Galerkin Least Square FEM for the European option price with CEV model

Transcription

1 Galerkin Least Square FEM for the European option price with CEV model A Major Qualifying Project Submitted to the Faculty of Worcester Polytechnic Institute In partial fulfillment of requirements for the Degree of Bachelor of Science By Tuan Nhat Le Date: January 4th, 014 Advisor: Professor Marcus Sarkis

2 Abstract For the evaluation of European options with constant local volatility, a general closed-form analytical solution was given by the classic Black-Scholes formula. In practice, such local volatility may vary, and in those situations, the Black-Scholes formula does not work eciently. A common way to deal with such problem is to apply numerical methods, particularly the Finite Element Method (FEM). Furthermore, the extensive use of Galerkin Least Square (GLS) stabilization method combined with adaptive mesh renements is explored for bad scenarios having large local volatility. Such local volatility was described by the Constant Elasticity of Variance (CEV) model. We implement our numerical schemes in Matlab and observe the accurracy of our numerical solutions. Finally, we take advantage of better ways to discretize our domain with geometric partition to achieve high accuracy despite of large local volatility. Keywords: Option, Black-Scholes equation, Finite Element Method, Hat function, Hilbert space. 1

3 Contents 1 Introduction European Option Black-Scholes Formula Constant Elasticity of Variance (CEV) Model Remarks about the Black-Scholes equation Variational Formulation The Finite Element Method 13.1 Subdivision of the Domain The Ritz - Galerkin Approach Finite Element Method applied to European Vanilla Option Galerkin Least squares stabilization Method Mesh Renements Numerical results 43 4 Conclusions 55

4 List of Figures 1 Partition of Ω into rectangle S T -lattice Setup of the hat functions Values of a put option with S b = 100, K = 50, σ = 0.5, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, geometric partition in S and uniform partition in T Errors by uniform partition in [0, 100] and in T with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 1500, N 3 = 1000, σ = Errors by uniform partition in [0, 100] and geometric partition in [100, 1500], uniform partition in T with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 1500, N 3 = 1000, σ = Errors by uniform partition in [0, 100] and in T with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 1500, N 3 = 1000, σ = Errors by uniform partition in [0, 100] and geometric partition in [100, 1500], uniform partition in T with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 1500, N 3 = 1000, σ = Solutions by Black-Scholes formula with K = 50, γ = 0, r = 0.03, T = 1, σ = 0.4, N t = 00, N s = 1000, S = 1500, N 3 = Solutions by Black-Scholes formula with K = 50, γ = 0, r = 0.03, T = 1, σ = 0.8, N t = 00, N s = 1000,S = 1500, N 3 = Errors with GLS when K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, σ = 0.4, geometric partition in S and uniform partition in T Errors with classical FEM when K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, σ = 0.4, geometric partition in S and uniform partition in T Errors with GLS when K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, σ = 0.8, geometric partition in S and uniform partition in T Errors with classical FEM when K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, σ = 0.8, geometric partition in S and uniform partition in T. 51 3

5 14 Numerical solutions in the case γ = 0.03, S = 1500, N 3 = 1000, σ 0 = 0.3, K = 50, S b = 100, N t = 00, N s = 1000, r = 0.03, T = 1, geometric partition in S and uniform partition in T Solutions by Black-Scholes in the case γ = 0, σ = 0.3, r = 0.03, T = 1, K = 50, S b = 100, N t = 00, N s = 1000, S = 1500, N 3 = Numerical solutions in the case γ = 0.07, S = 1500, N 3 = 1000, σ 0 = 0.3, K = 50, S b = K, N t = 00, N s = 1000, r = 0.03, T = 1, geometric partition in S and uniform partition in T Solutions by Black-Scholes in the case γ = 0, σ = 0.3, r = 0.03, T = 1, K = 50, S b = 100, N t = 00, N s = 1000, S = 1500, N 3 = Numerical solutions in the case γ = 0, S = 1500, N 3 = 1000, σ 0 = 0.3, K = 50, S b = K, N t = 00, N s = 1000, r = 0.03, T = 1, geometric partition in S and uniform partition in T Solutions by Black-Scholes in the case γ = 0, σ = 0.3, r = 0.03, T = 1, K = 50, S b = 100, N t = 00, N s = 1000, S = 1500, N 3 = Numerical solutions in the case γ = 0.07, S = 1500, N 3 = 500, σ 0 = 0.3, K = 50, S b = 100, N t = 00, N s = 1000, r = 0.03, T = 1, geometric partition in S and uniform partition in T 54 1 Solutions by Black-Scholes in the case γ = 0, σ = 0.3, r = 0.03, T = 1, K = 50, S b = 100, N t = 00, N s = 1000, S = 1500, N 3 =

6 List of Tables 1 L (D)- norm and L (D)- norm errors when σ = 0.4 and σ = 0.8 with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 30K, N 3 = L (Ω)- norm and L (W )- norm errors when σ = 0.4 and σ = 0.8 with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 30K, N 3 = L (D)- norm and L (D)- norm errors for uniform partitions in both S T directions with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, T = 1, S = 1500, N 3 = L (D)- norm and L (D)- norm errors for geometric partition in S, uniform partition in T with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, T = 1, S = 1500, N 3 = L (D)- norm and L (D)- norm errors for geometric partition in both S-T directions with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, T = 1, S = 1500, N 3 = L (Ω)- norm and L (Ω)- norm errors for uniform partition in both S T directions with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, T = 1, S = 1500, N 3 = L (Ω)- norm and L (Ω)- norm errors for geometric partition in S, uniform partition in T with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, S = 1500, N 3 = L (Ω)- norm and L (Ω)- norm errors for geometric partition in both S-T directions with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, S = 1500, N 3 = L (D)- norm and L (D)- norm errors for σ = 0.4 and σ = 0.8 with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 1500, N 3 = 1000, geometric partition in S and uniform partition in T L (Ω)- norm and L (Ω)- norm errors for σ = 0.4 and σ = 0.8 with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, geometric partition in S and uniform partition in T

7 Acknowledgement I want to dedicate special thanks to Professor Marcus Sarkis for his carefully weekly guidance, his continual support and his availability throughout the four terms. I am very grateful to receive his recommendation for an interesting book and good problems that he brought up in most of our meetings. Without his help, this project would not be possible.

8 1 Introduction In our nancial world nowadays, option is a very important nancial instrument used for hedging, arbitrage-free investing and highly speculative trading by giving the buyer the right to buy (call) or sell (put) a security or other nancial asset at an agreed-upon price (the strike price) during a certain period of time or on a specic date (exercise date). The three most popular types of options are European, American, and Asian options. For this project, we only concern about the pricing of European option. Pricing this type of option requires the use of Black-Scholes model, which assumes the price of a risky asset (underlying asset) is a solution to the stochastic dierential equation: ds = s( r td t + σ tw t) (1) where W t is a standard Brownian motion, σ t is a volatility, r t is an instantaneous interest rate, and s is a price of a risky asset at time t. 1.1 European Option When σ t = σ t (s, t) and r t = r( t) (0 t T ) are continuous functions such that s s σ(s, t) is a Lipschitz regular function of s with a Lipschitz constant independent of t, and is bounded from above and away from 0 uniformly in t, the Black-scholes formula for an European call and put option at time t are following (note that F t is the ltration and s T ( t) = s e r t (T t) ): C(s, t) = E (e T t r(τ)dτ (s T ( t) K) + F t ) P (s, t) = E (e T t r(τ)dτ (K s T ( t)) + F t ) () We can also rewrite equation (1) in the form of a partial dierential equation (PDE) as follows P (s, t) t + s σ (s, t) P (s, t) + r( t) s P (s, t) r( t) P (s, t) = 0 (3) s s 7

9 with the boundary conditions P (s, T ) = Ke r(t t). { K s if s < K 0 if s > K and P (0, t) = 1. Black-Scholes Formula When σ(s, t) = σ and r( t) = r are constants, by using the transformation t = T t, x = log s, and φ(x, t) = P (e x, T t), equation (3) becomes the Black-Scholes equation φ t 1 φ σ σ (r x ) φ + rφ = 0 as (x,t) R [0,T] (4) x φ(x, 0) = (K e x ) + (5) φ(x, t) Ke rt as x - () φ(x, t) = 0 as x (7) The advantage of equation (4) is that it has constant coecients, and by set- σ (r+ ting φ(x, t) = Ψ(x, t) e b )t+( 1 r σ )x, we obtain a following one-dimensional heat equation Ψ t σ Ψ = 0 as (x, t) R [0, T ] x Ψ(x, 0) = (K e ( 1 + r σ )x e ( 1 + r σ )x ) + Ψ(x, t) = 0 as x Solving the heat equation above with the given boundary conditions, we obtain the explicit Black-Scholes formula for the price of vanilla European put (or call option by using the same transformation but with some modications to the boundary conditions (5) (7)) 8

10 C(s, t) = s N(d 1 ) Ke r(t t) N(d ) P (s, t) = s N( d 1 ) + Ke r(t t) N( d ) (8) where N(d) = 1 π d e y dy, K is the exercise price or strike price of a call or put option, d 1 = log( s K )+(r+ σ σ T t )(T t) and d = d 1 σ T t. Remark 1.. If r( t) is not a constant, d 1 = log( s K )+ T t d is the same as above r(τ)dτ+( r+ σ )(T t) σ T t and 1.3 Constant Elasticity of Variance (CEV) Model Although pricing derivatives under the assumption of constant volatility, as in the Black-Scholes model (equation (3) with constant σ and r) for an European put option pricing, is well-known to give results which cannot be reconciled with market observations, such problem did not widely manifest themselves until the 1987 market crash. After this event, many stochastic volatility models, such as Heston model, Stochastic Alpha-Beta- Rho model, were introduced as ideal approaches to resolve a shortcoming of the Black- Scholes model. In particular, since the Black-Scholes model assumes that the underlying volatility is constant over the life of the derivative, and unaected by changes in the price level of the underlying security, such model cannot explain long-observed features of the volatility smile and skew, which indicate that implied, or local, volatility does tend to vary with respect to the strike price and expiry time. By assuming that the volatility of the underlying price is a stochastic process rather than a constant, it becomes possible to model prices of the derivatives more accurately. One of the most popular stochastic volatility models, which is widely used in practice, is the Constant Elasticity of Variance (CEV) model. It was rst proposed by Cox & Ross (see [4]) as an alternative to the Black-Scholes model of underlying asset price movements. The CEV model describes the following relationship between the volatility and price, ds = µs d t + σ 0 s γ dw t where σ 0, γ are constant parameters satisfying σ 0 0 and γ > 1, W t is a standard Brownian motion, s is the price of an underlying asset at time t and µ is the expected return. The term σ 0 s γ denotes an instantaneous, or local, volatility of our option. We use the Finite Element method introduced in 9

11 Section () to solve numerically the following PDE, given the two boundary conditions, of an European put option whose local volatility is σ(s, t) = σ 0 s γ, u(s, t) t σ 0s +γ u (s, t) u (s, t) rs + ru (s, t) = 0 (9) s s u (0, t) = Ke rt (10) u(s, 0) = { K s if s < K 0 if s > K (11) 1.4 Remarks about the Black-Scholes equation On the other hand, when σ(s, t) and r( t) are not constants, since t = T t, we let u(s, t) = P (s, T t), σ(s, t) = σ(s, T t) and r(t) = r(t t). Assume that for each time t, dene q(t) = q(t t) as the dividend yield, and the underlying asset pays out a dividend q(t)s dt in dt, equation (3) becomes (note that s > 0 and t [0, T ]) u(s, t) t σ (s, t) s u(s, t) u(s, t) (r(t) q(t))s s s +r(t) u(s, t) = 0 (1) with the Cauchy data u(s, 0) = u 0 (s) where s R + and u 0 is the payo function. If q is sucently well-behaved, then equation (1) does not have any additional mathematical diculties compared to equation (3). Thus, we assume that q(t) = 0, which implies there are no discretely paid dividends. This means the equation of an American vanilla call option is exactly the same as that of an European one, which is equation (3). The Cauchy problem (1) can then be proved, with additional conditions: The function (s, t) sσ(s, t) is Holder regular on R + [0, T ]. 10

12 The function σ(s, t) is bounded on R + [0, T ] and bounded below by a positive constant. The function t r(t) is bounded and Lipschitz continuous. The Cauchy data P 0 satises 0 u 0 (s) C(1 + s) for a given constantc. then there exists an unique function u C 0 (R + [0, T ]), C 1 -regular with respect to t and C -regular with respect to s, which is solution to the boundaryvalue problem (1) and satises 0 u(s, t) C (1 + s) for a given constant C. 1.5 Variational Formulation Since our underlying asset does not pay any dividends, by letting g(s, t) = C(s, T t), we introduce the well-known put-call parity as follows g (s, t) u (s, t) = s Ke rt (13) We would like to solve the following equation to determine the approximated price of an European vanilla put option with one underlying asset for which the annual interest rate r(t) = r is constant and the local volatility σ(s, t) is variable, with the boundary conditions derived from the put-call parity, equation (13), given the time to maturity t = 1, and g (0, t) = 0 when s = 0. u(s, t) t σ (s, t) s u (s, t) u (s, t) rs + ru (s, t) = 0 (14) s s u (0, t) = Ke rt (15) u(s, 0) = { K s if s < K 0 if s > K (1) 11

13 By dening the space V = {v L (R + ) : v dv ds L (R + )}, C 0 ([0, 1]; L (R + )) as the space of continuous function on[0, 1] with values in L (R + ), and L (0, 1; V ) as the space of square integrable functions on(0, 1) with values in V, we can write the variational formulation of the Boundary-Value Problem(14) (1) as follows Weak Variational Problem. Find u C 0 ([0, 1]; L (R + )) L (0, 1; V ) such that u t L (0, 1; V ), where V is a topological dual space of V, satisfying v V, a t (u, v) + v u ds = 0 (17) R + t u (0, t) = Ke rt u(s, 0) = { K s if s < K 0 if s > K where a t (u, v) is called the bilinear form, a t (u, v) = R + σ (s, t)s +r R + uv ds u v ds s s + R + ( r + σ (s, t) + sσ (s, t) σ s (s, t)) s u s v ds We now introduce the Finite Element Method (FEM) and apply this method using Ritz-Galerkin approach, together with Crank-Nicolson implicit scheme for dierent discretizations in the S-T directions (i.e, dierent S-T-lattices) to nd the approximated prices of an European put option with constant or variable local volatility. Such problem is equivalent to nd the numerical solution to the weak variational problem (equation (17)). We then compare our numerical solutions to the exact solutions 1

14 given by the classic Black-scholes formula to measure the errors of our numerical scheme in L - norm and L - norm. We compare the errors to see which mesh renement gives the most accurate numerical solutions (that is, the closest one to the exact solution). Finally, for an European put option with variable local volatility in which the Black-Scholes formula is not a reliable tool, we use FEM combining with the most ecient mesh renements determined from previous experiments to obtain the approximated prices for each time t and to understand the behavior of the solution when we consider dierent values of σ. The Finite Element Method Compared to the Finite Dierences Method (FDM) whose lattice is basically rectangular and adaption to non-trivial geometric domain is dicult, the FEM is far more exible due to the following typical properties: Division of the domain into simple geometric subdomains, such as rectangles (for 1D domain), triangles and/or quadrilaterals (for D domain), or cubes (for 3D domain) Setup of test-functions (continuous piecewise polynomials) on subdomains Global assembling of test functions FEM can be applied to the variational formulation of a PDE, such as equation (17), or the variational inequality, which is often derived from the freeboundary conditions of American option. 13

15 .1 Subdivision of the Domain Let Ω R denote the domain. We want to nd a partition P of Ω, which consists of dividing the two intervals in S and T directions,[0, S b ] and [0, T ], into N s and N t subintervals Γ i = [s i 1, s i ] and T k = [t k 1, t k ] such that h i = s i s i 1 and t k = t k t k 1, respectively (i = 1,,..., N s and k = 1,,..., N t ). We can either use an equidistant or nonequidistant partition for the S and T direction. Figure 1 gives an example for a S T -lattice with uniform partitions in both S and T directions. Figure 1: Partition of Ω into rectangle S T -lattice We introduce the following denition for an uniform and geometric parition on the interval [0, S b ], also dened as the S-direction. Partition of the interval [0, S b ] into subintervals Γ i = [s i 1, s i ], 1 i N s such that 0 = s 0 < s 1 < < s Ns = S b and h 1 = h =... = h Ns. Let Γ q = max i=1,...,ns Γ i, and the mesh Γ of [0, S b ] be the set {Γ 1, Γ,..., Γ Ns }. Realistically, we will assume that the strike price K of our put option coincides with some nodes of Γ, which means there exists z such that s z = K. This partition is dened as the uniform partition in the s-direction. 14

16 Partition of the interval [0, S b ] into two intervals [0, K] and [K, S b ]. Each of these intervals is partitioned as follows: we divide the intervals [0, K] and [K, S b ] into N 1 and N subintervals, Γ i = [s i 1, s i ] and Γ j = [s j 1, s j ] (1 i N 1, N j N 1 + N ), such that 0 = s 0 < s 1 < < s N1 = K < < s N = S b, h N1 = r N h 1 = h and h N1 +1 = r N 1 h N1 +N = h 4 3 N1 +N for 0 r 1, r 1 (the reason for the choice h and h 4 3 N1 +N is because when taking a small step in the S-direction, and plotting the L -norm errors with respect to dierent number of timesteps in the T-direction, the convergence order is less than, and in fact, the errors decay slower than h r 1, r 1. and h 3 4 N1 +N ) for We now show that if h 1 is given, the values of r 1 and N 1 are uniquely determined (same arguments for nding r and N in the interval [K, S b ]). Notice that since h i=1,,...,n1 follows a geometric series in the interval [0, K], h 1 + h 1 r h 1 r 1 N 1 1 = h 1(1 r N 1 1 ) 1 r 1 = K. Since r N h 1 = h 4 3 1, we obtain h 1 h r 1 1 r 1 = K. Solving for r 1, we get r 1 = K h 1 K h Furthermore, since r N = h 1 3 1, solving for N 1 gives N 1 = 1 log h 1 3 log r 1 value of h 1, there exists an unique pair log h 1 log r 1 + 1) (r 1, N 1 ) = ( K h 1, 1 K h satisfying all the given restriction. Similarly, an unique pair (r, N ) = ( S b K h N1 +N, 1 log h N1 +N + 1) S b K h log r N1 +N + 1. Therefore, given the exists for the interval [K, S b ]. Finally, let Γ q = max i=1,...,n1 +N Γ i, and the mesh Γ of [0, S b ] as the set {Γ 1, Γ,..., Γ N1 +N }. Realistically, we will assume that the strike price K of our put option coincides with some nodes of Γ, which means there exists N 1 such that s N1 = K. This partition is dened as a geometric partition in the s-direction. Our main purpose for discretizing the s-direction in this way is to increase signicantly the number of points s i near K where the singularity occurs, so that the accuracy of our numerical solutions is increased. For the accuracy conditions of our numerical schemes, the similar partitions are used for the interval [0, T ]. Specically, we will choose ( t 1 ) 8 3 same reasons used to choose h for the

17 . The Ritz - Galerkin Approach For numerical computation, Ritz and Galerkin suggested to treat the variational problem on a nitely-dimensional Hilbert subspace Vh b H1 0(the Hilbert space of functions with square-integrable value and derivatives in Ω with zero value on the boundary Ω). V b h is called trial space. Therefore, we consider the discrete variational problem to nd u h Vh b, which is the solution of a (u h (s), v h (s)) = f(s)v h (s)ds Ω for all v h V b h (18) where Ω f(s)v h(s)ds = (f, v h ) and a (u h (s), v h (s)) is a bilinear form stated in the weak variational problem. Let ϕ 1, ϕ,..., ϕ k be a basis of V b h, with k = dim ( V b h ). Then the uh in (18) can be interpolated by the basis elements with corresponding weights ϕ j R in the following way: u h = k u h, j ϕ j (19) j=1 By means of the Finite Element method, the ϕ j are call basis or test functions, which we will use later on as piecewise well-denied polynomials. The representation of u h as a nite sum of weighted test functions, in fact, gives the name Finite Element Method. Using (19), we can rewrite the discrete variational problem (18) as: Find a u h V b h with a(u h, v h ) = (f, v h ) v h Vh b a(u h, ϕ i ) = (f, ϕ i ) i = 1,..., k a( k j=1 u h, jϕ j ϕ i ) = (f, ϕ i ) i = 1,..., k k j=1 a(ϕ j, ϕ i )u h, j = (f, ϕ i ) i = 1,..., k Au h = B (0) 1

18 with A := (a (ϕ j, ϕ i )) i,,j R k k, u h := (u h, 1,..., u h, k )) T and B := (f, ϕ i ) i, i = 1,..., k. Hence, the Ritz-Galerkin approach is equivalent to a linear system of equations, and the computation of vector u h, applied to equation (0), gives an approximation for u. Due to its application to mechanics, A is called the stiness matrix. It is positive denite and symmetric for an arbitrary basis ϕ 1,..., ϕ k, which implies A is invertible. Note that for certain choices of the basis, the matrix A is sparse, which means that only a few elements (A) i, j are non-zero. This would reduce the cost for the computation of (0) signicantly..3 Finite Element Method applied to European Vanilla Option We now dene the discrete space Vh b as follows to ensure that the boundary condition u (s, 0) = 0 for s > K belongs to Vh b, V b h = { ϕ(s) C 0 [0, S b ], ϕ (S b ) = 0 s Γ, ϕ s is ane } In the real world, the option value is dependent mainly on the underlying asset price s whose local volatility varies based on the changes in the values of s. Therefore, we consider the general case when the local volatility σ, which is a function of s and t, is described through the CEV model σ(s, t) = σ 0 s γ for constant parameters σ 0 0, γ > 1 and s is the price of an underlying asset at time t. We rewrite equation (14) in the following form where α = s(1+γ) σ 0, β = (1 + γ)σ 0s γ+1 rs, u t s (α u s ) + β u + ru = 0 (1) s 17

19 , and by using inte- ds (since Multiplying both sides by the test function ϕ(s) Vh b gration by parts for the term S b ϕ(s) u (α ) ds = S b 0 s s ϕ(s b ) = 0), equation (1) becomes ( u, ϕ) + (α u t s, ϕ s α u 0 s ϕ(s) s u ) + (β, ϕ) + (ru, ϕ) = 0 () s where (, ) is a broken L inner product with respect to all partitions Γ i of the interval [0, S b ]. Let u t denote u t, and uk h denote u h(, t k ). After discretizing equation () in space and time (dividing the interval [0, 1] into N t subintervals [t k, t k 1 ], k = 1,..., N t with lengths t k = t k t k 1 ) and rewriting u h (s, t k ) = i=1 u h, i(t k )ϕ i (s), where ϕ i is the nodal basis of Vh b, we obtain the fully discrete variational problem with two boundary conditions: Find u h (s, t k ) Vh b such that u 0 h = max(k s, 0) for s is the price of the underlying asset at time t = 0 (3) and for 1 k N t, u h (0, t k ) = Ke rt k (4) M(u h (s, t k ), ϕ) + A(u h (s, t k ), ϕ) = 0 ϕ V b h and t k [0, 1] (5) where m ( u k h, ϕ) is the inertial form of the mass matrix M, m(u h (s, t k ), ϕ) = (u h (s, t k ), ϕ) () and a(u k h, ϕ) is the Galerkin form of the stiness matrix A, a(u h (s, t k ), ϕ) = (α uk h s, ϕ s ) + (β uk h s, ϕ) + (ruk h, ϕ) (7) Let (α uk h ) = s a1, k, (β uk h, ϕ) = s a, k, (ru k h, ϕ) = a3, k, the equation (7) can be rewritten as follows s, ϕ 18

20 a(u h (s, t k ), ϕ) = a 1, k + a, k + a 3, k (8) Given u 0 h = max(k s, 0), we can use either the Euler explicit/implicit scheme or the second-order stable Crank-Nicolson scheme corresponding to the values of θ = 0, 1, 1, respectively, to nd uk h for k = 1,,..., N t from the equation 1 m(u k h u k 1 h, ϕ) + a(θu k h + (1 θ)u k 1 h, ϕ) = 0 ϕ Vh b (9) t k In order to obtain the most accurate and stable numerical solutions for large local volatility, we choose the second-order stable Crank-Nicolson scheme (that is, θ = 1). Now, for each nite element dened over the nodes (s i, s j ), the mass matrix M has its entries m i, j = (ϕ j, ϕ i ), and let A be the stiness matrix dened by a i,,j = a (ϕ j, ϕ i ) = a 1 i, j + a i, j + a 3 i, j, 0 i, j N s By letting u k h = ( u k h (s 0),..., u k h (s N) )T and u 0 h = (u 0 h (s 0),..., u 0 h (s N)) T, which is the boundary condition, we can rewrite equation (9) as follows M ( ) u k h u k 1 t k h + ( ) Au k h + Au k 1 h = 0 (M + t k A)uk h = (M t k A)uk 1 h We now choose the nodal basis ϕ i to be the hat functions, which were dened below over two consecutive nodes s i,s i 1 and s i, s i+1. Denition 1. (Hat Functions) With i = 0, 1,..., N s, the hat functions are ϕ i (s) = s s i 1 h i s (s i 1, s i ) ϕ i (s) = s i+1 s h i+1 s (s i, s i+1 ) 19

21 Figure : Setup of the hat functions Note that the rst and last element ϕ 0 and ϕ Ns need to be cut o to t into the discretized domain, as shown in Figure. These hat functions ϕ i correspond to the nodes s i and are supported in [s i 1, s i+1 ]. When i j > 1, the intersection of the support ϕ i and ϕ j has measure 0. This implies the matrices M and A are tridiagonal. We now compute the actual assembling of matrix M as follows: by denition, m i,,j with (ϕ j, ϕ i ) = S b 0 ϕ i ϕ j ds. =(ϕ j, ϕ i ) i, j=0,..., Ns The rst element of M, m 0, 0 and m Ns, N s just contains one of the two summands of the equation (34), since it is dened by the cut boundary nite element ϕ 0 and ϕ Ns. The calculations for m 0, 0 and m Ns, N s are follows, m 0, 0 = Sb 0 ϕ 0 ϕ 0 ds = s1 s 0 (ϕ 0 ) ds = s1 s 0 1 h 1 (s 1 s) ds = 1 ( s 3h 1 + s 0 ) 3 = h (30) m Ns, N s = = Sb 0 sns ϕ Ns ϕ Ns ds = sns s Ns 1 (ϕ Ns ) ds s Ns 1 1 h N s (s Ns s Ns 1) ds = 1 ( s 3h Ns + s Ns 1) 3 = h N s N s 3 (31) 0

22 The matrix M is symmetric, since both subdiagonals m i, i 1 and m i, i+1 have the following formulas for its entries m i, i+1 = Sb 0 ϕ i+1 ϕ i ds = si+1 1 (s s s i h i ) (s i+1 s) ds i+1 = 1 ( s i+1s s h i+1 s i s s3 i s is ) s i+1 s i = 1 ( s 3 i+1 h i+1 s i+1s i + s i+1s i ) s3 i = 1 (s h i+1 s i ) 3 = h i+1 i+1 for i = 0,..., N s 1 (3) m i, i 1 = Sb 0 = 1 h i = 1 h i ϕ i 1 ϕ i ds = si s i 1 1 h i ( s is s i 1s i s s3 (s s i 1 ) (s i s) ds 3 + s i 1s ) s i s i 1 ( s 3 i s i s i 1 + s ) is i 1 s3 i 1 = 1 (s h i s i 1 ) 3 = h i i for i = 1,..., N s (33) On the diagonal of M, the elements m i, i are m i, i = Sb 0 ϕ i ϕ i ds = si+1 + s i si s i 1 ( ) 1 (s s i 1 ) ds h i ( ) 1 (s i+1 s) ds h i+1 = 1 (s 3h i s i 1 ) 3 1 ( (s i 3h i+1 s i ) 3 ) i+1 1

23 = h i + h i+1 3 = (m 1 i+1 + m 1 i 1) for i = 1,..., N s 1 (34) Now, for assembling matrix A, we need to compute the sum a 1 i, j + a i, j + a 3 i, j for j {0, i 1, i, i + 1, N s }. For example, similar to m 0, 0 and m Ns, N s, a 0, 0 and a Ns, N s needs special treatment and requires the following computations a 1 0, 0 = σ 0 Sb 0 s (1+γ) ( ϕ 0 s ) ds = σ s1 0 s (1+γ) ds 0 h 1 = σ 0s 1+γ 1 (3 + γ) (35) a 0, 0 = (1 + γ)σ 0 Sb 0 = (1 + γ)σ 0 h 1 s γ+1 ϕ Sb 0 ϕ 0 s ds r ϕ 0 sϕ 0 0 s ( s3+γ 1 + γ s3+γ γ ) + rs3 1 h 1 = σ 0s 1+γ 1 (3 + γ) + rs 1, since s 0 = 0 and h 1 = s 1 (3) a 3 0, 0 = r Sb 0 s1 (ϕ 0 ) (s 1 s) ds = r ds = rh 1 0 h 1 3 (37) a 1 N s, N s = σ 0 Sb 0 s (1+γ) ( ϕ N s s ) ds = σ 0 sns s Ns 1 s (1+γ) h N s ds ( s3+γ N s s 3+γ N s 1 ) h N s 3 + γ = σ 0 a N s, N s = (1 + γ)σ 0 Sb Sb ϕ Ns r sϕ Ns 0 s 0 s γ+1 ϕ Ns ϕ Ns s ds

24 = (1 + γ)σ 0 h N s ( sγ+3 N s s Ns 1 γ + 3 sγ+ N s γ + ) + σ 0s γ+3 N s 1 h N s (γ + 3) r ( s3 N s h N s 3 + s3 N s 1 ) = (1 + γ)σ 0 h N s + rs N s s Ns 1 h N s ( sγ+3 N s γ + 3 ) σ 0s γ+ N s h N s s Ns 1 s γ+3 N s 1 h N s (γ + 3) r (s N s + s Ns 1) + σ 0 (38) a 3 N s, N s = r Sb 0 sns (ϕ Ns ) (s s Ns 1) ds = r ds S Ns 1 h N s = rh N s 3 For the elements a i, i 1, we need to calculate the following elements (39) a 1 i, i 1 = σ 0 Sb 0 s (γ+1) ϕ i s ϕ i 1 s ds = σ 0 si s i 1 s (γ+1) h i ds a i, i 1 = (1 + γ)σ 0 = σ h i Sb 0 = (1 + γ)σ 0 h i i s γ+3 i 1 ) (40) γ + 3 ( sγ+3 s γ+1 ϕ i ϕ i 1 s + r h i ( sγ+3 i Sb ϕ i 1 ds r sϕ i ds 0 s s i 1 γ + 3 sγ+ i γ + ) ( s3 i 3 s i s i 1 + s3 i 1 ) 3

25 = (1 + γ)σ 0 h i σ 0s γ+3 i 1 h i (γ + 3) s γ+3 i γ + 3 σ 0s γ+ i h i s i 1 σ 0 h i s γ+3 i 1 (γ + 3) + r(s i + s i 1 ) (41) a 3 i, i 1 = rm 1 i, i 1 = r h i For the elements a i, i+1, we need to calculate the elements (4) a 1 i, i+1 = σ 0 Sb 0 s γ+ ϕ i s ϕ i+1 s ds = σ 0 si+1 s i s γ+ h i+1 ds a i, i+1 = (1 + γ)σ 0 = σ h i+1 Sb 0 i+1 s γ+3 i ) (43) γ + 3 ( sγ+3 sϕ i ϕ i+1 s Sb ϕ i ds r sϕ i+1 0 s ds = (1 + γ)σ 0 h i+1 ( sγ+ i s i+1 γ + + sγ+3 i γ + 3 ) r h i+1 ( s3 i+1 s i s i+1 + s3 i 3 ) = (1 + γ)σ 0 h i+1 + σ 0 s γ+3 i+1 h i+1 (γ + 3) s γ+3 i i s i+1 h i+1 γ + 3 σ 0s γ+ + σ 0 s γ+3 i+1 h i+1 (γ + 3) r(s i+1 + s i ) (44) 4

26 a 3 i, i+1 = rm 1 i+1 = rh i+1 (45) Its diagonal elements a i, i will contain the following integrals (note that ϕ i 1 h i s (s i 1, s i ) and ϕ i = 1 s h i+1 s (s i, s i+1 )) s = a 1 i, i = σ 0 Sb 0 s γ+ ( ϕ i s ) ds = σ si 0 ( s γ+ ds) s i 1 h i + σ 0 si+1 s i s γ+ h i+1 ds = σ 0 h i a i, i = (1 + γ)σ 0 = (1 + γ)σ 0 h i i s γ+3 i 1 ) + σ 0 γ + 3 h i+1 ( sγ+3 Sb 0 ( sγ+3 i i+1 s γ+3 i ) (4) γ + 3 ( sγ+3 s γ+1 ϕ Sb i ϕ i s ds r ϕ i sϕ i 0 s ds γ + 3 sγ+ i s i 1 γ + + s γ+3 i 1 (γ + )(γ + 3) ) (1 + γ)σ 0 h i+1 s γ+3 i+1 i s i+1 ( (γ + 3)(γ + ) sγ+ γ + + sγ+3 i γ + 3 ) r h i = (1 + γ)σ 0 h i ( s3 i 3 s i s i 1 + s3 i 1 ) + r ( s3 i+1 h i+1 s i s i+1 + s3 i 3 ) ( sγ+3 i s i 1 γ + 3 sγ+ i γ + ) + σ 0 (1 + γ)σ 0 h i+1 h i ( sγ+ i s i+1 γ + s γ+3 i 1 (γ + 3) + sγ+3 i γ + 3 ) σ 0 s γ+3 i+1 h i+1 (γ + 3) + r(h i + h i+1 ) (47) a 3 i, i = rm i,i = r( h i + h i+1 ) (48) 3 From equations (3) (4), we can calculate the entries of A as follows (remember that A is a tridiagonal matrix) 5

27 a i, i 1 = a 1 i, i 1 + a i, i 1 + a 3 i, i 1 = σ 0 h i (1 + γ)σ 0 h i i s γ+3 i 1 ) γ + 3 ( sγ+3 s γ+3 i γ σ 0 h i s γ+ i s i 1 σ 0s γ+3 i 1 h i (γ + 3) + r(s i + s i 1 ) + r h i = σ 0s (γ+1) i h i + rs i 1 i N s (49) a i, i+1 = a 1 i, i+1 + a i, i+1 + a 3 i, i+1 = σ 0 h i+1 + σ 0 h i+1 s γ+3 i+1 i+1 s γ+3 i ) γ + 3 ( sγ+3 i s i+1 h i+1 (γ + 3) σ 0s γ+ + (1 + γ)σ 0s γ+3 i h i+1 (γ + 3) r(s i+1 + s i ) + r h i+1 = s(γ+1) i σ 0 h i+1 rs i 0 i N s 1 (50) a 0,0 = a 1 0, 0 + a 0, 0 + a 3 0, 0 = σ 0s 1+γ 1 (3 + γ) σ 0s 1+γ 1 (3 + γ) + rs 1 + rh 1 3 = rs 1 (51) a Ns, N s = a 1 N s, N s + a N s, N s + a 3 N s, N s = (1 + γ)σ 0 h N s + σ 0 γ + 3 sγ+ N s γ + ) ( sγ+3 N s ( s3+γ N s s 3+γ h N s 3 + γ N s 1 s Ns 1 ) + rh N s 3

28 + σ 0s γ+3 N s 1 h N s (γ + 3) r(s N s + s Ns 1) = s(γ+1) N s σ 0 h Ns rs N s 1 (5) a i, i = a 1 i, i + a i, i + a 3 i, i = σ 0 h i + (1 + γ)σ 0 h i + σ 0 h i+1 ( sγ+3 (1 + γ)σ 0 h i+1 ( sγ+3 i i s γ+3 i 1 ) γ + 3 ( sγ+3 s i 1 γ + 3 sγ+ i γ + ) i+1 s γ+3 i ) + σ 0s γ+3 i 1 γ + 3 h i (γ + 3) ( sγ+ i s i+1 γ + + sγ+3 i γ + 3 ) + r(h i + h i+1 ) + r(h i + h i+1 ) 3 + h i+1 σ0s γ+3 i+1 (γ + 3) = s(γ+1) i σ0 ( 1 h i + 1 h i+1 ) + r(h i + h i+1 ) 1 i N s 1 (53) In summary, we get the form of the stiness and mass matrices, A and M, rspectively (note that the entries of matricesa and M don't depend on time, which is k, in this case) : A := a 0,0 a 0, a 1, a Ns 1, N s a Ns, N s 1 a Ns, N s where a 0,0, a 0, 1, a Ns 1, N s, a 1, 0, a Ns, N s 1 and a Ns, N s are given by the formulas (49) (53) with i = 1 and N s 1 7

29 M := h 1 h 1 3 h 1 h 1 +h 3 h h h Ns 1 +h Ns h Ns h Ns h Ns 3 For an equidistant grid in the S-direction (that is, uniform partition), implying h := h 1 =... = h Ns, the matrices can be simplied further, which further on will reduce the cost of calculation. On the non-equidistant grid, with A and M dened as in (3) and (4), the required time of calculation increases proportionally for higher neness of the mesh (larger k). Thus when a non-equidistant lattice is used, it should be designed in a manner that good accuracy will be achieve for a relatively small amount of grid points. Note that matrix M + t k A can be further reduced by omitting the rst row and rst column due to the two boundary conditions u 0 h = max(k s, 0) and u h (0, t k ) = Ke rt k. These boundary values are each interpolated by one hat k=0,1,..., Nt function with the corresponding coecients uh, 0 and u 0 h, i=1,..., N s. This means these coecients are known and we only have N s unknown coecients left. Hence, we may cancel the rst column and row of matrix M + t k A, the rst row of matrix (M t k A) uk 1 h and the rst element of vectors u k h. The matrix (M + t k A) now has size N s N s, (M t k A) uk 1 h has size N s 1 and vector u k h now has size N s 1. The boundary terms that were dropped during this reduction process are put into a vector c k with size N s 1, and the nal system of equations will have the following form: in which c k = (M + t k A)uk h = (M t k ( h 1 + t k a 1, 0) u k h, A)uk 1 and u k h = 8 h c k (54) u k h, 1 u k h,... u k h, N s 1 u k h, N s.

30 The linear system of equations (54) is equivalent to one with homogeneous boundary conditions where the function vanishes on the boundary. We can rewrite the linear of system of equations (54) in the form Eu k h = F uk 1 h c k, and the entries of matrices E and F are below, respectively e i,i 1 = t k 0s (γ+1) ( σ i h i + rs i ) + h i, i N s e i,i = t k (s(γ+1) i σ0 ( ) + r(h i + h i+1 ) ) h i h i+1 + h i + h i+1, 1 i N s 1 3 e i,i+1 = t k ( s(γ+1) i σ0 h i+1 rs i ) + h i+1, 1 i N s 1 e Ns, N s = t k (s(γ+1) N s σ0 h Ns rs N s 1 ) + h N s 3 f i,i 1 = t k ( σ 0s (γ+1) i h i + rs i ) + h i, i N s f i,i = t k ( s(γ+1) i σ0 ( ) + r(h i + h i+1 ) ) h i h i+1 + h i + h i+1, 1 i N s 1 3 f i,i+1 = t k ( s(γ+1) i σ 0 h i+1 rs i ) + h i+1, 1 i N s 1 e Ns, N s = t k ( s(γ+1) N s σ0 h Ns rs N s 1 ) + h N s 3 With those formulas and the given value of u 0 h = max(k s, 0), we can compute iteratively the unknown variables u k h, i for i = 1,,..., N s, and k = 1,,..., N t, and obtain the prices of our put option for any given values of both time t [0, 1] and the underlying asset s [0, S b ]. 9

31 Note that although we choose geometric partition in the S-direction, we decide to employ two ways of partition for the time interval [0, 1] into N t subintervals with t k = t k t k 1 for k = 1,..., N t : uniform or geometric partition, in order to see which partition gives a higher accuracy. Both partitions are exactly the same as those used in the S-direction, which means for the geometric partition, the number of points t i around t = 0 increase signicantly. However, although the geometric partition increases the cost of computation, the errors of our numerical solutions in this case are much larger than those obtained in the case of uniform partition. Therefore, for discretizing our mesh in the T-direction, it is best to choose the uniform partition to increase the accurracy of our numerical solutions while reducing the computation time required, especially for cases when σ 0 and γ are suciently small (for example, σ 0 = 0.3 and γ = 0.03).4 Galerkin Least squares stabilization Method Theoretically, the local volatility σ can be any positive values, but in practice, the local volatility of an option may be abnormally large due to some unexpectedly horrible news or events that occur (for example, in 008, after Lehman Brothers and Bear Sterns collapsed, the Down Jones closed down just over 500 points at the time the largest drop by points in a single day since the days following the attacks on September 11, 001, or in Japan, banks and insurers announced a combined 49 billion yen ($.4 billion) in losses due to this collapse). The impacts of such events will increase volatility of option prices signicantly, and such cases correspond to suciently large vales of σ 0 and γ (for example, σ 0 = 0.7 and γ = 0.05). We realize that the presented Galerkin FEM does not work well for large local volatility due to its high sensitivity to the large values of σ 0. One way to x this problem is to apply the Least Square Regression (LSR) method, but it has one major drawback: the solution is much harder to compute by iterative methods and more sensitive to roundo errors since the number of 30

32 matrices in the equation () scales as the square of the number of the matrix in the Ritz - Galerkin method. The LSR method is also less accurate in the regions where the solution is smooth. Thus, we decide to combine the Least Square and Ritz- Galerkin methods together, called the Galerkin Least Square (GLS) method, by adding an additional Least Square term to the LHS of equation (), which is equivalent to one of the two options belows: 1. ( u t, ϕ) + (α u ru, ρ i ( s. ( u t, ϕ) + (α u ru, ρ i ( s, ϕ s s (α ϕ s ) + β ϕ, ϕ s s (α ϕ s ) + β ϕ u ) + (β, ϕ) + (ru, ϕ) + s τ T h ( u u (α ) + β u t s s + rϕ)) = 0 s s + u ) + (β, ϕ) + (ru, ϕ) + s τ T h ( u u (α ) + β u t s s )) = 0 s s + in which the stability parameter ρ i is dened locally based on the Peclet number P e i ( s i+s i+1 ) = h s T h b( i +s i+1 ) p 4ɛ on each partition τ as follows, ρ i (s, P e i ( s h i + s T i i+1 )) = 48ɛ i( s i+s i+1 ) < 1 h Ti, P e b( s i +s i+1 i ( s i + s i+1 ) 1 ) p (55) { b( s i + s i+1 ( N i=1 ) p = i( s i+s i+1 ) p ) 1 p, 1 p < max i=1,..,n b i ( s i+s i+1 ), p = (5) After discretizing the equation () in space, time (dividing the interval [0, 1] into N t subintervals [t k, t k 1 ], k = 1,..., N t with lengths t k = t k t k 1 ) and rewriting u h (s, t k ) = again the linear test function of Vh b discrete variational problem: i=1 u h, i(t k )ϕ i (s), where ϕ i is, we again obtain the stabilized fully Find u h (s, t k ) V b h such that M LQ (u k h, ϕ) + A LQ (u k h, ϕ) = 0 ϕ V b h and t k [0, 1], (57) 31

33 where m LQ (u k h, ϕ) = (u k h, ϕ) + ( uk h t, ρ i( ε s (α ϕ s ) + β ϕ + rϕ)) (58) s τ T h is an augmented inertial form of the mass matrix M LQ, and a LQ (u k h, ϕ) = (α uk h s, ϕ s ) + (β uk h s, ϕ) + (ruk h, ϕ) + ( s (α uk h s ) + β uk h s + ruk h, ρ i ( ε s (α ϕ s ) + β ϕ + rϕ)) (59) s τ T h is a stabilized Galerkin form of the stinex matrix A LQ. In this formula, ε are either 0,1 or 1, which corresponds to the SUPG, GLS and MS method, respectively. We will fully solve the problem using option 1's formulation (ε = 1), which is the most intense computational case. Without repetition, we only present the general formula to form matrices M LQ and A LQ in the case of option (). Given u 0 h = max(k s, 0), by using the Crank-Nicolson scheme corresponding to the values of θ = 1, we can nd uk+1 h for k = 0, 1,..., N t 1 from the following equation 1 m LQ (u k h u k 1 h, ϕ) + a LQ (θu k h + (1 θ)u k 1 h, ϕ) = 0 ϕ Vh b (0) t k Let (ϕ i ) i=0,...,ns be the basis, or hat functions in V h, and let M LQ be the mass matrix whose entries are dened by i, j = (ϕ j, ϕ i ) + ρ i [(ϕ j, rϕ i ) (ϕ j, rs ϕ i )] (1) s τ T h m LQ since ϕ is the hat functions and α ϕ s ϕ = 0, (α ) + β ϕ s s s = rs. Futher- ) = s mlq, i, j, we more, let τ T h ρ i (ϕ j, rϕ i ) = m LQ, 1 i, j and τ T h ρ i (ϕ j, rs ϕi can rewrite equation (59) as 3

34 m LQ i, j = (ϕ j, ϕ i ) + m LQ, 1 i, j m LQ, i, j () Since a (ϕ j, ϕ i ) = (α ϕ j, ϕ i ) + (β ϕ j, ϕ s s s i) + (rϕ j, ϕ i ) as dened in equation (), the entries of the stiness matrix A LQ are given as a LQ i,,j = a (ϕ j, ϕ i ) + τ T h ρ i [(rϕ j, rϕ i ) (rs ϕ j s, rϕ i) + (rs ϕ j s, rs ϕ i s )] From the following identities,, τ T h ρ i (rϕ j, rs ϕ i s ), 0 i, j N s. (3) ρ i (rϕ j, rϕ i ) = rm LQ, 1 i, j τ T h ρ i (rs ϕ j s, rs ϕ i s ) = r a LQ, 1 i, j ρ i σ τ T h τ T h ρ i (rs ϕ j s, rϕ i) = r a LQ, i, j ρ i r + σ τ T h τ T h the equation (59) is equivalent to τ T h ρ i (rϕ j, rs ϕ i s ) = rmlq, i, j. a LQ i, j = a (ϕ j, ϕ i ) + rm LQ, 1 i, j + ρ i [ r a LQ, i, j τ T h r + σ + r a LQ, 1 i, j σ ] rm LQ, i, j By letting u k h = ( u k h (s 0),..., u k h (s N) )T and u 0 h = (u0 h (s 0),..., u 0 h (s N)) T which is the boundary condition, the fully discrete problem (0) can be rewritten as a system of linear equations (M LQ + t k ALQ )u k h = (M LQ t k ALQ )u k 1 h (4) 33

35 We now compute the following general formula for the stiness matrix M LQ. By using equation (), we need to compute the following elements m LQ, i, i 1 = r Sb 0 sϕ i 1 ϕ i s ds = ρ i 1r( s i + s i 1 3 ) m LQ, i, i+1 = r Sb 0 sϕ i+1 ϕ i s ds = ρ i 1r( s i s i ) Sb m LQ, ϕ 0 0, 0 = r sϕ 0 0 s ds = ρ 0rh 1 m LQ, N s, N S Sb ϕ Ns = r sϕ Ns 0 s ds = ρ Ns 1r( s N s 1 + s N s 3 ) Sb m LQ, ϕ i i, i = r sϕ i 0 s ds = r(ρ i 1(s i + s i 1 ) ) r ρ i(s i + s i+1 ) Together with the formulas (30) (34), the entries of M are given as follows m LQ i, i 1 = (ϕ i 1, ϕ i ) + m LQ, 1 i, i 1 mlq, i, i 1 = h i + ρ i 1r h i ρ i 1 r( s i + s i 1 3 ) = h i ρ i 1s i 1 r, 1 i N s m LQ i, i+1 = (ϕ i+1, ϕ i ) + m LQ, 1 i, i+1 mlq, i, i+1 = h i+1 + ρ ir h i+1 ρ i r( s i+1 3 s i ) 34

36 = h i+1 + ρ is i+1 r, 0 i N s 1 m LQ 0, 0 = (ϕ 0, ϕ 0 ) + m LQ, 1 0, 0 m LQ, 0, 0 = h ρ 0r h ρ 0rh 1 = h ρ 0h 1 r (5) m LQ N s, N s = (ϕ Ns, ϕ Ns ) + m LQ, 1 N s, N s m LQ, N s, N s = h N s 3 + ρ Ns 1r h N s 3 ρ Ns 1r( s N s 1 + s N s 3 ) = h N s 3 ρ N s 1rs Ns 1 () m LQ i, i = (ϕ i, ϕ i ) + m LQ, 1 i, i m LQ, i, i = h i + h i+1 + r ρ i 1h i + ρ i h i r ρ i 1(s i + s i 1 ) r ρ i 1(s i + s i 1 ) + r ρ i(s i + s i+1 ) = ( rs i 1ρ i 1 Similarly, the entries of A are given as + h i 3 ) +(rs i+1ρ i + h i+1 3 ), 1 i N s 1 (7) a LQ i, i 1 = a (ϕ i 1, ϕ i ) + rm LQ, 1 i, i 1 + ρ i [ r a LQ, i, i 1 τ T h r + σ + r a LQ, 1 i, i 1 ] σ rm LQ, i, i 1 = σ 0s (γ+1) i h i + rs i 35

37 + r h i ρ i 1 + ρ i 1r s i 1 + ρ i 1 r s i 3 ρ i 1 r ( s i s i s is i 1 ) ρ i 1 r ( s i h i + s i 1 3 ) = ρ i 1r s i s i 1 h i σ 0s (γ+1) i h i + rs i, 1 i N s (8) a LQ i, i+1 = a (ϕ i+1, ϕ i ) + rm LQ, 1 i, i+1 + ρ i [ r a LQ, i, i+1 τ T h r + σ + r a LQ, 1 i, i+1 ] σ rm LQ, i, i+1 = s(γ+1) i σ0 h i+1 rs i + r ρ i h i+1 r ρ i ( s i+1 + s i 3 ) r ρ i ( s i+1 s i 3 + s is i+1 h i+1 ) + r ρ i ( s i s i ) = ρ ir s i s i+1 h i+1 s(γ+1) i σ 0 h i+1 rs i, 0 i N s 1 a LQ 0, 0 = a (ϕ 0, ϕ 0 ) + rm LQ, 1 0, 0 + ρ i [ r a LQ, 0, 0 τ T h rm LQ, 0, 0 = rs 1 + h 1 r ρ h 1 r ρ 0 r + σ + r a LQ, 1 0, 0 ] σ + r h 1 ρ h 1 r ρ 0 = p 0h 1 r + rs 1 (9) a LQ N s, N s σ N s, N s = ρ i [ r a LQ, N s, N s r + σ + r a LQ, 1 ] τ T h +rm LQ, 1 N s, N s rm LQ, N s, N s 3

38 +a (ϕ Ns, ϕ Ns ) = s(γ+1) N s σ 0 h Ns rs N s 1 + r h Ns ρ Ns 1 3 r ρ Ns 1( s N s 1 + s N s 3 ) + r ρ Ns 1( h N s 3 + s N s s Ns 1 h Ns ) ρ Ns 1r ( s N s 1 + s N s 3 ) = ρ N s 1r s Ns s Ns 1 h Ns + s(γ+1) N s σ 0 h Ns rs N s 1 r ρ Ns 1s Ns 1 (70) i, i = a (ϕ i, ϕ i ) + ρ i [ r a LQ, i, i r + σ + r a LQ, 1 i, i ] τ T h a LQ +rm LQ, 1 i, i rm LQ, i, i σ = s(γ+1) i σ0 ( ) + r(h i + h i+1 ) h i h i+1 + r ( ρ i 1h i + ρ i h i+1 ) r ρ i 1(s i + s i 1 ) 3 + r ρ i(s i + s i+1 ) + r ρ i 1 ( h i 3 + s is i 1 h i ) + r ρ i ( h i s is i+1 ) r ( ρ i 1(s i + s i 1 ) ) h i+1 + r ρ i(s i + s i+1 ) = (s i+1 ρ i r + s i r ρ is i+1 h i+1 ) + (s i r ρ i 1s i 1 h i r s i 1 ρ i 1 ) 37

39 + s(γ+1) i σ0 ( 1 h i + 1 h i+1 ) + r(h i + h i+1 ), 1 i N s 1 (71) Note that matrix M LQ + t k ALQ can be further reduced by omitting the rst row and rst column due to the two boundary conditions u 0 h = max(k s, 0) and u h (0, t k ) = Ke rt k. These boundary values are each interpolated by one k=0,1,..., Nt hat function with the corresponding coecients uh, 0 and u 0 h, i=1,..., N s. This means these coecients are known and we only have N s unknown coecients left. Hence, we may cancel the rst column and row of matrix M LQ + t k ALQ, the rst row of matrix (M LQ t k ALQ ) u k 1 h and the rst element of vectors u k h. The matrix (M LQ + t k ALQ ) then has size N s N s and (M LQ t k ALQ )u k 1 h then has dimension N s 1 and vector u k h now has size N s 1. The boundary terms that were dropped during this reduction process are put into a vector c k with sizes N s 1, and the nal system of equations will have the following form: (M LQ + t k ALQ )u k h = (M LQ t k ALQ )u k 1 h c k (7) (m LQ 1,0 + t k alq 1,0 ) u k u k h, 0 h, 1 0 in which c k =... 0 and uk h = u k h,... u k. h, N s 1 0 u k h, N s The linear system of equations (7) is equivalent to one with homogeneous boundary conditions, where the function vanishes on the boundary. We can rewrite the linear system of equations (7) in the form E LQ u k h = F LQ u k 1 h c k, and the entries of matrices E LQ and F LQ are below, respectively e LQ i, i = ( rs i 1ρ i 1 + h i 3 ) + (rs i+1ρ i + h i+1 3 ) + t k (s i+1ρ i r + s i r ρ is i+1 h i+1 ) 38

40 + t k (s ir ρ i 1s i 1 h i r s i 1 ρ i 1 ) + t k s (γ+1) i σ0 + t k ( 1 h i + 1 h i+1 ) r(h i + h i+1 ), 1 i N s 1 e LQ i, i 1 = h i ρ i 1s i 1 r t k ρ i 1 r s i s i 1 h i + t k 0s (γ+1) ( σ i h i + rs i ), i N s e LQ i, i+1 = h i+1 + ρ is i+1 r + t k ρ i r s i s i+1 h i+1 + t k (s(γ+1) i σ0 h i+1 + rs i ), 1 i N s 1 e LQ N s, N s = t k ρ Ns 1r s Ns s Ns 1 h Ns + t k (s(γ+1) N s σ0 h Ns rs N s 1 ) t k r ρ Ns 1s Ns 1 + h N s 3 ρ N s 1rs Ns 1 f LQ i, i = ( rs i 1ρ i 1 + h i 3 ) + (rs i+1ρ i + h i+1 3 ) t k (s i+1ρ i r + s i r ρ is i+1 h i+1 ) 39

41 t k (s ir ρ i 1s i 1 h i r s i 1 ρ i 1 ) t k t k s (γ+1) i σ0 ( 1 h i + 1 h i+1 ) r(h i + h i+1 ), 1 i N s 1 f LQ i, i 1 = h i ρ i 1s i 1 r t krs i 4 + t k ρ i 1 r s i s i 1 h i t k 0s (γ+1) ( σ i ), i N s h i f LQ i, i+1 = h i+1 + ρ is i+1 r t k ρ i r s i s i+1 h i+1 + t krs i 4 t ks (γ+1) i σ 0 4h i+1, 1 i N s 1 f LQ N s, N s = t k ρ Ns 1r s Ns s Ns 1 h Ns t k (s(γ+1) N s σ0 h Ns rs N s 1 ) + t k r ρ Ns 1s Ns 1 + h N s 3 ρ N s 1rs Ns 1 With these formulas and the given value of u 0 h = max(k s, 0), we compute iteratively the unknown variables u k h, i for i = 1,,..., N s, and k = 1,,..., N t, and obtain the prices of our put option for any given values of both time t [0, 1] and the underlying asset s [0, S b ]. For option 's formulation with Crank-Nicolson scheme, using exactly the same calculations as in option 1, the entries for the mass and stiness matrix M and A are 40

42 m LQ i, i 1 = h i p i 1r (s i + s i 1 ), 1 i N s m LQ i, i = ( rs i 1ρ i 1 + h i 3 ) + rs i+1ρ i + h i+1 3, 1 i N s 1 m LQ 0, 0 = h 1 3 ρ 0rh 1 m LQ N s, N s = h N s 3 + ρ Ns 1r( s N s 1 + s N s 3 ) m LQ i, i+1 = h i+1 + ρ ir (s i + s i+1 ), 0 i N s 1 i, i 1 = σ 0s (γ+1) i h i a LQ + rs i, 1 i N s a LQ i σ0 h i+1 i, i+1 = s(γ+1) rs i, 0 i N s 1 a LQ 0,0 = rs 1 a LQ N s, N s = s(γ+1) N s σ0 h Ns rs N s 1 a LQ i, i = s(γ+1) i σ0 ( ) + r(h i + h i+1 ), 1 i N s 1 h i h i+1 41

43 .5 Mesh Renements For improving the accuracy of our numerical solutions to the equation (14) with boundary conditions (15) (1), especially when s is near the strike price K = 50 and S b, we extend the mesh in the S-direction (that is, add another interval [S b, S ], where S is a very large number, 30K) and divide the new interval [0, S ] into three subintervals [0, K], [K, S b ] and [S b, S ]. Then we discretize each of the subintervals as follows: First, we use either the same uniform partition or geometric partition, which was described at the beginning of Section (3.4), with N 1 and N subintervals for [0, K] and [K, S b ], respectively. Second, we decide to employ the geometric partition for the interval [S b, S ] with N 3 subintervals h N +N 1 +1, h N +N 1 +,..., h N3 +N +N 1 followed by the similar geometric partition used for [K, S b ] but now the geometric ratio r 3 is determined by the equation N h N +N 1 +1 = r h N3 +N +N 1 = h 1+γ N 3 +N +N 1. Our discrete space Vh b H1 0([0, S b ]) now becomes Vh H0([0, 1 S ]). The reason we choose h max( 1, 1+γ ) N 3 +N +N 1 can be explained as follows: we pick u in equation () as a function that will decay quickly when s, such as u = 1 for some parameters c. Let ϕ = u, we want to nd the best constant s c c such that each of the following integral in equation () converges: (α ϕ s, ϕ c(c + 1) ) = s 0 s γ c ds = c(c + 1) s γ c+1 γ c 0 converges only when c 1 + γ (β ϕ s, ϕ) = [(1 + γ)σ0s γ+1 rs] ds 0 s c+1 = (1 + γ)σ 0s (γ c)+1 rs c+1 c converges only when c max( 1, 1 + γ ) 4

44 1 rs c+1 (rϕ, ϕ) = r ds = 0 sc c ds converges only when c 1 Therefore, since γ > 1, we need to choose c = max( 1+γ, 1 ) since the convergence rate of our error will decay slower than h max( 1, 1+γ ) N 3 +N +N 1. Now we can nd the unique pair (r 3, N 3 ) in the same way used to nd the pair (r 1, N 1 ), and the formula is (r 3, N 3 ) = S S b h N3 +N ( +N 1 S S b (h N3 +N +N 1 ), (max(1 max( 1, 1+γ ), 1 + γ ) 1) log h N 3 +N +N 1 + 1) log r 3 Combining the new geometric partition for [S b, S ] with the uniform partition or geometric partition for each of the subinterval [0, K] and [K, S b ], we have four options to choose for discretizations in the S-direction. We employ a non-equidistant partition only in the S-direction, since having a mesh with an uniform parition in the T-direction gives the most accurate numerical solutions, especially when the local volatility is suciently large ( 0.7). Our purpose for discretizing the S-direction nonuniformly is to increase the number of points near the point u(k, 0) and u(s b, 1). This method helps eliminate the errors when the underlying asset price is near S b. Therefore, this non-equidistant mesh gives a much better accuracy, even for the cases when the local volatility is very high (that is, when σ 0.8). However, the trade-o is that this mesh requires a large number of points, and the required time of calculation increases proportionally for higher neness of the mesh (increase value of either N 1,N or N 3 ). 3 Numerical results A feature of the FEM method when applying it to solve the PDE representation of option pricing is the knowledge about the development of the option value function for each time step, which is denied as the complete term structure of the option. We can easily visualize the whole term structure of the European put option as in Figure 3 below. At the present time, t = 0, which is the front edge of the surface, one can clearly see the shape of the 43

45 non-smooth payo function u(s, 0) = (K s) +. By solving the functions u(s, t) for each s over the life-time of the option (that is, for each t runs from 0 to 1), the complete term structure is obtained. Towards the maturity time, t = 1, after N t time steps, we approach the function u(s, 1). Of the whole option surface, the special points of interest are near the points u(k, 0) and u(s b, 1) in which the errors of our numerical solutions were mainly distributed. Figure 3: Values of a put option with S b = 100, K = 50, σ = 0.5, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, geometric partition in S and uniform partition in T We set our parameters as follows: K = 50, γ = 0, S b = K, r = 0.03, N t = 00, N s = 1000,T = 1, S = 30K, N 3 = We will run our numerical schemes in two cases with σ = 0.4 and σ = 0.8. For each of these cases, we use either the uniform partitions for the interval [0, S b ] or extend the mesh by adding another interval [S b, S ] and for the new interval [0, S ], the geometric partition described in Section 4.5 is applied for [S b, S ] and the same uniform partition for [0, S b ]. In addition, for the T-direction, we employ the uniform partition in [0, T ]. For the two mesh renements, dene a domain D = [0, T ] [0, S b ] and compute the L (D)- norm and L (D)- norm errors (see Table 1) when comparing our numerical solutions to the exact solutions f(s, t) given the Black-Scholes formula, and compute the reduction factors of our errors in each case. The formulas for the L (D)- norm and L (D)- norm errrors are, L (D)-norm error = i=n s 1, j=n t 1 [ i, j=0 ( f(s i, t j ) + f(s i+1, t j ) + f(s i, t j+1 ) + f(s i+1, t j+1 ) 4 44

46 u(s i, t j ) + u(s i+1, t j ) + u(s i, t j+1 ) + u(s i+1, t j+1 ) ) 4 (t j+1 t j )(s i+1 s i )] 1 (73) L (D)-norm error = max 1 i Ns, 1 j N t u(s i, t j ) f(s i, t j ) (74) Based on the errors, we determine if adding the new interval [S b, S ] with geometric partition improves the accuracy of our numerical solutions. Figures 4-7 are the graphs of the errors for each mesh renement when σ = 0.4 and σ = 0.8. The exact solutions f(s, t) for these two cases given by the Black- Scholes formula are also shown in Figures 8-9. Since the investor mainly concerns about the errors of our numerical solutions at time t = 1, dene a domain Ω = [T ] [0, S b ], the L (Ω)- norm and L (Ω)- norm of those errors and the reduction factors are also shown in Table for cases when σ = 0.4. The formulas for the errors in L (Ω)- norm and L (Ω)- norm at time t = 1 are: L (Ω)-norm error at time 1 = i=n s 1 [ i=0 ( f(s i+1, t Nt ) + f(s i, t Nt ) u(s i+1, t Nt ) u(s i, t Nt ) (s i+1 s i )] 1 (75) ) L (Ω)-norm error at time 1 = max 1 i Ns u(s i, t Nt ) f(s i, t Nt ) (7) 45

47 Figure 4: Errors by uniform partition in [0, 100] and in T with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 1500, N 3 = 1000, σ = 0.4 Figure 5: Errors by uniform partition in [0, 100] and geometric partition in [100, 1500], uniform partition in T with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 1500, N 3 = 1000, σ = 0.4 Figure : Errors by uniform partition in [0, 100] and in T with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 1500, N 3 = 1000, σ = 0.8 Figure 7: Errors by uniform partition in [0, 100] and geometric partition in [100, 1500], uniform partition in T with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 1500, N 3 = 1000, σ = 0.8 4

48 Figure 8: Solutions by Black- Scholes formula with K = 50, γ = 0, r = 0.03, T = 1, σ = 0.4, N t = 00, N s = 1000, S = 1500, N 3 = 1000 Figure 9: Solutions by Black-Scholes formula with K = 50, γ = 0, r = 0.03, T = 1, σ = 0.8, N t = 00, N s = 1000,S = 1500, N 3 = 1000 Type of partition L (D)-norm error (σ = 0.4) L (D)-norm error (σ = 0.8) L (D)-norm error (σ = 0.4) L (D)-norm error (σ = 0.8) Uniform partition in [0,100] Geometric partition in [100,1500] Reduction Factor Table 1: L (D)- norm and L (D)- norm errors when σ = 0.4 and σ = 0.8 with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 30K, N 3 = 1000 Type of partition L (Ω)-norm error (σ = 0.4) L (Ω)-norm error (σ = 0.8) L (Ω)-norm error (σ = 0.4) L (Ω)-norm error (σ = 0.8) Uniform partition in [0,100] Geometric partition in [100,1500] Reduction Factor Table : L (Ω)- norm and L (Ω)- norm errors when σ = 0.4 and σ = 0.8 with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 30K, N 3 = 1000 Looking at Figures 4 7 and Table 1 above, we see that by adding the new interval [S b, S ] with geometric partition in the S- direction, the major errors near the point u(s b, 1) are eliminated, although the errors near the point u(k, 0) slightly increase. However, by using an extended mesh renement on [0, S ], the errors in L (D)- norm and L (D)- norm are reduced by a factor larger than compared to the errors obtained when using only the 47

49 mesh [0, S b ]. The reduction factors of those errors at time t = 1 are quite signicant (greater than 14, see Table ), which shows the advantage of an extended mesh renement by eliminating most of the errors at that time. In general, this extended mesh renement indeed helps improve the accuracy of our numerical solutons. Therefore, from this point, we will always include the new interval [S b, S ] with geometric partition when doing the renement in the S- direction. We also run our numerical schemes when S b = 1.K (that is, using less number of points in the S-direction), but the result is not suciently good. With the same set of parameters K = 50, γ = 0, S b = K, r = 0.03, σ = 0.4, S = 30K, N 3 = 1000, we now compare the errors in L (D)- norm and L (D)- norm (formulas (73) and (74)) when using dierent partitions for the intervals [0, S b ] and [0, T ] with dierent number of subintervals N s and N t to determine which partition best improves the accuracy of our numerical solutions. Note that we always use the geometric partition for [S b, S ] because it helps eliminate the major errors near the points u(s b, 1). There are exactly four possibilities of partitions in [0, S b ] [0, T ]: geometric partition for [0, S b ] and uniform partition for [0, T ], uniform partition for [0, S b ] and geometric partition for [0, T ], uniform partition for [0, S b ] and uniform partition for [0, t], geometric partition for [0, S b ] and geometric partition for [0, T ]. However, we tried our numerical schemes using uniform partition for [0, S b ] and geometric partition for [0, T ] and got bad results. We also computed the reduction factor of our errors in L (D)- norm and L (D)- norm for each of the other three renements to determine the approximated convergence rate of our reduction factor. The complete details are shown in Tables 3-5 below. Since the investor mainly concerns about the errors of our numerical solutions at time t = 1, the L (Ω)- norm and L (Ω)- norm of our errors and the reduction factors are also shown in Tables - 8 for cases when σ = 0.4. (N s, N t ) Max h i=1,...,ns, max t j=1,...,nt Min h i=1,...,ns, min t j=1,...,nt L (D)-norm error (σ = 0.4) L (D)-norm error (σ = 0.4) Reduction Factor in L (D)-norm Reduction Factor in L (D)-norm (000,400) 0.05, , (1000,00) 0.1, , (500,100) 0., , (50,50) 0.4, , Table 3: L (D)- norm and L (D)- norm errors for uniform partitions in both S T directions with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, T = 1, S = 1500, N 3 =

50 (N s, N t ) Max h i=1,...,ns, max t j=1,...,nt Min h i=1,...,ns, min t j=1,...,nt L (D)-norm error (σ = 0.4) L (D)-norm error (σ = 0.4) Reduction Factor in L (D)-norm Reduction Factor in L (D)-norm (000,400) , , (1000,00) , , (500,100) 0.35, , (50,50) 0.457, , Table 4: L (D)- norm and L (D)- norm errors for geometric partition in S, uniform partition in T with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, T = 1, S = 1500, N 3 = 1000 (N s, N t ) Max h i=1,...,ns, max t j=1,...,nt Min h i=1,...,ns, min t j=1,...,nt L (D)-norm error (σ = 0.4) L (D)-norm error (σ = 0.4) Reduction Factor in L (D)-norm Reduction Factor in L (D)-norm (000,400) , , (1000,00) 0.30, , (500,100) 0.35, , (50,50) 0.457, , Table 5: L (D)- norm and L (D)- norm errors for geometric partition in both S-T directions with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, T = 1, S = 1500, N 3 = 1000 (N s, N t ) L (Ω)-norm error (σ = 0.4) L (Ω)-norm error (σ = 0.4) Reduction Factor in L (Ω)-norm Reduction Factor in L (Ω)-norm (000,400) (1000,00) (500,100) (50,50) Table : L (Ω)- norm and L (Ω)- norm errors for uniform partition in both S T directions with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, T = 1, S = 1500, N 3 = 1000 (N s, N t ) L (Ω)-norm error (σ = 0.4) L (Ω)-norm error at t = 1 (σ = 0.4) Reduction Factor in L (Ω)-norm Reduction Factor in L (Ω)-norm (000,400) (1000,00) (500,100) (50,50) Table 7: L (Ω)- norm and L (Ω)- norm errors for geometric partition in S, uniform partition in T with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, S = 1500, N 3 =

51 (N s, N t ) L (Ω)-norm error (σ = 0.4) L (Ω)-norm error (σ = 0.4) Reduction Factor in L (Ω)-norm Reduction Factor in L (Ω)-norm (000,400) (1000,00) (500,100) (50,50) Table 8: L (Ω)- norm and L (Ω)- norm errors for geometric partition in both S-T directions with K = 50, γ = 0, S b = 100, r = 0.03, σ = 0.4, S = 1500, N 3 = 1000 From Tables 3-8, we nd that for suciently large values of N s and N t, the geometric partition in S and uniform partition in T for the intervals [0, S b ] and [0, T ] gives the smallest L (D)- norm and L (D)- norm errors and the highest reduction factors in both L (D)-norm and L (D)-norm. The approximated rate of convergence for each type of partition, in the order above, is: 1.84, 3.3,. in L (D)-norm and 1.53,.53, 1.5 in L (D)-norm. At time t = 1, such rate of convergence in the same order is:.3, 3.1,.83 in L (Ω)-norm and 1.43,.4,.89 in L (Ω)-norm. Due to these results, we determine to use the geometric partition in S and uniform partition in T for the interval [0, S b ] for all the following tests, since out of the four partitions, that partition improves the accuracy of our solutions the most. Although the geometric partition in S, uniform partition in T for the interval [0, S b ] [0, T ] and geometric partition for [S b, S ] is a good choice, we realize that for suciently large σ (that is, σ 0.7), our numerical schemes can reduce the errors near the point u(k, 0) by applying the Galerkin Least Square (GLS) stabilization method presented in Section 4.4. With the same set of parameters K = 50,γ = 0, S b = K, r = 0.03,N t = 00, N s = 1000, S = 30K, N 3 = 1000, we run our numerical schemes with GLS and the classical FEM for wo cases when σ = 0.4 (suciently small value) and σ = 0.8 (suciently large value). The errors of our numerical solutions are shown in Figure We also compute the L (D)- norm and L (D)- norm errors using formulas (73) and (74) to see the advantage of applying the GLS (see Table 9). The errors in L (Ω)-norm and L (Ω)-norm at time t = 1, which was the main concern of investors, are also shown in Table 10 for two cases of σ = 0.4 and

52 Figure 10: Errors with GLS when K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, σ = 0.4, geometric partition in S and uniform partition in T Figure 11: Errors with classical FEM when K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, σ = 0.4, geometric partition in S and uniform partition in T Figure 1: Errors with GLS when K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, σ = 0.8, geometric partition in S and uniform partition in T Figure 13: Errors with classical FEM when K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, σ = 0.8, geometric partition in S and uniform partition in T Numerical scheme L (D)-norm error (σ = 0.4) L (D)-norm error (σ = 0.4) L (D)-norm error (σ = 0.8) L (D)-norm error (σ = 0.8) Classical FEM GLS Reduction factor Table 9: L (D)- norm and L (D)- norm errors for σ = 0.4 and σ = 0.8 with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, T = 1, S = 1500, N 3 = 1000, geometric partition in S and uniform partition in T 51

53 Numerical scheme L (Ω)-norm error(σ = 0.4) L (Ω)-norm error (σ = 0.4) L (Ω)-norm error (σ = 0.8) L (Ω)-norm error (σ = 0.8) Classical FEM GLS Reduction factor Table 10: L (Ω)- norm and L (Ω)- norm errors for σ = 0.4 and σ = 0.8 with K = 50, γ = 0, S b = 100, r = 0.03, N t = 00, N s = 1000, S = 1500, N 3 = 1000, geometric partition in S and uniform partition in T As we can see from Table 9 and 10, the GLS did a great job in reducing the errors near the singular point u(k, 0). The reduction factors of our L (D)- norm and L (D)- norm errors are almost as large as and 1.8 even when σ is sucient large (0.8), which is demonstrated in the real-world situations in which the option prices are highly volatile. At time t = 1, the reduction factors in L (Ω)- norm and L (Ω)- norm are a little bit smaller, 1.9 and 1.4. The feature of improving the accuracy of our numerical solutions even when σ is large is exactly the advantage of adding the GLS into our numerical schemes. Finally, we consider applying the GLS on the same mesh renements used in Figures 10 and 1 with the set of parameters σ 0 = 0.3, K = 50, S b = K, N t = 00, N s = 1000, r = 0.03, S = 30K, N 3 = 1000 to run our numerical schemes in three scenarios: γ is negative (γ = 0.03), γ is positive (γ = 0.07) and γ = 0. We then obtain the numerical solutions for these three cases. Since there are no formulas to compute the exact solutions for γ 0, we expect that our numerical solutions are approximately close to the correct solution, thus we compare our numerical solutions to those given by the Black-Scholes formula when γ = 0. 5

54 Figure 14: Numerical solutions in the case γ = 0.03, S = 1500, N 3 = 1000, σ 0 = 0.3, K = 50, S b = 100, N t = 00, N s = 1000, r = 0.03, T = 1, geometric partition in S and uniform partition in T Figure 15: Solutions by Black-Scholes in the case γ = 0, σ = 0.3, r = 0.03, T = 1, K = 50, S b = 100, N t = 00, N s = 1000, S = 1500, N 3 = 1000 Figure 1: Numerical solutions in the case γ = 0.07, S = 1500, N 3 = 1000, σ 0 = 0.3, K = 50, S b = K, N t = 00, N s = 1000, r = 0.03, T = 1, geometric partition in S and uniform partition in T Figure 17: Solutions by Black-Scholes in the case γ = 0, σ = 0.3, r = 0.03, T = 1, K = 50, S b = 100, N t = 00, N s = 1000, S = 1500, N 3 =

55 Figure 18: Numerical solutions in the case γ = 0, S = 1500, N 3 = 1000, σ 0 = 0.3, K = 50, S b = K, N t = 00, N s = 1000, r = 0.03, T = 1, geometric partition in S and uniform partition in T Figure 19: Solutions by Black-Scholes in the case γ = 0, σ = 0.3, r = 0.03, T = 1, K = 50, S b = 100, N t = 00, N s = 1000, S = 1500, N 3 = 1000 Figure 0: Numerical solutions in the case γ = 0.07, S = 1500, N 3 = 500, σ 0 = 0.3, K = 50, S b = 100, N t = 00, N s = 1000, r = 0.03, T = 1, geometric partition in S and uniform partition in T Figure 1: Solutions by Black-Scholes in the case γ = 0, σ = 0.3, r = 0.03, T = 1, K = 50, S b = 100, N t = 00, N s = 1000, S = 1500, N 3 = 1000 Comparing the two pairs of gures side by side (Figures and Figures 18-19), we nd that for the case when γ= 0.3 or γ = 0, the graph of our numerical solutions are very much similar to that of the exact solutions given by Black-Scholes formula. The L (D)- norm and L (D)- norm errors, which were computed using formulas (73) - (74), when γ = 0.3, 0, 0.07 are 0.075, , and 0.141, ,1.39, respectively. However, 54