1 3

2014 c 5

u(g,)Journal of East China Normal University (Natural Science)

No. 3

May 2014

Article ID: 1000-5641(2014)03-0008-06

Perpetual American straddle option

CEN Yuan-jun1, YI Fa-huai2

(1. Shunde Polytechnic, Foshan Guangdong 528333, China;

2. School of Mathematical Sciences, South China Normal University, Guangzhou 510631, China)

Abstract: By appplying the comparison principle for the variational inequality, we ana-

lyzed the behavior of exercise boundaries for the perpetual American straddle option. We

found that it is a free boundary problem. Different from the standard perpetual American

option, it has two exercise boundary points with dividends and only one free boundary

point without dividends. These results can be understood very well from the financial point

of view. We will present a rigorous mathematical proof, and find the bounds of exercise

boundaries for the American straddle option with finite expiry.

Key words: straddle option; exercise boundary; variational inequality

CLC number: O175.26 Document code: A

DOI: 10.3969/j.issn.1000-5641.2014.03.002

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0 Introduction

American-style options can be exercised at any time before or up to the expiration date.

This feature often leads to a free boundary problem.

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1 3 , [{(=) 9

The price of American straddle option V (S, t) satisfies ([1, 2])

min{tV 2

2 S2SSV (r q)SSV + rV,

V |S K|} = 0, S > 0, 0 < t < T,

V (S, T ) = |S K|, S > 0;

(0.1)

where S is the stock price at time t; , r and K are all positive constants representing the

volatility of the stock, the risk-free interest rate and the strike price, respectively; q is a non-

negative constant representing the dividend; T is the expiry of the option.

In [1] the authors derived a pair of integral equations, giving the location of the optimal

exercise boundaries for an American straddle option. The integral equations close to the expiry

were solved in [2]. Some other properties were studied in [3], such as it has only one C-smooth

exercise boundary if dividend q = 0 but two C-smooth exercise boundaries if q > 0.

In system (0.1), if |S K| is replaced by (K S)+ or (S K)+, then V (S, t) is the

price of a standard American put (or call) option. In order to analyze the asymptotic behavior

as T + of the exercise boundary for a standard American option, one should research

the corresponding perpetual option first. Because the price of the perpetual option is always

higher than the American option with finite expiry. It is technical in mathematics to show the

existence of exercise boundaries of the perpetual American straddle option when q > 0.

The perpetual American straddle option is studied in this paper. Its price V0(S) satisfies

min{2

2S2V

0 (r q)SV

0 + rV0, V0 |S K|} = 0,S > 0. (0.2)

(0.2) is equivalent to the variational inequality

2

2S2V

0 (r q)SV

0 + rV0 > 0, V0 > |S K|, S > 0,[2

2S2V

0 (r q)SV

0 + rV0

][V0 |S K|

]= 0.

(0.3)

In the next section, we focus our attention on the case q = 0, where system (0.3) has

only one exercise boundary point S0 =rK

2r+2 . In Section 2, we consider the case q > 0 and

prove that it has two exercise boundary points S1 (0,K), S2 (K,+). Finally we show

the bounds of exercise boundary point for the problem (0.1).

1 Dividend q = 0

Denote

Lv = 2

2S2v

rSv

+ rv.

Lemma 1.1 Suppose V0(S) is the solution of the problem (0.3) with q = 0. Then

S K 6 V0(S) 6 S +K. (1.1)

Proof Since

L(S +K) = rS + r(S +K) = rK > 0 and (S +K) |S K| > 0,

10 u(g,) 2014 c

so S +K is the supersolution of problem (0.3). On the other hand, V0(S) > |S K| > S K,

hence S K 6 V0(S) 6 S +K.

Theorem 1.2 Problem (0.3) with q = 0 has only one optimal exercise point S0. More-

over

S0 =rK

2r + 2, (1.2)

V0(S) =

S + (K 2S0)(

S0S)

2r2 , S > S0,

K S, 0 6 S 6 S0.

Proof Rewrite system (0.3) as a free boundary problem

2

2S2V

0 rSV

0 + rV0 = 0, S0 < S S0; S0 =rK

2r + 2.

2 Dividend q > 0

When q > 0, the problem (0.3) is equivalent to (V0(S), S1, S2) satisfying

2

2S2V

0 (r q)SV

0 + rV0 = 0, S1 < S < S2,

V0(S1) = K S1, V

0 (S1) = 1,

V0(S2) = S2 K, V

0 (S2) = 1.

(2.1)

V0(S) = AS+BS is the general solution for the ordinary differential equation in (2.1), where

> 0 > ,

= +

2 +

2r

2, =

2 +

2r

2, =

r + q + 2

2

2. (2.2)

Notice that

q=

q=

(1 +

2 +

2r

2

)1

2> 0,

1 3 , [{(=) 11

and = 1 if q = 0, hence > 1. Applying boundary conditions in (2.1), we have

AS1 +BS1 = K S1,

AS11 +BS11 = 1,

AS2 +BS2 = S2 K,

AS12 +BS12 = 1.

Eliminating A and B, we obtain that

S1 + (K S1)

S1

=(S2 K) S2

S2

, (2.3)

S1 + (K S1)

S1=

(S2 K) S2S2

. (2.4)

Lemma 2.1 If 0 < S1 < K, then K < S2 < +.

Proof If 0 < S1 < K, then the left hand side in the equality (2.3) fits

0 K, and then

f

S2= S12 [( 1)(1 )S2 + K]

> S12 K[( 1)(1 ) + ].

Combining (2.2), we find

( 1)(1 ) + = + 2 = 2 +4r

2=

2r + 2q + 2

2> 1.

Hence fS2

> 0. Thus (2.3) has a unique solution S2 (K,+) if S1 (0,K).

Theorem 2.2 If q > 0, the problem (0.3) has two optimal exercise points S1 (0,K)

and S2 (K,+).

Proof According to Lemma 2.1, we only have to prove S1 (0,K). It is clear that

S1 < K by V0(S) > 0 for S > 0 and |S K| = 0 if S = K. Next, we divide the proof of S1 > 0

into three cases.

Case 1 r > q > 0. Suppose U0(S) is the solution of2

2S2U

0 (r q)SU

0 + (r q)U0 > 0, U0 > |S K|, S > 0,[2

2S2U

0 (r q)SU

0 + (r q)U0

][U0 |S K|

]= 0.

(2.5)

12 u(g,) 2014 c

Replacing r by r q in (1.3), from Theorem 1.2, we get the exercise boundary

S =(r q)K

2(r q) + 2, (2.6)

U0(S) =

S + (K 2S)(

SS)

2(rq)

2 , S > S,

K S, 0 6 S 6 S.(2.7)

It turns out, by(0.3), (2.5) and U0(S) > 0,

U0(S) > V0(S), S > 0.

Especially when S < S ,

K S = U0(S) > V0(S) > K S

is obtained. It follows that V0(S) = K S for 0 6 S < S, hence S1 > S. Notice that S > 0

if r > q, therefore S1 > 0.

Case 2 q > r > 0. Let

S =K2

x, V0(S) =

K

yv0(x).

Then dxdS = x2

K2,

V 0(S) =[Kxv0(x)

K

x2v0(x)

](

x2

K2

)=

x

Kv0(x) +

1

Kv0(x),

V 0 (S) =[

x

Kv0 (x)

1

Kv0(x) +

1

Kv0(x)

](

x2

K2

)=

x3

K3v0 (x).

Denote

x1 =K2

S2, x2 =

K2

S1. (2.8)

Then (2.1) become

2

2x2v

0 (q r)xv

0 + qv0 = 0, x1 < x < x2,

v0(x1) = K x1, v

0(x1) = 1,

v0(x2) = x2 K, v

0(x2) = 1.

(2.9)

It is the (2.1) by exchanging r and q. Applying the result of r > q in Case 1, we see that

0 < x1 < K, K < x2 < +.

So

0 < S1 =K2

x2< K, K < S2 =

K2

x1< +.

1 3 , [{(=) 13

Case 3 r = q > 0. In this case, problems (2.1) and (2.9) are the same, so that S1 =

x1,S2 = x2. Obviously, K2 = S1S2 is indicated by (2.8). Suppose

=S1

K=

K

S2.

It is clear that 0 6 6 1. On the other hand, (2.4) can be rewritten as

S1 + (K S1)

(S2 K) S2=

S1S2

. (2.10)

Divided the numerator and denominator by K, (2.10) can be transformed into

+ (1 )

( 1 1) 1

=

( 1).

Thus+ (1 )

(1 ) 1= 21.

Denote

f() =+ (1 )

(1 ) 1 21 = 1 +

1 +

(1 ) 1 21.

Then

f(0) =

1> 0, f(1) =

1

1= 1 < 0.

f () =(1 + )

[(1 ) 1]2 (2 1)22 < 0.

Hence f() = 0 has a unique solution

=S1

K=

K

S2 (0, 1).

As an application of Theorems 1.2 and 2.2, we can deduce the following bounds of exercise

boundaries for the problem(0.1).

Theorem 2.3 If q = 0, the problem (0.1) has only one exercise boundary h0(t) > S0,

where S0 is defined in (1.2). While q > 0, the problem(0.1) has two exercise boundaries

h1(t) > S1, h2(t) 6 S2, where S1 (0,K) and S2 (K,+) are defined in Theorem 2.2.

The American straddle option with finite expiry (0.1) and the properties of its free bound-

aries have been discussed in reference [3].

[ References ]

[ 1 ] ALOBAIDI G, MALLIER R. Laplace transformations and the American straddle [J]. Journal of Applied Math-ematics, 2002(2): 121-129.

[ 2 ] ALOBAIDI G, MALLIER R. The American straddle close to expiry [J]. Boundary Value Problems, 2006, (ArticleID 32835): 121-129.

[ 3 ] YI F H, CEN Y J. American straddle Option [J]. Chinese Journal of Engineering Mathematics, 2012, 29: 787-790.

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