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Chapter 7

A Foreign Exchange Model

7.1 The assets and the risk-neutral measures

Steve E. ShreveOctober 10, 2005

In this section we consider an N -period binomial model, but with twocurrencies. In particular, there is a domestic interest rate r ≥ 0 and a foreigninterest rate rf ≥ 0. The domestic interest rate leads to a domestic moneymarket account whose price at each time n is

Mn = (1 + r)n, n = 0, 1, . . . , N. (7.1.1)

The foreign interest rate leads to a foreign money market account whoseprice at each time n is

Mfn = (1 + rf )n, n = 0, 1, . . . , N. (7.1.2)

The domestic money market account is denominated in domestic currencyand the foreign money market account is denominated in foreign currency.

We also have an exchange rate process Qn, where Qn is the price of oneunit of foreign currency in units of domestic currency at time n. The initialexchange rate Q0 is assumed to be positive, and Qn+1 is defined in terms ofQn by the recursive equation

Qn+1(ω1 . . . ωnωn+1) ={

uQQn(ω1 . . . ωn) if ωn+1 = H,dQQn(ω1 . . . ωn) if ωn+1 = T.

(7.1.3)

The up factor uQ and down factor dQ for the exchange rate satisfy

0 < dQ <1 + r

1 + rf< uQ. (7.1.4)

1

2 CHAPTER 7. A FOREIGN EXCHANGE MODEL

Although it is not really necessary for our analysis, we shall assume thatuQ > 1 and dQ < 1 so the exchange rate really does increase when a cointoss results in H and really does decrease when a coin toss results in T .

In the domestic currency there are two assets in which an agent caninvest. One is the domestic money market account, and the other is theforeign money market account converted to domestic currency. The price ofthe latter asset at time n is

Sn = MfnQn = (1 + rf )nQn. (7.1.5)

Whereas the exchange rate has an up factor uQ and a down factor dQ, theasset S has an up and down factors

u = (1 + rf )uQ, d = (1 + rf )dQ. (7.1.6)

Even if dQ < 1, it is not necessarily the case that d < 1; thus, the term“down factor” should not be taken literally. We are only assured that d < ubecause of (7.1.4), so a coin toss of a tail results in a worse performancefor S than a coin toss of a head, although this worse performance may notactually be a decrease in price.

From (7.1.4) and (7.1.5) we see that

0 < d < 1 + r < u, (7.1.7)

which is the no-arbitrage condition for the binomial model. In particular,we may define risk-neutral probabilities

p =1 + r − d

u− d, q =

u− 1− r

u− d, (7.1.8)

and define the domestic risk-neutral measure P to correspond to independentcoin tosses with probability p for H on each toss and probability q for T oneach toss. Under P, the discounted process S is a martingale:

En

[Sn+1

Mn+1

]=

Sn

Mn, n = 0, 1, . . . , N − 1, (7.1.9)

Furthermore, P is the only probability measure under which discounted Sis a martingale. Recalling the definition (7.1.5) of S, we see that

En

[Mf

n+1Qn+1

Mn+1

]=

MfnQn

Mn, n = 0, 1, . . . , N − 1. (7.1.10)

7.1. THE ASSETS AND THE RISK-NEUTRAL MEASURES 3

In other words,

MfnQn

Mn=

(1 + rf

1 + r

)n

Qn, n = 0, 1, . . . , N, (7.1.11)

is a martingale under P.

In the foreign currency there are also two assets in which an agentcan invest. One is the foreign money market account, and the other is thedomestic money market account converted to foreign currency. The price ofthe latter asset at time n is

Wn =Mn

Qn=

(1 + r)n

Qn. (7.1.12)

When a coin toss results in a head, the exchange rate Q increases by thefactor uQ, so the reciprocal exchange rate decreases by the factor 1

uQ. There-

fore, W “decreases” by the factor dW = 1+ruQ

. When the coin toss resultsin a tail, the exchange rate Q decreases by the factor dQ, so the reciprocalexchange rate increases by the factor 1

dQ. Therefore, W increases by the

factor uW = 1+rdQ

. We do not really know that dW is less than one, but wecan see that dW < uW . Moreover, from (1.4), we see that with the “down”and up factors for W defined by the above formulas, which we repeat here,

uW =1 + r

dQ, dW =

1 + r

uQ, (7.1.13)

satisfy the no-arbitrage condition in the foreign currency

0 < dW < 1 + rf < uW . (7.1.14)

Because a T results in an up move of W and an H results in a down move,which is opposite to the effect of H and T on S, we define the foreignrisk-neutral probabilities opposite to the formula (7.1.8) for the domesticrisk-neutral probabilities. In particular, we define

qf =1 + rf − dW

uW − dW, pf =

uW − 1− rf

uW − dW(7.1.15)

and define the foreign risk-neutral measure Pf to correspond to indepen-dent coin tosses with probability pf for H (a “down” move of W ) on eachtoss and probability qf for T (an up move of W ) on each toss. Under Pf ,


the discounted (using the foreign money market account) process W is amartingale,

Efn

[Wn+1

Mfn+1

]=

Wn

Mfn

, n = 0, 1, . . . , N − 1, (7.1.16)

as we now show. If the first n coin tosses result in w = ω1 . . . ωn, then

Efn

[Wn+1

Mfn+1

](w)

= pf Wn+1(wH)

Mfn+1

+ qf Wn+1(wT )

Mfn+1

=uW − 1− rf

uW − dW· dW Wn(w)

(1 + rf )Mfn

+1 + rf − dW

uW − dW· uW Wn(w)

(1 + rf )Mfn

=(uW − 1− rf )dW + (1 + rf − dW )uW

(uW − dW )(1 + rf )· Wn(w)

Mfn

=uW dW − dW − rfdW + uW + rfuW − dW uW

(uW − dW )(1 + rf )· Wn(w)

Mfn

=uW − dW + rfuW − rfdW

(uW − dW )(1 + rf )· Wn(w)

Mfn

=Wn(w)

Mfn

.

Since w = ω1 . . . ωn is an arbitrary sequence of coin tosses, (7.1.16) holdsregardless of the outcome of the coin tosses. Recalling the definition (7.1.12)of W , we see that

Efn

[Mn+1

Mfn+1Qn+1

]=

Mn

MfnQn

, n = 0, 1, . . . , N − 1. (7.1.17)

In other words,

Mn

MfnQn

=(

1 + r

1 + rf

)n 1Qn

, n = 0, 1, . . . , N, (7.1.18)

is a martingale under Pf . Furthermore, Pf is the only probability measureunder which this process is a martingale.

7.2. EXAMPLE 5

��

ZZ

��

ZZ

��

ZZ

Q0 = 2

Q1(T ) = 43

Q1(H) = 3

Q2(TT ) = 89

Q2(HT ) = Q2(TH) = 2

Q2(HH) = 92

Figure 7.2.1: A two-period model.

7.2 Example

We consider a two-period example in which the initial exchange rate is Q0 =2 and the up and down factors for the exchange rate are

uQ =32, dQ =

23. (7.2.1)

The resulting exchange rate process is shown in Figure 7.2.1.

We take the domestic and foreign interest rates to be

r =14, rf =

13, (7.2.2)

so that

1 + r =54,

11 + r

=45, 1 + rf =

43,

11 + rf

=34. (7.2.3)

Note that

dQ =23

<1 + r

1 + rf=

1516

< uQ =32, (7.2.4)

as required by (7.1.4), so the no-arbitrage conditions (7.1.8) and (7.1.14) aresatisfied.

The up and down factors for S, the foreign money market denominatedin domestic currency, are given by (7.1.6). In this example these are

u = (1 + rf )uQ =43· 32

= 2, d = (1 + rf )dQ =43· 23

=89. (7.2.5)


��

ZZ

��

ZZ

��

ZZ

S0 = 2

S1(T ) = 169

S1(H) = 4

S2(TT ) = 12881

S2(HT ) = Q2(TH) = 329

S2(HH) = 8

Figure 7.2.2: The process Sn = MfnQn.

The process S is shown in Figure 7.2.2. The risk-neutral probabilities, givenby (7.1.8), are

p =1− r − d

u− d=

1 + 14 −

89

2− 89

=1340

, q = 1− p =2740

. (7.2.6)

The up and down factors for the domestic money market denominatedin foreign currency, given by (7.1.13), are

dW =1 + r

uQ=

54· 23

=56, uW =

1 + r

dQ=

54· 32

=158

. (7.2.7)

The price process for the domestic money market denominated in foreigncurrency, given by (7.1.12), is shown in Figure 7.2.3. The risk-neutral prob-abilities under the domestic currency, given by (7.1.15), are

qf =1 + rf − dW

uW − dW=

1 + 13 −

56

158 − 5

6

=1225

, pf = 1− pf =1325

. (7.2.8)

One can verify that Wn

Mfn

is a martingale under Pf , the probability measure

corresponding to independent coin tosses with probability pf for each H andqf for each tail. For example,

Ef1

[W2

Mf2

](H) = pf W2(HH)

Mf2 (HH)

+ qf W2(HT )

Mf2 (HT )

=1325

·2572(43

)2 +1225

·2532(43

)2

=916

[1325

· 2572

+1225

· 2532

]=

916

· 59

=516

=34· 512

=W1(H)

Mf1

.

7.3. RELATIONS BETWEEN RISK-NEUTRAL MEASURES 7

��

ZZ

��

ZZ

��

ZZ

W0 = 12

W1(T ) = 1516

W1(H) = 512

W2(TT ) = 225128

W2(HT ) = W2(TH) = 2532

W2(HH) = 2572

Figure 7.2.3: The process Wn = MnQn

.

7.3 Relations between risk-neutral measures

To understand the relationship between the risk-neutral measure P for thedomestic currency and the risk-neutral measure Pf for the foreign currency,we shall rely on the following theorem.

Theorem 7.3.1 (Quotient of martingales) In an N -period binomial model,suppose that Xn and Zn, n = 0, 1, . . . , N are martingales under a probabilitymeasure P such that P(ω1 . . . ωN ) > 0 for every ω1 . . . ωN . Suppose furtherthat ZN (ω1 . . . ωN ) is strictly positive for every ω1 . . . ωN and EZN = 1.Then we can define a new measure P by

P(ω1 . . . ωN ) = ZN (ω1 . . . ωN )P(ω1 . . . ωN ) for all ω1 . . . ωN . (7.3.1)

We have P(ω1 . . . ωN ) > 0 for every ω1 . . . ωN . We can also define Yn = XnZn

.Under P, the process Y0, Y1, . . . , YN is a martingale.

Proof: Because Zn is a martingale, if we set Z = ZN , then we are in thesetting of Theorem 3.2.1 and Lemmas 3.2.5 and 3.2.6 apply. Let n be anonnegative integer less than N . Since Yn+1 depends on only the first n + 1tosses, Lemma 3.2.6 implies

En[Yn+1] =1

ZnEn[Yn+1Zn+1] =

1Zn

En[Xn+1].

Since X0, X1, . . . , XN is a martingale under P, we have

1Zn

En[Xn+1] =1

ZnXn = Yn.


Therefore,En[Yn+1] = Yn,

which is the martingale property for Y0, Y1, . . . , YN under P. �

Remark 7.3.2 Let Zn, n = 0, 1, . . . , N , be a martingale under a probabilitymeasure P as described in Theorem 7.3.1. It is typically not the case that1

Zn, n = 0, 1, . . . , N , is also a martingale under P. In fact, from Jensen’s

inequality applied to the convex function ϕ(x) = 1x , defined for x > 0, and

from the fact that the expected value of martingale does not vary with timeso that Z0 = E[ZN ], we see that

1Z0

= ϕ(Z0) = ϕ(E[ZN ]

)≤ E

[ϕ(ZN )

]= E

[ 1ZN

]. (7.3.2)

Because ϕ′′(x) > 0 for all x > 0, the function ϕ has no “flat spots,” soinequality (7.3.2) is strict unless ZN (ω1 . . . ωn) = Z0 = 1 for every ω1 . . . ωN .But if this were the case, then we would have Zn = En[ZN ] = 1 for alln = 0, 1, . . . , N , and the martingale Zn would just be the constant 1 for allvalues of n and all outcomes of the coin tossing. Except for this case, theinequality in (7.3.2) is strict, and because the expected value of a martingalecannot vary with n, we conclude that 1

Zn, n = 0, 1, . . . , N , is not a martingale

under P. However, 1Zn

, n = 0, 1, . . . , N , is a martingale under the probabilitymeasure P given by (7.3.1). This is just a special case of Theorem 7.3.1 inwhich the martingale Xn is 1 for all values of n and all outcomes of the cointossing. �

We now apply Remark 7.3.2 to the foreign exchange model of Section7.1. In that section, the risky asset denominated in domestic currency wasthe foreign money market account converted to domestic currency, whichwe called Sn = Mf

nQn. We then discounted at the domestic interest rate toobtain Sn

Mn= Mf

nQn

Mn. We constructed the risk-neutral probability measure

P, so thatSn

Mn=

MfnQn

Mn, n = 0, 1, . . . , N, (7.3.3)

is a martingale under P. This was done using the risk-neutral probabilitiesp and q defined by (7.1.8).

The risky asset dominated in foreign currency was the domestic moneymarket converted to foreign currency, which we called Wn = Mn

Qn. We then

7.3. RELATIONS BETWEEN RISK-NEUTRAL MEASURES 9

discounted at the foreign interest rate to obtain Wn

Mfn

= Mn

MfnQn

. We con-

structed the risk-neutral probability measure Pf so that

Wn

Mfn

=Mn

MfnQn

, n = 0, 1, . . . , N, (7.3.4)

is a martingale under Pf .

The expression in (7.3.4) is the reciprocal of the expression in (7.3.3).According to Remark 7.3.2, the reciprocal of a martingale is not a martingaleunless we change the probability measure. We can apply this idea to changefrom the measure P under which Sn

Mnis a martingale to the measure Pf under

which Wn

Mfn

is a martingale. In fact, we use as the Zn process the martingaleSnMn

itself, except that we must normalize it so that it has expected value 1at time N , as required by Theorem 7.3.1. We thus define

Zn =Sn

S0Mn=

MfnQn

MnQ0, n = 0, 1, . . . , N. (7.3.5)

Because Zn, n = 0, 1, . . . , N , is a martingale under P, its expected valueunder P does not depend on n, and hence we have

E[ZN ] = Z0 = 1 (7.3.6)

because M0 = 1 and Mf0 = 1.

Lemmas 3.2.5 and 3.2.6 now take the form of the following lemmas.Note that here we are using the martingale Zn to change from P to Pf ,whereas in Lemmas 3.2.5 and 3.2.6 we were changing from P to P, butotherwise the statements of the following lemmas are just those of Lemmas3.2.5 and 3.2.6.

Lemma 7.3.3 Let n be a positive integer between 0 and N , and let Y be arandom variable depending only on the first n coin tosses. Then

Ef [Y ] = E[ZnY ], (7.3.7)

where Ef is the expected value using the probability measure Pf .

Lemma 7.3.4 Let n ≤ m be positive integers between 0 and N , and let Ybe a random variable depending only on the first m coin tosses. Then

Efn[Y ] =

1Zn

En[ZmY ], (7.3.8)

where Efn is the conditional expected value using the probability measure Pf .


We could of course make the change of measure in the reverse direction.The process 1

Zn, n = 0, 1, . . . , N , is a martingale under Pf , and so we can

use it as the “Zn” process in Remark 7.3.2, and thereby change from theprobability measure Pf to the probability measure P. The resulting formulasare given by the following lemmas.

Lemma 7.3.5 Let n be a positive integer between 0 and N , and let Y be arandom variable depending only on the first n coin tosses. Then

E[Y ] = Ef

[Y

Zn

]. (7.3.9)

Lemma 7.3.6 Let n ≤ m be positive integers between 0 and N , and let Ybe a random variable depending only on the first m coin tosses. Then

En[Y ] = ZnEfn

[Y

Zm

], (7.3.10)

Example 7.3.7 We illustrate Lemma 7.3.6 using the numerical example ofSection 7.3. We take n = 1, m = 2 and

Y (HH) = 8, Y (HT ) = 4, Y (TH) = 16, Y (TT ) = 8.

We compute E1[Y ](H). Under P, the probability of H on the second toss is(see (7.2.6)) p = 13

40 and the probability of T is q = 2740 . Therefore,

E1[Y ](H) = pY (HH) + qY (HT ) =1340

· 8 +2740

· 4 =2610

+2710

=5310

.

To compute the right-hand side of (7.3.10), we need some values of Z1 andZ2. The relevant values are

Z1(H) =Mf

1 Q1(H)M1Q0

=43 · 354 · 2

=85,

Z2(HH) =Mf

2 Q2(HH)M2Q0

=169 · 9

22516 · 2

=6425

,

Z2(HT ) =Mf

2 Q2(HT )M2Q0

=169 · 22516 · 2

=256225

.

7.4. APPLICATIONS 11

Under Pf , the probability of H on the second toss is (see (7.2.8)) pf = 1325

and the probability of T is qf = 1225 . Therefore,

Z1(H)Ef2

[Y

Zm

]= Z1(H)

[pf Y (HH)

Z2(HH)+ qf Y (HT )

Z2(HT )

]=

85·[1325

· 8 · 2564

+1225

· 4 · 225256

]=

85·[138

+2716

]=

85· 5316

=5310

.

The two-sides of (7.3.10) agree.

7.4 Applications

7.4.1 Forward Exchange Rates

Let n and m be integers satisfying 0 ≤ n ≤ m ≤ N . The time-n forwardexchange rate for foreign currency per unit of domestic currency for deliveryat time m is a random variable Forn,m depending on only the first n cointosses. This random variable is chosen so that a contract to receive oneunit of foreign currency in exchange for a payment of Forn,m in domesticcurrency at time m has value zero at time n. According to the risk-neutralpricing formula, the contract just described has time-n value

En

[1

(1 + r)m−n

(Qm − Forn,m

)]. (7.4.1)

Because Forn,m depends on only the first n tosses, En[Forn,m] = Forn,m. Inorder for the expression in (7.4.1) to be zero, we must therefore have

Forn,m = En[Qm].

Because MfnQn

Mnis a martingale under P (see (7.1.11)), we have

Forn,m = En[Qm] =Mm

Mfm

En

[Mf

mQm

Mm

]=

Mm

Mfm

·MfnQn

Mn=

(1 + r

1 + rf

)m−n

Qn.

We summarize with the following definition.


Definition 7.4.1 Let n and m be integers satisfying 0 ≤ n ≤ m ≤ N .The time-n forward exchange rate for foreign currency per unit of domesticcurrency for delivery at time m is

Forn,m =(

1 + r

1 + rf

)m−n

Qn. (7.4.2)

The time-n forward exchange rate for domestic currency per unit ofdomestic currency for delivery at time m is a random variable Forf (n, m)depending on only the first n coin tosses. This random variable is chosen sothat the contract to receive one unit of domestic currency in exchange for apayment of Forf

n,m in foreign currency at time m has value zero at time n.According to the risk-neutral pricing formula, the contract just describedhas time-n value (priced in foreign currency)

Efn

[1

(1 + rf )m−n

(1

Qm− Forf

n,m

)]. (7.4.3)

It is left as an exercise to show that this contract has value zero at time nif and only if Forf

n,m is given by the following definition.

Definition 7.4.2 Let n and m be integers satisfying 0 ≤ n ≤ m ≤ N .The time-n forward exchange rate for domestic currency per unit of foreigncurrency for delivery at time m is

Forfn,m =

(1 + rf

1 + r

)m−n 1Qn

. (7.4.4)

Example 7.4.3 Let 0 ≤ n ≤ N be given. What is the time-n value indomestic currency of a contract to receive one unit of foreign currency attime N in exchange for a payment of F0,N?

To solve this problem, we apply the risk-neutral pricing formula. Wealso use the fact that

(1+rf

1+r

)nQn is a martingale under P (see (7.1.11)),

which means that the average growth rate of Qn under P is 1+r1+rf per period.

7.4. APPLICATIONS 13

The time-n value of the contract described is

Vn = En

[1

(1 + r)N−n(QN − For0,N )

]=

1(1 + r)N−n

En[QN ]− 1(1 + r)N−n

For0,N

=1

(1 + r)N−n

(1 + r

1 + rf

)N−n

Qn −1

(1 + r)N−n

(1 + r

1 + rf

)N

Q0

=1

(1 + rf )N−nQn −

(1 + r)n

(1 + rf )NQ0.

If n = 0, this is zero, as it should be, since the forward price in this exampleis the one set at time zero. However, if n is not zero, then this contracttypically has a nonzero value, and this value could be either positive ornegative.

Example 7.4.4 Let 0 ≤ n < m ≤ N be given. Consider a contract toreceive Forf

m−1,m paid in foreign currency at time m. What is the value inforeign currency of this contract at time n?

To solve this problem, we apply the risk-neutral pricing formula underPf . We also use the fact that

(1+r1+rf

)n 1Qn

is a martingale under Pf (see

(7.1.18)), which means that the average growth rate of 1Qn

under Pf is 1+rf

1+r .The time-zero value of the contract described is

V fn = Ef

n

[1

(1 + rf )m−nForf

m−1,m

]= Ef

n

[1

(1 + rf )m−n· 1 + rf

1 + r· 1Qm−1

]=

1(1 + rf )m−n−1(1 + r)

Efn

[1

Qm−1

]=

1(1 + rf )m−n−1(1 + r)

·(

1 + rf

1 + r

)m−n−1 1Qn

=1

(1 + r)m−nQn.

7.4.2 Put-Call Duality

In this subsection we consider call and put options on currency.


We begin with a contract that grants the option to buy one unit offoreign currency in exchange for K units of domestic currency at time N .In other words, this contract is a call paying (QN −K)+ units of of domesticcurrency at time N . For 0 ≤ n ≤ N , the time-n domestic currency value ofthis contract is

Cn = En

[1

(1 + r)N−n(QN −K)+

](7.4.5)

We also consider a contract that grants the option to sell one unit ofdomestic currency in exchange of 1

K units of foreign currency at time N . Inother words, this second contract is a put paying

(1K − 1

QN

)+ units of foreigncurrency at time N . For 0 ≤ n ≤ N , the time-n foreign currency value ofthis contract is

Pn = Efn

[1

(1 + rf )N−n

(1K− 1

QN

)]. (7.4.6)

Theorem 7.4.5 We have

Cn = KQnPn. (7.4.7)

Proof: We recall the process Zn = MfnQn

MnQ0of (7.3.5) that permits us to

change from the probability measure P to the probability measure Pf . Usingthis process and Lemma 3.2.6, we may write

KQnPn = KQnEfn

[1

(1 + rf )N−n

(1K− 1

QN

)+]

= KQn1

ZnEn

[ZN

Mfn

MfN

(1K− 1

QN

)+]

= KQnMnQ0

MfnQn

En

[Mf

NQN

MNQ0· Mf

n

MfN

(1K− 1

QN

)+]

= Mn En

[KQN

MN

(1K− 1

QN

)+]

. (7.4.8)

There are two cases to consider. If 1K − 1

QN≥ 0, then

KQN

(1K− 1

QN

)+

= KQN

(1K− 1

QN

)= QN −K,

7.5. REPLICATING PORTFOLIO FOR A FORWARD EXCHANGE CONTRACT15

and this expression is also nonnegative, so QN −K = (QN −K)+. In otherwords,

KQN

(1K− 1

QN

)+

= (QN −K)+. (7.4.9)

On the other hand, if 1K − 1

QN< 0, then QN − K < 0 and we again have

(7.4.9), this time with both sides equal to zero. Substituting (7.4.9) into(7.4.8), we obtain

KQnPn = MnEn

[1

MN(QN −K)+

]= En

[1

(1 + r)N−n(QN −K)+

]= Cn.

�

We can understand Theorem 7.4.5 in the following way. The call isthe right to exchange K units of domestic currency for one unit of foreigncurrency, and this would be exercised when one unit of foreign currency ismore valuable than K units of domestic currency. We could think of thecall as K contracts, each granting the right to exchange 1 unit of domesticcurrency for 1

K units of foreign currency, and this would be exercised when 1K

units of foreign currency is more valuable than one unit of domestic currency.This is therefore the same as K contracts granting the right to sell one unitof domestic currency for 1

K units of foreign currency, which is K puts onthe domestic currency. Thus, one call on foreign currency with strike priceK is worth K puts on domestic currency with strike price 1

K . However,the call is naturally denominated in domestic currency and the K puts arenaturally denominated in foreign currency. To write an equation relatingthese two types of contracts, we need to also convert the currency at thetime of pricing. The term Qn in (7.4.7) is present to make this conversion.

7.5 Replicating portfolio for a forward exchangecontract

In this section we return to the forward exchange rates of Subsection 7.4.1and discuss how to construct a portfolio that replicates a forward exchangecontract. To begin this discussion, we must first find an equation thatdescribes the evolution of the value of a portfolio that trades in the domesticmoney market account and the foreign money market account. We beginwith an example.


Example 7.5.1 Consider the exchange rate process in Figure 7.2.1. As inSection 7.4.1, we take the domestic and foreign interest rates to be r = 1

4 andrf = 1

3 , respectively. To make matters more specific, we assume that thedomestic currency is dollars (denoted $) and the foreign currency is Britishpounds (denoted £). The time-zero exchange rate is $Q0/£ = $2/£, whichwe read as “two dollars per pound.” We begin with initial capital $X0 = $0and purchase £1 at time zero, which we invest in the British money market.To do this, we must borrow $2 from the U.S. money market. When we goto time one, the investment in the British money market grows to

(1 + rf ) ·£1 =43·£1 = £

43.

The debt in the U.S. money market, which we call −$2 because it is a debt,grows to

(1 + r) · $(−2) =54· $(−2) = $

(−5

2

).

To determine the value of the portfolio at time one, we must convert to acommon currency. Since we are recording the portfolio value in dollars, weconvert the pounds to dollars. But the exchange rate depends on the firstcoin toss, and we thus get two possible portfolio values at time one, whichare

X1(H) = £43· $3/£ + $

(−5

2

)= $4 + $

(−5

2

)= $

32,

X1(T ) = £43· $4

3/£ + $

(−5

2

)= $

169

+ $(−5

2

)= $

(−13

18

).

We see that even though the British interest rate is higher than the U.S.interest rate, borrowing dollars in the U.S. in order to invest in pounds inthe UK can lead to a loss. �

In general, the computation of X1 proceeds as follows. If we beginwith initial capital $X0 and purchase ∆0 pounds, then we will need to pay£∆0·$Q0/£ = $∆0Q0, and this is subtracted from our initial capital, leavingus with $(X0 −∆0Q0). This might be negative, in which case it representsborrowing in the U.S. money market and paying the U.S. interest rate r.It could also be positive, in which case it represents investing in the U.S.money market and receiving the U.S. interest rate r. In either case, whenwe go to time one, the cash in dollars changes from $(X0 −∆0Q0) to

$(1 + r)(X0 −∆0Q0). (7.5.1)


This is the change from $(−2) to $(− 5

2

)in the numerical case above where

X0 = 0, ∆0 = 1, Q0 = 2, and 1 + r = 54 . On the other hand, the ∆0 pounds

purchased at time zero is invested at the foreign interest rate and grows tobe (1 + rf )∆0 pounds at time one. This is the £4

3 in the example above,where ∆0 = 1 and 1 + rf = 4

3 . We convert this to dollars by multiplying by$Q1/£, which results in

$(1 + rf )∆0Q1. (7.5.2)

The sum of (7.5.1) and (7.5.2) is the value in dollars of the portfolio at timeone. In other words, the value in dollars of the portfolio at time one is

X1 = (1 + rf )∆0Q1 + (1 + r)(X0 −∆0Q0). (7.5.3)

Because Q1 depends on the first coin toss, the value of the portfolio at timeone also depends on the first coin toss (unless ∆0 = 0, in which case Q1

drops out of (7.5.3)).

An equation analogous to (7.5.3) holds when we move from time n totime n+1. Suppose at time n we decide to hold ∆n pounds, selling or buyingpounds as necessary to achieve this when we arrive at time n. Any moneyleft over is invested at the U.S. interest rate r, and if there is insufficientmoney to pay for ∆n pounds, we borrow at the U.S. interest rate r. Thenthe value Xn+1 of the portfolio at time n + 1, denominated in dollars, andthe value Xn of the portfolio at time n, also denominated in dollars, arerelated by the equation

Xn+1 = (1 + rf )∆nQn+1 + (1 + r)(Xn −∆nQn)= ∆n

((1 + rf )Qn+1 − (1 + r)Qn

)+ (1 + r)Xn. (7.5.4)

In (7.5.4), ∆n is permitted to depend on the first n coin tosses. Then Xn

will depend on only the first n coin tosses and Xn+1 will depend on onlythe first n + 1 coin tosses. In other words, both ∆0,∆1, . . . ,∆N−1 andX0, X1, . . . , XN are adapted processes.

Theorem 7.5.2 The process Xn(1+r)n , n = 0, 1, . . . , N , is a martingale under

the domestic risk-neutral measure P.

Proof: Recall from (7.1.11) that(

1+rf

1+r

)nQn, n = 0, 1, . . . , N , is a martin-

gale under P. Therefore, for 0 ≤ n ≤ N − 1,

En

[(1 + rf

1 + r

)n+1

Qn+1

]=

(1 + rf

1 + r

)n

Qn.


We can factor the nonrandom quantity (1 + rf )n out of the conditionalexpectation on the left-hand side to obtain

(1 + rf )nEn

[1 + rf

(1 + r)n+1Qn+1

]=

(1 + rf

1 + r

)n

Qn,

which simplifies to

En

[1 + rf

(1 + r)n+1Qn+1

]=

Qn

(1 + r)n. (7.5.5)

From the first line of (7.5.4) we have

Xn+1

(1 + r)n+1=

1 + rf

(1 + r)n+1∆nQn +

Xn

(1 + r)n−∆n

Qn

(1 + r)n.

Using linearity of conditional expectations, taking out what is known, andusing (7.5.5), we obtain

En

[Xn+1

(1 + r)n+1

]= En

[1 + rf

(1 + r)n+1∆nQn

]+ En

[Xn

(1 + r)n

]− En

[∆n

Qn

(1 + r)n

]= ∆nEn

[1 + rf

(1 + r)n+1Qn

]+

Xn

(1 + r)n−∆n

Qn

(1 + r)n

= ∆nQn

(1 + r)n+

Xn

(1 + r)n−∆n

Qn

(1 + r)n

=Xn

(1 + r)n.

This is the martingale property for Xn(1+r)n . �

Consider a forward contract to purchase one unit of foreign currency for

For0,N =(

1+r1+rf

)NQ0 units of domestic currency at time N . The formula

for For0,N is (7.4.2), and the formula is chosen so that at time zero thiscontract has value zero. If we take a short position in this contract (i.e.,we agree to deliver a unit of foreign currency at time N in exchange for the

payment(

1+r1+rf

)NQ0 in domestic currency), we receive

X0 = 0 (7.5.6)


at time zero and we are obligated to pay QN −(

1+r1+rf

)NQ0 at time N .

(This payment may be negative, in which case we actually receive moneyat time N . If there were no possibility that we would receive money attime N , we would not accept the short position in this contract withoutdemanding an initial payment.) Starting at X0 = 0, we want to determinerandom variables ∆0,∆1, . . . ,∆N−1, each ∆n depending on only the first ncoin tosses, so that regardless of how the tossing turns out, the final valueof our portfolio is

XN = QN −(

1 + r

1 + rf

)N

Q0. (7.5.7)

In other words, we want to replicate the payoff of the forward contract.

The method we use to determine ∆n is to first use the martingaleproperty of Theorem 7.5.2 to obtain a formula for Xn. Once this formula isobtained, we can replace n by n + 1 to obtain a formula for Xn+1. We thensubstitute both these formulas into (7.5.4) and solve for ∆n.

We now use the martingale property of Theorem 7.5.2. We proved “theone-step ahead” version of the martingale property in that theorem, butthis implies the “multiple-step ahead” version. In particular, the martingaleproperty proved in Theorem 7.5.2 implies that for all n = 0, 1, . . . , N ,

Xn

(1 + r)n= En

[XN

(1 + r)N

]. (7.5.8)

But XN is given by (7.5.7) and(

1+rf

1+r

)nQn is a martingale under P, so

Xn

(1 + r)n= En

[QN

(1 + r)N− Q0

(1 + rf )N

]= En

[QN

(1 + r)N

]− Q0

(1 + rf )N

=1

(1 + rf )NEn

[(1 + rf

1 + r

)N

QN

]− Q0

(1 + rf )N

=1

(1 + rf )N·(

1 + rf

1 + r

)n

Qn −Q0

(1 + rf )N.

Solving this equation for Xn yields

Xn =Qn

(1 + rf )N−n− (1 + r)nQ0

(1 + rf )N, n = 0, 1, . . . , N. (7.5.9)


Note that if we substitute n = 0 into (7.5.9), we obtain (7.5.6), and if wesubstitute n = N , we obtain (7.5.7).

Equation (7.5.9) with n + 1 replacing n is

Xn+1 =Qn+1

(1 + rf )N−n−1− (1 + r)n+1Q0

(1 + rf )N, n = 0, 1, . . . , N − 1. (7.5.10)

We substitute this formula for Xn+1 into the left-hand side of (7.5.4) andsubstitute formula (7.5.9) for Xn into the second line of the right-hand sideof (7.5.4). This results in the equation

Qn+1

(1 + rf )N−n−1− (1 + r)n+1Q0

(1 + rf )N

= ∆n

((1 + rf )Qn+1 − (1 + r)Qn) +

(1 + r)Qn

(1 + rf )N−n− (1 + r)n+1Q0

(1 + rf )N.

We cancel the term − (1+r)n+1Q0

(1+rf )N , which appears on both sides, and thenmove all the terms involving Qn+1 to the left-hand side of the equation.This results in

(1 + rf )Qn+1

(1

(1 + rf )N−n−∆n

)= (1 + r)Qn

(1

(1 + r)N−n−∆n

).

(7.5.11)We now argue from this equation that

1(1 + rf )N−n

−∆n = 0. (7.5.12)

If this were not the case, then the left-hand side of (7.5.11) would dependon the first n+1 coin tosses while the right-hand side would depend on onlythe first n tosses. In such a case, the two sides cannot be equal. The onlyway for the two sides to be equal is for (7.5.12) to hold, in which case bothsides are equal to zero. We conclude that (7.5.12) holds, or equivalently,

∆n =1

(1 + rf )N−n, n = 0, 1, . . . , N − 1. (7.5.13)

It turns out that (7.5.13) mandates a “buy and hold” strategy. At timezero, we should buy

∆0 =1

(1 + rf )N


units of the foreign currency. This costs us

Q0

(1 + rf )N

units of the domestic currency, an amount that we borrow at the domesticinterest rate. At time one, our ∆0 units of the foreign currency have grownat the foreign interest rate to

(1 + r)f∆0 =1

(1 + rf )N−1= ∆1

units of foreign currency. Therefore, we arrive at time one already holdingthe right amount of foreign currency in order to proceed to time two; wedo not need to make any adjustment. At time two, the ∆1 units of foreigncurrency we had at time one have grown to

(1 + rf )∆1 =1

(1 + rf )N−2= ∆2

units of foreign currency. Again we do not need to make any adjustment.We continue on like this, allowing our position in the foreign currency togrow at the foreign interest rate and making no adjustments in the position,until at time N , we have

∆N = 1

unit of foreign currency, which has value QN in domestic currency. On theother hand, our initial debt of Q0

(1+rf )N in domestic currency has grown fromtime zero to time N at the domestic interest rate to become a debt of(

1 + r

1 + rf

)N

Q0.

We thus have a portfolio valued at

XN = QN −(

1 + r

1 + rf

)N

Q0, (7.5.14)

and this is the value of XN we desired (see (7.5.7)). Whereas (7.5.7) wassimply a desire, a statement of what we wanted our trading strategy to leadto, (7.5.14) is the fulfillment of that desire. We have confirmed that thetrading strategy (7.5.13), starting with initial capital X0 = 0, leads to thefinal portfolio value (7.5.14).

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