The Black-Scholes Equation using Heat Equation

Transcription

1 The Black-Scholes Equation using Heat Equation Peter Cassar May 0, 05 Assumptions of the Black-Scholes Model We have a risk free asset given by the price process, dbt = rbt The asset price follows a geometric Brownian motion, ds t = αs t dt + σs t dw t. Trading can take place continuously without any transaction costs. Short Selling is permitted. The assets are perfectly divisible. The continuously compounded risk-free interest rate is constant. Investors can borrow or lend at the same risk-free rate of interest. There are no arbitrage opportunities => All risk-free portfolios must have the same return. Set Up We are interested in deriving the Black-Scholes PDE for a simple T-claim where the underlying stock has a constant dividend yield. Our stock process will be given by the following stochastic differential equation : ds t = µ δ S t dt + σs t dw t S 0 = s

2 Where µ is the mean return of our stock S t, δ is the continuous dividend rate σ is the volatility. The dividend process D t will be defined as D t = δs t 3 dd t = δs t dt 4 Lastly we want our portfolio to be entirely self financing deterministic. Since we are operating in an arbitrage free market the return of our portfolio must be equivalent to that of the risk free asset. In other words if we want our portfolio to be risk free it has to satisfy the following differential equation dπ t = rπ t dt 5 3 Constructing a Hedging Portfolio In order to price the T-claim, we will construct a portfolio that would hedge the option exactly. We will do this by : Buy an amount of Stock S t Sell the option The option price will be denoted by the function C t, T, S t, σ, r. To avoid confusion we will denote the pricing function as C the first derivative of the pricing function with respect to time as C t. But it is important to note that neither nor C are constants but continuous processes. Our goal now is to find an appropriate choice for that will make our portfolio deterministic. Our portfolio can be given by: π t = S t C 6 The value of the portfolio will go up or down according to the stock price, the dividends received on the shares owned the value of the option. This give us the stochastic differential equation: dπt = ds t + dd t dc 7 The change in the value of the option, also depends of the movements of the stock price. If we apply Itô s Formula to the option price we will get.

3 dc = C t dt + C s ds t + C ss ds t 8 dc = C t dt + C s S t µ δ dt + σdw t + C ss S t σ dt 9 dc = C t + C s S t µ δ + C ss S t σ dt + C s S t σdw t 0 If we plug everything into equation 7 we get: dπt = ds t + dd t dc dπt = St µ δdt + σdw t + δstdt dc dπt = St µ δ + δ dt + StσdW t dc 3 dπt = St µ δ + δ C t + C s S t µ δ + C ss S t σ dt + Stσ C s dw t Our goal is to remove the stochastic element from our portfolio completely. Hence the choice for is now immediately clear, = C s. Substituting back into our portfolio dynamics we are left with: dπt = 4 C s Stδ C t C ssst σ dt 5 We notice here that, after our substitution to eliminate the stochastic element in our portfolio, we have also eliminated the effect of µ, the mean return of our stock. This means that the mean return of the stock plays no role in our hedging portfolio. By our assumptions for the Black-Scholes model, any risk free portfolio will have the same return r will have dynamics given by equation 5. Using equation 5 6 we get: rπtdt = C s Stδ C t C ssst σ dt 6 rπt = C s Stδ C t C ssst σ 7 r C s St C = C s Stδ C t C ssst σ 8 0 = C t + C ssst σ + C s St r δ rc 9 Hence we have concluded that the arbitrage free price of a simple T-claim must be the solution to the above equation subject to some terminal condition. Which for the European call option will be given by: C t + C ssst σ + C s St r δ rc = 0, t, s [0, T ] 0, CT, s = Φs s 0, 3

4 4 Pricing a European Call Option We will solve the previous PDE analytically by transforming our problem into the heat equation, which has a well know solution. We will being by making this substitution u = e rt C Using the proct rule we can find the derivative of C in term of our new variable u. dc dlt = u te rt + rue rt 0 dc ds = ert u s d C ds = ert u ss Hence we can now derive the PDE that u must satisfy by substituting everything into our PDE for a call option. u t e rt + rue rt + ert u ss St σ + e rt u s St r δ rue rt = 0 3 = u t + u ssst σ + u s St r δ = 0 4 Here we make two further substitutions: y = lns, x = T t 5 Recall S describes a geometric Brownian motion, so lns describes a Brownian motion, hence y should satisfy some sort of diffusion equation. The partial derivatives are given by ds = S u y 6 d u ds = S u y + S u yy 7 dt = u x 8 Hence our PDE becomes u x + = u x + S u y + S u yy St σ + St r δ = 0 9 S u y r δ σ u y + σ u yy 30 Here we will make one last change of variables to transform our problem into the heat equation. Let z = y + r δ σ x, τ = x 3 4

5 Under our new co-ordinate system we have the following relations dy = u z z y + u τ τ y 3 = u z 33 d u dy = d dy u z 34 = u z z z y + u z τ τ y 35 = u z z 36 dx = u z z x + u τ τ x = u τ + u z r δ σ And substituting back to our PDE we get u τ u z r δ σ + u z r δ σ + σ u zz = 0 39 u τ = σ u zz This is one form of the diffusion equation with z, τ [0, T ]. Under the transformation the initial condition is 40 u0, z = e rt Φe z 4 The fundamental solution to this PDE under that initial condition is know is given by G τ z = πσ τ e z σ τ 4 And the solution is given by the convolution Let uτ, z = u0, z G τ z 43 uτ, z = ρ = z ω σ τ = e rt πσ τ e rt Φe ω πσ τ e z ω σ τ dω 44 Φe ω e z ω σ τ dω = dρ = σ dω 47 τ Hence uτ, z = e rt πσ τ = e rt π Φe z σ τρ e ρ σ τdρ 48 Φe z σ τρ e ρ dρ 49 Now recall that for a European Call option, Φx = x K + = max[x K, 0]. And we have 5

6 Φ = 0 = e z σ τρ < K = ρ > z lnk σ τ Φ 0 = e z σ τρ > K = ρ < z lnk σ τ 50 = d 5 Hence uτ, z = e rt π d [ ] e z σ τρ K e ρ e rt d dρ π 0.e ρ dρ 5 = e rt [ ] e z σ τρ K e ρ dρ 53 π d π = e rt = e rt e z π d d = e rt e z e σ τ π e z e ρ +σ τ dρ e rt K π d e ρ dρ 54 e ρ+σ τ σ τ dρ e rt KN[ d ] 55 d e ρ+σ τ dρ e rt KN[ d ] 56 = e rt e z e σ τ N[ d + σ τ] e rt KN[ d ] 57 uτ, z = e rt e z e σ τ N[ d ] e rt KN[ d ] 58 Recalling that Ct, s = e rt ut t, lns + Ct, s = e rt e rt e lns+ r δ σ r δ σ T t our solution becomes T t σ T t e N[ d ] e rt KN[ d ] 59 = se δt t N[ d ] e rt t KN[ d ] 60 Where d d are d = z lnk σ τ = σ T t = σ T t lns + ln s K r δ σ + r δ σ T t lnk T t And d = d + σ T t 64 = σ ln s + r δ + σ T t T t K 65 6