Partial Differential Equations The Black-Scholes-Merton (BSM or just BS) PDE for the value Ct, S) of a derivative security as a function of time t and underlying asset price level S(t) = S takes the form: ac 028 ac t + (ry)sas += sas2 = rC, where o is the logarithmic) volatility of the asset price S(t), r is the risk-free interest rate, and y is the dividend or convenience yield of S. This PDE, when combined with a terminal payoff condition at some maturity time T, suffices to determine the derivative's value Ct, S) Vt 0. For example, the payoff of a European call option is: C(t=T, S)= max(S(T) - K,0) = S(T) V0= (S(T) - K]+, where K is called the exercise price or strike/striking price. Remember that the BSM PDE is a so-called backward equation: we work backward from time t, when the payoff is known with certainty as a function of S to today (time t), incorporating the uncertainty due to the stochastic dynamics of S along the way. To put these equations into a more recognizable (Cauchy problem) form, we change to the new time variable T = T-t. Then: ac 02 20- C ac rzs asa + (r y)sac-rc, C(r=0, S)= max [S - K,0) = SVO= [S - K], where we can now interpret the payoff formula as an initial condition. (a) A common procedure for modelling problems like this one is to non-dimensionalize the variables, grouping them together with related parameters to eliminate any de pendence on specific choices of measurement units. Thereby, we are able to reflect the greatest possible generality of the solution. For example, the only fixed asset price is the strike price K, which we therefore use to scale all other asset prices, defining: = f and $ = Similarly, the variance rate ois a natural scale for time, suggesting that we define: *=or, = 2, and y = Use these definitions to rewrite the BSM PDE and payoff condition in terms of C, S, , r, and y. Given that these are all dimensionless variables or parameters while T was originally expressed in time units, what does this imply about the original) units of r, y, and o? (b) As we have previously seen, a change to log co-ordinates converts an equi-dimensional (in S) differential equation like this one into a constant-coefficients equation (at least in terms of S or the variable which replaces it). Use the definition r = log(S) to write your non-dimensional PDE and payoff condition in terms of C(1,). (c) From here, our objective is to reduce the PDE to a form as close to the "heat equation" as possible. Begin by making the substitution: C(1,2)= e-PD(1,1). Can you find a value of p that eliminates the term proportional to fC in your PDE? What effect does this substitution have on your payoffinitial condition? Given that the PDE and payoff represent only a single, random cash flow at time T, can you use what you learned about bond calculations and discounting to interpret this substitu- tion (and D and p) in terms of a discounted expectation of the random payoff? What represents the expectation and what does the discounting? aD (d) Now, we want to eliminate the term proportional to Define a new "spatial" variable z = 1+ it and substitute to find a PDE for D(1, ) instead of D(1,2). (If it helps avoid confusion, feel free to define a proxy G(t, ) and only set G(,)= D(1, 2) at the end of the calculation). What choice of j causes the term proportional to t o vanish? How do you interpret this substitution? What effect does it have on the initial condition? At this point, you should have a PDE for D(1, ) that closely resembles the "heat" equa- tion, with the exception of a simple coefficient multiplying the spatial second partial deriva- tive. Now, we can work on incorporating the payoff condition. (e) From the preceding analysis in this problem as well as in problem 2, it should be clear that the PDE for D(1, 2) represents an expectation for a standard Brownian motion, 2(1) W(+), albeit with some non-trivial initial condition. For example, if z(f=0)=zo, then z() ~ N(20,t). Verify that the probability density D(1, 2; 2o) - 9 A ) exp(-(2 - 0)/21) 211 PDE and that this approaches D( 0,2) = 5(2 Zo) as 10. This is the Green's function or "fundamental solution" to your PDE. (f) Next, make use of the superposition principle and the properties of the delta function to show that the solution to your PDE for an arbitrary continuous, integrable initial condition D(1=0,2)= D. (2) can be written as: D(1,2)= |dzo Dolzo)(7,2; zo), where 20 is a dummy integration variable representing values of z at t=0. (g) Substitute the initial condition you have derived for D(f =0,2) = Do() along with the Green's function D(1, 2; zo) into the integral representation of D(1, 2) and evaluate the integral(s). You may find it useful to apply the formula: /* are }(2+hoto - petroleon (var + ) T e -1/2 where N (X) =/ dx6 is the cumulative distribution function (CDF) for the is the cumulative distribution 21 -20 standard normal distribution. (h) Finally, reverse your steps in parts (d) to (a), substituting to find a formula for the value of a European call option, C(T-t, S;K). If you have successfully made it this far, congratulations! You have reproduced the Nobel-prize winning work of Black, Scholes, and Merton. Partial Differential Equations The Black-Scholes-Merton (BSM or just BS) PDE for the value Ct, S) of a derivative security as a function of time t and underlying asset price level S(t) = S takes the form: ac 028 ac t + (ry)sas += sas2 = rC, where o is the logarithmic) volatility of the asset price S(t), r is the risk-free interest rate, and y is the dividend or convenience yield of S. This PDE, when combined with a terminal payoff condition at some maturity time T, suffices to determine the derivative's value Ct, S) Vt 0. For example, the payoff of a European call option is: C(t=T, S)= max(S(T) - K,0) = S(T) V0= (S(T) - K]+, where K is called the exercise price or strike/striking price. Remember that the BSM PDE is a so-called backward equation: we work backward from time t, when the payoff is known with certainty as a function of S to today (time t), incorporating the uncertainty due to the stochastic dynamics of S along the way. To put these equations into a more recognizable (Cauchy problem) form, we change to the new time variable T = T-t. Then: ac 02 20- C ac rzs asa + (r y)sac-rc, C(r=0, S)= max [S - K,0) = SVO= [S - K], where we can now interpret the payoff formula as an initial condition. (a) A common procedure for modelling problems like this one is to non-dimensionalize the variables, grouping them together with related parameters to eliminate any de pendence on specific choices of measurement units. Thereby, we are able to reflect the greatest possible generality of the solution. For example, the only fixed asset price is the strike price K, which we therefore use to scale all other asset prices, defining: = f and $ = Similarly, the variance rate ois a natural scale for time, suggesting that we define: *=or, = 2, and y = Use these definitions to rewrite the BSM PDE and payoff condition in terms of C, S, , r, and y. Given that these are all dimensionless variables or parameters while T was originally expressed in time units, what does this imply about the original) units of r, y, and o? (b) As we have previously seen, a change to log co-ordinates converts an equi-dimensional (in S) differential equation like this one into a constant-coefficients equation (at least in terms of S or the variable which replaces it). Use the definition r = log(S) to write your non-dimensional PDE and payoff condition in terms of C(1,). (c) From here, our objective is to reduce the PDE to a form as close to the "heat equation" as possible. Begin by making the substitution: C(1,2)= e-PD(1,1). Can you find a value of p that eliminates the term proportional to fC in your PDE? What effect does this substitution have on your payoffinitial condition? Given that the PDE and payoff represent only a single, random cash flow at time T, can you use what you learned about bond calculations and discounting to interpret this substitu- tion (and D and p) in terms of a discounted expectation of the random payoff? What represents the expectation and what does the discounting? aD (d) Now, we want to eliminate the term proportional to Define a new "spatial" variable z = 1+ it and substitute to find a PDE for D(1, ) instead of D(1,2). (If it helps avoid confusion, feel free to define a proxy G(t, ) and only set G(,)= D(1, 2) at the end of the calculation). What choice of j causes the term proportional to t o vanish? How do you interpret this substitution? What effect does it have on the initial condition? At this point, you should have a PDE for D(1, ) that closely resembles the "heat" equa- tion, with the exception of a simple coefficient multiplying the spatial second partial deriva- tive. Now, we can work on incorporating the payoff condition. (e) From the preceding analysis in this problem as well as in problem 2, it should be clear that the PDE for D(1, 2) represents an expectation for a standard Brownian motion, 2(1) W(+), albeit with some non-trivial initial condition. For example, if z(f=0)=zo, then z() ~ N(20,t). Verify that the probability density D(1, 2; 2o) - 9 A ) exp(-(2 - 0)/21) 211 PDE and that this approaches D( 0,2) = 5(2 Zo) as 10. This is the Green's function or "fundamental solution" to your PDE. (f) Next, make use of the superposition principle and the properties of the delta function to show that the solution to your PDE for an arbitrary continuous, integrable initial condition D(1=0,2)= D. (2) can be written as: D(1,2)= |dzo Dolzo)(7,2; zo), where 20 is a dummy integration variable representing values of z at t=0. (g) Substitute the initial condition you have derived for D(f =0,2) = Do() along with the Green's function D(1, 2; zo) into the integral representation of D(1, 2) and evaluate the integral(s). You may find it useful to apply the formula: /* are }(2+hoto - petroleon (var + ) T e -1/2 where N (X) =/ dx6 is the cumulative distribution function (CDF) for the is the cumulative distribution 21 -20 standard normal distribution. (h) Finally, reverse your steps in parts (d) to (a), substituting to find a formula for the value of a European call option, C(T-t, S;K). If you have successfully made it this far, congratulations! You have reproduced the Nobel-prize winning work of Black, Scholes, and Merton