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THE CHINESE UNIVERSITY OF HONG KONG Department of Mathematics MATH4210 (2016/17 Term 1) Financial Mathematics Assignment 3 solution Note: If you have any questions about the solution or assignment score, please let me know by sending an Email to kckchan@math.cuhk.edu.hk 1. ([2;p55]) Draw the expiry payoff diagrams for each of the following portfolios: (a) Short one share, long two calls with exercise price E (this combination is called a straddle); (b) Long one call and one put, both with exercise price E (this is also a straddle: why?); (c) Long one call and two puts, all with exercise price E (a strip); (d) Long one put and two calls, all with exercise price E (a strap); (e) Long one call with exercise price E1 and one put with exercise E2. Compare the three cases E1 > E2 (known as a strangle), E1 = E2, E1 < E2. (f) As (e) but also short one call and one put with exercise price E (when E1 < E < E2, this is called a butterfly spread). Solution: { −S S ≤ E (a) Λ = S − 2E S > E (b) Λ = |E − S| 1
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E2 − S S ≤ E2 (e) (i) If E1 > E2, Λ = 0 E2 < S < E1 S − E1 S > E1 (ii) If E1 = E2 = E, Λ = |E − S| 3
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E2 − S S ≤ E1 (iii) If E1 > E2, Λ = E2 − E1 E2 < S < E1 S − E1 S > E2 (f) We have Λ = max(S − E1,0) + max(E2 − S,0) − max(S − E,0) − max(E − S,0) Here we have 9 cases: (a) E1 > E2 > E (b) E1 > E2 = E (c) E1 > E > E2 (d) E2 > E1 > E (e) E2 > E1 = E 4
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(f) E2 > E > E1 (g) E > E2 > E1 (h) E > E2 = E1 (i) E > E1 > E2 By explicitly writing down the equation, we can obtain the 9 graphs. Remark: When you are asked to draw a diagram, please don’t just show me the line without indicating any information. You have to do more than stating the equation of the graph. Please state additional information, such as x-intercept, y-intercept and slope, to let people know what you are drawing. This time marks are given to you, but please make sure you draw a graph with sufficient information provided in exams. 2. Derive the price formula of an European put based on the Black-Scholes model. Solution: Observe that the rescaled payoff function of an European option is Λ(z) = max{1 − ez,0} So we have ∫ 0 1 σ2 2 2 −(x−z+(r− 2 )τ ) v(x, τ) = e−rτ √ e 2 (1 − ez) dz 2σ τ −∞ 2πσ2τ By direct integration, we get p(S, t) = Ee−r(T−t)N(−d ) − SN(−d ) 2 1 3. Show that the payoff function of a portfolio c−p is S −E. From this and the Black-Scholes formula, show the formula of the put-call parity. Solution: Π = max{S − E, 0} − max{E − S, 0} = S − E Next consider the rescaled payoff function of the above portfolio: Λ(z) = ez − 1 By using equation 3.15 in lecture notes, we have ∫ ∞ 2 1 (x−z+(r− σ )τ2)2 v(x, τ) = e−rτ √ e− 2 (ez − 1) dz 2σ2τ −∞ 2πσ2τ By direct integration and change of variable, we get c(s, t) − p(s, t) = S(t) − Ee−r(T−t) which is the put-call parity. 4. ([6;p56]) Find the most general solution of the Black-Scholes equation that has the special form (a) V = V (S) (b) V = A(t)B(S) Solution: 5
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∂V (a) Note that = 0. So we obtain the following ODE: ∂t 1 d2S dV σ2S + rS − rV = 0 2 dS2 dS The above equation has solutions of the form V (S) = Sp for some p, we will then solve for the p. By putting V = Sp, we have 1 2 σ p(p01) + rp − r = 0 2 −2r which gives p = 1 or p = . Hence the solution is a linear combination of these σ2 two. i.e. V (S) = AS + BS(−2r)/σ2 where A and B are constants, which could be determined if initial conditions are given. (b) By simple differentiation we could obtain the following equation: A′(t) 1 B′′(S) B′(S) − = σ2S2 + rS − r A(t) 2 B(S) B(S) Since left hand side depends only on t and right hand side depends only on S, we then have A′(t) 1 B′′(S) B′(S) − = λ = σ2S2 + rS − r A(t) 2 B(S) B(S) where λ is a constant. For the t-related ODE, we have lnA(t) = −λt + C′ and A(t) = Ce−λt where C is a constant. For the S-related ODE, using similar idea as (a), we have √ −(r − σ2 ) ± (r + σ2 )2 + 2σ2λ p = 2 2 ± σ2 Hence B(S) = Ae(p+)S + Be(p−)S and V (S) = A(t)B(S) = e−λt(Ae(p+)S + Be(p−)S) 5. ([9;p57]) Suppose that a share price S is currently $100, and that tomorrow it will be either $101, with probability p, or $99 , with probability 1 − p. A call option, with value c, has exercise price $100. (a) Set up a Black-Scholes hedged portfolio and hence find the value of c. (Ignore interest rates.) (b) Now repeat the calculation for a cash-or-nothing call option with payoff $100 if the final asset price is above $100, zero otherwise. What difference do you notice? Solution: (a) Construst the following portfolio: Π = c − ∆S. Observe that { 1 − 101∆ with probaility p dΠ = −99∆ with probaility 1 − p 6
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and hence we should have 1 − 101∆ = 0 = −99∆ Solving gives ∆ = 0.5 and Π = −49.5 tomorrow. Ignoring interest rate we have −49.5 = c − 0.5(100) and hence c = $0.5. (b) Using the same construction as (a), we have { 100 − 101∆ with probaility p dΠ = −99∆ with probaility 1 − p Solving gives ∆ = 50 and Π = −4950. Ignoring interet rate we have −4950 = c − 50(100) and hence c = $50. 6. ([5;p86])Suppose that in the Black-Scholes equation, r(t) and (t) are both non-constant but known functions of t. Show that the the following procedure reduces the Black-Scholes equation to the diffusion equation. Solution: (a) Observe that ∂ ∂t′ ∂ ∂ = = − ∂t ∂t ∂t′ ∂t′ ∂ 1 ∂ = ∂S Eex ∂x ∂2 e−2x ∂2 e−2x ∂ = − ∂S2 E2 ∂x2 E2 ∂x Result follows by subsituting the above formulas into the Black-Scholes equation (b) Note that dτˆ = 1σ2(t′)dt′, hence, 2 1 ∂v 1 ∂2v 1 ∂v σ2(t′) = σ2(t′) + (r(t′) − σ2(t′)) − r(t′)v 2 ∂τˆ 2 ∂x2 2 ∂x and result directly follows by dividing the above equation by 1σ2(t′). 2 (c) By direct differentiation, we have ∂v ′ −B(ˆτ) −B(ˆτ) = F (x + A(ˆτ)) · a(ˆτ) · e − F(x + A(ˆτ)) · b(ˆτ) · e ∂τˆ ∂v ′ −B(ˆτ) = F (x + A)e ∂x Then we can verify that ∂v ∂v = a(ˆτ) − b(ˆτ)v ∂τˆ ∂x 7
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which confirms that v = F e−B is the general solution. (d) Define v(x,τˆ) = e−B(ˆτ)V (x + A(ˆτ),τˆ), we obtain v = e−B(ˆτ)V x xˆ v = e−B(ˆτ)V xx xˆxˆ v = −b(ˆτ)e−B(ˆτ)V + e−B(ˆτ)V a(ˆτ) + e−B(ˆτ)V τˆ xˆ τˆ Put back into the equation we obtain Vτˆ = Vxˆxˆ, and hence we have the following close form solution: ∫ 1 ∞ V (ˆx,τˆ) = √ V (s)e−(ˆx−s)2/4ˆτds 4πτˆ 0 −∞ and hence ∫ e−B(ˆτ) ∞ v(ˆx,τˆ) = √ V (s)e−(x+A(ˆτ)−s)2/4ˆτds 4πτˆ 0 −∞ (e) They have same initial condition. i.e. V0(s) = v0(s) 7. ([1;p104]) What is the put-call parity relation for options on an asset that pays a constant continuous dividend yield? Solution: Note that both options satisfies ∂V 1 ∂2V ∂V + σ2S2 + (r − D )S − rV = 0 ∂t 2 ∂S2 0 ∂S By linearity, V = P − C also satisfies the above equation, with terminal condition V (S, T) = E − S. In order to make use of BS equation, we let V (S, t) = eD0(T−t)V (S, t) 1 Hence V1 satisfies ∂V 1 ∂2V ∂V + σ2S2 + (r − D )S − (r − D )V = 0 ∂t 2 ∂S2 0 ∂S 0 which gives V (S, t) = Ee−(r−D0)(T−t) − S and P − C = Ee−r(T−t) − Se−D0(T−t) 1 9. ([10;p105])Derive the put-call parity result for the forward/futures price in the form C − P = (F − E)e−r(T−t) What is the corresponding version when the asset pays a constant continuous dividend yield? Solution: Consider two portfolio given by Λ (F, t) = C(F, t) + Ee−r(T−t) and Λ (F, t) = P(F, t) + F e−r(T−t) 1 2 8
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At t = T, we have Λ1(F, T) = max{F − E,0} + E = max{F, E} = Λ2(F, T) By no arbitrage assumption, we have C − P = (F − E)e−r(T−t) For the dividend case, using simliar idea we can obtain C − P = (F − E)e−(r−D0)(T−t) 10. ([11;p105])What is the forward price for an asset that pays a single dividend dyS(td) at time td? Solution: For T ≥ t ≥ t+, we have V (S, t) = S(t) − F e−r(T−t). By continuity, we d have [ ] F e−r(T−t−) − + d V (S, td ) = V (S(1 − dy), td ) = (1 − dy) S − 1 − dy Assuming zero cost in entering the forward contract, we obtain [ ] F e−r(T−td−) V (S(1 − d ), t+) = (1 − d ) S − = 0 and F = (1 − d )Se−r(T−t) y d y 1 − d y y 9
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