Valuation of American Strangle Option: Variational Inequality Approach

DISCRETE AND CONTINUOUS doi:10.3934/dcdsb.2018206 DYNAMICAL SYSTEMS SERIES B Volume 24, Number 2, February 2019 pp. 755–781

VALUATION OF AMERICAN STRANGLE OPTION: VARIATIONAL INEQUALITY APPROACH

Junkee Jeon Department of Mathematical Sciences Seoul National University Seoul 08826, Republic of Korea Jehan Oh∗ FakultätfürMathematik UniversitätBielefeld Postfach 100131, D-33501 Bielefeld, Germany

(Communicated by Bei Hu)

Abstract. In this paper, we investigate a parabolic variational inequality problem associated with the American strangle option pricing. We obtain the 2,1 existence and uniqueness of Wp,loc solution to the problem. Also, we analyze the smoothness and monotonicity of two free boundaries. Finally, numerical results of the model based on this problem are described and used to show the boundary properties and the price behavior.

1. Introduction. Various strategies have been studied to reduce the risk of options since the financial crisis. According to Chaput and Ederington [1], strangle and straddle account for more than 80% of options strategies. A strangle option is a strategy that holds a position at the same time in both a call and a put with different strike prices but with the same expiry. If we are expecting large movements in underlying assets, but are not sure which direction the movement will be, we can buy or sell them to reduce the risk exposed by a European call or put option. In particular, a straddle option is one of strangle options where the strike price of the call portion is the same as the strike price of the put portion. Here we focus on American strangle options. The definition of an American type option is an option contract that allows option holders to exercise their rights at any time before expiry. Since option holders in the American option can exercise their rights at any time before expiry, the pricing of such options is often categorized as the optimal stopping problem or the free boundary problem. In this paper, we

2010 Mathematics Subject Classiﬁcation. Primary: 35R35; Secondary: 91G80. Key words and phrases. American strangle options, free boundary problem, option pricing, variational inequality. The ﬁrst author gratefully acknowledges the support of the National Research Foundation of Korea grant funded by the Korea government (Grant No. NRF-2017R1C1B1001811), BK21 PLUS SNU Mathematical Sciences Division and the POSCO Science Fellowship of POSCO TJ Park Foundation. ∗ Corresponding author: Jehan Oh.

755 756 JUNKEE JEON AND JEHAN OH study the parabolic variational inequality associated with the model of American strangle options pricing. In other words, we will investigate V (t, s) satisfying

 2 σ 2  ∂tV + s ∂ssV + (r − q)s∂sV − rV = 0,  2  + +  if V > (s − K2) + (K1 − s) , (t, s) ∈ (0,T ] × (0, +∞),   σ2 ∂ V + s2∂ V + (r − q)s∂ V − rV ≤ 0, (1.1)  t 2 ss s  + +  if V = (s − K2) + (K1 − s) , (t, s) ∈ (0,T ] × (0, +∞),   + +  V (T, s) = (s − K2) + (K1 − s) , s ∈ [0, +∞), where r, σ, K1,K2 are positive constants with K1 < K2 and q is a constant with q ≥ 0. In AppendixA, we present the formulation and the financial background of the problem (1.1). There are various studies on the American strangle option. Chiarella and Ziogas derived the integral equation satisfying the American strangle option in [3] by using the incomplete Fourier transform method. Qiu [10] gave an alternative method to derive the EEP representation of the American strangle option value and analyzed the properties of the option value and the early exercise boundary. Ma et al. [9] construct tight lower and upper bounds for the price of an American strangle. In addition, there are a variety of studies on parabolic variational inequality arising in option pricing. Yang et al. [12], [15] considered parabolic variational inequalities associated with European-style installment call or put option pricing and obtained 2,1 the existence and uniqueness of Wp,loc solution to the problem and the monotonicity, smoothness and boundedness properties of free boundaries. Also, Chen et al. [2] proved existence and uniqueness of weak solution in variational inequality in the case of American lookback option with fixed strike price. However, the parabolic variational inequalities in the above researches, have only one free boundary. Of course, Yang and Yi [13] already considered a parabolic variational inequality problem associated with the American-style continuous-installment options with two free boundaries, the lower obstacle of the variational inequality is a monotone function in spatial variables. In the present paper, variational inequality with two free boundaries does not have monotonicity condition on the lower obstacle function in (1.1). The novelty of this paper is that we analyze more general case of this problem. The contributions of the paper are threefold: (i) we prove the existence and 2,1 uniqueness of Wp,loc ((0,T ] × (0, +∞)) ∩ C([0,T ] × (0, +∞)) solution to the parabolic variational inequality (1.1). (ii) We prove that two free boundaries (1.1) are monotone and C∞-regular. (iii) We prove the existence and uniqueness of 2 Wp,loc ((0, +∞)) solution to the stationary problem of (1.1) and utilize it to show that the free boundaries are bounded. The rest of this paper is organized as follows. In section2, we prove the exis- 2,1 tence and uniqueness of Wp,loc solution to problem (1.1). In section3, we show that the monotonicity and C∞-regularity of two free boundaries based on the results in section2. Moreover, we will prove the starting points of the free boundaries. In section4, we conduct comparative static analysis of variational inequality (1.1). In section5, we solve the stationary problem arising from American strangle option and use it to show that two free boundaries are bounded. In section6, we describe the numerical result applying the finite difference scheme. AppendixA is AMERICAN STRANGLE OPTION 757 the formulation of the model. AppendixB shows that the unique solution to the problem (1.1) coincides with the expected value of the American strangle option.

2. Existence and uniqueness of a solution. We ﬁrst transform the degenerate backward parabolic problem (1.1) into a familiar forward non-degenerate parabolic problem. Setting s V (t, s) τ = T − t, x = ln ,Y (τ, x) = , K2 K2 we have  ∂ Y − LY = 0, if Y > (ex − 1)+ + (κ − ex)+, (τ, x) ∈ [0,T ) × ,  τ R  x + x + ∂τ Y − LY ≥ 0, if Y = (e − 1) + (κ − e ) , (τ, x) ∈ [0,T ) × R, (2.1)  x + x +  Y (0, x) = (e − 1) + (κ − e ) , x ∈ R, where K κ := 1 ∈ (0, 1) K2 and σ2 σ2 LY := ∂ Y + r − q − ∂ Y − rY. (2.2) 2 xx 2 x We now consider the problem in the bounded domain [0,T ) × (−n, n): x + x +  ∂τ Yn − LYn = 0, if Yn > (e − 1) + (κ − e ) , (τ, x) ∈ [0,T ) × (−n, n),   x + x +  ∂τ Yn − LYn ≥ 0, if Yn = (e − 1) + (κ − e ) , (τ, x) ∈ [0,T ) × (−n, n), −n n (2.3) ∂xYn(τ, −n) = −e , ∂xYn(τ, n) = e , τ ∈ [0,T ],   x + x +  Yn(0, x) = (e − 1) + (κ − e ) , x ∈ [−n, n],

2 where n ∈ with n > ln . N κ The lower obstacle and terminal condition function of variational inequality (2.1) are not monotonic for spatial variable x. Thus, we appropriately transform the value function to have monotonicity. The following lemma provides the existence, uniqueness and properties of a solution to the above problem. 2 Lemma 2.1. For each ﬁxed n ∈ with n > ln , there exists a unique solution N κ 2,1 Yn ∈ C ([0,T ] × [−n, n]) ∩ Wp (([0,T ) × (−n, n)) \ (Bρ(0, 0) ∪ Bρ(0, ln κ))) to the 2 2 problem (2.3), where 1 < p < ∞, ρ > 0 and Bρ(0, x0) = (τ, x): τ + (x − x0) ≤ 2 ρ . Furthermore, if n ∈ N is large enough, then we have x + x + x (e − 1) + (κ − e ) ≤ Yn ≤ e + κ, (2.4)

x x ∂τ Yn ≥ 0, −e ≤ ∂xYn ≤ e . (2.5) ∞ Proof. We ﬁrst deﬁne a penalty function βε ∈ C (R) (0 < ε < 1) satisfying  0 00 βε(t) ≤ 0, βε(t) ≥ 0, βε (t) ≤ 0, ∀t ∈ R;   n βε(t) = 0 if t ≥ ε; βε(0) = −C0 for C0 = 3(r + q)e + 3r; (2.6)  lim βε(t) = 0 if t > 0; lim βε(t) = −∞ if t < 0.  ε→0 ε→0 758 JUNKEE JEON AND JEHAN OH

Since the functions (ex − 1)+ and (κ − ex)+ are not smooth enough, we also deﬁne ∞ a function ϕε ∈ C (R) satisfying

 0 00 ϕε(t) ≥ 0, 0 ≤ ϕε(t) ≤ 1, ϕε (t) ≥ 0, ∀t ∈ R;   ϕε(t) = t if t ≥ ε; ϕε(t) = 0 if t ≤ −ε; (2.7)  +  lim ϕε(t) = t , ∀t ∈ R.  ε→0

We then consider the following approximation of the problem (2.3):

x x  ∂τ Yn,ε − LYn,ε + βε (Yn,ε − ϕε(e − 1) − ϕε(κ − e )) = 0,    (τ, x) ∈ [0,T ) × (−n, n), −n n (2.8) ∂xYn,ε(τ, −n) = −e , ∂xYn,ε(τ, n) = e , τ ∈ [0,T ],   x x  Yn,ε(0, x) = ϕε(e − 1) + ϕε(κ − e ), x ∈ [−n, n].

2,1 By Schauder’s ﬁxed point theorem, the above problem (2.8) has a unique Wp solution, see [14]. We next claim that

x x x ϕε(e − 1) + ϕε(κ − e ) ≤ Yn,ε ≤ e + κ (2.9) for suﬃciently large n. Observe from (2.7) that

+ + t ≤ ϕε(t) ≤ (t + ε) , ∀t ∈ R. (2.10) Then we deduce from (2.7) and (2.10) that

x x x x ∂τ [ϕε(e − 1) + ϕε(κ − e )] − L [ϕε(e − 1) + ϕε(κ − e )] x x x x + βε (ϕε(e − 1) + ϕε(κ − e ) − ϕε(e − 1) − ϕε(κ − e )) x x = −L [ϕε(e − 1) + ϕε(κ − e )] + βε(0) σ2 = − [ϕ00(ex − 1) + ϕ00(κ − ex)] e2x + (q − r)[ϕ0 (ex − 1) − ϕ0 (κ − ex)] ex 2 ε ε ε ε x x + r [ϕε(e − 1) + ϕε(κ − e )] − C0 x x + x + ≤ 2(q + r)e + r (e − 1 + ε) + (κ − e + ε) − C0 x x ≤ 2(q + r)e + r [(e + ε) + (κ + ε)] − C0 x n ≤ (2q + 3r)e + 3r − C0 ≤ 3(q + r)e + 3r − C0 = 0. κ Furthermore, we see from the boundary conditions in (2.8) that if 0 < ε < , 2  ϕ (ex − 1) + ϕ (κ − ex) = Y (τ, x), τ = 0,  ε ε n,ε  x x −n ∂x [ϕε(e − 1) + ϕε(κ − e )] = −e = ∂xYn,ε(τ, x), x = −n,  x x n  ∂x [ϕε(e − 1) + ϕε(κ − e )] = e = ∂xYn,ε(τ, x), x = n.

By the comparison principle, we get

x x ϕε(e − 1) + ϕε(κ − e ) ≤ Yn,ε. (2.11) AMERICAN STRANGLE OPTION 759

κ 1 On the other hand, it follows from (2.6) and (2.7) that if 0 < ε < < , 2 2 x x x x x ∂τ [e + κ] − L [e + κ] + βε (e + κ − ϕε(e − 1) − ϕε(κ − e )) x x x x = −L [e + κ] + βε (e + κ − ϕε(e − 1) − ϕε(κ − e )) x x x x = qe + rκ + βε (e − ϕε(e − 1) + κ − ϕε(κ − e )) x x ≥ qe + rκ + βε(1 − ε) = qe + rκ ≥ 0. Also, from the boundary conditions in (2.8), we get  ex + κ ≥ ϕ (ex − 1) + ϕ (κ − ex) = Y (τ, x), τ = 0,  ε ε n,ε  x −n −n ∂x [e + κ] = e ≥ −e = ∂xYn,ε(τ, x), x = −n,  x n  ∂x [e + κ] = e = ∂xYn,ε(τ, x), x = n. By the comparison principle, we obtain x e + κ ≥ Yn,ε. (2.12) Therefore, the claim (2.9) follows from (2.11) and (2.12). x x We see from (2.9) that −C0 ≤ βε (Yn,ε − ϕε(e − 1) − ϕε(κ − e )) ≤ 0. Employ- ing a Cα estimate (see [8, Theorem 6.33]), we get

||Yn,ε||Cα,α/2([0,T ]×[−n,n]) ≤ c for some constant c > 0 which is independent of ε. We then apply the method in [5] to deduce that 2,1 Yn,ε *Yn in Wp (([0,T ) × (−n, n)) \ (Bρ(0, 0) ∪ Bρ(0, ln κ))) and

Yn,ε → Yn in C ([0,T ] × [−n, n]) as ε → 0+, where Yn is the solution to the problem (2.3). Hence, (2.4) follows immediately from (2.9). Now let us prove (2.5). For any small δ > 0, we see from (2.3) and (2.4) that the function Yn(τ + δ, x) satisﬁes

 ∂τ Yn(τ + δ, x) − LYn(τ + δ, x) = 0,  x + x +  if Yn(τ + δ, x) > (e − 1) + (κ − e ) , (τ, x) ∈ [0,T − δ) × (−n, n)    ∂τ Yn(τ + δ, x) − LYn(τ + δ, x) ≥ 0,  x + x + if Yn(τ + δ, x) = (e − 1) + (κ − e ) , (τ, x) ∈ [0,T − δ) × (−n, n)  −n n  ∂xYn(τ + δ, −n) = −e , ∂xYn(τ + δ, n) = e , τ ∈ [0,T − δ],   x + x +  Yn(0 + δ, x) ≥ (e − 1) + (κ − e ) = Yn(0, x), x ∈ [−n, n].

Applying the monotonicity of solution of variational inequality with respect to initial value (see [5]), we obtain

Yn(τ + δ, x) ≥ Yn(τ, x), ∀(τ, x) ∈ [0,T − δ) × (−n, n), which yields the ﬁrst inequality in (2.5). For the second inequalities in (2.5), we diﬀerentiate (2.8) with respect to x, then we have 760 JUNKEE JEON AND JEHAN OH

 0 0 0 x 0 x x ∂τ Z − LZ + βε(··· )Z = βε(··· )[ϕε(e − 1) − ϕε(κ − e )] e ,    (τ, x) ∈ [0,T ) × (−n, n), (2.13) Z(τ, −n) = −e−n,Z(τ, n) = en, τ ∈ [0,T ],   0 x 0 x x  Z(0, x) = [ϕε(e − 1) − ϕε(κ − e )] e , x ∈ [−n, n],

0 0 x x where Z := ∂xYn,ε and βε(··· ) = βε (Yn,ε − ϕε(e − 1) − ϕε(κ − e )). Letting x Z1 := Z + e , we see from (2.13) that Z1 satisﬁes  ∂ Z − LZ + β0 (··· )Z  τ 1 1 ε 1  = −L [ex] + β0 (··· )[ϕ0 (ex − 1) − ϕ0 (κ − ex) + 1] ex, (τ, x) ∈ [0,T ) × (−n, n),  ε ε ε n Z1(τ, −n) = 0,Z1(τ, n) = 2e , τ ∈ [0,T ],   0 x 0 x x  Z1(0, x) = [ϕε(e − 1) − ϕε(κ − e ) + 1] e , x ∈ [−n, n]. Since x x 0 0 −L [e ] = qe , βε(··· ) ≥ 0, 0 ≤ ϕε(t) ≤ 1, we get  ∂ Z − LZ + β0 (··· )Z ≥ 0, (τ, x) ∈ [0,T ) × (−n, n),  τ 1 1 ε 1  Z1(τ, −n) = 0,Z1(τ, n) ≥ 0, τ ∈ [0,T ],   Z1(0, x) ≥ 0, x ∈ [−n, n].

Then we conclude from the maximum principle that Z1 ≥ 0 in [0,T ] × [−n, n], that x x is, ∂xYn,ε ≥ −e in [0,T ] × [−n, n]. Similarly, we set Z2 := Z − e and then deduce that  ∂ Z − LZ + β0 (··· )Z  τ 2 2 ε 2  = −qex + β0 (··· )[ϕ0 (ex − 1) − ϕ0 (κ − ex) − 1] ex ≤ 0, (τ, x) ∈ [0,T ) × (−n, n),  ε ε ε −n Z2(τ, −n) = −2e ≤ 0,Z2(τ, n) = 0, τ ∈ [0,T ],   0 x 0 x x  Z2(0, x) = [ϕε(e − 1) − ϕε(κ − e ) − 1] e ≤ 0, x ∈ [−n, n]. x It follows from the maximum principle that Z2 ≤ 0, and hence that ∂xYn,ε ≤ e . Thus we obtain (2.5). To prove the uniqueness, suppose that Yn and Yen are two solutions to the problem (2.3) and that the set n o N := (τ, x) ∈ [0,T ] × [−n, n]: Yn(τ, x) < Yen(τ, x) is non-empty. Observe from (2.4) that if (τ, x) ∈ N , x + x + Yen(τ, x) > (e − 1) + (κ − e ) , and hence ∂τ Yen − LYen = 0. Therefore, the function U := Yen − Yn satisfies  ∂ U − LU ≤ 0, in N ,  τ  ∂xU = 0, on ∂pN ∩ ([0,T ] × {−n} ∪ [0,T ] × {n})   U = 0 on ∂pN\ ([0,T ] × {−n} ∪ [0,T ] × {n}), AMERICAN STRANGLE OPTION 761 where ∂pN is the parabolic boundary of the domain N . Then it follows from the ABP maximum principle (see [11]) that U ≤ 0 in N , which contradicts the definition of the set N . Hence N = ∅ and Yn ≥ Yen. Similarly, we conclude that Yn ≤ Yen, and finally that Yn = Yen. Theorem 2.2. There exists a unique solution 2,1 Y ∈ C ([0,T ] × R) ∩ Wp (([0,T ) × (−R,R)) \ (Bρ(0, 0) ∪ Bρ(0, ln κ))) 2 to the problem (2.1) for all R > ln , ρ > 0 and 1 < p < ∞. Moreover, we have κ (ex − 1)+ + (κ − ex)+ ≤ Y ≤ ex + κ, (2.14)

x x ∂τ Y ≥ 0, −e ≤ ∂xY ≤ e . (2.15) 2,1 Proof. Since the solution Yn of the problem (2.3) belongs to Wp,loc ([0,T ) × (−n, n)), we can rewrite the problem (2.3) as  ∂ Y − LY = f(τ, x), in [0,T ) × (−n, n),  τ n n  −n n ∂xYn(τ, −n) = −e , ∂xYn(τ, n) = e , τ ∈ [0,T ], (2.16)  x + x +  Yn(0, x) = (e − 1) + (κ − e ) , x ∈ [−n, n], p where f ∈ Lloc([0,T ) × (−n, n)) and x x f(τ, x) = χ{Yn=e −1}(τ, x) · (qe − r) x x + χ{Yn=κ−e }(τ, x) · (−qe + rκ) a.e. in [0,T ) × (−n, n). (2.17)

Here, χA denotes the characteristic function of the set A. Then we see that |f(τ, x)| ≤ c(R) for (τ, x) ∈ [0,T ) × [−R,R], where the constant c(R) depends on R, but is independent of n. Therefore, it 2 follows from the W 2,1 estimates (see [8]) and (2.4) that for n > R > ln , p κ

||Y || 2,1 n Wp (([0,T )×(−R,R))\(Bρ(0,0)∪Bρ(0,ln κ))) x + ≤ c ||Yn||L∞([0,T )×(−R,R)) + (e − 1) C2([−R,−ρ]∪[ρ,R]) x + + (κ − e ) C2([−R,ln κ−ρ]∪[ln κ+ρ,R]) + ||f||L∞([0,T )×(−R,R)) ≤ c(R) for some constant c(R) which is independent of n. Letting n → ∞, we deduce that, up to a subsequence, R 2,1 Yn *Y in Wp (([0,T ) × (−R,R)) \ (Bρ(0, 0) ∪ Bρ(0, ln κ))) . In addition, we obtain from the Sobolev embedding theorem that R R Yn → Y in C([0,T ) × (−R,R)) and ∂xYn → ∂xY in C([0,T ) × (−R,R)). We now deﬁne Y := Y R if x ∈ [−R,R]. Then it is obvious that Y is well-deﬁned R R and that Y is the solution to the problem (2.1). Since Y , ∂xY ∈ C([0,T ) × α (−R,R)), we have Y, ∂xY ∈ C([0,T ) × R). Furthermore, the C estimate yields Y ∈ C([0,T ] × R) and the inequalities (2.14) and (2.15) follow from (2.4) and (2.5), respectively. The proof of the uniqueness is the same as that of Lemma 2.1. 762 JUNKEE JEON AND JEHAN OH

3. Analysis of the free boundaries. In this section we analyze the free boundaries of variational inequality (2.1). If q 6= 0, we will show that the variation inequality (2.1) has two free boundaries. Since the two free boundaries interact with each other, it is not easy to prove the monotonicity and smoothness of the free boundaries. Let D = [0,T ] × R be the whole region. We denote E = (τ, x) ∈ D : Y (τ, x) = (ex − 1)+ + (κ − ex)+ (the exercise region), C = (τ, x) ∈ D : Y (τ, x) > (ex − 1)+ + (κ − ex)+ (the continuation region). The exercise region E is a disjoint union of three subregions E0, EA, EB of E, where E0 = {(τ, x) ∈ D : Y (τ, x) = 0} , EA = (τ, x) ∈ D : Y (τ, x) = (κ − ex)+ > 0 , EB = (τ, x) ∈ D : Y (τ, x) = (ex − 1)+ > 0 . x x Theorem 2.2 shows that ∂−x [Y − (κ − e )] ≤ 0 and ∂x [Y − (e − 1)] ≤ 0. Hence we can define the free boundaries A(τ) = sup {x : x ≤ ln κ, Y (τ, x) = κ − ex} , τ > 0 B(τ) = inf {x : x ≥ 0,Y (τ, x) = ex − 1 > 0} , τ > 0. We note that the free boundary A(τ) separates EA from C and that the free boundary B(τ) separates EB from C. Let us denote A A(τ) B B(τ) S (τ) = e · K2 and S (τ) = e · K2. Then, the SA(τ) and SB(τ) are the free boundaries of the variational inequality (1.1). When the underlying asset hits below SA(τ) or above SB(τ), the option holder can benefit by exercising his/her rights. Also, the region surrounded by free boundaries SA(τ) and SB(τ) is the continuation region C of American strangle options. Lemma 3.1. For any τ ∈ (0,T ], Y (τ, ln κ) > 0 and Y (τ, 0) > 0. Accordingly, we have {0 < τ ≤ T, x = ln κ} , {0 < τ ≤ T, x = 0} ⊂ C. Proof. Let us first show that Y (τ, 0) > 0 for all τ ∈ (0,T ]. Suppose the assertion x + x + is false. Since Y (τ, 0) ≥ (e − 1) + (κ − e ) |x=0 = 0, it follows that there exists τ0 > 0 such that Y (τ0, 0) = 0. Then we deduce from Y (0, 0) = 0 and ∂τ Y ≥ 0 that Y (τ, 0) = 0 for all 0 ≤ τ ≤ τ0. By Theorem 2.2, we obtain that for x ≥ 0, x x x Y − (e − 1) ≥ 0 and ∂x(Y − (e − 1)) = ∂xY − e ≤ 0. x x This forces Y (τ, x) = e − 1 and, in consequence, ∂xY (τ, x) = e for all (τ, x) ∈ (0, τ0) × (0, ∞). On the other hand, by Theorem 2.2 and the Sobolev embedding 1,α theorem, there exists α ∈ (0, 1) such that Y (τ, ·) ∈ Cloc (R) for a.e. τ ∈ (0,T ). Therefore, there exists τ1 ∈ (0, τ0) such that 1 ∂ Y (τ , x) ≥ , ∀x ≥ −ε, x 1 2 for some small ε > 0. Then it follows from Y (τ1, 0) = 0 that Z 0 ε Y (τ1, −ε) = − ∂xY (τ1, x) dx ≤ − < 0. −ε 2 AMERICAN STRANGLE OPTION 763

This contradicts the fact that Y ≥ (ex − 1)+ + (κ − ex)+ ≥ 0. We next show that Y (τ, ln κ) > 0 for all τ ∈ (0,T ]. If the assertion is not true, x + x + then we deduce from Y (τ, ln κ) ≥ (e − 1) + (κ − e ) |x=ln κ = 0 that there exists τ0 > 0 such that Y (τ0, ln κ) = 0. Hence we have Y (τ, ln κ) = 0 for all 0 ≤ τ ≤ τ0. Observe from Theorem 2.2 that for x ≤ ln κ, x x x x Y − (κ − e ) ≥ 0 and ∂−x(Y − (κ − e )) = −∂x(Y − (κ − e )) = −∂xY − e ≤ 0. x x We thus get Y (τ, x) = κ−e and ∂xY (τ, x) = −e for all (τ, x) ∈ (0, τ0)×(−∞, ln κ). 1,α Since Y (τ, ·) ∈ Cloc (R) for a.e. τ ∈ (0,T ), there exist τ2 ∈ (0, τ0) and small ε > 0 such that κ −2κ ≤ ∂ Y (τ , x) ≤ − , ∀x ∈ (ln κ − ε, ln κ + ε). x 2 2 Therefore, we obtain Z ln κ+ε κε Y (τ2, ln κ + ε) = Y (τ2, ln κ + ε) − Y (τ2, ln κ) = ∂xY (τ2, x) dx ≤ − < 0, ln κ 2 a contradiction. Lemma 3.2. If τ > 0, Y is always positive, i.e., Y (τ, x) > 0 for all τ > 0 and x ∈ R. Proof. We observe from (2.1) and Lemma 3.1 that   ∂τ Y − LY ≥ 0, (τ, x) ∈ (0,T ] × (ln κ, 0),   Y (τ, ln κ) > 0,Y (τ, 0) > 0, τ ∈ (0,T ],  x + x +  Y (0, x) = (e − 1) + (κ − e ) ≥ 0, x ∈ [ln κ, 0]. Therefore, it follows from the strong maximum principle that Y (τ, x) > 0 for (τ, x) ∈ (0,T ] × (ln κ, 0). Moreover, (2.14) and Lemma 3.1 show that Y (τ, x) > 0 for (τ, x) ∈ (0,T ] × ((−∞, ln κ] ∪ [0, ∞)) , which completes the proof. We now prove the regularity and the monotone property of the free boundaries. In addition, we describe the limiting behavior of the free boundaries as time to maturity goes to zero. Theorem 3.3. (1) If q = 0, then EB is the empty set, and hence B(τ) does not exist. (2) If q > 0, then B(τ) is smooth and strictly increasing in (0,T ]. Moreover, r r B(0) := lim B(τ) = max 0, ln = ln max 1, . (3.1) τ→0+ q q Proof. (1) Suppose that EB 6= ∅. We observe from the deﬁnition of EB that x x x B (∂τ − L)Y = (∂τ − L)[e − 1] = −L[e − 1] = qe − r in E . B When q = 0, it follows that (∂τ − L)Y = −r < 0 in E . On the other hand, (2.1) B leads to (∂τ − L)Y ≥ 0 in E . This is a contradiction. Therefore, we conclude that EB = ∅, and hence that B(τ) does not exist. 764 JUNKEE JEON AND JEHAN OH

(2) From the deﬁnition of B(τ), it is clear that B(τ) ≥ 0. Since ∂τ Y ≥ 0, we see that B(τ) is increasing in (0,T ]. Therefore, B(0) := lim B(τ) exists. Let us prove τ→0+ (3.1). We ﬁrst consider the case q ≥ r. Suppose that B(0) = x0 > 0. Then

B(τ) ≥ B(0) = x0 > 0, ∀τ ≥ 0. It follows from the deﬁnition of B(τ) that x Y (τ, x) > e − 1, ∀(τ, x) ∈ (0,T ] × (0, x0), and hence x x (∂τ − L)[Y − (e − 1)] = 0 + L[e − 1] = −qex + r ≤ −q(ex − 1) < 0 in (0,T ) × (0, x0), where we have used the fact that r ≤ q. Then we obtain x ∂τ Y (0, x) < L[Y (0, x) − (e − 1)] = 0, ∀x ∈ (0, x0), a contradiction. Therefore, we conclude that B(0) = 0 in the case q ≥ r. r We now turn to the case q < r. Suppose that B(0) = x0 > ln q . In the same manner we can see that r Y (τ, x) > ex − 1, ∀(τ, x) ∈ (0,T ] × ln , x , q 0 and that x x x (∂τ − L)[Y − (e − 1)] = 0 + L[e − 1] = −qe + r < 0 r in (0,T ) × ln q , x0 . We thus get r ∂ Y (0, x) < L[Y (0, x) − (ex − 1)] = 0, ∀x ∈ ln , x , τ q 0 r r a contradiction. Therefore, B(0) ≤ ln q . To show that B(0) = ln q , we now suppose r 1 r that B(0) = x0 < ln q . Let x1 := 2 x0 + ln q . We deduce that there exists τ0 > 0 such that B(τ0) ≤ x1, and hence r Y = ex − 1 in (0, τ ) × x , ln . 0 1 q Then it is clear that r (∂ − L)[Y − (ex − 1)] = 0 in (0, τ ) × x , ln . τ 0 1 q On the other hand, we discover from (2.1) that r (∂ − L)[Y − (ex − 1)] ≥ L[ex − 1] = −qex + r > 0 for x < ln . τ q r We have arrived at a contradiction which proves B(0) = ln q in the case q < r. Thus (3.1) is proved. We now show that B(τ) is strictly increasing in (0,T ]. Conversely, suppose that B(τ) is not strictly increasing in (0,T ]. Then B(τ1) = B(τ2) = x0 for some 0 < τ1 < τ2 ≤ T . We note from Lemma 3.1 that x0 > 0. It follows from the deﬁnition of B(τ) that

x0 Y (τ, x0) = e − 1, ∀τ ∈ [τ1, τ2], and hence ∂τ Y (τ, x0) = 0, ∀τ ∈ (τ1, τ2). AMERICAN STRANGLE OPTION 765

Moreover, we have

(∂τ − L)Y = 0 in (τ1, τ2) × (0, x0), and, in consequence,

(∂τ − L)(∂τ Y ) = 0 in (τ1, τ2) × (0, x0).

Since ∂τ Y ≥ 0, the strong maximum principle implies that either ∂τ Y ≡ 0 or ∂xτ Y (τ, x0) < 0 for any τ ∈ (τ1, τ2). On the other hand, we deduce from the x0 deﬁnition of B(τ) that ∂xY (τ, x0) = e for any τ ∈ [τ1, τ2], and so ∂xτ Y (τ, x0) = 0 for any τ ∈ (τ1, τ2). This is a contradiction. Hence B(τ) is strictly increasing in (0,T ]. Let us show that B(τ) is continuous in (0,T ]. Conversely, suppose that B(τ) is not continuous in (0,T ]. Then there exist τ0 ∈ (0,T ], x0 ≥ 0 and small ε0, δ0 > 0 such that

B(τ0 − ε) ≤ x0 and B(τ0 + ε) ≥ x0 + δ0 for all ε ∈ (0, ε0). It follows from the deﬁnition of B(τ) that for any ε ∈ (0, ε0), x Y (τ0 − ε, x) = e − 1 for all x ≥ x0 (3.2) and x Y (τ0 + ε, x) > e − 1 for all 0 ≤ x < x0 + δ0. (3.3) In addition, (3.2) and the continuity of Y yield x Y (τ0, x) = e − 1 for all x ≥ x0. (3.4)

n r o Since B(τ) is strictly increasing in (0,T ], we observe that x0 > B(0) = max 0, ln q . We deduce from (3.3) that in (τ0, τ0 + ε0) × (x0, x0 + δ0),

x x x x0 (∂τ − L)[Y − (e − 1)] = 0 + L[e − 1] = −qe + r ≤ −qe + r < 0 (3.5) We thus obtain from (3.4) and (3.5) that x ∂τ Y (τ0, x) < L[Y (τ0, x) − (e − 1)] = 0, ∀x ∈ (x0, x0 + δ0), which is a contradiction. Therefore, B(τ) is continuous in (0,T ]. Moreover, since x + x + ∂τ Y ≥ 0 and (e − 1) + (κ − e ) is the lower obstacle, it can be proved that B(τ) ∈ C0,1(0,T ] by the method developed by Friedman in [4]. At this point, it follows from the bootstrap argument (see also [6]) that B(τ) ∈ C∞(0,T ]. This completes the proof.

Remark 3.1. If q 6= 0, Theorem 3.3 implies that when the underlying asset increase suﬃciently, if it is optimal to exercise the long-term American strangle, it is never optimal to leave un-exercised the short-term American strangle. Theorem 3.4. A(τ) is smooth and strictly decreasing in (0,T ]. Moreover, ( n o ln κ min 1, r , if q > 0, A(0) := lim A(τ) = q (3.6) τ→0+ ln κ, if q = 0.

Proof. We observe from the deﬁnition of A(τ) that A(τ) ≤ ln κ. Since ∂τ Y ≥ 0, A(τ) is decreasing in (0,T ]. Hence A(0) := lim A(τ) exists. We ﬁrst consider the τ→0+ case 0 ≤ q ≤ r. Suppose that A(0) = x0 < ln κ. Since A(τ) is decreasing, we get

A(τ) ≤ A(0) = x0 < ln κ, ∀τ ≥ 0. 766 JUNKEE JEON AND JEHAN OH

Then it follows from the definition of A(τ) that x Y (τ, x) > κ − e , ∀(τ, x) ∈ (0,T ] × (x0, ln κ), and hence x x x (∂τ − L)[Y − (κ − e )] = 0 + L[κ − e ] = qe − rκ. x in (0,T ) × (x0, ln κ). If q = 0, then (∂τ − L)[Y − (κ − e )] = −rκ < 0. If 0 < q ≤ r, x x then (∂τ − L)[Y − (κ − e )] = qe − rκ < qκ − rκ ≤ 0 in (0,T ) × (x0, ln κ). Thus we deduce that in (0,T ) × (x0, ln κ), x (∂τ − L)[Y − (κ − e )] < 0. From this we conclude that x ∂τ Y (0, x) < L[Y (0, x) − (κ − e )] = 0, ∀x ∈ (x0, ln κ), which is a contradiction. Therefore, we obtain A(0) = ln κ in the case 0 ≤ q ≤ r. rκ We now consider the case q > r. Suppose that A(0) = x0 < ln q . We obtain similarly that rκ Y (τ, x) > κ − ex, ∀(τ, x) ∈ (0,T ] × x , ln , 0 q and that x x x (∂τ − L)[Y − (κ − e )] = 0 + L[κ − e ] = qe − rκ < 0 rκ in (0,T ) × x0, ln q . Thus we get rκ ∂ Y (0, x) < L[Y (0, x) − (κ − ex)] = 0, ∀x ∈ x , ln , τ 0 q rκ rκ a contradiction. Therefore, A(0) ≥ ln q . We now suppose that A(0) = x0 > ln q . As in the proof of the previous theorem, there exists τ0 > 0 such that rκ 1 rκ (∂ − L)[Y − (κ − ex)] = 0 in (0, τ ) × ln , ln + x . (3.7) τ 0 q 2 q 0 On the other hand, we deduce from (2.1) that rκ (∂ − L)[Y − (κ − ex)] ≥ L[κ − ex] = qex − rκ > 0 for x > ln , τ q rκ rκ which contradicts (3.7). This yields A(0) ≤ ln q , and hence A(0) = ln q . Analysis similar to that in the proof of Theorem 3.3 shows that A(τ) is strictly decreasing in (0,T ]. Moreover, the smoothness of A(τ) follows by the same method as Theorem 3.3. Remark 3.2. Contrary to Remark 3.1, Theorem 3.4 means that when the underlying asset decrease sufficiently, if it is to exercise the short-term American strangle, the long-term American strangle already has been exercised. Remark 3.3. If q 6= 0, Theorem 3.3 and Theorem 3.4 imply that the free boundaries SA(τ) and SB(τ) are well-defined. If q = 0, (1) in Theorem 3.3 implies that the free boundary B(τ) or SB(τ) does not exist. That is, for the American strangle option written on an underlying asset without dividends, although the underlying asset increases enough, the option holder does not exercise his/her right. This phenomena is consistent with American call option with non-dividend. Since the price of the American call option on a non-dividend-paying stock always exceeds its intrinsic value prior to expiration, the early exercise is never optimal. AMERICAN STRANGLE OPTION 767

4. Comparative static analysis. In this section we conduct comparative static analysis with respect to important model parameters. First, we analyze the eﬀect of value σ, the volatility of the American strangle option. Lemma 4.1. Let Y be the solution to (2.1). Then we have

∂xxY − ∂xY ≥ 0. (4.1) Proof. We denote C = (τ, x) ∈ D : Y (τ, x) > (ex − 1)+ + (κ − ex)+ , E = (τ, x) ∈ D : Y (τ, x) = (ex − 1)+ + (κ − ex)+ , where D = (0,T ] × R. In the continuation region C, it follows from (2.1) that

∂τ (∂xY ) − L(∂xY ) = 0, and hence ∂τ (∂xY − Y ) − L(∂xY − Y ) = 0. (4.2) On the other hand, we observe from Lemma 3.2 that E ⊂ ((−∞, ln κ) ∪ (0, ∞)) × (0,T ]. Hence, we see that in the exercise region E,  x  κ − e if x < ln κ, Y (τ, x) = x  e − 1 if x > 0, and so   −κ if x < ln κ, (∂xY − Y )(τ, x) =  1 if x > 0. We thus get   −rκ if x < ln κ, ∂τ (∂xY − Y ) − L(∂xY − Y ) = (4.3)  r if x > 0, in the exercise region E. Combining (4.2) and (4.3) gives   0 in C,   −rκ in E ∩ {x < ln κ}, ∂τ (∂xY − Y ) − L(∂xY − Y ) = (4.4)   r in E ∩ {x > 0}.

For δ > 0, we now set Yδ(τ, x) := Y (τ, x − δ). We also denote x−δ + x−δ + Cδ = (τ, x) ∈ D : Yδ(τ, x) > (e − 1) + (κ − e ) , x−δ + x−δ + Eδ = (τ, x) ∈ D : Yδ(τ, x) = (e − 1) + (κ − e ) . Then we obtain similarly that   0 in Cδ,   −rκ in E ∩ {x < ln κ + δ}, ∂τ (∂xYδ − Yδ) − L(∂xYδ − Yδ) = δ (4.5)   r in Eδ ∩ {x > δ}. 768 JUNKEE JEON AND JEHAN OH

In the region E ∩ {x < ln κ}, we know that Y (τ, x) = κ − ex. Then we deduce x−δ from Section3 that Yδ(τ, x) = Y (τ, x − δ) = κ − e , and so

∂τ (∂xYδ − Yδ) − L(∂xYδ − Yδ) = −rκ. Therefore, we have

∂τ [(∂xY − Y ) − (∂xYδ − Yδ)] − L [(∂xY − Y ) − (∂xYδ − Yδ)] = −rκ − (−rκ) = 0 in E ∩ {x < ln κ}. (4.6) In addition, we notice that

C ∩ (Eδ ∩ {x > δ}) = C ∩ ((E + (0, δ)) ∩ {x > δ}) = C ∩ ((E ∩ {x > 0}) + (0, δ)) ⊂ C ∩ (E ∩ {x > 0}) = ∅. (4.7) We now combine (4.4)–(4.7) to discover that

∂τ [(∂xY − Y ) − (∂xYδ − Yδ)] − L [(∂xY − Y ) − (∂xYδ − Yδ)] ≥ 0 (4.8) x + x + in the whole domain D = (0,T ] × R. Letting Y0(x) := (e − 1) + (κ − e ) , we see that ∂xY0 − Y0 is increasing in R, and hence

[(∂xY − Y ) − (∂xYδ − Yδ)] (0, x) ≥ 0, ∀x ∈ R. (4.9) Then we conclude from (4.8) and (4.9) that

[(∂xY − Y ) − (∂xYδ − Yδ)] (τ, x) ≥ 0, ∀(τ, x) ∈ D. It follows that (∂xY − Y )(τ, x) ≥ (∂xY − Y )(τ, x − δ) for all (τ, x) ∈ D and δ > 0. This yields the desired inequality (4.1). Using the above Lemma, we prove the following theorem, the behavior of the free boundaries according to σ. Theorem 4.2. (1) A(τ) is decreasing with respect to σ. (2) B(τ) is increasing with respect to σ.

Proof. Let σ1, σ2 be positive constants such that σ1 > σ2. Let Y1 be the solution to (2.1) with σ = σ1 and let Y2 be the solution to (2.1) with σ = σ2. Then it follows from (2.1), (2.2) and Lemma 4.1 that Y1 satisﬁes  2 2 σ2 σ2 1 2 2  ∂τ Y1 − ∂xxY1 − r − q − ∂xY1 + rY1 = (σ − σ )(∂xxY1 − ∂xY1) ≥ 0,  2 2 2 1 2  x + x +  if Y1 > (e − 1) + (κ − e ) , (τ, x) ∈ [0,T ) × R,   2 2 σ2 σ2 1 2 2 ∂τ Y1 − 2 ∂xxY1 − r − q − 2 ∂xY1 + rY1 ≥ 2 (σ1 − σ2)(∂xxY1 − ∂xY1) ≥ 0,  x + x +  if Y1 = (e − 1) + (κ − e ) , (τ, x) ∈ [0,T ) × R,   x + x +  Y1(0, x) = (e − 1) + (κ − e ) = Y2(0, x), x ∈ R.

A A By the comparison principle, we have Y1 ≥ Y2, From this we deduce that E1 ⊂ E2 B B and E1 ⊂ E2 , where A x + Ej := (τ, x) ∈ D : Yj(τ, x) = (κ − e ) > 0 , B x + Ej := (τ, x) ∈ D : Yj(τ, x) = (e − 1) > 0 , AMERICAN STRANGLE OPTION 769 for j = 1, 2. Therefore, we obtain A1(τ) ≤ A2(τ) and B1(τ) ≥ B2(τ), where Aj(τ) and Bj(τ) are the free boundaries of Yj for j = 1, 2. This completes the proof.

For ﬁxed r, q, σ and maturity T , let us deﬁne Fp(τ) and Fc(τ) as the free boundaries of the degenerate backward parabolic problem arising from the American put option with strike price K1 and the American call option with strike price K2, respectively. Then, by the following theorem, we can compare A(τ) and Fp(τ), B(τ) and Fc(τ), respectively.

Theorem 4.3. For any τ ∈ (0,T ], we have A(τ) ≤ Fp(τ) and B(τ) ≥ Fc(τ).

Proof. We ﬁrst prove that B(τ) ≥ Fc(τ). Let Yc be the solution to the problem  ∂ Y − LY = 0, if Y > (ex − 1)+, (τ, x) ∈ [0,T ) × ,  τ c c c R  x + ∂τ Yc − LYc ≥ 0, if Yc = (e − 1) , (τ, x) ∈ [0,T ) × R,  x +  Yc(0, x) = (e − 1) , x ∈ R, which is the forward non-degenerate parabolic problem arising from the model of American call option. In view of

x + x + x + Y (0, x) = (e − 1) + (κ − e ) ≥ (e − 1) = Yc(0, x), the monotonicity of solution of variational inequality with respect to initial value yields Y ≥ Yc in [0,T ] × R. This implies that B(τ) ≥ Fc(τ) for all τ ∈ (0,T ]. To prove A(τ) ≤ Fp(τ), setting s V (t, s) τ = T − t, x = ln , Ye(τ, x) = , K1 K1 we have  x + x +  ∂τ Ye − LYe = 0, if Ye > (e − κe) + (1 − e ) , (τ, x) ∈ [0,T ) × R,   x + x + ∂τ Ye − LYe ≥ 0, if Ye = (e − κe) + (1 − e ) , (τ, x) ∈ [0,T ) × R, (4.10)  x + x +  Ye(0, x) = (e − κe) + (1 − e ) , x ∈ R, where K2 1 κe := = ∈ (1, ∞). K1 κ

Now let Yec be the solution to the forward non-degenerate parabolic problem arising from the model of American put option:

 x +  ∂τ Yec − LYec = 0, if Yec > (1 − e ) , (τ, x) ∈ [0,T ) × R,   x + ∂τ Yec − LYec ≥ 0, if Yec = (1 − e ) , (τ, x) ∈ [0,T ) × R,  x +  Yec(0, x) = (1 − e ) , x ∈ R.

Since x + x + x + Ye(0, x) = (e − κe) + (1 − e ) ≥ (1 − e ) = Yec(0, x), we have Ye ≥ Yec in [0,T ] × R, and hence A(τ) ≤ Fp(τ) for all τ ∈ (0,T ]. 770 JUNKEE JEON AND JEHAN OH

5. Stationary problem for American strangle option. In this section, we consider a stationary problem arising in the American strangle option. According to Theorem 3.3, the problem (2.1) is divided into two cases according to the range of q: (1) There exist two free boundaries as q > 0, (2) There is only one free boundary as q = 0. Therefore, we also need to consider about the stationary problems corresponding to problem (2.1) separately. (1) For q > 0, the stationary problem is ( x + x + −LV = 0, if V > (e − 1) + (κ − e ) , x ∈ R, (5.1) −LV ≥ 0, if V = (ex − 1)+ + (κ − ex)+, x ∈ R, where L is deﬁned in (2.2).

2 Theorem 5.1. The variational inequality (5.1) has a unique Wp,loc solution, which is  x  κ − e , x < sp,  n1x n2x V (x) = c1e + c2e , x ∈ [sp, sc], (5.2)  x  e − 1, x > sc, where (n − 1)esc − n (n − 1)esc − n 2 2 −n1sc 1 1 −n2sc c1 = e , c2 = e , n2 − n1 n1 − n2 and n1, n2 are the positive and negative roots of the algebraic equation σ σ2 n2 + r − q − n − r = 0. (5.3) 2 2

The constants sc and sp are deﬁned as n κyn1 + 1 sp 2 sc sp e = · n , e = e · y. n2 − 1 y 1 + y Here, y ∈ (1, +∞) is the unique solution of the algebraic equation f(y) = 0 with

n1 1−n1 n2−n1 n2 1−n1 f(y) = −(n1 − 1)n2(1 + κy )(y + y ) − n1(1 − n2)(1 + κy )(y + 1). (5.4) 1 In fact, y > κ . Proof. We ﬁrst consider the following free boundary problem:   LV = 0, sp < x < sc,  sp 0 sp V (sp) = κ − e ,V (sp) = −e , (5.5)   sc 0 sc  V (sc) = e − 1,V (sc) = e . Then we extend the solution V onto R by x x V (x) = κ − e if x ∈ (−∞, sp) and V (x) = e − 1 if x ∈ (sc, +∞). (5.6) Next, we show that V is the unique solution to the variational inequality (5.1). We can let the general solution for (5.5) in the form of

n1x n2x V (x) = c1e + c2e , x ∈ [sp, sc]. (5.7)

It is easy to check that n1 = 1 if q = 0 and n1 > 1 if q > 0. AMERICAN STRANGLE OPTION 771

From (5.5) and (5.7),

n1sc n2sc sc V (sc) = c1e + c2e = e − 1, (5.8) 0 n1sc n2sc sc V (sc) = c1n1e + c2n2e = e .

Therefore, c1 and c2 are given by (n − 1)esc − n (n − 1)esc − n 2 2 −n1sc 1 1 −n2sc c1 = e , c2 = e . (5.9) n2 − n1 n1 − n2 Similarly,

n1sp n2sp sp V (sp) = c1e + c2e = κ − e , (5.10) 0 n1sp n2sp sp V (sp) = c1n1e + c2n2e = e , and (1 − n )esp + κn (1 − n )esp + κn 2 2 −n1sp 1 1 −n2sp c1 = e , c2 = e . (5.11) n2 − n1 n1 − n2 From (5.9) and (5.11), we obtain (n − 1)esc − n (1 − n )esp + κn 2 2 = 2 2 , n s n1sp e 1 c e (5.12) (n − 1)esc − n (1 − n )esp + κn 1 1 = 1 1 , en2sc en2sp and

n1sp n1sc sc+n1sp sp+n1sc n2e + κn2e = (n2 − 1)e − (1 − n2)e , (5.13) n2sc n2sp sc+n2sp sp+n2sc n1e + κn1e = (n1 − 1)e − (1 − n1)e . Therefore, we have n1(sc−sp) (1−n1)(s −s ) (n2−n1)(sc−sp) 0 = − (n1 − 1)n2 1 + κe e c p + e (5.14) n2(sc−sp) (1−n1)(sc−sp) − n1(1 − n2) 1 + κe e + 1 .

Let us deﬁne y = esc−sp and f(y) as (5.4). Then, we prove that f(y) = 0 possesses 1 a unique solution y ∈ (1, +∞) and that y > κ . We observe that

0 −n1 n2−n1−1 n2−1 f (y) = − (n1 − 1)n2 (1 − n1)y + κ + (n2 − n1)y + κn2y

−n1 n2−n1 n2−1 − n1(1 − n2) (1 − n1)y + κ(n2 − n1 + 1)y + κn2y

n2−n1−1 =y (−(n1 − 1)n2(n2 − n1) − κn1(1 − n2)(n2 − n2 + 1)y)

−n1 n2−1 + (1 − n1)(n2 − n1)y + κ(1 − n1)n2 + κn2(n2 − n1)y .

Since n1 > 1 and n2 < 0,

−n1 n2−1 (1 − n1)(n2 − n1)y > 0, κ(1 − n1)n2 ≥ 0, κn2(n2 − n1)y > 0, 1 and for y ∈ κ , +∞ ,

n2−n1−1 y (−(n1 − 1)n2(n2 − n1) − κn1(1 − n2)(n2 − n1 + 1)y)

n2−n1−1 ≥ y (−(n1 − 1)n2(n2 − n1) − n1(1 − n2)(n2 − n1 + 1))

n2−n1−1 2 = y ((n2 − n1) + n1(n2 − 1))

n2−n1−1 2 n2−n1−1 ≥ y ((n2 − n1) + n1(n2 − n1)) = y (n2 − n1)n2 > 0. 772 JUNKEE JEON AND JEHAN OH

0 1 Therefore, f (y) > 0 for κ < y < +∞. Moreover, we see that 1 f = (n − n )(1 + κ1−n2 )(1 + κn1−1) < 0, κ 2 1 f(+∞) = +∞.

1 Hence, f(y) = 0 has a unique solution in y ∈ κ , +∞ . 1 Also, for 1 < y < κ ,

n1 1−n1 n2−n1 n2 1−n1 f(y) < − n1n2(1 + κy )(y + y ) + n1n2(1 + κy )(y + 1)

n2−n1 = n1n2(1 − ky)(1 − y ) ≤ 0, 1 then, in y ∈ 1, κ , there is no solution of f(y) = 0. Hence f(y) = 0 possesses a 1 unique solution y ∈ (1, +∞) with y > κ .

Since z = esc−sp in (5.12), we obtain n κyn1 + 1 sp 2 sc sp e = · n , e = e · y. n2 − 1 y 1 + y 2 Next, we show that V deﬁned as (5.2) is the unique Wp,loc solution to the variational inequality (5.1). By a method similar to Section2, we can get the uniqueness 2 of the solution of the problem (5.1). Also, it is not hard to check that V (x) ∈ Wp,loc. Therefore, we only need to show that V satisﬁes the variational inequality (5.1). Firstly, for f(y) = 0, we obtain

n1 n2 n1 − 1 y + κy n2 − 1 y + κy · n1 = · n n1 y + y n2 y + y 2 and n 1 + κyn1 n 1 + κyn2 n sc 2 1 1 e = · n −1 = n −1 > > 1. n2 − 1 1 + y 1 n1 − 1 1 + y 2 n1 − 1 Also, it is not hard to check that sc > 0 and sp < ln κ < 0. Therefore,

c1, c2 > 0 and

V (x) ≥ 0 for x ∈ [sp, sc]. Secondly, by (5.5), we obtain that  −L(V 0 − ex) = Lex = −qex < 0, s < x < s ,  p c  0 x x x V (x) − e = e − e ≤ 0, x = sc,  0 x x x  V (x) − e = −e − e ≤ 0, x = sp, and  −L(−V 0 − ex) = Lex = −qex < 0, s < x < s ,  p c  0 x x x −V (x) − e = −e − e ≤ 0, x = sc,  0 x x x  −V (x) − e = e − e ≤ 0, x = sp. x 0 x Applying the maximum principle, we have −e ≤ V (x) ≤ e , x ∈ [sp, sc]. Since (V (x) − (ex − 1))0 = V 0(x) − ex ≤ 0, (V (x) − (κ − ex))0 = V 0(x) + ex ≥ 0, AMERICAN STRANGLE OPTION 773

sc sp and V (sc) = e − 1,V (sp) = κ − e , we get V (x) ≥ ex − 1,V (x) ≥ κ − ex. We see from V ≥ 0 that V (x) ≥ (ex − 1)+ + (κ − ex)+.

Finally, if x > sc, −LV = −L(ex − 1) = qex − r. It is easy to check that n n r 1 · 2 = . n1 − 1 n2 − 1 q Therefore, for x > sc, n 1 + κyn1 r 1 − n 1 + κyn2 r sc 2 2 e = · n −1 = n −1 > n2 − 1 1 + y 1 q −n2 1 + y 2 q and −LV = qex − r ≥ qesc − r ≥ 0.

If x < sp, −LV = −L(κ − ex) = −qex + κr.

For x < sp, n 1 + κyn1 r n − 1 1 + κyn1 r sp 2 1 e = · n = n < κ n2 − 1 y + y 1 q n1 y + y 1 q and −LV = −qex + κr ≥ −qesp + κr ≥ 0. 2 Therefore, V is the unique Wp,loc solution to problem (5.1).

(2) As q = 0, the subregion EB of exercise region E does not exist and for the uniqueness of the solution to the corresponding stationary problem, we need to impose an additional condition, which comes from the properties of strangle options. The problem is  −LV = 0, if V > (ex − 1)+ + (κ − ex)+, x ∈ ,  R  −LV ≥ 0, if V = (ex − 1)+ + (κ − ex)+, x ∈ R, (5.15)   lim e−xV (x) = 1. x→∞ 2 Theorem 5.2. The variational inequality (5.15) has a unique Wp,loc solution, which is  x  κ − e , x < sp, V (x) = κ (5.16) ex + en(x−sp), x ∈ [s , +∞],  1 − n p

2 sp n κ where n = −2r/σ and e = n−1 2 . Proof. In the ﬁrst, we solve the free boundary problem (5.17)  LV = 0, sp < x,   sp 0 sp V (sp) = κ − e ,V (sp) = −e , (5.17)   lim e−xV (x) = 1. x→∞ 774 JUNKEE JEON AND JEHAN OH

The general solution form of (5.17) is given by

x nx V (x) = c1e + c2e , (5.18)

2 −x where n = −2r/σ . Since lim e V (x) = 1, we obtain c1 = 1 and x→∞

x nx V (x) = e + c2e , x ∈ [sp, +∞).

sp 0 sp From V (sp) = κ − e and V (sp) = −e , we have

sp nsp sp sp nsp sp e + c2e = κ − e , e + nc2e = −e .

It is easily seen that n κ κ esp = , c = e−nsp , n − 1 2 2 1 − n

x then (5.16) follows if we extend V by V (x) = κ − e for any x < sp. We note that sp e < κ and c2 > 0, and so V (x) ≥ 0. 2 Next, we prove that V possessing the form (5.16) is the unique Wp,loc solution to the variational inequality (5.15). Similarly in case (1), we only need to prove that V satisﬁes (5.15). First, for x < sp,

−LV = −L(κ − ex) = rκ ≥ 0. (5.19)

Secondly, from (5.16), we see that for any x > sp, there holds κn V 0(x) = ex + en(x−sp). 1 − n

Moreover, for x ≥ sp, κn n − 1 2 V 0(x) =ex + · · en(x−sp)+sp 1 − n n κ =ex − 2ex · e(n−1)(x−sp) ≥ −ex.

0 sp x Since V (sp) = κ − e , we have V (x) ≥ κ − e , x ∈ [sp, +∞). We also have κ V (x) = ex + en(x−sp) > ex − 1, x ∈ [s , +∞). 1 − n p

x + x 2 Thus, V ≥ (e − 1) + (κ − e ) and V is the Wp,loc solution to variational inequality (5.15).

Applying the comparison principle with respect to initial value of variational inequality, we can obtain the following theorem.

Theorem 5.3. There exist constants sp < 0 and sc > 0 such that

r r s ≤ A(τ) ≤ ln κ min 1, , ln max 1, ≤ B(τ) ≤ s . p q q c

In fact, the constants sp and sc are deﬁned in Theorem 5.1 and Theorem 5.2. AMERICAN STRANGLE OPTION 775

6. Numerical methods and results. In this section, we obtain the numerical solution of the value function and free boundaries of the American strangle option by applying the ﬁnite diﬀerence scheme.

Starting from (2.1), we have  σ2 σ2  min ∂τ Y − ∂xxY − r − q − ∂xY + rY,  2 2  x + x + Y − (e − 1) − (κ − e ) = 0, for x ∈ R, τ ∈ (0,T ], (6.1)  x + x +  Y (0, x) = (e − 1) + (κ − e ) , for x ∈ R.

n Given mesh size ∆τ, ∆x > 0, Yj = Y (n∆τ, j∆x) represents the value of numerical approximation at (n∆τ, j∆x), then the PDE is converted to the following diﬀerence equation:

 ( n n−1 2 n−1 n−1 n−1 Yj − Vj σ Yj+1 − 2Yj + Yj−1  min −  ∆τ 2 ∆x2  )  σ2 Y n−1 − Y n−1  − r − q − j+1 j−1 + rY n,Y n − (ex − 1)+ − (κ − ex)+ = 0, 2 2∆x j j (6.2)     0 j∆x + j∆x +  Yj = (e − 1) + (κ − e ) .

It means  1 σ2∆τ σ2 + (r − q − σ2/2)∆x Y n = max 1 − Y n−1 + ∆τY n−1  j 2 j 2 j+1  1 + r∆τ ∆x 2∆x  σ2 − (r − q − σ2/2)∆x + ∆τY n−1 , ej∆x − 1, κ − ej∆x , (6.3) 2∆x2 j−1   0 j∆x + j∆x +  Yj = (e − 1) + (κ − e ) . σ∆τ Choosing = 1, we have ∆x2  1 1 r − q − σ2/2√  n n−1  Yj = max + ∆τ Yj+1  1 + r∆τ 2 2σ  2 √ 1 r − q − σ /2 n−1 j∆x j∆x + − ∆τ Yj−1 , e − 1, κ − e , (6.4)  2 2σ   0 j∆x + j∆x +  Yj = (e − 1) + (κ − e ) . √ We denote u = eσ ∆τ , d = u−1, ρ = er∆τ , p = (ρe−q∆τ − d)(u − d)−1. Applying the Taylor expansion, we see that as ∆τ → 0+, there holds 1 r − q − σ2/2√ √ 1 1 + ∆τ = p + o( ∆τ), = + O(∆τ 2). 2 2σ 1 + r∆τ ρ √ Ignoring a higher order of ∆τ, we get  n 1 n−1 n−1 j j  Y = max pV + (1 − p)V , u − 1, κ − u ,  j ρ j+1 j−1 (6.5)  0 j + j +  Yj = (u − 1) + (κ − u ) . 776 JUNKEE JEON AND JEHAN OH

Consider the point (τ, x) = (n∆τ, j∆x), then n Yj = Y (τ, x), n−1 Yj = V (τ − ∆τ, x), n−1 Yj+1 = V (τ − ∆τ, x + ∆x), n−1 Yj−1 = V (τ − ∆τ, x − ∆x), and get Figures1–5.

1.5 T-t=1 T-t=0

1 V option price 0.5

0 0 0.5 1 1.5 2 2.5 3 s

Figure 1. The change of the option value function V (t, s) with respect to stock price s where r = 0.05, q = 0.1, σ = 0.3,K1 = 1 and K2 = 1.5.

0.8 σ=0.1 0.6 σ=0.2 σ=0.3

0.4

0.2

0 x

-0.2

-0.4

-0.6

-0.8 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 τ = T − t time to matuarity

Figure 2. The change of the free boundaries A(τ) and B(τ) with respect to σ where r = 0.05, q = 0.05,K1 = 1 and K2 = 1.1.

Figure1 shows the value function of the American strangle option using the numerical method described above. Since the American strangle option has both the properties of the call option and the put option, the value of the option increases as the underlying asset increases or decreases suﬃciently. That is, the value function V does not hold monotonicity with respect to s. This intuition is consistent with the numerical result in Figure1. AMERICAN STRANGLE OPTION 777

0.45

0.4

0.35

0.3

0.25 x 0.2

0.15 B(τ) F (τ) 0.1 c

0.05

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 τ = T − t time to matuarity

Figure 3. Compare the free boundary B(τ) and the free boundary Fc(τ) with r = 0.05, q = 0.05, σ = 0.2,K1 = 1 and K2 = 1.1.

-0.05 A(τ) F (τ) -0.1 p

-0.15

-0.2

-0.25

x -0.3

-0.35

-0.4

-0.45

-0.5

-0.55 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 τ = T − t time to matuarity

Figure 4. Compare the free boundary A(τ) and the free boundary Fp(τ) with r = 0.05, q = 0.05, σ = 0.2,K1 = 1 and K2 = 1.1.

0.8

B(τ) 0.6 A(τ)

0.4

0.2

0 x -0.2

-0.4

-0.6

-0.8

-1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 τ = T − t time to matuarity

Figure 5. Upper and lower bounds of A(τ) and the free boundary B(τ), respectively, with r = 0.05, q = 0.05, σ = 0.2,K1 = 1 and K2 = 1.1. 778 JUNKEE JEON AND JEHAN OH

In Figure2, it can be seen that as the volatility σ of the underlying asset increases, A(τ) decreases and B(τ) increases. As mentioned in the introduction, the American strangle option can obtain more profits as the underlying asset fluctuates more. That is, as the volatility σ of the underlying assets increases, the price of the option increases, so it should be exercised when the underlying asset price is higher or lower than when the volatility is lower. This is consistent with the result of Theorem 4.2. In addition, the area enclosed by the two free boundaries A(τ) and B(τ) is the continuation region C of Problem (2.1) and the complement of this region is the exercise region E of Problem (2.1). Figure3 plots the behavior of free boundaries B(τ) and Fc(τ) according to τ. Observe that B(0) = Fc(0) ,B(τ) ≥ Fc(τ), B(τ) and Fc(τ) are increasing. Like- wise, Figure4 shows the behavior of free boundaries A(τ) and Fp(τ) according to τ. Observe that A(0) = Fp(0),A(τ) ≤ Fp(τ), A(τ) and Fp(τ) are decreasing. Therefore, Figure4 and Figure5 illustrate the result of Theorem 4.3 numerically. The American strangle option is more expensive than the American call option and the American put option because it can obtain benefits both cases in which the underlying asset increases or decreases. Therefore, the American strangle option should be exercised at a higher price than the corresponding American call option and at a lower price than the corresponding American put option. This intuition is also consistent with the numerical results in Figure3-4. Figure5 shows the upper and lower bounds of free boundaries A(τ) and B(τ), r r respectively. In other words, sp ≤ A(τ) ≤ ln(κ max(1, q )) and ln(max(1, q )) ≤ B(τ) ≤ sc. The numerical results coincide with Theorem 5.3.

Appendix A. Formulation of the model. An American strangle option whose underlying asset is the stock which has the price St given by

dSt = (r − q)Stdt + σStdWt, (A.1) where t is calender time, r represents the risk-free interest rate, q(≥ 0) is the continuous dividend rate, and σ > 0 is the constant volatility of St. Also, (Wt)t≥0 is a standard Brownian motion on a filtered probability space (Ω, (Ft)t≥0 , P) with a risk-neutral measure P, where (Ft)t≥0 is the natural filtration generated by (Wt)t≥0. Let Tt,T be the set of all stopping time in [t, T ], and the value of American strangle option price at time t is defined by

h −r(θ−t) + + i V (t, St) = max E e (Sθ − K2) + (K1 − Sθ) | Ft , (A.2) θ∈Tt,T where K1 and K2 denote the strike price of put and call options, respectively, and we assume K1 < K2. It is obvious that + + V (t, St) ≥ (St − K2) + (K1 − St) . (A.3) By the deﬁnition of (A.2) and the law of iterated expectations, −rt −rθ + + e V (t, St) = max E e (Sθ − K2) + (K1 − Sθ) | Ft θ∈Tt,T −rθ + + = max E E e (Sθ − K2) + (K1 − Sθ) | Ft+h | Ft θ∈Tt,T −rθ + + ≥ E max E e (Sθ − K2) + (K1 − Sθ) | Ft+h | Ft θ∈Tt+h,T AMERICAN STRANGLE OPTION 779

−r(t+h) = E e V (t + h, St+h) | Ft , −rt which yields e V (t, St) is a supermatingale under P. By the Itˆoformula, we can obtain σ2 ∂ V (t, S ) + S2∂ V (t, S ) + (r − q)S ∂ V (t, S ) − rV (t, S ) ≤ 0. (A.4) t t 2 t SS t t S t t + + When V (t, St) > (St − K2) + (K1 − St) , there exists h > 0 small enough such ∗ that the optimal stopping time θ ∈ Tt+h,T . Thus, −rt −rθ + + e V (t, St) = max E e (Sθ − K2) + (K1 − Sθ) | Ft θ∈Tt,T h −rθ∗ + + i = E e (Sθ∗ − K2) + (K1 − Sθ∗ ) | Ft + + = max E E (Sθ − K2) + (K1 − Sθ) | Ft+h | Ft θ∈Tt+h,T h −r(t+h) i = E e V (t + h, St+h) | Ft ,

−r(t∧θ∗) ∗ which implies that e V (t ∧ θ ,St∧θ∗ ) is a martingale under P. We then apply the Itôformula to get σ2 ∂ V (t, S ) + S2∂ V (t, S ) + (r − q)S ∂ V (t, S ) − rV (t, S ) = 0. (A.5) t t 2 t SS t t S t t By the definition of V (t, St) in (A.2), + + V (T, s) = (s − K2) + (K1 − s) . (A.6) It follows from (A.3)–(A.5) that V (t, s) satisfies  2 σ 2  ∂tV + s ∂ssV + (r − q)s∂sV − rV = 0,  2  + +  if V > (s − K2) + (K1 − s) , (t, s) ∈ (0,T ] × (0, +∞),   σ2 ∂ V + s2∂ V + (r − q)s∂ V − rV ≤ 0, t 2 ss s (A.7)   + +  if V = (s − K2) + (K1 − s) , (t, s) ∈ (0,T ] × (0, +∞),  + +  V (T, s) = (s − K2) + (K1 − s) , s ∈ [0, +∞). 

Appendix B. Veriﬁcation. In this section, we show that the unique solution V (t, s) to the problem (1.1) coincides with the expected value of the American stran- 2,1 gle option price, i.e., (A.2) holds. By Theorem 2.2, we know v(·, ·) ∈ Wp,loc((0,T )× (0, +∞), so V (·, ·) and ∂sV (·, ·) are continuous in (0,T ) × (0, +∞) by embedding + + theorem [7]. Since ∂tV ≤ 0 and (S − K2) + (K1 − s) is the lower obstacle, we can obtain from a standard method as explained in Friedman [4] that ∂tV (·, ·) is continuous across s = SA(t) and s = SB(t), where SA(t),SB(t) are the free boundaries of problem (1.1). Therefore ∂tV (·, ·) is continuous in (0,T ) × (0, +∞). Moreover, ∂tV (·, ·) and ∂ssV (·, ·) are locally bounded in (0,T ) × (0, +∞). It follows from Itˆo’slemma that for any θ ∈ Tt,T , −r(θ−t) e V (θ, Sθ) Z θ 2 −r(u−t) σ 2 =V (θ, Sθ) + e ∂tV (u, Su) + Su∂ssV (u, Su) t 2 (B.1) Z θ +(r − q)Su∂uV (u, Su) − rV (u, Su)) du + σSu∂sV (u, Su)dWu. t 780 JUNKEE JEON AND JEHAN OH

+ + Since V (t, St) ≥ (St − K2) + (K1 − St) , we get

h −r(θ−t) i V (t, St) ≥E e V (θ, Sθ) | Ft

h −r(θ−t) + + i ≥E e (Sθ − K2) + (K1 − Sθ) | Ft . Therefore,

h −r(θ−t) + + i V (t, St) ≥ max E e (Sθ − K2) + (K1 − Sθ) | Ft . (B.2) θ∈Tt,T We deﬁne the stopping time θ∗ as follows: + + ∗ min{t ∈ [0,T ) | V (t, St) = (s − K2) + (K1 − s) }, θ = + + (B.3) T, if V (t, St) > (s − K2) + (K1 − s) , t ∈ [0,T ).

A B ∗ If S (t) ≥ St or St ≥ S (t), then θ = t, and + + h −r(θ∗−t) + + i V (t, St) = (St−K2) +(K1−St) = E e (Sθ − K2) + (K1 − Sθ) | Ft .

A B ∗ If S (t) < St < S (t), choose θ deﬁned in (B.3), then,  σ2  ∂ V (u, S ) + S2∂ V (u, S ) + (r − q)S ∂ V (u, S ) − rV (u, S ) = 0,  t u 2 u ss u u s u u for u ∈ (t, θ∗), (B.4)  ∗ + +  V (θ ,Sθ∗ ) = (Sθ∗ − K2) + (K1 − Sθ∗ ) .

By (B.1) and (B.4), we can obtain

h −r(θ∗−t) ∗ i V (t, St) = E e V (θ ,Sθ∗ ) | Ft

h −r(θ∗−t) + + i = E e (Sθ∗ − K2) + (K1 − Sθ∗ ) | Ft . Then we conclude from (B.2) that

h −r(θ−t) + + i V (t, St) = max E e (Sθ − K2) + (K1 − Sθ) | Ft . θ∈Tt,T

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