(Solution Manual) Introduction to Stochastic Finance with Market Examples

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  5α+β= 0 2α+β= 6,with solution  α=−2 β= +10. The hedging strategy at t= 0is toshortsell−α= +2units of the asset SpricedS0= 4, and to put β= $10on the savings account. The price V0=αS0+βof the initial portfolio at time t= 0is V0=αS0+β=−2×4 + 10 = $2 , which yields the price of the claim at time t= 0. In order to hedge then option, one should: i) At time t= 0, a. Charge the $2 option price. b. Shortsell−α= +2units of the stock priced S0= 4, which yields $8. c. Putβ= $8 + $2 = $10 on the savings account. ii) At time t= 1, a. IfS1= $5, spend $10 from savings to buy back −α= +2stocks. b. IfS1= $2, spend $4 from savings to buy back −α= +2stocks, and deliver a $10 - $4 = $6 payoff. Pricing the option by the expected value IE∗[C]yields the equality $2 =IE∗[C] = 0×P∗(C= 0) + 6×P∗(C= 6) = 0×P∗(S1= 2) + 6×P∗(S1= 5) = 6×q∗, 1Solutions Manual for Introduction to Stochastic Finance with Market Examples, 2e by Nicolas Privault (All Chapters) Chapter 1 Exercise 1.1 The payoff C is that of a put option with strike price K = $3. Exercise 1.2 Each of the two possible scenarios yields one equation:Solutions Manual hence the risk-neutral probability measure P∗is given by p∗=P∗(S1= 5) =2 3andq∗=P∗(S1= 2) =1 3. Exercise 1.3 a) Each of the stated conditions yields one equation, i.e.   4α+β= 1 5α+β= 3,with solution  α= 2 β=−7. Therefore, the portfolio allocation at t= 0consists to purchase α= 2 unit of the asset SpricedS0= 4, and to borrow−β= $7in cash. We can check that the price V0=αS0+βof the initial portfolio at timet= 0is V0=αS0+β= 2×4−7 = $1. b) This loss is expressed as α×$2 +β= 2×2−7 =−$3. Note that the $1 received when selling the option is not counted here be- cause it has already been fully invested into the portfolio. Exercise 1.4 a) i) Does this model allow for arbitrage? Yes |✓No| ii) If this model allows for arbitrage opportunities, how can they be real- ized? By shortselling |By borrowing on savings |✓N.A.| b) i) Does this model allow for arbitrage? Yes |No|✓ ii) If this model allows for arbitrage opportunities, how can they be real- ized? By shortselling |By borrowing on savings |N.A.|✓ c) i) Does this model allow for arbitrage? Yes |✓No| ii) If this model allows for arbitrage opportunities, how can they be real- ized? By shortselling |✓By borrowing on savings |N.A.| Exercise 1.5 2 2 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition a) We need to search for possible risk-neutral probability measure(s) P∗such that IE∗/bracketleftbig S(1) 1/bracketrightbig = (1 +r)S(1) 0. Letting   p∗=P∗/parenleftbig S(1) 1=S(1) 0(1 +a)/parenrightbig =P∗/parenleftigg S(1) 1−S(1) 0 S(1) 0=a/parenrightigg , θ∗=P∗/parenleftbig S(1) 1=S(1) 0(1 +b)/parenrightbig =P∗/parenleftigg S(1) 1−S(1) 0 S(1) 0=b/parenrightigg , q∗=P∗/parenleftbig S(1) 1= (1 +c)S(1) 0/parenrightbig =P∗/parenleftigg S(1) 1−S(1) 0 S(1) 0=c/parenrightigg , We have   (1 +a)p∗S(1) 0+ (1 +b)θ∗S(1) 0+ (1 +c)q∗S(1) 0= (1 +r)S(1) 0 p∗+θ∗+q∗= 1, from which we obtain   p∗a+θ∗b+q∗c=r, p∗+θ∗+q∗= 1.=⇒  p∗=(1−θ∗)c+θ∗b−r c−a∈(0,1), q∗=r−(1−θ∗)a−θ∗b c−a∈(0,1). In order for p∗andq∗to belong to the interval (0,1)we should have   0<(1−θ∗)c+θ∗b−r<c−a, 0<r−(1−θ∗)a−θ∗b<c−a, i.e.  r−a b−c<θ∗<c−r c−b, r−c b−a<θ∗<r−a b−a. Therefore there exists an infinity of risk-neutral probability measures de- pending on the value of θ∗∈/parenleftbigg max/parenleftbiggr−a b−c,r−c b−a/parenrightbigg ,min/parenleftbiggc−r c−b,r−a b−a/parenrightbigg/parenrightbigg , in which case the market is without arbitrage but not complete. " 3 September 15, 2022Solutions Manual b) Hedging a claim with possible payoff values Ca,Cb,Ccwould require to solve   (1 +a)αS(1) 0+ (1 +r)βS(0) 0=Ca (1 +b)αS(1) 0+ (1 +r)βS(0) 0=Cb (1 +c)αS(1) 0+ (1 +r)βS(0) 0=Cc, forαandβ, which is not possible in general due to the existence of three conditions with only two unknowns. Exercise 1.6 a) The risk-neutral condition IE∗[R1] = 0reads bP∗(R1=b) + 0×P∗(R1= 0) + (−b)×(R1=−b) =bp∗−bq∗= 0, hence p∗=q∗=1−θ∗ 2, sincep∗+q∗+θ∗= 1. b) We have Var∗/bracketleftigg S(1) 1−S(1) 0 S(1) 0/bracketrightigg =IE∗/bracketleftbig R2 1/bracketrightbig −(IE∗[R1])2 =IE∗/bracketleftbig R2 1/bracketrightbig =b2P∗(R1=b) + 02×P∗(R1= 0) + (−b)2×(R1=−b) =b2(p∗+q∗) =b2(1−θ∗) =σ2, henceθ∗= 1−σ2/b2, and therefore p∗=q∗=1−θ∗ 2=σ2 2b2, provided that σ2≤2b2. Exercise 1.7 a) We denote the risk-neutral measure by p∗=P∗(S1= 2),q∗=P∗(S1= 1). i) Yes|No|✓Comment : No loss is possible, while a 100% profit is possible with non-zero probability 1/3. 4 4 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition ii) Yes|No|✓Comment : The (unique) risk-neutral measure (p∗,q∗) = (0,1)is given by $2×p∗+ $1×q∗= $1×(1 +r) = $1andp∗+q∗= 1, and is not equivalent to Pgiven by (p,q) = (1/3,2/3). b) We denote the risk-neutral measure by p∗=P∗(S1= 2),θ∗=P∗(S1= 1), q∗=P∗(S1= 0). i) Yes|✓No|Comment : The risk-neutral measure (p∗,θ∗,q∗) is given by the equations $2×p∗+ $1×θ∗+ $0×q∗= $1×(1 +r) = $1andp∗+θ∗+q∗= 1, (S.1.1) which clearly admit solutions, see (biv) below. ii) Yes|✓No|Comment : Realizing arbitrage would mean building a portfolio achieving no strictly negative return with prob- ability one, which is impossible since the probability of 100% loss is P(S1= 0) = 1−1/4−1/9 = 23/36>0. iii) Yes|No|✓Comment : Examples of claims that cannot be attained can be easily constructed in this market. For example, the claim 1{S1>0}cannot be attained since there is no portfolio allocation (α,β)satisfying  $2×α+β= $1 $1×α+β= $1 $0×α+β= $0. iv) Yes|No|✓Comment : The risk-neutral measure is clearly not unique, as for example (p∗,θ∗,q∗) = (1/4,1/2,1/4)and (p∗,θ∗,q∗) = (1/3,1/3,1/3) are both solutions of (S.1.1). Exercise 1.8 a) The possible values of Rareaandb. b) We have IE∗[R] =aP∗(R=a) +bP∗(R=b) " 5 September 15, 2022Solutions Manual =ab−r b−a+br−a b−a =r. c) By Theorem 1.6, there do not exist arbitrage opportunities in this market since from Question (b) there exists a risk-neutral probability measure P∗ whenevera<r<b . d) The risk-neutral probability measure is unique hence the market model is complete by Theorem 1.13. e) Taking η=α(1 +b)−β(1 +a) π1(b−a)andξ=β−α S0(b−a), we check that   ηπ1+ξS0(1 +a) =αifR=a, ηπ1+ξS0(1 +b) =βifR=b, which shows that ηπ1+ξS1=C in both cases R=aandR=b. f) We have π0(C) =ηπ0+ξS0 =α(1 +b)−β(1 +a) (1 +r)(b−a)+β−α b−a =α(1 +b)−β(1 +a)−(1 +r)(α−β) (1 +r)(b−a) =αb−βa−r(α−β) (1 +r)(b−a). (S.1.2) g) We have IE∗[C] =αP∗(R=a) +βP∗(R=b) =αb−r b−a+βr−a b−a. (S.1.3) h) Comparing (S.1.2) and (S.1.3) above we do obtain π0(C) =1 1 +rIE∗[C] i) The initial value π0(C)of the portfolio is interpreted as the arbitrage- free price of the option contract and it equals the expected value of the discounted payoff. j) We have 6 6 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition C= (K−S1)+= (11−S1)+=  11−S1ifK >S 1, 0 ifK≤S1. k) We have S0= 1,a= 8,b= 11,α= 2,β= 0, hence ξ=β−α S0(b−a)=0−2 11−8=−2 3, and η=α(1 +b)−β(1 +a) π1(b−a)=24 3×1.05. l) The arbitrage-free price π0(C)of the contingent claim with payoff Cis π0(C) =ηπ0+ξS0= 6.952. Exercise 1.9 Letting Rdenote the price of one right, it will require 10R/3to purchase one stock at €6.35, hence absence of arbitrage tells us that 10 3R+ 6.35 = 8, from which it follows that R=3 10(8−6.35) = €0.495. Note that the actual share right was quoted at €0.465according to market data. See also Exercise 17.8 for the pricing of convertible bonds. Exercise 1.10 Let a:= (152−180)/180 =−7/45andb:= (203−180)/180 = 23/180denote the potential market returns, with r= 0.03. From the strike priceKand the risk-neutral probabilities p∗ r=r−a b−a= 0.6549andq∗ r=b−r b−a= 0.3451, the price of the option at the beginning of the year is given from Proposi- tion 1.16 as the discounted expected value 1 1 +rIE∗[(K−S1)+] =1 1 +r/parenleftbig p∗ r(K−203)++q∗ r(K−152)+/parenrightbig . Equating this price with the intrinsic value (K−180)+of the put option yields the equation (K−180)+=1 1 +r/parenleftbig p∗ r(K−203)++q∗ r(K−152)+/parenrightbig " 7 September 15, 2022Solutions Manual which requires K > 180(the caseK≤152is not considered because both the option price and option payoff vanish in this case). Hence we consider the equation K−180 =1 1 +r/parenleftbig p∗ r(K−203)++q∗ r(K−152)+/parenrightbig , with the following cases. i) IfK∈[180,203]we get (1 +r)(K−180) =q∗ r(K−152), hence K=(1 +r)180−q∗ r152 1 +r−q∗r=(1 +r)180−q∗ r152 p∗r+r= 194.11. ii) IfK≥203we find K=180(1 +r)−203p∗ r−152q∗ r r<203, which is out of range and leads to a contradiction. We note that the above formula K=(1 +r)180−q∗ r152 p∗r+r=28b−180a+r(180(b−a) + 152) (b+ 1−a)r−a yields a decreasing function K(r)ofrin the interval [0,100%], although the function is not monotone over R+. 60 70 80 90 100 110 120 130 140 150 0 0.5 1 1.5 2 r Fig. S.1: Strike price as a function of risk-free rate r. 8 8 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition Chapter 2 Exercise 2.1 Let m:= $2,550denote the amount invested each year. a) By (2.1), the value of the plan after N= 10years becomes mN/summationdisplay k=1(1 +r)k=m(1 +r)(1 +r)N−1 r, which in turns becomes (1 +r)NmN/summationdisplay k=1(1 +r)k=m(1 +r)N+1(1 +r)N−1 r, afterNadditional years without further contributions to the plan. Equat- ing A= 30835 = m(1 +r)N+1(1 +r)N−1 r shows that (1 +r)2N+1−(1 +r)N+1 r=A m, withm= 2550, or (1 +r)21−(1 +r)11 r=30835 2550≃12.09215, hencer≃1.23%according to Figure S.2, which is typical of an annual fixed deposit interest rate. 10 10.5 11 11.5 12 12.5 13 13.5 14 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 r in % Fig. S.2: Graph ofr∝⇕⊣√∫⊔≀−→((1 +r)21−(1 +r)11)/r. In the hypothesis r= 3.25%we would find A=m(1 +r)N+1(1 +r)N−1 r= 42040.42. " 9 September 15, 2022Solutions Manual b) TakingN= 10,m= 2,550andr= 0.0325, we find A2N: =m(1 +r)NN/summationdisplay k=1(1 +r)N−k+1 =m(1 +r)NN/summationdisplay k=1(1 +r)k =m(1 +r)N+1(1 +r)N−1 r = $42,040.42. c) In this case, with N= 10,m= 2,550andr= 0.0325, we find A2N=AN=mN/summationdisplay k=1(1 +r)N−k+1=m(1 +r)(1 +r)N−1 r= $30,532.79. Exercise 2.2 a) Letm:= $3,581denote the amount invested each year. After multiplying (2.1) by (1 +r)Nin order to account for the compounded interest from year 11until year 20, we get the equality A=m(1 +r)N+1(1 +r)N−1 r shows that (1 +r)21−(1 +r)11=r50862 3581≃14.2033r, showing that r≃2.28%according to Figure S.3. 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3 r in % Fig. S.3: Graph ofr∝⇕⊣√∫⊔≀−→((1 +r)21−(1 +r)11)/r. 10 10 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition b) TakingN= 10,m= 3,581andr= 0.0325, we find A2N: =m(1 +r)NN/summationdisplay k=1(1 +r)N−k+1 =m(1 +r)NN/summationdisplay k=1(1 +r)k =m(1 +r)N+1(1 +r)N−1 r = $59,037.94. c) In this case, we find A2N=mN/summationdisplay k=1(1 +r)N−k+1=m(1 +r)(1 +r)N−1 r= $42,877.61. Exercise 2.3 a) We find m= $10,000. b) TakingN= 10,A= 100,000andr= 0.02, Proposition 2.1 shows that m=r 1−(1 +r)−NA=0.02×100,000 1−1.02−10= $11,132.65. Exercise 2.4 a) The discounted value of the loan after Nmonths is m(1 +r)−NN−1/summationdisplay k=0(1 +r)k=m(1 +r)−N1−(1 +r)N 1−(1 +r)=m1−(1 +r)−N r, which should match A= $3,000withm= $275andN= 12, hence 1−(1 +r)−12 r=A m= 10.909090909, see also Equation (3.1.2) in Proposition 3.1 of MH4514_Slides.pdf, hence r≃1.49767% per month. b) The yearly interest rate is given by (1 +r)N−1 = (1 +r)12−1 = 1.014976712−1≃19.529%per year. Remark: Computing the interest rate as " 11 September 15, 2022Solutions Manual 12×$275 $3000−1 = 0.1 = 10% is not correct because this implicitly means that the 12×$275 = $3,300 are repaid as one lump sum at the end of the 12 thmonth, which is not the case. c) The analysis of replies to Question (c)shows that “All of the above” was the most popular answer, followed by “Block”. Fig. S.4: Histogram of replies to Question c). Exercise 2.5 We check that for any P∗of the form P∗(Rt=−1) :=p∗,P∗(Rt= 0) := 1−2p∗,P∗(Rt= 1) :=p∗, we have IE∗[S1] =S0(2p∗+ 1−2p∗) =S0, and similarly IE∗[S2|S1] =S1(2p∗+ (1−2p∗)) =S1, hence the probability measure P∗is risk-neutral. Exercise 2.6 a) In order to check for arbitrage opportunities we look for a risk-neutral probability measure P∗which should satisfy IE∗/bracketleftbig S(1) k+1|Fk/bracketrightbig = (1 +r)S(1) k, k = 0,1,...,N−1, withr= 0. Rewriting IE∗/bracketleftbig S(1) k+1|Fk/bracketrightbig as IE∗/bracketleftbig S(1) k+1|Fk/bracketrightbig 12 12 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition = (1−b)S(1) kP∗(Rk+1=a|Fk) +S(1) kP∗(Rk+1= 0|Fk) + (1 +b)S(1) kP∗(Rk+1=b|Fk) = (1−b)S(1) kP∗(Rk+1=a) +S(1) kP∗(Rk+1= 0) + (1 + b)S(1) kP∗(Rk+1=b), k= 0,1,...,N−1, it follows that any risk-neutral probability measure P∗should satisfy the equations   (1 +b)S(1) kP∗(Rk+1=b) +S(1) kP∗(Rk+1= 0) + (1−b)S(1) kP∗(Rk+1=a) =S(1) k P∗(Rk+1=b) +P∗(Rk+1= 0) + P∗(Rk+1=−b) = 1, k= 0,1,...,N−1,i.e.   bP∗(Rk=b)−bP∗(Rk=−b) = 0, P∗(Rk=b) +P∗(Rk=−b) = 1−P∗(Rk= 0), k= 1,2,...,N, with solution P∗(Rk=b) =P∗(Rk=−b) =1−θ∗ 2, k= 1,2,...,N. b) We have Var∗/bracketleftigg S(1) k+1−S(1) k S(1) k/vextendsingle/vextendsingle/vextendsingle/vextendsingleFk/bracketrightigg =IE∗ /parenleftigg S(1) k+1−S(1) k S(1) k/parenrightigg2/vextendsingle/vextendsingle/vextendsingle/vextendsingleFk −/parenleftigg IE∗/bracketleftigg S(1) k+1−S(1) k S(1) k/vextendsingle/vextendsingle/vextendsingle/vextendsingleFk/bracketrightigg/parenrightigg2 =IE∗ /parenleftigg S(1) k+1−S(1) k S(1) k/parenrightigg2/vextendsingle/vextendsingle/vextendsingle/vextendsingleFk  =b2P∗ σ(Rk+1=−b|Fk) +b2P∗ σ(Rk+1=b|Fk) =b21−P∗ σ(Rk+1= 0) 2+b21−P∗ σ(Rk+1= 0) 2 =b2(1−θ∗) =σ2, k= 0,1,...,N−1, hence P∗ σ(Rk= 0) =θ∗= 1−σ2 b2, " 13 September 15, 2022Solutions Manual and therefore P∗ σ(Rk=b) =P∗ σ(Rk=−b) =1−P∗ σ(Rk= 0) 2=σ2 2b2, k= 0,1,...,N−1, under the condition 0<σ2<b2. Exercise 2.7 a) The possible values of Rtareaandb. b) We have IE∗[Rt+1|Ft] =aP∗(Rt+1=a|Ft) +bP∗(Rt+1=b|Ft) =ab−r b−a+br−a b−a=r. c) Lettingp∗= (r−a)/(b−a)andq∗= (b−r)/(b−a)we have IE∗[St+k|Ft] =k/summationdisplay i=0(p∗)i(q∗)k−i/parenleftbiggk i/parenrightbigg (1 +b)i(1 +a)k−iSt =Stk/summationdisplay i=0/parenleftbiggk i/parenrightbigg (p∗(1 +b))i(q∗(1 +a))k−i =St(p∗(1 +b) +q∗(1 +a))k =St/parenleftbiggr−a b−a(1 +b) +b−r b−a(1 +a)/parenrightbiggk = (1 +r)kSt. Assuming that the formula holds for k= 1, its extension to k≥2can also be proved recursively from the tower property of conditional expectations, as follows: IE∗[St+k|Ft] =IE∗[IE∗[St+k|Ft+k−1]|Ft] = (1 +r)IE∗[St+k−1|Ft] = (1 +r)IE∗[IE∗[St+k−1|Ft+k−2]|Ft] = (1 +r)2IE∗[St+k−2|Ft] = (1 +r)2IE∗[IE∗[St+k−2|Ft+k−3]|Ft] = (1 +r)3IE∗[St+k−3|Ft] =··· = (1 +r)k−2IE∗[St+2|Ft] = (1 +r)k−2IE∗[IE∗[St+2|Ft+1]|Ft] = (1 +r)k−1IE∗[St+1|Ft] = (1 +r)kSt. 14 14 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition Exercise 2.8 a) We check that IE∗[Rt+1|Ft] =aP∗(Rt+1=a|Ft) +bP∗(Rt+1=b|Ft) =ab−r b−a+br−a b−a=r. b) We have IE∗/bracketleftbig/tildewideSt+1/vextendsingle/vextendsingleFt/bracketrightbig =1 A0(1 +r)t+1E∗[St+1|Ft] =1 A0(1 +r)t+1/parenleftbig (1 +a)StP∗(Rt+1=a|Ft) + (1 +b)StP∗(Rt+1=b|Ft)/parenrightbig =1 A0(1 +r)t+1/parenleftbigg (1 +a)Stb−r b−a+ (1 +b)Str−a b−a/parenrightbigg =/tildewideStb−a+ (b−a)r (1 +r)(b−a) =/tildewideSt, t = 0,1,...,N−1. c) We have IE∗/bracketleftbig (SN)β/bracketrightbig =S0IE∗ /parenleftiggN/productdisplay k=1(1 +Rk)/parenrightiggβ  =S0IE∗/bracketleftiggN/productdisplay k=1(1 +Rk)β/bracketrightigg =S0N/productdisplay k=1IE∗/bracketleftbig (1 +Rk)β/bracketrightbig , after using the independence of the returns (Rk)k=1,2,...,N, with IE∗/bracketleftbig (1 +Rk)β/bracketrightbig = (1 +a)βb−r b−a+ (1 +b)βr−a b−a, k = 0,1,...,N, hence we find IE∗/bracketleftbig (SN)β/bracketrightbig =Sβ 0/parenleftbigg (1 +a)βb−r b−a+ (1 +b)βr−a b−a/parenrightbiggN . d) We have P∗/parenleftbig St≥αAtfor somet∈{0,1,...,N}/parenrightbig =P∗/parenleftbigg max t=0,1,...,nSt At≥α/parenrightbigg " 15 September 15, 2022Solutions Manual ≤IE/bracketleftbig (MN)β/bracketrightbig αβ =/parenleftbiggS0 (1 +r)NαA0/parenrightbiggβ/parenleftbigg (1 +a)βb−r b−a+ (1 +b)βr−a b−a/parenrightbiggN , since the discounted price process (Mt)t=0,1,...,N :=/parenleftbiggSt At/parenrightbigg t=0,1,...,N is a nonnegative martingale by part (b). e) Since (Mt)t=0,1,...,Nis a nonnegative martingale, we have IE[St+1|Ft] =IE[Mt+1At+1|Ft] =At+1IE[Mt+1|Ft] =At+1Mt ≥AtMt =St, t = 0,1,...,N−1, becauser≥0, hence (St)t=0,1,...,Nis a nonnegative submartingale. There- fore, we have P∗/parenleftig max t=0,1,...,nSt≥x/parenrightig ≤IE/bracketleftbig (MN)β/bracketrightbig xβ ≤/parenleftbiggS0 x/parenrightbiggβ/parenleftbigg (1 +a)βb−r b−a+ (1 +b)βr−a b−a/parenrightbiggN . Chapter 3 Exercise 3.1 (Exercise 2.5 continued). We consider the following trinomial tree. 16 16 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition S0= 1 S1= 0 S2= 0,C= 1p∗S2= 0,C= 11−2p∗S2= 0,C= 1 p∗p∗S1= 1 S2= 0,C= 1p∗S2= 1,C= 01−2p∗S2= 2,C= 0 p∗ 1−2p∗S1= 2 S2= 0,C= 1p∗S2= 2,C= 01−2p∗S2= 4,C= 0 p∗ p∗ At timet= 0, we find π0(C) =1 (1 +r)2IE∗[(K−S2)+] =p∗(p∗+ (1−2p∗) +p∗) + (1−2p∗)p∗+ (p∗)2 =p∗+ (1−2p∗)p∗+ (p∗)2 = 2p∗−(p∗)2. At timet= 1, we find π1(C) =1 1 +rIE∗[(K−S2)+|S1] =  p∗ifS1= 2S0, p∗ifS1=S0, 1ifS1= 0. Exercise3.2 Wehave p∗= (r−a)/(b−a) = 1/2andq∗= (b−r)/(b−a) = 1/2, and the following underlying asset price tree: " 17 September 15, 2022Solutions Manual S2= 4, C= 3 S1= 2 S0= 1 S2= 2, C= 1 S1= 1 S2= 1, C= 3.p∗ q∗p∗ q∗ p∗ q∗ We first price, and then hedge. At time t= 1, by Theorem 3.5 we have π1(C) =V1=  3p∗+q∗ 1 +r=4 3ifS1= 2 p∗+ 3q∗ 1 +r=4 3ifS1= 1,andπ0(C) =V0=4 3p∗+q∗ 1 +r=8 9. This leads to the following option pricing tree: S2= 4, V2= 3 V1= 4/3 S1= 2 V0= 8/9 S0= 1S2= 2V2= 1 V1= 4/3 S1= 1 S2= 1V2= 3.p∗ q∗p∗ q∗ p∗ q∗ Regarding hedging, if S1= 2the condition ξ2·S2=ξ2S2+η2A2=V2reads S1= 2 =⇒  4ξ2+η2(1 +r)2= 3 2ξ2+η2(1 +r)2= 1, hence (ξ2,η2) = (1,−4/9). On the other hand, if S1= 1we have 18 18 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition S1= 1 =⇒  2ξ2+η2(1 +r)2= 1 ξ2+η2(1 +r)2= 3, hence (ξ2,η2) = (−2,20/9). Finally, at time t= 0withS0= 1the condition ξ1·S1=ξ1S1+η1A1=V1yields   2ξ1+η1(1 +r) =4 3 ξ1+η1(1 +r) =4 3, hence (ξ1,η1) = (0,8/9). The results can be summarized in the following table: S1= 2,V1= 4/3S2= 4 S0= 1ξ2= 1,η2=−4/9V2= 3 V0= 8/9 S2= 1 ξ1= 0 V2= 3 η1= 8/9S1= 1,V1= 4/3S2= 1 ξ2=−2,η2= 20/9V2= 3 Table S.1: CRR pricing and hedging table. In addition, it can be checked that the portfolio strategy (ξk,ηk)k=1,2is self- financing, as we have ξ1S1+η1A1=8 9×3 2 =  2−4 9×3 2 −2 +20 9×3 2 =ξ2S1+η2A1. Exercise 3.3 a) We have IE∗[St+1|Ft] =IE∗[St+1|St] =St 2P∗(Rt=−0.5) +StP∗(Rt= 0) + 2StP∗(Rt= 1) =/parenleftbiggr∗ 2+q∗+ 2p∗/parenrightbigg St =St, t = 0,1, " 19 September 15, 2022Solutions Manual withr= 0. b) We have the following graph: S0= 1 S1= 0.5 S2= 0.25,C= 0r∗= 1/2S2= 0.5,C= 0q∗= 1/4S2= 1,C= 0 p∗= 1/4r∗= 1 /2S1= 1 S2= 0.5,C= 0r∗= 1/2S2= 1,C= 0q∗= 1/4S2= 2,C= 0 p∗= 1/4 q∗= 1/4S1= 2 S2= 1,C= 0r∗= 1/2S2= 2,C= 0.5q∗= 1/4S2= 4,C= 2.5 p∗= 1/4 p∗= 1/4 c) The down-an-out barrier call option is priced at time t= 0as V0=IE∗[C] = 2.5×(p∗)2+ 0.5×p∗q∗=3 16. At timet= 1we have V1= 2.5×p∗+ 0.5×q∗= 2.5×1 4+ 0.5×1 4=3 4 ifS1= 2, andV1= 0in both cases S1= 1andS1= 0.5. d) This market is not complete, and not every contingent claim is attainable, because the risk-neutral probability measure P∗is not unique, for example (r∗,q∗,p∗) = (1/4,5/8,1/8)and (r∗,q∗,p∗) = (1/2,1/4,1/4)are both risk-neutral probability measures. Exercise 3.4 The CRR model can be described by the following binomial tree. 20 20 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition (1 +b)2S0 (1 +b)S0 S0= 1 (1 +a)(1 +b)S0 (1 +a)S0 (1 +a)2S0p∗ q∗p∗ q∗ p∗ q∗ a) By the formulas V1=1 1 +rIE∗[V2|F1] =1 1 +rIE∗[V2|S1] =S0(1 +b)2−8 1 +rP∗(S2=S0(1 +b)2|S1) =p∗(S0(1 +b)2−8) 1 +r1{S1=S0(1+b)}, and V0=1 1 +rIE∗[V1|F0] =1 1 +r/parenleftbigg p∗(S0(1 +b)2−8) 1 +r×P∗(S1=S0(1 +b)) + 0×P∗(S1=S0(1 +a))/parenrightbigg = (p∗)2(S0(1 +b)2−8) (1 +r)2, we find the table S2= 9 S1= 3,V1= 1/4V2= 1 S0= 1 S2= 3 V0= 1/16 V2= 0 S1= 1,V1= 0S2= 1 V2= 0 Table S.2: CRR pricing tree. Note that we could also directly compute V0from " 21 September 15, 2022Solutions Manual V0=1 (1 +r)2IE∗[V2|F0]. b) WhenS1=S0(1 +b), the equation ξ2S2+η2A2=V2reads   ξ2S0(1 +b)2+η2A0(1 +r)2=S0(1 +b)2−8 ξ2S0(1 +b)(1 +a) +η2A0(1 +r)2= 0, which yields ξ2=S0(1 +b)2−8 S0(b−a)(1 +b)andη2=−(S0(1 +b)2−8)(1 +a) (b−a)A0(1 +r)2.(S.3.4) WhenS1=S0(1 +a), the equation ξ2S2+η2A2=V2reads   ξ2S0(1 +a)2+η2A0(1 +r)2= 0 ξ2S0(1 +b)2+η2A0(1 +r)2= 0, which has the unique solution (ξ2,η2) = (0,0). Next, the equation ξ1S1+ η1A1=V1reads   ξ1S0(1 +b) +η1A0(1 +r) =p∗(S0(1 +b)2−8) 1 +r, ξ1S0(1 +a) +η1A0(1 +r) = 0, which yields ξ1=p∗S0(1 +b)2−8 S0(b−a)(1 +r)andη1=−p∗(1 +a)(S0(1 +b)2−8) (b−a)A0(1 +r)2. (S.3.5) This can be summarized in the following table: S1= 3,V1= 1/4S2= 9 S0= 1 V2= 1 V0= 1/16ξ2= 1/6,η2=−1/8S2= 3 ξ1= 1/8S1= 1,V1= 0V2= 0 η1=−1/16 S2= 1 ξ2= 0,η2= 0V2= 0 Table S.3: CRR pricing and hedging tree. WhenS1=S0(1 +a)at timet= 1the option price is V1= 0and the hedging strategy is to cut all positions: ξ2=η2= 0. On the other hand, 22 22 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition ifS1=S0(1 +b)then there is a chance of being in the money at maturity and we need to increase our position in the underlying asset from ξ1= 1/8 toξ2= 1/6. Note that the self-financing condition ξ1S1+η1A1=ξ2S1+η2A1, (S.3.6) is verified. For example when S1=S0(1 +a)we have 1 8×S1−1 16A1= 0×S1+ 0×A1= 0, while when S1=S0(1 +b)we find 1 8×S1−1 16A1=1 6×S1−1 8×A1=1 4. c) We can also use the self-financing condition (S.3.6) to recover (S.3.5) by rewriting the system of equations as   ξ1S0(1 +b) +η1A0(1 +r) =ξ2S0(1 +b) +η2A0(1 +r) ξ1S0(1 +a) +η1A0(1 +r) = 0, with (ξ2,η2)given by (S.3.4), which recovers V1=ξ1S1+η1A1=  3 8−2 16=1 4ifS1= 3, 1 8−2 16= 0ifS1= 1. Exercise 3.5 a) We build a portfolio based at times t= 0,1onαt+1units of stock and $βt+1in cash. When S1= 2, we should have /braceleftbigg4α2+β2= 0 2α2+β2= 1, hence (α2,β2) = (−1/2,2). On the other hand, when S1= 1we should have /braceleftbigg2α1+β1= 1 α1+β1= 0, hence (α2,β2) = (1,−1). b) WhenS1= 2, the price of the claim at t= 1is " 23 September 15, 2022Solutions Manual α2S1+β2= (−1/2)×2 + 2×1 = 1. WhenS1= 1,thepriceoftheclaimat t= 1isα2S1+β2= 1×1−1×1 = 0. c) At time t= 1we build a portfolio using α1units of stock and $ β1in cash. We should have /braceleftbigg 2α1+β1= 1 α1+β1= 0, hence (α1,β1) = (1,−1). d) The price of the claim Cat timet= 0isα1S0+β1= 1×1+(−1)×1 = 0. e) The probabilities (p∗,q∗) = ((r−a)/(b−a),(b−r)/(b−a)) = (0,1)are clearly risk-neutral in the sense of Definition 2.12, as they yield IE∗[S2|S1] =S1and IE∗[S1|S0] =S0. with the risk-free rate r= 0. However, this does not form a risk-neutral probabilitymeasure P∗equivalentto PinthesenseofDefinition2.14when q=P(R1= 0) = P(R2= 0)>0. In case (p,q) = (0,1), the probabilities (p∗,q∗) = (0,1)would yield an equivalent risk-neutral probability measure. f) According to Theorem 2.15 this model allows for arbitrage opportuni- ties as the unique available risk-neutral probability measure P∗are not be equivalent to the historical probability measure Pwhenq=P(R1= 0) = P(R2= 0)>0. In this case, arbitrage opportunities are easily implemented by purchasing the option at the price 0of part (d) while receiving a strictly positive payoff at maturity. More generally, arbitrage opportunities exist when the underlying price may increase with nonzero probability, without a possibility of strict decrease. In case (p,q) = (0,1), no arbitrage is possible as prices remain constant. Exercise 3.6 We have the model-free answer πk(C) =1 (1 +r)N−kIE∗[h(SN)|Fk] =1 (1 +r)N−kIE∗[α+βSN|Fk] =α (1 +r)N−k+β (1 +r)N−kIE∗[SN|Fk] =α (1 +r)NAk+βSk, k = 0,1,...,N. The hedging portfolio strategy is to hold βunits of the underlying asset pricedSkandα/(1 +r)Nunits of the riskless asset priced Ak= (1 +r)kat 24 24 September 15, 2022Introduction to Stochastic Finance with Market Examples, Second Edition timek= 0,1,...,N. Note that in the particular case of the CRR model, this answer is also com- patible with (3.19)-(3.20). Exercise 3.7 Call-put parity. a) The relation (x−K)+=x−K+ (K−x)+can be verified by successively checking the cases x≤Kandx≥K. b) Respectively denoting by C(k)andP(k)the call and put prices at time k= 0,1,...,N, by part (a) we have C(k) = (1 +r)−(N−k)IE∗/bracketleftbig (SN−K)+/vextendsingle/vextendsingleFk/bracketrightbig = (1 +r)−(N−k)IE∗/bracketleftbig SN−K+ (K−SN)+/vextendsingle/vextendsingleFk/bracketrightbig = (1 +r)−(N−k)IE∗[SN|Fk]−(1 +r)−(N−k)K +(1 +r)−(N−k)IE∗/bracketleftbig (K−SN)+/vextendsingle/vextendsingleFk/bracketrightbig =Sk−(1 +r)−(N−k)K+ (1 +r)−(N−k)IE∗/bracketleftbig (K−SN)+/vextendsingle/vextendsingleFk/bracketrightbig =Sk−(1 +r)−(N−k)K+P(k). Exercise 3.8 a) Takingq∗= 1−p∗= 1/4, we find the binary tree 6.25 = (1 +b)2 2.5 = 1 +b S0= 1 1.25 = (1 +a)(1 +b) 0.5 = 1 +a 0.25 = (1 +a)(1 +b)p∗ q∗p∗ q∗ p∗ q∗ b) We find the binary tree " 25 September 15, 2022