MAFS550 - Computational Methods for Pricing Structured Products Solution to Homework Two Course instructor: Prof YK Kwok 1 Expand f(x 0 ) and f(x 0 x) at x 0 into Taylor series, where f(x 0 ) = f(x 0 ) f (x 0 ) + 4 x f (x 0 ) + O( x 3 ) f(x 0 x) = f(x 0 ) xf (x 0 ) + x f (x 0 ) + O( x 3 ) We determine α, α 1 and α 0 such that α f(x 0 ) + α 1 f(x 0 x) + α 0 f(x 0 ) = f (x 0 ) + O( x ) Collecting like terms, we obtain (α + α 1 + α 0 )f(x 0 ) + ( α α 1 ) xf (x 0 ) + (4α + α 1 ) x f (x 0 ) = f (x 0 ) + O( x ) The corresponding linear system of equations for α, α 1 and α 0 is 1 1 1 α 0 1 0 α 1 = 1/ x 4 1 0 α 0 0 The solution of the system gives α = 1, α 1 = x and α 0 = 3 The corresponding finite difference formula is the one-sided backward difference formula for the first order derivative, namely f (x 0 ) f(x 0 ) 4f(x 0 x) + 3f(x 0 ) By changing x to x, we deduce that the one-sided forward difference formula for the first order derivative is given by f (x 0 ) f(x 0 + ) + 4f(x 0 + x) 3f(x 0 ) Local truncation error of the Crank-Nicolson scheme V ( x, (n + 1)) V ( x, n) = σ V (( + 1) x, (n + 1)) V ( x, (n + 1)) + V (( 1) x, (n + 1)) 4 x V (( + 1) x, n) V ( x, n) + V (( 1) x, n)) + x 1 ) (r σ V (( + 1) x, (n + 1)) V (( 1) x, (n + 1)) V (( + 1) x, n) V (( 1) x, n) + + r V ( x, (n + 1)) + V ( x, n) 1
Expanding ( ( each term using the Taylor series expansion at the intermediate time level x, n + 1 ) ), we obtain = V ( x, (n + 1)τ) V ( x, n) { V + V τ + 1 ( ) V + 1 τ 6 V V τ + 1 ( V τ = V τ + 1 3 V 4 τ 3 () + O( 4 ) 3 V τ 3 ) 1 3 V 6 τ 3 ( ) 3 + ( ) } 3 / + Here, we adopt the convention that any derivative ( with no ( argument specified would implicitly implies that the derivative is evaluated at x, n + 1 ) ) Note that and V (( + 1) x, n) V ( x, n) + V (( 1) x, n)/( x) = V x 4 V ( x, n) + x 1 x ( x, n) + 4 O( x4 ) = V x 3 V x τ + 1 ( ) 4 V x τ + + x 4 V 1 x 5 V 4 x 4 τ + 1 ( ) 6 V x 4 τ + + O( x 4 ), V (( + 1) x, (n + 1)) V ( x, (n + 1)) + V (( 1) x, (n + 1))/( x) = V x + 3 V x τ + 1 ( ) 4 V x τ + + x V 1 x + 5 V 4 x 4 τ + 1 ( ) 6 V x 4 τ + + O( x 4 ) Combining the results, we have σ V (( + 1) x, (n + 1)) V ( x, (n + 1)) + V (( 1) x, (n + 1)) 4 x V (( + 1) x, n) V ( x, n) + V (( 1) x, n) + x = σ V x + O( ) + O( x )
Similarly, we obtain 1 = 1 (r σ ) V (( + 1) x, (n + 1)) V (( 1) x, (n + 1)) V (( + 1) x, n) V (( 1) x, n) +, ) (r σ V x + O( ) + O( x ) and r V ( x, (n + 1)) + V ( x, n) = rv + O() ( ( By noting that V x, n + 1 ) ) satisfies the Black-Scholes equation, we obtain the local truncation error of the Crank-Nicholson scheme = O( ) + O( x ) 3 Instead of imposing artificial boundary conditions for the bond price at r = r min and r = r max, we enforce the condition that the option values along the boundary nodes of the computational domain remain to be governed by the bond price equation We apply backward difference operators to approximate the continuous differential operators: B r rmax=(n+1) r B r rmax=(n+1) r The corresponding explicit FTCS becomes B N+1 Bn N+1 = σ r max B N+1 5B N + 4B N 1 B N r, 3B N+1 4B N + B N 1 r B n N+1 5Bn N + 4Bn N 1 Bn N r + α(β r max ) 3Bn N+1 4Bn N + Bn N 1 r r max B n N+1 In a similar manner, we now apply the forward difference operators to approximate the differential operators: B r B 0 5B 1 + 4B B 3, rmin r B r 3B 0 + 4B 1 B rmin r We obtain the following explicit FTCS at = 0: B 0 B n 0 = σ r min B n 0 5B n 1 + 4B n B n 3 r + α(β r min ) 3Bn 0 + 4B n 1 B n r 3 r min B n 0
4 For the given pricing formulation of the floating strike lookback put, the binomial parameters are determined by equating the mean and variance of the discrete random walk and the continuous price process up to O( t): ) (1 α) x α x = (q r σ t, (1 α) x + α x = σ t We obtain x = σ t and This gives (1 α) x = ) (q r σ t = α = 1 + x ( r q + 1 ) σ n 1 V 1 ) (q r σ x σ 1 x n V x Here, α denotes the probability of down-move in the binomial tree The binomial scheme takes the form: V n = αv 1 + (1 α)v+1 qv n t, V n 1 = αv 1 + (1 α)v+1, 1 1 + q t The term qv in the governing equation for V is similar to the discount term rv in the 1 usual Black-Scholes equation This gives rise to the discount factor in the above 1 + q t binomial pricing formula When = 0, the binomial formula involves the grid point at = 1, which is outside the computational domain We then approximate the Neumann V boundary condition: (0, t) = 0 using the one-sided finite difference formula, that is, V 1 = V n 1 V 1 x 0 The numerical boundary value is given by V n 0 = 1 αv 0 + (1 α)v1 1 + q t This binomial scheme is similar to the Cheuk-Vorst scheme To complete the construction of the binomial scheme, we adopt the terminal payoff condition: V N = e x 1, 0 4
5 We write the FTCS scheme in the form of the standard -level-4-point scheme: ( ) σ V = S S + rs V+1 n S ( ) + 1 σ S S r V n ( ) σ + S S rs V n S 1 In order to avoid spurious oscillations, it suffices to have all the coefficients to be positive That is, σ S S + rs S > 0 1 σ S S r > 0 σ S S rs S > 0 The first inequality is always satified The last two inequalities are satisfied provided that S < σ S r and < 1 r + σ S S 6 Let the first barrier be an up-stream barrier B H while the second barrier be the downstream barrier B L The sequential barrier option is equivalent to a one-sided up-and-out barrier with a rebate paid upon knock-out The rebate is a one-sided down-and-out barrier option We assume a call payoff of the sequential barrier option Let B n and R n denote the numerical option value of the sequential barrier option and the rebate barrier option at the (, n) th node, respectively Design of the computational domain For the sequential barrier option, we set the right boundary to coincide with the upstream barrier B H The left boundary must lie sufficiently far to the left end For the down-and-out barrier option (treated as the rebate upon breaching the up-barrier in the sequential barrier option), we set the left boundary to coincide with the downstream barrier B L The right boundary must lie sufficiently far to the right end Here, we choose M such that M >> N 5
Boundary conditions (i) At = 0, the sequential barrier option is deep out-of-the-money so that the option value is close to zero We set B n 0 = 0, for all n (ii) At = N, which corresponds to the up-barrier B H, the sequential barrier option is apparently knocked out, receiving the down-and-out barrier option as the rebate Therefore, we have B n N = R n N, for all n (iii) At = M, the (rebate) down-and-out barrier option is deep in-the-money so that it is almost like a forward contract Accordingly, the second order derivative of the price function with respect to the stock price is close to zero That is, RM n = 5Rn M 1 4Rn M + Rn M 3, for all n (iv) At = L, the rebate barrier option is knocked out at x = B L with zero value That is R n L = 0, for all n Both price functions of the sequential barrier option and the rebate barrier option satisfy the Black-Scholes equation We start with computation of the rebate barrier option, then continue with the sequential barrier option The explicit finite difference schemes in both option calculations take an identical form: R = B = ( where µ = σ and c = x µ + c Rn +1 + (1 µ)r n + µ c Rn 1 e r, = L + 1,, M 1, n = 0, 1,, µ + c Bn +1 + (1 µ)b n + µ c Bn 1 e r, = 1,, N 1, n = 0, 1,,, ) r q σ x 6
7 In the continuation region where V > h, the penalty term vanishes However, if the option remains unexercised when it should be optimally exercised (in the region of optimal stopping), we have V < h As a result, the penalty term becomes highly dominant since the penalty parameter ρ is typically chosen to be a very large positive parameter Intuitively, we observe that V should assume the exercise payoff h in the limit ρ Suppose we use S as the state variable and let V n denote the numerical option value at the node ( S, n), where S = S + S 0 and S 0 is the initial stock price The Crank-Nicolson scheme takes the form: V V +1 V n = σ S V + V 1 + V n S V +1 + (r q)s V 1 + V n S r V + V n +1 V n S +1 V 1 n S + ρ max{h(s ) V + V n + V n 1 Due to the non-linearity exhibited in the penalty term, we cannot apply the Thomas algorithm Indeed, we need to use an iteration method to solve the resulting non-linear algebraic system of equations 8 The exponential distribution is the distribution of the random time between two successive umps of a Poisson process with rate 1/θ Inverting the exponential distribution gives This can be implemented as X = θ ln(1 U) X = θ ln(u) since U and 1 U have the same distribution As a summary, with U U(0, 1), we have X = θ ln U exp(θ) 9 Write ϵ 1 α 11 0 0 x 1 ϵ = α 1 α 0 x ϵ 3 α 31 α 3 α 33 x 3 where α 11 0 0 M = α 1 α 0 α 31 α 3 α 33 The entries in M are determined by the relation: MM T = Σ This yields α MM T 11 α 11 α 1 α 31 α 11 1 06 05 = α 11 α 1 α1 + α α 1 α 31 + α α 3 = 06 1 07, α 31 α 11 α 1 α 31 + α α 3 α31 + α3 + α33 05 07 1 7, 0}
and accordingly, α 11 = 1, α 11 α 1 = 06, α 31 α 11 = 05 α 1 + α = 1, α 1 α 31 + α α 3 = 07, α 31 + α 3 + α 33 = 1 We take α 11 = 1 (we choose the positive root without loss of generality), then Also, α 31 = 05, then α 1 = 06, α = 1 06 = 08 (choosing the positive root) (06)(05) + 08α 3 = 07 so that 07 03 α 3 = = 05 08 Lastly, α 33 = 1 α 31 α 3 = 1/ (choosing the positive root) Hence 10 Since c AV = c + c so that ( ) c + c var(c AV ) = var ϵ 1 1 0 0 x 1 ϵ = 06 08 0 ϵ 3 05 05 1/ x x 3 = 1 4 var(c) + 1 4 var( c) + 1 cov(c, c) As c is generated using ϵ (i) while c is generated using ϵ (i), we expect to have var(c) = var( c) so that var(c AV ) = 1 var(c) + 1 cov(c, c) (A) Now, we apply the following criterion of determining the trade-off between computational work units and variances: σ1/σ < W /W 1 Since the amount of computational work to compute c AV is about twice that of c, the control variate is preferred in terms of computational efficiency provided that Based on Eq (A), Ineq (B) is equivalent to var(c AV ) < var(c) (B) cov(c, c) < 0 Since we have chosen ϵ (i) for computing c i, the chances are high that c i and c i are negatively correlated Hence, the antithetic variates method improves computational efficiency 8
11 E Ŷ = E t ( Z 1 + 1 3 Z ) = 0; ( ) E Ŷ = te Z 4 1 + 1 3 Z = t 4 E Ŷ Ŵ = t E Z 1 (Z 1 + 1 3 Z 3 ) = t Correlation coefficient between Ŷ t E Ŷ Ŵ E Ŷ E Ŵ = = var( Ŷ ) var( Ŵ ) 1 We consider t 3 var(x t ) = E = 3 t ( ) 1 + 1 3 = t ; 3 and Ŵ ( W t t ) T W T E W t t T W T = EW t t T EW tw T + t T EW T = t t T EW t(w T W t ) + W t + t T T Since the Brownian increments over non-overlapping time intervals are independent, so Hence Let Y t = EW t (W T W t ) = 0 var(x t ) = t t ( T t + t T T = t t T = t 1 t ) T t ( ) 1 t T Z, Z N(0, 1), we have ( EY t = EX t = 0 and var(y t ) = t 1 t ) = var(x t ) T Also, Y 0 = X 0 = 0 and Y T X T = 0; so t ( ) 1 t T Z, with Z N(0, 1), is a realization of X t 13 (a) Note that var( X Y ) = 1 N var(x Y ) = 1 N var(x) + var(y ) cov(x, Y ) We achieve a reduction of the variance for the control variate estimator provided that var(x) > var(x Y ), which is equivalent to cov(x, Y ) > var(y ) If Y is very close to X, then we have the satisfaction of the inequality This would imply the significant reduction of the variance of the crude Monte Carlo estimator by using its control variate variable (b) The efficiency of the control variate hinges on the fact that we can directly simulate the difference X i Y i as one random variable It is supposed that we know its exact distribution, so one needs to generate one random variable 9
To obtain the confidence interval for the control variate Monte Carlo estimator, one starts with that of the crude Monte Carlo estimator for EX Y and add EY to the interval For example, we obtain an approximate 95%-confidence interval by X Y 196 ˆσ X Y, XY + 196 ˆσ X Y, N N where ˆσ X Y = 1 N 1 N X i Y i 1 N i=1 N (X Y ) =1 10