that \(S^{\nu - S^\nu(\bar{\mu}_x^y)}(\mu_y^{y'})(\mathbb{R})=\mu_y^{y'}(\mathbb{R}) \downarrow 0\) and hence \((\nu-\mu)_{w(x,y)}^{w(x,y')}(\mathbb{R}) \downarrow 0\) and \((\nu-\mu)_{z(x,y')}^{z(x,y)}(\mathbb{R}) \downarrow 0\). Since, by the Dispersion Assumption (recall Definition 36), \((\nu-\mu)\) admits a strictly positive density on \((\alpha_\nu,e^L)\cup(e^R,\beta_\nu)\), we find that \(w(x,y') \downarrow w(x,y)\) and \(z(x,y') \uparrow z(x,y)\). For the left-continuity we can let \(y \uparrow y'\) and apply a similar argument. Hence \(w(x, \cdot)\) and \(z(x,\cdot)\) are continuous.

Finally, still fixing \(x\), note that if \(y \downarrow x\) then \((\nu-S^\nu(\bar\mu_x^y))(\mathbb{R})={\mu}_x^y(\mathbb{R})\downarrow0\). But \((\nu-S^\nu(\bar\mu_x^y))=(\nu-\mu)_{\alpha_\nu}^{w(x,y)} + (\nu-\mu)_{z(x,y)}^{\beta_\nu}\), and therefore \((\nu-\mu)_{\alpha_\nu}^{w(x,y)}(\mathbb{R})\downarrow0\) and \((\nu-\mu)_{z(x,y)}^{\beta_\nu}(\mathbb{R})\downarrow 0\). Then, again using that \((\nu-\mu)\) admits a strictly positive density on \((\alpha_\nu,e^L)\cup(e^R,\beta_\nu)\), we get that \(w(x,y) \downarrow \beta_\nu\) and \(z(x,y) \uparrow \beta_\nu\).

Proof of Proposition 48↩︎

Fix \((x,y')\in\mathcal{A}^<\) such that \(\Lambda(x,y')=0\), and note that \(T_{a,a}^{x,y'}\) is well-defined and \(T_{a,a}^{x,y'}(z(x,y'))\leq b(z(x,y'))\). Furthermore, the convexity of \(a\) ensures that \(a\leq T_{a,a}^{x,y'}\) on \([x,y']\) and \(a\geq T_{a,a}^{x,y'}\) on \(\mathbb{R}\setminus [x,y']\). Similarly, by the convexity of \(a\) and \(b\), and the fact that \(a\geq b\), we have that \(b\geq T^{a,a}_{x,y'}\) on \((z(x,y'),\infty)\); also, if \(\Lambda(x,y')=0\), then \(b\leq T^{a,a}_{x,y'}\) on \([x,z(x,y')]\).

We first prove the monotonicity of \(y\to\Lambda(x,y)\) on \((x,y')\). Since \(b\) is convex, \(a\geq b\), and \(T_{a,a}^{x,y'}(z(x,y'))\leq b(z(x,y'))\), we have that \(b\geq T_{a,a}^{x,y'}\) and (more importantly) \((T_{a,a}^{x,y'}-b)\) is non-increasing on \((z(x,y'),\infty)\). Then since \(z(x,\cdot)\) non-increasing on \((x,y')\) (recall Lemma 45) it follows that \(y\mapsto\Lambda(x,y)=T_{a,a}^{x,y'}(z(x,y))-b(z(x,y))\) is non-decreasing on \((x,y')\). (Here, in the case \(y'=\beta_\mu\), we use \(T_{a,a}^{x,y'}(z)=a(\beta_\mu-)+\frac{z-\beta_\mu}{\beta_\mu-x}(a(\beta_\mu-)-a(x))\) and \(\hat{\Lambda}(x,\beta_\mu,z)=T_{a,a}^{x,y'}(z)-b(z)\) as in 47 .)

We now suppose that \(\Lambda(x,y')=0\). We will only prove that \(\Lambda(x,y)>0\) for \(y \in (y',B(x)]\); that \(\Lambda(x,y)<0\) for \(y<y'\) can be proved in an identical fashion.

Assume that \(y'<B(x)\), else there is nothing to prove. Fix \(y\in(y',B(x)]\) and note that \(x<z(x,y)<z(x,y')\). Then \(a(y)\geq T_{a,a}^{x,y'}(y)\) (from which it follows that \(T_{a,a}^{x,y}\geq T_{a,a}^{x,y'}\) on \([x,\infty)\)) and \(b(z(x,y))\leq T_{a,a}^{x,y'}(z(x,y))\). It follows that \(\Lambda(x,y)=T_{a,a}^{x,y}(z(x,y))-b(z(x,y))\geq 0\). It is left to prove that \(\Lambda(x,y)=0\) cannot happen.

Suppose that \(\Lambda(x,y)=0\). Then we must have that \(T_{a,a}^{x,y}(z(x,y))=T_{a,a}^{x,y'}(z(x,y))=b(z(x,y))\), and therefore \(T_{a,a}^{x,y}=T^{a,a}_{x,y'}\). However, due to the convexity of \(a\) (resp., convexity of \(b\) and the fact that \(a\geq b\)) this is only possible if \(a=T_{a,a}^{x,y'}\) on \([x,y]\) (resp., \(b=T_{a,a}^{x,y'}\) on \([x,z(x,y')]\)). It follows that \(a=T_{a,a}^{x,z'}=b\) on \((x,y)\cap(x,z(x,y'))\neq\emptyset\), which contradicts Standing Assumption 2. We conclude that \(\Lambda(x,y)>0\).

Proof of Proposition 49↩︎

For \(\bar{x} \leq x \leq (g^L)^{-1}(\beta_\mu)\) we have \(z(x,B(x))=B(x)\) and \(\Lambda(x,B(x)) = \hat{\Lambda}(x,B(x),B(x))= a(B(x)) - b(B(x)) \geq 0\). In order to complete the proof it is sufficient to consider \(x \in [x_0, \bar{x})\).

Observe that, for \(x\in[x_0,\bar x)\) we have that \(x<B(x)<e^R<z(x,B(x))\), and therefore both \(T_{a,a}^{x,B(x)}\) and \(T_{a,b}^{B(x),z(x,B(x))}\) are well-defined. Then \(\Lambda(x,B(x))=0\) if and only if \(T_{a,a}^{x,B(x)}=T_{a,b}^{B(x),z(x,B(x))}\) (or equivalently, \(\tilde{S}_{a,a}^{x,B(x)}=\tilde{S}_{a,b}^{B(x),z(x,B(x))}\)), and \(\Lambda(x,B(x))>0\) if and only if \(T_{a,a}^{x,B(x)}>T_{a,b}^{B(x),z(x,B(x))}\) on \([B(x),\infty)\) (or equivalently, \(\tilde{S}_{a,a}^{x,B(x)}>\tilde{S}_{a,b}^{B(x),z(x,B(x))}\)).

By construction, since we are in either Case \((a)\) or Case \((b)\), we have that \(\tilde{S}_{a,a}^{x_0,y_0}\geq\tilde{S}_{a,b}^{y_0,z(x_0,y_0)}\), and thus \(\Lambda(x_0,y_0=B(x_0))\geq0\). We are left to prove that \(\Lambda(x,B(x)) \geq 0\) for \(x\in(x_0,\bar x)\).

By the definition of \(B\), we have that \(w(x,B(x))=f^L(x)\) and \(z(x,B(x))=g^R(B(x))\), for all \(x\in(\alpha_\mu,(g^L)^{-1}(\beta_\mu))\). In particular, \(w(\cdot,B(\cdot))\) is increasing (resp., decreasing) on \((\alpha_\mu,e^L)\) (resp., \((e^L,\beta_\mu)\)), while \(z(\cdot,B(\cdot))\) is decreasing (resp., increasing) on \((\alpha_\mu,B^{-1}(e^R))\) (resp., \((B^{-1}(e^R),\beta_\mu)\))

We show that \(\tilde{S}_{a,a}^{x,B(x)} \geq \tilde{S}_{a,b}^{B(x),z(x,B(x))=g^R(B(x))}\) for \(x\in(x_0,\bar x)\). It then follows that \(\Lambda(x,B(x))\geq 0\) for \(x\in(x_0,\bar x)\). Since \(B(\cdot)\) is non-decreasing, by the convexity of \(a(\cdot)\) we have that \(\tilde{S}_{a,a}^{x_0,y_0=B(x_0)}\leq\tilde{S}_{a,a}^{x,B(x)}\) for all \(x\in[x_0,(g^L)^{-1}(\beta_\mu))\). On the other hand, on \([x_0,\bar x)\), \(x\mapsto z(x,B(x))=g^R(B(x))\) is non-increasing, and therefore, using that \(a\geq b\) and the convexity of \(a(\cdot)\) and \(b(\cdot)\), we have that, \(\tilde{S}_{a,b}^{y_0=B(x_0),z(x_0,y_0)=g^R(B(x_0))}\geq \tilde{S}_{a,b}^{B(x),z(x,B(x))=g^R(B(x))}\) on \([x_0,\bar x)\). It follows that, for all \(x\in[x_0,\bar x)\), \(\tilde{S}_{a,b}^{B(x),z(x,B(x))=g^R(B(x))}\leq \tilde{S}_{a,b}^{y_0=B(x_0),z(x_0,y_0)=g^R(B(x_0))}\leq\tilde{S}_{a,a}^{x_0,y_0=B(x_0)}\leq\tilde{S}_{a,a}^{x,B(x)}\), and thus \(\Lambda(x , B(x))\geq 0\).

Proof of Corollary 50↩︎

We have that \(\Lambda(x,B(x))\geq 0\) for all \(x \in \mathcal{A}^\Lambda_{>}\cup\mathcal{A}^\Lambda_{=}=([x_1, \tilde{x})\setminus\mathcal{A}^\Lambda_{<})\). Then the existence of \(y^*(x)\) follows from the definition of \(\tilde{x}\) and the intermediate value theorem, while the uniqueness follows from Proposition 48. We are left to prove the continuity.

We first claim that \(\Lambda(x,y)<0\) for all \(x\in[x_1,\tilde{x})\) and \(y\in(x,y^*(x))\). Indeed, \(\Lambda(x,y^*(x))\leq0\) for all \(x\in[x_1,\tilde{x})\), and thus the claim follows by Proposition [prop:Lambday].

Fix \(x\in[x_1,\tilde{x})\) and \(y\in(x,y^*(x))\). Then \(\Lambda(x,y)<0\). By the continuity of \(\Lambda(\cdot,y)\), there exists a small enough \(\epsilon\in(0,y\wedge\tilde{x}-x)\) such that \(\Lambda(x',y)<0\) for all \(x'\in[x,x+\epsilon)\). Recall that \(B\) is non-decreasing. If \(x'\in\mathcal{A}^\Lambda_>\cup\mathcal{A}^\Lambda_=\), then \(y<y^*(x')\leq B(x')\). If \(x'\in\mathcal{A}^\Lambda_<\), then \(y<y^*(x)\leq B(x)\leq\beta_\mu=y^*(x')\). In either case we have that \(y<y^*(x')\) for all \([x,x+\epsilon)\), and therefore \(\liminf_{x'\downarrow x}y^*(x')\geq y\). Since \(y\in(x,y^*(x))\) was arbitrary, we conclude that \(\liminf_{x'\downarrow x}y^*(x')\geq y^*(x)\).

We now show that \(\limsup_{x'\downarrow x}y^*(x')\leq y^*(x)\). Fix \(x\in\mathcal{A}^\Lambda_>\) so that \(y^*(x) < B(x)\). Choose \(y\in(y^*(x),B(x))\) so that \(\Lambda(x,y)>0\). Then by the continuity of \(\Lambda(\cdot,y)\), there exists a small enough \(\epsilon\in(0,(y\wedge\tilde{x}-x))\) such that \(\Lambda(x',y)>0\) for all \(x'\in[x,x+\epsilon)\). Then \(y^*(x')<y\) for all \(x'\in[x,x+\epsilon)\) and thus \(\limsup_{x' \downarrow x} y^*(x') \leq y\). But \(y\in (y^*(x),B(x))\) was arbitrary, and thus \(\limsup_{x' \downarrow x} y^*(x') \leq y^*(x)\). Now suppose that \(x\in\mathcal{A}^\Lambda_=\cup\mathcal{A}^\Lambda_<\), so that \(y^*(x)=B(x)\). Then, for \(x'\in(x,\tilde{x})\), we have \(y^*(x') \leq B(x')\), and since \(B\) is continuous, \(\limsup_{x' \downarrow x} y^*(x') \leq \limsup_{x' \downarrow x} B(x') = B(x) = y^*(x)\).

We conclude that \(\liminf_{x' \downarrow x} y^*(x')=\limsup_{x' \downarrow x} y^*(x')=y^*(x)\), which proves the right-continuity on \([x_1,\tilde{x})\). The argument for the left-continuity on \((x_1,\tilde{x})\) is similar.

Proof of Lemma 51↩︎

Since \(x< y^*(x)\leq B(x)\) for all \(x\in[x_1,\tilde{x})\), the claim immediately follows if \(\tilde{x}=\beta_\mu\). In the rest of the proof we assume that \(\tilde{x}<\beta_\mu\).

By Corollary 50, \(x<y^*(x)\) for \(x\in[x_1,\tilde{x})\), and therefore \(\limsup_{x \uparrow \tilde{x}}y^*(x)\geq\liminf_{x \uparrow \tilde{x}}y^*(x)\geq\tilde{x}\). To prove that we have equalities throughout, we argue by contradiction. Suppose \(y=\limsup_{x \uparrow \tilde{x}}y^*(x)>\tilde{x}\). Note that, since \(\tilde{x}<\beta_\mu\) and \(y^*\leq B\) on \([x_1,\tilde{x})\), we have that \(y\leq\lim_{x\uparrow\tilde{x}}B(x)=B(\tilde{x})<\infty\).

Let \((x_n)_{n\geq 1} \uparrow \tilde{x}\) be such that \((y^*(x_n))_{n\geq 1} \to y\). Then \[0 \geq \Lambda(x_n,y^*(x_n))=T_{a,a}^{x_n, y^*(x_n)}(z(x_n,y^*(x_n))) - b(z(x_n,y^*(x_n)));\] note that strict inequality is possible in the case \(x_n\in\mathcal{A}^\Lambda_<\), since then \(y^*(x_n)=B(x_n)=\beta_\mu\) and, by definition, \(\Lambda(x,B(x))<0\) for all \(x\in\mathcal{A}^\Lambda_<\). Also, \(x_n<y^*(x_n)\) for each \(n\geq 1\) (resp., \(\tilde{x}<y\)), and thus \(T_{a,a}^{x_n,y^*(x_n)}\) (resp., \(T_{a,a}^{\tilde{x},y}\)) is well-defined. Taking limits, and using that \(z(x_n,y^*(x_n)) \rightarrow z(\tilde{x},y)\) by the joint continuity of \(z(\cdot,\cdot)\), we get \[0 \geq T_{a,a}^{\tilde{x}, y}(z(\tilde{x},y)) - b(z(\tilde{x},y)) = \Lambda(\tilde{x},y).\] Since \(y>\tilde{x}\), using Proposition 48 we have that \(\Lambda(\tilde{x},y') <0\) for all \(y' \in (\tilde{x},y)\). By the continuity of \(\Lambda(\cdot,y')\), there exists (small enough) \(\epsilon>0\) such that \(\Lambda(x,y') <0\) for all \(x \in [\tilde{x},\tilde{x}+\epsilon)\subset[\tilde{x},\beta_\mu)\). Then (again by Proposition 48) \(\Lambda(x,x+) <0\) for all \(x \in [\tilde{x},\tilde{x}+\epsilon)\), contradicting the definition of \(\tilde{x}\).

References↩︎

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1. Department of Statistics, University of Warwick d.hobson@warwick.ac.uk↩︎

2. Department of Mathematics, NC State University dnorgil@ncsu.edu↩︎

3. Two integrable (Borel) measures \(\eta,\chi\) on \(\mathbb{R}\), with \(\eta(\mathbb{R})=\chi(\mathbb{R})<\infty\), are in convex order (\(\eta\leq_{cx}\chi\)) if \(\int fd\eta\leq\int f d\chi\) for all convex \(f:\mathbb{R}\to\mathbb{R}\).↩︎

4. As in the discussion after the initial introduction of Cases (C1), (C2) and (C3), it must be the case that \(\Psi\) is not convex, but we do not show this.↩︎