Passport Option

J. Math. Sci. Univ. Tokyo5 (1998), 747785.

Pricing of Passport Option

By Izumi Nagayama

Abstract. Passport options are derivatives on actively managedfunds. There is no known explicit formula for the price of passportoptions in general. We estimate and analyze the value function, andgive the condition that the optimal strategy becomes trivial. We alsoshow an approximation theorem which enables us to compute the valuefunction numerically. Finally we give the explicit formula for the valuefunction in the case that the interest rate is zero by using stochasticanalytic approach. This is somehow done in symmetric case by Hyer,Lipton-Lifschitz, and Pugachevsky using partial dierential equationapproach.

1. Introduction

Many traded securities are managed in the trading account. The value

of the trading account is the wealth accumulated by trading the securities

following to some trading strategies. The derivative security called Pass-

port option is the contract that an option holder has the right to gain the

positive part of trading account at maturity. The option holder executes

a trading strategy (ctitiously) which is chosen by himself in the class of

strategies specied in the contract, and whenever the option holder changes

the position, he has to report the amount of security to the option writer.

In this paper we suppose that there is one traded security whose value

process St follows Black & Scholes model [1], i.e. under the risk neutral

measure P , St satises

dSt = St(rdt+ dBt),(1.1)

where r and are constants and B is a P -Brownian motion. The value

X,xt of trading account at time t under the strategy with initial value x

1991 Mathematics Subject Classication. Primary 90A09; Secondary 60J55.Key words: passport option, Black & Scholes model, local time, Skorohods Theorem,

stochastic control.

747

748 Izumi Nagayama

is expressed as

X,xt = x+

t0sdSs.(1.2)

Let 1 < 2. We denote by (1, 2) the set of predictable processes

satisfying 1 t 2 for any t 0. The passport option whose pay-o isa positive part of trading account at maturity T is evaluated by

C(T, x) = sup(1,2)

E[erT (X,xT )+].(1.3)

We dene Zt as a process which satises

dZt = Zt(rdt+ dBt), Z0 = 1

and dene

(t, y) = r,1,2(t, y)def= sup

(1,2)E[(y +

t0udZu)

+].

Then, we have C(T, x) = S0erT(T,

x

S0).

The optimal strategy which attain the supremum in (1.3) does not al-

ways exist in the strategy class (1, 2), and the explicit formula of (1.3)

is not known in general.

In Section 2 we give the denition of passport option and show that its

price is given by (1.3).

In Section 3 we assume that r 0, 1 = 1, and 2 = 1. And letting

0(t, y) = r,0 (t, y)

def= E[(y +

t0dZu)

+] = E[(y + Zt 1)+],

we estimate (t, y) and examine the conditions for = 0. Let (x) =12

ex2

2 and (x) =

x

(y)dy. We have the following theorem as the

estimation of (t, y).

Theorem 1.1. For any c > 0, t > 0, and y R,(t, y) 0(t, y)

(e2t 1) 122

( c

t

)+ 2

|y|

ec

(1 e2t)

14

e3r2

22t.

Pricing of Passport Option 749

For each T > 0, we dene

R(T )def= inf{r > 0 : r,(t, y) = r,0 (t, y)

for all r r, t [0, T ], and y R}.We show the following results.

Theorem 1.2. Let c be the solution of c(

1 c) = 1 c (

1 c). Then

R(T ) = 2c+3c(1 32c)

1 cT

4c{c2(1 2c)16(1 c)5/2 +

131c3 270c2 + 168c 3248(1 c)2

}T + o(T ),

as T 0.

Theorem 1.3. R(T ) = 2 2 log(22T )

2(2T + 1)+ o

(1

T 2

), as T .

Theorem 1.4. For any r, there exists T0 = T0(r, 2) > 0 such that

0(t, y) = (t, y), for any t [0, T0] and y 1.

In Section 4, we prove that the value function of passport option can be

derived as the limit of discrete time approximation.

In Section 5, we consider the case that r = 0. Hyer, Lipton-Lifschitz

and Pugachevsky [2] derived the explicit formula in the case that 1 =

2 using partial dierential equation approach. We derive the explicitformula in general case that 1 2 using stochastic analytic approach.By introducing the Equivalent Martingale Measure P with respect to the

numeraire St, we show that the problem of passport option valuation comes

down to the problem of calculation of an expected value E[(Y yT )+] where

Y yt is the solution of

dYt = (1 + |Yt|)dBt, Y0 = y.

750 Izumi Nagayama

Using Skorohods Theorem, we derive the explicit formula for valuation of

passport options (see Theorem 5.6).

In Section 6, by numerical calculations we illustrate the functions r(t)and (r(t), t) introduced in Section 3. Furthermore numerical experiencesin the discrete time framework discussed at Section 4 are reported.

The author expresses her sincere thanks to Professors Shigeo Kusuoka

and Freddy Delbaen for their kind advice and discussion.

2. Notation and Problems

Let (Bt)0t


To prove this proposition, we need some preparations. Let Zt be a

process given by the following S.D.E.

dZt = Zt(rdt+ dBt), Z0 = 1,

i.e. Zt = exp(Bt 122t+rt). Let U ,xt = sup

E

[(X,xt +

TtudSu)

+|Ft]

for , x R, and t [0, T ]. Let

(t, y) = r,1,2(t, y)def= sup

(1,2)E[(y +

t0udZu)

+].

Then it is easy to see that 0 (t, y2) (t, y1) y2 y1 for y1 y2 andthat U ,xt = St(T t, X

,xtSt

). Then we have the following.

Lemma 2.2. {U ,xt }0tT is a non-negative supermartingale.

Proof. E[|U ,xT |

]= E

[|X,xT |

]

752 Izumi Nagayama

Therefore (T t, X,xt

St) = lim

nE

(X,xt

St+

Ttns dZs

)+Ft, a.s. Then

by the dominated convergence theorem we have

E[U ,xt |Fs] = limnE[(X,xt +

TtnudSu)

+

Fs]

sup

E

[(X,xt +

TtudSu)

+

Fs].

Proof of Proposition 2.1. Suppose that C(T, x, S, 1, 2) sup

E

[erT (X,xT )

+S0=S

]. Then we

sell the passport option by value C(T, x, S, 1, 2). Suppose the option

holder takes a trading strategy . From Lemma 2.2, {erTU ,xt }0tT is anon-negative supermartingale, so it has a unique decomposition erTU ,xt =M ,xt A,xt , where {M ,xt } is a square integrable martingale and {A,xt } isa non-decreasing predictable process with A0 = 0. Let St = e

rtSt, thenwe have dSt = StdBt and S is a P -martingale. From the representation

theorem of Brownian martingales, there exists an adapted process {Ht} suchthat E[

T0 H

2s S

2sds]


U ,xT +erTA,xT = (X

,xT )

++erTA,xT . Then we will have the gain erTA,xT 0

at time T . Since the initial gain was C(T, x, S, 1, 2) M ,x0 > 0, it is anarbitrage.

For any positive constant , by considering the passport option

whose underlying is times the original underlying, we have that

C(T, x, S, 1, 2) = C(T, x, S, 1, 2). And by changing time, we have

Cr,(T, x, S, 1, 2) = Cr/2,1(2T, x, S, 1, 2).

Proposition 2.3. (t, y) is convex, increasing, and Lipschitz with re-

spect to y. Moreover if 1 0 2, (t, y) is increasing in t.

Proof. For any convex function g and 0 1, we have

sup

E[g(x+ (1 )y + t0udZu)]

sup

[E[g(x+

t0udZu)] + (1 )E[g(y +

t0udZu)]

]

sup

E[g(x+

t0udZu)] + (1 ) sup

E[g(y +

t0udZu)].

Therefore (t, ) is convex. We can easily see that (t, y) is increasing andLipschitz with respect to y. Let 1 0 2 and s < t, then

(t, y) = sup

E[(y +

t0udZu)

+] sup

E[(y +

s0udZu)

+] = (s, y).

3. The Estimation of the Value Function

In this section, we assume r 0, 1 = 1, and 2 = 1. By Proposition2.1, C(t, x, S) = Sert(t, xS ). Let

0(t, y) = r,0 (t, y)

def= E[(y+

t0dZu)

+] = E[(y+Zt1)+] , t 0, y R.

Then obviously we have (t, y) 0(t, y). Let

(x) =12

ex2

2 and (x) =

x

(y)dy.

754 Izumi Nagayama

Theorem 3.1. For any c > 0,

(t, y) 0(t, y)

(e2t 1) 122

( c

t

)+ 2

|y|

ec

(1 e2t)

14

e3r2

22t.

Proof. For each , let = inf{t > 0 : y + t0 sdZs = 0}. First,let us assume that y 0. Then we have

(t, y) = sup

(E[y +

t0sdZs] + E[(y +

t0sdZs)

, t > ])

0(t, y) + sup

E[(y +

t0sdZs)

, t > ].

Recall that dZt = Zt(rdt+dBt). Dene a P -equivalent probability measure

Q by (dQdP )t = tdef= exp( rBt r

2

22t). Since Wt

def= r t+Bt is a Q-Brownian

motion, Z is a Q-square integrable martingale. Then, for each , wehave

E

[(y +

t0sdZs)

, t > ] E

[ t0sdZs

I{t>}]

= EQ[ t0sdZs

I{t>} 1t]

EQ[ t0sdZs

2] 1

2

EQ[I{t>}]14EQ

[1

4t

] 14

.

Here we have EQ[

1

4t

]= E

[1

3t

]= E

[exp

(3r

Bt +

3r2

22t

)]= e

6r2

2t. Let

Mt =

t0sdZs, then we have

[M,M ]t =

t02sd[Z,Z]s

t02Z2sds =

2 t0

exp(2Ws 2s)ds.

Therefore we have

EQ[ t0sdZs

2] 2

t0e

2sds = (e2t 1).


By Knights theorem (see Ikeda and Watanabe [3]) there exists a Q-

Brownian motion {Wt}t[0,) such that Ms = W[M,M ]s . Therefore we havefor any c > 0

EQ[I{t>}] = Q[

inf0st

Ms < y]

= Q

[inf

0s[M,M ]tWs < y

]

Q[inf{Ws : 0 s (1 e2t) exp(2 max

0utWu)} < y

]

Q[

max0ut

Wu c]

+Q

[inf

0s(1e2t)e2cWs < y

].

It is well known that Q[ max0ut

Wu a] = 2( a

t

)for a > 0. So we have

EQ[I{t>}] 2( c

t

)+ 2

y

ec

(1 e2t)

. Therefore we have

our assertion in the case that y 0.We can prove the case of y < 0 similarly.

By Theorem 3.1, we can easily prove that for any p > 0, there exists

cp > 0 such that (t, y) 0(t, y) + cp|y|p/2, |y| 1. Therefore we have

limy(t, y) = 0, limy

1

y(t, y) = 1, and

limy

+

y+(t, y) = lim

y

y(t, y) = 1.

In the rest of this section, we examine the condition such that = 0is satised.

Remark 3.2. We can easily see the following.

0(t, y) =

(r,y,t,)

(y + etx 1

22t+rt 1)e

x22

2dx if y < 1

y + ert 1 if y 1

where (r, y, t, ) = log(1 y) 122t+ rt

t

for t > 0, y < 1.

756 Izumi Nagayama

For each T > 0 and > 0, dene

R(T ) = inf{r > 0 : r,(t, y) = r,0 (t, y)for all r r, t [0, T ], and y R}.

Let V ,xt = V,x,r,t

def= St

r,0

(T t, X

,xt

St

), for , x R, 0 t T .

Then {V 1,xt }0tT is a martingale. Let

F (r, y, t, ) = r0y

(t, y) y220y2

(t, y).

Lemma 3.3. For any T > 0 and > 0,

R(T ) = inf {r : F (r, y, t, ) 0, for any y (0, 1), t (0, T ]} .

Proof. Noting that V ,xT =

(x+

T0tdSt

)+, we have the following

from the proof of Proposition 2.1.

R(T ) = inf{r > 0 : {V ,x,r,t }0tT is a P -supermartingale,for all r r, x R, and }.

So it is sucient to show that {V ,x,r,t }0tT is a P -supermartingale forall and x R if and only if F (r, y, t, ) 0 for all y (0, 1) andt (0, T ]. Let

Y ,ytdef=

X,xtSt

, x = yS0.(3.1)

Then by Itos formula we have

dY ,yt = (t Y ,yt )(rdt+ dBt 2dt)

and

dV ,x,r,t = V (St, Y

,yt , t, T t)dBt + V (St, Y ,yt , t, T t)dt,


where

V (S, y, , t) = S( y)0y

(t, y) + S0(t, y)

V (S, y, , t) = rS( y)0y

(t, y) S0t

(t, y)

+1

2S( y)22

20y2

(t, y) + rS0(t, y).

Since {V 1,x,r,t }0tT is a martingale, we have V (S, y, 1, t) = 0 for allt (0, T ]. Note that {V ,x,r,t }0tT is supermartingale if and only ifV (S, y, , t) 0 = V (S, y, 1, t) for any y R, S > 0, t (0, T ], || 1.This is equivalent to

r0y

(t, y) +2

2( + 1 2y)

20y2

(t, y) 0,(3.2)for y R, t (0, T ], || 1.

Since 0(t, y) is convex in y, we have20y2

0, and so the condition (3.2)is equivalent to that

F (r, y, t, ) = r0y

(t, y) y220y2

(t, y) 0, t (0, T ](3.3)

for all y R. Here we have0y

(t, y) =

{1 if y 1((r, y, t, )) if y < 1

20y2

(t, y) =

{0 if y 1

1(1y)t((r, y, t, )) if y < 1.

So in the case that y 1 or y 0, condition (3.3) is satised for allt (0, T ]. This proves our assertion.

Remark 3.4. From Remark 3.2, we have the following.

F (r, y, t, ) = r((r, y, t, )) y(1 y)t((r, y, t, ))(3.4)

F

r(r, y, t, ) = ((r, y, t, ))(3.5)

758 Izumi Nagayama

+((r, y, t, ))

{t

r +

y

1 y(r, y, t, )}

F

y(r, y, t, ) =

((r, y, t, ))

(1 y)2t3

{(r 2)t 1

22ty y log(1 y)

}(3.6)

F

t(r, y, t, ) = ((r, y, t, ))

{y

2(1 y)t3

(3.7)

+

(r +

y(r, y, t, )

(1 y)t

)log(1 y) 122t+ rt

2t3

}.

Lemma 3.5. R(T ) 2.

Proof. From (3.5), we haveF

r(r, y, t, ) 0 for all r 2, y (0, 1)

and t (0, T ]. So it is sucient to show that F (2, y, t, ) 0 for ally (0, 1) and t (0, T ]. From (3.6), we have

F

y(2, y, t, )

{ 0 if 0 < y 1 e 122t 0 if 1 e 122t y < 1.

It is sucient to show that F0(t)def= F (2, 1 e 122t, t, ) 0 for all t > 0.

Since F 0(t) =(e

122t 1)2t3

(t) 0, F0(t) is an increasing function.

Also,

limt0

F0(t) = limt0

2(t) lim

t0(e

122t 1)t

(t) =

2

2.

Therefore F (2, 1 e 122t, t, ) 22 for all t > 0.

Lemma 3.6. Let r < 2. Then for each t > 0, there exists a unique

(r, t) (0, 1) such that F (r, (r, t), t, ) = miny(0,1)

F (r, y, t, ).

Proof. Note that((r, y, t, ))

(1 y)2t3

> 0. So from (3.6),F

y(r, y, t, ) > 0

(< 0) if and only if f(y) = (r 2)t 122ty y log(1 y) > 0 (< 0,


respectively). Note that f (y) = 122t log(1 y) + y

1 y . So f(y) is

increasing in y (0, 1). Also we have f (0) = 122t < 0, lim

y1f (y) = ,

f(0) = rt 2t < 0, and limy1

f(y) = . So we see that the equationF

y(r, y, t, ) = 0 in y has a unique solution and it gives the minimum of

F (r, , t, ).

Remark 3.7. The proof of Lemma 3.6 shows that (r, t) is a solution

ofF

y(r, y, t, ) = 0. Therefore (r, t) is characterized as a solution of

log(1 (r, t)) = rt 2t

(r, t) 1

22t, 0 < (r, t) < 1.(3.8)

Then we have

(r, (r, t), t, ) =2 r(r, t)

t+

rt

.(3.9)

By (3.8) we see that limt0{(r, t) log(1 (r, t))} and lim

t{(r, t)

log(1 (r, t))} = . This implies

limt0

(r, t) = 0 and limt

(r, t) = 1.(3.10)

Lemma 3.8. For r (0, 2), (r, t) =t(2 r) + r

4t+ o(t) as t 0.

Proof. Note that we have log(1 x) = x(1 +R(x)), where R(x) =i=1

1

i+ 1xi, for |x| < 1. From Equation (3.8), we have

(r, t)2(1 +R((r, t))) 122t(r, t) (2 r)t = 0.

Since (r, t) (0, 1), we have

(r, t) =t2t+

4t+ 16(2 r)(1 +R((r, t)))

4(1 +R((r, t))).(3.11)

760 Izumi Nagayama

So limt0

(r, t)t

=2 r. Let f(t) = (r, t)2 rt . Then from (3.8)

we have

(t(2 r) + f(t)) log(1

t(2 r) f(t))

+1

22t(

t(2 r) + f(t)) + (2 r)t = 0.

Since we have

log(1t(2 r)f(t)) = log(1

t(2 r)) f(t)

1t(2 r)(1+R(t))

where R(t) = R

(f(t)

1t(2 r))

, f(t) is the solution of the following

equation.

Atf(t)2 +Btf(t) + Ct = 0,

where At = 1 + R(t)1t(2 r)

Bt = log(1t(2 r))

t(2 r)(1 + R(t))1t(2 r) +

1

22t

Ct =t(2 r) log(1

t(2 r))

+1

22t

t(2 r) + (2 r)t.

Since limt0At = 1, limt0

Btt

= 22 r, and lim

t0Ctt3

= 12(2 r)3/2 +

1

222 r, we have lim

t0f(t)

t=r

4.

We can show the following lemma similarly to Lemma 3.8.

Lemma 3.9. For r (0, 2),

(r, t) =t(2 r) + r

4t+

3r2 12r(2 r) 16(2 r)2962 r t

3/2

+o(tt) as t 0.


Let G(r, t) = F (r, (r, t), t, ). Then we have the following.

Proposition 3.10. For each t > 0, the equation G(r, t) = 0 in r has a

unique solution r(t) in (0, 2). Moreover r(t) is continuous in t (0,).

Proof. Note that by (3.5) we have

G

r(r, t) = ((r, (r, t), t, ))

+((r, (r, t), t, ))

t

r +

(r, t)

1 (r, t) r2(r,t) t+ rt

t

> 0 for 0 r < 2.

From (3.8) we have (r, t) = 1exp{

(r 2)t(r, t)

122t

}. Since 0 < (r, t) 0 from (3.8), so limr2

(r, t) = 1 e 122t. This

implies that limr2

G(r, t) = F (2, 1 e 122t, t, ) 2

2(see the proof of

Lemma 3.5). So we have the rst assertion and we complete the proof by

implicit function theorem.

Corollary 3.11. R(T ) = max0tT

r(t). In particular, R(T ) < 2.

Proposition 3.12. limt0

r(t) = c2 , where c is the solution of

c(

1 c) = 1 c(1 c).

Proof. Let G0(r) = limt0

G(r, t). Then we have

G0(r) = r

(1

2 r

)

2 r

(1

2 r

),

and

G0(c2) = c2(

1 c) 21 c(1 c) = 0.

762 Izumi Nagayama

Since G0(r) = (

1

2 r

)> 0, we have G0(c

2 8) < 0 < G0(c2 + 8),for any 8 > 0. So by Proposition 3.10 lim

t0r(t) = c2.

We get c = 0.2945... by numerical calculation.

Proposition 3.13. R(T ) = 2c+3c(1 32c)

1 cT+o(

T ), as T 0.

Proof. Let H(t, a) = G(c2 + at, t), where c is the value dened in

Proposition 3.12. By Lemma 3.8 we have

(c2 + at, t) =

1 ct

(a

2

1 c c2

4

)t+ o(t).

Therefore we have from (3.9)

(c2 + at, (c2 + a

t, t), t, )

=2(1 c) at

2

1 c(

a2

1c c3

4

)t

+ ct+ o(

t)

=

1 c+ h(a)t+ o(t),

where h(a) = a22

1 c +

3

4c. And we have

(c2 + at, t)

(1 (c2 + at, t))t =2

1 c(

a2

1c c3

4

)t

1 1 ct+(

a2

1c c2

4

)t

+ o(t)

= 2

1 c+ k(a)t+ o(t),

where k(a) = a2

1 c + 3 3c

3

4. Since (x + z) = (x) + (x)z +

o(z) as z 0 and (x+ z) = (x) x(x)z + o(z) as z 0 , we have

H(t, a) = (c2 + at){(1 c) + (1 c)h(a)t+ o(t)}

(21 c+ k(a)t){(1 c)1 c(1 c)h(a)t+ o(t)}.


Therefore from the denitions of c, we have

H0(a) = limt0

H(t, a)t

= (

1 c){c2h(a)+a

1 cc

+2(1c)h(a)k(a)}.

Noting that H 0(a) = (

1 c) 1 2cc

1 c > 0 and H0(3c(1 32c)

1 c

)= 0,

we see H0(3c(1 3

2c)

1c 8) < 0 < H0(3c(1 3

2c)

1c + 8) for any 8 > 0. So r(t) =

c2 +3c(1 32c)

1 ct+ o(

t). By Corollary 3.11, we have our assertion.

We can also show the following. Although we need ner computation,

the idea of the proof is similar and so we omit the proof.

Theorem 3.14.

R(T ) = 2c+3c(1 32c)

1 cT

4c{c2(1 2c)16(1 c)5/2 +

131c3 270c2 + 168c 3248(1 c)2

}T

+o(T ), as T 0.

We also have the following theorem.

Theorem 3.15. R(T ) = 2 2 log(22T )

2(2T + 1)+ o

(1

T 2

), as T .

Proof. For each h R, set r0(h, t) = 2 2 log(22t)

2(2t+ 1)+h

t2. Then,

(3.8) yields

0(h, t) = (r0(h, t), t)) = 1 e2

2t exp

(

2t log(22t)

2(2t+ 1)0(h, t)+

h

t0(h, t)

)

= 1 o(e2

2t) as t ,

764 Izumi Nagayama

and we have from (3.9)

0(h, t)def= (r0(h, t), 0(h, t), t, )

= t+

(2t log(22t)

2(2t+ 1) htt

)(1

0(h, t) 1

)

as t .Then we have

(0(h, t))t(1 0(h, t))

= exp

( log(2

2t)

2(2t+ 1) h

t

)

exp12

(2t log(22t)

2(2t+ 1) htt

)2 (1 0(h, t)0(h, t)

)2 .

Therefore we have

G(r0(h, t), t) = r0(h, t)(0(h, t)) 0(h, t) (0(h, t))t(1 0(h, t))

=

{2

2 log(22t)

2(2t+ 1)+

h

t2

}(1

0(h,t)

(x)dx

)

20(h, t) exp( log(2

2t)

2(2t+ 1) h

t

)

exp12

(2t log(22t)

2(2t+ 1) htt

)2 (1 0(h, t)0(h, t)

)2

=h

t2+2h

t+ o

(1

t

)as t .

So we have tG(r0(h, t), t) 2h as t . Consequently, we have r(t) =2

2 log(22t)

2(2t+ 1)+ o

(1

t2

)as t , since G is increasing in r by Propo-

sition 3.10. So Corollary 3.11 implies our assertion.

Theorem 3.16. For any r > 0, there exists T0 = T0(r, 2) > 0 such

that

0(t, y) = (t, y), 0 t T0, y 1.


In order to prove this theorem, we show some propositions. Let Y ,yt be

the process given by (3.1). Note that we have the following by Itos formula.

d(ertS1t ) = ertS1t (dBt 2dt).

Let tdef= exp

(Bt 122t

)and dene the probability measure P by(

dP

dP

)t

= t. Then Btdef= t + Bt is a P - Brownian motion. Since

ertS1t is a martingale under the measure P , P is EMM with respect to thenumeraire St. By applying Itos formula to (3.1), we have

dY ,yt = (t Y ,yt )(dBt + rdt), Y ,y0 = y.(3.12)

Moreover by denition of P , we have E[erTX] = S0E[S1T X] for any FTmeasurable random variable X, where E[ ] denotes the expectation under

P . So we have

C(T, x, S) = S sup

E[(Y

,x/ST )

+].(3.13)

Let (t, y) = sup

E[(Y ,yT )

+]

and 0(t, y) = E[(Y 1,yT )

+]. Then (t, y) =

ert(t, y) and 0(t, y) = ert0(t, y), so (t, y) = 0(t, y) if and only if

(t, y) = 0(t, y).

Proposition 3.17. Y 1,yt 1, P -a.s, for all y 1.

Proof. From (3.12) we have that Y 1,yt = 1+(y1)eBtrt122t 1

if y 1.

Proposition 3.18. (1) E[Y 1,yt Y ,yt ] = r t0 e

r(st)E[(1 s)]ds, forall and t > 0.(2) E[(Y 1,yt Y ,yt )2] 22e2

2t t0 E[(1 s)2]ds, for all and t > 0.

Proof. (1) Let Ut = Y1,yt Y ,yt . Then{

dUt = {(1 t) Ut}(dBt + rdt)U0 = 0.

766 Izumi Nagayama

Therefored

dtE[Ut] = rE[1 t] rE[Ut] . So E[Ut] = r

t0er(st)E[(1

s)]ds.

(2) Similarly we have by Itos formula,

d

dtE[U2t ] = 2rE[{(1 t) Ut}Ut] + 2E[{(1 t) Ut}2]

= (2 2r)E[U2t ] + 2E[(1 t)2] + 2(r 2)E[Ut(1 t)] {(2 2r) + |2 r|}E[U2t ] + (2 + |2 r|)E[(1 t)2].

Therefore we have from Gronwalls inequality

E[U2t ] (2 + |2 r|) exp({(2 2r) + |2 r|}t) t0E[(1 s)2]ds

22e22t t0E[(1 s)2]ds.

Proposition 3.19. There exists a constant K = K(r, 2) such that

P ( sup0st

|Y 1,ys Y ,ys | > 1) KtE[ t0

(1 s)2ds],for any and t [0, 1].

Proof. Letting Ut = Y1,yt Y ,yt , we have

Ut =

t0

(1 s)dBs t0UsdBs + r

t0

(1 s)ds r t0Usds.

So we have

P ( sup0st

|Us| > 1) P( sup

0st| s0

(1 u)dBu| > 14

)

+P

( sup

0st| s0UudBu| > 1

4

)

+P

(r

t0|1 s|ds > 1

4

)+ P

(r

t0|Us|ds > 1

4

).


Note that

P

(r

t0|1 s|ds > 1

4

) 16r2E

[( t0|1 s|ds

)2]

16r2tE[ t

0(1 s)2ds

].

Also by Proposition 3.18 (2)

P

(r

t0|Us|ds > 1

4

) 16r2tE

[ t0U2s ds

]

32r22e22ttE[ t

0(1 s)2ds

].

By Doobs inequality

P

( sup

0st| s0UudBu| > 1

4

) 162E

[sup

0st| s0UudBu|2

]

642E[| t0UsdBs|2

]

= 642E

[ t0U2s ds

]

1284e22ttE[ t

0(1 s)2ds

].

By Burkholders inequality there exists some constant C4 such that

P

( sup

0st| s0

(1 u)dBu| > 14

) 444E

[sup

0st| s0

(1 u)dBu|4]

C44E[(

t0

(1 s)2ds)2] 4C44tE

[ t0

(1 s)2ds].

Proof of Theorem 3.16. If x 1 and y R, then

x+ y+ = x y 1{y0}|y| x y 1{|xy|1}|x y|.

Therefore for each , if ertr (82Ke22t)1/2t1/2 0

E[(Y 1,yt )+] E[(Y ,yt )+]

768 Izumi Nagayama

E[Y 1,yt Y ,yt ] E[|Y 1,yt Y ,yt |, |Y 1,yt Y ,yt | 1] E[Y 1,yt Y ,yt ] E[|Y 1,yt Y ,yt |2]1/2P (|Y 1,yt Y ,yt | 1)1/2

rE[ t0er(st)(1 s)ds]

{

22E[e22t t0

(1 s)2ds]}1/2 {

KtE[

t0

(1 s)2ds]}1/2

(ertr (82Ke22t)1/2t1/2)E[ t0

(1 s)ds] 0.

We can obtain similar results in the case of r < 0 similarly.

4. Discrete Time Approximation

Let us x T > 0 and assume r 0. We introduce the discrete timeframework. Let N be any positive integer, and let

= N =T

N,

N = { : t = n, for t [n, (n+ 1)), n = 0, 1, , N 1},

Y N,N ,y

t = y +N1n=0

(Nn Y N,N ,y

n )

{(Bt(n+1) Btn) + r(t (n+ 1) t n)},

and

N (m, y) = supNN

E[(Y N,N ,y

m )+].

We will consider the optimal trading strategy in the case of discrete time

framework.

Proposition 4.1. N (m, ) is convex and

N (m, y) = max{E[N

(m 1, y + (2 y)(B + r)

)],

E[N

(m 1, y + (1 y)(B + r)

)]}.


Proof. This can be shown inductively. Let us assume that N (m 1, y) is convex in y. Then we have

N (m, y) = sup1k2

E[ supN ,0=k

E[(y + (k y)(B + r)

+m1n=1

(n Y N,,yn )((B(n+1) Bn) + r))+|F]]

= sup1k2

E[N (m 1, y + (k y)(B + r))]

max{E[N (m 1, y + (1 y)(B + r))],

E[N (m 1, y + (2 y)(B + r))]}.

The opposite side of the inequality is trivial. And so N (m, y) is convex in

y.

Lemma 4.2. sup

E[(Y ,yt )2] (y2 + 22 max{21, 22}t)e2

2t.

Proof. Since d(Y ,yt )2 = 2Y ,yt (tY ,yt )(dBt+rdt)+2(tY ,yt )2dt

for any , we have the following similarly to Proposition 3.18 (2).d

dtE[|Y ,yt |2] {(2 2r) + |2 r|}E[(Y ,yt )2] + (2 + |2 r|)E[2t ].

Therefore we have

E[|Y ,yt |2] (y2 + 22 t0E[2s ]ds) exp(2

2t)

(y2 + 22 max(21, 22)t)e22t.

Lemma 4.3. Let b 0, c 0 and let {an}n0 be a sequence of numberssuch that a0 = 0 and

an b+ cn1k=0

ak, n 1.

Then an b(1 + c)n1.

770 Izumi Nagayama

Proof. We show our assertion by induction. It is easy to verify it in

case of n = 1. Let an b+ cn1k=0

ak b(1 + c)n1. Then we have

an+1 b+ cn

k=0

ak b(1 + c)n1 + can b(1 + c)n.

Lemma 4.4. There exists a sequence {CN} such that CN 0 as N such that

E[|Y ,yT Y N,,yT |2

](

1 + 4(2 + r2T )T

N

)N1CN ,

for any N .

Proof. Let N . From the denitions of Y N,,yT and Y ,yT we have

E[|Y ,yN Y N,,yN |2

] 2(2 + r2T )E

[N1n=0

(n+1)n

|Y ,yt Y N,,yn |2dt]

= 4(2 + r2T )N1n=0

(n+1)n

E[|Y ,yt Y ,yn |2

]dt

+4(2 + r2T )N1n=0

E[|Y ,yn Y N,,yn |2

].

Since we have Y ,yt = Y,yn +(

NnY ,yn )(1 e(BtBn)(

2

2+r)(tn)) for

n t < (n+ 1), we have (n+1)n

E[|Y ,yt Y ,yn |2

]dt

=

(n+1)n

E

[(Nn Y ,yn )2(1 e(BtBn)(

2

2+r)(tn))2

]dt

K(

2r(1 er) + 12r + 2 (e

(2r+2) 1)),

where K = 2(y2 +22 max{21, 22}T )e22T +2 supN E[(Y ,yn Nn)2].

Therefore

E[|Y ,yN Y N,,yN |2

]


4(2 + r2T )KN(

2(1 er)

r+e(2r+2) 12r + 2

)

+4(2 + r2T )N1n=0

E[|Y ,yn Y N,,yn |2

].

So by Lemma 4.3

E[|Y ,yN Y N,,yN |2

]

4(2 + r2T )KN(

2(1 er)

r+e(2r+2) 12r + 2

)

(1 + 4(2 + r2T )

)N1.

Since 2(1 er)

r+e(2r+2) 12r + 2 = O

(2)

, we have our assertion.

Proposition 4.5. limN

N (t, y) = (t, y).

Proof. Lemma 4.4 and N imply thatlim supN

supN

E[(Y N,,yt )+] = lim sup

NsupN

E[(Y ,yt )+] (t, y).

On the other hand, for each , there exists a sequence of strategies{N}N=1 such that N N and lim

NE

[ T0

(t Nt )2dt]

= 0. Then,

E[|Y ,yt Y

N ,yt |2

]= 2(2 + r2T )E

[ t0|(s Ns ) (Y ,ys Y

N ,ys )|2ds

]

4(2 + r2t) t0E[(s Ns )2

]ds

+4(2 + r2t)

t0E[|Y ,ys Y

N ,ys |2

]ds.

By Gronwalls inequality

E[|Y ,yt Y

N ,yt |2

] 4(2 + r2t)E

[ t0

(s Ns )2ds]e4(

2+r2t)t 0,as N .

772 Izumi Nagayama

Lemma 4.2 implies that for each N there exists a sequence {(N,n) n}nsuch that E

[|Y (N),yT Y n,

(N,n),yT |2

] 0 as n uniformly with respect

to N . Therefore we have subsequence {(N,n(N))}N such that E[|Y ,yT

Yn(N),(N,n(N)),yT |2

] 0 as N . Therefore lim

NE[(Y

n(N),(N,n(N)),yT )

+] =

E[(Y ,yT )+]. Therefore lim inf

NN (T, y) (T, y).

5. Closed Form of Valuation in the Case of r = 0

In this section we assume that r = 0. Then, under the measure P , we

have

dY ,yt = (t Y ,yt )dBt.Letting 1 < 2, we derive the closed form for valuation of passport call

option. Let =2 + 1

2, =

2 12

and x T > 0. We consider the

discrete time approximation dened in the previous section.

Lemma 5.1. For any convex function g, we have

sup(1,2)

E

[g

( t0sdBs

)]= E[g(iBt)],

where i =

{2 if |1| |2|1 if |1| > |2|.

Proof. Let B be a Brownian motion independent of B. Then for any

E

[g

( t0sdBs

)]= E

[g

(E

[ t0

(sdBs + (2i 2s)1/2dBs)|Ft

])]

E[g

( t0

(sdBs + (2i 2s)1/2dBs)

)]= E [g (iBt)] .

Corollary 5.2.

N (m, y) =

E[N

(m 1, y + (2 y)B

)]if y

E[N

(m 1, y + (1 y)B

)]if y > .


Proof. This follows from Lemma 5.1 and the following.

N (m, y) = sup12

E[N (m 1, y + ( y)B)]

= sup1y2y

E[N (m 1, y + B)].

By Corollary 5.2 the optimal strategy N in the discrete framework isinductively given by the following.

Nt ={1 if Y

N,N,yn >

2 if YN,N,yn ,

n t < (n+ 1),

where

Y N,N,y

t = y +N1n=0

(Nn Y N,N,y

n )(Bt(n+1) Btn)

= y N1n=0

[sign

(Y N,

N,yn

) (+

Y N,N,yn )

(Bt(n+1) Btn)].

Let BN,$t be given inductively by

BN,$t = BN,$n sign(Y N,

N,yn )(Bt Bn), n < t (n+ 1).

Then {BN,$t } is another P -Brownian motion and we have

Y N,N,y

t = y +N1n=0

((+

Y N,N,yn )(BN,$t(n+1) BN,$tn)).

Now let

Y N,$,yt = y +N1n=0

((+

Y N,$,yn )(Bt(n+1) Btn)) .

774 Izumi Nagayama

Then Y N,N,y

t and YN,$,yt have the same distribution and we have

E[(Y N,N,y

t )+] = E[(Y N,$,yt )

+]. And let Y $,yt to be the solution of the fol-

lowing S.D.E.

dY $,yt =(+

Y $,yt )dBt, Y $,y0 = y.(5.1)Then we have lim

NE[|Y N,$,yT Y $,yT |2] = 0 by Euler-Maruyama approxima-

tion (see Kloeden and Platen [5]), and so limN

E[(Y N,$,yT )+] = E[(Y $,yT )

+].

Therefore by Proposition 4.5 we have (T, y) = E[(Y $,yT )+].

Remark 5.3. From the form of N, one may guess that the optimalstrategy is given by the following.

t ={1 if Yt >

2 if Yt , n t < (n+ 1)

where Y is given by

dYt = sign(Yt )(+ |Yt |)dBt, Y0 = y.For simplicity, let 1 = 1 and 2 = 1. Then recalling that Bt = Bt + t isP -Brownian motion, we see that Xt = YtSt satises

dXt = sign(Xt)dSt, X0 = y,(5.2)under the measure P . It is well known as Tanakas counter-example that

S.D.E. (5.2) does not have any strong solutions. Therefore such an adapted

process does not exist.

Now we derive the formula for pricing the passport option. Let Y ytdef=

1

(Y $,y+t

). Then we have

dY yt = (1 + |Y yt |)dBt.Let f Y

0(y : t)dy = P [Y 0t dy] and q(a, b, t;)dadb = P [Bt + t da,

min0st

(Bs + s) db]. It is well known(see Karatzas & Shreve [4]) that

q(a, b, t;) =2(a 2b)

2t3exp

((a 2b t)

2

2t+ 2b

).


Skorohods Theorem implies the following lemma.

Lemma 5.4.

f Y0(y : t) =

1

(1 + |y|)2{

1

t

( log(1 + |y|)

t

+1

2t

)

+1

2

( log(1 + |y|)

t

+1

2t

)}.

Proof. We have

d(log(1 + |Y 0t |)) = sign(Y 0t )dBt + dL0t 1

22dt = dBt + dL

0t

1

22dt,

where L0t is the local time of Y0 at 0, and Bt =

t0 sign(Y

0s )dBs. Dene a P -

equivalent probability measure Q by (dQ

dP)t = t

def= exp(

2Bt

2

8t). Then

BQtdef= Bt

2t is Q-Brownian motion, and we have d

(1

log(1 + |Y 0t |)

)=

dBQt + dL0t . Let Wt

def=

1

log(1 + |Y 0t |). From Skorohods Theorem, we have

Wt = BQt min

0utBQu (See Ikeda and Watanabe [3]). Since

Q[Wt dw,BQt db] = q(b, b w, t, 0)dwdb

=2(2w b)

2t3exp

((2w b)

2

2t

)dwdb,

we have

g(w : t)def=

d

dwP [Wt w] = d

dwEQ[I{Wtw}

1t ]

=

w

e182t 1

2b d

2

dwdbQ[Wt w,BQt b]db

= 2ew{

1

t

( w

t+

1

2t

)+

1

2

( w

t+

1

2t

)},

t, w 0.Then,

f Y0(y : t) =

1

2(1 + |y|)g(1

log(1 + |y|) : t)

776 Izumi Nagayama

=1

(1 + |y|)2{

1

t

( log(1 + |y|)

t

+1

2t

)

+1

2

( log(1 + |y|)

t

+1

2t

)}.

Lemma 5.5.

C(T, S, S, 1, 2)

= S

[

2

{(1 +

Td1(T ))(d1(T )) +

T(d1(T ))

}

++ + ||2

(d2(T ))

]

where d1(T ) = 1T

log+ ||

+T

2and d2(T ) = d1(T )

T .

Proof. Note that C(T, S, S, 1, 2) = SE[(Y $,T )

+]. Then we have

E[(Y $,T )

+]

= E

[(Y 0T +

)+]=

(y + ) f Y0(y : T )dy

=

||

(y + ) f Y0(y : T )dy + 2+

||

0f Y

0(y : T )dy

=

d1(T )

(e

Tz+ 1

22T +

)eTz 1

22T

((z) +

T

2(z)

)dz

+2+ 1

2T

d1(T )eTz 1

22T

((z) +

T

2(z)

)dz

=

2

{(1 +

Td1(T ))(d1(T )) +

T(d1(T ))

}

++ + ||2

(d2(T )).

Theorem 5.6.

C(T, x, S, 1, 2)


= SeTf2(x,T )

(f2(x, T ) f3(x, T )+ +

2

T

)

S( sign(x S))(f2(x, T ) f3(x, T )+

2

T

)

+SeTf2(x,T )( sign(x S))

(f2(x, T ) f3(x, T )+

2

T

)

S(f2(x, T ) f3(x, T )+ +

2

T

)+ x1{x0} 1{x 0

if sign(x S) 0gx(t) = f2(x, t)

t

(f2(x, t) +

1

2t

).

Proof. Note that C(T, x, S, 1, 2) = SE[(Y

$,x/ST )

+]

and let y =

inf{t > 0 : Y $,yt =

}= inf

{t > 0 : Y yt = 0

}, where y =

y

. Then, we

778 Izumi Nagayama

have

E[(Y $,yT )

+]

= E[(Y $,yT )

+, T y]+ E

[(Y $,yT )

+, T < y].

And

dY yty ={(1 + Y yty)dBty if y 0(1 Y yty)dBty if y < 0.

Therefore

Y yty = sign(y)[(1 + |y|) exp

{sign(y)Bty 1

22(t y)

} 1

].

Then, we have

y =

inf{t > 0 : Bt = 1( log(1 + y) + 122t

)} if y 0

inf{t > 0 : Bt = 1( log(1 y) + 122t

)} if y < 0

= inf

{t > 0 : Bt = sign(y)

1

( log(1 + |y|) + 1

22t

)}.

Then

E[(Y $,yT )

+, T < y]

= E

[(Y yT +

)+, T < y

]

=

0 1

log(1+|y|)

( b

q(a, b, T ;2)

{sign(y)((1 + |y|)ea 1) + }+da

)db

=

0f

( b(f+h)

q(a, b, T ;2)

{(1 + y)ea

}da

)db

if y > 0

0f

( b(f+h)b

q(a, b, T ;2)

{(y 1)ea + +

}da

)db

if y < 0

where

f = 1

log(1 + |y|)

h =

{1 log

sign(y) if sign(y) > 0

if sign(y) 0.


Since

( eaq(a, b, T ;

2)da

)db

= e[

(2

T+

2

T

)

(2

T+

2

T

)]

e[

(2

T+

2

T

)

(2

T+

2

T

)],

( beaq(a, b, T ;

2)da

)db

=

(T

2

T

)

(T

2

T

)

+e[

(2

T+

2

T

)

(T

+

2

T

)]

+e[

(T

+

2

T

)

(2

T+

2

T

)], for ,

and

q(a, b, T ;2) = eaq(a, b, T ;

2),

we have

E[(Y $,yT )

+, T < y]

= (+ |y |)(f + h

+

T

+

2

T

)

( sign(y))(f + h

+

T

2

T

)

+ sign(y)

(+ |y |)

(f h+

T

2

T

)

(f h+

T+

2

T

)

1{y

780 Izumi Nagayama

Next, we have

E[(Y $,yT )

+, T y]

=

T0g

y(T t)E

[(Y $,t )

+]dt

=

T0g

y(T t)

[

2

{(1 +

td1(t))(d1(t)) +

t(d1(t))

}

++ + ||2

(d2(t))

]dt

=

T0g

y(T t)

[

2

{(1 log + ||

+2t

2)(d1(t)) +

t(d1(t))

}

++ + ||2

(d2(t))

]dt

where

gy(t)

def=

d

dtP [y t] = log(1 + |y|)

t3

( log(1 + |y|)

t

+1

2t

)

= ft3

(ft

+1

2t

).

Since

T0g

y(T t)dt =

(fT

+

2

T

)+ (1 + |y|)

(fT

2

T

) T0g

y(T t)(d1(t))dt =

(d1(T ) +

fT

) T0g

y(T t)(d2(t))dt = (1 + |y|)

(d2(T ) +

fT

),

we have

E[(Y $,yT )

+, T y]

= +[

(fT

+

2

T

)+ (1 + |y|)

(fT

2

T

)]

+

2

(1 log + ||

+2t

2

)

(d1(T ) +

fT

)


2

+ ||2

(1 + |y|)(d2(T ) +

fT

)

+

2

T0g

y(T t)

{2t

2(d1(t)) +

t(d1(t))

}dt.

6. Remarks and Numerical Calculations

We examined the functions r(t) and (r(t), t) with = 1by numerical calculations based on Equation (3.8) and the denition

of r(t) in Proposition 3.10. Figure 1 depicts the function r(t) and its

approximation functions r0(t) = c +c(1 32c)

1 ct c

{c2(1 2c)

16(1 c)5/2 +131c3 270c2 + 168c 32

48(1 c)2}t and r(t) = 1

log(2)

2(t+ 1)which are exam-

ined in Theorem 3.14 and 3.15. Figures 2 and 3 show these approximation

errors respectively. In Figure 4, we illustrate the function (r(t), t) with its

approximation function 0(r(t), t) =

t(1 r(t))+ r(t)4 t+

{r(t)2

32

1r(t)

Figure 1.

782 Izumi Nagayama

Figure 2.

Figure 3.

r(t)8

1 r(t) 16(1 r(t))3/2

}tt which are examined in Lemma 3.8.

Figure 5 shows approximation error.

As numerical experiences we also examined the area where the optimal

trading strategy is 1 in the discrete time framework discussed in Propo-sition 4.1, when 1 = 1, 2 = 1, and = 1. We calculate the value ofN by using numerical integration technique inductively. Figures 6 and 7

show the case of r = 0.35(> c) and r = 0.25(< c) respectively. The sym-

bol in these gures indicates the points where the optimal strategy is


Figure 4.

Figure 5.

784 Izumi Nagayama

Figure 6.

Figure 7.

1, in the coordinate (y, t) where y is the value of trading strategy dividedby the current value of the security, and t is the remaining life time of


passport option. By numerical calculation we have r(0.075) 0.35 and(0.35, 0.075) 0.225. The symbol in Figure 6 indicates the pointwhere y = 0.225 and t = 0.075.

References

[1] Black, F. and M. Scholes, The Pricing of Options and Corporate Liabilities,J. Political Economy 81 (1973), 637654.

[2] Hyer, T., Lipton-Lifschitz, A. and D. Pugachevsky, Passport to Success, Risk10 (1997), September, 127131.

[3] Ikeda, N. and S. Watanabe, Stochastic dierential equations and diusionprocesses, 2nd ed, North Holland/Kodansha, 1988.

[4] Karatzas, I. and S. E. Shreve, Brownian Motion and Stochastic Calculus,Springer-Verlag, 1988.

[5] Kloden, P. E. and E. Platen, Numerical Solution of Stochastic DierentialEquations, Springer-Verlag, 1991.

(Received May 21, 1998)(Revised July 21, 1998)

Graduate School of Mathematical ScienceThe University of Tokyo3-8-1 Komaba, Meguro-kuTokyo 153-8914, Japan, andThe Bank of Tokyo-Mitubishi Ltd.Derivative and Structured Products Division2-7-1 Marunouchi, Chiyoda-kuTokyo 100-0005, JapanE-mail: izumi [email protected]

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