Queens College, CUNY, Department of Computer Science Computational Finance CSCI 365 / 765 Fall 2017 Instructor: Dr. Sateesh Mane.

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1 Queens College, CUNY, Department of Computer Science Computational Finance CSCI 365 / 765 Fall 217 Instructor: Dr. Sateesh Mane c Sateesh R. Mane Lecture 13 November 15, 217 Derivation of the Black-Scholes-Merton formula In this lecture we derive the Black-Scholes-Merton formula for European call and put options. We do it by solving the Black-Scholes-Merton partial differential equation. We also derive the probability density function for geometric Brownian motion. 1

2 13.1 Black-Scholes-Merton equation First recall the Black-Scholes-Merton equation (which includes continuous dividends) V t σ2 S 2 2 V V + (r q)s S2 S rv =. (13.1.1) For a European call, the terminal condition on V at time T is V call (S T, T ) = max(s T K, ). (13.1.2) For a European put, the terminal condition on V at time T is V put (S T, T ) = max(k S T, ). (13.1.3) We solve eq. (13.1.1) by integrating backwards in time from expiration t = T to today t = t. Actually we can just solve eq. (13.1.1) by integrating backwards in time from expiration to an arbitrary value of the time t. 2

3 13.2 Change of variables We can solve eq. (13.1.1) as is but it is simpler to change variable to x = ln(s/k). Note that < S < so < x <. It is also convenient to define τ as the time to expiration τ = T t. Then τ = at expiration and we solve to obtain the solution for τ >. Define U(x, τ) = V (S, t) and solve for U. With this change of variables, eq. (13.1.1) is transformed to U τ σ2 2 U x 2 + (r q 1 2 σ2 ) U x ru =. (13.2.1) Then eq. (13.2.1) is a linear partial differential equation with constant coefficients. This makes eq. (13.2.1) simpler to solve than eq. (13.1.1). For a European call, the terminal condition on U (i.e. τ = ) is (put y = x T ) U call (y, ) = max(ke y K, ) = K max(e y 1, ). (13.2.2) Hence the domain of integration at expiration is y <. For a European put, the terminal condition on U is U put (y, ) = max(k Ke y, ) = K max(1 e y, ). (13.2.3) Hence the domain of integration at expiration is < y. 3

4 13.3 Discount factor There are many ways to solve eq. (13.2.1). First let us factor out the discount factor e r(t t) = e rτ. Let us express U(x, t) in the form Substitution into eq. (13.2.1) yields the following U(x, τ) = G(x, τ) e rτ. (13.3.1) G τ e rτ + rg e rτ σ2 2 G x 2 e rτ + (r q 1 2 σ2 ) G x e rτ rg e rτ =. (13.3.2) Simplifying and factoring out e rτ (which never equals zero) yields the following G τ = 1 2 σ2 2 G x 2 + (r q 1 2 σ2 ) G x. (13.3.3) Then eq. (13.3.3) has the form of a diffusion equation with a drift term. Technically, eq. (13.3.3) is a diffusion-convection equation. Mathematically, eq. (13.3.3) is a parabolic-hyperbolic partial differential equation. We may or may not have time to discuss what that means, in these lectures. 4

5 13.4 Green s function Let us try the following solution for eq. (13.3.3), including y = x T as a parameter: G(x, τ; y) = e (x y+(r q σ2 /2)τ) 2 /(2σ 2 τ) 2σ 2 τ. (13.4.1) The partial derivative G/ x is given by The partial derivative 2 G/ x 2 is given by G x = x y + (r q 1 2 σ2 )τ σ 2 G. (13.4.2) τ 2 G x 2 = G σ 2 τ x y + (r q 1 2 σ2 )τ σ 2 τ G x = G σ 2 τ + (x y + (r q 1 2 σ2 )τ) 2 σ 4 τ 2 G. (13.4.3) The partial derivative G/ τ is given by Substituting into eq. (13.3.3) yields G τ = G 2τ + (x y + (r q 1 2 σ2 )τ) 2 2σ 2 τ 2 G (r q 1 2 σ2 ) x y + (r q 1 2 σ2 )τ σ 2 τ G. (13.4.4) 1 2 σ2 2 G x 2 + (r q 1 2 σ2 ) G x = G 2τ + (x y + (r q 1 2 σ2 )τ) 2 2σ 2 τ 2 G = G τ. Hence G(x, τ; y) in eq. (13.4.1) is a solution of eq. (13.3.3). The function G(x, τ; y) is called a Green s function. (r q 1 2 σ2 ) x y + (r q 1 2 σ2 )τ σ 2 τ G (13.4.5) Grammatically, it should really be called a Green function (named for George Green). However everyone writes Green s function and ignores grammar. The function G(x, τ; y) is the solution of eq. (13.3.3) with the terminal boundary condition Here δ(x y) is the Dirac delta function. lim G(x, τ; y) = δ(x y). (13.4.6) τ 5

6 13.5 General solution for arbitrary terminal payoff Let the terminal payoff function be P(y). The solution of eq. (13.2.1) with terminal payoff P(y) is V (S, t) = U(x, τ) = e rτ G(x, τ; y) P(y) dy. (13.5.1) Note that eq. (13.5.1) is the general solution of the Black-Scholes-Merton equation eq. (13.2.1), for the terminal payoff function P(y), where y = ln(s T /K). The structure of eq. (13.5.1) is a discount factor e rτ = e r(t t) multiplying the expectation value of the terminal payoff. For a European call option, the terminal payoff function is P call (y) = max(s T K, ) = max(ke y K, ) = K max(e y 1, ). (13.5.2) This is nonzero in the interval y <. For a European put option, the terminal payoff function is P put (y) = max(k S T, ) = max(k Ke y, ) = K max(1 e y, ). (13.5.3) This is nonzero in the interval < y. 6

7 13.6 Geometric Brownian Motion: probability density function The Green s function G(x, τ; y) in eq. (13.4.1) is actually a probability density function, but for standard Brownian motion (with a drift term). Let us derive the probability density function for geometric Brownian motion. Hence we want a function in terms of S and t (with parameters S T and T ). Define the function p GBM (S, t; S T, T ) via Then dx/ds = 1/S, hence we obtain p GBM (S, t; S T, T ) = G(x, τ; y) dx ds. (13.6.1) p GBM = 1 S e (ln(s/k) ln(s T /K)+(r q σ 2 /2)(T t)) 2 /(2σ 2 (T t)) 2σ 2 (T t). (13.6.2) The probability density function for geometric Brownian motion is given by p GBM. The function p GBM in eq. (13.6.2) is the probability density that, if the stock price is S at the time t, the random walk of the stock price will have the value S T at the future time T, where T > t. 7

8 13.7 Solution for call option using terminal boundary condition The solution of eq. (13.2.1) which satisfies the terminal boundary condition is given by U(x, τ) = e rτ G(x, τ; y)p call (y) dy = Ke rτ G(x, τ; y)(e y 1) dy = Ke rτ G(x, τ; y)e y dy Ke rτ G(x, τ; y) dy. Hence we must evaluate the following integrals I 1 = G(x, τ; y)e y dy, I 2 = (13.7.1) G(x, τ; y) dy. (13.7.2) We recognize the second integral I 2 immediately as a cumulative Normal distribution: e (x y+(r q σ2 /2)τ) 2 /(2σ 2 τ) G(x, τ; y) dy = dy 2σ 2 τ ( x + (r q 1 2 = N σ2 )τ ) σ τ = N(d 2 ). Recall the definitions of d 1 and d 2 from previous lectures (and recall x = ln(s/k)): (13.7.3) d 1 = d 2 = ln(s/k) + (r q)(t t) σ T t σ T t, (13.7.4a) ln(s/k) + (r q)(t t) σ 1 2 T t σ T t = d 1 σ T t. (13.7.4b) To evaluate I 1 we require the following identity Ke rτ e (x y+(r q σ2 /2)τ) 2 /(2σ 2 τ) e y = Se qτ e (x y+(r q+σ2 /2)τ) 2 /(2σ 2 τ). (13.7.5) The proof of eq. (13.7.5) consists basically of completing the square in the exponent. Then the integral for I 1 also yields a cumulative Normal distribution: Ke rτ I 1 = Ke rτ G(x, τ; y)e y dy = Se qτ e (x y+(r q+σ2 /2)τ) 2 /(2σ 2 τ) 2σ 2 τ ( x + (r q + = Se qτ 1 2 N σ2 )τ ) σ τ = Se qτ N(d 1 ). dy (13.7.6) Summing the integrals yields the Black-Scholes-Merton formula for a European call option. 8

9 13.8 Solution for put option using terminal boundary condition The discount factor and the Green s function is the same for a European put option. Only the terminal boundary condition (terminal payoff) is different. The solution of eq. (13.2.1) which satisfies the terminal boundary condition for a put is U(x, τ) = e rτ G(x, τ; y)p put (y) dy = Ke rτ G(x, τ; y)(1 e y ) dy = Ke rτ G(x, τ; y) dy Ke rτ G(x, τ; y)e y dy. (13.8.1) Hence we must evaluate the following integrals I 3 = G(x, τ; y) dy, I 4 = The evaluation of these integrals is similar to that for the call option. G(x, τ; y)e y dy. (13.8.2) The results are I 3 = N( d 2 ), Ke rτ I 4 = Se qτ N( d 1 ). (13.8.3) Summing the integrals yields the Black-Scholes-Merton formula for a European put option. 9

10 13.9 Black-Scholes-Merton formula The Black-Scholes-Merton formulas for a European call and put option are respectively c(s, t) = Se q(t t) N(d 1 ) Ke r(t t) N(d 2 ), p(s, t) = Ke r(t t) N( d 2 ) Se q(t t) N( d 1 ). (13.9.1a) (13.9.1b) The definitions of d 1 and d 2 were displayed above, but for ease of reference they are d 1 = d 2 = ln(s/k) + (r q)(t t) σ T t σ T t, (13.9.2a) ln(s/k) + (r q)(t t) σ 1 2 T t σ T t = d 1 σ T t. (13.9.2b) 1