Stochastic Calculus for Finance I: TheBinomial Asset Pricing ModelSolution of Exercise Problems

Yan Zeng

Version 1.1, last revised on 2014-10-26

Abstract

This is a solution manual for Shreve [6]. If you find any typos/errors or have any comments, pleaseemail me at zypublic@hotmail.edu.

Contents1 The Binomial No-Arbitrage Pricing Model 2

2 Probability Theory on Coin Toss Space 9

3 State Prices 21

4 American Derivative Securities 30

5 Random Walk 37

6 Interest-Rate-Dependent Assets 46

1

1 The Binomial No-Arbitrage Pricing Model⋆ Comments:

1) Example 1.1.1 illustrates the essence of arbitrage: buy low, sell high. Since concrete numbers oftenobscure the nature of things, we review Example 1.1.1 in abstract symbols.

First, the possibility of replicating the payoff of a call option, (S1 −K)+, and its reverse, −(S1 −K)+.To replicate the payoff (S1 −K)+ of a call option at time 1, we at time 0 construct a portfolio (X0 −

∆0S0,∆0S0). At the operational level, we borrow X0, buy ∆0 shares of stock, and invest the residualamount (X0−∆0S0) into money market account. The result is a net cash flow of X0 into the portfolio. Thereplication requirement at time 1 is

(1 + r)(X0 −∆0S0) + ∆0S1 = (S1 −K)+.

Plug in S1(H) = uS0 and S1(T ) = dS0, we obtain a system of two linear equations for two unknowns (X0

and ∆0) and it has a unique solution as long as u = d. This is how we obtain the magic number X0 = 1.20and ∆0 = 1

2 in Example 1.1.1.To replicate the reverse payoff −(S1−K)+ of a call option at time 1, we at time 0 construct a portfolio

(−X0 + ∆0S0,−∆0S0). At the operational level, we short sell ∆0 shares of stock, invest the income ∆0S0

into money market account, and withdraw a cash amount of X0 from the money market account. The resultis a net cash flow of X0 out of the portfolio.

Second, the realization of arbitrage opportunity through buy low, sell high.If the market price C0 of the call option is greater than X0, we just sell the call option for C0 (sell high),

spend X0 on constructing the synthetic call option (buy low), and take the residual amount (C0 − X0) asarbitrage profit. At time 1, the payoff of short position in the call option will cancel out with the payoff ofthe synthetic call option.

If the market price C0 of the call option is less than X0, we buy the call option for C0 (buy low) and setup the portfolio (−X0 + ∆0S0,−∆0S0) at time 0, which allows us to withdraw a cash amount of X0 (sellhigh). The net cash flow (X0 − C0) at time 0 is taken as arbitrage profit. At time 1, the payoff of longposition in the call option will cancel out with the payoff of the portfolio (−X0 +∆0S1,−∆0S1).

2) The essence of Definition 1.2.3 is that we can find a replicating portfolio (∆0, · · · , ∆N−1) and “define”the portfolio’s value at time n as the price of the derivative security at time n. The rationale is that ifP (ω1ω2 · · ·ωN ) > 0 and the price of the derivative security at time n does not agree with the portfolio’svalue, we can make an arbitrage on sample path ω1ω2 · · ·ωN , starting from time n. For a formal presentationof this argument, we refer to Delbaen and Schachermayer [3], Chapter 2.

Also note the definition needs the uniqueness of the replicating portfolio. This is guaranteed in thebinomial model as seen from the uniqueness of solution of equation (1.1.3)-(1.1.4).

Finally, we note the wealth equation (1.2.14) can be written as

Xn+1

(1 + r)n+1=

Xn

(1 + r)n+∆n

[Sn+1

(1 + r)n+1− Sn

(1 + r)n

]This leads to a representation by discrete stochastic integral:

XT = X0 + (∆ · S)T ,

where Xn = Xn

(1+r)n and Sn = Sn

(1+r)n , n = 1, 2, · · · , N .

I Exercise 1.1. Assume the one-period binomail market of Section 1.1 that both H and T have positiveprobability of occurring. Show that condition (1.1.2) precludes arbitrage. In other words, show that ifX0 = 0 and

X1 = ∆0S1 + (1 + r)(X0 −∆0S0),

then we cannot have X1 strictly positive with positive probability unless X1 is strictly negative with positiveprobability as well, and this is the case regardless of the choice of the number ∆0.

2

Proof. Note the random stock price has only two states at time 1, H and T . So “we cannot have X1 strictlypositive with positive probability unless X1 is strictly negative with positive probability as well” can besuccinctly summarized as

‘‘X1(H) > 0 ⇒ X1(T ) < 0 and X1(T ) > 0 ⇒ X1(H) < 0”.

We prove a slightly more general version of the problem to expose the nature of no-arbitrage:

‘‘X1(H) > (1 + r)X0 ⇒ X1(T ) < (1 + r)X0 and X1(T ) > (1 + r)X0 ⇒ X1(H) < (1 + r)X0”.

Note this is indeed a generalization since the original problem assumes X0 = 0.For a formal proof, we write X1 as

X1 = ∆0S0

(S1 − S0

S0− r

)+ (1 + r)X0.

ThenX1(H)− (1 + r)X0 = ∆0S0[u− (1 + r)]

andX1(T )− (1 + r)X0 = ∆0S0[d− (1 + r)].

Given the condition d < 1 + r < u, the factor S0[u− (1 + r)] is positive and the factor S0[d− (1 + r)] isnegative. So

X1(H) > (1 + r)X0 ⇒ ∆0 > 0 ⇒ X1(T ) < (1 + r)X0

andX1(T ) > (1 + r)X0 ⇒ ∆0 < 0 ⇒ X1(H) < (1 + r)X0.

This concludes our proof.

Remark 1.1. The textbook (page 2-3) has shown “negation of d < 1 + r < u ⇒ arbitrage”, or equivalently,“no arbitrage ⇒ d < 1 + r < u”. This exercise problem asks us to prove “d < 1 + r < u ⇒ no arbitrage”.

Remark 1.2. In the equation X1 = ∆0S1 + (1 + r)(X0 − ∆0S0), the first term ∆0S1 is the value of thestock position at time 1 while the second term (1 + r)(X0 −∆0S0) is the value of the money market accountat time 1.

Remark 1.3. The condition X0 = 0 in the original problem formulation is not really essential, as far as aproper definition of arbitrage can be given. Indeed, for the one-period binomial model, we can define arbitrageas a trading strategy such that P (X1 ≥ X0(1 + r)) = 1 and P (X1 > X0(1 + r)) > 0. That is, arbitrage isa trading strategy whose return beats the risk-free rate. See Shreve [7, page 254] Exercise 5.7 for this moregeneral definition.

Remark 1.4. Note the condition d < 1 + r < u is just S1(T )−S0

S0< r < S1(H)−S0

S0: the return of the stock

investment is not guaranteed to be greater or less than the risk-free rate. This agrees with the intuition that“arbitrage is a trading strategy that is guaranteed to beat the risk-free investment”.

I Exercise 1.2. Suppose in the situation of Example 1.1.1 that the option sells for 1.20 at time zero.Consider an agent who begins with wealth X0 = 0 and at time zero buys ∆0 shares of stock and Γ0 options.The numbers ∆0 and Γ0 can be either positive or negative or zero. This leaves the agent with a cash positionof −4∆0−1.20Γ0. If this is positive, it is invested in the money market; if it is negative, it represents moneyborrowed from the money market. At time one, the value of the agent’s portfolio of stock, option, and moneymarket assets is

X1 = ∆0S1 + Γ0(S1 − 5)+ − 5

4(4∆0 + 1.20Γ0).

Assume that both H and T have positive probability of occurring. Show that if there is a positive probabilitythat X1 is positive, then there is a positive probability that X1 is negative. In other words, one cannot findan arbitrage when the time-zero price of the option is 1.20.

3

Proof. X1(u) = ∆0× 8+Γ0× 3− 54 (4∆0+1.20Γ0) = 3∆0+1.5Γ0, and X1(d) = ∆0× 2− 5

4 (4∆0+1.20Γ0) =−3∆0 − 1.5Γ0. That is, X1(u) = −X1(d). So if there is a positive probability that X1 is positive, then thereis a positive probability that X1 is negative. This finishes the proof of the original problem.

Since the use of numbers often obscures the nature of a problem, and since the notation of this exerciseproblem is rather confusing (given the notation in Example 1.1.1), we shall re-state the problem in differentnotation and rewrite the proof in abstract symbols.

First, a re-statement of the problem in different notation: “If the option is priced at the replication costC0 (which is denoted by X0 in Example 1.1.1), we cannot find arbitrage by investing in tradable securities(stock, option, and money market account). Formally, suppose an investor borrows money to buy α sharesof stock and β options at time 0, the portfolio thus constructed is (−αS0 − βC0, αS0 + βC0). Its value attime 1 becomes

X1 = αS1 + βV1 − (1 + r)(αS0 + βC0)

where V1 is the option payoff (S1 −K)+. Prove P (X1 > 0) > 0 ⇒ P (X1 < 0) > 0.”Second, our proof in abstract symbols. Recall for the replication cost C0, there exists some δ > 0 such

that(1 + r)(C0 − δS0) + δS1 ≡ V1.

Plug this into the expression of X1, we have

X1 = αS1 + βV1 − (1 + r)(αS0 + βC0)

= αS1 + β[(1 + r)(C0 − δS0) + δS1]− (1 + r)(αS0 + βC0)

= αS1 + β(1 + r)C0 − βδS0(1 + r) + δβS1 − (1 + r)αS0 − (1 + r)βC0

= (α+ δβ)S1 − S0(1 + r)(α+ βδ)

= (α+ βδ)S0

[S1

S0− (1 + r)

].

Since S1/S0 = u or d and d < 1+ r < u, we conclude X1(H) and X1(T ) have opposite signs. This concludesour proof.

Remark 1.5. Example 1.1.1 has shown that “no arbitrage ⇒ the time-zero price of the option is 1.20”by proving “the time-zero price of the option is not 1.20 ⇒ there exists arbitrage via linear combination ofinvestments in stock, option, and money market account”.

This exercise problem asks us to prove “the time-zero price of the option is 1.20 ⇒ there exists no arbitragevia the linear combination of investments in stock, option and money market account”. Although logically,“linear combination of tradable securities” is only a special way of constructing portfolios, in practice it isthe only way. So it’s all right to use this special form of arbitrage to stand for general arbitrage.

I Exercise 1.3. In the one-period binomial model of Section 1.1, suppose we want to determine the priceat time zero of the derivative security V1 = S1 (i.e., the derivative security pays off the stock price.) (Thiscan be regarded as a European call with strike price K = 0). What is the time-zero price V0 given by therisk-neutral pricing formula (1.1.10)?

Solution. V0 = 11+r

[1+r−du−d S1(H) + u−1−r

u−d S1(T )]= S0

1+r

(1+r−du−d u+ u−1−r

u−d d)= S0. This is not surprising,

since this is exactly the cost of replicating S1, via the buy-and-hold strategy.

I Exercise 1.4. In the proof of Theorem 1.2.2, show under the induction hypothesis that

Xn+1(ω1ω2 · · ·ωnT ) = Vn+1(ω1ω2 · · ·ωnT ).

4

Proof.

Xn+1(T ) = ∆ndSn + (1 + r)(Xn −∆nSn)

= ∆nSn(d− 1− r) + (1 + r)Vn

=Vn+1(H)− Vn+1(T )

u− d(d− 1− r) + (1 + r)

pVn+1(H) + qVn+1(T )

1 + r= p[Vn+1(T )− Vn+1(H)] + pVn+1(H) + qVn+1(T )

= pVn+1(T ) + qVn+1(T )

= Vn+1(T ).

I Exercise 1.5. In Example 1.2.4, we considered an agent who sold the look-back option for V0 = 1.376and bought ∆0 = 0.1733 shares of stock at time zero. At time one, if the stock goes up, she has a portfoliovalued at V1(H) = 2.24. Assume that she now takes a position of ∆1(H) = V2(HH)−V2(HT )

S2(HH)−S2(HT ) in the stock.Show that, at time two, if the stock goes up again, she will have a portfolio valued at V2(HH) = 3.20,whereas if the stock goes down, her portfolio will be worth V2(HT ) = 2.40. Finally, under the assumptionthat the stock goes up in the first period and down in the second period, assume the agent takes a positionof ∆2(HT ) = V3(HTH)−V3(HTT )

S3(HTH)−S3(HTT ) in the stock. Show that, at time three, if the stock goes up in the thirdperiod, she will have a portfolio valued at V3(HTH) = 0, whereas if the stock goes down, her portfolio willbe worth V3(HTT ) = 6. In other words, she has hedged her short position in the option.

Proof. First, on the path ω1 = H, the investor’s portfolio is worth of

X1(H) = (1 + r)(X0 −∆0S0) + ∆0S1(H) = (1 + 0.25)(1.376− 0.1733 · 4) + 0.1733 · 8 = 2.24 = V1(H).

If on the path ω1 = H, the investor takes a position of

∆1(H) =V2(HH)− V2(HT )

S2(HH)− S2(HT )=

3.20− 2.40

16− 4= 0.0667

in the stock, then on the path ω1ω2 = HH, the investor’s portfolio is worth of

X2(HH) = (1 + r)[X1(H)−∆1(H)S1(H)] + ∆1(H)S2(HH)

= (1 + 0.25)(2.24− 0.0667 · 8) + 0.0667 · 16= 3.2

= V2(HH),

while on the path ω1ω2 = HT , the investor’s portfolio is worth of

X2(HT ) = (1 + r)[X1(H)−∆1(H)S1(H)] + ∆1(H)S2(HT )

= (1 + 0.25)(2.24− 0.0667 · 8) + 0.0667 · 4= 2.4

= V2(HT ).

Second, if on the path ω1ω2 = HT , the agent takes a position

∆2(HT ) =V3(HTH)− V3(HTT )

S3(HTH)− S3(HTT )=

0− 6

8− 2= −1

in the stock, then on the path ω1ω2ω3 = HTH, the investor’s portfolio is worth of

X3(HTH) = (1 + r)[X2(HT )−∆2(HT )S2(HT )] + ∆2(HT )S3(HTH)

= (1 + 0.25)[2.4− (−1) · 4] + (−1) · 8= 0

= V3(HTH),

5

while on the path ω1ω2ω3 = HTT , the investor’s portfolio is worth of

X3(HTT ) = (1 + r)[X2(HT )−∆2(HT )S2(HT )] + ∆2(HT )S3(HTT )

= (1 + 0.25)[2.4− (−1) · 4] + (−1) · 2= 6

= V3(HTT ).

I Exercise 1.6. (Hedging a long position-one period). Consider a bank that has a long position inthe European call written on the stock price in Figure 1.1.2. The call expires at time one and has strikeprice K = 5. In Section 1.1, we determined the time-zero price of this call to be V0 = 1.20. At time zero,the bank owns this option, while ties up capital V0 = 1.20. The bank wants to earn the interest rate 25% onthis capital until time one (i.e., without investing any more money, and regardless of how the coin tossingturns out, the bank wants to have

5

4· 1.20 = 1.50

at time one, after collecting the payoff from the option (if any) at time one). Specify how the bank’s tradershould invest in the stock and money markets to accomplish this.

Solution. The bank’s trader should set up a replicating portfolio whose payoff is the opposite of the option’spayoff. More precisely, we solve the equation

(1 + r)(X0 −∆0S0) + ∆0S1 = −(S1 −K)+.

Then X0 = −1.20 and ∆0 = −12 since this equation is a linear equation of X0 and ∆0. The solution means

the trader should sell short 0.5 share of stock, put the income 2 into a money market account, and thentransfer 1.20 into a separate money market account. At time one, the portfolio consisting of a short positionin stock and 0.8(1+ r) in money market account will cancel out with the option’s payoff. In the end, we endup with 1.20(1 + r) in the separate money market account.

Remark 1.6. This problem illustrates why we are interested in hedging a long position. In case the stockprice goes down at time one, the option will expire worthless. The initial amount of money 1.20 paid attime zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) whichguarantees a sure payoff at time one. As to why we hedge a short position (as a writer), see Wilmott [8,page 11-13], 1.4 What are Options For?

I Exercise 1.7. (Hedging a long position-multiple periods). Consider a bank that has a long positionin the lookback option of Example 1.2.4. The bank intends to hold this option until expiration and receivethe payoff V3. At time zero, the bank has capital V0 = 1.376 tied up in the option and wants to earn theinterest rate of 25% on this capital until time three (i.e., without investing any more money, and regardlessof how the coin tossing turns out, the bank wants to have(

5

4

)3

· 1.376 = 2.6875

at time three, after collecting the payoff from the lookback option at time three). Specify how the bank’strader should invest in the stock and the money market account to accomplish this.

Solution. The idea is the same as Exercise 1.6. The bank’s trader only needs to set up the reverse of thereplicating trading strategy described in Example 1.2.4. More precisely, he should short sell 0.1733 share ofstock, invest the income 0.6933 into money market account, and transfer 1.376 into a separate money marketaccount. The portfolio consisting a short position in stock and 0.6933-1.376 in money market account willreplicate the opposite of the option’s payoff. After they cancel out, we end up with 1.376(1 + r)3 in theseparate money market account.

6

I Exercise 1.8. (Asian option). Consider the three-period model of Example 1.3.1,1 with S0 = 4, u = 2,d = 1

2 , and take the interest rate r = 14 , so that p = q = 1

2 . For n = 0, 1, 2, 3, define Yn =∑n

k=0 Sk to be thesum of the stock prices between times zero and n. Consider an Asian call option that expires at time threeand has strike K = 4 (i.e., whose payoff at time three is

(14Y3 − 4

)+). This is like a European call, exceptthe payoff of the option is based on the average stock price rather than the final stock price. Let vn(s, y)denote the price of this option at time n if Sn = s and Yn = y. In particular, v3(s, y) =

(14y − 4

)+.(i) Develop an algorithm for computing vn recursively. In particular, write a formula for vn in terms of vn+1.

Solution. By risk-neutral pricing formula,

vn(s, y) =1

1 + r[pvn+1(us, y + us) + qvn+1 (ds, y + ds)] =

2

5

[vn+1(2s, y + 2s) + vn+1

(s2, y +

s

2

)].

(ii) Apply the algorithm developed in (i) to compute v0(4, 4), the price of the Asian option at time zero.

S0 = 4 Y0 = 4v0(4, 4) = 1.216

S1(T ) = 2Y1(T ) = 6

v1(2, 6) = 0.08

S2(TT ) = 1Y2(TT ) = 7v2(1, 7) = 0 S3(TTT ) = 0.5

Y3(TTT ) = 7.5v3(0.5, 7.5) = 0

S3(TTH) = 2Y3(TTH) = 9v3(2, 9) = 0

S2(TH) = 4Y2(TH) = 10v2(4, 10) = 0.2 S3(THT ) = 2

Y3(THT ) = 12v3(2, 12) = 0

S3(THH) = 8Y3(THH) = 18v3(8, 18) = 0.5

S1(H) = 8Y1(H) = 12

v1(8, 12) = 2.96

S2(HT ) = 4Y2(HT ) = 16v2(4, 16) = 1 S3(HTT ) = 2

Y3(HTT ) = 18v3(2, 18) = 0.5

S3(HTH) = 8Y3(HTH) = 24v3(8, 24) = 2

S2(HH) = 16Y2(HH) = 28v2(16, 28) = 6.4 S3(HHT ) = 8

Y3(HHT ) = 36v3(8, 36) = 5

S3(HHH) = 32Y3(HHH) = 60v3(32, 60) = 11

Exercise 1.8. Asian option.

Solution.1The textbook said “Example 1.2.1” by mistake.

7

(iii) Provide a formula for δn(s, y), the number of shares of stock that should be held by the replicatingportfolio at time n if Sn = s and Yn = y.

Solution.δn(s, y) =

vn+1(us, y + us)− vn+1(ds, y + ds)

(u− d)s.

I Exercise 1.9. (Stochastic volatility, random interest rate). Consider a binomial pricing model,but at each time n ≥ 1, the “up factor” un(ω1ω2 · · ·ωn), the “down factor” dn(ω1ω2 · · ·ωn), and the interestrate rn(ω1ω2 · · ·ωn) are allowed to depend on n and on the first n coin tosses ω1ω2 · · ·ωn. The initial upfactor u0, the initial down factor d0, and the initial interest rate r0 are not random. More specifically, thestock price at time one is given by

S1(ω1) =

u0S0 if ω1 = H,d0S0 if ω1 = T ,

and, for n ≥ 1, the stock price at time n+ 1 is given by

Sn+1(ω1ω2 · · ·ωnωn+1) =

un(ω1ω2 · · ·ωn)Sn(ω1ω2 · · ·ωn) if ωn+1 = H,dn(ω1ω2 · · ·ωn)Sn(ω1ω2 · · ·ωn) if ωn+1 = T .

One dollar invested in or borrowed from the money market at time zero grows to an investment or debt of1 + r0 at time one, and, for n ≥ 1, one dollar invested in or borrowed from the money market at time ngrows to an investment or debt of 1 + rn(ω1ω2 · · ·ωn) at time n+ 1. We assume that for each n and for allω1ω2 · · ·ωn, the no-arbitrage condition

0 < dn(ω1ω2 · · ·ωn) < 1 + rn(ω1ω2 · · ·ωn) < un(ω1ω2 · · ·ωn)

holds. We also assume that 0 < d0 < 1 + r0 < u0.(i) Let N be a positive integer. In the model just described, provide an algorithm for determining the priceat time zero for a derivative security that at time N pays off a random amount VN depending on the resultof the first N coin tosses.

Solution. Similar to Theorem 1.2.2, but replace r, u and d everywhere with rn, un and dn. More precisely,set pn = 1+rn−dn

un−dnand qn = 1− pn. Then

Vn =pnVn+1(H) + qnVn+1(T )

1 + rn.

(ii) Provide a formula for the number of shares of stock that should be held at each time n (0 ≤ n ≤ N − 1)by a portfolio that replicates the derivatives security VN .

Solution. ∆n = Vn+1(H)−Vn+1(T )Sn+1(H)−Sn+1(T ) =

Vn+1(H)−Vn+1(T )(un−dn)Sn

.

(iii) Suppose the initial stock price is S0 = 80, with each head the stock price increases by 10, and with eachtail the stock price decreases by 10. In other words, S1(H) = 90, S1(T ) = 70, S2(HH) = 100, etc. Assumethe interest rate is always zero. Consider a European call with strike price 80, expiring at time five. Whatis the price of this call at time zero?

Solution. un = Sn+1(H)Sn

= Sn+10Sn

= 1 + 10Sn

and dn = Sn+1(T )Sn

= Sn−10Sn

= 1 − 10Sn

. So the risk-neutralprobabilities at time n are pn = 1−dn

un−dn= 1

2 and qn = 12 .

To price the European call, we only need to focus on the nodes at time 5 of the binomial tree since pnand qn are constants. In order to have non-zero payoffs, a node at time 5 must have at least 3 H’s. For sucha node, we have

8

number of H’s option payoff risk-neutral probability of paths3 (80 + 30− 20)− 80 = 10 1

25 · 5!3!2! =

1032

4 (80 + 40− 10)− 80 = 30 125 · 5 = 5

32

5 (80 + 50)− 80 = 50 125 = 1

32

So the price of the call at time zero is

1

32· 50 + 5

32· 30 + 10

32· 10 =

300

32= 9.375.

2 Probability Theory on Coin Toss Space⋆ Comments:

1) The second proof of Theorem 2.4.4 also works for the random interest rate model of Exercise 1.9, asfar as Sn+1

(1+r)n+1 is replaced by Sn+1

(1+r0)···(1+rn). The requirement on the risk-neutral probability P is that

P(wn+1 = H|ω1, · · · , ωn) := pn =1 + rn − dnun − dn

andP(wn+1 = T |ω1, · · · , ωn) := 1− pn = qn.

Results in measure theory guarantee the existence and uniqueness of such a probability measure P on thesample space Ω = w : (w1, · · · , wN ) for a given family of conditional probabilities (pn, qn)

Nn=1.2 With this

setup, we haveEn

[Sn+1

Sn

]= unpn + dnqn = (1 + rn).

In other words, the stock price process, when discounted by the random interest rates, is a martingale underP. When rn’s are deterministic, we are back to the case where ωn’s are independent of each other. Thisunifies the deterministic and random interest rate models.

2) On Theorem 2.4.8 (Cash flow valuation): by Theorem 2.4.7, it is natural to “conjecture” that theno-arbitrage price process of the derivative security is given by the risk-neutral pricing formula (2.4.13).However, pricing always needs to be justified by hedging/replication. We need to find a portfolio processwhich replicates the cash flows while equals to (Vn)

Nn=0 in value. The key insight to the construction of such

a replicating portfolio is the wealth equation (2.4.16)

Xn+1 = ∆nSn+1 + (1 + r)(Xn − Cn −∆nSn),

which clearly indicates that the wealth process (Xn)Nn=1 produces a cash outflow of Cn at step n.

Note the presentation of Theorem 2.4.8 follows the following flow of logic:

define Vn’s ⇒ define ∆n’s ⇒ define Xn’s ⇒ prove Xn = Vn.

We could have taken a different, yet more natural path of logic, namely

define Xn’s and ∆n’s simultaneously⇒ prove formula (2.4.14) holds with Vn’s replaced by Xn’s⇒ prove formula (2.4.13) holds with Vn’s replaced by Xn’s⇒ define Vn = Xn.

2See, for example, Shiryaev [5, page 249], Theorem 2 (Ionescu Tulcea’s Theorem on Extending a Measure and the Existenceof a Random Sequence).

9

Indeed, we define XN = CN and solve for XN−1 and ∆N−1 in the following equation

XN = ∆N−1SN + (1 + r)(XN−1 − CN−1 −∆N−1SN−1).

By imitating the trick of (1.1.3)-(1.1.8) (page 6), we obtain

∆N−1(ω1 · · ·ωN−1) =CN (ω1 · · ·ωN−1H)− CN (ω1 · · ·ωN−1T )

SN (ω1 · · ·ωN−1H)− SN (ω1 · · ·ωN−1T )

XN−1 = CN−1 +1

1 + r[pXN (ω1 · · ·ωN−1H) + qXN (ω1 · · ·ωN−1T )]

= CN−1 + En

[XN

1 + r

]= En

[N∑

k=N−1

Ck

(1 + r)k−(N−1)

].

Working backward by induction, we can provide definitions for each Xn and ∆n, and prove for each n

Xn = Cn + En

[Xn+1

1 + r

]= En

[N∑

k=n

Ck

(1 + r)k−n

].

Finally, we comment that Theorem 1.2.2 is a special case of Theorem 2.4.8, with C0 = C1 = · · · =CN−1 = 0.

3) The textbook gives readers the impression that Markov property is preserved under the risk-neutralprobability, at least in the setting of binomial model. A general result in this regard is recently announcedin Schmock [4]:

Theorem 1 (Uwe Schmock and Ismail Cetin Gülüm). Let (Ω,F , Ftt=0,··· ,T ,P) be a (general) filteredprobability space and let S = Stt∈0,··· ,T be an adapted, Rd-valued discounted asset price process, whichhas the k-multiple Markov property w.r.t. P. Then the following properties are equivalent:(a) The financial market model is free of arbitrage.(b) There exists a probability measure P∗ on (Ω,F ,P) such that

• P∗ ∼ P with ϱ := dP∗/dP ∈ L∞(Ω,FT ,P).• Integrability: St ∈ L1(Ω,Ft,P∗) for all t ∈ 0, · · · , T.• Martingale property w.r.t. P∗:

EP∗ [St|Ft−1]a.s.= St−1 for all t ∈ 1, · · · , T.

• k-multiple Markov property: For all B ∈ E and t ∈ k, · · · , T

P∗(St ∈ B|Ft−1)a.s.= P∗(St ∈ B|St−1, St−2, · · · , St−k).

For continuous time financial models, it is well-known that for Lipschitz continuous f , g, the stochasticdifferential equation of K. Itô

Xt = X0 +

∫ t

0

f(s,Xs)dWs +

∫ t

0

g(s,Xs)ds

has a unique solution which is a Markov process with continuous paths. Moreover if f and g satisfy f(t, x) =f(x), g(t, x) = g(x), then X is a time homogenous strong Markov process. These results combined withGirsanov’s Theorem (Shreve [7] Chapter 5) will preserve Markov property of discounted asset process underthe risk-neutral measure.

I Exercise 2.1. Using Definition 2.1.1, show the following.(i) If A is an event and Ac denotes its complement, then P(Ac) = 1− P(A).

10

Proof. P(Ac) + P(A) =∑

ω∈Ac P(ω) +∑

ω∈A P(ω) =∑

ω∈Ω P(ω) = 1.

(ii) If A1, A2, · · · , AN is a finite set of events, then

P(∪Nn=1An) ≤

N∑n=1

P(An). (2.8.1)

If the events A1, A2, · · · , AN are disjoint, then equality holds in (2.8.1).

Proof. By induction, it suffices to work on the case N = 2. When A1 and A2 are disjoint, P(A1 ∪ A2) =∑ω∈A1∪A2

P(ω) =∑

ω∈A1P(ω) +

∑ω∈A2

P(ω) = P(A1) + P(A2). When A1 and A2 are arbitrary, usingthe result when they are disjoint, we have P(A1 ∪ A2) = P((A1 − A2) ∪ A2) = P(A1 − A2) + P(A2) ≤P (A1) + P(A2).

I Exercise 2.2. Consider the stock price S3 in Figure 2.3.1.(i) What is the distribution of S3 under the risk-neutral probabilities p = 1

2 , q = 12 .

Solution. Under the risk-neutral probability, the distribution of ωn+1 conditioning on ω1, · · · , ωn is deter-ministic

P(ωn+1 = H|ω1ω2 · · ·ωn) := p =1 + r − d

u− d, P(ωn+1 = T |ω1ω2 · · ·ωn) := q =

u− 1− r

u− d.

So ωn’s are independent of each other under P and we have

P(S3 = 32) = p3 =1

8, P(S3 = 8) = 3p2q =

3

8, P(S3 = 2) = 3pq2 =

3

8, P(S3 = 0.5) = q3 =

1

8.

(ii) Compute ES1, ES2, and ES3. What is the average rate of growth of the stock price under P?

Solution. E[S1] = 8P(S1 = 8) + 2P(S1 = 2) = 8p+ 2q = 5

E[S2] = 16p2 + 4 · 2pq + 1 · q2 = 6.25

E[S3] = 32 · 18 + 8 · 3

8 + 2 · 38 + 0.5 · 1

8 = 7.8125.

So the average rates of growth of the stock price under P are, respectively:r0 = E[S1]

S0− 1 = 5

4 − 1 = 0.25,

r1 = E[S2]

E[S1]− 1 = 6.25

5 − 1 = 0.25,

r2 = E[S3]

E[S2]− 1 = 7.8125

6.25 − 1 = 0.25.

Remark 2.1. An alternative solution is to use martingale property:

E[

Sn

(1 + r)n

]= S0 ⇒ E [Sn] = (1 + r)nS0,

E [Sn]

E [Sn−1]= (1 + r).

(iii) Answer (i) and (ii) again under the actual probabilities p = 23 , q = 1

3 .

Solution. P(S3 = 32) = ( 23 )3 = 8

27 , P(S3 = 8) = 3 · ( 23 )2 · 13 = 4

9 , P(S3 = 2) = 2 · 19 = 29 , and P(S3 = 0.5) = 1

27 .Accordingly, E[S1] = 6, E[S2] = 9 and E[S3] = 13.5. So the average rates of growth of the stock price

under P are, respectively: r0 = 64 − 1 = 0.5, r1 = 9

6 − 1 = 0.5, and r2 = 13.59 − 1 = 0.5.

11

Remark 2.2. An alternative solution is to use Markov property: En[Sn+1] = SnEn[Sn+1/Sn] = Sn(pu+qd).

I Exercise 2.3. Show that a convex function of a martingale is a submartingale. In other words, let M0,M1, · · · , MN be a martingale and let φ be a convex function. Show that φ(M0), φ(M1), · · · , φ(MN ) is asubmartingale.

Proof. Apply conditional Jensen’s inequality, Theorem 2.3.2 (v).

I Exercise 2.4. Toss a coin repeatedly. Assume the probability of head on each toss is 12 , as is the

probability of tail. Let Xj = 1 if the jth toss results in a head and Xj = −1 if the jth toss results in a tail.Consider the stochastic process M0, M1, M2, · · · defined by M0 = 0 and

Mn =n∑

j=1

Xj , n ≥ 1.

This is called a symmetric random walk; with each head, it steps up one, and with each tail, it steps downone.(i) Using the properties of Theorem 2.3.2, show that M0, M1, M2, · · · is a martingale.

Proof. En[Mn+1] =Mn + En[Xn+1] =Mn + E[Xn+1] =Mn.

(ii) Let σ be a positive constant and, for n ≥ 0, define

Sn = eσMn

(2

eσ + e−σ

)n

.

Show that S0, S1, S2, · · · is a martingale. Note that even though the symmetric random walk Mn has notendency to grow, the “geometric symmetric random walk” eσMn does have a tendency to grow. This is theresult of putting a martingale into the (convex) exponential function (see Exercise 2.3). In order to againhave a martingale, we must “discount” the geometric symmetric random walk, using the term 2

eσ+e−σ as thediscount rate. This term is strictly less than one unless σ = 0.

Proof.En

[Sn+1

Sn

]= En

[eσXn+1

2

eσ + e−σ

]=

2

eσ + e−σE[eσXn+1

]= 1.

I Exercise 2.5. Let M0, M1, M2, · · · be the symmetric random walk of Exercise 2.4, and define I0 = 0and

In =n−1∑j=0

Mj(Mj+1 −Mj), n = 1, 2, · · · .

(i) Show thatIn =

1

2M2

n − n

2.

Proof.

2In = 2

n−1∑j=0

Mj(Mj+1 −Mj) = 2

n−1∑j=0

MjMj+1 +M2n −

n−1∑j=0

M2j+1 −

n−1∑j=0

M2j

= M2n −

n−1∑j=0

(Mj+1 −Mj)2 =M2

n −n−1∑j=0

X2j+1 =M2

n − n.

12

Remark 2.3. This is the discrete version of the integration-by-parts formula for stochastic integral:

M2T −M2

0 = 2

∫ T

0

MtdMt + [M,M ]T .

with IT =∫ T

0MtdMt.

(ii) Let n be an arbitrary nonnegative integer, and let f(i) be an arbitrary function of a variable i. In termsof n and f , define another function g(i) satisfying

En[f(In+1)] = g(In).

Note that although the function g(In) on the right-hand side of this equation may depend on n, the onlyrandom variable that may appear in its argument is In; the random variable Mn may not appear. You willneed to use the formula in part (i). The conclusion of part (ii) is that the process I0, I1, I2, · · · is a Markovprocess.

Solution.

En[f(In+1)]

= En[f(In +Mn(Mn+1 −Mn))]

= En[f(In +MnXn+1)]

=1

2[f(In +Mn) + f(In −Mn)]

= g(In),

where g(x) = 12 [f(x+

√2x+ n) + f(x−

√2x+ n)], since

√2In + n = |Mn|.

I Exercise 2.6 (Discrete-time stochastic integral). Suppose M0, M1, · · · , MN is a martingale, andlet ∆0, ∆1, · · · , ∆N−1 be an adapted process. Define the discrete-time stochastic integral (sometimes calleda martingale transform) I0, I1, · · · , IN by setting I0 = 0 and

In =n−1∑j=0

∆j(Mj+1 −Mj), n = 1, · · · , N.

Show that I0, I1, · · · , IN is a martingale.

Proof. En[In+1 − In] = En[∆n(Mn+1 −Mn)] = ∆nEn[Mn+1 −Mn] = 0.

I Exercise 2.7. In a binomial model, give an example of a stochastic process that is a martingale but isnot Markov.

Solution. We denote by Xn the result of the nth coin toss:Xn = 1 if ωn = H

Xn = −1 if ωn = T

We also suppose P(X = 1) = P(X = −1) = 12 . Define S1 = X1 and Sn+1 = Sn + bn(X1, · · · , Xn)Xn+1,

where bn(·) is a bounded function on −1, 1n, to be determined later on. Clearly (Sn)n≥1 is an adaptedstochastic process, and we can show it is a martingale. Indeed,

En[Sn+1 − Sn] = bn(X1, · · · , Xn)En[Xn+1] = 0.

For any arbitrary function f , En[f(Sn+1)] =12 [f(Sn+ bn(X1, · · · , Xn))+f(Sn− bn(X1, · · · , Xn))]. Then

intuitively, En[f(Sn+1] cannot be solely dependent upon Sn when bn’s are properly chosen. Therefore ingeneral, (Sn)n≥1 cannot be a Markov process.

13

Remark 2.4. If Xn is regarded as the gain/loss of n-th bet in a gambling game, then Sn would be the wealthat time n. bn is therefore the wager for the (n+ 1)th bet and is devised according to past gambling results.

I Exercise 2.8. Consider an N -period binomial model.(i) Let M0, M1, · · · , MN and M ′

0, M ′1, · · · , M ′

N be martingales under the risk-neutral measure P. Show thatif MN =M ′

N (for every possible outcome of the sequence of coin tosses), then, for each n between 0 and N ,we have Mn =M ′

n (for every possible outcome of the sequence of coin tosses).

Proof. Note Mn = En[MN ] = En[M′N ] =M ′

n, n = 0, 1, · · · , N .

(ii) Let VN be the payoff at time N of some derivative security. This is a random variable that can dependon all N coin tosses. Define recursively VN−1, VN−2, · · · , V0 by the algorithm (1.2.16) of Chapter 1. Showthat

V0,V1

1 + r, · · · , VN−1

(1 + r)N−1,

VN(1 + r)N

is a martingale under P.

Proof. In the proof of Theorem 1.2.2, we proved by induction that Xn = Vn where Xn is defined by (1.2.14)of Chapter 1: Xn+1 = ∆nSn+1 + (1 + r)(Xn − ∆nSn). Since

Xn

(1+r)n

0≤n≤N

is a martingale under P

(Theorem 2.4.5),

Vn

(1+r)n

0≤n≤N

is also a martingale under P.

(iii) Using the risk-neutral pricing formula (2.4.11) of this chapter, define

V ′n = En

[VN

(1 + r)N−n

], n = 0, 1, · · · , N − 1.

Show thatV ′0 ,

V ′1

1 + r, · · · ,

V ′N−1

(1 + r)N−1,

VN(1 + r)N

is a martingale.

Proof. This is obvious by iterated conditioning.

(iv) Conclude that Vn = V ′n for every n (i.e., the algorithm (1.2.16) of Theorem 1.2.2 of Chapter 1 gives the

same derivative security prices as the risk-neutral pricing formula (2.4.11) of Chapter 2).

Proof. Combine (ii) and (iii), then use (i).

I Exercise 2.9 (Stochastic volatility, random interest rate). Consider a two-period stochastic volatil-ity, random interest rate model of the type described in Exercise 1.9 of Chapter 1. The stock prices andinterest rates are shown in Figure 2.8.1.(i) Determine risk-neutral probabilities

P(HH), P(HT ), P(TH), P(TT ),

such that the time-zero value of an option that pays off V2 at time two is given by the risk-neutral pricingformula

V0 = E[

V2(1 + r0)(1 + r1)

].

14

S0 = 4r0 = 1

4

S1(T ) = 2r1(T ) =

12

S2(TT ) = 2

S2(TH) = 8

S1(H) = 8r1(H) = 1

4

S2(HT ) = 8

S2(HH) = 12

Fig. 2.8.1. A stochastic volatility, random interest rate model.

Solution.

u0 =S1(H)

S0= 2, d0 =

S1(H)

S0=

1

2,

u1(H) =S2(HH)

S1(H)= 1.5, d1(H) =

S2(HT )

S1(H)= 1,

u1(T ) =S2(TH)

S1(T )= 4, d1(T ) =

S2(TT )

S1(T )= 1.

Therefore

p0 =1 + r0 − d0u0 − d0

=1

2, q0 = 1− p0 =

1

2

p1(H) =1 + r1(H)− d1(H)

u1(H)− d1(H)=

1

2, q1(H) = 1− p1(H) =

1

2,

p1(T ) =1 + r1(T )− d1(T )

u1(T )− d1(T )=

1

6, q1(T ) = 1− p1(T ) =

5

6.

and

P(HH) = p0p1(H) =1

4, P(HT ) = p0q1(H) =

1

4, P(TH) = q0p1(T ) =

1

12, P(TT ) = q0q1(T ) =

5

12.

The proofs of Theorem 2.4.4, Theorem 2.4.5 and Theorem 2.4.7 still work for the random interest ratemodel, with proper modifications (i.e. P would be constructed according to the family of conditional proba-bilities P(ωn+1 = H|ω1, · · · , ωn) := pn and P(ωn+1 = T |ω1, · · · , ωn) := qn. See Footnote 2 for comments.).So the time-zero value of an option that pays off V2 at time two is given by the risk-neutral pricing formulaV0 = E

[V2

(1+r0)(1+r1)

].

(ii) Let V2 = (S2 − 7)+. Compute V0, V1(H), and V1(T ).

Solution. V2(HH) = 5, V2(HT ) = 1, V2(TH) = 1 and V2(TT ) = 0. So

V1(H) =p1(H)V2(HH) + q1(H)V2(HT )

1 + r1(H)= 2.4

V1(T ) =p1(T )V2(TH) + q1(T )V2(TT )

1 + r1(T )=

1

9

V0 =p0V1(H) + q0V1(T )

1 + r0= 1.00444.

15

(iii) Suppose an agent sells the option in (ii) for V0 at time zero. compute the position ∆0 she should takein the stock at time zero so that at time one, regardless of whether the first coin toss results in head or tail,the value of her portfolio is V1.

Solution. ∆0 = V1(H)−V1(T )S1(H)−S1(T ) =

2.4− 19

8−2 = 0.4− 154 ≈ 0.3815.

(iv) Suppose in (iii) that the first coin toss results in head. What position ∆1(H) should the agent now takein the stock be sure that, regardless of whether the second coin toss results in head or tail, the value of herportfolio at time two will be (S2 − 7)+?

Solution. ∆1(H) = V2(HH)−V2(HT )S2(HH)−S2(HT ) =

5−112−8 = 1.

I Exercise 2.10 (Dividend-paying stock).3 We consider a binomial asset pricing model as in Chapter1, except that, after each movement in the stock price, a dividend is paid and the stock price is reducedaccordingly. To describe this in equations, we define

Yn+1(ω1 · · ·ωnωn+1) =

u, if ωn+1 = H,d, if ωn+1 = T .

Note that Yn+1 depends only on the (n + 1)st coin toss. In the binomial model of Chapter 1, Yn+1Sn wasthe stock price at time n + 1. In the dividend-paying model considered here, we have a random variableAn+1(ω1 · · ·ωnωn+1), taking values in (0, 1), and the dividend paid at time n+ 1 is An+1Yn+1Sn. After thedividend is paid, the stock price at time n+ 1 is

Sn+1 = (1−An+1)Yn+1Sn.

An agent who begins with initial capital X0 and at each time n takes a position of ∆n shares of stock,where ∆n depends only on the first n coin tosses, has a portfolio value governed by the wealth equation (see(2.4.6))4

Xn+1 = ∆nSn+1 + (1 + r)(Xn −∆nSn) + ∆nAn+1Yn+1Sn = ∆nYn+1Sn + (1 + r)(Xn −∆nSn). (2.8.2)

(i) Show that the discounted wealth process is a martingale under the risk-neutral measure (i.e., Theorem2.4.5 still holds for the wealth process (2.8.2)). As usual, the risk-neutral measure is still defined by theequations

p =1 + r − d

u− d, q =

u− 1− r

u− d.

Proof.

En

[Xn+1

(1 + r)n+1

]= En

[∆nYn+1Sn

(1 + r)n+1+

(1 + r)(Xn −∆nSn)

(1 + r)n+1

]=

∆nSn

(1 + r)n+1En[Yn+1] +

Xn −∆nSn

(1 + r)n

=∆nSn

(1 + r)n+1(up+ dq) +

Xn −∆nSn

(1 + r)n

=∆nSn +Xn −∆nSn

(1 + r)n=

Xn

(1 + r)n.

3Compare this problem with §5.5 of Shreve [7].4Note the assumption of the equation is that the dividends are reinvested. Also note the equation can be written as

Xn+1 −Xn = ∆n(Sn+1 − Sn) + r(Xn −∆nSn) + ∆nAn+1Yn+1Sn,

which gives three sources of wealth change: change of stock price, interest earned from money market account, and stockdividends.

16

(ii) Show that the risk-neutral pricing formula still applies (i.e., Theorem 2.4.7 holds for the dividend-payingmodel).

Proof. From (2.8.2), we have∆nuSn + (1 + r)(Xn −∆nSn) = Xn+1(H)∆ndSn + (1 + r)(Xn −∆nSn) = Xn+1(T ).

So∆n =

Xn+1(H)−Xn+1(T )

uSn − dSn, Xn = En

[Xn+1

1 + r

].

To make the portfolio replicate the payoff at time N , we must have XN = VN . So Xn = En

[XN

(1+r)N−n

]=

En

[VN

(1+r)N−n

]. Since (Xn)0≤n≤N is the value process of the unique replicating portfolio (uniqueness is

guaranteed by the uniqueness of the solution to the above linear equations), the no-arbitrage price of VN attime n is Vn = Xn = En

[VN

(1+r)N−n

].

(iii) Show that the discounted stock price is not a martingale under the risk-neutral measure (i.e., Theorem2.4.4 no longer holds). However, if An+1 is a constant a ∈ (0, 1), regardless of the value of n and the outcomeof the coin tossing ω1 · · ·ωn+1, then Sn

(1−a)n(1+r)n is a martingale under the risk-neutral measure.

Proof.

En

[Sn+1

(1 + r)n+1

]=

1

(1 + r)n+1En[(1−An+1)Yn+1Sn]

=Sn

(1 + r)n+1p[1−An+1(ω1 · · ·ωnH)]u+ q[1−An+1(ω1 · · ·ωnT )]d

<Sn

(1 + r)n+1[pu+ qd]

=Sn

(1 + r)n.

If An+1 is a constant a, then

En

[Sn+1

(1 + r)n+1

]=

Sn

(1 + r)n+1(1− a)(pu+ qd) =

Sn

(1 + r)n(1− a).

So En

[Sn+1

(1+r)n+1(1−a)n+1

]= Sn

(1+r)n(1−a)n , which implies Sn

(1−a)n(1+r)n is a martingale under the risk-neutralmeasure.

I Exercise 2.11 (Put-call parity). Consider a stock that pays no dividend in an N -period binomialmodel. A European call has payoff CN = (SN −K)+ at time N . The price Cn of this call at earlier timesis given by the risk-neutral pricing formula (2.4.11):

Cn = En

[CN

(1 + r)N−n

], n = 0, 1, · · · , N − 1.

Consider also a put with payoff PN = (K − SN )+ at time N , whose price at earlier time is

Pn = En

[PN

(1 + r)N−n

], n = 0, 1, · · · , N − 1.

17

Finally, consider a forward contract to buy one share of stock at time N for K dollars. The price of thiscontract at time N is FN = SN −K, and its price at earlier times is

Fn = En

[FN

(1 + r)N−n

], n = 0, 1, · · · , N − 1.

(Note that, unlike the call, the forward contract requires that the stock be purchased at time N for K dollarsand has a negative payoff if SN < K.)(i) If at time zero you buy a forward contract and a put, and hold them until expiration, explain why thepayoff you receive is the same as the payoff of call; i.e., explain why CN = FN + PN .

Solution.

FN + PN = SN −K + (K − SN )+ =

SN −K +K − SN if K > SN

SN −K if K ≤ SN

= (SN −K)+ = CN .

(ii) Using the risk-neutral pricing formulas given above for Cn, Pn, and Fn and the linearity of conditionalexpectations, show that Cn = Fn + Pn for every n.

Proof. Cn = En[CN

(1+r)N−n ] = En[FN

(1+r)N−n ] + En[PN

(1+r)N−n ] = Fn + Pn.

(iii) Using the fact that the discounted stock price is a martingale under the risk-neutral measure, show thatF0 = S0 − K

(1+r)N.

Proof. F0 = E[ FN

(1+r)N] = 1

(1+r)NE[SN −K] = S0 − K

(1+r)N.

(iv) Suppose you begin at time zero with F0, buy one share of stock, borrowing money as necessary to dothat, and make no further trades. Show that at time N you have a portfolio valued at FN . (This is calleda static replication of the forward contract. If you sell the forward contract for F0 at time zero, you can usethis static replication to hedge your short position in the forward contract.)

Proof. At time zero, the trader has F0 − S0 in money market account and one share of stock. At time N ,the trader has a wealth of (F0 − S0)(1 + r)N + SN = −K + SN = FN .

(v) The forward price of the stock at time zero is defined to be that value of K that causes the forwardcontract to have price zero at time zero. The forward price in this model is (1 + r)NS0. Show that, at timezero, the price of a call struck at the forward price is the same as the price of a put struck at the forwardprice. This fact is called put-call parity.

Proof. By (iii), F0 = S0 − (1+r)NS0

(1+r)N= 0. So by (ii), C0 = F0 + P0 = P0.

Remark 2.5. The forward price can be easily determined by the static replication: to hedge a short positionin forward contract, we can buy and hold one share of stock until time N . The cost of this strategy is S0 attime zero and S0(1 + r)N at time N . At time N , we receive K dollars for compensation. So we must haveK = S0(1 + r)N to eliminate arbitrage.

(vi) If we choose K = (1 + r)NS0, we just saw in (v) that C0 = P0. Do we have Cn = Pn for every n?

Proof. By (ii), Cn = Pn if and only if Fn = 0. Note Fn = En[SN−K

(1+r)N−n ] = Sn − (1+r)NS0

(1+r)N−n = Sn − S0(1 + r)n.So Fn is zero if and only if n = 0 or r = 0.

18

I Exercise 2.12 (Chooser option). Let 1 ≤ m ≤ N − 1 and K > 0 be given. A chooser option is acontract sold at time zero that confers on its owner the right to receive either a call or a put at time m.The owner of the chooser may wait until time m before choosing. The call or put chosen expires at timeN with strike K. Show that the time-zero price of a chooser option is the sum of the time-zero price of aput, expiring at time N and having strike price K, and a call, expiring at time m and having strike price

K(1+r)N−m . (Hint: Use put-call parity (Exercise 2.11).)

Proof. First, the no-arbitrage price of the chooser option at time m must be max(Cm, Pm), where

Cm = Em

[(SN −K)+

(1 + r)N−m

]and Pm = Em

[(K − SN )+

(1 + r)N−m

].

That is, Cm is the no-arbitrage price of a call option at time m and Pm is the no-arbitrage price of a putoption at time m. Both of the call and the put have maturity date N and strike price K. Suppose the marketis liquid, then the chooser option is equivalent to receiving a payoff of max(Cm, Pm) at time m. Therefore,its no-arbitrage price at time 0 should be E[max(Cm,Pm)

(1+r)m ].

By the put-call parity, Cm = Sm − K(1+r)N−m + Pm. So max(Cm, Pm) = Pm +

[Sm − K

(1+r)N−m

]+.

Therefore, the time-zero price of a chooser option is

E[

Pm

(1 + r)m

]+ E

(Sm − K

(1+r)N−m

)+(1 + r)m

= E[(K − SN )+

(1 + r)N

]+ E

(Sm − K

(1+r)N−m

)+(1 + r)m

.The first term stands for the time-zero price of a put, expiring at time N and having strike price K, and thesecond term stands for the time-zero price of a call, expiring at time m and having strike price K

(1+r)N−m .

I Exercise 2.13 (Asian option). Consider an N -period binomial model. An Asian option has a payoffbased on the average stock price, i.e.,

VN = f

(1

N + 1

N∑n=0

Sn

),

where the function f is determined by the contractual details of the option.(i) Define Yn =

∑nk=0 Sk and use the Independence Lemma 2.5.3 to show that the two-dimensional

process Sn, Yn, n = 0, 1, · · · , N is Markov.

Proof. Note the conditional distribution

P(ωn+1 = i|ω1, · · · , ωn) =

1+r−du−d if i = H

u−1−4u−d if i = T

is deterministic (i.e., independent of ω1, · · · , ωn). So ωn’s are i.i.d. under P, as well as under P. Hence,without loss of generality, we can work under P. For any function g,

En[g(Sn+1, Yn+1)]

= En

[g

(Sn+1

SnSn, Yn +

Sn+1

SnSn

)]= pg(uSn, Yn + uSn) + qg(dSn, Yn + dSn),

which is a function of (Sn, Yn). This shows (Sn, Yn)0≤n≤N is Markov under P.

(ii) According to Theorem 2.5.8, the price Vn of the Asian option at time n is some function vn of Sn

and Yn; i.e.Vn = vn(Sn, Yn), n = 0, 1, · · · , N.

Give a formula for vN (s, y), and provide an algorithm for computing vn(s, y) in terms of vn+1.

19

Proof. Set vN (s, y) = f( yN+1 ). Then vN (SN , YN ) = f

(∑Nn=0 Sn

N+1

)= VN . Suppose vn+1 is given, then

Vn = En

[Vn+1

1 + r

]= En

[vn+1(Sn+1, Yn+1)

1 + r

]=

1

1 + r[pvn+1(uSn, Yn + uSn) + qvn+1(dSn, Yn + dSn)].

So the formula for computing vn(s, y) in terms of vn+1 is

vn(s, y) =pvn+1(us, y + us) + qvn+1(ds, y + ds)

1 + r.

I Exercise 2.14 (Asian option continued). Consider an N -period binomial model, and let M be a fixednumber between 0 and N − 1. Consider an Asian option whose payoff at time N is

Vn = f

(1

N −M

N∑n=M+1

Sn

),

where again the function f is determined by the contractual details of the option.(i) Define

Yn =

0, if 0 ≤ n ≤M ,∑n

k=M+1 Sk, if M + 1 ≤ n ≤ N .

Show that the two-dimensional process (Sn, Yn), n = 0, 1, · · · , N is Markov (under the risk-neutral measureP).

Proof. For n ≤ M , (Sn, Yn) = (Sn, 0). Since coin tosses ωn’s are i.i.d. under P, (Sn, Yn)0≤n≤M is Markovunder P . More precisely, for any function h, En[h(Sn+1)] = ph(uSn) + qh(dSn), for n = 0, 1, · · · ,M − 1.

For any function g of two variables, we have

EM [g(SM+1, YM+1)] = EM [g(SM+1, SM+1)] = pg(uSM , uSM ) + qg(dSM , dSM ).

And for n ≥M + 1,

En[g(Sn+1, Yn+1)] = En

[g

(Sn+1

SnSn, Yn +

Sn+1

SnSn

)]= pg(uSn, Yn + uSn) + qg(dSn, Yn + dSn).

So (Sn, Yn)0≤n≤N is Markov under P.

(ii) According to Theorem 2.5.8, the price Vn of the Asian option at time n is some function vn of Sn

and Yn, i.e.Vn = vn(Sn, Yn), n = 0, 1, · · · , N.

Of course, when n ≤M , Yn is not random and does not need to be included in this function. Thus, for suchn we should seek a function vn of Sn alone and have

Vn =

vn(Sn), if 0 ≤ n ≤M ,vn(Sn, Yn), if M + 1 ≤ n ≤ N .

Give a formula for vN (s, y), and provide an algorithm for computing vn in terms of vn+1. Note that thealgorithm is different for n < M and n > M , and there is a separate transition formula for vM (s) in termsof vM+1(·, ·).

20

Proof. Set vN (s, y) = f( yN−M ). Then vN (SN , YN ) = f

(∑NK=M+1 Sk

N−M

)= VN .

Suppose vn+1 is already given.a) If n > M , then

En[vn+1(Sn+1, Yn+1)] = pvn+1(uSn, Yn + uSn) + qvn+1(dSn, Yn + dSn).

So vn(s, y) = pvn+1(us, y + us) + qvn+1(ds, y + ds).b) If n =M , then

EM [vM+1(SM+1, YM+1)] = pvM+1(uSM , uSM ) + vn+1(dSM , dSM ).

So vM (s) = pvM+1(us, us) + qvM+1(ds, ds).c) If n < M , then

En[vn+1(Sn+1)] = pvn+1(uSn) + qvn+1(dSn).

So vn(s) = pvn+1(us) + qvn+1(ds).

3 State Prices⋆ Comments:According to the Fundamental Theorem of Asset Pricing (FTAP), the no-arbitrage property is associated

with the existence of a probability measure called equivalent martingale measure (EMM) and a positiveprocess called numéraire, such that the price processes of tradable assets discounted by the numéraire aremartingales under the given probability measure (hence the name martingale measure).

Theorem 2.4.4 shows that the risk-neutral probability P defined by

p =1 + r − d

u− d, q =

u− 1− r

u− d

and the value process of the money market account (1+r)nNn=0 is such an EMM-numéraire pair. Theorem3.2.7 shows that the actual probability P and the inverse of the state price density process

1ζn

N

n=0is such

an EMM-numéraire pair. Note numéraire does not have to the price process of a tradable asset, as theinverse of the state price density process is just an abstract stochastic process.

For a survey of the various formulations of the two Fundamental Theorem of Asset Pricing, we refer toZeng [9].

I Exercise 3.1. Under the conditions of Theorem 3.1.1, show the following analogues of properties (i)-(iii)of that theorem:(i’) P

(1Z > 0

)= 1;

(ii’) E 1Z = 1;

(iii’) for any random variable Y ,

EY = E[1

Z· Y].

In other words, 1Z facilitates the switch from E to E in the same way Z facilitates the switch from E to E.

Proof. Define Z(ω) := P(ω)

P(ω)= 1

Z(ω) . Apply Theorem 3.1.1 with P, P, Z replaced by P, P, Z, we get theanalogues of properties (i)-(iii) of Theorem 3.1.1.

I Exercise 3.2. Let P be a probability measure on a finite probability space Ω. In this problem, weallow the possibility that P(ω) = 0 for some values of ω ∈ Ω. Let Z be a random variable on Ω with theproperty that P(Z ≥ 0) = 1 and EZ = 1. For ω ∈ Ω, define P(ω) = Z(ω)P(ω), and for events A ⊂ Ω, defineP(A) =

∑ω∈A P(ω). Show the following.

(i) P is a probability measure; i.e., P(Ω) = 1.

21

Proof. P(Ω) =∑

ω∈Ω P(ω) =∑

ω∈Ω Z(ω)P(ω) = E[Z] = 1.

(ii) If Y is a random variable, then EY = E[ZY ].

Proof. E[Y ] =∑

ω∈Ω Y (ω)P(ω) =∑

ω∈Ω Y (ω)Z(ω)P(ω) = E[Y Z].

(iii) If A is an event with P(A) = 0, then P(A) = 0.

Proof. P(A) =∑

ω∈A Z(ω)P(ω). Since P(A) = 0, P(ω) = 0 for any ω ∈ A. So P(A) = 0.

(iv) Assume that P(Z > 0) = 1. Show that if A is an event with P(A) = 0, then P(A) = 0.

Proof. If P(A) =∑

ω∈A Z(ω)P(ω) = 0, by P(Z > 0) = 1, we conclude P(ω) = 0 for any ω ∈ A. SoP(A) =

∑ω∈A P(ω) = 0.

When two probability measures agree which events have probability zero (i.e., P(A) = 0 if and onlyif P(A) = 0), the measure are said to be equivalent. From (iii) and (iv) above, we see that P and P areequivalent under the assumption that P(Z > 0) = 1.(v) Show that if P and P are equivalent, then they agree which events have probability one (i.e., P(A) = 1

if and only if P(A) = 1).

Proof. P(A) = 1 ⇐⇒ P(Ac) = 0 ⇐⇒ P(Ac) = 0 ⇐⇒ P(A) = 1.

(vi) Construct an example in which we have only P(Z ≥ 0) = 1 and P and P are not equivalent.

Solution. Pick ω0 such that 1 > P(ω0) > 0. Define

Z(ω) =

0 if ω = ω0

1P(ω0)

if ω = ω0.

Then P(Z ≥ 0) = 1 and E[Z] = 1P(ω0)

· P(ω0) = 1.Clearly

P(Ω \ ω0) =∑ω =ω0

Z(ω)P(ω) = 0.

But P(Ω \ ω0) = 1− P(ω0) > 0 since P(ω0) < 1. Hence P and P cannot be equivalent.

I Exercise 3.3. Using the stock price model of Figure 3.1.1 and the actual probabilities p = 23 , q = 1

3 ,define the estimates of S3 at various times by

Mn = En[S3], n = 0, 1, 2, 3.

Fill in the values of Mn in a tree like that of Figure 3.1.1. Verify that Mn, n = 0, 1, 2, 3, is a martingale.

22

M3(HHH) = 32

M2(HH) = 24

M1(H) = 18 M3(HHT ) =M3(HTH) =M3(THH) = 8

M0 = 13.5 M2(HT ) =M2(TH) = 6

M1(T ) = 4.5 M3(HTT ) =M3(THT ) =M3(TTH) = 2

M2(TT ) = 1.5

M3(TTT ) = .50

Exercise 3.3.

Solution. M3 = S3,

M2(HH) = E2[S3](HH) = pS3(HHH) + qS3(HHT ) =2

3· 32 + 1

3· 8 = 24

M2(HT ) = E2[S3](HT ) = pS3(HTH) + qS3(HTT ) =2

3· 8 + 1

3· 2 = 6

M2(TH) = E2[S3](TH) = pS3(THH) + qS3(THT ) =2

3· 8 + 1

3· 2 = 6

M2(TT ) = E2[S3](TT ) = pS3(TTH) + qS3(TTT ) =2

3· 2 + 1

3· 0.5 = 1.5

M1(H) = E1[S3](H) = p2S3(HHH) + pqS3(HHT ) + qpS3(HTH) + q2S3(HTT )

=

(2

3

)2

· 32 + 2

3· 13· (8 + 8) +

(1

3

)2

· 2 = 18

M1(T ) = E1[S3](T ) = p2S3(THH) + pqS3(THT ) + qpS3(TTH) + q2S3(TTT )

=

(2

3

)2

· 8 + 2

3· 13· (2 + 2) +

(1

3

)2

· 0.5 = 4.5

M0 = E0[S3] =

(2

3

)3

· 32 +(2

3

)2

· 13· (8 + 8 + 8) +

2

3·(1

3

)2

· (2 + 2 + 2) +

(1

3

)3

· 0.5 = 13.5

23

To verify (Mn)3n=0 is a martingale, we note M2 = E2[S3] = E2[M3], since M3 = S3. Moreover, we have

E1[M2](H) = pM2(HH) + qM2(HT ) =2

3· 24 + 1

3· 6 = 18 =M1(H)

E1[M2](T ) = pM2(TH) + qM2(TT ) =2

3· 6 + 1

3· 1.5 = 4.5 =M1(T )

E0[M1](T ) = pM1(H) + qM1(T ) =2

3· 18 + 1

3· 4.5 = 13.5 =M0.

Therefore, (Mn)3n=0 is a martingale.

I Exercise 3.4. This problem refers to the model of Example 3.1.2, whose Radon-Nikodým process Zn

appears in Figure 3.2.1.(i) Compute the state price densities

ζ3(HHH),

ζ3(HHT ) = ζ3(HTH) = ζ3(THH),

ζ3(HTT ) = ζ3(THT ) = ζ3(TTH),

ζ3(TTT )

explicitly.

Solution. By formula (3.1.5), we have

ζ3(HHH) =27

64· 1

(1 + 0.25)3= 0.216

ζ3(HHT ) = ζ3(HTH) = ζ3(THH) =27

32· 1

(1 + 0.25)3= 0.432

ζ3(HTT ) = ζ3(THT ) = ζ3(TTH) =27

16· 1

(1 + 0.25)3= 0.864

ζ3(TTT ) =27

8· 1

(1 + 0.25)3= 1.728.

(ii) Use the numbers computed in (i) in formula (3.1.10) to find the time-zero price of the Asian option ofExercise 1.8 of Chapter 1. You should get v0(4, 4) computed in part (ii) of that exercise.

Solution. Recall in Example 3.1.2, we have p = 23 and q = 1

3 . So, by formula (3.1.10), the time-zero price ofthe Asian option of Exercise 1.8 is

V0 = E

[ζ3

(1

4Y3 − 4

)+]=∑ω∈Ω

(1

4Y3(ω)− 4

)+

ζ(ω)P(ω)

=

(1

4· 60− 4

)+

· 0.216 ·(2

3

)3

+

(1

4· 36− 4

)+

· 0.432 ·(2

3

)2

·(1

3

)+(

1

4· 24− 4

)+

· 0.432 ·(2

3

)2

·(1

3

)+

(1

4· 18− 4

)+

· 0.864 ·(1

3

)2

·(2

3

)+(

1

4· 18− 4

)+

· 0.432 ·(2

3

)2

·(1

3

)= 1.216.

24

S0 = 4 Y0 = 4

S1(T ) = 2Y1(T ) = 6

S2(TT ) = 1Y2(TT ) = 7

S3(TTT ) = 0.5Y3(TTT ) = 7.5

S3(TTH) = 2Y3(TTH) = 9

S2(TH) = 4Y2(TH) = 10

S3(THT ) = 2Y3(THT ) = 12

S3(THH) = 8Y3(THH) = 18

S1(H) = 8Y1(H) = 12

S2(HT ) = 4Y2(HT ) = 16

S3(HTT ) = 2Y3(HTT ) = 18

S3(HTH) = 8Y3(HTH) = 24

S2(HH) = 16Y2(HH) = 28

S3(HHT ) = 8Y3(HHT ) = 36

S3(HHH) = 32Y3(HHH) = 60

Exercise 1.8. Asian option.

(iii) Compute also the state price densities ζ2(HT ) = ζ2(TH).

Solution. We note ζ2 = Z2

(1+r)2 = E2

[Z

(1+r)3

]· (1 + r). By formula (3.1.5),

ζ2(HT ) =1

(1 + r)2[pZ(HTH) + qZ(HTT )] =

23 · 27

32 + 13 · 27

16

1.252= 0.72

ζ2(TH) =1

(1 + r)2[pZ(THH) + qZ(THT )] =

1

(1 + r)2[pZ(HTH) + qZ(HTT )] = 0.72.

(iv) Use the risk-neutral pricing formula (3.2.6) in the form

V2(HT ) =1

ζ2(HT )E2[ζ3V3](HT ),

V2(TH) =1

ζ2(TH)E2[ζ3V3](TH)

to compute V2(HT ) and V2(TH). You should get V2(HT ) = v2(4, 16) and V2(TH) = v2(4, 10), where v2(s, y)was computed in part (ii) of Exercise 1.8 of Chapter 1. Note that V2(HT ) = V2(TH).

Solution. By Exercise 1.8 of Chapter 1, V2(HT ) = v2(4, 16) = 1, V2(TH) = v2(4, 10) = 0.2. Meanwhile, we

25

have

V2(HT ) =1

ζ2(HT )E2[ζ3V3](HT ) =

1

ζ2(HT )[pζ3(HTH)V3(HTH) + qζ3(HTT )V3(HTT )]

=1

0.72

[2

3· 0.432 ·

(1

4· 24− 4

)+

+1

3· 0.864 ·

(1

4· 18− 4

)+]

= 1,

and

V2(TH) =1

ζ2(TH)E2[ζ3V3](TH) =

1

ζ2(TH)[pζ3(THH)V3(THH) + qζ3(THT )V3(THT )]

=1

0.72

[2

3· 0.432 ·

(1

4· 18− 4

)+

+1

3· 0.864 ·

(1

4· 12− 4

)+]

= 0.2.

I Exercise 3.5 (Stochastic volatility, random interest rate). Consider the model of Exercise 2.9 ofChapter 2. Assume that the actual probability measure is

P(HH) =4

9, P(HT ) =

2

9, P(TH) =

2

9, P(TT ) =

1

9.

The risk-neutral measure was computed in Exercise 2.9 of Chapter 2.(i) Compute the Radon-Nikodým derivative Z(HH), Z(HT ), Z(TH), and Z(TT ) of P with respect to P.

Solution. In Exercise 2.9 of Chapter 2, we have

P(HH) =1

4, P(HT ) =

1

4, P(TH) =

1

12, P(TT ) =

5

12.

Therefore

Z(HH) =P(HH)

P(HH)=

9

16

Z(HT ) =P(HT )P(HT )

=9

8

Z(TH) =P(TH)

P(TH)=

3

8

Z(TT ) =P(TT )P(TT )

=15

4.

(ii) The Radon-Nikodým derivative process Z0, Z1, Z2 satisfies Z2 = Z. Compute Z1(H), Z1(T ), and Z0.Note that Z0 = EZ = 1.

Solution. From the specification of P(HH), P(HT ), P(TH), and P(TT ), we can conclude ω1 and ω2 are i.i.d.under P , with p := P(ω1 = H) = P(ω2 = H) = 2

3 and q := P(ω1 = T ) = P(ω2 = T ) = 13 . Therefore,

Z1(H) = E1[Z2](H) = Z2(HH)p+ Z2(HT )q =9

16· 23+

9

8· 13=

3

4,

Z1(T ) = E1[Z2](T ) = Z2(TH)p+ Z2(TT )q =3

8· 23+

15

4· 13=

3

2,

Z0 = E[Z1] = pZ1(H) + qZ1(T ) =2

3· 34+

1

3· 32= 1.

26

(iii) The version of the risk-neutral pricing formula (3.2.6) appropriate for this model, which does not usethe risk-neutral measure, is

V1(H) =1 + r0Z1(H)

E1

[Z2

(1 + r0)(1 + r1)V2

](H)

=1

Z1(H)(1 + r1(H))E1[Z2V2](H),

V1(T ) =1 + r0Z1(H)

E1

[Z2

(1 + r0)(1 + r1)V2

](T )

=1

Z1(H)(1 + r1(H))E1[Z2V2](T ),

V0 = E[

Z2

(1 + r0)(1 + r1)V2

].

Use this formula to compute V1(H), V1(T ), and V0 when V2 = (S2 − 7)+. Compare the result with youranswers in Exercise 2.9(ii) of Chapter 2.5

Solution.

V1(H) =Z2(HH)V2(HH)p+ Z2(HT )V2(HT )q

Z1(H)(1 + r1(H))=

916 · (12− 7)+ · 2

3 + 154 · (8− 7)+ · 1

334 ·(1 + 1

4

) = 2.4,

V1(T ) =Z2(TH)V2(TH)p+ Z2(TT )V2(TT )q

Z1(T )(1 + r1(T ))=

38 · (8− 7)+ · 2

3 + 154 · (2− 7)+ · 1

332 ·(1 + 1

2

) =1

9,

and

V0 =Z2(HH)V2(HH)

(1 + r0)(1 + r1(H))P(HH) +

Z2(HT )V2(HT )

(1 + r0)(1 + r1(H))P(HT ) +

Z2(TH)V2(TH)

(1 + r0)(1 + r1(T ))P(TH) + 0

=916 · (12− 7)+ · 4

9

(1 + 14 )(1 +

14 )

+98 · (8− 7)+ · 2

9

(1 + 14 )(1 +

14 )

+38 · (8− 7)+ · 2

9

(1 + 14 )(1 +

12 )

= 1.00444.

I Exercise 3.6. Consider Problem 3.3.1 in an N -period binomial model with the utility function U(x) =lnx. Show that the optimal wealth process corresponding to the optimal portfolio process is given byXn = X0

ζn, n = 0, 1, · · · , N , where ζn is the state price density process defined in (3.2.7).

Proof. U ′(x) = 1x , so I(x) = 1

x . (3.3.26) gives E[

Z(1+r)N

(1+r)N

λZ

]= X0. So λ = 1

X0. By (3.3.25), we have

XN = (1+r)N

λZ = X0

Z (1 + r)N . Hence

Xn = En

[XN

(1 + r)N−n

]= En

[X0(1 + r)n

Z

]= X0(1 + r)nEn

[1

Z

]= X0(1 + r)n

1

ZnEn

[Z · 1

Z

]=X0

ζn,

where the second to last “=” comes from Lemma 3.2.6.

I Exercise 3.7. Consider Problem 3.3.1 in an N -period binomial model with the utility function U(x) =1px

p, where p < 1, p = 0. Show that the optimal wealth at time N is

XN =X0(1 + r)NZ

1p−1

E[Z

pp−1

] ,

where Z is the Radon-Nikodým derivative of P with respect to P.5The textbook said “Exercise 2.6” by mistake.

27

Proof. U ′(x) = xp−1, so I(x) = x1

p−1 . By (3.3.26), we have E[

Z(1+r)N

(λZ

(1+r)N

) 1p−1

]= X0. Solve it for λ,

we get

λ =

X0

E[

Zp

p−1

(1+r)Npp−1

]

p−1

=Xp−1

0 (1 + r)Np(E[Z

pp−1

])p−1 .

So by (3.3.25),

XN =

[λZ

(1 + r)N

] 1p−1

=λ

1p−1Z

1p−1

(1 + r)N

p−1

=X0(1 + r)

Npp−1

E[Z

pp−1

] Z1

p−1

(1 + r)N

p−1

=(1 + r)NX0Z

1p−1

E[Z

pp−1

] .

I Exercise 3.8. The Lagrange Multiplier Theorem used in the solution of Problem 3.3.5 has hypothesesthat we did not verify in the solution of that problem. In particular, the theorem states that if the gradientof the constraint function, which in this case is the vector (p1ζ1, · · · , pmζm), is not the zero vector, thenthe optimal solution must satisfy the Lagrange multiplier equations (3.3.22). This gradient is not the zerovector, so this hypothesis is satisfied. However, even when this hypothesis is satisfied, the theorem does notguarantee that there is an optimal solution; the solution to the Lagrange multiplier equations may in factminimize the expected utility. The solution could also be neither a maximizer nor a minimizer. Therefore,in this exercise, we outline a different method for verifying that the random variable XN given by (3.3.25)maximizes the expected utility.

We begin by changing the notation, calling the random variable given by (3.3.25) X∗N rather than XN .

In other words,X∗

N = I

(λ

(1 + r)NZ

), (3.6.1)

where λ is the solution of equation (3.3.26). This permits us to use the notation XN for an arbitrary (notnecessarily optimal) random variable satisfying (3.3.19). We must show that

EU(XN ) ≤ EU(X∗N ). (3.6.2)

(i) Fix y > 0, and show that the function of x given by U(x)− yx is maximized by x = I(y).6 Conclude that

U(x)− yx ≤ U(I(y))− yI(y) for every x. (3.6.3)

Proof. ddx (U(x)−yx) = U ′(x)−y. So x = I(y) is an extreme point of U(x)−yx. Because d2

dx2 (U(x)−yx) =U ′′(x) ≤ 0 (U is concave), x = I(y) is a maximum point. Therefore U(x)− yx ≤ U(I(y))− yI(y) for everyx.

(ii) In (3.6.3), replace the dummy variable x by the random variable XN and replace the dummy variabley by the random variable λZ

(1+r)N. Take expectations of both sides and use (3.3.19) and (3.3.26) to conclude

that (3.6.2) holds.

Proof. Following the hint of the problem, we have

E[U(XN )]− E[XN

λZ

(1 + r)N

]≤ E

[U

(I

(λZ

(1 + r)N

))]− E

[λZ

(1 + r)NI

(λZ

(1 + r)N

)],

i.e.

E[U(XN )]− λX0(3.3.19)= E[U(XN )]− λE

[XN

(1 + r)N

]= E[U(XN )]− E

[XN

λZ

(1 + r)N

]≤ E[U(X∗

N )]− λE[

Z

(1 + r)NI

(λZ

(1 + r)N

)](3.3.26)= E[U(X∗

N )]− λX0.

6The textbook said “y = I(x)” by mistake.

28

So E[U(XN )] ≤ E[U(X∗N )].

I Exercise 3.9 (Maximizing probability of reaching a goal). (Kulldorf [30], Heath [19]) A wealthyinvestor provides a small amount of money X0 for you to use to prove the effectiveness of your investmentscheme over the next N periods. You are permitted to invest in the N -period binomial model, subject tothe condition that the value of your portfolio is never allowed to be negative. If at time N the value of yourportfolio XN is at least γ, a positive constant specified by the investor, then you will be given a large amountof money to manage for her. Therefore, your problem is the following:Maximize

P(XN ≥ γ),

where XN is generated by a portfolio process beginning with the initial wealth X0 and where the value Xn

of your portfolio satisfiesXn ≥ 0, n = 1, 2, · · · , N.

In the way that Problem 3.3.1 was reformulated as Problem 3.3.3, this problem may be reformulated asMaximize

P(XN ≥ γ)

subject toE

XN

(1 + r)N= X0, Xn ≥ 0, n = 1, 2, · · · , N.

(i) Show that if XN ≥ 0, then Xn ≥ 0 for all n.

Proof. By the martingale property, Xn = En

[XN

(1+r)N−n

]. So if XN ≥ 0, then Xn ≥ 0 for all n.

(ii) Consider the function

U(x) =

0, if 0 ≤ x < γ,1, if x ≥ γ.

Show that for each fixed y > 0, we have

U(x)− yx ≤ U(I(y))− yI(y) ∀x ≥ 0,

where

I(y) =

γ, if 0 < y ≤ 1

γ ,0, if y > 1

γ .

Proof. We consider the following four scenarios.a) If 0 ≤ x < γ and 0 < y ≤ 1

γ , then U(x)−yx = −yx ≤ 0 and U(I(y))−yI(y) = U(γ)−yγ = 1−yγ ≥ 0.So U(x)− yx ≤ U(I(y))− yI(y).

b) If 0 ≤ x < γ and y > 1γ , then U(x) − yx = −yx ≤ 0 and U(I(y)) − yI(y) = U(0) − y · 0 = 0. So

U(x)− yx ≤ U(I(y))− yI(y).c) If x ≥ γ and 0 < y ≤ 1

γ , then U(x)− yx = 1− yx and U(I(y))− yI(y) = U(γ)− yγ = 1− yγ ≥ 1− yx.So U(x)− yx ≤ U(I(y))− yI(y).

d) If x ≥ γ and y > 1γ , then U(x) − yx = 1 − yx < 0 and U(I(y)) − yI(y) = U(0) − y · 0 = 0. So

U(x)− yx ≤ U(I(y))− yI(y).

(iii) Assume there is a solution λ to the equation

E[

Z

(1 + r)NI

(λZ

(1 + r)N

)]= X0. (3.6.4)

Following the argument of Exercise 3.8, show that the optimal XN is given by

X∗N = I

(λZ

(1 + r)N

).

29

Proof. Using (ii) and set x = XN , y = λZ(1+r)N

, where XN is a random variable satisfying E[

XN

(1+r)N

]= X0,

we haveE[U(XN )]− E

[λZ

(1 + r)NXN

]≤ E[U(X∗

N )]− E[

λZ

(1 + r)NX∗

N

].

That is, E[U(XN )]− λX0 ≤ E[U(X∗N )]− λX0. So E[U(XN )] ≤ E[U(X∗

N )].

(iv) As we did to obtain Problem 3.3.5, let us list the M = 2N possible coin toss sequences, labeling themω1, · · · , ωM , and then define ζm = ζ(ωm), pm = P(ωm). However, here we list these sequences in ascendingorder of ζm i.e., we label the coin toss sequences so that

ζ1 ≤ ζ2 ≤ · · · ≤ ζM .

Show that the assumption that there is a solution λ to (3.6.4) is equivalent to assuming that for some positiveinteger K we have ζK < ζK+1 and

K∑m=1

ζmpm =X0

γ. (3.6.5)

Proof. Plug pm and ξm into (3.6.4), we have

X0 =

2N∑m=1

pmξmI(λξm) =

2N∑m=1

pmξmγ1λξm≤ 1γ .

So X0

γ =∑2N

m=1 pmξm1λξm≤ 1γ . Suppose there is a solution λ to (3.6.4), note X0

γ > 0, we then can concludem : λξm ≤ 1

γ = ∅. Let K = maxm : λξm ≤ 1γ , then λξK ≤ 1

γ < λξK+1. So ξK < ξK+1 andX0

γ =∑K

m=1 pmξm (Note, however, that K could be 2N . In this case, ξK+1 is interpreted as ∞. Also, notewe are looking for positive solution λ > 0). Conversely, suppose there exists some K so that ξK < ξK+1 and∑K

m=1 ξmpm = X0

γ . Then we can find λ > 0, such that ξK < 1λγ < ξK+1. For such λ, we have

E[

Z

(1 + r)NI(

λZ

(1 + r)N)

]=

2N∑m=1

pmξm1λξm≤ 1γ γ =

K∑m=1

pmξmγ = X0.

Hence (3.6.4) has a solution.

(v) Show that X∗N is given by

XN (ωm) =

γ, if m ≤ K

0, if m ≥ K + 1.

Proof.

X∗N (ωm) = I(λξm) = γ1λξm≤ 1

γ =

γ, if m ≤ K,0, if m ≥ K + 1.

4 American Derivative Securities⋆ Comments:

1) Before proceeding to the exercise problems, we first give a brief summary of pricing American derivativesecurities as presented in the textbook. We shall use the notation of the book.

From the buyer’s perspective. At time n, if the derivative security has not been exercised, then the buyercan choose a policy τ with τ ∈ Sn. The valuation formula for cash flow (Theorem 2.4.8) gives the fair price

30

of the derivative security exercised according to τ (the cash flow sequence Cn, · · · , CN at times n, · · · , Nare defined by Cn = 1τ=nGn, · · · , CN = 1τ=NGN ):

Vn(τ) =N∑

k=n

En

[1τ=k

Gk

(1 + r)k−n

]= En

[1τ≤N

Gτ

(1 + r)τ−n

].

The buyer wants to consider all the possible τ ’s, so that he can find the least upper bound of security value,which will be the maximum price of the derivative security acceptable to him. This is the price given byDefinition 4.4.1: Vn = maxτ∈Sn En

[1τ≤N

1(1+r)τ−nGτ

].

From the seller’s perspective. A price process (Vn)0≤n≤N is acceptable to him if and only if at each timen, (i) Vn ≥ Gn (selling price is sufficient to cover potential obligation) and (ii) he needs no further investinginto the portfolio as time goes by (no follow-up cost to maintain (i)).

Formally, the seller can find hedging process (∆n)0≤n≤N and cash flow process (Cn)0≤n≤N such thatCn ≥ 0 and Vn+1 = ∆nSn+1 + (1 + r)(Vn − Cn − ∆nSn) ≥ Gn+1. Since

Sn

(1+r)n

0≤n≤N

is a martingale

under the risk-neutral measure P, we conclude

En

[Vn+1

(1 + r)n+1

]− Vn

(1 + r)n= − Cn

(1 + r)n≤ 0,

i.e.

Vn

(1+r)n

0≤n≤N

is a supermartingale under P. This inspires us to check if the converse is also true.This is exactly the content of Theorem 4.4.4. So (Vn)0≤n≤N is the value process of a portfolio that needsno further investing if and only if

Vn

(1+r)n

0≤n≤N

is a supermartingale under P (note this is independent ofthe requirement Vn ≥ Gn). In summary, a price process (Vn)0≤n≤N is acceptable to the seller if and only if(i) Vn ≥ Gn; (ii)

Vn

(1+r)n

0≤n≤N

is a supermartingale under P.Theorem 4.4.2 shows the buyer’s upper bound is the seller’s lower bound. So it gives the price acceptable

to both. Theorem 4.4.3 gives a specific algorithm for calculating the price, Theorem 4.4.4 establishes theone-to-one correspondence between super-replication and supermartingale property, and finally, Theorem4.4.5 shows how to decide on the optimal exercise policy.

2) The definition of a stopping time typically seen in textbooks is

τ = n ∈ Fn := σ(ω1, · · · , ωn), n = 0, 1, 2, · · · , N.

To see Definition 4.3.1 agrees with this version of definition, note a typical element of Fn has the formω : ω1 ∈ A1, · · · , ωn ∈ An. So τ = n imposes conditions on ω1, ω2, · · · , ωn only.

3) Comment on Theorem 4.4.4 (Replication of path-dependent American derivatives): in the proof, thecondition that (Vn)

Nn=0 comes from Definition 4.4.1 is only used for Vn ≥ Gn. The proof applies to any

P-supermartingale

Vn

(1+r)n

n. This justifies the paragraph immediately after Theorem 4.4.2. Conversely, if

(Vn)n can be replicated via formula (4.4.16) with Cn ≥ 0, then

Vn

(1+r)n

n

is a P-supermartingale. That is,if we can replicate a stochastic process (Vn)n by money market account and stock without injecting morecash into the portfolio, then

Vn

(1+r)n

n

is a P-supermartingale.

I Exercise 4.1. In the three-period model of Figure 1.2.2 of Chapter 1, let the interest rate be r = 14 so

the risk-neutral probabilities are p = q = 12 .

(i) Determine the price at time zero, denoted V P0 , of the American put that expires at time three and has

intrinsic value gP (s) = (4− s)+.

31

Solution. At time three, the payoff of the put is

V P3 (HHH) = (4− 32)+ = 0

V P3 (HHT ) = V P

3 (HTH) = V P3 (THH) = (4− 8)+ = 0

V P3 (HTT ) = V P

3 (THT ) = V P3 (TTH) = (4− 2)+ = 2

V P3 (TTT ) = (4− 0.5)+ = 3.5.

At time two, the value of the put is calculated by

V P2 (·) = max

(4− S2)

+, E2

[V P3

1 + r

]= max

(4− S2)

+,pV P

3 (·H) + qV P3 (·T )

1 + r

.

Therefore,

V P2 (HH) = max

[4− S2(HH)]+,

2

5[V P

3 (HHH) + V P3 (HHT )]

= max

(4− 16)+,

2

5(0 + 0)

= 0

V P2 (HT ) = max

[4− S2(HT )]

+,2

5[V P

3 (HTH) + V P3 (HTT )]

= max

(4− 4)+,

2

5(0 + 2)

= 0.8

V P2 (TH) = max

[4− S2(TH)]+,

2

5[V P

3 (THH) + V P3 (THT )]

= max

(4− 4)+,

2

5(0 + 2)

= 0.8

V P2 (TT ) = max

[4− S2(TT )]

+,2

5[V P

3 (TTH) + V P3 (TTT )]

= max

(4− 1)+,

2

5(2 + 3.5)

= 3

A time one, the value of the put is calculated by

V P1 (·) = max

(4− S1)

+, E1

[V P2

1 + r

]= max

(4− S1)

+,pV P

2 (·H) + qV P2 (·T )

1 + r

Therefore

V P1 (H) = max

[4− S1(H)]+,

2

5[V P

2 (HH) + V P2 (HT )]

= max

(4− 8)+,

2

5(0 + 0.8)

= 0.32

V P1 (T ) = max

[4− S1(T )]

+,2

5[V P

2 (TH) + V P2 (TT )]

= max

(4− 2)+,

2

5(0.8 + 3)

= 2.

Finally, the value of the put at time zero is

V P0 = max

(4− S0)

+,2

5[V P

1 (H) + V P1 (T )]

= max

(4− 4)+,

2

5(0.32 + 2)

= 0.928.

(ii) Determine the price at time zero, denoted V C0 , of the American call that expires at time three and has

intrinsic value gC(s) = (s− 4)+.

Solution. By Theorem 4.5.1, at the time zero, the price of the American call is the same as the price of theEuropean call with the same strike. Therefore, at time three,

V C3 (HHH) = (32− 4)+ = 28,

V C3 (HHT ) = V C

3 (HTH) = V C3 (THH) = (8− 4)+ = 4,

V C3 (HTT ) = V C

3 (THT ) = V C3 (TTH) = (2− 4)+ = 0,

V C3 (TTT ) = (0.5− 4)+ = 0.

32

At time two,

V C2 (HH) =

2

5(28 + 4) = 12.8, V C

2 (HT ) = V C2 (TH) =

2

5(4 + 0) = 1.6, V C

2 (TT ) =2

5(0 + 0) = 0.

At time one,

V C1 (H) =

2

5(12.8 + 1.6) = 5.76, V C

1 (T ) =2

5(1.6 + 0) = 0.64.

And finally, at time zero,V C0 =

2

5(5.76 + 0.64) = 2.56.

(iii) Determine the price at time zero, denoted V S0 , of the American straddle that expires at time three and

has intrinsic value gS(s) = gP (s) + gC(s).

Solution. gS(s) = |4− s|. At time three,

V S3 (HHH) = |4− 32| = 28,

V S3 (HHT ) = V S

3 (HTH) = V S3 (THH) = |4− 8| = 4,

V S3 (HTT ) = V S

3 (THT ) = V S3 (TTH) = |4− 2| = 2,

V S3 (TTT ) = |4− 0.5| = 3.5.

At time two,

V S2 (HH) = max

|4− 16|, 2

5(28 + 4)

= 12.8,

V S2 (HT ) = V S

2 (TH) = max|4− 4|, 2

5(4 + 2)

= 2.4,

V S2 (TT ) = max

|4− 1|, 2

5(2 + 0.5)

= 3.

At time one,

V S1 (H) = max

|4− 8|, 2

5(12.8 + 2.4)

= 6.08,

V S1 (T ) = max

|4− 2|, 2

5(2.4 + 3)

= 2.16.

Finally, at time zero,V S0 = max

|4− 4|, 2

5(6.08 + 2.16)

= 3.296.

(iv) Explain why V S0 < V P

0 + V C0 .

Solution. In our previous calculations,

V S0 = 3.296 < V P

0 + V C0 = 0.928 + 2.56.

To explain this, we first note the simple inequality

max(a1, b1) + max(a2, b2) ≥ max(a1 + a2, b1 + b2),

33

where “>” holds if and only if b1 > a1 and b2 < a2 or b1 < a1 and b2 > a2.At time N , V S

N = gS(SN ) = gP (SN ) + gC(SN ) = V PN + V C

N . By induction, we can show

V Sn = max

gS(Sn),

pV Sn+1 + qV S

n+1

1 + r

≤ max

gP (Sn) + gC(Sn),

pV Pn+1 + qV P

n+1

1 + r+pV C

n+1 + qV Cn+1

1 + r

≤ max

gP (Sn),

pV Pn+1 + qV P

n+1

1 + r

+ max

gC(Sn),

pV Cn+1 + qV C

n+1

1 + r

= V P

n + V Cn ,

n = 0, 1, · · · , N − 1.To see when “<” holds, note for American call options, we always have gC(Sn) <

pV Cn+1+qV C

n+1

1+r forn = 0, 1, · · · , N − 1. In view of our observation of the simple inequality at the beginning of our solution,“<” holds whenever

gP (Sn) >pV P

n+1 + qV Pn+1

1 + r.

This is typically the case at the optimal exercise time of the put.Beside the mathematical argument, intuitively, a straddle is not an American put plus an American call,

since when exercised, a straddle has a payoff equal to the sum of the payoff of a put and the payoff of acall. However, the exercise time of a put typically differs from that of a call. So intuitively, we must haveV S0 < V P

0 + V C0 .

I Exercise 4.2. In Example 4.2.1, we computed the time-zero value of the American put with strike price5 to be 1.36. Consider an agent who borrows 1.36 at time zero and buys the put. Explain how this agentcan generate sufficient funds to pay off his loan (which grows by 25% each period) by trading in the stockand money markets and optimally exercising the put.

Solution. For this problem, we need Figure 4.2.1, Figure 4.4.1 and Figure 4.4.2. Then

∆1(H) =V2(HH)− V2(HT )

S2(HH)− S2(HT )= − 1

12, ∆1(T ) =

V2(TH)− V2(TT )

S2(TH)− S2(TT )= −1,

and∆0 =

V1(H)− V1(T )

S1(H)− S1(T )≈ −0.433.

The optimal exercise time is τ = infn : Vn = Gn. So

τ(HH) = ∞, τ(HT ) = 2, τ(TH) = τ(TT ) = 1.

Therefore, the agent borrows 1.36 at time zero and buys the put. At the same time, to hedge the longposition, he needs to borrow again and buy 0.433 shares of stock at time zero.

At time one, if the result of coin toss is tail and the stock price goes down to 2, the value of the portfoliois X1(T ) = (1+ r)(−1.36− 0.433S0)+0.433S1(T ) = (1+ 1

4 )(−1.36− 0.433× 4)+0.433× 2 = −3. The agentshould exercise the put at time one and get 3 to pay off his debt.

At time one, if the result of coin toss is head and the stock price goes up to 8, the value of the portfoliois X1(H) = (1 + r)(−1.36 − 0.433S0) + 0.433S1(H) = −0.4. The agent should borrow to buy 1

12 shares ofstock. At time two, if the result of coin toss is head and the stock price goes up to 16, the value of theportfolio is X2(HH) = (1+ r)(X1(H)− 1

12S1(H))+ 112S2(HH) = 0, and the agent should let the put expire.

If at time two, the result of coin toss is tail and the stock price goes down to 4, the value of the portfolio isX2(HT ) = (1 + r)(X1(H)− 1

12S1(H)) + 112S2(HT ) = −1. The agent should exercise the put to get 1. This

will pay off his debt.

34

I Exercise 4.3. In the three-period model of Figure 1.2.2 of Chapter 1, let the interest rate be r = 14 so

the risk-neutral probabilities are p = q = 12 . Find the time-zero price and optimal exercise policy (optimal

stopping time) for the path-dependent American derivative security whose intrinsic value at each time n,n = 0, 1, 2, 3, is

(4− 1

n+1

∑nj=0 Sj

)+. This intrinsic value is a put on the average stock price between time

zero and time n.

Solution. We need Figure 1.2.2 for this problem, and calculate the intrinsic value process and price processof the put as follows.

For the intrinsic value process, G0 = 0, G1(T ) = 1, G2(TH) = 23 , G2(TT ) = 5

3 , G3(THT ) = 1,G3(TTH) = 1.75, G3(TTT ) = 2.125. All the other outcomes of G is negative.

For the price process, V0 = 0.4, V1(T ) = 1, V1(TH) = 23 , V1(TT ) = 5

3 , V3(THT ) = 1, V3(TTH) = 1.75,V3(TTT ) = 2.125. All the other outcomes of V is zero.

Therefore the time-zero price of the derivative security is 0.4 and the optimal exercise time satisfies

τ(ω) =

∞ if ω1 = H,1 if ω1 = T .

I Exercise 4.4. Consider the American put of Example 4.2.1, which has strike price 5. Suppose at timezero we sell this put to a purchaser who has inside information about the stock movements and uses theexercise rule ρ of (4.3.2). In particular, if the first toss is going to result in H, the owner of the put exercisesat time zero, when the put has intrinsic value 1. If the first toss results in T and the second toss is going toresult in H, the owner exercises at time one, when the put has intrinsic value 3. If the first two tosses resultin TT , the owner exercises at time two, when the intrinsic value is 4. In summary, the owner of the put hasthe payoff random variable

Y (HH) = 1, Y (HT ) = 1, Y (TH) = 3, Y (TT ) = 4. (4.8.1)

The risk-neutral expected value of this payoff, discounted from the time of payment back to zero, is

E[(

4

5

)ρ

Y

]=

1

4

[1 + 1 +

4

5· 3 + 16

25· 4]= 1.74. (4.8.2)

The time-zero price of the put computed in Example 4.2.1 is only 1.36. Do we need to charge the insidermore than this amount if we are going to successfully hedge our short position after selling the put to her?Explain why or why not.

Solution. 1.36 is the cost of super-replicating the American derivative security. It enables us to construct aportfolio sufficient to pay off the derivative security, no matter when the derivative security is exercised. Soto hedge our short position after selling the put, there is no need to charge the insider more than 1.36.

To understand the paradox that the risk-neutral expected value 1.74 of the insider’s payoff is greater thanthe fair price 1.36, we note the rationale for risk-neutral pricing is the existence of a self-financing replicatingportfolio:

Xn+1 = ∆nSn+1 + (1 + r)(Xn − Cn −∆nSn), n = 0, 1, · · · , N − 1,

where the hedging process (∆n)n is assumed to be adapted, i.e. ∆n ∈ Fn. This is in contradiction of theuse of inside information. Therefore, it is fishy to use risk-neutral pricing formula when insider informationexists.

I Exercise 4.5. In equation (4.4.5), the maximum is computed over all stopping times in S0. List allthe stopping times in S0 (there are 26), and from among them, list the stopping times that never exercisewhen the option is out of the money (there are 11). For each stopping time τ in the latter set, computeE[1τ≤2

(45

)τGτ

]. Verify that the largest value for this quantity is given by the stopping time of (4.4.6),

the one that makes this quantity equal to the 1.36 computed in (4.4.7).

35

Solution. The stopping times in S0 are(1) τ ≡ 0;(2) τ ≡ 1;(3) τ(HT ) = τ(HH) = 1, τ(TH), τ(TT ) ∈ 2,∞ (4 different ones);(4) τ(HT ), τ(HH) ∈ 2,∞, τ(TH) = τ(TT ) = 1 (4 different ones);(5) τ(HT ), τ(HH), τ(TH), τ(TT ) ∈ 2,∞ (16 different ones).

When the option is out of money, the following stopping times do not exercise(i) τ ≡ 0;(ii) τ(HT ) ∈ 2,∞, τ(HH) = ∞, τ(TH), τ(TT ) ∈ 2,∞ (8 different ones);(iii) τ(HT ) ∈ 2,∞, τ(HH) = ∞, τ(TH) = τ(TT ) = 1 (2 different ones).

For (i), E[1τ≤2

(45

)τGτ

]= G0 = 1. For (ii), E

[1τ≤2

(45

)τGτ

]≤ E

[1τ∗≤2

(45

)τ∗

Gτ∗

], where

τ∗(HT ) = 2, τ∗(HH) = ∞, τ∗(TH) = τ∗(TT ) = 2. So

E

[1τ∗≤2

(4

5

)τ∗

Gτ∗

]=

1

4

[(4

5

)2

· 1 +(4

5

)2

(1 + 4)

]= 0.96.

For (iii), E[1τ≤2

(45

)τGτ

]has the biggest value when τ satisfies τ(HT ) = 2, τ(HH) = ∞, τ(TH) =

τ(TT ) = 1. This value is 1.36.

I Exercise 4.6 (Estimating American put prices). For each n, where n = 0, 1, · · · , N , let Gn be arandom variable depending on the first n coin tosses. The time-zero value of a derivative security that canbe exercised at nay time n ≤ N for payoff Gn but must be exercised at time N if it has not been exercisedbefore that time is

V0 = maxτ∈S0,τ≤N

E[

1

(1 + r)τGτ

]. (4.8.3)

In contrast to equation (4.4.1) in the Definition 4.4.1 for American derivative securities, here we consideronly stopping times that take one of the values 0, 1, · · · , N and not the value ∞.(i) Consider Gn = K − Sn, the derivative security that permits its owner to sell one share of stock forpayment K at any time up to and including N , but if the owner does not sell by time N , then she must doso at time N . Show that the optimal exercise policy is to sell the stock at time zero and that the value ofthis derivative security is K − S0.

Proof. The value of the put at time N , if it is not exercised at previous times, is K − SN . Hence

VN−1 = maxK − SN−1, EN−1

[VN1 + r

]= max

K − SN−1,

K

1 + r− SN−1

= K − SN−1.

The second equality comes from the fact that discounted stock price process is a martingale under risk-neutral probability. By induction, we can show Vn = K − Sn (0 ≤ n ≤ N). So by Theorem 4.4.5, theoptimal exercise policy is to sell the stock at time zero and the value of this derivative security is K−S0.

Remark 4.1. We cheated a little bit by using American algorithm and Theorem 4.4.5, since they are developedfor the case where τ is allowed to be ∞. But intuitively, results in this chapter should still hold for the caseτ ≤ N , provided we replace “maxGn, 0” with “Gn”.

(ii) Explain why a portfolio that holds the derivative security in (i) and a European call with strike K andexpiration time N is at least as valuable as an American put struck at K with expiration time N . Denotethe time-zero value of the European call by V EC

0 and the time-zero value of the American put by V AP0 .

Conclude that the upper boundV AP0 ≤ K − S0 + V EC

0 (4.8.4)

on V AP0 holds.

36

Proof. This is because at time N , if we have to exercise the put and K − SN < 0, we can exercise theEuropean call to set off the negative payoff. In effect, throughout the portfolio’s lifetime, the portfolio hasintrinsic values no less than that of an American put stuck at K with expiration time N . So, we must haveV AP0 ≤ K − S0 + V EC

0 .

(iii) Use put-call parity (Exercise 2.11 of Chapter 2) to derive the lower bound on V AP0 :

K

(1 + r)N− S0 + V EC

0 ≤ V AP0 . (4.8.5)

Proof. Let V EP0 denote the time-zero value of a European put with strike K and expiration time N . Then

V AP0 ≥ V EP

0 = V EC0 − E

[SN −K

(1 + r)N

]= V EC

0 − S0 +K

(1 + r)N.

I Exercise 4.7. For the class of derivative securities described in Exercise 4.6 whose time-zero price isgiven by (4.8.3), let Gn = Sn −K. This derivative security permits its owner to buy one share of stock inexchange for a payment of K at any time up to the expiration time N . If the purchase has not been made attime N , it must be made then. Determine the time-zero value and optimal exercise policy for this derivativesecurity.7

Solution. VN = SN −K,

VN−1 = maxSN−1 −K, EN−1

[VN1 + r

]= max

SN−1 −K,SN−1 −

K

1 + r

= SN−1 −

K

1 + r.

By induction, we can prove Vn = Sn − K(1+r)N−n (0 ≤ n ≤ N) and Vn > Gn for 0 ≤ n ≤ N − 1. So the

time-zero value is S0 − K(1+r)N

and the optimal exercise time is N .

5 Random WalkI Exercise 5.1. For the symmetric random walk, consider the first passage time τm to the level m. Therandom variable τ2 − τ1 is the number of steps required for the random walk to rise from level 1 to level 2,and this random variable has the same distribution as τ1, the number of steps required for the random walkto rise from level 0 to level 1. Furthermore, τ2−τ1 and τ1 are independent of one another; the latter dependsonly on the coin toss 1, 2, · · · , τ1, and the former depends only on the coin tosses τ1 + 1, τ1 + 2, · · · , τ2.(i) Use these facts to explain why

Eατ2 = (Eατ1)2 for all α ∈ (0, 1).

Proof. E[ατ2 ] = E[α(τ2−τ1)+τ1 ] = E[α(τ2−τ1)]E[ατ1 ] = E[ατ1 ]2.

Remark 5.1. The i.i.d. property of τ2−τ1 and τ1 can be proven by strong Markov property of random walk.

(ii) Without using (5.2.13), explain why for any positive integer m we must have

Eατm = (Eατ1)m for all α ∈ (0, 1). (5.7.1)

Proof. If we define M (m)n =Mn+τm −Mτm (m = 1, 2, · · · ), then (M

(m)· )m as random functions are i.i.d. with

distributions the same as that of M . So τm+1 − τm = infn : M(m)n = 1 are i.i.d. with distributions the

same as that of τ1. Therefore

E[ατm ] = E[α(τm−τm−1)+(τm−1−τm−2)+···+τ1 ] = E[ατ1 ]m.

7Compare with American call option, Theorem 4.5.1.

37

Remark 5.2. For those concerned with the meaning of “(M (m)· )m as random functions...”, note the function

space E := f : N → Z is a Polish space with metric ||f − g|| =∑∞

n=1|fn−gn|

2n . Then each M(m)· : Ω → E

is a random element from one probability space to another measurable space so that we can talk about itsdistribution.

(iii) Would equation (5.7.1) still hold if the random walk is not symmetric? Explain why or why not.

Solution. Yes, it would still hold since the strong Markov property of random walk, hence the i.i.d. propertyof (τm+1 − τm)∞m=0, still holds and the argument of (ii) still works for asymmetric random walk.

I Exercise 5.2 (First passage time for random walk with upward drift). Consider the asymmetricrandom walk with probability p for an up step and probability q = 1− p for a down step, where 1

2 < p < 1so that 0 < q < 1

2 . In the notation of (5.2.1), let τ1 be the first time the random walk starting from level 0reaches level 1. If the random walk never reaches this level, then τ1 = ∞.(i) Define f(σ) = peσ + qe−σ. Show that f(σ) > 1 for all σ > 0.

Proof. For all σ > 0, f ′(σ) = peσ − qe−σ > p− q > 0. So f(σ) > f(0) = 1 for all σ > 0.

(ii) Show that, when σ > 0, the process

Sn = eσMn

(1

f(σ)

)n

is a martingale.

Proof. En

[Sn+1

Sn

]= En

[eσXn+1 1

f(σ)

]= peσ 1

f(σ) + qe−σ 1f(σ) = 1.

(iii) Show that, for σ > 0,

e−σ = E[1τ1<∞

(1

f(σ)

)τ1].

Conclude that P(τ1 <∞) = 1.

Proof. By optional sampling theorem (Theorem 4.3.2), E[Sn∧τ1 ] = E[S0] = 1. Note

Sn∧τ1 = eσMn∧τ1

(1

f(σ)

)n∧τ1

≤ eσ

for all σ > 0, by bounded convergence theorem and the fact 0 < 1τ1=∞Sn∧τ1 ≤ eσ(

1f(σ)

)n→ 0 as n→ ∞,

we haveE[1τ1<∞Sτ1 ] = E[ lim

n→∞Sn∧τ1 ] = lim

n→∞E[Sn∧τ1 ] = 1.

That is, E[1τ1<∞e

σ(

1f(σ)

)τ1]= 1. So e−σ = E

[1τ1<∞

(1

f(σ)

)τ1]. Let σ ↓ 0, again by bounded

convergence theorem, 1 = E[1τ1<∞

(1

f(0)

)τ1]= P (τ1 <∞).

(iv) Compute Eατ1 for α ∈ (0, 1).

Solution. Set α = 1f(σ) = 1

peσ+qe−σ , then as σ varies from 0 to ∞, α can take all the values in (0, 1). Writeσ in terms of α, we have (note 4pqα2 < 4(p+q

2 )2 · 12 = 1)

eσ =1±

√1− 4pqα2

2pα.

38

We want to choose σ > 0, so we should take σ = ln(

1+√

1−4pqα2

2pα

). Therefore

E[ατ1 ] =2pα

1 +√1− 4pqα2

=1−

√1− 4pqα2

2qα.

(v) Compute Eτ1.

Solution. ∂∂αE[α

τ1 ] = E[ ∂∂αα

τ1 ] = E[τ1ατ1−1], and(1−

√1− 4pqα2

2qα

)′

=1

2q

[(1−

√1− 4pqα2)α−1

]′=

1

2q

[−1

2(1− 4pqα2)−

12 (−4pq2α)α−1 + (1−

√1− 4pqα2)(−1)α2

].

Therefore

E[τ1] = limα↑1

∂

∂αE[ατ1 ] =

1

2q

[−1

2(1− 4pq)−

12 (−8pq)− (1−

√1− 4pq)

]=

1

2p− 1=

1

p− q.

I Exercise 5.3 (First passage time for random walk with downward drift). Modify Exercise 5.2by assuming 0 < p < 1

2 so that 12 < q < 1.

(i) Find a positive number σ0 such that the function f(σ) = peσ + qe−σ satisfies f(σ0) = 1 and f(σ) > 1 forall σ > σ0.

Solution. Solve the equation peσ + qe−σ = 1 and a positive solution is

ln 1 +√1− 4pq

2p= ln 1− p

p= ln q − ln p.

Set σ0 = ln q − ln p, then f(σ0) = 1 and

f ′(σ) = peσ − qe−σ = qe−σ[e2(σ−

ln q−ln p2 ) − 1

]> 0

for σ > σ0. So f(σ) > f(σ0) = 1 for all σ > σ0.

(ii) Determine Pτ1 <∞. (This quantity is no longer equal to 1.)

Solution. As in Exercise 5.2, Sn = eσMn

(1

f(σ)

)nis a martingale, and

1 = E[S0] = E[Sn∧τ1 ] = E

[eσMn∧τ1

(1

f(σ)

)τ1∧n].

Suppose σ > σ0, then by bounded convergence theorem and f(σ) > 1, we have

1 = E

[lim

n→∞eσMn∧τ1

(1

f(σ)

)n∧τ1]= E

[1τ1<∞e

σ

(1

f(σ)

)τ1].

Let σ ↓ σ0, we get P(τ1 <∞) = e−σ0 = pq < 1.

39

(iii) Compute Eατ1 for α ∈ (0, 1).

Solution. Choose σ > σ0, then f(σ) > 1 and we have from (ii) E[1τ1<∞

(1

f(σ)

)τ1]= e−σ. Set α = 1

f(σ) ,then

E[ατ11τ1<∞

]= e−σ(α)

where σ(α) satisfies eσ(α) = 1±√

1−4pqα2

2pα . To make sure σ(α) > σ0, we choose σ(α) = ln(

1+√

1−4pqα2

2pα

). As

a consequence,

E[ατ11τ1<∞

]=

1−√

1− 4pqα2

2qα.

(iv) Compute E[1τ1<∞τ1

]. (Since P(τ1 = ∞) > 0, we have Eτ1 = ∞.)

Solution. Using result in (iii) and bounded convergence theorem, we have

E[τ11τ1<∞

]= E

[limα↑1

(τ1α

τ1−1)]

= limα↑1

E[(τ1α

τ1−1)]

= limα↑1

∂

∂αE[ατ1 ]

=1

2q

[4pq√1− 4pq

− (1−√

1− 4pq)

]=

1

2q

[4pq

2q − 1− 1 + 2q − 1

]=

p

q

1

q − p.

I Exercise 5.4 (Distribution of τ2). Consider the symmetric random walk, and let τ2 be the first timethe random walk, starting from level 0, reaches the level 2. According to Theorem 5.2.3,

Eατ2 =

(1−

√1− α2

α

)2

for all α ∈ (0, 1).

Using the power series (5.2.21), we may write the right-hand side as(1−

√1− α2

α

)2

=2

α· 1−

√1− α2

α− 1

= −1 +

∞∑j=1

(α2

)2j−2 (2j − 2)!

j!(j − 1)!

=∞∑j=2

(α2

)2j−2 (2j − 2)!

j!(j − 1)!

=

∞∑k=1

(α2

)2k (2k)!

(k + 1)!k!.

(i) Use the power series above to determine P τ2 = 2k, k = 1, 2, · · · .

Solution. E[ατ2 ] =∑∞

k=1 P(τ2 = 2k)α2k =∑∞

k=1

(α2

)2k P(τ2 = 2k)4k. So P(τ2 = 2k) = (2k)!4k(k+1)!k!

.

(ii) Use the reflection principle to determine Pτ2 = 2k, k = 1, 2, · · · .

40

Solution. P(τ2 = 2) = 14 . For k ≥ 2, P(τ2 = 2k) = P(τ2 ≤ 2k)− P(τ2 ≤ 2k − 2).

P(τ2 ≤ 2k) = P(M2k = 2) + P(M2k ≥ 4) + P(τ2 ≤ 2k,M2k ≤ 0)

= P(M2k = 2) + 2P(M2k ≥ 4)

= P(M2k = 2) + P(M2k ≥ 4) + P(M2k ≤ −4)

= 1− P(M2k = −2)− P(M2k = 0).

Similarly, P(τ2 ≤ 2k − 2) = 1− P(M2k−2 = −2)− P(M2k−2 = 0). So

P(τ2 = 2k) = P(M2k−2 = −2) + P(M2k−2 = 0)− P(M2k = −2)− P(M2k = 0)

=

(1

2

)2k−2 [(2k − 2)!

k!(k − 2)!+

(2k − 2)!

(k − 1)!(k − 1)!

]−(1

2

)2k [(2k)!

(k + 1)!(k − 1)!+

(2k)!

k!k!

]=

(2k)!

4k(k + 1)!k!

[4

2k(2k − 1)(k + 1)k(k − 1) +

4

2k(2k − 1)(k + 1)k2 − k − (k + 1)

]=

(2k)!

4k(k + 1)!k!

[2(k2 − 1)

2k − 1+

2(k2 + k)

2k − 1− 4k2 − 1

2k − 1

]=

(2k)!

4k(k + 1)!k!.

I Exercise 5.5 (Joint distribution of random walk and maximum-to-date). Let Mn be a symmetricrandom walk, and define its maximum-to-date process

M∗n = max

1≤k≤nMk. (5.7.2)

Let n and m be even positive integers, and let b be an even integer less than or equal to m. Assume m ≤ nand 2m− b ≤ n.(i) Use an argument based on reflected paths to show that

PM∗n ≥ m,Mn = b = PMn = 2m− b

=n!(

n−b2 +m

)!(n+b2 −m

)!

(1

2

)n

.

Proof. For every path that crosses level m by time n and resides at b at time n, there corresponds a reflectedpath that resides at time 2m− b. So

P(M∗n ≥ m, Mn = b) = P(Mn = 2m− b) =

(1

2

)nn!(

m+ n−b2

)!(n+b2 −m

)!.

(ii) If the random walk is asymmetric with probability p for an up step and probability q = 1− p for a downstep, where 0 < p < 1, what is PM∗

n ≥ m,Mn = b?

Solution. The argument is similar to (i), but we need to determine the powers of p and q from the followingequations

xp + xq = n

xp − xq = 2m− b.

Solving it gives us xp = m+ n−b

2

xq = n+b2 −m.

41

Therefore

P(M∗n ≥ m, Mn = b) = P(Mn = 2m− b) =

n!(m+ n−b

2

)!(n+b2 −m

)!pm+n−b

2 qn+b2 −m.

I Exercise 5.6. The value of the perpetual American put in Section 5.4 is the limit as n → ∞ of thevalue of an American put with the same strike price 4 that expires at time n. When the initial stock priceis S0 = 4, the value of the perpetual American put is 1 (see (5.4.6) with j = 2). Show that the value of anAmerican put in the same model when the initial stock price is S0 = 4 is 0.80 if the put expires at time 1,0.928 if the put expires at time 3, and 0.96896 if the put expires at time 5.

Proof. Instead of numerical examples, we give a rigorous proof.On the infinite coin-toss space, we define Mn = stopping times that takes values 0, 1, · · · , n,∞ and

M∞ = stopping times that takes values 0, 1, 2, · · · . Then the time-zero value V ∗ of the perpetual Ameri-can put as in Section 5.4 can be defined as

V ∗ = supτ∈M∞

E[1τ<∞

(K − Sτ )+

(1 + r)τ

].

For an American put with the same strike price K that expires at time n, its time-zero value V (n) is

V (n) = maxτ∈Mn

E[1τ<∞

(K − Sτ )+

(1 + r)τ

].

Clearly (V (n))n≥0 is nondecreasing and V (n) ≤ V ∗ for every n. So limn V(n) exists and limn V

(n) ≤ V ∗.

For any given τ ∈ M∞, we define τ (n) =

∞, if τ = ∞τ ∧ n, if τ <∞ , then τ (n) is also a stopping time, τ (n) ∈ Mn

and limn→∞ τ (n) = τ . By bounded convergence theorem,

E[1τ<∞

(K − Sτ )+

(1 + r)τ

]= lim

n→∞E[1τ(n)<∞

(K − Sτ(n))+

(1 + r)τ(n)

]≤ lim

n→∞V (n).

Take sup at the left hand side of the inequality, we get V ∗ ≤ limn→∞ V (n). Therefore V ∗ = limn V(n).

I Exercise 5.7 (Hedging a short position in the perpetual American put). Suppose you have soldthe perpetual American put of Section 5.4 and are hedging the short position in this put. Suppose that atthe current time the stock price is s and the value of your hedging portfolio is v(s). Your hedge is to firstconsume the amount8

c(s) = v(s)− 4

5

[1

2v(2s) +

1

2v(s2

)](5.7.3)

and then take a position

δ(s) =v(2s)− v

(s2

)2s− s

2

(5.7.4)

in the stock. (See Theorem 4.2.2 of Chapter 4. The processes Cn and ∆n in that theorem are obtained byreplacing the dummy variable s by the stock price Sn in (5.7.3) and (5.7.4); i.e. Cn = c(Sn) and ∆n = δ(Sn).)If you hedge this way, then regardless of whether the stock goes up or down on the next step, the value ofyour hedging portfolio should agree with the value of the perpetual American put.(i) Compute c(s) when s = 2j for the three cases j ≤ 0, j = 1, and j ≥ 2.

8The text missed a factor of 12

in fron of v(s2

).

42

Solution. By (5.4.6),

v(2j) =

4− 2j if j ≤ 142j if j ≥ 1.

For j ≤ 0,

c(2j) = v(2j)− 4

5

[1

2v(2j+1) +

1

2v(2j−1)

]= (4− 2j)− 2

5(4− 2j+1 + 4− 2j−1) =

4

5.

For j = 1,c(2) = v(2)− 4

5

[1

2v(4) +

1

2v(1)

]= 2− 2

5(1 + 4− 1) =

2

5.

For j ≥ 2,c(2j) = v(2j)− 2

5

[v(2j+1) + v(2j−1)

]=

4

2j− 2

5

(4

2j+1+

4

2j−1

)= 0.

(ii) Compute δ(s) when s = 2j for the three cases j ≤ 0, j = 1, and j ≥ 2.

Solution. For j ≤ 0,

δ(2j) =v(2j+1)− v(2j−1)

2j+1 − 2j−1=

(4− 2j+1)− (4− 2j−1)

2j+1 − 2j−1= −1.

For j = 1,δ(2) =

v(4)− v(1)

4− 1=

1− (4− 1)

4− 1= −2

3.

For j ≥ 2,

δ(2j) =v(2j+1)− v(2j−1)

2j+1 − 2j−1=

42j+1 − 4

2j−1

2j+1 − 2j−1= − 1

4j−1.

(iii) Verify in each of the three cases s = 2j for j ≤ 0, j = 1, and j ≥ 2 that the hedge works (i.e., regardlessof whether the stock goes up or down, the value of your hedging portfolio at the next time is equal to thevalue of the perpetual American put at that time).

Proof. The computation is too tedious; skipped for this version.

I Exercise 5.8 (Perpetual American call). Like the perpetual American put of Section 5.4, the per-petual American call has no expiration. Consider a binomial model with up factor u, down factor d, andinterest rate r that satisfies the no-arbitrage condition 0 < d < 1 + r < u. The risk-neutral probabilities are

p =1 + r − d

u− d, q =

u− 1− r

u− d.

The intrinsic value of the perpetual American call is g(s) = s − K, where K > 0 is the strike price. Thepurpose of this exercise is to show that the value of the call is always the price of the underlying stock, andthere is no optimal exercise time.(i) Let v(s) = s. Show that v(Sn) is always at least as large as the intrinsic value g(Sn) of the calland

(1

1+r

)nv(Sn) is a supermartingale under the risk-neutral probabilities. In fact,

(1

1+r

)nv(Sn) is a

martingale.9 These are the analogues of properties (i) and (ii) for the perpetual American put of Section5.4.

9The textbook said “supermartingale” by mistake.

43

Proof. v(Sn) = Sn ≥ Sn−K = g(Sn). Under risk-neutral probabilities, 1(1+r)n v(Sn) =

Sn

(1+r)n is a martingaleby Theorem 2.4.4.

(ii) to show that v(s) = s is not too large to be the value of the perpetual American call, we must find agood policy for the purchaser of the call. Show that if the purchaser of the call exercises at time n, regardlessof the stock price at that time, then the discounted risk-neutral expectation of her payoff is S0 − K

(1+r)n .Because this is true for every n, and

limn→∞

[S0 −

K

(1 + r)n

]= S0,

the value of the call at time zero must be at least S0. (The same is true at all other times; the value of thecall is at least as great as the current stock price.)

Proof. If the purchaser chooses to exercises the call at time n, then the discounted risk-neutral expectationof her payoff is E

[Sn−K(1+r)n

]= S0 − K

(1+r)n . Since limn→∞

[S0 − K

(1+r)n

]= S0, the value of the call at time

zero is at least supn

[S0 − K

(1+r)n

]= S0.

(iii) In place of (i) and (ii) above, we could verify that v(s) = s is the value of the perpetual Americancall by checking that this function satisfies the equation (5.4.16) and boundary conditions (5.4.18). Do thisverification.

Proof. maxg(s), pv(us)+qv(ds)

1+r

= maxs−K, pu+qv

1+r s = maxs−K, s = s = v(s), so equation (5.4.16) issatisfied. Clearly v(s) = s also satisfies the boundary condition (5.4.18).

(iv) Show that there is no optimal time to exercise the perpetual American call.

Proof. Suppose τ is an optimal exercise time, then according to result in (ii),

E[Sτ −K

(1 + r)τ1τ<∞

]≥ sup

n

[S0 −

K

(1 + r)n

]= S0,

which implies P(τ < ∞) = 0 and hence E[

K(1+r)τ 1τ<∞

]> 0. So E

[Sτ−K(1+r)τ 1τ<∞

]< E

[Sτ

(1+r)τ 1τ<∞

].

Since(

Sn

(1+r)n

)n≥0

is a martingale under the risk-neutral measure, by Fatou’s lemma,

E[

Sτ

(1 + r)τ1τ<∞

]≤ lim inf

n→∞E[

Sτ∧n

(1 + r)τ∧n1τ<∞

]≤ lim inf

n→∞E[

Sτ∧n

(1 + r)τ∧n

]= lim inf

n→∞E[S0] = S0.

Combined, we haveS0 ≤ E

[Sτ −K

(1 + r)τ1τ<∞

]< E

[Sτ

(1 + r)τ1τ<∞

]≤ S0.

Contradiction. So there is no optimal time to exercise the perpetual American call. Simultaneously, we haveshown E

[Sτ−K(1+r)τ 1τ<∞

]< S0 for any stopping time τ . Combined with (ii), we conclude S0 is the least

upper bound for all the prices acceptable to the buyer.

I Exercise 5.9. (Provided by Irene Villegas.) Here is a method for solving equation (5.4.13) for the valueof the perpetual American put in Section 5.4.(i) We first determine v(s) for large values of s. When s is large, it is not optimal to exercise the put, so themaximum in (5.4.13) will be given by the second term

4

5

[1

2v(2s) +

1

2v(s2

)]=

2

5v(2s) +

2

5v(s2

).

44

We thus seek solutions to the equation

v(s) =2

5v(2s) +

2

5v(s2

). (5.7.5)

All such solutions are of the form sp for some constant p or linear combination of functions of this form.Substitute sp into (5.7.5), obtain a quadratic equation for 2p, and solve to obtain 2p = 2 or 2p = 1

2 . Thisleads to the values p = 1 and p = −1, i.e., v1(s) = s and v2(s) =

12 are soltuions to (5.7.5).

Proof. Suppose v(s) = sp, then we have sp = 252

psp + 25sp

2p . So 1 = 2p+1

5 + 21−p

5 . Solve it for p, we get p = 1or p = −1.

(ii) The general solution to (5.7.5) is a linear combination of v1(s) and v2(s), i.e.,

v(s) = As+B

s. (5.7.6)

For large values of s, the value of the perpetual American put must be given by (5.7.6). It remains toevaluate A and B. Using the second boundary condition in (5.4.15), show that A must be zero.

Proof. Since lims→∞ v(s) = lims→∞(As+ Bs ) = 0, we must have A = 0.

(iii) We have thus established that for large values of s, v(s) = Bs for some constant B still to be determined.

For small values of s, the value of the put is its intrinsic value 4−s. We must choose B so these two functionscoincide at some point, i.e., we must find a value for B so that, for some s > 0,

fB(s) =B

s− (4− s)

equals zero. Show that, when B > 4, this function does not take the value 0 for any s > 0, but, when B ≤ 4,the equation fB(s) = 0 has a solution.

Proof. fB(s) = 0 if and only if B+s2−4s = 0. The discriminant ∆ = (−4)2−4B = 4(4−B). So for B ≤ 4,the equation has roots and for B > 4, this equation does not have roots.

(iv) Let B be less than or equal to 4, and let sB be a solution of the equation fB(s) = 0. Suppose sB isa stock price that can be attained in the model (i.e., sB = 2j for some integer j). Suppose further thatthe owner of the perpetual American put exercises the first time the stock price is sB or smaller. Then thediscounted risk-neutral expected payoff of the put is vB(S0), where vB(s) is given by the formula

vB(s) =

4− s, if s ≤ sB ,Bs , if s ≥ sB .

(5.7.7)

Which values of B and sB give the owner the largest option value?

Proof. Suppose B ≤ 4, then the equation s2 − 4s + B = 0 has solution 2 ±√4−B. By drawing graphs of

4− s and Bs , we can see the largest vB(s) is obtained by making 4− s and B

s tangent to each other, whichimplies

∆ = 4(4−B) = 0

4− sB = BsB.

Solving it gives us B = 4 and sB = 2.

(v) For s < sB , the derivative of vB(s) is v′B(s) = −1. For s > sB , this derivative is v′B(s) = − Bs2 . Show

that the best value of B for the option owner makes the derivative of vB(s) continuous at s = SB (i.e., thetwo formulas for v′B(s) give the same answer at s = sB).

Proof. To have continuous derivative, we must have −1 = − Bs2B

. Plug B = s2B back into 4 − sB = BsB

, weget sB = 2. This gives B = 4.

45

6 Interest-Rate-Dependent Assets⋆ Comments:1) In previous chapters, we started with a real probability measure and derived the risk-neutral measure

based on the no-arbitrage argument. The concrete form of the risk-neutral measure is known in that setting.In the current chapter, models are built under the risk-neutral measure, whose existence is assumed a prioriand whose concrete form (in terms of the real probability measure) is unknown. The justification for this isthe following result by Dalang et al. [2]:

Theorem 2 (Dalang-Morton-Willinger, Fundamental Theorem of Asset Pricing). Let (Ω,F , Ftt=0,··· ,T ,P)be a (general) filtered probability space and let S = Stt∈0,··· ,T be an adapted, Rd-valued stochastic processdescribing the discounted prices of d ∈ N financial assets. Then the following properties are equivalent:(a) The financial market model is free of arbitrage.(b) There exists a probability measure P∗ on (Ω,F ,P) such that

• P∗ is equivalent to P and the Radon-Nikodým density ϱ := dP∗/dP is bounded, i.e. in L∞(Ω,FT ,P),• Integrability: St ∈ L1(Ω,Ft,P∗) for all t ∈ 0, · · · , T,• Martingale property w.r.t. P:

EP∗ [St|Ft−1]a.s.= St−1 for all t ∈ 0, · · · , T.

According to the Fundamental Theorem of Asset Pricing (FTAP), the no-arbitrage property is associatedwith the existence of a probability measure called equivalent martingale measure (EMM) and a positiveprocess called numéraire, such that the price processes of tradable assets discounted by the numéraire aremartingales under the equivalent martingale measure. In the current chapter, the numéraire is the valueprocess of the money market account

Mn = (1 +R0) · · · (1 +Rn−1), n = 1, 2, · · · , N ;M0 = 1.

And the risk-neutral measure is the EMM associated with this numéraire. This justifies Definition 6.2.4(zero-coupon bond prices) since

DnBn,m = En [DmBm,m] ⇒ Bn,m = En

[Dm · 1Dn

].

2) The wealth equation (6.2.6) essentially assumes all coupons and other payouts of cash are reinvestedin the portfolio.

3) To see the intuition of the forward interest rate at time n for investing at time m (Definition 6.3.4),note at time n, we can invest $1 to buy 1

Bn,mshares of zero-coupon bond maturing at time m; at time m,

we reinvest the payoff 1Bn,m

at rate Fn,m to get paid at time (m+ 1). To eliminate arbitrage, this investingstrategy should have the same return as investing $1 at time n in the zero-coupon bond maturing at time(m+ 1):

1

Bn,m(1 + Fn,m) =

1

Bn,m+1.

This gives formula (6.3.3)Fn,m =

Bn,m

Bn,m+1− 1 =

Bn,m −Bn,m+1

Bn,m+1.

4) Whenever there is a hedge, static or not, the discounted value of the hedging portfolio is a martingaleunder the risk-neutral measure, and the risk-neutral pricing formula applies. This is regardless of whetheror not the market is complete.

i) Forward contract. a) Hedging of a short position: at time n, short Sn

Bn,mzero-coupon bond maturing

at time m and long 1 share of asset at price of Sn, hold the portfolio until maturity time m. b) Risk-neutralpricing:

En [Dm(K − Sm)] = KDnBn,m −DnSn = DnVn,

46

where Vn is the contract value at time n. By setting Vn = 0, we obtain m-forward price of the asset at timen (Theorem 6.3.2).

ii) Forward rate agreement (FRA).10 a) Hedging of a short position: the portfolio whose value is Rm attime (m + 1) can be replicated by a cash inflow of $1 at time m and a cash outflow of $1 at time (m + 1),since the inflow of $1 at time m can be invested in the money market account to return (1 + Rm) at time(m+1). The cash inflow of $1 at time m can be obtained by longing at time n a zero-coupon bond maturingat time m, while the cash outflow of $1 at time (m+1) can be obtained by shorting at time n a zero-couponbond maturing at time (m+ 1). So the value of the portfolio at time n is

Vn = Bn,m −Bn,m+1.

b) Risk-neutral pricing (Exercise 6.3):

Vn = En

[Dm+1Rm

Dn

]= En

[Dm −Dm+1

Dn

]= Bn,m −Bn,m+1.

This is Theorem 6.3.5.iii) Interest rate swap. A long swap position (“receive fixed”) is a long position in a coupon bond plus a

short position in a series of FRA’s. Therefore, an interest rate swap can be hedged and risk-neutral pricingformula applies. This is Theorem 6.3.7. More specifically,

Swapm = Km∑

n=1

+B0,m − 1

suggests a long position in a bond with coupon K and maturity m and a short position of $1 in the moneymarket account. A short position of $1 in the money market account at the beginning of each period willresult in −(1 +Rn) at the end of the period.

iv) Interest rate caps and floors. Suppose the replication uses the money market account and zero-couponbonds with various maturities. The wealth equation satisfies

Xn+1 = (1 +Rn)(Xn −∆nBn,m) + ∆nBn+1,m.

To successfully replicate Vn+1, it’s necessary and sufficient that Bn+1,m(H) = Bn+1,m(T ). See Exercise 6.4for illustration.

5) According to the Fundamental Theorem of Asset Pricing (FTAP), the no-arbitrage property is associ-ated with the existence of a probability measure called equivalent martingale measure (EMM) and a positiveprocess called numéraire, such that the price processes of tradable assets discounted by the numéraire aremartingales under the given probability measure (hence the name martingale measure).

The m-forward measure is the EMM associated with the zero-coupon bond (Bn,m)mn=0 as the numéraire.For a European contingent claim with maturity m, we have the risk-neutral pricing formula (note Bm,m ≡ 1)

Vn = Bn,mEmn [Vm] =

1

DnEn[DmVm].

If we denote the Radon-Nikodým derivative process by (Zn,m)mn=0, that is,

Zm,m :=dPm

dP, Zn,m := En[Zm,m],

by setting n = 0 in the risk-neutral pricing formula, we have Zm,m = Dm

B0,mand hence

Zn,m =En[Dm]

B0,m=DnBn,m

B0,m, n = 0, · · · ,m.

10A FRA is a contract involving three time instants: the current time t, the expiry time T > t, and the maturity time S > T .The contract gives its holder an interest-rate payment for the period between T and S. See Brigo and Mercurio [1, page 11] fora general definition.

47

This gives an alternative presentation of the m-forward measure Pm. A notable feature of working underthis measure is that any asset price process discounted by the zero-coupon bond (Bn,m)mn=0 is exactly itsm-forward price process (Theorem 6.3.2).

I Exercise 6.1. Prove Theorem 2.3.2 when conditional expectation is defined by Definition 6.2.2.

Proof. The proof is tedious and is not a lot simpler than the one using the most general definition ofconditional expectation (see, for example, Shiryaev [5]); skipped for this version.

I Exercise 6.2. Verify that the discounted value of the static hedging portfolio constructed in the proof ofTheorem 6.3.2 is a martingale under P.

Proof. The static hedging portfolio consists of shorting one forward contract, shorting Sn

Bn,mzero-coupon

bonds maturing at time m, and longing one share of the asset. So for any k = n, · · · ,m, the value of theportfolio at time k is

Xk = Sk − Ek[Dm(Sm −K)]D−1k − Sn

Bn,mBk,m.

Then

Ek−1[DkXk] = Ek−1

[DkSk − Ek[Dm(Sm −K)]− Sn

Bn,mBk,mDk

]= Dk−1Sk−1 − Ek−1[Dm(Sm −K)]− Sn

Bn,mEk−1

[Ek[Dm · 1]

]= Dk−1

Sk−1 − Ek−1[Dm(Sm −K)]D−1

k−1 −Sn

Bn,mBk−1,m

= Dk−1Xk−1.

This shows the discounted value of the static hedging portfolio constructed in the proof of Theorem 6.3.2 isa martingale under P.

I Exercise 6.3. Let 0 ≤ n ≤ m ≤ N − 1 be given. According to the risk-neutral pricing formula, thecontract that pays Rm at time m + 1 has time-n price 1

DnEn[Dm+1Rm]. Use the properties of conditional

expectations to show that this gives the same result as Theorem 6.3.5, i.e.,

1

DnEn[Dm+1Rm] = Bn,m −Bn,m+1.

Proof.

1

DnEn[Dm+1Rm] =

1

DnEn[Dm(1 +Rm)−1Rm] =

1

DnEn [Dm −Dm+1] = Bn,m −Bn,m+1.

I Exercise 6.4 (Short position hedge for caplets). Using the data in Example 6.3.9, this exerciseconstructs a hedge for a short position in the caplet paying

(R2 − 1

3

)+ at time three. We observe from thesecond table in Example 6.3.9 that the payoff at time three of this caplet is

V3(HH) =2

3, V3(HT ) = V3(TH) = V3(TT ) = 0.

Since this payoff depends on only the first two coin tosses, the price of the caplet at time two can bedetermined by discounting:

V2(HH) =1

1 +R2(HH)V3(HH) =

1

3, V2(HT ) = V2(TH) = V2(TT ) = 0.

48

Indeed, if one is hedging a short position in the caplet and has a portfolio valued at 13 at time two in the

event ω1 = H, ω2 = H, then one can simply invest this 13 in the money market in order to have the 2

3required to pay off the caplet at time three.

In Example 6.3.9, the time-zero price of the caplet is determined to be 221 (see (6.3.10)).

(i) Determine V1(H) and V1(T ), the price at time one of the caplet in the events ω1 = H and ω1 = T ,respectively.

Solution. D1V1 = E1[D2V2] = D2E1[V2]. So V1 = D2

D1E1[V2] =

11+R1

E1[V2]. In particular,

V1(H) =V2(HH)P (ω2 = H|ω1 = H) + V2(HT )P (ω2 = T |ω1 = H)

1 +R1(H)=

6

7

(1

3· 23+ 0 · 1

3

)=

4

21,

V1(T ) =V2(TH)P (ω2 = H|ω1 = T ) + V2(TT )P (ω2 = T |ω1 = T )

1 +R1(T )= 0.

(ii) Show how to begin with 221 at time zero and invest in the money market and the maturity two bond in

order to have a portfolio value X1 at time one that agrees with V1, regardless of the outcome of the first cointoss. Why do we invest in the maturity two bond rather than the maturity three bond to do this?

Proof. Let X0 = 221 . Suppose we buy ∆0 shares of the maturity two bond, then at time one, the value of

our portfolio is X1 = (1 +R0)(X0 −∆0B0,2) + ∆0B1,2. To replicate the value V1, we must haveV1(H) = (1 +R0)(X0 −∆0B0,2) + ∆0B1,2(H)V1(T ) = (1 +R0)(X0 −∆0B0,2) + ∆0B1,2(T ).

So∆0 =

V1(H)− V1(T )

B1,2(H)−B1,2(T )=

421 − 067 − 5

7

=4

3.

The hedging strategy is therefore to borrow 43B0,2 − 2

21 = 2021 and buy 4

3 share of the maturity two bond.The reason why we do not invest in the maturity three bond is that B1,3(H) = B1,3(T )(=

47 ) and the

portfolio will therefore have the same value at time one regardless the outcome of first coin toss. This makesimpossible the replication of V1, since V1(H) = V1(T ).

(iii) Show how to take the portfolio value X1 at time one and invest in the money market and the maturitythree bond in order to have a portfolio value X2 at time two that agrees with V2, regardless of the outcomeof the first two coin tosses. Why do we invest in the maturity three bond rather than the maturity two bondto do this?

Proof. Suppose we buy ∆1 share of the maturity three bond at time one, then to replicate V2 at time two,we must have V2 = (1 +R1)(X1 −∆1B1,3) + ∆1B2,3. So

∆1(H) =V2(HH)− V2(HT )

B2,3(HH)−B2,3(HT )= −2

3, ∆1(T ) =

V2(TH)− V2(TT )

B2,3(TH)−B2,3(TT )= 0.

The hedging strategy is as follows. If the outcome of first coin toss is T , then we keep everything in themoney market account. If the outcome of first coin toss is H, we short 2

3 shares of the maturity three bondand invest the income into the money market account. We do not invest in the maturity two bond, becauseat time two, the value of the bond is its face value and our portfolio will therefore have the same valueregardless outcomes of coin tosses. This makes impossible the replication of V2.

I Exercise 6.5. Let m be given with 0 ≤ m ≤ N − 1, and consider the forward interest rate

Fn,m =Bn,m −Bn,m+1

Bn,m+1, n = 0, 1, · · · ,m.

(i) Use (6.4.8) and (6.2.5) to show that Fn,m, n = 0, 1, · · · ,m, is a martingale under the (m + 1)-forwardmeasure Pm+1.

49

Proof. Before we start the computation, note Fn,m =Bn,m−Bn,m+1

Bn,m+1is a tradable asset discounted by

(Bn,m+1)n, so it must be a martingale under the equivalent martingale measure Pm+1 associated withthe numéraire (Bn,m+1)n. To verify this, suppose 1 ≤ n ≤ m, then

Em+1n−1 [Fn,m] = En−1[B

−1n,m+1(Bn,m −Bn,m+1)Zn,m+1Z

−1n−1,m+1]

= En−1

[(Bn,m

Bn,m+1− 1

)Bn,m+1Dn

Bn−1,m+1Dn−1

]=

Dn

Bn−1,m+1Dn−1En−1[Bn,m −Bn,m+1]

=Dn

Bn−1,m+1Dn−1En−1[D

−1n En[Dm]−D−1

n En[Dm+1]]

=En−1[Dm −Dm+1]

Bn−1,m+1Dn−1

=Bn−1,m −Bn−1,m+1

Bn−1,m+1

= Fn−1,m.

(ii) Compute F0,2, F1,2(H), and F1,2(T ) in Example 6.4.4 and verify the martingale property

E3[F1,2] = F0,2.

Solution.

F0,2 =B0,2

B0,3− 1 =

0.9071

0.8639− 1 = 0.05

F1,2(H) =B1,2(H)

B1,3(H)− 1 =

0.9479

0.8985− 1 = 0.055

F1,2(T ) =B1,2(T )

B1,3(T )− 1 =

0.9569

0.9158− 1 = 0.045

Therefore

E3[F1,2] = P3(H)F1,2(H) + P3(T )F1,2(T ) = 0.4952 · 0.055 + 0.5048 · 0.045 = 0.05 = F0,2.

I Exercise 6.6. Let Sm be the price at time m of an asset in a binomial interest rate model. Forn = 0, 1, · · · ,m, the forward price is Forn,m = Sn

Bn,mand the futures price is Futn,m = En[Sm].11

(i) Suppose that at each time n an agent takes a long forward position and sells this contract at time n+ 1.Show that this generates cash flow Sn+1 − SnBn+1,m

Bn,mat time n+ 1.

Proof. At time n+ 1, the forward contract has a value of

1

Dn+1En+1 [Dm(Sm − Futn,m)] = Sn+1 − Futn,mBn+1,m = Sn+1 −

SnBn+1,m

Bn,m.

So if the agent takes a long forward position at time n for no cost and sells this contract at time n + 1, hewill receive a cash flow of Sn+1 − SnBn+1,m

Bn,m.

11The textbook said “Sn” by mistake.

50

(ii) Show that if the interest rate is a constant r and at each time n an agent takes a long position of(1 + r)m−n−1 forward contracts, selling these contracts at time n + 1, then the resulting cash flow is thesame as the difference in the futures price Futn+1,m − Futn,m.

Proof. By (i), the cash flow generated at time n+ 1 is

(1 + r)m−n−1

(Sn+1 −

SnBn+1,m

Bn,m

)= (1 + r)m−n−1

(Sn+1 −

Sn

(1+r)m−n−1

1(1+r)m−n

)

= (1 + r)mSn+1

(1 + r)n+1− (1 + r)m

Sn

(1 + r)n

= (1 + r)mEn+1

[Sm

(1 + r)m

]− (1 + r)mEn

[Sm

(1 + r)m

]= Futn+1,m − Futn,m.

I Exercise 6.7. Consider a binomial interest rate model in which the interest rate at time n depends ononly the number of heads in the first n coin tosses. In other words, for each n there is a function rn(k) suchthat

Rn(ω1, · · · , ωn) = rn(#H(ω1, · · · , ωn)).

Assume the risk-neutral probabilities are p = q = 12 . The Ho-Lee model (Example 6.4.4) and the Black-

Derman-Toy model (Example 6.5.5) satisfy these conditions.Consider a derivative security that pays 1 at time n if and only if there are k heads in the first n tosses;

i.e., the payoff is Vn(k) = 1#H(ω1,··· ,ωn)=k. Define ψ0(0) = 1 and , for n = 1, 2, · · · , define

ψn(k) = E [DnVn(k)] , k = 0, 1, · · · , n,

to be the price of this security at time zero. Show that the functions ψn(k) can be computed by the recursion

ψn+1(0) =ψn(0)

2(1 + rn(0))

ψn+1(k) =ψn(k − 1)

2(1 + rn(k − 1))+

ψn(k)

2(1 + rn(k)), k = 1, · · · , n,

ψn+1(n+ 1) =ψn(n)

2(1 + rn(n)).

Proof.

ψn+1(0) = E [Dn+1Vn+1(0)]

= E[

Dn

1 + rn(0)1#H(ω1···ωn+1)=0

]= E

[Dn

1 + rn(0)1#H(ω1···ωn)=01ωn+1=T

]=

1

2(1 + rn(0))E[Dn1#H(ω1···ωn)=0

]=

ψn(0)

2(1 + rn(0)).

51

For k = 1, 2, · · · , n,

ψn+1(k) = E[

Dn

1 + rn(#H(ω1 · · ·ωn))1#H(ω1···ωn+1)=k

]= E

[Dn

1 + rn(k)1#H(ω1···ωn)=k1ωn+1=T

]+ E

[Dn

1 + rn(k − 1)1#H(ω1···ωn)=k−11ωn+1=H

]=

1

2

E[DnVn(k)]

1 + rn(k)+

1

2

E[DnVn(k − 1)]

1 + rn(k − 1)

=ψn(k)

2(1 + rn(k))+

ψn(k − 1)

2(1 + rn(k − 1)).

Finally,

ψn+1(n+ 1) = E[Dn+1Vn+1(n+ 1)] = E[

Dn

1 + rn(n)1#H(ω1···ωn)=n1ωn+1=H

]=

ψn(n)

2(1 + rn(n)).

Remark 6.1. In the above proof, we have used the independence of ωn+1 and (ω1, · · · , ωn) under P. This isguaranteed by the assumption that P(ωn+1 = H|ω1, · · · , ωn) = p and P(ωn+1 = T |ω1, · · · , ωn) = q, which aredeterministic. In case the binomial model has stochastic up- and down-factor un and dn, we have P(ωn+1 =

H|ω1, · · · , ωn) = pn and P(ωn+1 = T |ω1, · · · , ωn) = qn, where pn = 1+rn−dn

un−dnand qn = un−1−rn

un−dn. Then for

any X ∈ Fn = σ(ω1, · · · , ωn), we have E[Xf(ωn+1)] = E[XE[f(ωn+1)|Fn]] = E[X(pnf(H) + qnf(T ))].

References[1] D. Brigo and F. Mercurio. Interest rate models - Theory and practice: With smile, inflaiton and credit,

2nd Edition. Springer, 2007. 47

[2] R. C. Dalang, A. Morton, and W. Willinger. “Equivalent martingale measures and no-arbitrage instochastic securities market models”. Stochastics and Stochastics Reports 29, pp. 185-202, (1989). 46

[3] F. Delbaen and W. Schachermayer. The mathematics of arbitrage. Springer, 2006. 2

[4] U. Schmock. “Existence of an equivalent martingale measure in the Dalang-Morton-Willinger Theorem,which preserves the dependence structure”. 6th AMaMeF and Banach Center Conference, Warsaw, 2013.http://bcc.impan.pl/6AMaMeF/uploads/presentations/Folien_AMaMeF_2013.pdf. 10

[5] A. N. Shiryaev. Probability, 2nd Edition. Springer, 1995. 9, 48

[6] S. Shreve. Stochastic calculus for finance I: The binomial asset pricing model. Springer-Verlag, NewYork, 2004. 1

[7] S. Shreve. Stochastic calculus for finance II: Continuous-time models. Springer-Verlag, New York, 2004.3, 10, 16

[8] P. Wilmott. The mathematics of financial derivatives: A student introduction. Cambridge UniversityPress, 1995. 6

[9] Y. Zeng. “Fundamental Theorem of Asset Pricing in a nutshell: With a view toward numéraire change”,unpublished survey, 2009. http://opensrcsolutions.webs.com/.

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