The Early Exercise Premium Representation for American Options on Multiply Assets

Abstract

In the paper we consider the problem of valuation of American options written on dividend-paying assets whose price dynamics follow the classical multidimensional Black and Scholes model. We provide a general early exercise premium representation formula for options with payoff functions which are convex or satisfy mild regularity assumptions. Examples include index options, spread options, call on max options, put on min options, multiply strike options and power-product options. In the proof of the formula we exploit close connections between the optimal stopping problems associated with valuation of American options, obstacle problems and reflected backward stochastic differential equations.

1 Introduction

In the paper we study American options written on dividend-paying assets. We assume that the underlying assets dynamics follow the classical multidimensional Black and Scholes model. It is now well known that the arbitrage-free value of American options can be expressed in terms of the optimal stopping problem (Bensoussan [4], Karatzas [20]; see also [21] for nice exposition and additional references), in terms of variational inequalities (Jaillet et al. [19]) and in terms of solutions of reflected BSDEs (El Karoui and Quenez [14]). Although these approaches provide complete characterization of the option value (see Sect. 2 for a short review), the paper by Broadie and Detemple [7] shows that it is of interest to provide alternative representation, which expresses the value of an American option as the value of the corresponding European option plus the gain from early exercise. The main reason is that the representation, called the early exercise premium formula, gives useful information on the determinants of the option value. The formula was proved first by Kim [22] in the case of standard American put option on a single asset. Another important contributions in the case of single asset include Broadie and Detemple [6], El Karoui and Karatzas [13] and Jacka [18] (see also [10, 21] and the references therein). The case of options on multiply dividend-paying assets is more difficult and has received rather little attention in the literature. In the important paper [7] and next in Detemple et al. [11] (see also [10]) the early exercise premium formula was established for concrete classes of options on multiply assets. Note that in the last paper call on min option, i.e. option with nonconvex payoff function is investigated. A subclass of call on min options consisting of capped options is studied in [6–8] (see also [10]).

In the present paper we provide a unified way of treating a wide variety of seemingly disparate examples. It allows us to prove a general exercise premium formula for options with convex payoff functions satisfying the polynomial growth condition or payoff function satisfying quite general condition considered in Laurence and Salsa [26]. Verifying the last condition requires knowledge of the payoff function and the structure of the exercise set. Therefore it is a complicated task in general. Fortunately, in most interesting cases one can easily check convexity of the payoff function or check some simpler condition implying the general condition from [26]. The class of options covered by our formula includes index options, spread options, call on max options, put on min options, multiply strike options, power-product options and others.

In the proof of the exercise premium formula we rely on some results on reflected BSDEs and their links with optimal stopping problems (see [14]) and with parabolic variational inequalities established in Bally et al. [2]. We also use classical results on regularity of the solution of the Cauchy problem for parabolic operator with constant coefficients, and in case of convex payoffs, some fine properties of convex functions. Perhaps it is worth mentioning that we do not use any regularity results on the free boundary problem for an American option. The basic idea of the proof comes from our earlier paper [25] devoted to standard American call and put options on single asset.

2 Preliminaries

We will assume that under the risk-neutral probability measure the underlying assets prices $X^{s,x,1},\dots ,X^{s,x,n}$ evolve on the time interval $[s,T]$ according to stochastic differential equation of the form

$$\begin{aligned} X^{s,x,i}_{t}=x_i+\int _{s}^{t}(r-d_{i})X^{s,x,i}_{\theta }\,d\theta +\sum _{j=1}^n\int _{s}^{t}\sigma _{ij}X^{s,x,i}_{\theta }\,dW^j_{\theta }, \quad t\in [s,T]. \end{aligned}$$

(1)

Here $W=(W^1,\dots ,W^n)$ is a standard $n$-dimensional Wiener process, $r\ge 0$ is the rate of interest, $d_i\ge 0$ is the dividend rate of the asset $i$ and $\sigma =\{\sigma _{ij}\}$ is the $n$-dimensional volatility matrix. We assume that $a=\sigma \cdot \sigma ^*$ is positive definite. Since the distributions of the processes $X^{s,x,i}$ depend $\sigma$ only through $a$, we may and will assume that $\sigma$ is a symmetric square root of $a$. As for the payoff function $\psi$ we will assume that it satisfies the assumptions:

• (A1) $\psi$ is a nonnegative continuous function on ${\mathbb R}^n$ with polynomial growth,

• (A2) For every $t\in (0,T)$, $\psi$ is a smooth function on $\{\psi =u\}\cap \bar{Q}_{t}$, i.e. there exists an open set $U\subset {\mathbb R}^n$ such that $\{u=\psi \}\cap \bar{Q}_{t}\subset [0,t]\times U$ and $\psi$ is smooth on $U$ (Here $Q_t\equiv [0,t)\times {\mathbb R}^n$, $\bar{Q}_t\equiv [0,t]\times {\mathbb R}^n$ and $u$ is the value of an option with payoff $\psi$; see (5) and (9) below)

or

• (A3) $\psi$ is a nonnegative convex function on ${\mathbb R}^n$ with polynomial growth.

Note that convex functions are locally Lipschitz, so assumption (A3) implies (A1). Assumption (A2) is considered in [26]. It is satisfied for instance if

• (A2$'$) The region where $\psi$ is strictly positive is the union of several connected components in which $\psi$ is smooth.

Following [26] let us also note that unlike (A2$'$) or (A3), condition (A2) cannot be verified by appealing to the structure of the payoff alone. Verifying (A2) requires additional knowledge of the structure of the exercise set $\{u=\psi \}$.

Let $\Omega =C([0,T];{\mathbb R}^n)$ and let $X$ be the canonical process on $\Omega$. For $(s,x)\in Q_{T}$ let $P_{s,x}$ denote the law of the process $X^{s,x}=(X^{s,x,1},\dots ,X^{s,x,n})$ defined by (1) and let $\{{\mathcal {F}}^s_t\}$ denote the completion of $\sigma (X_{\theta };\theta \in [s,t])$ with respect to the family $\{P_{s,\mu };\mu$ a finite measure on ${\mathcal {B}}({\mathbb R}^n)\}$, where $P_{s,\mu }(\cdot )=\int _{{\mathbb R}^n}P_{s,x}(\cdot )\,\mu (dx)$. Then for each $s\in [0,T)$, ${\mathbb X}=(\Omega ,({\mathcal {F}}^s_t)_{t\in [s,T]},X,P_{s,x})$ is a Markov process on $[0,T]$.

Let $I=\{0,1\}^{n}$. For $\iota =(i_1,\dots ,i_n)\in I$ we set $D_{\iota }=\{x\in {\mathbb R}^n; (-1)^{i_{k}}x_{k}>0,k=1,\dots ,n\}$, $P=\bigcup _{\iota \in I} D_{\iota }$, $P_{T}=[0,T)\times P$. By Itô’s formula,

$$\begin{aligned} X^{s,x,i}_t=x^i\exp \left( (r-d_i-a_{ii})(t-s) +\sum ^n_{j=1}\sigma _{ij}(W^j_t-W^j_s)\right) ,\quad t\in [s,T]. \end{aligned}$$

(2)

Therefore if $s\in [0,T)$ and $x\in D_{\iota }$ for some $\iota \in I$ then $P_{s,x}(X_{t}\in D_{\iota },\, t\ge s)=1$. From this and the fact that $a$ is positive definite it follows that if $x\in P_T$ then $\det \sigma (X_t)>0$, $P_{s,x}$-a.s. for every $t\ge s$, where $\sigma (x)=\{\sigma _{ij}x_i\}_{i,j=1,\dots , n}$. Moreover, $[s,T]\ni t\mapsto \sigma ^{-1}(X_t)$ is a continuous process. Therefore, if $x\in P_T$ then by Lévy’s theorem the process $B_{s,\cdot }$ defined as $B_{s,t}=\int ^t_s\sigma ^{-1}(X_{\theta })\,dM_{\theta }$, where $M^i_t=X^i_t-X^i_0-\int ^t_0(r-d_i)X^i_{\theta }\,d\theta$, $t\in [s,T]$, is under $P_{s,x}$ a standard $n$-dimensional $\{{\mathcal {F}}^s_t\}$-Wiener process on $[s,T]$ and

$$\begin{aligned} X^i_t-x^i=\int _{s}^{t}(r-d_{i})X^{i}_{\theta }\,d\theta +\sum _{j=1}^n\int _{s}^{t}\sigma _{ij}X^{i}_{\theta }\,dB^j_{s,\theta }, \quad t\in [s,T],\quad P_{s,x}\text{-a.s. }, \end{aligned}$$

(3)

i.e.

$$\begin{aligned} X^{i}_t=x^i\exp \left( (r-d_i-a_{ii})(t-s) +\sum ^n_{j=1}\sigma _{ij}B_{s,t}\right) ,\quad t\in [s,T],\quad P_{s,x}\text{-a.s. } \end{aligned}$$

(4)

The above forms of the assets price dynamics will be more convenient for us than (1) or (2). Note that from the definition of the process $B_{s,\cdot }$ and (4) it follows that $\sigma (X_{\theta };\theta \in [s,t])=\sigma (B_{s,\theta };\theta \in [s,t])$ for $s\in [0,T)$, so for every $s\in [0,T)$ the filtration $\{{\mathcal {F}}^s_t\}$ is the completion of the Brownian filtration.

In Bensoussan [4] and Karatzas [20] (see also Sect. 2.5 in [21]) it is shown that under (A1) the arbitrage-free value $V$ of an American option with payoff function $\psi$ and expiration time $T$ is given by the solution of the stopping problem

$$\begin{aligned} V(s,x)=\sup _{\tau \in \mathcal {T}_{s}} E_{s,x}\big (e^{-r(\tau -s)}\psi (X_{\tau })\big ), \end{aligned}$$

(5)

where the supremum is taken over the set $\mathcal {T}_{s}$ of all $\{{\mathcal {F}}^s_t\}$-stopping times $\tau$ with values in $[s,T]$.

From the results proved in [12] it follows that under (A1) for every $(s,x)$ there exists a unique solution $(Y^{s,x},Z^{s,x},K^{s,x})$, on the space $(\Omega ,{\mathcal {F}}^s_T,P_{s,x})$, to the reflected BSDE with terminal condition $\psi (X_T)$, coefficient $f:{\mathbb R}\rightarrow {\mathbb R}$ defined as $f(y)=-ry$, $y\in {\mathbb R}$, and barrier $\psi (X)$ (RBSDE${}_{s,x}(\psi ,-r y,\psi )$ for short). This means that the processes $Y^{s,x},Z^{s,x},K^{s,x}$ are $\{{\mathcal {F}}^s_t\}$-progressively measurable, satisfy some integrability conditions and $P_{s,x}$-a.s.,

$$\begin{aligned} \left\{ \begin{array}{l} Y^{s,x}_t=\psi (X_T)-\int ^T_tr Y^{s,x}_{\theta }\,d\theta +K^{s,x}_T-K^{s,x}_t -\int ^T_tZ^{s,x}_{\theta }\,dB_{s,\theta },\quad t\in [s,T],\\ Y^{s,x}_t\ge \psi (X_t),\quad t\in [s,T], \\ K^{s,x} \text{ is } \text{ increasing, } \text{ continuous, } K^{s,x}_s=0,\,\, \int ^T_s(Y^{s,x}_t-\psi (X_t))\,dK^{s,x}_t=0. \end{array} \right. \end{aligned}$$

(6)

In [12] it is also proved that for every $(s,x)\in Q_T$,

$$\begin{aligned} Y^{s,x}_t=u(t,X_t),\quad t\in [s,T],\quad P_{s,x}\text{-a.s. }, \end{aligned}$$

(7)

where $u$ is a viscosity solution to the obstacle problem

$$\begin{aligned} \left\{ \begin{array}{ll} \min (u(s,x)-\psi (x),-u_s-L_{BS}u(s,x)+r u(s,x))=0, &{} (s,x)\in Q_T,\\ u(T,x)=\psi (x), &{} x\in {\mathbb R}^n \end{array} \right. \end{aligned}$$

(8)

with

$$\begin{aligned} L_{BS}u=\sum ^n_{i=1}(r-d_{i})x_{i}u_{x_{i}} +\frac{1}{2}\sum ^n_{i,j=1} a_{ij}x_ix_ju_{x_{i}x_{j}}. \end{aligned}$$

From [12, 14] we know that $V$ defined by (5) is equal to $Y^{s,x}_s$. Hence

$$\begin{aligned} V(s,x)=Y^{s,x}_s=u(s,x),\quad (s,x)\in [0,T]\times {\mathbb R}^n. \end{aligned}$$

(9)

In the next section we analyze $V$ via (9) but as a matter of fact instead of viscosity solutions of (8) we consider variational solutions which provide more information on the value function $V$.

3 Obstacle Problem for the Black and Scholes Equation

Assume that $\psi :{\mathbb R}^n\rightarrow {\mathbb R}_+$ is continuous and satisfies the polynomial growth condition. Let $L^2_{\varrho }=L^2({\mathbb R}^n;\varrho ^2\,dx)$, $H^{1}_{\varrho }=\{u\in L^2_{\varrho }:\sum ^n_{j=1} \sigma _{ij}x_iu_{x_j}\in L^{2}({\mathbb R}^n;\varrho ^2\,dx),\,i=1,\dots ,n\}$ and $W_{\varrho }=\{u\in L^2(0,T;H^1_{\varrho }):u_t\in L^2(0,T;H^{-1}_{\varrho })\}$, where $u_t,u_{x_i}$ denote the partial derivatives in the distribution sense, $\varrho (x)=(1+|x|^2)^{-\gamma }$ and $\gamma >0$ is chosen so that $\int _{{\mathbb R}^n}\varrho ^2(x)\,dx<\infty$ and $\int _{{\mathbb R}^n}\psi ^{2}(x)\varrho ^2(x)\,dx<\infty$. Following [2, 25] we adopt the following definition.

Definition

(a) A pair $(u,\mu )$ consisting of $u\in W_{\varrho }\cap C(\bar{Q}_T)$ and a Radon measure $\mu$ on $Q_T$ is a variational solution to (8) if

$$\begin{aligned} u(T,\cdot )=\psi ,\quad u\ge \psi ,\quad \int _{Q_T}(u-\psi )\varrho ^2\,d\mu =0 \end{aligned}$$

and the equation

$$\begin{aligned} u_t+L_{BS}u=ru-\mu \end{aligned}$$

is satisfied in the strong sense, i.e. for every $\eta \in C_{0}^{\infty }(Q_{T})$,

$$\begin{aligned} \langle u_t,\eta \rangle _{\varrho ,T}+ \langle L_{BS}u,\eta \rangle _{\varrho ,T} =r\langle u,\eta \rangle _{2,\varrho ,T}-\int _{Q_{T}}\eta \varrho ^{2}\,d\mu , \end{aligned}$$

where

$$\begin{aligned} \langle L_{BS}u,\eta \rangle _{\varrho ,T} =\sum ^n_{i=1}\langle (r-d_i)x_iu_{x_i},\eta \rangle _{2,\varrho ,T} -\frac{1}{2}\sum ^n_{i,j=1}a_{ij}\langle u_{x_i}, (x_ix_j\eta \varrho ^2)_{x_j}\rangle _{2,T}. \end{aligned}$$

Here $\langle \cdot ,\cdot \rangle _{\varrho ,T}$ stands for the duality pairing between $L^2(0,T;H^1_{\varrho })$ and $L^2(0,T;H^{-1}_{\varrho })$, $\langle \cdot ,\cdot \rangle _{2,\varrho ,T}$ is the usual scalar product in $L^2(0,T;L^2_{\varrho })$ and $\langle \cdot ,\cdot \rangle _{2,T} =\langle \cdot ,\cdot \rangle _{2,\varrho ,T}$ with $\varrho \equiv 1$.

(b) If $\mu$ in the above definition admits a density (with respect to the Lebesgue measure) of the form $\Phi _u(t,x)=\Phi (t,x,u(t,x))$ for some measurable $\Phi :\bar{Q}_T\times {\mathbb R}\rightarrow {\mathbb R}_+$, then we say that $u$ is a variational solution to the semilinear problem

$$\begin{aligned} u_t+L_{BS}u=ru-\Phi _u,\quad u(T,\cdot )=\psi ,\quad u\ge \psi . \end{aligned}$$

(10)

In our main theorems below we show that if $\psi$ satisfies (A1) and (A2) or (A3) then the measure $\mu$ is absolutely continuous with respect to the Lebesgue measure and its density has the form $\Phi _u(t,x)=\mathbf {1}_{\{u(t,x)=\psi (x)\}}\Psi ^{-}(x)$, where $\Psi ^{-}=\max \{-\Psi ,0\}$ and $\Psi$ is determined by $\psi$ and the parameters $r,d,a$. In the next section we compute $\Psi$ for some concrete options.

3.1 Payoffs Satisfying (A1), (A2)

Remark

One can check that if $u$ is a solution to (10) then $v$ defined as

$$\begin{aligned} v(t,x)=u(T-t,(-1)^{i_1}e^{x_1},\dots ,(-1)^{i_n}e^{x_n})\equiv u(T-t,e^x) \end{aligned}$$

for $t\in [0,T]$, $x=(x_1,\dots ,x_n)\in D_{\iota }$, $\iota \in I$ ($D_{\iota }$ is defined in Sect. 2) is a variational solution of the Cauchy problem

$$\begin{aligned} v_{t}-Lv=-r v+\bar{\Phi },\quad v\ge \bar{\psi },\quad v(0,\cdot )=\bar{\psi }, \end{aligned}$$

where

$$\begin{aligned} Lv=\sum ^n_{i=1}\left( r-d_{i}-\frac{1}{2} \sigma _{ii}^{2}\right) v_{x_{i}}+ \frac{1}{2}\sum ^n_{i,j=1}a_{ij}v_{x_{i}x_{j}} \end{aligned}$$

and $\bar{\Phi }(t,x)=\Phi _u(T-t,e^x)$, $\bar{\psi }(t,x)=\psi (T-t,e^x)$. Furthermore, a simple calculation shows that if $\eta$ is a smooth function on ${\mathbb R}^n$ with compact support and $U\subset {\mathbb R}^n$ is a bounded open set such that supp$[\eta ]\subset U$ then $\tilde{v}=v\eta$ is a solution of the Cauchy-Dirichlet problem

$$\begin{aligned} \tilde{v}_{t}-\tilde{L}\tilde{v}=-r\tilde{v}+f,\quad \quad \tilde{v}(0,\cdot )=\tilde{\psi },\quad \tilde{v}_{|[0,T]\times \partial U}=0, \end{aligned}$$

where $\tilde{\psi }=\bar{\psi }\eta$, $\tilde{L}$ is some uniformly elliptic operator with smooth coefficients not depending on $t$ and $f\in L^{2}(0,T;L^2(U))$. By classical regularity results (see, e.g., Theorem 5 in Sect. 7.1 in [15]), $\tilde{v}\in L^2(0,T;H^2(U))\cap L^{\infty }(0,T;H^1_0(U))$ and $\tilde{v}_t\in L^2(0,T;L^2(U))$. From this and the construction of $\tilde{v}$ we infer that the regularity properties of $\tilde{v}$ are retained by $u$. It follows in particular that

$$\begin{aligned} u_t+L_{BS}u=ru-\Phi _u\quad \text{ a.e. } \text{ on } P_T. \end{aligned}$$

(11)

Theorem 1

Assume (A1), (A2).

• (i) $u$ defined by (9) is a variational solution of the semilinear Cauchy problem

  $$\begin{aligned} u_{t}+L_{BS}u=r u-\Phi ^{-}_u,\quad u(T,\cdot )=\psi \end{aligned}$$

  (12)

  with

  $$\begin{aligned} \Phi _u(t,x)=\mathbf {1}_{\{u(t,x)=\psi (x)\}}\Psi (x),\quad (t,x)\in Q_T, \end{aligned}$$

  where for $x\in {\mathbb R}^n$ such that $(t,x)\in \{u=\psi \}$,

  $$\begin{aligned} \Psi (x)=-r\psi (x)+L_{BS}\psi (x). \end{aligned}$$

• (ii) Set $\sigma (x)=\{\sigma _{ij}x_i\}_{i,j=1,\dots ,n}$ and

  $$\begin{aligned} K_{s,t}=\int _{s}^{t}\Phi ^{-}_u(\theta ,X_{\theta })\,d\theta ,\quad t\in [s,T]. \end{aligned}$$

  (13)

  Then for every $(s,x)\in P_{T}$ the triple $(u(\cdot ,X),\sigma (X)u_x(\cdot ,X),K_{s,\cdot })$ is a unique solution of RBSDE$_{s,x}(\psi ,-r y,\psi )$.

Proof

Fix $(s,x)\in P_{T}$. Let $(Y^{s,x},Z^{s,x},K^{s,x})$ be a solution of RBSDE$_{s,x}(\psi ,-r y,\psi )$ and let $u$ be a viscosity solution of (8). For $t_0\in (s,T)$ let $U\subset {\mathbb R}^n$ be an open set of assumption (A2). Then there exists $\eta \in C^{\infty }({\mathbb R}^n)$ such that $\eta \ge 0$, $\eta \equiv 1$ on $\{u=\psi \}\cap Q_{t_0}$ and $\eta \equiv 0$ on $U^{c}$ (we make the convention that $\eta (t,x)=\eta (x)$). Of course $(Y^{s,x},Z^{s,x}, K^{s,x})$ is a solution of RBSDE$_{s,x}(Y^{s,x}_{t_{0}},-r y,\psi )$ on $[s,t_{0}]$. It is also a solution of RBSDE$_{s,x}(Y^{s,x}_{t_{0}},-ry, \tilde{\psi })$ on $[s,t_{0}]$ with $\tilde{\psi }(x)=\eta (x)\psi (x)$, because $\tilde{\psi }\le \psi$ and by (6) and (7),

$$\begin{aligned} \int ^{t_0}_s(Y^{s,x}_t-\tilde{\psi }(X_t))\,dK^{s,x}_t&=\int ^{t_0}_s(u(t,X_t)-\tilde{\psi }(X_t)) \mathbf {1}_{\{u(t,X_t)=\psi (X_t)\}}\,dK^{s,x}_t\\&=\int ^{t_0}_s(u(t,X_t)-\psi (X_t))\,dK^{s,x}_t=0. \end{aligned}$$

Since $\tilde{\psi }$ is smooth, applying Itô’s formula yields

$$\begin{aligned} \tilde{\psi }(X_{t}) =\tilde{\psi }(X_s)+ \sum ^n_{i=1}\int _{s}^{t}\tilde{\psi }_{x_i}(X_{\theta })\,dX^i_{\theta } +\frac{1}{2}\sum ^n_{i,j=1}\int _{s}^{t}a_{ij}X^i_{\theta }X^j_{\theta } \tilde{\psi }_{x_{i}x_{j}}(X_{\theta })\,d\theta . \end{aligned}$$

From the above, (7) and [12, Remark 4.3] it follows that there exists a predictable process $\alpha ^{s,x}$ such that $0\le \alpha ^{s,x}\le 1$ and

$$\begin{aligned} dK^{s,x}_{t}&=\alpha ^{s,x}_{t}\mathbf {1}_{\{u={\psi }\}}(X_{t}) \left( -r\tilde{\psi }(X_t) \right. +\sum ^n_{i=1}(r-d_{i})X^{i}_{t}\tilde{\psi }_{x_{i}}(X_{t})\\&\quad \left. +\,\frac{1}{2}\sum ^n_{i,j=1} a_{ij} X^i_tX^j_t\tilde{\psi }_{x_{i}x_{j}}(X_{t})\right) ^{-}\,dt \end{aligned}$$

on $[s,t_{0}]$. Thus

$$\begin{aligned} dK^{s,x}_{t}=\alpha ^{s,x}_{t}\mathbf {1}_{\{u(t,X_t)=\psi (X_t)\}} \Psi ^{-}(X_t)\,dt \end{aligned}$$

(14)

on $[s,t_{0}]$ for every $t_0\in [s,T)$. Consequently, the above equation is satisfied on $[s,T]$. Since the coefficients of the stochastic differential equation (3) satisfy the assumptions of the “equivalence of norm” result proved in [3] (see [3, Proposition 5.1]), it follows from [2, Theorem 3] that there exists a function $\alpha$ on $Q_{T}$ such that $0\le \alpha \le 1$ a.e. and for a.e. $(s,x)\in Q_{T}$,

$$\begin{aligned} \alpha ^{s,x}_{t}=\alpha (t,X_{t}),\,\quad dt\otimes P_{s,x}\text{-a.s. } \end{aligned}$$

(15)

Moreover, $u\in C(\bar{Q}_T)$ by [12, Lemma 8.4] and from [2, Theorem 3] it follows that $u\in W_{\varrho }$ and $u$ is a variational solution of the Cauchy problem

$$\begin{aligned} u_{t}+L_{BS}u=r u-\alpha \mathbf {1}_{\{u=\psi \}}\Psi ^{-},\quad u(T,\cdot )=\psi . \end{aligned}$$

By the above and (11),

$$\begin{aligned} u_{t}+L_{BS}u=r u -\alpha \mathbf {1}_{\{u=\psi \}}\Psi ^{-}\quad \text{ a.e. } \text{ on } Q_T, \end{aligned}$$

so by Lemma A.4 in Chapter II in [23],

$$\begin{aligned} \psi _{t}+L_{BS}\psi =r\psi -\alpha \Psi ^{-}\quad \text{ a.e. } \text{ on } \{u=\psi \}. \end{aligned}$$

On the other hand, by the definition of $\Psi$,

$$\begin{aligned} \psi _{t}+L_{BS}\psi =L_{BS}\psi =r\psi +\Psi \quad \text{ on } \{u=\psi \}. \end{aligned}$$

Thus $\Psi =-\alpha \Psi ^{-}$ a.e. on $\{u=\psi \}$, which implies that $\alpha \Psi =\Psi$ a.e. on $\{u=\psi \}$, and hence that

$$\begin{aligned} \mathbf {1}_{\{u=\psi \}}\alpha \Psi ^{-} =\mathbf {1}_{\{u=\psi \}}\Psi ^{-}\quad \text{ a.e. } \end{aligned}$$

(16)

Accordingly (12) is satisfied. From (2) it is clear that if $s\in [0,T)$ and $x\in D_{\iota }$ for some $\iota \in I$ then $P_{s,x}(X_{t}\in D_{\iota },\, t\ge s)=1$ and for every $t\in (s,T]$ the random variable $X_t$ has strictly positive density on $D_{\iota }$ under $P_{s,x}$. From this and (16) it follows that

$$\begin{aligned} \mathbf {1}_{\{u=\psi \}}(t,X_{t})\alpha (t,X_{t})\Psi ^{-}(X_t) =\mathbf {1}_{\{u=\psi \}}(t,X_{t})\Psi ^{-}(X_t),\quad dt\otimes P_{s,x}\text{-a.s. } \end{aligned}$$

(17)

for every $(s,x)\in P_{T}$. In [24] it is proved that the function $\mathbf {1}_{\{u=\psi \}}\alpha$ is a weak limit in $L^{2}(Q_{T})$ of some sequence $\{\alpha _{n}\}$ of nonnegative functions bounded by $1$ and such that $\alpha _{n}(t,X_{t})\rightarrow \alpha ^{s,x}_{t}$ weakly in $L^{2}([0,T]\times \Omega ;dt\otimes P_{s,x})$ for every $(s,x)\in Q_{T}$. Therefore using once again the fact that for every $(s,x)\in P_{T}$ the process $X$ has a strictly positive transition density under $P_{s,x}$ we conclude that (15) holds for every $(s,x)\in P_{T}$, which when combined with (17) implies (13). What is left is to prove that for every $(s,x)\in P_{T}$,

$$\begin{aligned} Z^{s,x}_{t}=\sigma (X_t)u_x (t,X_{t}),\quad dt\otimes P_{s,x}\text{-a.s. } \end{aligned}$$

(18)

From the results proved in [12, Sect. 6] it follows that for every $(s,x)\in Q_{T}$,

$$\begin{aligned} E_{s,x}\sup _{s\le t\le T} |Y^{s,x,n}_t-Y^{s,x}_{t}|^{2} +E_{s,x}\int _{s}^{T}|Z^{s,x,n}_{t}-Z^{s,x}_{t}|^{2}\,dt\rightarrow 0, \end{aligned}$$

(19)

where $(Y^{s,x,n},Z^{s,x,n})$ is a solution of the BSDE

$$\begin{aligned} Y^{s,x,n}_{t}&=\psi (X_{T})-\int _{t}^{T}r Y^{s,x,n}_{\theta }\,d\theta \nonumber \\&\quad +\int _{t}^{T}n\left( Y^{s,x,n}_{\theta }-\psi (X_{\theta })\right) ^{-}\,d\theta -\int _{t}^{T}Z^{s,x,n}_{\theta }\,dB_{s,\theta }. \end{aligned}$$

(20)

It is known (see [27]) that

$$\begin{aligned} Y^{s,x,n}_{t}=u_{n}(t,X_{t}),\quad t\in [s,T],\quad P_{s,x}\text{-a.s. }, \end{aligned}$$

(21)

where $u_{n}$ is a viscosity solution of the Cauchy problem

$$\begin{aligned} (u_n)_t+L_{BS}u_n=-r u_n+n(u_n-\psi )^{-},\quad u_n(T,\cdot )=\psi . \end{aligned}$$

We know that $P_{s,x}(X_{t}\in D_{\iota },\, t\ge s)=1$ if $x\in D_{\iota }$. Moreover, by classical regularity results (see, e.g., [17, Theorem 1.5.9] and Remark preceding Theorem 1), $u_{n}\in C^{1,2}(P_{T})$. Therefore applying Itô’s formula shows that (20) holds true with $Z^{s,x,n}_{\theta }$ replaced by $\sigma (X_{\theta })(u_n)_x(\theta ,X_{\theta })$. Since (20) has a unique solution (see [12, Corollary 3.7]), it follows that

$$\begin{aligned} Z^{s,x,n}_{t}=\sigma (X_t)(u_{n})_x(t,X_{t}), \quad dt\otimes P_{s,x}\text{-a.s. } \end{aligned}$$

(22)

for every $(s,x)\in P_{T}$. By (19) and (21), $u_n\rightarrow u$ pointwise in $Q_T$. Moreover, from (21), (22) and standard estimates for solutions of BSDEs (see, e.g., [12, Sect. 6]) it follows that there is $C>0$ such that for any $(s,x)\in P_T$,

$$\begin{aligned} E_{s,x}\sup _{s\le t\le T} |u_n(t,X_t)|^2+E_{s,x}\int ^T_s|\sigma (X_t)(u_n)_x(t,X_t)|^2\,dt \le CE_{s,x}|\psi (X_T)|^2, \end{aligned}$$

(23)

while from (19), (22) it follows that

$$\begin{aligned} E_{s,x}\sup _{s\le t\le T} \int ^T_s|\sigma (X_t)((u_n)_x-(u_m)_x)(t,X_t)|^2\,dt \rightarrow 0 \end{aligned}$$

(24)

as $n,m\rightarrow \infty$. From (23) one can deduce that $u_n\in L^2(0,T;H_{\varrho })$ and then, by using (24), that $u_n\rightarrow u$ in $L^2(0,T;H_{\varrho })$ (see the arguments following (2.12) in the proof of [25, Theorem 2.3]). From the last convergence and (19), (22) it may be concluded that

$$\begin{aligned} E_{s,x}\int ^T_s|\sigma (X_t)(u_n)_x(t,X_t)-Z^{s,x}_t|^2\,dt=0 \end{aligned}$$

for $(s,x)\in P_T$, which implies (18).

3.2 Convex Payoffs

Assume that $\psi :{\mathbb R}^n\rightarrow {\mathbb R}$ is convex. Let $m$ denote the Lebesgue measure on ${\mathbb R}^n$, $\nabla _i\psi$ denote the usual partial derivative with respect to $x_i$, $i=1,\dots ,n$, and let $E$ be set of all $x\in {\mathbb R}^n$ for which the gradient

$$\begin{aligned} \nabla \psi (x)=(\nabla _1\psi (x),\dots ,\nabla _n\psi (x)) \end{aligned}$$

exists. Since $\psi$ is locally Lipschitz function, $m(E^c)=0$ and $\nabla \psi =(\psi _{x_1},\dots ,\psi _{x_n})$ a.e. (recall that $\psi _{x_i}$ stands for the partial derivative in the distribution sense). Moreover, for a.e. $x\in E$ there exists an $n$-dimensional symmetric matrix $\{H(x)=\{H_{ij}(x)\}$ such that

$$\begin{aligned} \lim _{E\ni y\rightarrow x} \frac{\nabla \psi (y)-\nabla \psi (x) -H(x)(y-x)}{|y-x|}=0, \end{aligned}$$

(25)

i.e. $H_{ij}(x)$ are defined as limits through the set where $\nabla _i\psi$ exists (see, e.g., [1, Sect. 7.9]). By Alexandrov’s theorem (see, e.g., [1, Theorem 7.10]), if $x\in E$ is a point where (25) holds then $\psi$ has second order differential at $x$ and $H(x)$ is the hessian matrix of $\psi$ at $x$, i.e. $H(x)=\{\nabla ^2_{ij}\psi (x)\}$.

The second order derivative of $\psi$ in the distribution sense $D^2\psi =\{\psi _{x_ix_j}\}$ is a matrix of real-valued Radon measures $\{\mu _{ij}\}$ on ${\mathbb R}^n$ such that $\mu _{ij}=\mu _{ji}$ and for each Borel set $B$, $\{\mu _{ij}(B)\}$ is a nonnegative definite matrix (see, e.g., [16, Sect. 6.3]). Let $\mu _{ij}=\mu ^a_{ij}+\mu ^s_{ij}$ be the Lebesgue decomposition of $\mu _{ij}$ into the absolutely continuous and singular parts with respect to $m$. By Theorem 1 in Sect. 6.4 in [16],

$$\begin{aligned} \mu ^a_{ij}(dx)=\nabla ^2_{ij}\psi (x)\,dx. \end{aligned}$$

(26)

For $R>0$ set $D_R=P\cap \{x\in {\mathbb R}^n:|x|<R\}$ and $\tau _R=\inf \{t\ge s:X_t\notin D_R\}$. Let $\tilde{L}_{BS}$ denote the operator formally adjoint to $L_{BS}$. By [28, Theorem 4.2.5] for a sufficiently large $\alpha >0$ there exist the Green’s functions $G^{\alpha }_R$, $\tilde{G}^{\alpha }_R$ for $\alpha -L_{BS}$ and $\alpha -\tilde{L}_{BS}$ on $D_R$. Let $A$ be a continuous additive functional of ${\mathbb X}$ and let $\nu$ denote the Revuz measure of $A$ (see, e.g., [29]). By the theorem proved in Sect. V.5 of [29], for every nonnegative $f\in C_0({\mathbb R}^d)$,

$$\begin{aligned} E_{s,x}\int ^{\tau _R}_se^{-\alpha t}f(X_t)\,dA^{\nu }_t =\int _{{\mathbb R}^n}G^{\alpha }_R(x,y)f(y)\,\nu (dy). \end{aligned}$$

Since $G^{\alpha }_R(x,y)=\tilde{G}^{\alpha }_R(y,x)$ by [28, Corollary 4.2.6], it follows that

$$\begin{aligned} E_{s,g\cdot m}\int ^{\tau _R}_se^{-\alpha t} f(X_t)\,dA^{\nu }_t=\int _{{\mathbb R}^n}\tilde{G}^{\alpha }_Rg(y)f(y)\,\nu (dy) \end{aligned}$$

(27)

for any nonnegative $g\in C_0(D_R)$, where $E_{s,g\cdot m}$ denotes the expectation with respect to the measure $P_{s,g\cdot m}(\cdot )=\int P_{s,x}(\cdot )g(x)\,dx$ and

$$\begin{aligned} \tilde{G}^{\alpha }_Rg(y)=\int _{G_R}\tilde{G}^{\alpha }_R(y,x)g(x)\,dx. \end{aligned}$$

Note that if $g$ is not identically equal to zero then $\tilde{G}^{\alpha }_Rg$ is strictly positive (see [28, Theorem 4.2.5]).

Set

$$\begin{aligned} {\mathcal {L}}_{BS}=\sum ^n_{i=1}(r-d_{i})x_{i}\nabla _i +\frac{1}{2}\sum ^n_{i,j=1} a_{ij}x_ix_j\nabla ^2_{ij}\,. \end{aligned}$$

Theorem 2

Assume (A3). Then assertions (i), (ii) of Theorem 1 hold true with $L_{BS}$ replaced by ${\mathcal {L}}_{BS}$.

Proof

We use the notation of Theorem 1. Fix $s\in [0,T)$. Since $\psi$ is a continuous convex function, from Itô’s formula proved in [5] it follows that there exists a continuous increasing process $A$ such that for $x\in {\mathbb R}^n$,

$$\begin{aligned} \psi (X_{t})=\psi (X_{s})+A_{t} +\int _{s}^{t}\nabla \psi (X_{\theta })\, dX_{\theta },\quad t\in [s,T],\quad P_{s,x}\text{-a.s. } \end{aligned}$$

(28)

From (28) it follows that $A$ is a positive continuous additive functional (PCAF for short) of ${\mathbb X}$. Let $\nu$ denote the Revuz measure of $A$. We are going to show that $\mathbf {1}_P\cdot \nu =\mathbf {1}_P\cdot \mu$ where $\mu$ is the measure on ${\mathbb R}^n$ defined as

$$\begin{aligned} \mu (dx)=\sum ^n_{i,j=1}a_{ij}x_ix_j\,\mu _{ij}(dx). \end{aligned}$$

To this end, let us set

$$\begin{aligned} \mu ^{\varepsilon }_{ij}=\frac{\partial ^2\psi }{\partial x_i\partial x_j}*\rho _{\varepsilon }\,,\quad \mu ^{\varepsilon }(dx) =\sum ^n_{i,j=1}a_{ij}x_ix_j\,\mu ^{\varepsilon }_{ij}(dx), \end{aligned}$$

where $\{\rho _{\varepsilon }\}_{\varepsilon >0}$ is some family of mollifiers. Fix a nonnegative $g\in C_0(D_R)$ such that $g(x)>0$ for some $x\in D_R$ and denote by $A^{\varepsilon }$ the PCAF of ${\mathbb X}$ in Revuz correspondence with $\mu _{\varepsilon }$. Then for a sufficiently large $\alpha >0$,

$$\begin{aligned} E_{s,g\cdot m}\int ^{\tau _R}_se^{-\alpha t} f(X_t)\,dA^{\varepsilon }_t =\int _{{\mathbb R}^n}\tilde{G}^{\alpha }_Rg(y)f(y)\,\mu ^{\varepsilon }(dy) \end{aligned}$$

(29)

for all nonnegative $f\in C_0({\mathbb R}^d)$. By [9, Theorem 2], $E_{s,x}\sup _{t\ge s} |A^{\varepsilon }_{t\wedge \tau _R}-A_{t\wedge \tau _R}|\rightarrow 0$ as $\varepsilon \downarrow 0$ for every $x\in {\mathbb R}^d$. Hence $\int ^{\tau _R}_se^{-\alpha t}f(X_t)\,dA^{\varepsilon }_t \rightarrow \int ^{\tau _R}_se^{-\alpha t}f(X_t)\,dA_t$ weakly under $P_{s,x}$ for $x\in {\mathbb R}^d$. Since

$$\begin{aligned} A^{\varepsilon }_{t\wedge \tau _R}=\psi _{\varepsilon }(X_{t\wedge \tau _R}) -\psi _{\varepsilon }(X_s)-\int ^{t\wedge \tau _R}_s \nabla \psi _{\varepsilon }(X_{\theta })\,dX_{\theta } \end{aligned}$$

and $\sup _{\varepsilon >0}\sup _{|x|\le R}| \nabla \psi _{\varepsilon }(x)|\le C(R)<\infty$ by Lemma in [9], it follows that for every compact subset $K\subset {\mathbb R}^n$, $\sup _{x\in K} \sup _{\varepsilon >0}E_{s,x}|A^{\varepsilon }_{t\wedge \tau _R}|^2<\infty$. Therefore

$$\begin{aligned} E_{s,g\cdot m}\int ^{\tau _R}_se^{-\alpha t} f(X_t)\,dA^{\varepsilon }_t\rightarrow E_{s,g\cdot m} \int ^{\tau _R}_se^{-\alpha t}f(X_t)\,dA_t \end{aligned}$$

(30)

as $\varepsilon \downarrow 0$. On the other hand, since $\mu ^{\varepsilon }_{ij}\rightarrow \mu _{ij}$ weakly${}^{*}$ for $i,j=1,\dots ,n$ and, by [28, Theorem 4.2.5], $f\tilde{G}^{\alpha }_Rg\in C_0(G_R)$, we have

$$\begin{aligned} \sum ^n_{i,j=1}\int _{{\mathbb R}^n} \tilde{G}^{\alpha }_Rg(y)f(y)a_{ij}y_iy_j\, \mu ^{\varepsilon }_{ij}(dy) \rightarrow \sum ^n_{i,j=1}\int _{{\mathbb R}^n} \tilde{G}^{\alpha }_Rg(y)f(y)\, a_{ij}y_iy_j\mu _{ij}(dy). \end{aligned}$$

Combining this with (27), (29), (30) we see that for every $f\in C_0({\mathbb R}^n)$,

$$\begin{aligned} \int _{{\mathbb R}^n}\tilde{G}^{\alpha }_Rg(y)f(y)\mu (dy)=\int _{{\mathbb R}^n}\tilde{G}^{\alpha }_Rg(y)f(y)\,\nu (dy). \end{aligned}$$

Since $\tilde{G}^{\alpha }_Rg$ is strictly positive on $D_R$, we conclude from the above that $\mu =\nu$ on $D_R$ for each $R>0$. Consequently, $\mu =\nu$ on $P$. For $x\in P$, $P_{s,x}(X_t\in {\mathbb R}^n\setminus P)=0$ for $t\ge s$. Hence

$$\begin{aligned} A^{\nu }_t=\int ^t_s\mathbf {1}_P(X_s)\,dA^{\nu }_s=A^{\mathbf {1}_P\cdot \nu }_t =A^{\mathbf {1}_P\cdot \mu }_t,\quad t\ge s,\quad P_{s,x}\text{-a.s. } \end{aligned}$$

(31)

for $x\in P$. Let $\mu ^a$ denote the absolutely continuous part in the Lebesgue decomposition of $\mathbf {1}_P\cdot \mu$. By (26), $\mu ^a(dx)=\sum ^n_{i,j=1}\mathbf {1}_P(x)a_{ij}x_ix_j\nabla ^2_{ij}\psi (x)\,dx$. Hence

$$\begin{aligned} A^{\mu ^a}_t=\sum ^n_{i,j=1}\int ^t_sa_{ij}X^i_{\theta }X^j_{\theta } \nabla ^2_{ij}\psi (X_{\theta })\,d\theta , \quad t\ge s, \quad P_{s,x}\text{-a.s. } \end{aligned}$$

(32)

for $x\in P$. From (28), (31), (32) and [12, Remark 4.3] it follows that

$$\begin{aligned} dK^{s,x}_{t}&=\alpha ^{s,x}_{t}\mathbf {1}_{\{u\,=\,\psi \}} (X_{t})\left( -r\psi (X_{t}) +\sum _{i=1}^{n}(r-d_{i})X^{i}_{t}\nabla _i\psi (X_{t})\right. \\&\quad \left. +\,\frac{1}{2}\sum ^n_{i,j=1} a_{ij} X^i_tX^j_t\nabla ^2_{ij}\psi (X_{t})\right) ^{-}\,dt. \end{aligned}$$

Let $u$ be a viscosity solution of (8). From the above and the results proved in [2] (see the reasoning following (14)) we conclude that $u\in W_{\varrho }\cap C(\bar{Q}_{T})$ and there is a function $\alpha$ on $Q_T$ such that $0\le \alpha \le 1$ a.e., (15) is satisfied and $u$ is a variational solution of the Cauchy problem

$$\begin{aligned} u_{t}+L_{BS}u=r u -\alpha \mathbf {1}_{\{u=\psi \}}\Psi ^{-},\quad u(T,\cdot )=\psi \end{aligned}$$

(33)

with

$$\begin{aligned} \Psi =-r\psi +{\mathcal {L}}_{BS}\psi \quad \text{ on } \{u=\psi \}. \end{aligned}$$

(34)

By Remark preceding Theorem 1, $u(t,\cdot )\in H^2_{loc}({\mathbb R}^n)$. Therefore by Remark (ii) following Theorem 4 in Sect. 6.1 in [16] the distributional derivatives $u_{x_i}$, $u_{x_ix_j}$ are a.e. equal to the approximate derivatives $\nabla ^{ap}_iu$, $(\nabla ^{ap})^2_{ij}u$. Let ${\mathcal {L}}^{ap}_{BS}$ denote the operator defined as ${\mathcal {L}}_{BS}$ but with $\nabla _i$, $\nabla _{ij}$ replaced by $\nabla ^{ap}_i$, $(\nabla ^{ap})^2_{ij}$. Then $u$ is a variational solution of (33) with $L_{BS}$ replaced by ${\mathcal {L}}^{ap}_{BS}$ and (11) holds with $L_{BS}$ replaced by ${\mathcal {L}}^{ap}_{BS}$. Hence

$$\begin{aligned} u_{t}+{\mathcal {L}}^{ap}_{BS}u=ru -\alpha \mathbf {1}_{\{u=\psi \}}\Psi ^{-}\quad \text{ a.e. } \text{ on } Q_T. \end{aligned}$$

On the other hand, since $\psi$ is convex, $\psi \in BV_{loc}({\mathbb R}^n)$ as a locally Lipschitz continuous function and, by Theorem 3 in Sect. 6.3 in [16], $\psi _{x_i}\in BV_{loc}({\mathbb R}^n)$, $i=1,\dots ,n$. Therefore $\psi$ is twice approximately differentiable a.e. by Theorem 4 in Sect. 6.1 in [16]. It follows now from Theorem 3 in Sect. 6.1 in [16] that ${\mathcal {L}}^{ap}u={\mathcal {L}}^{ap}\psi$ a.e. on $\{u=\psi \}$. Consequently,

$$\begin{aligned} {\mathcal {L}}^{ap}_{BS}\psi =r\psi -\alpha \Psi ^{-}\quad \text{ a.e. } \text{ on } \{u=\psi \}. \end{aligned}$$

(35)

Moreover, since $\psi$ is convex, ${\mathcal {L}}_{BS}\psi ={\mathcal {L}}^{ap}_{BS}\psi$ a.e. on ${\mathbb R}^n$ by Remark (i) following Theorem 4 in Sect. 6.1 in [16]. Therefore combining (34) with (35) we see that $\Psi =-\alpha \Psi ^{-}$ a.e. on $\{u=\psi \}$ from which as in the proof of Theorem 1 we get (17). To complete the proof it suffices now to repeat step by step the arguments following (17) in the proof of Theorem 1. $\square$

4 The Early Exercise Premium Representation

Let $\xi$ denote the payoff process for an American option with payoff function $\psi$, i.e.

$$\begin{aligned} \xi _t=e^{-r(t-s)}\psi (X_t),\quad t\in [s,T], \end{aligned}$$

and let $\eta$ denote the Snell envelope for $\xi$, i.e. the smallest supermartingale which dominates $\xi$. It is known (see, e.g., Sect. 2.5 in [21]) that

$$\begin{aligned} \eta _t=e^{-r(t-s)}V(t,X_t),\quad t\in [s,T]. \end{aligned}$$

Assume (A1), (A2) or (A3). Applying Itô’s formula and using Theorem 1 or 2 we get

$$\begin{aligned} \eta _t=e^{-r(t-s)}Y^{s,x}_t&=e^{-r(T-s)}\psi (X_T) +\int ^T_t e^{-r(\theta -s)}\Phi ^{-}(X_{\theta },Y^{s,x}_{\theta })\,d\theta \\&\quad -\int ^T_te^{-r(\theta -s)}Z^{s,x}_{\theta }\,dW_{\theta },\quad t\in [s,T],\quad P_{s,x}\text{-a.s. }, \end{aligned}$$

which leads to the following corollary.

Corollary 3

For every $(s,x)\in Q_T$ the Snell envelope admits the representation

$$\begin{aligned} \eta _t=E_{s,x}\left( e^{-r(T-s)}\psi (X_T) +\int ^T_te^{-r(\theta -s)} \Phi ^{-}(X_{\theta },Y^{s,x}_{\theta })\,d\theta \,|{\mathcal {F}}_t\right) ,\quad t\in [s,T]. \end{aligned}$$

(36)

Taking $t=s$ in (36) and using (7) we get the early exercise premium representation for the value function.

Corollary 4

For every $(s,x)\in Q_T$ the value function $V$ admits the representation

$$\begin{aligned} V(s,x)=V^E(s,x)+E_{s,x}\int ^T_s e^{-r(t-s)} {\mathbf {1}}_{\{V(t,X_t)=\psi (X_t)\}}\Psi ^{-}(X_t)\,dt, \end{aligned}$$

where

$$\begin{aligned} V^E(s,x)=E_{s,x}\big (e^{-r(T-s)}\psi (X_T)\big ) \end{aligned}$$

is the value of the European option with payoff function $\psi$ and expiration time $T$.

In closing this section we show by examples that for many options $\Psi ^{-}$ can be explicitly computed. Using results of Sects. 4 and 5 in [30] one can check that the payoff functions $\psi$ in examples 1–4 below satisfy (A3). It is also easy to see that the payoff function $\psi$ in example 5 satisfies (A2$'$). Note that the payoff function in example 1 also satisfies (A2$'$) and, by [7, 26], the payoff functions in examples 2–4 satisfy (A2). We would like to stress that the last assertion is by no means evident. On the other hand, the convexity of $\psi$ in examples 2–4 is readily checked.

In all the examples we have computed the corresponding functions $\Psi ^{-}$ on the region $\{u=\psi \}$. When computing $\Psi$ we keep in mind that $\{u=\psi \}\subset [0,T]\times \{\psi >0\}$.

• 1. Index options and spread options

  $$\begin{aligned} \psi (x)=\left( \sum _{i=1}^{n} w_{i}x_i-K\right) ^{+},\quad \Psi ^-(x)=\left( \sum _{i=1}^{n} w_{i}d_{i}x_i-r K\right) ^{+}\quad \text{(call) } \end{aligned}$$
  $$\begin{aligned} \psi (x)=\left( K-\sum _{i=1}^{n} w_{i}x_i\right) ^{+},\quad \Psi ^-(x)=\left( r K-\sum _{i=1}^{n} w_{i}d_{i}x_i\right) ^{+}\quad \text{(put) } \end{aligned}$$

  (Here $w_i\in {\mathbb R}$ for $i=1,\dots ,n$).

• 2. Max options

  $$\begin{aligned} \psi (x)=(\max \{x_1,\dots ,x_n\}-K)^{+} \quad \text{(call } \text{ on } \text{ max) } \end{aligned}$$
  $$\begin{aligned} \Psi ^{-}(x)=\left( \sum _{i=1}^{n} d_{i}\mathbf {1}_{B_{i}}(x)x_i-r K\right) ^{+}, \end{aligned}$$

  where $B_{i}=\{x\in {\mathbb R}^n; x_{i}>x_{j},\, j\ne i\}$.

• 3. Min options

  $$\begin{aligned} \psi (x)=(K-\min \{x_1,\dots ,x_n\})^{+}\quad \text{(put } \text{ on } \text{ min) } \end{aligned}$$
  $$\begin{aligned} \Psi ^{-}(x)=\left( r K -\sum _{i=1}^{n}d_{i}\mathbf {1}_{C_{i}}(x)x_i\right) ^{+}, \end{aligned}$$

  where $C_{i}=\{x\in {\mathbb R}^n; x_{i}<x_{j},\, j\ne i\}$.

• 4. Multiple strike options

  $$\begin{aligned} \psi (x)=(\max \{x_1-K_{1},\dots , x_n-K_{n}\})^{+}, \end{aligned}$$
  $$\begin{aligned} \Psi ^{-}(x)=\left( \sum _{i=1}^{n} \mathbf {1}_{B_{i}}(x-K)(d_{i}x_i-r K_{i})\right) ^{+}, \end{aligned}$$

  where $K=(K_{1},\dots ,K_{n})$.

• 5. Power-product options

  $$\begin{aligned} \psi (x)=(|x_1\cdot \ldots \cdot x_n|^{\gamma }-K)^+\quad \text{ for } \text{ some } \gamma >0. \end{aligned}$$

  If $x\in D_{\iota }$ with $\iota =(i_1,\dots ,i_n)\in \{0,1\}^n$ then

  $$\begin{aligned} \Psi ^{-}(x)=\left( \left( r-\gamma \sum _{i=1}^n(r-d_i-a_{ii}) -\gamma ^2\sum ^n_{i,j=1}a_{ij}\right) f(x)-r K\right) ^{+}, \end{aligned}$$

  where $f(x)=((-1)^{|\iota |}\,x_1\cdot \ldots \cdot x_n)^{\gamma }$ and $|\iota |=i_1+\cdots + i_n$.