Numerical Challenge in Option Pricing - polytechnique · Numerical Challenge in Option Pricing University of Paris VI, Laboratoire J.-L. Lions ... Pricing Options: 3 numerical methods

Numerical Challenge in Option Pricing

University of Paris VI, Laboratoire J.-L. Lions

Olivier Pironneau1

1

with Yves Achdou (UP-VII), N. Lantos & A. Conze (NextGen-IXIS) + BP-IXIS and Zeliadewww.ann.jussieu.fr/pironneau

En l'honneur de Luc Tartar

O. Pironneau (LJLL-UPMC) Numerical Challenge in Option Pricing 7/07 in L.Tartar's Honor 1 / 33

Introduction to the Black & Scholes Model

A nancial asset with tendency µ and volatility σ

dSt = St(µdt + σdWt), S0 known

• µ = r(t) the interest rate under the risk-neutral probability law• (Wt) : a standard Brownian motion.

A European option on S is a contract giving the right to buy ( call) orsell ( put) this asset at date T (maturity) at price K (strike).

Its value at t is the expected prot, discounted at t:

Ct = e−r(T−t)E((ST − K )+

), Pt = e−r(T−t)E

((K − ST )+

)










((K − ST )+

)










((K − ST )+

)


Pricing Options: 3 numerical methods

1. Tree methods2. Montecarlo:


)dSt = St(µdt + σdWt), S(0) = S0. dWt ≈

√dtN (0, 1)

3. Ito Calculus :

∂tC +σ2x2

2∂2xxC + rx∂xC − rC = 0

C (T , x) = (x − K )+

C (0, t) = 0C (x , t) ∼ x − Ke−r(T−t) when x →∞

For a put P(x , t) = Ke−r(T−t) − x + C (x , t) just change the B.C.

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’result.dat’ using 1:2’result.dat’ using 1:3’result.dat’ using 1:4


Existence, Uniqueness and Regularity

∂tP −σ2x2

2∂2xxP − rx∂xP + rP = 0

P(0, x) = (K − x)+, limx→∞

P(x , t) = 0

Use a weighted Sobolev H1:

V = u ∈ L2(R+) : x∂xu ∈ L2(R+)(∂tP,w) + a(P,w) = 0, ∀w ∈ V , P(0) = (K − x)+,P ∈ L2(0,T ,V )

Theorem

Assume 0 < σm ≤ σ(x , t) < σM , x∂xσ ∈ L∞ and r , σ, x∂xσ Lipschitz in

t, then P exists, is continuous in time and x2∂xxP ∈ L2(R+). Furthermore

if x2∂xxσ ∈ L∞ then P is convex.


A posteriori estimates (Y.Achdou)

Proposition

[[u − uh,δt ]](tn) ≤ c(u0)δt

+µ

σ2min

n∑m=1

ηm2 +

δtmσ2min

g(ρδt)m−1∏i=1

(1− 2λδti )∑ω∈Tmh

ηm,ω2

1

2

where L, µ are the time-continuity constants of σ2, r , xσ∂xσ in L∞,c(u0) = (‖u0‖2 + δt‖∇u0‖2)1/2, g(ρδt) = (1 + ρδt)

2max(2, 1 + ρδt)

η2m = δtme−2λtm−1 σ

2min

2|umh − um−1h |2V ,

ηm,ω =hω

xmax(ω)‖umh − um−1h

δtm− rx

∂umh∂x

+ rumh ‖L2(ω)


Best Numerical Method

• Implicit in time, centered in space (upwind usually not necessary):

Pn+1 − Pn

δt− x2σ2

2∂2xxP

n+ 12 + rPn+ 1

2 − xr∂xPn+ 1

2 = 0

• FEM-P1 + LU factorization• mesh adaptivity + a posteriori estimates• Banks need 0.1% precision in split seconds ...within Excel

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PDE solutionBlack-Scholes formula


Numerical Example

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American Options

No arbitrage implies that if today's prot is greater than expected protthen the owner of the option should exercise his right and sell/buy beforematurity.

∂u

∂t− σ2x2

2

∂2u

∂x2− rx

∂u

∂x+ ru ≥ 0

u ≥ u

(∂u

∂t− σ2x2

2

∂2u

∂x2− rx

∂u

∂x+ ru)(u − u) = 0,

with u(t, 0) = Ke−rt , u(0, x) = u(x) := (K− x)+


Variational Setting

V =

v ∈ L2(R+), x

∂v

∂x∈ L2(R+)

,

K = v ∈ L2(0,T;V), v ≥ u a.e. in (0,T)×R+.

Find u ∈ K ∩ C 0([0,T ]; L2(R+)),∂u

∂t∈ L2(0,T ;V ′),

s.t. <∂u

∂t+ A(t)u, v − u > ≥ 0, ∀v ∈ K,

u(t = 0) = u, with

< A(t)v,w >=

∫R+

(σ2

2x2∂v

∂x

∂w

∂x+ (σ2 + x

∂σ2

∂x− r)x

∂v

∂xw + rvw

)dx.


Existence and Regularity Results (Y. Achdou)

If σ ∈ [σ, σ] and x |σ∂xσ| < M, then the solution exists and is unique.Assume that there exists M > 0 s.t.

|x2∂2σ2

∂x2|+ |∂σ

2

∂t|+ |x ∂

2σ2

∂x∂t| ≤ M,

• Then the free boundary t → γ(t)is C 0,• If t → γ(t) is Lipschitz in [τ,T ] then with L∞((τ,T )×R+)

‖(γ(σ′2)− γ(σ2))+‖3L3(τ,T ) ≤ cτ‖σ2 − σ′2‖2Q

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t

S

"fb0""fb6""fb12""fb19"


Semi-Smooth Newton Method (K.Kunisch)

Reformulate the problem at each time step as

a(u, v)− (λ, v) = (f , v) ∀v ∈ H1(R+), i .e.Au − λ = f

λ−min0, λ+ c(u − φ) = 0,

The last eq. is equivalent to λ ≤ 0, λ ≤ λ+ c(u − φ) i.e. u ≥ φ, λ ≤ 0,with equality on one of them for each x . This problem is equivalent for anyreal constant c > 0 because λ is the Lagrange multiplier of the constraint.

Newton's algorithm gives• Choose c > 0, , u0, λ0, set k = 0.• Determine Ak := x : λk(x) + c(uk(x)− φ(x)) < 0• Set uk+1 = argminu∈H1(R+)a(u, u)− 2(f , u) : u = φ on Ak• Set λk+1 = f − Auk+1


More Complex Options

Jump Processes In B&S use a Lévy instead of a Wiener process :

∂tC −σ2x2

2∂xxC − rx∂xC − rC

= −∫R

(C (xey , t)− C (x , t)− x(ey − 1)∂xC ) k(y)dy

Proposition (E.Voltchkova-R.Cont)The following converges if δt < C and (+FEM)

Cm − Cm−1

δt− σ2x2

2∂xxC

m − rx∂xCm − rCm

= −∫R

(Cm−1(xey )− Cm−1(x)− x(ey − 1)∂xC

m−1) kdyAsian Options

∂tC −σ2x2

2∂xxC − rx∂xC − rC +

x − y

t∂yC = 0

2D problem which requires upwind.O. Pironneau (LJLL-UPMC) Numerical Challenge in Option Pricing 7/07 in L.Tartar's Honor 12 / 33

Multidimensional problems

1. Basket Option d <WiWj >= σijdt

dSi = Si (rdt + σidWi ), i = 1..d , u = e−(T−t)rE(∑

Si − K )+

Ito calculus leads to a multidimensional Black-Scholes equation

∂tu −∑i

(σ2ijxixj

2∂xixiu − rxi∂xiu) + ru = 0 u(0) = (

∑i

xi − K )+

Stochastic Volatility models (Stein-Stein, Orstein-Uhlenbeck, Heston)

dSt = St(rdt + σtdWt), σt =√Yt ,

dYt = κ(θ − Yt)dt + βdZt , d <Wt ,Zt >= ρdt

∂tU + aU + bx∂xU −x2y

2∂xxU −

β2y

2∂yyU − ρβyx∂2xyU + c∂yU = 0

a = µ− κ− ρβ − y , b = (µ− 2y − ρβ), c = κ(θ − y)− ρβy − β2


www.freefem.org (F. Hecht, O.P. et al)

real T=5, K=120, r=0.0, mu =0.4,...; int n=1,L=500, LL=250, ...;mesh th = square(50*n,50*n,[x*L,y*LL]);fespace Vh(th,P1);Vh u=f,v,w,rhs;problem eq1(u,v,init=j) = int2d(th)( u*v/dt+ dx(u)*dx(v) *x*x*y*y*a11 + dy(u)*dy(v) *y*y*a22+ dy(u)*dx(v) *y*y*x*a12 + dx(u)*dy(v) *y*y*x*a21+ dx(u)*v *x*(y*( a1 + y/sc2)-r) + dy(u)*v *y*(y*a2+mu2))+ rhs[] + on(1,u=max(K-x,0.)) ;varf vrhs(u,v) = int2d(th)( v*u/dt );matrix M = rhs(Vh,Vh);for (int n=0; n<Nmax ; n++)w=u; rhs[] = M*u[]; eq1 ; t= t+dt; j++;if(n==20*(nM = vrhs(Vh,Vh); rhs =0; u=max(K-x,0.); t=0; j=0; Execute


SuperLU, Hypre... (N. Lantos)

Mesh size LU [s] Relative error superLU[s] Relative error

101× 101 10.094 3.532 % 2.39 3.076 %

126× 126 14.547 1.338 % 4.016 1.797 %

151× 151 22.313 0.751 % 6.203 0.489 %

176× 176 31.985 1.131% 8.735 0.790 %

201× 201 43.938 0.432 % 12.109 0.670%

Comparison of CPU time for the LU algorithm and superLU for a product

put option on a uniform mesh and 200 step time.


Numerical Results: adapted meshes

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"exercise_250"

Basket Euro option Exercise region for an American basket option.


Calibration: How to choose σ(S , t)?

... by trying to reproduce the prices observable on the market:

Every day, one can observe (in the news)• The underlying asset So• The values (ci )i∈I of calls (CKi ,Ti

(S0, 0))i∈I .

Inverse Problem : nd σ(x , t) s.t. for all , Cii∈I given by

∂tCi +σ2x2

2∂2xxCi + rx∂xCi − rCi = 0,

Ci (x ,Ti ) = (x −Ki)+, for all t ∈ [0,Ti[, x ∈ R+,

such that Ci (x0, 0) = ci ...it involves as many B&S as dierent Ki , i ∈ I .

min∑|Ci (x0, 0)− ci |2 : subject to all PDEs, i=1..I


Dupire's Equation

Theorem Dupire's equation for CS,t(T ,K ) = CT ,K (S , t)

∂TC−σ2(T ,K )K 2

2∂2KKC+ rK∂KC = 0 C(0,K ) = ( S − K )+

Proof (r = 0) Let ∂tu + µ(x , t)∂xxu = 0; ∂tp − ∂xx(µp) = 0

Then, with appropriate boundary conditions∫R+

u(0)p0 =

∫R+

uTp(T ) = u(x0, 0) when p(x , 0) = δ(x − x0)

Let ∂xxv = p then∂tv − µ∂xxv = ax + b, v(0) = c + d(x − x0)

+ in R× (0,T )

uT = (K − x)+, ⇒ ∂xxuT = −δ(x − K ) ⇒ u(x0, 0) = −v(K ,T )d Thisproof does not use Kolmogorov's eq. nor the PDE's Green function and itsets Dupire's variable in the S , t-space.


Bruno Dupire

Inverse Problem : nd σ(S , t) s.t.

min∑|u(Ki ,Ti )− ci |2

subject to Cii∈I given by

∂tu − ∂2xxu + rx∂xu = 0, .u(x , 0) = (x − S0)

+

Only one PDE.


Dupire identities Basket Options

An option on two assets would be modeled by

∂tC + A : ∇∇C + rxi∂xiC − rC = 0

A =1

2

(σ21x

21 2qx1x2

2qx1x2 σ22x22

), C (T ) = (x1 + x2 − K )+

So let p be solution of the adjoint equation with

p(t0) = δ(x1 − S01)δ(x2 − S02).

As before

CK ,T (S01, S02, t0) =

∫R+2

p(x1, x2,T )(x1 + x2 − K )+dx1dx2


To remove the Dirac singularities

Let p = ∂x1x2w .Then, when σi does not depend on xj , j 6= i ,

∂tw −1

2∇ · (A∇w) + rx1∂x1w + rx2∂x2w + rw = 0

w(x1, x2, t0) = (1− H(x1 − S01))(1− H(x2 − S02))

It corresponds to a special choice for the integration constant which givesan exponential decay at innity for w ; H is the Heaviside function. Finally :Proposition If r is function of t only and σi does not depend on Sj , j 6= i

CK ,T (S01, S02, t0) =

∫x1+x2=K

w(x1, x2,T )

+

∫ ∞K

w(x1, 0,T )dx1 +

∫ ∞K

w(0, x2,T )dx2


Comparison

Plots of S → C computed by B&S and S → C computed by Dupire for abarrier option.

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"u.txt" using 1:2"u.txt" using 1:3"u.txt"using 1:4S1\S2 20 50 80 110 140

20:c -0.097 -0.089 5.61 31.57 61.4920:d 1.1e-07 0.0065 5.88 32.11 61.67

50:c 4.32 31.50 61.49 91.49 31.5750:d 4.39 31.52 61.81 91.44 32.11

80:c 61.49 91.49 121.49 61.49 5.6180:d 61.70 91.99 121.62 61.81 5.88

110:c 121.49 151.49 91.5 31.50 -0.089110:d 122.23 151.86 91.99 31.52 0.0065

140:c 181.49 121.5 61.49 4.3 -0.097140:d 181.60 122.2 61.69 4.3 0

Comparison between a direct calculation of a basket call with several solutions of

Dupire's


The General Case

Use only the adjoint equation (which is Kolmogorov's or Fokker-Planck)

∂tp−σ2(x , t)x2

2∂2xxp+ rx∂xp = 0∫

R+

u(0)p(0) =

∫R+

uTp(T ) = u(S0, 0) when p(x , 0) = δ(x − S0)

Approximate δ by µ−1e−(

x−S0µ

)2, use mesh adaption:


Precision

adjoint Analytical relative error

ES0 94.1552 94.1548 0.00042%

ES20 11044.8 11048.3 0.03168%

ES1 96.9252 96.9219 0.00340%

ES21 12932.3 12939.4 0.05487%

ES0S1 9766.74 9768.25 0.01546%

E(S0 − K0)+ 4.53898 4.54716 0.17989%

E(S1 − K1)+ 11.426 11.4235 0.02188%

E(K0 − S0)+(K1 − S1)

+ 211.865 211.82057 0.02098%Comparison between direct calculation of C , the basket call on S1, S2,based on duality lines :c and C computed by solving the Dupireequation) lines :d.


Duality at the Discrete Level

FEM + Euler implicit with time step δt ⇒

(B + A)Cn − BCn+1 = 0 CN = CT

where Cn is the vector of values of Ch(qi , nδt) and A,B are

Bij =1

δt

∫Ωh

W iW j , Aij = a(W i ,W j)

the bilinear form of the PDE. Let Pn+1 be

(A + B)TPn+1 − BTPn = 0 P0 = P0

Then 0 = PnT (A + B)Un − PnTBUn+1 = Pn−1TBUn − PnTBUn+1

Summing up over all n gives P0TBU1 = PN−1TBUN

Choosing P0j = δij gives (

I∑j=1

bij)U1i ≈ (BU1)i = PN−1TBUN


Results on a Basket Call

S1\S2 20 50 80 110 140

20:c -0.097 -0.089 5.61 31.57 61.49

20:d 1.1e-07 0.0065 5.88 32.11 61.67

50:c 4.32 31.50 61.49 91.49 31.57

50:d 4.39 31.52 61.81 91.44 32.11

80:c 61.49 91.49 121.49 61.49 5.61

80:d 61.70 91.99 121.62 61.81 5.88

110:c 121.49 151.49 91.5 31.50 -0.089

110:d 122.23 151.86 91.99 31.52 0.0065

140:c 181.49 121.5 61.49 4.3 -0.097

140:d 181.60 122.2 61.69 4.3 0

Comparison between direct calculation of C based on B&S lines :c and U

computed by solving the Discrete Kolmogorov-Dupire equation lines :d.

at T=1, σ1 = 0.2σ2 = 0.3, q = −0.6, rate = µ1 = µ2 = 5%, K= 100.


Part II Calibration

Program: Build a robust set of tools, for all models (Americans?).Start with European vanilla options.• Automatic Dierentiation• Conjugate Gradient Methods (and more)• Spline representations of the volatility surface

Build all the constraints within the parametrization.Ex: σ(x , t) bilinear inside an (x,t) parallelogram, grows large outside and isC 1 ⇒ 8 parameters: Sij , σij but 0 < S1(t) < S2(t) is needed so set

z0 =√S11, z1 =

√S21, z2 =

√S21 − S11, z3 =

√S12 − S22,

z2∗i+j+1 =

√σij

1 +√σij, i , j = 1, 2

so as to work with an unconstrained set of parameters zi70;


The spline smile

σ(S , t) =

a + (2σ1−a

S1− σ2−σ1

S2−S1 )S + (σ2−σ1S2−S1 −

σ1−aS1

)S2

S1if (S < S1)

σ2S−S1S2−S1 + σ1

S2−SS2−S1 if S1 ≤ S ≤ S2

σ2 + (S − S2)σ2−σ1S2−S1 +

((S − S2)

σ2−σ1S2−S1

)2if S > S2

where Si , σi , i = 1, 2 are linear in t:

Si = Si1(1−t

T) + Si2

t

T, σi = σi1(1−

t

T) + σi2

t

T


Observation Data

Strike 1 Month 2 Months 6 Months 12 Months 24 Months 36 Months

700 733800 650.6900 569.81000 467.8... ... ... ... ... ... ...

1215 253.41225 219 2451250 196.6 224.2 269.21275 174.5 203.9 2511300 152.9 184.1 233.21325 131.9 164.9 215.81350 111.7 146.3 198.91365 1001375 50.6 60 92.5 139 182.61380 46.1 55.81385 41.8 51.8... ... ... ... ... ... ...

1700 32.7 75,71800 15.51900 5.2

index SPX on 21.12.2006) at spot price 1418.3, r=3/100


Results I

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Perspective

• The future of PDE in nance is both bright and uncertain.• Monte-Carlo still much in use• Calibration in 1D is a challenge what if d > 1?• Yet the demand is for d > 1.• applies also to interest rates, risk,• insurance, resource management etc.• Handbook of Numerical Analysis 2008.


Sparse Grids

If f is analytic then ∫D

f ≈∑i

f (qi )ωi

where most of the points are on ∂D. The argument is recursive.S.A. Smolyak: Quadrature and interpolation formulas for tensorproducts of certain classes of functions Dokl. Akad. Nauk SSSR 4 pp240-243, 1963.

Michael Griebel: The combination technique for the sparse gridsolution of PDEs on multiprocessor machine. Parallel ProcessingLetters 2 1(61-70), 1992.In polynomial approximations of f most of the mixed terms x i1x

j2... are

not needed.


Sparse Grids

If f is analytic then ∫D

f ≈∑i

f (qi )ωi

where most of the points are on ∂D. The argument is recursive.S.A. Smolyak: Quadrature and interpolation formulas for tensorproducts of certain classes of functions Dokl. Akad. Nauk SSSR 4 pp240-243, 1963.

Michael Griebel: The combination technique for the sparse gridsolution of PDEs on multiprocessor machine. Parallel ProcessingLetters 2 1(61-70), 1992.In polynomial approximations of f most of the mixed terms x i1x

j2... are

not needed.


Sparse Grids

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-0.1 0

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1

"Bsk_Put_SparseG"

Computed by D. Pommier (CIFRE at Credit Lyonnais)see also C. Schwab et al.


Documents

Numerical Challenge in Option Pricing - polytechnique · Numerical Challenge in Option Pricing University of Paris VI, Laboratoire J.-L. Lions ... Pricing Options: 3 numerical methods