Hi,

I need some assistance for a question in a quiz from a course I take in Coursera.

Compute the fair value of achooseroption which expires aftern=10periods. At

expiration the owner of the chooser gets to choose (at no cost) a European call option

or a European put option. The call and put each have strikeK=100and they mature

5 periods later, i.e. atn=15.

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Quiz instruction :

Questions 1-8should be answered by building a 15-period binomial model whose parameters should be calibrated to a Black-Scholes geometric Brownian motion model with:T=.25years,S0=100,r=2%,?=30%

and a dividend yield ofc=1%.

Hint

Your binomial model should use a value ofu=1.0395.... (This has been rounded to four decimal places but you should not do any rounding in your spreadsheet calculations.)

Submission Guidelines

Round all your answers to 2 decimal places. So if you compute a price of 12.9876 you should submit an answer of 12.99.

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For this quiz, the instructor gives us an excel sheet which contains somes lattices examples to resolve this question (I attached it and the pdf of the lesson).

Is it very IMPORTANT that the result is correct otherwise I didnt get the point and so it s useless for me.

Thanks in advance for all tutors who help me.

Zeksou

Financial Engineering & Risk Management The Multi-Period Binomial Model M. Haugh G. Iyengar Department of Industrial Engineering and Operations Research Columbia University A 3-period Binomial Model Recall R = 1.01 and u = 1/d = 1.07. 122.5 \u0010 114.49 \u0010 \u0010 \u0010\u0010PPP \u0010 \u0010\u0010 PP 107 107 \u0010 P \u0010 \u0010\u0010P PP 100 \u0010\u0010\u0010 \u0010\u0010 100 \u0010 PP \u0010 \u0010\u0010P PP PP 93.46 PP93.46 \u0010\u0010\u0010 P P\u0010 \u0010 PP PP87.34 \u0010\u0010\u0010 P\u0010 PP PP 81.63 P t=0 t=1 t=2 t=3 Just a series of 1-period models spliced together! - all the results from the 1-period model apply - just need to multiply 1-period probabilities along branches to get probabilities in multi-period model. 2 Pricing a European Call Option Assumptions: expiration at t = 3, strike = $100 and R = 1.01. 22.5 122.50 \u0010 ? \u0010\u0010 114.49 \u0010 \u0010 \u0010PP 7 PP 107 \u0010\u0010 107 \u0010 P \u0010 \u0010 \u0010PP PP 100 \u0010\u0010\u0010 \u0010\u0010 P\u0010 0 \u0010PP PP PP 93.46 PP93.46 \u0010\u0010\u0010 \u0010 P PP PP 87.34 \u0010\u0010\u0010 PP\u0010\u0010 0 PP PP 81.63 P t=0 t=1 t=2 t=3 100 \u0010 \u0010 3 Pricing a European Call Option 22.5 122.50 \u0010 15.48 \u0010 114.49 \u0010\u0010 \u0010 P 7 P 10.23 \u0010\u0010 PP 107 107 \u0010 P \u0010 \u0010 3.86 6.57 \u0010\u0010 PPP 100 \u0010\u0010\u0010 100 \u0010 \u0010 0 PP 2.13 \u0010PPP \u0010\u0010 P P93.46 \u0010 93.46 P PP \u0010 P 0 \u0010 PP 87.34 \u0010\u0010 PP \u0010 \u0010 0 PP 81.63 P t=0 t=1 t=2 P t=3 Q q3 3q 2 (1 q) 3q(1 q)2 (1 q)3 We can also calculate the price as C0 = 1 Q E [max(ST 100, 0)] R3 0 (1) - this is risk-neutral pricing in the binomial model - avoids having to calculate the price at every node. How would you nd a replicating strategy? - to be dened and discussed in another module. 4 Financial Engineering & Risk Management What's Going On? M. Haugh G. Iyengar Department of Industrial Engineering and Operations Research Columbia University What's Going On? Stock ABC \u0010a 110 p = .99 \u0010 \u0010 S0 = 100 \u0010\u0010 \u0010hh a h \u0010 \u0010\u0010 hhhh h a 90 h 1 p = .01 t=0 Stock XYZ t=1 \u0010a 110 p = .01 \u0010 \u0010 S0 = 100 \u0010\u0010 \u0010 \u0010\u0010 a \u0010hh h h hhhh ha 1 p = .99 t=0 90 t=1 Question: What is the price of a call option on ABC with strike K = $100? Question: What is the price of a call option on XYZ with strike K = $100? 2 What's Going On? Saw earlier C0 = = = 1 Rd uR Cu + Cd R ud ud 1 [qCu + (1 q)Cd ] R 1 Q E [C1 ] R 0 So it appears that p doesn't matter! This is true ... ... but it only appears surprising because we are asking the wrong question! 3 Another Surprising Result? R = 1.02 Stock Price 100.00 t=0 106.00 94.34 t=1 European Option Price: K = 95 112.36 100.00 89.00 119.10 106.00 94.34 83.96 11.04 t=2 t=3 t=0 14.76 4.56 19.22 7.08 0.00 24.10 11.00 0.00 0.00 t=1 t=2 t=3 R = 1.04 Stock Price 100.00 t=0 106.00 94.34 t=1 European Option Price: K = 95 112.36 100.00 89.00 119.10 106.00 94.34 83.96 15.64 t=2 t=3 t=0 18.19 6.98 21.01 8.76 0.00 24.10 11.00 0.00 0.00 t=1 t=2 t=3 Question: So the option price increases when we increase R. Is this surprising? (See \"Investment Science\" (OUP) by D. G. Luenberger for additional examples on the binomial model.) 4 Existence of Risk-Neutral Probabilities No-Arbitrage Recall our analysis of the binomial model: no arbitrage d 0, 1 q > 0 and n = # of periods. (If t is the length of a period, then T = n t.) In fact for any model if there exists a risk-neutral distribution, Q, such that (2) holds, then arbitrage cannot exist. Why? Reverse is also true: if there is no arbitrage then a risk-neutral distribution exists. Together, these last two statements are often called the rst fundamental theorem of asset pricing. 5 Financial Engineering & Risk Management Pricing American Options M. Haugh G. Iyengar Department of Industrial Engineering and Operations Research Columbia University Pricing American Options Can also price American options in same way as European options - but now must also check if it's optimal to early exercise at each node. But recall never optimal to early exercise an American call option on non-dividend paying stock. e.g. Price American put option: expiration at t = 3, K = $100 and R = 1.01. 0 122.50 114.49 \u0010 \u0010 \u0010 \u0010\u0010 \u0010\u0010PPP 0 107 P P PP \u0010 \u0010 PP \u0010\u0010\u0010 100 \u0010\u0010 100 \u0010 P 6.54 PP \u0010\u0010PP PP \u0010\u0010\u0010 93.46 93.46 PP PP \u0010 PP 87.34\u0010\u0010\u0010 P\u0010 18.37 PP PP 81.63 P t=0 t=1 t=2 t=3 107 \u0010\u0010 \u0010 2 Pricing American Options 0 122.50 \u0010 \u0010 0 \u0010\u0010 114.49 \u0010 \u0010 \u0010PP 0 1.26 \u0010\u0010 PP 107 107 \u0010\u0010 P P \u0010 2.87 \u0010 PPP 3.82 \u0010\u0010 \u0010\u0010 PP \u0010\u0010\u0010 100 100 \u0010 \u0010 \u0010 P 7.13 PP 6.54 PP 93.46 \u0010\u0010\u0010 PPP 93.46 \u0010 PP PP \u0010 12.66 \u0010 PP \u0010 PP 87.34 \u0010\u0010 \u0010 P\u0010 PP PP 18.37 P81.63 P t=0 t=1 t=2 t=3 Price option by working backwards in binomial the lattice. e.g. 12.66 = max 12.66, 1 (q 6.54 + (1 q) 18.37) R 3 A Simple Die-Throwing Game Consider the following game: 1. You can throw a fair 6-sided die up to a maximum of three times. 2. After any throw, you can choose to 'stop' and obtain an amount of money equal to the value you threw. e.g. if 4 thrown on second throw and choose to 'stop', then obtain $4. Question: If you are risk-neutral, how much would you pay to play this game? Solution: Work backwards, starting with last possible throw: 1. You have just 1 throw left so fair value is 3.5. 2. You have 2 throws left so must gure out a strategy determining what to do after 1st throw. We nd fair value = 1 (4 + 5 + 6) + 1 3.5 = 4.25. 6 2 3. Suppose you are allowed 3 throws. Then ... Question: What if you could throw the die 1000 times? 4 Financial Engineering & Risk Management Replicating Strategies in the Binomial Model M. Haugh G. Iyengar Department of Industrial Engineering and Operations Research Columbia University Trading Strategies in the Binomial Model Let St denote the stock price at time t. Let Bt denote the value of the cash-account at time t - assume without any loss of generality that B0 = 1 so that Bt = Rt - so now explicitly viewing the cash account as a security. Let xt denote # of shares held between times t 1 and t for t = 1, . . . , n. Let yt denote # of units of cash account held between times t 1 and t for t = 1, . . . , n. Then t := (xt , yt ) is the portfolio held: (i) immediately after trading at time t 1 so it is known at time t 1 (ii) and immediately before trading at time t. t is also a random process and in particular, a trading strategy. 2 Trading Strategies in the Binomial Model 3 \u0010\u0010u S0 \u0010\u0010 \u0010\u0010 \u0010\u0010PPP u 2 S0 PP \u0010\u0010 P \u0010\u0010 \u0010PP \u0010\u0010 0 \u0010 uS0 uS PP \u0010\u0010 \u0010\u0010 PP\u0010\u0010 \u0010\u0010 \u0010P P S0 PP \u0010\u0010 S0 PPP PP \u0010\u0010 PP P\u0010 P \u0010\u0010dS0 dS0 PP \u0010\u0010 PP P\u0010\u0010 PP d 2 S0 P P PP 3 d S0 t=0 t=1 t=2 t=3 3 Self-Financing Trading Strategies Denition. The value process, Vt (), associated with a trading strategy, t = (xt , yt ), is dened by x1 S0 + y1 B0 for t = 0 Vt = xt St + yt Bt for t 1. (3) Denition. A self-nancing trading strategy is a trading strategy, t = (xt , yt ), where changes in Vt are due entirely to trading gains or losses, rather than the addition or withdrawal of cash funds. In particular, a self-nancing strategy satises Vt = xt+1 St + yt+1 Bt , t = 1, . . . , n 1. (4) The denition states that the value of a self-nancing portfolio just before trading is equal to the value of the portfolio just after trading - so no funds have been deposited or withdrawn. 4 Self-Financing Trading Strategies Proposition. If a trading strategy, t , is self-nancing then the corresponding value process, Vt , satises Vt+1 Vt = xt+1 (St+1 St ) + yt+1 (Bt+1 Bt ) so that changes in portfolio value can only be due to capital gains or losses and not the injection or withdrawal of funds. Proof. For t 1 we have Vt+1 Vt = (xt+1 St+1 + yt+1 Bt+1 ) (xt+1 St + yt+1 Bt ) = xt+1 (St+1 St ) + yt+1 (Bt+1 Bt ) and for t = 0 we have V1 V0 = (x1 S1 + y1 B1 ) (x1 S0 + y1 B0 ) = x1 (S1 S0 ) + y1 (B1 B0 ). 5 Risk-Neutral Price Price of Replicating Strategy We have seen how to price derivative securities in the binomial model. The key to this was the use of the 1-period risk neutral probabilities. But we rst priced options in 1-period models using a replicating portfolio - and we did this without needing to dene risk-neutral probabilities. In the multi-period model we can do the same, i.e., can construct a self-nancing trading strategy that replicates the payo of the option - this is called dynamic replication. The initial cost of this replicating strategy must equal the value of the option - otherwise there's an arbitrage opportunity. The dynamic replication price is of course equal to the price obtained from using the risk-neutral probabilities and working backwards in the lattice. And at any node, the value of the option is equal to the value of the replicating portfolio at that node. 6 The Replicating Strategy For Our European Option Key: Replicating strategy [xt , yt ] Option price Ct Stock price St 22.5 122.50 [1, 97.06] \u0010\u0010 \u0010\u0010 15.48 \u0010 114.49 \u0010 \u0010 \u0010\u0010PPP 7 10.23 \u0010\u0010\u0010 PP PP 107 107 \u0010 \u0010 [.52, 46.89] \u0010PP \u0010 PP \u0010\u0010 \u0010\u0010 6.57 PP 3.86 \u0010\u0010\u0010 \u0010\u0010 100 100 \u0010 P\u0010 \u0010 PP \u0010\u0010PPP PP 2.13 0 \u0010\u0010 PP PP93.46 \u0010\u0010 93.46 PP P\u0010 PP \u0010 PP \u0010\u0010 0 PP 87.34 \u0010\u0010 P\u0010\u0010 P [0, 0] PPP PP 0 81.63 P t=0 t=1 t=2 t=3 7 The Replicating Strategy For Our European Option Key: Replicating strategy [xt , yt ] Option price Ct Stock price St 22.5 122.50 [1, 97.06] \u0010\u0010 \u0010\u0010 15.48 \u0010 114.49 \u0010 \u0010 [.802, 74.84] \u0010\u0010PPP 7 10.23 \u0010\u0010\u0010 PP PP 107 107 \u0010 \u0010 [.517, 46.89] \u0010PP \u0010 [.598, 53.25] PP \u0010\u0010 \u0010\u0010 6.57 PP 3.86 \u0010\u0010\u0010 \u0010\u0010 100 100 \u0010 P\u0010 \u0010 PP \u0010\u0010PPP PP 2.13 0 \u0010\u0010 PP PP93.46 \u0010\u0010 93.46 PP P\u0010 PP \u0010 [.305, 26.11] \u0010\u0010 PP 0 PP 87.34 \u0010\u0010 P\u0010\u0010 P [0, 0] PPP PP 0 81.63 P t=0 t=1 t=2 t=3 e.g. .802 107 + (74.84) 1.01 = 10.23 at upper node at time t = 1 8 Financial Engineering & Risk Management Including Dividends M. Haugh G. Iyengar Department of Industrial Engineering and Operations Research Columbia University Including Dividends C1 (S1 ) S0 \u0010 \u0010\u0010h ahh a \u0010\u0010 uS0 + cS0 Cu p\u0010\u0010 \u0010 \u0010\u0010 hhhh 1p t=0 hhha dS0 + cS0 Cd t=1 Consider again 1-period model and assume stock pays a proportional dividend of cS0 at t = 1. No-arbitrage conditions are now d + c