Unformatted text preview: STAT/ACTSC 446/846
Assignment #3 (due November 9, 2007)
Note:When handing in your assignment, please use a cover page showing only your UWID number and section (lecture)
number. Please write your name on the ﬁrst actual page of your assignment. ACTSC/STAT 846 students: please indicate on the cover page that you are a graduate student by writing “846”. Recall that assignments must be handed in at MC 6028 before 3pm. If you want to use Excel for some of the problems (e.g., (4) and (5)), you are welcome to do so. In that case, hand in a copy of your spreadsheet and explain which approach/formulas you used. 1. (This is a variant on Exercise 5.5 of Financial Economics (FE).) A one-period arbitrage-free model has two assets with the following price evolution: for asset 1, S1 (0) = 10.8 and S1 (1, ω1 ) = 12, while S1 (1, ω2 ) = 8; for asset 2, S2 (0) = 10 and S2 (1, ω1 ) = a, where a > 0 is a constant to be determined, while S2 (1, ω2 ) = 3. In addition, you are told that a call option on asset 1 with a strike price of 9 has a price of 2.1. a) Compute the state-price vector implicit in this model and determine a. b) Show that the implied interest rate for this model is 0 (hint: construct a riskless portfolio whose payoﬀ is always 1 and ﬁnd its time 0 value). c) Determine the risk-neutral probabilities for this model. d) The price of a call option on asset 2 is 2.8. What is the strike price of this option? ———————————————————————————————————————————Answer: 12 3 a) We use the asset 1 and the call on asset 1. So that S (0) = [10.8 2.1]. S (1, Ω) = . We 80 want ψ such that S (0) = ψS (1, Ω). So then ψ = S (0)[S (1, Ω)]−1 = [10.8 2.1] . [0.7 0.3]. Then we can ﬁnd a so that 10 = 0.7 × a + 0.3 × 3 that is a = 13. b) Let’s ﬁnd θ1 , θ2 so that 1 12 3 θ1 = 1 θ2 80 After solving the two equations, one ﬁnds that θ1 = 1/8 and θ2 = −1/6. The value at time 0 is θ1 × 10.8 − θ2 × 2.1 = 1.35 − 0.35 = 1. So the interest rate must be r = 0%. c)Then we can compute the risk neutral probabilities:Q(ω1 ) = ψ (ω1 )(1 + r) = ψ (ω1 ) = 0.7 and Q(ω2 ) = ψ (ω2 )(1 + r) = ψ (ω2 ) = 0.3. d) The price of a call option on the second asset is 2.8, then that means: 2.8 = ψ1 × (13 − K )+ + ψ2 × (3 − K )+ * In the ﬁrst case, if K < 3 then K could have to satisfy: 2.8 = 0.7(13 − K ) + 0.3(3 − K ) = 10 − K So that K = 7.2 which is in contradiction with K < 3. * In the second case, if K > 13 then we would have a call option with a payoﬀ always equal to 0. Its price will be 0 and not 2.8! * In the third case, 3 ≤ K ≤ 13, so that K satisﬁes: 2.8 = 0.7(13 − K ) 0 −3 −8 12 .
−1 24 = It then implies K= 0.7 × 13 − 2.8 = 13 − 4 = 9 0.7 ———————————————————————————————————————————12 . Determine all state-price vectors for the one1 10 period securities market model. (Exercise 5.9 of FE replacing 0.3 by 3). ———————————————————————————————————————————Answer: We want to ﬁnd all vectors ψ such that S (0) = ψS (1, Ω). Since S (1, Ω) is invertible, there is a unique
−1 2. Assume S (0) = [0.7 , 3.0] and S (1, Ω) = 12 = [ 0.5 0.2]. 1 10 Remark: It proves in particular that this market is arbitrage free (there exists a state-price vector and that it is complete since it is unique. ———————————————————————————————————————————state-price vector. So, then ψ = S (0)[S (1, Ω)]−1 = [0.7 3] . 3. (This is more or less a reformulation of Exercise 5.2 of FE.) Consider a one-period security market model. There are two assets, a bond and a stock. The interest rate is assumed to be 0 so that the bond sells for $1 and pays $1 at the end of the period. The stock ր 30 price is assumed to have the following trinomial structure: 20 → 20 ց 15 a) Is the market arbitrage-free? Justify your answer. b) Is it possible to calculate the exact price of a call option with strike K = $23 by arbitrage arguments? Justify your answer. c) Introduce the Arrow-Debreu security associated with ω1 in this market and determine the range of possible prices for this security that preserve the arbitrage-free condition. d) Use your answer to (c) to derive a lower-bound and an upper-bound for the call option price that preserve the arbitrage-free condition. ———————————————————————————————————————————Answer: a) First Method: The market is arbitrage-free because there is no arbitrage opportunities. If we have θ1 and θ2 such that 20θ1 + θ2 ≤ 0 (which is the value at time 0 of the portfolio) then: (i) if θ1 ≤ 0 then 30θ1 + θ2 < 0 (outcome ω1 ). It can’t be an arbitrage. (ii) if θ1 > 0 then 15θ1 + θ2 < 0 (outcome ω2 ).It can’t be an arbitrage. (iii) if θ1 = 0 then the portfolio at 1 is equal to: which is not > 0 since the condition at 0 is 20θ1 + θ2 ≤ 0, so that θ2 ≤ 0 when θ1 = 0. Second Method: The market is arbitrage-free because there exists one state price vector. You can give at least one. To ﬁnd a state price vector ψ = [ψ1 , ψ2 , ψ3 ]. such that ψ1 + ψ2 + ψ3 = 1 and such that 30ψ1 + 20ψ2 + 15ψ3 = 20. There is one free variable since there are two equations and three unknown variables! Let’s express everything with respect to θ3 . Then ψ = ( 1 θ3 , 1 − 3 θ3 , θ3 ) For instance for θ3 = 1/2, 2 2 then ψ = [1/4 1/4 1/2] is a possible state-price vector (with three positive coordinates). θ2 θ2 θ2 b)If we have a call option with strike 23, then its ﬁnal payoﬀ is: We try to ﬁnd a portfolio (θ1 , θ2 )t , that replicates the ﬁnal payoﬀ of the call option. Then 7 = θ1 +30θ2 , 0 = θ1 + 20θ2 and 0 = θ1 + 15θ2 . There is no solution that satisﬁes the three equations. So that it is impossible to ﬁnd the price of this call option using arbitrage arguments. c) We introduce the ﬁrst arrow Debreu security in the market. It cannot be replicated for the same reason as in b). The possible prices for this Arrow Debreu security is given by the range: L<x<U where L = max(20θ1 + θ2 ) and U = min(20θ1 + θ2 ) s.t. s.t. 30θ1 + θ2 ≤ 1 30θ1 + θ2 ≥ 1 1 2 15θ + θ ≥ 0 1 2 1 2 15θ + θ ≤ 0 1 2 7 0 0 20θ + θ ≤ 0 20θ + θ ≥ 0 The intuition is that if x ≤ L or if x ≥ U then there is a portfolio θ1 , θ2 that dominates the security in every states and creates an arbitrage opportunity. Thus, we must have L < x < U . To solve these two programming problems, you can either do it “by hand” using a graph and represent graphically possible regions of θ1 and θ2 because it is a simple case. Or you can simply use Excel solver. Note that you are not expected to be able to do it in limited time during an exam. One ﬁnd L = 0 and U = 1/3. So that 1 0<x< 3 d) The option is equal to 7 times the Arrow Debreu security so that the range of the price of the option is: 7 0, 3 Note that you can also apply the same techniques as the one described above and solve the two linear programming problems for the new security. ———————————————————————————————————————————4. (This is Exercise 6.5 from FE.) A European derivative pays the square of the asset price in 3 months. The current price of the underlying asset is $ 20, and the asset price has a volatility σ of 15% per year. The continuously compounded interest rate is 7% per year. Use the binomial model with three time intervals to calculate the value of this derivative. Determine the self-ﬁnancing portfolio for the ﬁrst time interval and an estimate of the delta. (Use u = exp(σ T /3) and d = 1/u.) ———————————————————————————————————————————Answer: rh − q = eu−dd = 0.5567. The standard backward recursive method, we get: 409.3 442.9 479.3 518.7 372.5 403.1 436.2 338.97 366.8 308.5 The delta of an option is the diﬀerentiate of the price with respect to the underlying: ∆= ∂V ∂S Here, we are considering a discrete model (binomial model), we can thus only obtain an approximation of the delta: The self-ﬁnancing portfolio is given at 0 by: 409.3 = θ1 S1 (0) + θ2 S2 (0)
442.9−372.5 S1 (0) = 1, S2 (0) = 20 where θ2 = 20.88−19.15 ≈ 40.7 and θ1 ≈ −404. The estimate of the delta corresponds to θ2 = ∆V which is approximately equal to 40.7. ∆S The strategy is given at time 0 and is correct for the ﬁrst period, it replicates the payoﬀ 372.5 (if the underlying goes down) and 442.9 (if the underlying goes up). It was not asked in the exercise, but then you should reorganize the portfolio diﬀerently if you are in the state “up” and in the state “down”. In the binomial model, the portfolio is self-ﬁnancing, at any time you reallocate the portfolio and you neither add any money to the portfolio, nor subtract anything from the portfolio. The replicating portfolio obtained in the binomial model is automatically a self-ﬁnancing strategy. (To see other examples, have a look for instance at the problem 10.6 in assignment 2 and the problem on the equity-linked product in the ﬁrst midterm.) ———————————————————————————————————————————- 5. (This is Exercise 6.9 from FE but with maturity 3 months and three time intervals.) Determine the value of an European put option on the arithmetic average of the price of a stock with current price $20, strike price $20, continuously compounded risk-free rate 7.5% per year, volatility 25% per year, and time to maturity 3 months. The arithmetic average uses the current value plus the value at the end of each month to the date of calculation of the average (sampling frequency is monthly). This means the payoﬀ at the end of 3 months is: Payoﬀ = max 0, K − S (1) + S (2) + S (3) 3 where S (i) denotes the value of the underlying asset at the end of the ith month and K denotes the strike. (It is called an Asian put option). No dividends are due on the stock over the next 3 months. Use a monthly time interval and a binomial model (with u = exp(σ T /3) and d = 1/u). (a) Calculate the price of this European Asian put option. Hint:You need to consider a nonrecombining tree because the option is path-dependent. (b) Calculate the value if the Asian put option were American style.(Note: At the end of the ﬁrst month, the payoﬀ of the American option is max(0, K − S (1)) at the end of the second month, the payoﬀ is max 0, K − S (1)+S (2) .) 2 (c) Calculate the value of the European Asian option if there is a dividend of $1 paid at the end of the ﬁrst month and a dividend of $1 at the end of the second month. (d) Calculate the value of the American Asian option if there is a dividend of $1 paid at the end of the ﬁrst month and a dividend of $1 at the end of the second month. ———————————————————————————————————————————Answer: This exercise is an example where the price could not be studied from the recombining tree since the payoﬀ at the end depends on the path! For a put, it is more interesting to exercise right after the dividends are paid. For a call, it is optimal to exercise right before. The exercise value for the American put option should thus be the exercise value calculated with the stock price right after the payment of dividends. ———————————————————————————————————————————6. Binomial Model with Short-selling Constraints. Assumptions: we assume a binomial model with a riskless asset S1 and a risky asset S2 , with S (0) = 1.1 1.2 [1 , 1], and S (1, Ω) = . 1.1 0.8 a) Assume there is no short selling for the risky asset. That means it is not allowed to have a short position in the risky asset (S2 ). Show that the contingent claim S3 such that S3 (1, ω1 ) = 1.15 and S3 (1, ω2 ) = 1.35 cannot be replicated. b) Is the binomial model with the no-short selling constraint complete? c) Without short-selling constraints, what should be the price of S3 in an arbitrage-free market? Do we have a unique possible price? Why? d) Assume now that it is not possible to take a short position in S2 and in S3 . Let x denote the price of S3 at time 0. (i) Show that the price of S3 should be strictly greater than 115/110 to avoid arbitrage opportunities, that is x > 115/110. Note: (ii) and (iii) are bonus questions for 446 students - Required questions for 846 students. (ii) Show that it is equivalent: • to verify that there is no arbitrage opportunity θ = (θ1 , θ2 , θ3 )T in the market with θ2 ≥ 0 and θ3 ≥ 0. • to verify that it is not possible to ﬁnd θ1 ∈ R and θ2 ≥ 0 such that: θ1 + θ2 + x = 0 and 1.1θ1 + 1.2θ2 + 1.15 1.1θ1 + 0.8θ2 + 1.35 > 0, where the notation > in the second inequality is as seen in class (at least one component is > 0 and all components are ≥ 0). (iii) Show that the market (with short-selling constraints on S2 and S3 ) is arbitrage-free if and only if x > 120/110. (Use (ii) with θ1 = −x − θ2 ). Do we have a unique possible price? Why? ———————————————————————————————————————————Answer: 1.15 θ1 be a replicating portfolio for the asset S3 = , then S (1, Ω)θ = S3 , that is: (a) Let θ = 1.35 θ2 1.1 1.2 1.1 0.8 . θ1 θ2 = 1.1θ1 + 1.2θ2 1.1θ1 + 0.8θ2 = 1.15 1.35 The only solution is θ2 = −0.5 and θ1 = 1.15+0..15×1.2 ≈ 1.59. It is an impossible portfolio because of 1 the short selling constraints. (b) The binomial model with the no-short selling constraint is incomplete because there exists payoﬀs that can’t be replicated. (c) Without short-selling constraints, the price of S3 is the value of its replicating portfolio at time 0: 1θ1 + 1θ2 ≈ 1.09 The price is unique because the binomial model is complete. θ1 (d)-(i) Let θ = θ2 be a portfolio such that θ2 = 0 and θ3 = 1, then θ3 V θ (1) = 1.1 1.2 1.15 1.1 0.8 1.35 .θ = 1.1θ1 + 1.15 1.1θ1 + 1.35 115 If θ1 ≥ −115 , then V θ (1) > 0. (the ﬁrst component can be equal to 0 when θ1 = − 110 but the second 110 component is always positive). So, the price at the beginning of this portfolio should be positive, otherwise it is an arbitrage opportunity since you never lose money at the end. Let us compute the value at time 0: V θ (0) = θ1 + S3 (0) If the market is arbitrage-free, then for all θ1 ≥ −115 110 , θ1 + S3 (0) ≥ 0, so that S3 (0) ≥ 115 110 . (d)-(ii) Assume the ﬁrst part of the equivalence. If there is no arbitrage opportunity θ = (θ1 , θ2 , θ3 )T in the market with θ2 ≥ 0 and θ3 ≥ 0, then by deﬁnition of an arbitrage opportunity, in the special case when θ3 = 1, we ﬁnd the second part of the equivalence. Now let us assume the second part. ∀θ1 ∈ R, ∀θ2 ≥ 0, It is equivalent to have: ˜ ˜ ∀θ1 ∈ R, ∀θ2 ≥ 0, θ3 ≥ 0 1.1θ1 + 1.2θ2 + 1.15 < 0 or 1.1θ1 + 0.8θ2 + 1.35 < 0 , θ1 + θ2 + x = 0 ⇒ or both equal to 0 ˜ ˜ θ1 + θ2 + θ3 x = 0 ⇒ ˜ ˜ 1.1θ1 + 1.2θ2 + 1.15θ3 < 0 ˜ ˜ or 1.1θ1 + 0.8θ2 + 1.35θ3 < 0 , or both equal to 0 ˜ ˜ because it is possible to multiply the previous result by θ3 ≥ 0 (where θ1 = θ3 θ1 and θ2 = θ3 θ2 but they are still arbitrary parameters). This last equation says that there is no arbitrage opportunity (θ1 , θ2 , θ3 )T such that θ2 ≥ 0 and θ3 ≥ 0. We indeed don’t need to have a negative portfolio at the beginning of the period to have an arbitrage opportunity. If there exists an arbitrage opportunity so that V θ (0) ≤ 0 then there exists another arbitrage strategy β such that V β (0) = 0. You remove the money received at 0:V θ (0) and you invest it in a riskless asset. You still have an arbitrage opportunity because the payoﬀ is still positive (it is even greater now since you received the interest of your investment). (d)-(iii) We then use (ii). We want to avoid arbitrage opportunities. To do that, we only need to check ˜ ˜ 1.1θ1 + 1.2θ2 + 1.15θ3 >0 that the condition θ1 + θ2 + θ3 x = 0 with θ2 ≥ 0 and the condition ˜ ˜ 1.1θ1 + 0.8θ2 + 1.35θ3 will never happen at the same time. From the ﬁrst condition θ1 = −x − θ2 . Then, you replace in both equations: v1 v2 = (−x − θ2 )1.1 + 1.2θ2 + 1.15 (−x − θ2 )1.1 + 0.8θ2 + 1.35 One then analyses when each component is positive. v1 ≥ 0 ⇔ x ≤ 1.15 + 0.1θ2 1.1 1.35 − 0.3θ2 1.1 So that, to avoid arbitrage opportunities, v1 and v2 can’t be positive at the same time. −0 One method to solve this question it is to represent the two lines on a same graph (y = 1.351.1.3θ2 and 1.15+0.1θ2 y= ). If x is below the intersection point, it is possible to ﬁnd θ2 ≥ 0 such that both are 1.1 positive. We thus need that x is above the intersection point (obtained at θ2 = 0.5). Then x = 120 . 110 120 The price x such that v1 > 0 and v2 > 0 are not possible should satisfy: x ≥ 110 . Note that when 120 x = 110 , it is impossible to ﬁnd an arbitrage opportunity because both components of the portfolio are zero. ———————————————————————————————————————————v2 ≥ 0 ⇔ x ≤ and ...
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This note was uploaded on 03/02/2011 for the course ACTSC 446 taught by Professor Adam during the Fall '09 term at Waterloo.
- Fall '09