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Stochastic Calculus for Finance I: TheBinomial Asset Pricing ModelSolution of Exercise ProblemsYan ZengVersion 1.1, last revised on 2014-10-26AbstractThis is a solution manual for Shreve [6]. If you find any typos/errors or have any comments, pleaseemail me at [email protected] The Binomial No-Arbitrage Pricing Model22 Probability Theory on Coin Toss Space93 State Prices214 American Derivative Securities305 Random Walk376 Interest-Rate-Dependent Assets461
1The 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 0 S0 , 0 S0 ). At the operational level, we borrow X0 , buy 0 shares of stock, and invest the residualamount (X0 0 S0 ) 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 0 S0 ) 0 S1 (S1 K) .Plug in S1 (H) uS0 and S1 (T ) dS0 , we obtain a system of two linear equations for two unknowns (X0and 0 ) and it has a unique solution as long as u ̸ d. This is how we obtain the magic number X0 1.20and 0 12 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 0 S0 , 0 S0 ). At the operational level, we short sell 0 shares of stock, invest the income 0 S0into 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 0 S0 , 0 S0 ) 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 0 S1 , 0 S1 ).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 (ω1 ω2 · · · 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 ω1 ω2 · · · 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 1XnSn 1Sn n(1 r)n 1(1 r)n(1 r)n 1(1 r)nThis leads to a representation by discrete stochastic integral:eT X0 ( · S)e T,Xen where XXn(1 r)nand Sen 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 andX1 0 S1 (1 r)(X0 0 S0 ),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()S1 S0X1 0 S0 r (1 r)X0 .S0ThenX1 (H) (1 r)X0 0 S0 [u (1 r)]andX1 (T ) (1 r)X0 0 S0 [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. SoX1 (H) (1 r)X0 0 0 X1 (T ) (1 r)X0andX1 (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 0 S1 (1 r)(X0 0 S0 ), the first term 0 S1 is the value of thestock position at time 1 while the second term (1 r)(X0 0 S0 ) 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.00Remark 1.4. Note the condition d 1 r u is just S1 (TS) S r S1 (H) S: the return of the stockS00investment 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 is5X1 0 S1 Γ0 (S1 5) (4 0 1.20Γ0 ).4Assume 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 54 (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 becomesX1 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 suchthat(1 r)(C0 δS0 ) δS1 V1 .Plug this into the expression of X1 , we haveX1 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)[]S1 (α βδ)S0 (1 r) .S0Since 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)?[]()S011 r du 1 r1 r du 1 rSolution. V0 1 ru d S1 (H) u d S1 (T ) 1 ru d u 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 thatXn 1 (ω1 ω2 · · · ωn T ) Vn 1 (ω1 ω2 · · · ωn T ).4
Proof.Xn 1 (T ) n dSn (1 r)(Xn n Sn ) n Sn (d 1 r) (1 r)VnVn 1 (H) Vn 1 (T )p̃Vn 1 (H) q̃Vn 1 (T ) (d 1 r) (1 r)u d1 r p̃[Vn 1 (T ) Vn 1 (H)] p̃Vn 1 (H) q̃Vn 1 (T ) p̃Vn 1 (T ) q̃Vn 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 portfolio2 (HT )valued at V1 (H) 2.24. Assume that she now takes a position of 1 (H) SV22 (HH) V(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 positionH) V3 (HT T )of 2 (HT ) SV33 (HT(HT H) S3 (HT T ) in the stock. Show that, at time three, if the stock goes up in the thirdperiod, she will have a portfolio valued at V3 (HT H) 0, whereas if the stock goes down, her portfolio willbe worth V3 (HT T ) 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 ofX1 (H) (1 r)(X0 0 S0 ) 0 S1 (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 )3.20 2.40 0.0667S2 (HH) S2 (HT )16 4in the stock, then on the path ω1 ω2 HH, the investor’s portfolio is worth ofX2 (HH) (1 r)[X1 (H) 1 (H)S1 (H)] 1 (H)S2 (HH) (1 0.25)(2.24 0.0667 · 8) 0.0667 · 16 3.2V2 (HH),while on the path ω1 ω2 HT , the investor’s portfolio is worth ofX2 (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 ω1 ω2 HT , the agent takes a position 2 (HT ) V3 (HT H) V3 (HT T )0 6 1S3 (HT H) S3 (HT T )8 2in the stock, then on the path ω1 ω2 ω3 HT H, the investor’s portfolio is worth ofX3 (HT H) (1 r)[X2 (HT ) 2 (HT )S2 (HT )] 2 (HT )S3 (HT H) (1 0.25)[2.4 ( 1) · 4] ( 1) · 8 0 V3 (HT H),5
while on the path ω1 ω2 ω3 HT T , the investor’s portfolio is worth ofX3 (HT T ) (1 r)[X2 (HT ) 2 (HT )S2 (HT )] 2 (HT )S3 (HT T ) (1 0.25)[2.4 ( 1) · 4] ( 1) · 2 6 V3 (HT T ).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 have5· 1.20 1.504at 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 0 S0 ) 0 S1 (S1 K) .Then X0 1.20 and 0 21 since this equation is a linear equation of X0 and 0 . The solution meansthe 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( )35· 1.376 2.68754at 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 n S0 4, u 2,d 21 , and take the interest rate r 14 , so that p̃ q̃ 12 . For n 0, 1, 2, 3, define Yn k 0 Sk to be thesum of the stock prices between times zero and n. Consider an Asian call option that expires at time three() and has strike K 4 (i.e., whose payoff at time three is 41 Y3 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) 14 y 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) (s12[s )][p̃vn 1 (us, y us) q̃vn 1 (ds, y ds)] vn 1 (2s, y 2s) vn 1,y .1 r522(ii) Apply the algorithm developed in (i) to compute v0 (4, 4), the price of the Asian option at time zero.S2 (HH) 16Y2 (HH) 28v2 (16, 28) 6.4S1 (H) 8Y1 (H) 12v1 (8, 12) 2.96S2 (HT ) 4Y2 (HT ) 16v2 (4, 16) 1S3 (HHH) 32Y3 (HHH) 60v3 (32, 60) 11S3 (HHT ) 8Y3 (HHT ) 36v3 (8, 36) 5S3 (HT H) 8Y3 (HT H) 24v3 (8, 24) 2S3 (HT T ) 2Y3 (HT T ) 18v3 (2, 18) 0.5S0 4 Y0 4v0 (4, 4) 1.216S2 (T H) 4Y2 (T H) 10v2 (4, 10) 0.2S1 (T ) 2Y1 (T ) 6v1 (2, 6) 0.08S2 (T T ) 1Y2 (T T ) 7v2 (1, 7) 0Exercise 1.8. Asian option.Solution.1 Thetextbook said “Example 1.2.1” by mistake.7S3 (T HH) 8Y3 (T HH) 18v3 (8, 18) 0.5S3 (T HT ) 2Y3 (T HT ) 12v3 (2, 12) 0S3 (T T H) 2Y3 (T T H) 9v3 (2, 9) 0S3 (T T T ) 0.5Y3 (T T T ) 7.5v3 (0.5, 7.5) 0
(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)sI Exercise 1.9. (Stochastic volatility, random interest rate). Consider a binomial pricing model,but at each time n 1, the “up factor” un (ω1 ω2 · · · ωn ), the “down factor” dn (ω1 ω2 · · · ωn ), and the interestrate rn (ω1 ω2 · · · ωn ) are allowed to depend on n and on the first n coin tosses ω1 ω2 · · · 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{u0 S0 if ω1 H,S1 (ω1 ) d0 S0 if ω1 T ,and, for n 1, the stock price at time n 1 is given by{un (ω1 ω2 · · · ωn )Sn (ω1 ω2 · · · ωn ) if ωn 1 H,Sn 1 (ω1 ω2 · · · ωn ωn 1 ) dn (ω1 ω2 · · · ωn )Sn (ω1 ω2 · · · 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 (ω1 ω2 · · · ωn ) at time n 1. We assume that for each n and for allω1 ω2 · · · ωn , the no-arbitrage condition0 dn (ω1 ω2 · · · ωn ) 1 rn (ω1 ω2 · · · ωn ) un (ω1 ω2 · · · 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,n dnset p̃n 1 ren . Thenun dn and q̃n 1 pVn p̃n Vn 1 (H) q̃n Vn 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?(H)(T )Solution. un Sn 1 SnS 10 1 S10n and dn Sn 1 SnS 10 1 S10n . So the risk-neutralSnSnnnn 12 and q̃n 12 .probabilities at time n are p̃n u1 dn dnTo price the European call, we only need to focus on the nodes at time 5 of the binomial tree since p̃nand q̃n 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 have8
number of H’s345option payoff(80 30 20) 80 10(80 40 10) 80 30(80 50) 80 50risk-neutral probability of paths15!1025 · 3!2! 321525 · 5 321125 32So the price of the call at time zero is1510300· 50 · 30 · 10 9.375.323232322Probability 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, asSn 1Sn 1efar as (1 r)n 1 is replaced by (1 r )···(1 r ) . The requirement on the risk-neutral probability P is that0ne n 1 H ω1 , · · · , ωn ) : p̃n 1 rn dnP(wun dnande n 1 T ω1 , · · · , ωn ) : 1 p̃n q̃n .P(we on theResults in measure theory guarantee the existence and uniqueness of such a probability measure PN2sample space Ω {w : (w1 , · · · , wN )} for a given family of conditional probabilities (p̃n , q̃n )n 1 . With thissetup, we have[]e n Sn 1 un p̃n dn q̃n (1 rn ).ESnIn other words, the stock price process, when discounted by the random interest rates, is a martingale undere When rn ’s are deterministic, we are back to the case where ωn ’s are independent of each other. ThisP.unifies 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 sucha replicating portfolio is the wealth equation (2.4.16)Xn 1 n Sn 1 (1 r)(Xn Cn n Sn ),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 simultaneouslyprove formula (2.4.14) holds with Vn ’s replaced by Xn ’s prove formula (2.4.13) holds with Vn ’s replaced by Xn ’sdefine Vn Xn .2 See, 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 equationXN N 1 SN (1 r)(XN 1 CN 1 N 1 SN 1 ).By imitating the trick of (1.1.3)-(1.1.8) (page 6), we obtain N 1 (ω1 · · · ωN 1 ) XN 1CN (ω1 · · · ωN 1 H) CN (ω1 · · · ωN 1 T )SN (ω1 · · · ωN 1 H) SN (ω1 · · · ωN 1 T )1[p̃XN (ω1 · · · ωN 1 H) q̃XN (ω1 · · · ωN 1 T )]1 r[]e n XN CN 1 E1 r][ N Cken. E(1 r)k (N 1)k N 1 CN 1 Working backward by induction, we can provide definitions for each Xn and n , and prove for each n][ N[] XCn 1kenenXn Cn E. E1 r(1 r)k nk nFinally, 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, {Ft }t 0,··· ,T , P) be a (general) filteredprobability space and let S {St }t {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 :a.s.EP [St Ft 1 ] St 1 for all t {1, · · · , T }. k-multiple Markov property: For all B E and t {k, · · · , T }P (St B Ft 1 ) P (St B St 1 , St 2 , · · · , St k ).a.s.For continuous time financial models, it is well-known that for Lipschitz continuous f , g, the stochasticdifferential equation of K. Itô t tXt X0 f (s, Xs )dWs g(s, Xs )ds00has 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 AcP(ω) ω AP(ω) ω ΩP(ω) 1.(ii) If A1 , A2 , · · · , AN is a finite set of events, thenP( Nn 1 An ) N P(An ).(2.8.1)n 1If the events A1 , A2 , · · · , AN are disjoint, then equality holds in (2.8.1).Proof.By induction,work on the case N 2. When A1 and A2 are disjoint, P(A1 A2 ) it suffices to ω A1 A2 P(ω) ω A1 P(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 p̃ 12 , q̃ 12 .Solution. Under the risk-neutral probability, the distribution of ωn 1 conditioning on ω1 , · · · , ωn is deterministice n 1 H ω1 ω2 · · · ωn ) : p̃ 1 r d , P(ωe n 1 T ω1 ω2 · · · ωn ) : q̃ u 1 r .P(ωu du de and we haveSo ωn ’s are independent of each other under Pe 3 32) p̃3 1 , P(Se 3 8) 3p̃2 q̃ 3 , P(Se 3 2) 3p̃q̃ 2 3 , P(Se 3 0.5) q̃ 3 1 .P(S8888e 1 , ESe 2 , and ESe 3 . What is the average rate of growth of the stock price under P?e(ii) Compute ESSolution. eee E[S1 ] 8P(S1 8) 2P(S1 2) 8p̃ 2q̃ 52eE[S2 ] 16p̃ 4 · 2p̃q̃ 1 · q̃ 2 6.25 eE[S3 ] 32 · 18 8 · 38 2 · 38 0.5 · 18 7.8125.e are, respectively:So the average rates of growth of the stock price under P e 1]5 r̃ E[S S0 1 4 1 0.25, 0e 2]E[S6.25r̃1 E[Se 1 ] 1 5 1 0.25, e 3] r̃ E[S 1 7.8125 1 0.25.2e 2]E[S6.25Remark 2.1. An alternative solution is to use martingale property:eE[]eSne [Sn ] (1 r)n S0 , E [Sn ] (1 r). S0 Ene [Sn 1 ](1 r)E(iii) Answer (i) and (ii) again under the actual probabilities p 23 , q 13 .81Solution. P(S3 32) ( 32 )3 27, P(S3 8) 3 · ( 23 )2 · 13 94 , P(S3 2) 2 · 19 29 , and P(S3 0.5) 27.Accordingly, E[S1 ] 6, E[S2 ] 9 and E[S3 ] 13.5. So the average rates of growth of the stock priceunder P are, respectively: r0 64 1 0.5, r1 69 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 ] Sn En [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 theprobability 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 andMn n Xj , n 1.j 1This 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()n2Sn eσMn.eσ e σShow 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 again2have a martingale, we must “discount” the geometric symmetric random walk, using the term eσ e σ as thediscount rate. This term is strictly less than one unless σ 0.Proof.[En][][]Sn 122σXn 1 En e σE eσXn 1 1.σ σ σSne ee eI Exercise 2.5. Let M0 , M1 , M2 , · · · be the symmetric random walk of Exercise 2.4, and define I0 0andn 1 In Mj (Mj 1 Mj ), n 1, 2, · · · .j 0(i) Show thatIn 1 2 nM .2 n 2Proof.2In 2n 1 Mj (Mj 1 Mj ) 2j 0 Mn2 n 1 Mj Mj 1 Mn2 j 0n 1 j 0j 0122Mj 1 j 0n 1 (Mj 1 Mj )2 Mn2 n 1 2Xj 1 Mn2 n.n 1 j 0Mj2
Remark 2.3. This is the discrete version of the integration-by-parts formula for stochastic integral: TMT2 M02 2Mt dMt [M, M ]T .0with IT T0Mt dMt .(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) satisfyingEn [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 Mn Xn 1 )]1 [f (In Mn ) f (In Mn )]2 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
Oct 26, 2014 · Binomial Asset Pricing Model Solution 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, please email me at [email protected] Contents 1 The Binomial No-Arbitrage Pricing M
