On perpetual American options in a multidimensional Black-Scholes model
aa r X i v : . [ m a t h . P R ] J a n On perpetual American options in a multidimensionalBlack-Scholes model
Andrzej Rozkosz
Faculty of Mathematics and Computer Science, Nicolaus Copernicus UniversityChopina 12/18, 87-100 Toru´n, PolandE-mail address: [email protected]
Abstract
We consider the problem o pricing perpetual American options written ondividend-paying assets whose price dynamics follow a multidimensional Black andScholes model. For convex Lipschitz continuous functions, we give a probabilisticcharacterization of the fair price in terms of a reflected BSDE, and an analyticalone in terms of an obstacle problem. We also provide the early exercise premiumformula.
Keywords:
Perpetual American option, backward stochastic differential equation, obstacleproblem . . Primary 91G20, Secondary 60H10, 60H30. In this paper, we consider the problem of pricing perpetual American options writtenon dividend-paying assets whose price dynamics follow the classical multidimensionalBlack and Scholes model. In this model, under the risk-neutral measure P , the assetprices X s,x, , . . . , X s,x,d on [ s, ∞ ) evolve according to the stochastic differential equation X s,x,it = x i + Z ts ( r − δ i ) X s,x,iθ dθ + n X j =1 Z ts σ ij X s,x,iθ dW jθ , t ≥ s. (1.1)In (1.1), W is a standard d -dimensional Wiener process, x i , i = 1 , . . . , d , are the initialprices at time s , r ≥ δ i ≥ i = 1 , . . . , d , are dividendrates and σ = { σ ij } i,j =1 ,...,d is the volatility matrix. We assume that a = σ · σ ∗ , where σ ∗ is the transpose of σ , is strictly positive definite.Let T > ψ : R d → R be a nonnegative continuous functions with polynomialgrowth. Under the measure P , the value at time s of the American option with payofffunction ψ and expiration time T is given by V T ( s, x ) = sup s ≤ τ ≤ T Ee − r ( τ − s ) ψ ( X s,xτ ) , (1.2)and the value of the perpetual option with payoff function ψ is V ( s, x ) = sup τ ≥ s Ee − r ( τ − s ) ψ ( X s,xτ ) . (1.3)1see [9, 10, 19]). In (1.2), the supremum is taken over the set of all stopping times withvalues in [ s, T ], and in (1.3), over the set of stopping times in [ s, ∞ ]. In the event that τ = ∞ , we interpret e − r ( τ − s ) ψ ( X s,xτ ) to be zero.At present, properties of V T are quite well investigated. It is known (see [5, 6, 7])that V T can be represented by a solution of a reflected backward stochastic differentialequation (RBSDE). A detailed study of the structure of this RBSDE, which in partic-ular leads to the early exercise premium formula, is given in [12] (also see Section 3.1).The value V T can also be characterized analytically as a solution of some obstacle prob-lem (or, in different terminology, variational inequality) (see [5, 6, 7, 12] and Section3.2). It is worth pointing here that the analytical characterization relies heavily on thecharacterization via solutions of RBSDEs.In case of perpetual options less in known, except for put and call options in case d = 1, which were thoroughly investigated as early as in [16, 17]. For a nice presentationof these results as well as some newer results and historical comments see the books[10, 19]. Presumably, the main reason that less attention has been paid to V thanto V T is that perpetual options are not traded. On the other hand, in our opinion,perpetual American options are interesting from historical reasons and from a purelytheoretical point of view. This motivated us to ask whether in the multidimensionalcase, for a wide class of payoffs functions one can represent V in terms of BSDEs orsolutions of obstacle problems. Another reason for writing this paper is that the desiredrepresentations of V can be derived in a quite elegant way from those for V T . The mainidea is as follows. Intuitively, V is the limit of V T as T → ∞ (in fact this is true; seeSection 3.1). This suggests that properties of V we are interested in can be derivedby studying the behaviour, as T → ∞ , of the solution of the RBSDE with terminalcondition at time T , which is used to represent V T . By modifying some results fromthe recent paper [14], we show that the idea sketched above is indeed realizable. Asa result we show that for convex and Lipschitz continuous ψ the value function V isrepresented by a solution of some RBSDE with terminal condition 0 at infinity and weget the exercise premium formula. We also show that V is a solution a unique of someobstacle problem. Finally, we estimate that rate of convergence of V T to V . It seemsthat some of our results (the representation in terms of RBSDEs, rate of convergence)are new even in the case of classical call/put option and d = 1. In our considerations only the distribution of the processes X s,x,i will be important.Since they depend on σ only through a , we may and will assume that σ is a symmetricsquare root of a . From the same reason (only the distributions are important), as in[12], we will use a slightly different from (1.1) form of the price dynamics. It appearsto be more convenient for us than (1.1).Let Ω = C ([0 , T ]; R d ) and let X be the canonical process on Ω. For ( s, x ) ∈ [0 , T ] × R d let P s,x denote the law of the process X s,x = ( X s,x, , . . . , X s,x,d ) definedby (1.1) and let {F st } denote the completion of σ ( X θ ; θ ∈ [ s, t ]) with respect to thefamily { P s,µ ; µ a finite measure on B ( R n ) } , where P s,µ ( · ) = R R d P s,x ( · ) µ ( dx ). Then foreach s ∈ [0 , T ), X = (Ω , ( F st ) t ∈ [ s,T ] , X, P s,x ) is a Markov process on [0 , T ]. Using Itˆo’sformula and L´evy’s characterization of the Wiener process motion one can check (see212, Section 2] for details) that X it = X i + Z ts ( r − δ i ) X iθ dθ + d X j =1 Z ts σ ij X iθ dB js,θ , t ≥ s, P s,x -a.s. , (2.1)where { B s,t , t ≥ s } is under P s,x a standard d -dimensional {F st } -Wiener process on[ s, ∞ ). It is well known that the unique solution of (2.1) is of the form X it = X i exp (cid:0) ( r − δ i − a ii / t − s ) + d X j =1 σ ij B js,t (cid:1) , t ≥ s, P s,x -a.s. (2.2)Let σ i = √ a ii . Since ˜ B i := P dj =1 σ ij B js, · is a continuous martingale with the quadraticvariation h ˜ B is, · i t = a ii ( t − s ), t ≥ s , the process X i has the form X it = X i e ( r − δ i )( t − s ) N is,t , t ≥ s, (2.3)where N is,t = exp( − ( t − s ) a ii / B is,t ), t ≥ s , is an ( F st )-martingale under P s,x . Let D = { x = ( x , . . . , x d ) : x i > , i = 1 , . . . , d } . From (2.2) it follows that if x ∈ D , then P s,x ( X t ∈ D, t ≥ s ) = 1.Below we recall some known results on the pricing of American options with finiteexpiration time T >
0. They will be needed in the next section.In this paper, we assume that the payoff function satisfies the following condition:(A1) ψ : R d → R is a nonnegative convex function which is Lipschitz continuous, i.e.there is L > | ψ ( x ) − ψ ( y ) | ≤ L | x − y | for all x, y ∈ R d .In particular, ψ ( x ) ≤ C (1 + | x | ) with C = max { L, ψ (0) } . Furthermore, since ψ isconvex, for a.e. x ∈ R d there exist the usual partial derivatives ∇ ψ ( x ) , . . . , ∇ d ψ ( x )of ψ at x . Furthermore, by Alexandrov’s theorem (see, e.g., [1, Theorem 7.10]), ψ hassecond order derivatives at x for a.e. x ∈ R d , which we denote by ∇ ij ψ ( x ).Let T s,T denote the set of all ( F st )-stopping times with values in [ s, T ]. The fairprice (or value) V T ( s, x ) of the American option with expiration time T and payofffunction ψ is given by V T ( s, x ) = sup τ ∈T s,T E s,x e − r ( τ − s ) ψ ( X τ ) . (2.4)Since ψ is continuous with linear growth, from [5, Theorem 5.2] it follows thatfor every ( s, x ) ∈ [0 , T ] × R d there exists a unique solution ( Y T,s,x , K
T,s,x , Z
T,s,x ), onthe space (Ω , F sT , P s,x ), of the RBSDE with coefficient f ( y ) = − ry , y ∈ R , terminalcondition ψ ( X T ) and barrier ψ ( X ), that is RBSDE of the form Y T,s,xt = ψ ( X T ) − R Tt rY T,s,xθ dθ + R Tt dK T,s,xθ − R Tt Z T,s,xθ dB s,θ , t ∈ [ s, T ] ,Y T,s,xt ≥ ψ ( X t ) , t ∈ [ s, T ] ,K T,s,x = 0 , K T,s,x is continuous and increasing, and satisfiesthe minimality condition R Ts ( Y T,s,xt − ψ ( X t )) dK T,s,xt = 0 . (2.5)For the precise definition of a solution we defer the reader to [5]. Here let us only notethat E s,x R Ts | Z T,s,xθ | dθ < ∞ , so the process M T,s,xt = Z ts Z T,s,xθ dB s,θ , t ∈ [ s, T ] ,
3s a martingale under P s,x . Let L BS denote the Black-Scholes operator defined by L BS = d X i =1 ( r − δ i ) x i ∂ x i + 12 d X i,j =1 a ij x i x j ∂ x i x j , where ∂ x i , ∂ x i x j denote the partial derivatives in the distribution sense. In [5, Theorem8.5] it is also proved that for every ( s, x ) ∈ [0 , T ] × R d , Y T,s,xt = u T ( t, X t ) , t ∈ [ s, T ] , P s,x -a.s. , (2.6)where u T is a (unique) viscosity solution to the obstacle problem ( min { u T − ψ, − ∂ s u T − L BS u T + ru T } = 0 in [0 , T ] × R d ,u T ( T, · ) = ψ on x ∈ R d . (2.7)The process ¯ Y T,s,x defined as ¯ Y T,s,xt = e − r ( t − s ) Y T,s,xt , t ∈ [ s, T ], is the first componentof the solution of RBSDE with coefficient f = 0, terminal condition e − rT ψ ( X T ) andbarrier e − rt ψ ( X t ), t ∈ [ s, T ]. Therefore from (2.6) with t = s and [5, Proposition 2.3](or [6, Proposition 3.3]) it follows that V T = u T . Let L BS = d X i =1 ( r − δ i ) x i ∇ i + 12 d X i,j =1 a ij ∇ ij . In [12, Theorem 2] it is proved that under (A1), for every ( s, x ) ∈ [0 , T ] × D , K T,s,xt = Z ts Φ( X θ , u T ( θ, X θ )) dθ, t ∈ [ s, T ] , P s,x -a.s.where Φ( x, y ) = Ψ − ( x ) ( −∞ ,ψ ( x )] ( y ) , Ψ( x ) = − rψ ( x ) + L BS ψ ( x ) (2.8)and Ψ − = max {− Ψ , } . Since u T ( s, x ) ≥ ψ ( x ) ≥
0, we haveΦ( x,
0) = Ψ − ( x ) , Φ( x, u T ( s, x )) = Ψ − ( x ) { u T ( s,x )= ψ ( x ) } , ( s, x ) ∈ [0 , T ] × D. To shorten notation, in this section we set V ( x ) = V (0 , x ), F t = F t , P x = P ,x , andwe denote by E x the expectation with respect to P x . With this notation, (1.3) takesthe form V ( x ) = sup τ ∈T E x e − rτ ψ ( X τ ) , (3.1)where T is the set of all ( F t )-stopping times. Assume (A1) and let Y Tt = u T ( t, X t ) , K Tt = Z t Φ( X s , u T ( s, X s )) ds, t ∈ [0 , T ] . Y T and K T are independent of x versions of Y T, ,x and K T, ,x , respectively. Since V T = u T , we have V T ( t, X t ) = Y Tt = u T ( t, X t ) , t ∈ [0 , T ] , P x -a.s. (3.2)From the first equation in (2.5) it follows that M T, ,x also has a version independentof x , which we denote by M T . Set¯ Y Tt = e − rt Y Tt , ¯ K Tt = Z t e − rs dK Ts , ¯ M Tt = Z t e − rs dM Ts , t ∈ [0 , T ] . Since Y Tt = ψ ( X T ) − Z Tt rY Ts ds + Z Tt dK Tθ − Z Tt dM Ts , t ∈ [ s, T ] , integrating by parts we obtain¯ Y Tt = e − rT ψ ( X T ) + Z Tt d ¯ K Ts − Z Tt d ¯ M Ts , t ∈ [0 , T ] . (3.3)We will also need the following condition.(A2) For every x ∈ D ,(a) lim t →∞ E x e − rt ψ ( X t ) = 0 , (b) E x Z ∞ e − rt Ψ − ( X t ) dt < ∞ . (3.4) Remark 3.1. (i) Condition (3.4) can be equivalently stated as(a) lim t →∞ e − rt P t ψ ( x ) = 0 , (b) R r Ψ − ( x ) < ∞ , where ( P t ) t> (resp. ( R α ) α> ) is the semigroup (resp. resovent) associated with X .(ii) Assume that r >
0. Clearly (3.4)(a) is satisfied if ψ is bounded. By (2.3), E x X it = x i e ( r − δ i ) t , t ≥
0. Therefore (3.4)(a) is satisfied for general Lipschitz-continuous ψ if δ i > i = 1 , . . . d . Similarly, (3.4)(b) is satisfied if Ψ − is bounded or Ψ − satisfies thelinear growth condition and δ i > i = 1 , . . . , d .We are going to show that if (3.4) is satisfied for some x ∈ D , then ¯ Y T convergesas T → ∞ to a process ¯ Y x being the first component of the solution ( ¯ Y x , ¯ K x , ¯ M x ) ofthe reflected BSDE which informally can written as¯ Y xt = Z ∞ t d ¯ K xs − Z ∞ t d ¯ M xs , t ≥ . (3.5)We will also show that ¯ K x has the representation¯ K xt = Z t e − rs Φ( X s , e rs ¯ Y xs ) ds, t ≥ , (3.6)so in fact ( ¯ Y x , ¯ M x ) is a solution of the usual BSDE¯ Y xt = Z ∞ t e − rs Φ( X s , e rs ¯ Y xs ) ds − Z ∞ t d ¯ M xs , t ≥ . (3.7)5efore giving the definition of solutions of (3.5) and (3.7) let us recall that a con-tinuous ( F t )-adapted process Y is said to be of class (D) under the measure P x if thecollection { Y τ : τ ∈ T } is uniformly integrable under P x . Let L ( P x ) denote the spaceof continuous processes with finite norm k Y k x, = sup { E x | Y τ | : τ ∈ T } . It is knownthat L ( P x ) is complete (see [3, Theorem VI.22]). Moreover, if Y n are of class (D) and Y n → Y in L ( P x ), then Y is of class (D) (see [14, Section 3]). Definition. (i) We say that a triple ( ¯ Y x , ¯ K x , ¯ M x ) of adapted continuous processes is asolution of the reflected BSDE (3.5) with lower barrier ¯ L t = e − rt ψ ( X t ) if ¯ Y x is of classD, ¯ M x is a local martingale with ¯ M x = 0, ¯ K x is an increasing process with ¯ K = 0,and for every T > ¯ Y xt = ¯ Y xT + R Tt d ¯ K xs − R Tt d ¯ M xs , t ≥ , ¯ Y xt ≥ ¯ L t , t ∈ [0 , T ] , R T ( ¯ Y xt − ¯ L t ) d ¯ K xt = 0 , ¯ Y xT → P x -a.s. as T → ∞ . (3.8)(ii) We say that a pair ( ¯ Y x , ¯ M x ) of adapted continuous processes is a solution of theBSDE (3.7) if ¯ Y x is of class D, ¯ M x is a local martingale with ¯ M x = 0 and for every T > R T e − rt Φ( X t , e rt ¯ Y xt ) dt < ∞ P x -a.s. ( ¯ Y xt = ¯ Y xT + R Tt e − rs Φ( X s , e rs ¯ Y xs ) ds − R Tt d ¯ M xs , t ≥ , ¯ Y xT → P x -a.s. as T → ∞ . (3.9) Remark 3.2.
Let ( ¯ Y x , ¯ M x , ¯ K x ) be a solution of (3.8). Then for every t ∈ T , E x ¯ Y x ≥ E x Y xτ ≥ E x e − rτ ψ ( X τ ) . To see this, consider a localizing sequence { τ n } for ¯ M x . Since¯ Y xt = ¯ Y x − Z t d ¯ K xs + Z t d ¯ M xs , t ≥ , we have E x ¯ Y x ≥ lim inf n →∞ E x Y xτ ∧ τ n . Applying Fatou’s lemma yields the desiredinequalities. Proposition 3.3.
Assume that ψ satisfies (A1) and (3.4) for some x ∈ D . Then thereis at most one solution of (3.8) . Similarly, there is at most one solution of (3.7) .Proof. Suppose that ( ¯ Y i , ¯ K i , ¯ M i ), i = 1 ,
2, are solutions of (3.8). Write ¯ Y = ¯ Y − ¯ Y ,¯ K = ¯ K − ¯ K , ¯ M = ¯ M − ¯ M . Then¯ Y t = ¯ Y − Z t d ¯ K s + Z t d ¯ M s , t ≥ . By the Meyer-Tanaka formula (see, e.g., [18, Theorem IV.68]),¯ Y + t ≤ ¯ Y + T + Z Tt { ¯ Y s > ¯ Y s } d ¯ K s − Z Tt { ¯ Y s > ¯ Y s } d ¯ M s . (3.10)Since e − rt ψ ( X t ) L t ≤ ¯ Y t ∧ ¯ Y t ≤ ¯ Y t , we have Z Tt { ¯ Y s > ¯ Y s } d ¯ K s = Z Tt { ¯ Y s > ¯ Y s } ( ¯ Y s − ¯ Y s ) − ( ¯ Y s − ¯ Y s ∧ ¯ Y s ) d ¯ K s ≤ .
6y the above inequality and (3.10), E x ¯ Y + t ≤ E x ¯ Y + T . Since E x ¯ Y + T → T → ∞ , wesee that ¯ Y + t = 0, t ≥ P x -a.s. In the same way we show that ( − ¯ Y t ) + = 0, t ≥ P x -a.s. Thus ¯ Y = ¯ Y . That ¯ M = ¯ M and ¯ K = ¯ K now follows from uniqueness ofthe Doob-Meyer decomposition of ¯ Y .The proof of the second assertion is similar. Suppose that ( ¯ Y , ¯ M ), ( ¯ Y , ¯ M ) aresolutions of (3.7). Let ¯ Y = ¯ Y − ¯ Y , ¯ M = ¯ M − ¯ M . Applying the Meyer-Tanakaformula yields¯ Y + t ≤ ¯ Y + T + Z Tt { ¯ Y s > ¯ Y s } e − rs (Ψ( X s , e rs ¯ Y s ) − Ψ( X s , e rs ¯ Y s )) ds − Z Tt { ¯ Y s > ¯ Y s } d ¯ M s . But (Φ( x, y ) − Φ( x, y ))( y − y )= Ψ − ( x )( ( −∞ ,ψ ( x )] ( y ) − ( −∞ ,ψ ( x )] ( y ))( y − y ) ≤ , (3.11)so ¯ Y + t ≤ ¯ Y + T − R Tt { ¯ Y s > ¯ Y s } d ¯ M s . To prove that ¯ Y = ¯ Y and ¯ M = ¯ M it suffices nowto repeat the argument from the proof of the first assertion.By (3.3) with T = n ,¯ Y nt = e − rn ψ ( X n ) + Z nt e − rs Φ( X s , e rs ¯ Y ns ) ds − Z nt d ¯ M ns , t ∈ [0 , n ] . (3.12)We put ˜ Y nt = ¯ Y nt , ˜ M nt = ¯ M nt , t < n, ˜ Y nt = 0 , ˜ M nt = ¯ M nn , t ≥ n. The proof of the following theorem is a modification of the proof of [14, Propositions4.1, 4.2].
Theorem 3.4.
Assume that ψ satisfies (A1) and (3.4) for some x ∈ D . Then thereexists a unique solution ( ¯ Y x , ¯ M x ) of (3.7) on (Ω , F , P x ) . Moreover, E x Z ∞ e − rt Φ( X t , e rt ¯ Y xt ) dt ≤ E x Z ∞ e − rt Ψ − ( X t ) dt, (3.13)lim n →∞ k ¯ Y n − ¯ Y x k x, = 0 (3.14) and for every q ∈ (0 , , lim n →∞ E x sup t ≥ | ¯ Y nt − ¯ Y xt | q = 0 . (3.15) Proof.
Uniqueness follows from Proposition 3.3. The proof of the existence and (3.13)–(3.15) we divide into two steps.Step 1. We shall prove some a priori estimates for the process ¯ Y n and the difference δ ˜ Y := ˜ Y m − ˜ Y n . Specifically, we shall prove that k δ ˜ Y k x, ≤ E x (cid:16) e − rm ψ ( X m ) + e − rn ψ ( X n ) + Z mn e − rt Ψ − ( X t ) dt (cid:17) , (3.16) E x sup t ≥ | δ ˜ Y t | q ≤ − q E x (cid:16) e − rm ψ ( X m ) + e − rn ψ ( X n )) + Z mn e − rt Ψ − ( X t ) dt (cid:17) q (3.17)7or every q ∈ (0 , t ≥ E x Z t e − rs Φ( X s , e rs ¯ Y ns ) ds ≤ E x (cid:16) ¯ Y nt + 2 Z t e − rs Ψ − ( X s ) ds (cid:17) . (3.18)By (3.12),¯ Y nt = ¯ Y n − Z t [0 ,n ] ( s ) e − rs Φ( X s , e rs ¯ Y ns ) ds + Z t [0 ,n ] ( s ) d ¯ M ns , t ∈ [0 , n ] . (3.19)Moreover,˜ Y nt = ˜ Y n − Z t [0 ,n ] ( s ) e − rs Φ( X s , e rs ˜ Y ns ) ds + Z t dV ns + Z t [0 ,n ] ( s ) d ˜ M ns , t ≥ , where V nt = 0 , t < n, V nt = − ¯ Y nn , t ≥ n. Hence δ ˜ Y t = δ ˜ Y + R t + Z t ( [0 ,m ] ( s ) d ˜ M ms − [0 ,n ] ( s ) d ˜ M ns ) , t ≥ R t = − Z t [0 ,n ] ( s ) e − rs (Φ( X s , e rs ˜ Y ms ) − Φ( X s , e rs ˜ Y ns )) ds − Z t ( n,m ] ( s ) e − rs Φ( X s , e rs ˜ Y ms ) ds + Z t d ( V ms − V ns ) . By the Meyer-Tanaka formula, for t < m we have | δ ˜ Y m | − | δ ˜ Y t | ≥ Z mt sign( δ ˜ Y s − ) d ( δ ˜ Y ) s , where sign( x ) = 1 if x > x ) = − x ≤
0. Therefore, for t < m , | δ ˜ Y t | = E x ( | δ ˜ Y t | |F t ) ≤ E x (cid:16) | δ ˜ Y m | − Z mt sign( δ ˜ Y s − ) dR s (cid:12)(cid:12) F t (cid:17) . From this it follows that for t ∈ [0 , m ], | δ ˜ Y t | ≤ E x (cid:16) | δ ˜ Y m | + Z mt [0 ,n ] ( s ) e − rs sign( δ ˜ Y s )(Φ( X s , e rs ˜ Y ms ) − Φ( X s , e rs ˜ Y ns )) ds + Z mt ( n,m ] ( s ) e − rs sign( δ ˜ Y s )Φ( X s , e rs ˜ Y ms ) ds + | V mm | + | V nn | (cid:12)(cid:12) F t (cid:17) . By (3.11), Z mt [0 ,n ] ( s ) e − rs sign( δ ˜ Y s )(Φ( X s , e rs ˜ Y ms ) − Φ( X s , e rs ˜ Y ns )) ds ≤ . (3.20)Furthermore, since ˜ Y nt = 0 for t ≥ n , it follows from (3.11) that Z mt ( n,m ] ( s ) e − rs sign( δ ˜ Y s )Φ( X s , e rs ˜ Y ms ) ds ≤ Z mt ( n,m ] ( s ) e − rs sign( δ ˜ Y s )Ψ − ( X s ) ds ≤ Z mn e − rs Ψ − ( X s ) ds. δ ˜ Y m = 0 and | V mm | + | V nn | = | ¯ Y mm | + | ¯ Y nn | = e − rm ψ ( X m ) + e − rn ψ ( X n ).Therefore, for t ∈ [0 , m ] we have | δ ˜ Y t | ≤ E x (cid:16) e − rm ψ ( X m ) + e − rn ψ ( X n ) + Z mn e − rs Ψ − ( X s ) ds (cid:12)(cid:12) F t (cid:17) =: N t , from which (3.16) follows. By the above inequality and [2, Lemma 6.1], E x sup ≤ t ≤ m | δ ˜ Y t | q ≤ (1 − q ) − ( E x N m ) q , which shows (3.17). To prove (3.18), we first observe that by the Meyer-Tanaka formula, E x | ¯ Y nt | − E x | ¯ Y n | ≥ E x Z t sign( ¯ Y ns − ) d ¯ Y ns . By the above inequality and (3.19), for t < n we have E x | ¯ Y nt | − E x | ¯ Y n | ≥ − E x Z t [0 ,n ] ( s )sign( ¯ Y ns ) e − rs Φ( X s , e rs ¯ Y ns ) ds. On the other hand, for every t ≥ Z t e − rs Φ( X s , e rs ¯ Y ns ) ≤ Z t e − rs | Φ( X s , e rs ¯ Y ns ) − Φ( X s , | ds + Z t e − rs Φ( X s , ds = − Z t sign( ¯ Y ns ) e − rs (Φ( X s , e rs ¯ Y ns ) − Φ( X s , ds + Z t e − rs Φ( X s , ds ≤ − Z t sign( ¯ Y ns ) e − rs Φ( X s , e rs ¯ Y ns ) ds + 2 Z t e − rs Ψ − ( X s ) ds, which when combined with (3.20) proves (3.18).Step 2. We will prove the existence of a solution of (3.7) and (3.14), (3.15). From(3.4) and (3.16) it follows that k ¯ Y n − ¯ Y m k x, → n, m → ∞ . Hence there existsa process Y x ∈ L ( P x ) of class D such that (3.14) is satisfied. By (3.4) and (3.16),lim n,m →∞ E x sup t ≥ | ¯ Y nt − ¯ Y mt | q →
0. Since the space D q ( P x ) is complete, the lastconvergence and (3.14) imply that ¯ Y x ∈ D q ( P x ) and (3.15) is satisfied. By (2.4) and(3.2), ¯ Y nt ≤ ¯ Y n +1 t , t ≥ P x -a.s. By this and (3.15),lim n →∞ { e rt ¯ Y nt ≤ ψ ( X s ) } = { e rt ¯ Y t ≤ ψ ( X s ) } , t ≥ , P x -a.s.Hence lim n →∞ Φ( X t , e rt ¯ Y nt ) = Φ( X t , e rt ¯ Y t ) , t ≥ , P x -a.s. , (3.21)so applying Fatou’s lemma we conclude from (3.18) that for every T > E x Z T e − rt Φ( X t , e rt ¯ Y xt ) dt ≤ E x (cid:16) ¯ Y xT + 2 Z T e − rt Ψ − ( X t ) dt (cid:17) . (3.22)From (3.15) it follows that ¯ Y xT → P x as T → ∞ . As a consequence,since ¯ Y x is of class D, E x ¯ Y xT →
0. Letting T → ∞ in (3.22), we therefore get (3.13).By (3.12), ¯ Y nt = ¯ Y nT + Z Tt e − rs Φ( X s , e rs ¯ Y ns ) ds − Z Tt d ¯ M ns , t < T ≤ n. M n is a martingale, it follows that¯ Y nt = E x (cid:16) ¯ Y nT + Z Tt e − rs Φ( X s , e rs ¯ Y ns ) ds (cid:12)(cid:12) F t (cid:17) , t < T ≤ n. (3.23)By Doob’s inequality and (3.14),lim n →∞ P x ( sup ≤ t ≤ T | E x ( ¯ Y nT − ¯ Y T |F t ) | > ε ) ≤ ε − lim n →∞ E x | ¯ Y nT − ¯ Y xT | = 0 . (3.24)By (3.4), (3.21) and the dominated convergence theorem,lim n →∞ E x Z T e − rs | Φ( X s , e rs ¯ Y ns ) − Φ( X s , | ds = 0 . (3.25)From (3.23)–(3.25) we deduce that¯ Y xt = E x (cid:16) ¯ Y xT + Z Tt e − rs Φ( X s , e rs ¯ Y s ) ds (cid:12)(cid:12) F t (cid:17) . Letting T → ∞ and using (3.13) and the fact that lim T →∞ E x ¯ Y T = 0 yields¯ Y xt = E x (cid:16) Z ∞ t e − rs Φ( X s , e rs ¯ Y xs ) ds (cid:12)(cid:12) F t (cid:17) . Let ¯ M x be a c`adl`ag version of the martingale t E x (cid:16) Z ∞ e − rs Φ( X s , e rs ¯ Y xs ) ds (cid:12)(cid:12) F t (cid:17) − ¯ Y . (3.26)One can check that ( ¯ Y x , ¯ M x ) is a solution of (3.7). Remark 3.5.
Since ¯ M x is a version of (3.26), it follows from (3.13) and (A2)(b) that itis a closed martingale. Hence (see, e.g., [18, Theorem I.12]), ¯ M x ∞ = lim t →∞ ¯ M xt exists P x -a.s. and ¯ M x is a martinagale on [0 , ∞ ]. Therefore (3.5) is satisfied P x -a.s. and E x ¯ M x ∞ = E x ¯ M x = 0. As a result, E x ¯ Y x = E x Z ∞ Φ( X t , e rt ¯ Y xt ) dt. (3.27) Corollary 3.6.
Let the assumption of Theorem 3.4 hold. (i) If ( ¯ Y x , ¯ M x ) is a solution of (3.7) , then ( ¯ Y x , ¯ K x , ¯ M x ) with ¯ K x defined by (3.6) isa solution of (3.5) . (ii) Conversely, if ( ¯ Y x , ¯ K x , ¯ M x ) is a solution of (3.5) , then ¯ K x admits the represen-tation (3.6) .Proof. To prove (i), we only have to show that ¯ Y x , ¯ K x have the properties formulatedin the second line of (3.8). By (3.15), ¯ Y xt ≥ ¯ L t , t ≥
0, since ¯ Y nt ≥ L t , t ∈ [0 , n ], forevery n ≥
1. Clearly ¯ K x = 0 and ¯ K x is continuous and increasing. Since we knowthat ¯ Y xt ≥ ¯ L t , t ≥
0, directly from the definition of Φ it follows that ¯ K x satisfies theminimality condition. Part (ii) follows from (i) and the first part of Proposition 3.3. Corollary 3.7.
Assume that (A1), (A2) are satisfied. Then V ( x ) = E x ¯ Y x , x ∈ D . Moreover, e rt ¯ Y xt = V ( X t ) , t ≥ , P x -a.s. for every x ∈ D . (ii) lim T →∞ V T ( t, x ) = V ( x ) for all t ≥ and x ∈ D . Moreover, for every x ∈ D , V ( x ) − V T (0 , x ) ≤ e (cid:16) e − rT ψ ( X T ) + Z ∞ T e − rt Ψ − ( X t ) dt (cid:17) , T > . (3.28) Proof.
By (2.4) and (3.1), V n (0 , x ) ≤ V ( x ), n ≥
1, whereas by (3.2) and Theorem 3.4, V n (0 , x ) = E x ¯ Y n ր E x ¯ Y x . Hence E x ¯ Y x ≤ V ( x ). On the other hand, by Remark 3.2, E x ¯ Y x ≥ V ( x ), which proves the first part of (i). From (2.2) and (2.4) it follows that V T ( t, x ) = V T − t (0 , x ), t ∈ [0 , T ], x ∈ D . By (3.2) and (3.14), lim T →∞ V T − t (0 , x ) =lim T →∞ E x ¯ Y T − t = E x ¯ Y x , which equals V ( x ). This proves the first part of (ii). By(3.15) and (3.17), for every q ∈ (0 , | V T (0 , x ) − V ( x ) | ≤ (1 − q ) − /q E x (cid:16) e − rT ψ ( X T ) + Z ∞ T e − rt Ψ − ( X t ) dt (cid:17) , T > . Letting q ↓ x ∈ D , e rt ¯ Y Tt = Y Tt = V T ( t, X t ) → V ( X t ) P x -a.s. as T → ∞ . On the other hand, by (3.14) again, e rt ¯ Y Tt → e rt ¯ Y xt P x -a.s.as T → ∞ . Hence e rt ¯ Y xt = V ( X t ) P x -a.s. for every t ≥
0, which proves the second partof (i) because the processes t e rt ¯ Y xt and V ( X ) are continuous. Remark 3.8. (i) From Corollary 3.6(ii) and Corollary 3.7(i) it follows that the solution( ¯ Y x , ¯ K x , ¯ M x ) of (3.7) has a version ( ¯ Y , ¯ K, ¯ M ) independent of x .(ii) The argument from the proof of [13, Proposition 5.6] shows that if ψ ( x ) > x ∈ D , then { x ∈ D : V ( x ) = ψ ( x ) } ⊂ { x ∈ D : ψ ( x ) > } . Therefore ¯ K can bewritten in the form K t = Z t e − rs Ψ − ( X s ) { V ( X s )= ψ ( X s ) , ψ ( X s ) > } ds, t ≥ . The value of “perpetual European option” with payoff function ψ is defined as V E ( x ) = lim T →∞ E x e − rt ψ ( X T ). Under the assumption (A2) it is equal to zero. There-fore the next result can be called the early exercise premium formula for perpetualAmerican options. This formula extends the corresponding formula for call option inone-dimensional model (see [10, (6.31)]). Corollary 3.9.
Assume that (A1), (A2) are satisfied. Then for every x ∈ D , V ( x ) = E x Z ∞ e − rt Ψ − ( X t ) { V ( X t )= ψ ( X t ) , ψ ( X t ) > } dt. (3.29) Proof.
Follows immediately from (3.27) and Corollary 3.7(i) and Remark 3.8(ii).
Lemma 3.10.
Assume (A1) . Then (i) D ∋ x V T ( x ) , D ∋ x V ( x ) are Lipschitz continuous with constant L . (ii) For all x ∈ D , T > and t ∈ [0 , T ] , V T ( t, x ) ≤ C (1 + | x | ) with C = max { L, ψ (0) } . roof. (i) For y ∈ D set ˜ X = ( ˜ X , . . . , ˜ X d ), where ˜ X i , i = 1 , . . . , d , is defined by (2.2)with x i replaced by y i . Let x, y ∈ D . By (2.4), | V T (0 , x ) − V T (0 , y ) | ≤ sup τ ∈T T E x e − rτ | ψ ( X τ ) − ψ ( ˜ X τ ) | ≤ LE x e − rτ | X τ − ˜ X τ | . Define N i as in (2.3). Since | X iτ − ˜ X iτ | ≤ | x i − y i | E x N i ,τ = | x i − y i | , it follows that | V T (0 , x ) − V T (0 , y ) | ≤ L | x − y | for all T >
0. This and Corollary 3.7 imply that wealso have | V ( x ) − V ( y ) | ≤ L | x − y | for x, y ∈ D .(ii) Since ψ ( x ) ≤ C (1 + | x | ), x ∈ D , for all T > t ∈ [0 , T ] we have V T ( t, x ) ≤ V T (0 , x ) ≤ C + C sup τ ∈T ,T E x e − rτ | X τ | . Since δ i ≥ i = 1 , . . . , d , for any τ ∈ T ,T wealso have | X τ | ≤ P di =1 X i e rτ N i ,τ . Since E x N i ,τ = 1, i = 1 , . . . , d , this proves (ii). Let ̺ ( x ) = (1+ | x | ) − γ with γ > (2+ d ) /
4. By an elementary calculation, R R d ̺ ( x ) dx < ∞ and R R d | x | ̺ ( x ) dx < ∞ . In particular, Z R d | ψ ( x ) | ̺ ( x ) dx < ∞ , Z R d | Ψ − ( x ) | ̺ ( x ) dx < ∞ if ψ satisfies (A1) and Ψ − ( x ) ≤ c (1 + | x | ) , x ∈ R d , (3.30)for some c >
0. Define L ̺ ( D ) = L ( D ; ̺ dx ) , H ̺ ( D ) = { u ∈ L ̺ ( D ) : d X j =1 σ ij x i u x j ∈ L ̺ , i = 1 , . . . , d } , and for φ, ϕ ∈ C ∞ ( D ) set B BS̺ ( φ, ϕ ) = d X i =1 Z D ( r − δ i ) x i ∂ x i φ ( x ) ϕ ( x ) ̺ ( x ) dx − d X i,j =1 Z D a ij ∂ x i φ ( x ) ∂ x j ( x i x j ϕ ( x ) ̺ ( x )) dx. One can check that B BS̺ ( φ, ϕ ) ≤ c k φ k H ̺ ( D ) k ϕ k H ̺ ( D ) for some c >
0. Therefore theform B BS̺ can be extended to a bilinear form on on H ̺ ( D ) × H ̺ ( D ), which we stilldenote by B BS̺ . For an open set U ⊂ R D , we define the spaces H ( U ), H ( U ) in theusual way. Definition.
We say that v ∈ H ̺ ( D ) is a variational solution of the semilinear problem L BS v = rv − Φ( · , v ) , v ≥ ψ (3.31)if v ( x ) ≥ ψ ( x ) for x ∈ D , Φ( · , v ) ∈ L ̺ ( D ) and the equation in (3.31) is satisfied in theweak sense, i.e. for every ϕ ∈ H ̺ ( D ), B BS̺ ( v, ϕ ) = ( rv − Φ( · , v ) , ϕ ) L ̺ ( D ) . (3.32)12 roposition 3.11. Assume that (A1), (A2) and (3.30) are satisfied. If v is a varia-tional solution of (3.31) then v ∈ H loc ( D ) . In particular, L BS v ( x ) = rv ( x ) − Φ( x, v ( x )) for a.e. x ∈ D. (3.33) Proof.
Fix a bounded open set U such that U ⊂ ¯ U ⊂ D . Let ξ ∈ C ∞ ( U ) and ϕ = ξ/̺ .Then ϕ ∈ H ̺ , so from (3.32) it follows that B BS ( v, ξ ) = ( rv − Φ( · , v ) , ξ ) L ( R d ; dx ) , where B BS is defined as B BS̺ but with ̺ = 1. Therefore v is a weak solution, in thespace H ( U ), of the problem L BS v = rv − Φ( · , v ) in U . Write e x = ( e x , . . . , e x d ) for x = ( x , . . . , x d ) ∈ R d , and then define ˜ v ( x ) = v ( e x ), ˜Φ( x ) = Φ( e x , ˜ v ( x )), ˜ U = { x ∈ R d : e x ∈ U } and ˜ L = d X i =1 ( r − δ i − a ii / ∂ x i + 12 d X i,j =1 a ij ∂ x i x j . One can check that ˜ v ∈ H ( ˜ U ) and ˜ v is a weak solution of the problem ˜ L ˜ v = r ˜ v − ˜Φ in˜ U . By [4, Theorem 1, Section 6.3], ˜ v ∈ H ( ˜ U ), from which it follows that v ∈ H ( U ).Because of arbitrariness of U , v ∈ H loc ( D ). The equality (3.33) now follows by astandard argument (see Remark (ii) following [4, Section 6.3, Theorem 1]). Theorem 3.12.
Assume that (A1), (A2) and (3.30) are satisfied. Then V is a varia-tional solution of (3.31) .Proof. Let W ̺ = { u ∈ L (0 , T ; H ̺ ) : u t ∈ L (0 , T ; H − ̺ ) } . In [12] it is proved that forevery T > V T ∈ W ̺ and V T is a variational solution of the Cauchy problem ∂ t V T + L BS V T = rV T − Φ( · , V T ) , V T ( T, · ) = ψ, (3.34)i.e. V T ≥ ψ and (3.34) is satisfied in the weak sense. In particular, for any η ∈ C ∞ ((0 , T ) × D ) we have Z T h ∂ t V T ( t ) , η ( t ) i dt + Z T B BS̺ ( V T ( t ) , η ( t )) dt = Z T ( rV T ( t ) − Φ( · , V T ( t )) , η ( t )) L ̺ dt, where V t ( t ) = V T ( t, · ), η ( t ) = η ( t, · ) and h· , ·i denotes the duality pairing between L (0 , T ; H − ̺ ) and L (0 , T ; H ̺ ). From this one can deduce that for every ϕ ∈ C ∞ ( D ), Z h ∂ t V T ( t ) , ϕ i dt + Z B BS̺ ( V T ( t ) , ϕ ) dt = Z ( rV T ( t ) − Φ( · , V T ( t )) , ϕ ) L ̺ dt. (3.35)By Corollary 3.7(ii), for every x ∈ D , V T (0 , x ) → V ( x ) and V T (1 , x ) → V ( x ), soapplying the dominated convergence theorem we getlim T →∞ Z h ∂ t V T ( t ) , ϕ i dt = lim T →∞ ( V T (1) − V T (0) , ϕ ) L ̺ = 0 . (3.36)Suppose that supp[ ϕ ] ⊂ U for some relatively compact open set U ⊂ D . By Lemma3.10, | ∂ x i V T | ≤ L a.e. for all i = 1 , . . . , d and T >
0, and V T are bounded on (0 , × U T >
0. By this and Corollary 3.7(ii), V T → V weakly in L (0 , H ( U )).Therefore lim T →∞ Z B BS̺ ( V T ( t ) , ϕ ) dt = B BS̺ ( V, ϕ ) . (3.37)Since V T ≤ V T ′ if T ≤ T ′ , in fact V T ր V as T → ∞ . Therefore Φ( · , V T ) → Φ( · , V )pointwise, so applying the dominated convergence theorem we get (jakies ograniczeniana Ψ − ). lim T →∞ Z ( rV T ( t ) − Φ( · , V T ( t )) , ϕ ) L ̺ dt = ( rV − Φ( · , V ) , ϕ ) L ̺ . (3.38)From (3.35)–(3.38) it follows that V satisfies (3.32) for ϕ ∈ C ∞ c ( D ), and hence for ϕ ∈ H ̺ by an approximation argument. Clearly V ≥ ψ , so V is a solution of (3.31).Before stating the uniqueness result, we note that under the assumptions on ψ and δ , . . . , δ d stated in Remark 3.1(ii), e − rt P t V ( x ) → → ∞ . Proposition 3.13.
Under the assumptions of Theorem 3.12 there is at most one vari-ational solution v of (3.31) such that lim t →∞ e − rt P t v ( x ) = 0 for every x ∈ D .Proof. Let v , v be two solutions of (3.31) such that lim t →∞ e − rt P t v k ( x ) → x ∈ D , k = 1 ,
2, and let v = v − v . Define ˜ L as in Proposition 3.11 and set ˜ v ( x ) = v ( e x ).Then v ( X ) = ˜ v ( Z ), where Z = ( Z , . . . , Z d ), Z it = ln x i +( r − δ i − a ii / t + P dj =1 σ ij B j ,t , t ≥
0. Choose an increasing sequence { U n } of bounded open sets such that ¯ U n ⊂ U n +1 and S n ≥ U n = D and set τ n = inf { t > X t / ∈ U n } = inf { t > Z t / ∈ ˜ U n } , where˜ U n = { x ∈ R d : e x ∈ U } . Since ˜ v ∈ H ( ˜ U n ), by the extension of Itˆo’s formula provedby Krylov (see [15, Chapter II, §
10, Theorem 1]) we have˜ v ( Z t ∧ τ n ) = ˜ v ( Z ) + d X i,j =1 Z t ∧ τ n ∂ x i ˜ v ( Z s ) σ ij dB j ,s + Z t ∧ τ n ˜ L ˜ v ( Z s ) ds, t ≥ . Define Y t = v ( X t ), t ≥
0. Since v ( X ) = ˜ v ( Z ), it follows that Y t ∧ τ n = Y + d X i,j =1 Z t ∧ τ n L BS v ( X s ) ds + R t ∧ τ n , t ≥ , (3.39)where R t = P di,j =1 R t σ ij X is ∂ x i v ( X s ) dB j ,s . Since P x ( X t ∈ D, t ≥
0) = 1, τ n → ∞ P x -a.s. as n → ∞ . Therefore letting n → ∞ in (3.39) shows that it holds true with t ∧ τ n replaced by t . Let ¯ Y t = e − rt Y t . Integrating by parts we obtain¯ Y t = ¯ Y + Z t ( − re − rs Y s ds + Z t e − rs dY s = ¯ Y + Z t e − rs ( − rv + L BS v )( X s ) ds + Z t e − rs dR s = ¯ Y − Z t e − rs (Φ( X s , v ( X s )) − Φ( X s , v ( X s )) ds + Z t e − rs dR s . Repeating now the argument from the proof of Proposition 3.3 we show that E x ¯ Y +0 ≤ E x ¯ Y + t , t ≥
0. In much the same way we show that E x ¯ Y − ≤ E x ¯ Y − t , t ≥
0. Hence E x | ¯ Y | ≤ E x | ¯ Y t | = e − rt E x | v ( X t ) | = e − rt P t | v | ( x ), which converges to zero as t → ∞ .Thus | v ( x ) | = E x ¯ Y = 0.Note that in case of American call and American put on single asset explicit formulasfor the solution of (3.31) are known (see, e.g., [8, 10, 16, 19]).14 Examples
Below we give examples of payoff functions satisfying (A1), (A2) and (3.30). In all theexamples Ψ − is computed in the subset D ∩ { ψ > } (see Remark 3.8(ii)). Example 4.1.
Let d = 1. ψ ( x ) = ( x − K ) + , Ψ − ( x ) = ( δx − rK ) + (call) ψ ( x ) = ( K − x ) + , Ψ − ( x ) = ( rK − δx ) + (put)The assumptions (A1) and (A2) are satisfied if r > r > , δ > V ( x ) − V T (0 , x ) ≤ e (cid:16) Ke − rT + rK Z ∞ T e − rt dt (cid:17) = 2 eKe − rT , x > K. For call option, V ( x ) − V T (0 , x ) ≤ eKe − δT , T > x ∈ (0 , K ). Example 4.2.
In the examples below d ≥
2. In all the cases where ψ is bounded, (A1)and (A2) are satisfied if r >
0. In the other cases they are satisfied if r > δ i > i = 1 , . . . , d .1. Index options and spread options. ψ ( x ) = (cid:0) d X i =1 w i x i − K (cid:1) + , Ψ − ( x ) = (cid:0) d X i =1 w i δ i x i − rK (cid:1) + (call) ψ ( x ) = (cid:0) K − d X i =1 w i x i (cid:1) + , Ψ − ( x ) = (cid:0) rK − d X i =1 w i δ i x i (cid:1) + (put)2. Call on max option. ψ ( x ) = (max { x , . . . , x d } − K ) + , Ψ − ( x ) = (cid:0) d X i =1 δ i B i ( x ) x i − rK (cid:1) + , where B i = { x ∈ R d : x i > x j , j = i } .3. Put on min option. ψ ( x ) = ( K − min { x , . . . , x d } ) + , Ψ − ( x ) = (cid:0) rK − d X i =1 δ i C i ( x ) x i (cid:1) + , where C i = { x ∈ R d : x i < x j , j = i } .4. Multiple strike options. ψ ( x ) = (max { x − K , . . . , x d − K d } ) + , Ψ − ( x ) = (cid:0) d X i =1 B i ( x − K )( δ i x i − rK i ) (cid:1) + with K = ( K , . . . , K d ) . cknowledgements This work was supported by the Polish National Science Centre under Grant2016/23/B/ST1/01543).
References [1] G. Alberti and L. Ambrosio,
A geometrical approach to monotone functions in R n ,Math. Z. (1999) 259–316.[2] P. Briand, B. Delyon, Y. Hu, Y., E. Pardoux, and L. Stoica, L p solutions ofbackward stochastic differential equations , Stochastic Process. Appl. (2003)109–129.[3] C. Dellacherie and P.-A. Meyer, Probabilities and Potential B. Theory of Martin-gales . North-Holland Publishing Co., Amsterdam, 1982.[4] L.C. Evans,
Partial Differential Equations . American Mathematical Society, Prov-idence, RI 1998.[5] N. El Karoui, C. Kapoudjian, E. Pardoux, S. Peng, and M.C. Quenez,
Reflectedsolutions of backward SDEs, and related obstacle problems for PDE’s , Ann. Probab. (1997) 702–737.[6] N. El Karoui, E. Pardoux, and M.C. Quenez, Reflected backward SDEs and Amer-ican options , Numerical methods in finance, Publ. Newton Inst., 13, CambridgeUniv. Press, Cambridge, 1997, pp. 215–231[7] N. El Karoui and M.C. Quenez,
Non-linear pricing theory and backward stochasticdifferential equations , Lecture Notes in Math. , Springer, Berlin, 1997.[8] S. Jacka,
Optimal stopping and the American put , Math. Finance (2) (1991) 1–14.[9] I. Karatzas, On the pricing of American options , Appl. Math. Optim. (1988)37-60.[10] I. Karatzas and S.E. Shreve, Methods of Mathematical Finance , Springer, NewYork, 1998.[11] T. Klimsiak and A. Rozkosz,
On backward stochastic differential equations ap-proach to valuation of American options , Bull. Polish Acad. Sci. Math. (2011)275–288.[12] T. Klimsiak and A. Rozkosz, The early exercise premium representation for Amer-ican options on multiply assets , Appl. Math. Optim. (2016) 99–114.[13] T. Klimsiak and A. Rozkosz, The valuation of American options in a multidimen-sional exponential L´evy model , Math. Finance (2018) 1107–1142.[14] T. Klimsiak and A. Rozkosz, Large time behaviour of solutions to parabolic equa-tions with Dirichlet operators and nonlinear dependence on measure data , PotentialAnal. DOI: 10.1007/s11118-018-9711-9.1615] N.V. Krylov,
Controlled Diffusion Processes , Springer, New York, 1980.[16] H.P. McKean, Jr.,
A free-boundary problem for the heat equation arising from aproblem in mathematical economics , Industr. Manag. Rev. (1965) 32–39.[17] R.C. Merton, Theory of rational option pricing , Bell J. Econom. and ManagementSci. (1973) 141–183.[18] P. Protter, Stochastic Integration and Differential Equations . Second Edition.Springer, Berlin, 2004.[19] S.E. Shreve,