computational methods for quantitative finance

PreliminariesPricing by binomial tree method

Computational Methods for Quantitative FinanceModels, Algorithms, Numerical Analysis

Christoph SchwabN. Hilber, C. Winter

ETHZ-UNIZ Master of Advanced Studies in Finance

Spring Term 2008

Christoph Schwab Computational Methods for Quantitative Finance


Scope of the course

Analysis and implementation of numerical methods forpricing options.

Models: Black-Scholes, stochastic volatility, exponential Levy.

Options: European, American, Asian, barrier, compound . . .

In this course: Focus on deterministic (PDE based) methods

I binomial and multinomial trees

I Finite Difference methods

I Finite Element methods

This course will be continued by the course Monte Carlo methodsin autumn 2008.



Organization of the course

13 lectures (2 hours) + 12 exercise classes (1 hour).

I No lectures on March, 27. and May, 1.

I No exercise class on March, 25.

I Testat: 75% of solved homework assignments(theoretical exercises + MATLAB programming).

Examination: On Thursday, May 29, 10–12.Written, closed-book examination includes theoretical andMATLAB programming problems.Examination takes place on ETH-workstations running MATLABunder LINUX. Own computer will NOT be required.



Details of the course

I Elementary Pricing

1. Pricing by binomial trees2. Pricing by Finite Difference Solution of PDEs3. American options4. Barrier and Asian contracts

II Pricing by Finite Element methods

5. Variational Formulation of BS model6. Finite Element Methods for infinite horizon problems7. Time stepping for finite horizon European vanillas8. Finite horizon American contracts

III Pricing in incomplete markets

9. Stochastic volatility models10. Levy models



OptionsStochastic processesBlack-Scholes model

Outline

PreliminariesOptionsStochastic processesBlack-Scholes model

Pricing by binomial tree methodBinomial tree modelMatlab implementation: Euro1.m–Euro9.m




Definition

An option is a contract which gives the right (but not theobligation) to buy or to sell a risky asset at a specified date T inthe future (or within a specified period) at a specified price K.

T : maturity or expiry date (after T the option is worthless)

K: strike or exercise price

Options are derivative contracts.

Examples of underlying assets (or underlyings): stocks (shares ofa company), indices (weighted means of several stocks), bonds,currencies, other options . . .




Examples

Call: right to buy the underlying

Put: right to sell the underlying

Exercise styles

I European: exercise only at maturity T

I American: exercise at any date before the maturity T(the holder chooses when exercise the option)

More complicated, exotic (as opposed to simple vanilla) options:

I Barrier: depend on particular asset price level(s) attained over a period

I Asian: depend on the average price of the underlying over a period

I Lookback: depend on the minimum or maximum asset price over a period

(⇒ path-dependent options)




Value of an option

The right without obligations is not given for free: option has avalue called premium (do not confuse with the exercise price K!)

What is the fair price of an option? It depends on:

I Assumptions on the market

I Model for the underlying price

Note: the price of standard liquid options is defined by the market(offer-demand equilibrium). Models are mostly needed forvaluation of exotic options.




Modelling assumptions on the market

I There are no arbitrage opportunities: no gain without initialinvestment nor risk (no free lunch).

I There exists a riskless bank account with riskless continuouslycompounded interest rate r ≥ 0:

time 0 t

$ X ertX

I Market is frictionless: there are no transaction costs, interestrates for lending and borrowing are equal, all securities andcredits are available at any time and in any volume.

I All market participants have immediate access to anyinformation; individual trading does not influence the price.




Payoff function

The payoff of an option is its value at the time of exercise.The payoff is a function of the underlying price S at this time.

Examples:

I Call: max(S −K, 0) ≡ (S −K)+.

I Put: max(K − S, 0) ≡ (K − S)+.

Note: the payoff function can be obtained by no arbitrageargument. It does not depend on the model for the underlyingprice.On the contrary, to find values of the option before the exercise, wehave to do some assumptions on the evolution of the asset price.




Stochastic modelling of the stock price

Stock prices in the future are not known: they are modelled bystochastic processes.

A stochastic process (St)t∈[0,T ] is a random function of t.

Filtered probability space: (Ω,F ,P, (Ft)t∈[0,T ]).

I Ω: set of elementary events ω

I F : σ-algebra of all events (i.e. subsets of Ω)

I P: probability measure on FI (Ft)t∈[0,T ] (filtration): increasing family of σ-algebras.Ft represents the information available up to time t.

(St(ω))t∈[0,T ] is a trajectory or sample path of the process.




Important definitions

A stochastic process X is adapted (w.r.t. the filtration) if Xt isFt-measurable (notation: Xt ∈ Ft).It means that the value of Xt is revealed at time t.

X is a Markov process if, for any bdd Borel function f and ∀s ≥ t,E[f(Xs) | Ft] = E[f(Xs) |Xt]:the future values of X do not depend on the past but only on thepresent value.

An adapted process X is a martingale if

(i) E[|Xt|] <∞ for all t ≥ 0(ii) E[Xs | Ft] = Xt, for all s ≥ t

(the best prediction for future values is the present value).

X is cadlag if its trajectories Xt(ω) are almost surely rightcontinuous and have left limits at all t ∈ [0, T ].French: continu a droite, limite a gauche.




Brownian motion

Normal distribution N (µ, σ2) has the density

f(x) = 1√2πσ2

exp(− (x−µ)2

2σ2

).

A stochastic process W is a Wiener process or Brownian motion if

I W0 = 0 a.s.

I W has independent increments: for all t1 ≤ t2 ≤ t3 ≤ t4,Wt4 −Wt3 ⊥⊥Wt2 −Wt1 .

I increments of W are normally distributed:Wt+s −Wt ∼ N (0, s).

Properties:

I Wiener process is a Markov process and a martingale.

I Wt ∼ N (0, t), E[Wt] = 0, VarWt = t.

I W has continuous paths.




Black-Scholes model

In the Black-Scholes model, the stock price is given by

St = S0e(r−σ2/2)t+σWt

where Wt is a Brownian motion (under the risk-neutral probability).

Risk-neutral means the discounted price e−rtSt is a martingale.

The value at time t ≤ T of a European option with maturity Tand payoff H(S) is defined as

Vt = e−r(T−t) E[H(ST ) | Ft].Due to the Markov property,

Vt = V (S, t) = e−r(T−t) E[H(ST ) |St = S].




Example: European call price

European Call: payoff function H(S) = (S −K)+.

Let Xt = (r − σ2/2)t+ σWt. Call price at time t is:

C(S, t) = e−r(T−t) E[(ST −K)+ |St = S]= e−r(T−t) E[(S0e

XT −K)+ |St = S]= e−r(T−t) E[(S0e

Xt+(XT−Xt) −K)+ |St = S]

= e−r(T−t) E[(Ste(r−σ2/2)(T−t)+σ(WT−Wt) −K)+ |St = S]

= e−r(T−t) E[(Se(r−σ2/2)(T−t)+σWT−t −K)+]

= E[(Se−σ2/2(T−t)+σWT−t −Ke−r(T−t))+]

Let τ = T − t (time to maturity). Since Wτ ∼ N (0, τ), we obtain

C(S, t) =∫ ∞−∞

(Se−σ2/2τ+σx −Ke−rτ )+

1√2πτ

exp(−x

2

2τ

)dx.




Black-Scholes formula

Calculation gives the Black-Scholes formula for European call price:

C(S, t) = S N(d1)−Ke−rτN(d2),

τ = T − t, d1 =log(S/K) + (r + σ2/2)τ

σ√τ

, d2 = d1 − σ√τ

where N(x) is the cumulative distribution function of N (0, 1):

N(x) =∫ x

−∞1√2π

exp(−y

2

2

)dy.

Black-Scholes formula in Matlab:

[Call, Put] = blsprice(S, K, r, τ , σ, DividendRate)



Binomial tree modelMatlab implementation: Euro1.m–Euro9.m

Outline

PreliminariesOptionsStochastic processesBlack-Scholes model

Pricing by binomial tree methodBinomial tree modelMatlab implementation: Euro1.m–Euro9.m




Model setting

We construct a simple discrete time model:

I M time steps: 0 = t1 < t2 < · · · < tM < tM+1 = T ,ti = (i− 1)∆t, i = 1, . . . ,M + 1, ∆t = T/M .

I At each time step there are only two possibilities for the price:S Su with probability p or S Sd with probability 1− p,d < 1 < u.

Parameters u, d, and p will be chosen such that the binomialmodel converges to the Black-Scholes model as M →∞.




Choice of parameters u, d, p

We equate expectations and variances in binomial and BS models:

Binomial BS modelE[Si+1 |Si] pSiu+ (1− p)Sid = Sier∆t

Var [Si+1 |Si] p(Siu)2 + (1− p)(Sid)2

−(pSiu+ (1− p)Sid)2 = (Si)2e2r∆t(eσ2∆t − 1)

To close the system we choose a third condition. For instance,ud = 1

(another popular choice is p = 1/2). Solving these equations yields:

u = A+√A2 − 1, d = 1/u, p =

er∆t − du− d ,

where A = 12(e−r∆t + e(r+σ2)∆t).




Option valuation using the binomial tree

Forward phase: initializing the treeStarting from S1

1 = S we can find all nodes Sin, i = 2, . . . ,M + 1,n = 1, . . . , i of the tree.

Backward phase: calculating the option valuesThe values of the option at tM+1 = T are given by the payoff:

VM+1n = H(SM+1

n )

We use the martingale property of the discounted option price:

V i = e−r∆t E[V i+1 |V i]which leads to

V in = e−r∆t (pV i+1

n+1 + (1− p)V i+1n )

So, we can go back through the tree to find V 11 = V (0, S).




% EURO1 Binomial method for a European Put.

% Unvectorized version.

W = zeros(M+1,1);

W(1) = S;

% Asset prices at time T

for i = 1:M

for n = i:-1:1

W(n+1) = u*W(n);

end

W(1) = d*W(1);

end

% Option values at time T

for n = 1:M+1

W(n) = max(K-W(n),0);

end

% Re-trace to get option value at time zero

for i = M:-1:1

for n = 1:i

W(n) = exp(-r*dt)*(p*W(n+1) + (1-p)*W(n));

end

end

disp(’Option value is’), disp(W(1))




Avoid nested pair of for loops, which requires O(M2) operations.Use instead SM+1

n = dM+1−nun−1S, 1 ≤ n ≤M + 1.


% Initial tree computation improved.

W = zeros(M+1,1);

% Asset prices at time T

for n = 1:M+1

W(n) = S*(d^(M+1-n))*(u^(n-1));

end


for n = 1:M+1

W(n) = max(K-W(n),0);

end


for i = M:-1:1

for n = 1:i

W(n) = exp(-r*dt)*(p*W(n+1) + (1-p)*W(n));

end

end





Eliminate the first two for loops by elementwise operations onvectors.


% Partially vectorized version.


W = max(K-S*d.^([M:-1:0]’).*u.^([0:M]’),0);


for i = M:-1:1

for n = 1:i

W(n) = exp(-r*dt)*(p*W(n+1) + (1-p)*W(n));

end

end





Remove unnecessary computations. The scaling factor e−r∆t ineach time step can be accumulated into e−rT . Precompute 1− pas q.


% Partially vectorized version.

% Redundant scaling removed.


W = max(K-S*d.^([M:-1:0]’).*u.^([0:M]’),0);


q = 1-p;

for i = M:-1:1

for n = 1:i

W(n) = p*W(n+1) + q*W(n);

end

end

W(1) = exp(-r*T)*W(1);





The final O(M2) loop is still dominating. Remove one for loop byworking directly with arrays rather than scalars.


% Vectorized version, uses shifts via colon notation.


W = max(K-S*d.^([M:-1:0]’).*u.^([0:M]’),0);


q = 1-p;

for i = M:-1:1

W = p*W(2:i+1) + q*W(1:i);

end

W = exp(-r*T)*W;

disp(’Option value is’), disp(W)

value = W;




Use the binomial expansion

V 11 = e−rT

M+1∑k=1

CMk−1pk−1(1−p)M+1−kVM+1

k , CNk =N !

k!(N − k)!.


% Uses explicit solution based on binomial expansion.

% Unvectorized, subject to overflow.


W = max(K-S*d.^([M:-1:0]’).*u.^([0:M]’),0);

% Binomial coefficients

v = zeros(M+1,1);

for k = 1:M+1

v(k) = nchoosek(M,k-1);

end

value = exp(-r*T)*sum(v.*(p.^([0:M]’)).*((1-p).^([M:-1:0]’)).*W)’;

disp(’Option value is’), disp(value)




Individual coefficients CNk take on some extremely large values:the largest element in v is of order 1075! ⇒ rounding errors mayarise. Compute V 1

1 =∑M+1

k=1 akVM+1k . The coefficients ak satisfy

the following recurrence

ak =p

1− pM − k + 2k − 1

ak−1, a1 = (1− p)M

Vectorize the computation of v using cumprod.



% Vectorized, subject to underflow.


W = max(K-S*d.^([M:-1:0]’).*u.^([0:M]’),0);

% Recursion for coefficients

a = cumprod([(1-p)^M,(p/(1-p))*[M:-1:1]./[1:M]]);

value = exp(-r*T)*sum(a’.*W);





The first element in a is (1− p)M . If this value drops below 10−325

which is Matlab’s realmin*eps threshold, then it underflows tozero. To overcom this underflow problem use logarithms:

log(ak) =M∑i=1

log(i)−k−1∑i=1

log(i)−M−k+1∑i=1

log(i)

+(k − 1) log(p) + (M + 1− k) log(1− p).% EURO8 Binomial method for a European Put.


% Vectorized and based on logs to avoid overflow.


W = max(K-S*d.^([M:-1:0]’).*u.^([0:M]’),0);

% log/cumsum version

csl = cumsum(log([1;[1:M]’]));

tmp = csl(M+1) - csl - csl(M+1:-1:1) + log(p)*([0:M]’) + log(1-p)*([M:-1:0]’);

value = exp(-r*T)*sum(exp(tmp).*W);





Certain terms in V 11 = e−rT

∑M+1k=1 CMk−1p

k−1(1− p)M+1−kVM+1k

are predictably zero: ∃z ∈ 1, . . . ,M + 1 such that

VM+1k = 0, z + 1 ≤ k ≤M + 1.



% Vectorized, based on logs to avoid overflow, and avoids computing with zeros.

% cut-off index

z = max(1,min(M+1,floor( log((E*u)/(S*d^(M+1)))/(log(u/d)) )));


W = K - S*d.^([M:-1:M-z+1]’).*u.^([0:z-1]’);

% log/cumsum version using cut-off index z

tmp1 = cumsum(log([1;[M:-1:M-z+2]’])) - cumsum(log([1;[1:z-1]’]));

tmp2 = tmp1 + log(p)*([0:z-1]’) + log(1-p)*([M:-1:M-z+1]’);

value = exp(-r*T)*sum(exp(tmp2).*W);





Reference:

Desmond J. Higham, Nine Ways to Implement the BinomialMethod for Option Valuation in MATLAB, SIAM Review, 2002.

(See lecture web site for the link on this paper and accompanyingMatlab files)




Improving execution time in Matlab

Some useful general rules:

I Eliminate redundant computations (e.g. precompute aconstant expression if it is used many times);

I Avoid for loops;

I Work directly on entire arrays or subarrays instead ofindividual array elements:

I supply array-valued arguments to Matlab functions;

I apply elementwise operations such as .* and .ˆ to arrays;

I exploit the colon notation to access subarrays: W(1:i).


Pricing by Finite Difference Solution of PDEs

Computational Methods for Quantitative FinancePricing by Finite Difference Solution of PDEs

Christoph SchwabN. Hilber, Ch. Winter

Lecture 2


Pricing by Finite Difference Solution of PDEsOption prices as solutions of PDEs: Fokker-Planck equationFinite Difference method for the heat equation

Outline




Setup

Consider the SDE

dXt = b(t,Xt)dt+ σ(t,Xt)dWt, X0 = x. (2.1)

Assume ∃K > 0 such that ∀x, y ∈ R, ∀0 ≤ t ≤ T(A1) |b(t, x)− b(t, y)|+ |σ(t, x)− σ(t, y)| ≤ K|x− y|(A2) |b(t, x)|+ |σ(t, x)| ≤ K(1 + |x|)(A3) the RV x is independent of the σ−algebra generated by W

and satisfies E[|x|2] <∞(A1)–(A3) ⇒ for any T ≥ 0 (2.1) admits an unique solution(Xt)0≤t≤T with the properties

I (Xt)0≤t≤T is t–continuous,

I (Xt)0≤t≤T is adapted to the filtration generated by x and W ,

I E[∫ T0 |Xt|2dt] <∞.



Setup

I Assume the risky underlying (Xt)0≤t≤T evolves according tothe SDE

dXt = b(t,Xt)dt+ σ(t,Xt)dWt, X0 = x.

I Assume r(t, x) is a bounded and continuous functionmodelling the riskless interest rate.

Objective: Compute the value of an option with payoff h(·) whichis the conditional expectation

Vt = u(t,Xt) = E[e−

R Tt r(s,Xt,x

s )dsh(Xt,xT )|Ft

], (2.2)

where Xt,xs is the solution of the SDE (2.1) starting from x at time

t.



Infinitesimal generator

Solution: characterize u(t, x) as solution of a partial differentialequation.

We need some auxiliary results.

Assume that σ, b in (2.1) are independent of t. Denote by A thedifferential operator

(Af)(x) =12σ2(x)f ′′(x) + b(x)f ′(x), (2.3)

acting on functions f ∈ C2.




Proposition

Assume that σ, b in (2.1) are independent of t. Then the process

Mt := f(Xt)−∫ t

0(Af)(Xs) ds (2.4)

is a martingale with respect to the filtration of Wt. In particular,for all t,

E[f(Xt)

]= f(x) + E

[∫ t

0Af(Xs)ds

]. (2.5)

Proof. Ito’s Lemma, stochastic integral w.r. to W is a martingale.




Based on this Proposition, we have

TheoremDenote by Xx

t the solution of the SDE (2.1) with initial conditionXx

0 = x. Then for any function f ∈ C2, the functiont→ E[f(Xx

t )] is continuously differentiable and

ddt

E[f(Xx

t )]|t=0 = (Af)(x)

with A as in (2.3)

RemarkThe operator A is called infinitesimal generator of the diffusionprocess Xx

t .



Prices of European contracts are moments of the price process Xxt

of the formu(x, t) = E

[e−

R t0 r(X

xs )dsh(Xx

t )]

(2.6)

where h(x) and r(x) are continuous functions of x ∈ R.

Such moments are solutions of deterministic PDEs, as thefollowing theorem, the so-called Feynman-Kac theorem, shows.



Feynman-Kac

TheoremLet u(x, t) ∈ C2,1(R× R+) be a solution of the parabolic Cauchyproblem

∂tu+Au− ru = 0 in R× R+

u(x, 0) = h(x) in R (2.7)

with A as in (2.3). Then u(x, t) can be represented as in (2.6), i.e.

u(x, t) = E[e−

R t0 r(X

xs )dsh(Xx

t )].

Viceversa, any u(x, t) as in (2.6) which is in C2,1(R× R+) solvesthe deterministic Fokker Planck equation (2.7).



This result is still true in a multi-dimensional model and for thecase when the coefficient functions b, σ and the discounting factorc depend also on t.For p ∈ N consider the system of SDEs

dXkt = bk(t,Xt)dt+

p∑j=1

σkj(t,Xt)dWjt , 1 ≤ k ≤ n. (2.8)

Denote by A the operator

(Au)(x, t) =12

tr[σσ>D2u] + 〈b,∇u〉, (2.9)

where D2 = (∂xixj )1≤i,j≤n is the Hessian, σ = (σij)1≤i≤n,1≤j≤pand 〈·, ·〉 denotes the standard inner product in Rn.



Feynman-Kac (baskets)

Theorem(baskets)

(i) If (Xt)t≥0 is a solution of (2.8) and u(x, t) : Rn × R+ → Rhas bounded, second order derivatives in x and bounded firstorder derivatives in t and if r(t, x) : R+ × Rn → R is acontinuous, bounded function on R+ × Rn, then the process

Mt := e−R t0 r(s,Xs)dsu(Xt, t)

−∫ t

0e−

R s0 r(τ,Xτ )dτ (∂tu+Au− ru) (Xs, s)ds

is a martingale.



(ii) Let u(x, t) ∈ C2,1(Rn × [0, T ]) be a solution of the parabolicCauchy problem

∂tu+Au− ru = 0 in Rn × R+

u(x, T ) = h(x) in Rn (2.10)

with A as in (2.9). Then u(x, t) can be represented as

u(x, t) = E[e−

R Tt r(s,Xt,x

s )dsh(Xt,xT )]. (2.11)



Application to the Black Scholes model

Assume for simplicity that no dividends are paid.Dynamics of risky underlying: Geometric Brownian motion

dSt = St(rdt+ σdWt), S0 = S (2.12)

i.e. we have b(S) = rS, σ(S) = σS. The infinitesimal generator Ais

A =: ABS =12σ2S2∂SS + rS∂S .



The discounted price of a European contract with payoff h(S), i.e.

V (S, t) := E[e−r(T−t)h(ST ) | St = S],

is equal to a regular solution V (S, t) of the Black Scholes (BS)equation

∂tV +ABSV − rV = 0 in R+ × [0, T )V (S, T ) = h(S) in R+

, (2.13)

where

ABS =12σ2S2∂SS + rS∂S .



Heat equation

By change of variables

y(x, τ) := e−αx−βτV (ex, T − 2τ/σ2), α, β ∈ R

the Black-Scholes equation can be transformed to the heatequation (see exercise):

∂τy − ∂xxy = 0 in R× (0, 1/2σ2T ]y(x, 0) = y0(x) in R

We first study finite difference method on this simple example.Boundary conditions If the equation is satisfied for x ∈ (a, b), weneed conditions at x = a and x = b, for all τ . Examples:

I Dirichlet conditions: y(a, τ) = ga(τ), y(b, τ) = gb(τ).I Neumann conditions: ∂xy(a, τ) = ga(τ), ∂xy(b, τ) = gb(τ).

Remark If x ∈ R, we choose a bounded computational domain(a, b) and impose artificial boundary conditions.



Discretization of the domain

Computational domain [a, b]× [0, τmax] is replaced by discrete grid:

(xi, τm), i = 0, . . . , N + 1, m = 0, . . . ,M,

where xi are equidistant space grid points with space step size h:

xi = a+ ih, h ≡ ∆x =b− aN + 1

,

and τm the time levels with time step size k:

τm = mk, k ≡ ∆τ =τmax

M.

Remark It is possible to use variable step sizes ∆xi and ∆τm:for example, to refine the grid near the irregularities of the solution.



Solution on the grid

The exact solution y(x, t) is a surface on [a, b]× [0, τmax]:infinitely dimensional object.

Computer can work only with finite quantities.

Therefore, we represent the solution by its values on the grid:the surface is replaced by (N + 2)× (M + 1) points:

y(x, t) −→ ymi = y(xi, τm), i = 0, . . . , N +1, m = 0, . . . ,M.

The goal is to approximate the values ymi . Values of the solutionbetween grid points are then found by some interpolation.



Difference Quotients (= Finite Differences)

We want to approximate the derivatives of y using only its valueson the grid. First, let us consider a function f(x) of one variable.

Assume that f is twice differentiable with continuous derivatives;we write f ∈ C2. Then, using Taylor’s formula,

f ′(x) =f(x+ h)− f(x)

h+h

2f ′′(ξ), ξ ∈ [x, x+ h] .

If fi = f(xi) are the values of f on the grid xi, we obtain

f ′(xi) ≈ fi+1 − fih

with a remainder that goes to zero as h→ 0.



Notations

f ′(xi) =fi+1 − fi

h+O(h) =: (δ+x f)i +O(h)

f ′(xi) =fi − fi−1

h+O(h) =: (δ−x f)i +O(h)

(δ+x f)i and (δ−x f)i are called one-sided difference quotients of fwith respect to x at xi. They are accurate of first order (O(h)).

Other examples (valid for more regular functions):

f ′(xi) =fi+1 − fi−1

2h+O(h2) =: (δxf)i +O(h2) f ∈ C3 ,

f ′′(xi) =fi+1 − 2fi + fi−1

h2+O(h2) =: (δ2xxf)i +O(h2) f ∈ C4 ,

f ′(xi) =−fi+2 + 4fi+1 − 3fi

2h+O(h2) =: (δ+x f)i +O(h2) f ∈ C3 .



Explicit (Euler) finite difference scheme

We replace PDE ∂y∂τ − ∂2y

∂x2 = 0 by the set of algebraic equations

ym+1i − ymi

k− ymi+1 − 2ymi + ymi−1

h2= 0

for m = 0, . . . ,M − 1, i = 1, . . . , N .

Initialization:

y0i = y0(xi), i = 0, . . . , N + 1.

For m = 0, . . . ,M− 1:

ym+1i = k

h2 ymi−1 + (1− 2 k

h2 )ymi + kh2 y

mi+1, i = 1, . . . , N

ym+10 = ga(τm+1), ym+1

N+1 = gb(τm+1) (if Dirichlet boundary conditions)



Neumann boundary conditions

If boundary conditions are specified in Neumann form:

∂y

∂x(a, τ) = ga(τ),

∂y

∂x(b, τ) = gb(τ), τ ∈ [0, τmax]

we can approximate the derivatives by one-sided finite differences:

ym1 − ym0h

= ga(τm),ymN+1 − ymN

h= gb(τm), m = 0, . . . ,M

which gives boundary conditions for the discrete scheme:

ym0 = ym1 −h ga(τm), ymN+1 = ymN +h gb(τm), m = 0, . . . ,M.

Remark We can also use more accurate approximations of ∂y∂x as,

for example, (δ+x y).



Implicit finite difference scheme

Here, the space derivative is taken on the time level m+ 1:

ym+1i − ymi

k− ym+1

i+1 − 2ym+1i + ym+1

i−1

h2= 0

for m = 0, . . . ,M − 1, i = 1, . . . , N .

Initialization:

y0i = y0(xi), i = 0, . . . , N + 1.

For m = 0, . . . ,M− 1:

solve

ym+10 = ga(τm+1),

− kh2 y

m+1i−1 + (1 + 2 k

h2 )ym+1i − k

h2 ym+1i+1 = ymi , i = 1, . . . , N

ym+1N+1 = gb(τm+1).



θ-scheme

For a parameter θ ∈ [0, 1], θ-scheme is a combination of explicitand implicit schemes:

Emi : =ym+1i − ymi

k−[(1− θ)(δ2xxy)mi + θ(δ2xxy)m+1

i

]=ym+1i − ymi

k−[(1− θ)y

mi+1 − 2ymi + ymi−1

h2

+θym+1i+1 − 2ym+1

i + ym+1i−1

h2

]= 0

θ = 0 =⇒ explicit scheme.θ = 1 =⇒ implicit scheme.θ = 1/2 =⇒ Crank-Nicolson scheme.



θ-scheme in matrix form

Let us introduce column vectors

ym = (ym1 , . . . , ymN )>, Em = (Em1 , . . . , EmN )>

and the tridiagonal N ×N matrix

G =

2 −1

−1. . .

. . .. . .

. . . −1−1 2

Then the FD θ-scheme Em = 0 becomes, in matrix form,(

I + θk

h2G)ym+1 =

(I − (1− θ) k

h2G)ym.

/here, we assume homogeneous Dirichlet boundary conditions/



Notation

To shorten the notation, let us introduce matrices

B = I + θk

h2G and C = I − (1− θ) k

h2G.

The scheme becomes

Bym+1 = Cym.

To find ym+1, we have to solve this linear system:

ym+1 = B−1Cym.



θ-scheme for some modifications of the PDE

Inhomogeneous heat equation: ∂y∂τ = ∂2y

∂x2 + f(x, τ) =⇒Bym+1 = Cym + k[θfm+1 + (1− θ)fm].

where fm = (fm1 , . . . , fmN ) and fmi = f(xi, τm).

Inhomogeneous Dirichlet BC: y(a, τ) = ga(τ), y(b, τ) = gb(τ) ⇒Bym+1 = Cym +

k

h2[θgm+1 + (1− θ)gm].

where gm =(ga(τm), 0, . . . , 0, gb(τm)

)>.

Neumann BC: ∂y∂x(a, τ) = ga(τ), ∂y

∂x(b, τ) = gb(τ) =⇒

(B−θ kh2

F )ym+1 = (C+(1− θ) kh2

F )ym+k

h[θgm+1 + (1− θ)gm].

where gm =(− ga(τm), 0, . . . , 0, gb(τm)

)>, and F contains only

two non-zero elements: F ij = 0 except F 11 = FNN = 1.


Pricing by Finite Difference Solution of PDEs

Computational Methods for Quantitative FinancePricing by Finite Difference Solution of PDEs (part 2)


Lecture 3


Pricing by Finite Difference Solution of PDEsAnalysis of the Finite Difference methodFinite Difference solution of the Black-Scholes equation

Outline




Consistency of the scheme

We have replaced continuous PDE by the following FD equations:

Emi = 0, i = 1, . . . , N, m = 0, . . . ,M − 1 ,

y0i = y0(xi), i = 1, . . . , N

ym0 = 0, ymN+1 = 0, m = 0, . . . ,M (homogeneous Dirichlet BC)

DefinitionThe consistency error Emi at (xi, τm) is the difference operator Emiwith ymi replaced by y(xi, τm) where y is the exact solution.

Note Consistency error measures the quality of the approximationof the differential operator by finite differences. A scheme is calledconsistent if its consistency error goes to zero as h, k → 0.



Consistency error

We want to estimate the consistency errors Emi in terms of meshwidth h and time step size k.

Proposition

If the exact solution y(x, τ) of the heat equation is sufficientlysmooth, then, for m = 1, ...,M − 1 and i = 1, ..., N , we have

|Emi | ≤ C(h2 + k) 0 ≤ θ ≤ 1 ,

|Emi | ≤ C(h2 + k2) θ =12.

where C depends on the exact solution y and its derivatives.



Discretization error

Define εm, the error vector at time τm:

εmi = y(xi, τm)− ymi , 1 ≤ i ≤ N, 0 ≤ m ≤M .

The error vectors εmMm=0 satisfy the difference equation

1k

B εm+1 − 1k

C εm = Em

or, in explicit form,

εm+1 = Aθ εm + kB−1Em

where Aθ ≡ B−1C is called amplification matrix.

Note: the scheme is convergent (in L∞(0, τmax;L2(a, b)) norm) ifsupm

√h‖εm‖l2 → 0 as h, k → 0.



Recall on matrix and vector norms

Let v = (v1, . . . , vN )> and M be an N ×N matrix. We define

‖v‖`2 := (v21 + · · ·+ v2

N )1/2

‖M‖2 := supv 6=0

‖Mv‖`2‖v‖`2

.

Remark If M is symmetric with eigenvalues λl, 1 ≤ l ≤ N , then

‖M‖2 = max1≤l≤N

|λl|.

Property For all v and M , we have ‖Mv‖`2 ≤ ‖M‖2‖v‖`2 .

Remark Let f be a function on (a, b) with f(a) = f(b) = 0, andlet f = (f1, . . . , fN ) be the vector of its values on the grid. Then

‖f‖2L2(a,b) =∫ b

af2(x)dx =

N∑i=0

∫ xi+1

xi

f2(x)dx ≈N∑i=1

f2i h = h‖f‖2`2 .



Consistency + Stability = Convergence

Proposition

For all M ∈ N, 1 ≤ m ≤M , holds

‖εm‖`2 ≤ ‖Aθ‖m2 ‖ε0‖`2 + km−1∑n=0

‖Aθ‖m−1−n2 ‖En‖`2 .

Corollary

If the stability condition ‖Aθ‖2 ≤ 1 holds then, as M →∞,N →∞, the FD scheme converges and, if ε0 = 0,

supm‖εm‖`2 ≤ τmax sup

m‖Em‖`2 .



Stability

When the stability condition ‖Aθ‖2 ≤ 1 is satisfied?

We need some auxiliary results.

LemmaThe eigenvalues of the tridiagonal matrix X ∈ RN×N :

X =

α β

γ α. . .

. . .. . . βγ α

.

are given by

λX` = α+ 2

√βγ cos

( `π

N + 1

), ` = 1, . . . , N.



Stability (cont.)

I Eigenvalues of B = I + θkh−2G are given by

λB` = 1+2θ

k

h2+2θ

k

h2cos( `π

N + 1

)= 1+4θ

k

h2sin2

( `π

2(N + 1)

)I Eigenvalues of C = I − (1− θ)kh−2G are given by

λC` = 1− 4(1− θ) k

h2sin2

( `π

2(N + 1)

)I Eigenvalues of Aθ = B−1C are given by

λAθ` =

λC`

λB`

=1− 4(1− θ) kh−2 sin2

( `π

2(N + 1)

)1 + 4θ kh−2 sin2

( `π

2(N + 1)

) .



Stability of the θ-scheme

‖Aθ‖2 ≤ 1 ⇐⇒ ∀l = 1, . . . , N, |λAθl | ≤ 1.

Denoting x` = 4kh−2 sin2(`π/2(N + 1)) ≥ 0 we can write

λAθl =

1− (1− θ)x`1 + θx`

= 1− x`1 + θx`

≤ 1.

It remains to check that λAθl ≥ −1, ∀l = 1, . . . , N .

Proposition

If 12 ≤ θ ≤ 1, the θ-scheme is stable for all k and h.

If 0 ≤ θ < 12 , the θ-scheme is stable if the CFL-condition holds:

k

h2≤ 1

2(1− 2θ).



Convergence of the θ-scheme

TheoremIf y(x, τ) is sufficiently smooth (y ∈ C4([a, b]× [0, τmax])), then

i) for 12 < θ ≤ 1 or for 0 ≤ θ < 1

2 and CFL-condition,

supm

h12 ‖εm‖`2 ≤ C(y)(h2 + k) ,

ii) for θ = 12 (Crank-Nicolson scheme),

supm

h12 ‖εm‖`2 ≤ C(y)(h2 + k2) .

Conclusion If the scheme is consistent and stable, then it isconvergent with the same order that the consistency error.



Black-Scholes equation in real price

Recall that in the BS model European option prices satisfy

∂V

∂t+

12σ2S2∂

2V

∂S2+ rS

∂V

∂S− rV = 0, in R+ × [0, T )

V (S, T ) = H(S), in R+

We denote

ABS V =12σ2S2∂

2V

∂S2+ rS

∂V

∂S.

I The coefficients of this Black-Scholes operator arenon-constant.

I The operator degenerates near S = 0.



Black-Scholes equation in log-price

To obtain a non-degenerate operator with constant coefficients, wechange to log-price:

x = logS.

It is also convenient to work with time to maturity instead of time:

τ = T − t.Define v(x, τ) := V (S, t) = V (ex, T − τ). Then v solves

∂τv −ABS−logv + rv = 0 in R× (0, T )

v(0, x) = H(ex) in R

where ABS−log denotes the BS operator in log-price:

ABS−logv =σ2

2∂2v

∂x2+(r − σ2

2

)∂v

∂x.



Initial condition

Terminal condition for V becomes initial conditions for v.

European call:

European put:



Truncation of the domain

In the European case, v(x, τ) is defined for −∞ < x <∞.

For numerical solution, we truncate the domain to (−R,R):

∂τvR −ABS−logvR + rvR = 0 in (−R,R)× (0, T ),

vR(x, 0) = H(ex) |x| < R

We then need artificial boundary conditions at x = ±R.

The simplest choice are homogeneous Dirichlet conditions:

vR(±R, τ) = 0 0 < τ < T .



Estimation of the truncation error

Let us estimate the difference between v and vR.

TheoremAssume that the payoff function is bounded, i.e.

there exists M such that |H(ex)| ≤M ∀x ∈ R.

Then with µ := r − σ2

2 , for all x ∈ (−R+ |µ|τ,R− |µ|τ),

|v(x, τ)− vR(x, τ)| ≤ M

exp(− |R− |µ|τ − x|

2

2σ2τ

)+ exp

(− |R− |µ|τ + x|2

2σ2τ

)In particular, |v(x, τ)− vR(x, τ)| → 0 as R→∞.



Boundary layer

The error is large near the boundary:

We have to choose R much larger than the values x of interest.



Asymptotics of the solution

How to reduce the error near the boundary? We have to chooseboundary conditions accordingly to the asymptotics of the solution.

For European calls and puts, we have:

∀t, VC(S, t)|S=0 = 0, VP (S, t)|S→∞ = 0.

At the complementary sides, we use Call-Put parity:

S + VP − VC = Ke−r(T−t).

This implies, for all t:

VC(S, t) ≈ S −K e−r(T−t) S →∞ ,

VP (S, t) ≈ K e−r(T−t) − S S → 0.

Note Asymptotics for v(x, τ) may be obtained by change of variables.



Asymptotic boundary conditions

We obtain the following boundary conditions at Smin and Smax:

Dirichlet or Neumann

VC(Smin, t) = 0 ∂SVC(Smin, t) = 0

VP (Smin, t) = Ke−r(T−t) − Smin ∂SVP (Smin, t) = −1

VC(Smax, t) = Smax −Ke−r(T−t) ∂SVC(Smax, t) = 1

VP (Smax, t) = 0 ∂SVP (Smax, t) = 0



Discretization

∂τvR −ABS−logvR + rvR = 0 in (−R,R)× (0, T ),

vR(x, 0) = H(ex) |x| < R

For N,M ∈ N define (equidistant) grid (xi, τm) in[−R,R]× [0, T ]:

I space grid with mesh width h: xi = −R+ ih, h = 2RN+1 ,

i = 0, . . . , N + 1I time grid with mesh width k: τm = mk, k = T

M ,m = 0, . . . ,M



Discretization (cont.)

Discretize ABS−log by second order finite differences: Fori = 1, . . . , N

(ABS−logh v

)i=

12σ2

h2(vi−1−2vi+vi+1)+

(r− σ

2

2

)12h

(vi+1−vi−1),

where v(τ) = (v1(τ), . . . , vN (τ))> and vi(τ) is an approximationin space to vR(xi, τ). For simplicity, adopt homogeneous Dirichletboundary conditions:

v0(τ) = vN+1(τ) = 0.



Discretization (cont.)

Let Ah = ABS−logh − rI ∈ RN×N . Then (where ˙= ∂τ )

v = Ahv, v(0) = H(ex),

where x = (x1, . . . , xN )>.

This is a N ×N -system of ordinary differential equations.

To discretize it, use the θ-scheme. Let wm be an approximation tov(τm). We obtain

1k(wm+1 − wm) = Ah(θwm+1 + (1− θ)wm).



θ-scheme for Black-Scholes equation

We obtain the following scheme:

Given w0 = H(ex) ∈ RN , for m = 0, . . . ,M − 1 find wm+1 ∈ RN

such that (I − kθAh

)wm+1 =

(I + k(1− θ)Ah

)wm,

where Ah is given by

Ah = tridiag(α−, β, α+),

with

α± =σ2

2h2± 1

2h

(r − σ2

2

), β = −σ

2

h2− r.



Stability analysis

Denote by λi,N < 0, 1 ≤ i ≤ N , the eigenvalues of Ah ∈ RN×N .The scheme is stable if the amplification factors

Ri(λi,Nk) := 1 +λi,Nk

1− θλi,Nk , 1 ≤ i ≤ N

satisfy|Ri(λi,Nk)| ≤ 1,∀i.

These inequalities always hold if 12 ≤ θ ≤ 1 (unconditionally

stable).

If, however, 0 ≤ θ < 12 , we must have the CFL-condition



Stability analysis (cont.)

k ≤ −2λi,N (1− 2θ)

, 1 ≤ i ≤ N.

Since λN,N < · · · < λ1,N < 0, this is equivalent to

k ≤ −2λN,N (1− 2θ)

.

By Lemma 2.12 of the lecture notes we have

λN,N = −σ2

h2− r +

1h

√σ4

h2−(r − σ2

2

)2

cos(

Nπ

N + 1

).



Stability analysis (cont.)

We may estimate

λN,N ≥ −2σ2

h2− r

and obtain the CFL-condition

k ≤ 2h2

(1− 2θ)(2σ2 + rh2).

Example. θ = 0 (explicit Euler).Let T = 1, σ = 0.3, r = 0.05, R = 5, N = 500. Then

M ≥ T

k=

2(N + 1)2σ2 + 4R2r

8R2≈ 226

time steps are needed to fullfill the CFL-condition.



Stability of θ-scheme



Summary

θ CFL-condition Convergence Remark

0 k ≤ 2|λN,N | O(h2 + k) no linear system has to be solved

12 none O(h2 + k2) linear systems have to be solved

1 none O(h2 + k) linear systems have to be solved


Pricing American options using Finite Difference method

Computational Methods for Quantitative FinancePricing American options using Finite Difference method


Lecture 4



Black-Scholes Variational InequalityDomain truncationFinite Difference discretizationSolution of matrix linear complementarity problems (LCPs)

Outline

Pricing American options using Finite Difference methodBlack-Scholes Variational InequalityDomain truncationFinite Difference discretizationSolution of matrix linear complementarity problems (LCPs)




Assumption

Black-Scholes market, no dividends (δ = 0):

dSt = St(rdt+ σ dWt) t > 0 .

American options are options that can be exercised at any time upto the expiration date T . The problem of optimal exercising isequivalent to an optimal stopping problem: denote by Tt,T the setof all stopping times with values in the interval (t, T ). Then, withthe payoff H(S) = (S −K)+ (call) or H(S) = (K − S)+ (put),

V am(S, t) = supτ∈Tt,T

E[e−r(τ−t)H(Se(r−

σ2

2)(τ−t)+σ(Wτ−Wt))

].




Basic properties of the Value V am(S, t) of an American contract(δ = 0):

V amC = V eur

C

V amP ≥ V eur

P

V amP (S, t) ≥ (K − S)+, V am

C (S, t) ≥ (S −K)+ .

As for the European vanilla, there is a close connection betweenthe probabilistic representation of the option’s value and adeterministic, PDE based representation of this value.




American option as Free Boundary Problem

There exists a point Sf = Sf (t) such that

V amP (S, t) > (K − S)+ for S > Sf (t)V amP (S, t) = (K − S)+ for S ≤ Sf (t)




American option as Free Boundary Problem (cont.)

A priori the boundary Sf (t) is not known. Therefore, the problemof calculating V am

P (S, t) for S > Sf (t) is called freeboundary-value problem and Sf (t) is called free boundary.

The free boundary Sf (t) separates R+ × [0, T ] into thecontinuation region (hold the option) and the stopping region(exercise the option).

As soon the price of the asset reaches Sf (t) the holder shouldexercise the option. The corresponding time instant τ is calledstopping time.




American option as Free Boundary Problem (cont.)

Calculated free boundary for a put with maturity T = 1 and strikeK = 20.




A simple obstacle problem

Let −1 < α < β < 1. Assume an obstacle ψ such that ψ(x) > 0for α < x < β, ψ ∈ C2, ψ′′ < 0.The obstacle problem reads: Find u ∈ C1[−1, 1] such that

u′′ = 0 (then u > ψ) for x ∈ (−1, α) ∪ (β, 1)

u = ψ (then u′′ = ψ′′ < 0) for x ∈ (α, β)




A simple obstacle problem (cont.)

This manifests a complementarity in the sense

if u > ψ, then u′′ = 0if u = ψ, then u′′ < 0

The obstacle problem can be reformulated as: Find u ∈ C1[−1, 1]such that u(±1) = 0 and such that

−u′′ ≥ 0 in (−1, 1)u ≥ ψ in (−1, 1)

u′′(u− ψ) = 0 in (−1, 1)




Linear complementarity problem (LCP)

The finite difference discretization of the obstacle problemproceeds in the usual fashion:

−1 = x0 < x1 < . . . < xi < . . . < xN < xN+1 = 1

with

xi = −1 + i∆x, i = 0, 1, . . . , N + 1 , ∆x = 2/(N + 1),

yi ≈ u(xi) : y = (y1, . . . , yN )> ∈ RN .




Linear complementarity problem (LCP) (cont.)

We approximate the obstacle problem by the LCP

Ay ≥ b = 0

y ≥ ψ(Ay − b) · (y − ψ) = 0

where

A = tridiag(−1, 2,−1)

ψ =(ψ(x1), . . . , ψ(xN )

)>.




Projected Successive Over Relaxation (PSOR)

We solve the LCP by the projected SOR iteration:

Definition (Projected SOR/Cryer 1971)

Given y0 ≥ ψ, for k = 0, 1, 2, . . . , compute yk+1 from yk via:

i) for i = 1, . . . , n, do

yk+1i :=

1Aii

(bi −

i−1∑j=1

Aij yk+1j −

N∑j=i+1

Aij ykj

).

ii) yk+1i := max

(ψi, y

ki + ω(yk+1

i − yki ))

if ‖yk+1 − yk‖2 < ε stop; else: k → k + 1.




Remark1) if ii) is replaced by

yk+1i = yki + ω(yk+1

i − yki ) ,

then we get the classical SOR for the solution of Ay = b.

Step ii) ensures the inequality condition (y ≥ ψ).

2) 0 < ω < 2 is a relaxation parameter - the PSOR convergesfor all these ω. Convergence speed can be optimized for1 < ω < 2.

We turn back to American options.

In analogy to the simple obstacle problem, we have the followingTheorem.




TheoremLet v(x, t) be a regular solution of the BS-inequality(g(x) := H(ex)):

∂tv +ABS−log v − rv ≤ 0, in R× (0, T ) ,

v(x, t) ≥ H(ex) =: g(x) in R× (0, T ) ,

(∂tv +ABS−log v − rv)(g − v) = 0, in R× (0, T ) ,

v(x, T ) = g(x) in R .

ThenV am(S, t) = v(logS, t) .




To prepare the FD discretization, we truncate the domain: werestrict x ∈ R to |x| < R <∞, where R is a truncation parameter,and ΩR = (−R,R). Then vR(x, t) solves

∂tvR +ABS−logvR − rvR ≤ 0 in ΩR × (0, T )

vR ≥ g in ΩR × (0, T )

(vR − g)(∂tvR +ABS−logvR − rvR) = 0 in ΩR × (0, T )

vR(x, T ) = g(x) in ΩR

∂xvR(±R, t) = 0 in (0, T )




As in the European case, we choose the time step ∆t = k = TM ,

and the space step ∆x = h = 2RN+1 .

Then

tm = mk, xi = −R+ ih, i = 0, . . . , N + 1, k = 0, . . . ,M.

The vector of approximate option prices at time level m is

w = (w1, . . . , wN )>; we expect that w(m)i ∼ v(xi, tm).




Applying the FD discretization in space and the θ-scheme in timeyields the following sequence of Matrix Inequality Problems:

w(M) = gh

:=(g(x1), . . . , g(xN )

)>.

For m = M − 1, . . . , 0: find w(m) ∈ RN s.t.

w(m+1) − w(m) + k[θAhw

(m) + (1− θ) Ahw(m+1)

]≤ 0

w(m) ≥ gh(

w(m+1) − w(m) + k[θAhw

(m) + (1− θ) Ahw(m+1)

])·(w(m) − g

h) = 0 .

Note: the problem is posed in real time t rather than in time tomaturity τ , hence the time-stepping is backward in time.




Here, as in the European case

Ah := ABS−logh − rI,

where, due to homogeneous Neumann boundary conditions

0 = ∂xvR(±R, t)

=1

2h(± 3vR(±R, t)∓ 4vR(±R∓ h)± vR(±R∓ 2h, t)

)+O(h2)

the tridiagonal matrix ABS−logh ∈ RN×N is given by




ABS−logh =

β+4

3α− α+−1

3α−

α− β α+

α−. . .

. . .. . . β α+

α−−13α

+ β+43α

+

with

β = −σ2

h2, α± =

σ2

2h2± 1

2h

(r − σ2

2

).




We write the matrix inequality problem succinctly asLinear Complementarity Problem (LCP):

Ay ≥ b

y ≥ g

(y − g) · (Ay − b) = 0

where

A := 1− kθAh, y = w(m), b = (1 + k(1− θ)Ah)w(m+1)




Transformation to normal form

With the same transformation that we used in the European case,the problem transforms as follows: the exercise condition

V amP (S, t) ≥ (K − S)+ = K max1− ex, 0

becomes (where q := 2rσ2 )

y(x, τ) ≥ e 12(q−1)x+ 1

4(q+1)2τ max1− ex, 0

= e14(q+1)2τ max

e

12(q−1)x − e 1

2(q+1)x, 0

=: ψ(x, τ) .




Hence, the problem becomes (with R > 0 sufficiently large and

J := (0, σ2T2 )):

Find y(·, τ) ∈ C1(−R,R) such that

∂τy − ∂xxy ≥ 0 in (−R,R)× Jy ≥ ψ in (−R,R)× J

(y − ψ)(∂τy − ∂xxy) = 0 in (−R,R)× Jy(x, 0) = ψ(x, 0) in (−R,R)

y(±R, τ) = ψ(±R, τ) in J

RemarkThe condition y ∈ C1 is called smooth fit or smooth pastingcondition.




Inserting finite differences in the parabolic variational inequality(assumed to hold pointwise), we get at each time level the linearsystem of algebraic inequalities for the unknown values

y(m)i ≈ y(xi, τm) at the nodes (xi, τm):

y(m+1)i −λθ(y(m+1)

i+1 − 2y(m+1)i + y

(m+1)i−1

)− y

(m)i −λ(1− θ)(y(m)

i+1 − 2y(m)i + y

(m)i−1

) ≥ 0,

with

λ :=∆t

∆x2=

k

h2.




This is equivalent to the following LCP: Given

y(0) = ψ(0) =(ψ(x1, 0), . . . , ψ(xN , 0)

)> ∈ RN ,

for m = 0, . . . ,M − 1, find y(m+1) ∈ RN such that

Ah,ky(m+1) ≥ b(m)

y(m+1) ≥ ψ(m+1)(Ah,ky

(m+1) − b(m)) · (y(m+1) − ψ(m+1))

= 0




whereAh,k := I − θλG ∈ RN×N

sym

and

G := tridiag(1,−2, 1)

b(m) = (I + (1− θ)λG)y(m) + λ(θF (m+1) + (1− θ)F (m)

)F (m) :=

(ψ(−R, τm), 0, . . . , 0, ψ(R, τm)

)>ψ(m) :=

(ψ(x1, τm), . . . , ψ(xN , τm)

)>.




We focus now on the matrix LCP obtained from the discretizationof the BS inequality, i.e.

Ay ≥ b

y ≥ ψ

(y − ψ) · (Ay − b) = 0

where inequalities between vectors are to be read componentwise.

Assume that A is SPD (symmetric positive definite):

RN×N 3 A = A> and ∃γ > 0 : ∀x ∈ RN : x>Ax ≥ γx>x.




It is useful to rewrite the LCP into the following standard form:Put

x := y − ψ, b := b−Aψ, z := Ay − b.Then

x ≥ 0, z = Ax− b ≥ 0, x · z = 0 .

This LCP is equivalent to

x = arg miny∈RN+

12y>Ay − b>y

.




Generalized Cryer algorithm

Let H ∈ RN×N be any symmetric positive definite matrix and letC ∈ RN×N be any invertible matrix.

Consider the minimization problem

minuu>Hu− 2f>u subject to Cu ≥ 0 element-wise,

which corresponds to the LCP

Find u ∈ RN such that Cu ≥ 0

C−>(Hu− f) ≥ 0

u>(Hu− f) = 0.




We solve this LCP approximately by the Cryer algorithm

Algorithm

(Cryer) Choose ω ∈ (0, 2). Set si := C−1ei, i = 1, . . . , N .0) Choose a starting vector u0 with Cu0 ≥ 0.1) For k = 1, . . . do (until convergence):

1.1) Set uk0 := uk−1.1.2) For i = 1, . . . , N do:

1.2.1) Set rki = (s>i Hsi)−1s>i[f −Huki−1

]si.

1.2.2) Choose the maximal ωki with ωki ≤ ω,such that C(uki−1 + ωki r

ki ) ≥ 0.

1.2.3) Set uki = uki−1 + ωki rki .

1.3) Next i.1.4) Set uk := ukN .

2) Next k.




TheoremUnder the above assumptions, the Generalized Cryer Algorithmconverges to the solution of the minimization problem.


Pricing barrier and Asian contracts by Finite Difference method

Computational Methods for Quantitative FinanceFinite Difference methods for barrier and Asian contracts

Prof. Ch. SchwabN. Hilber, Ch. Winter

Lecture 5


Pricing barrier and Asian contracts by Finite Difference methodBarrier OptionsAsian OptionsMatlab implementation

Outline




Barrier options differ from vanillas in the sense that the optioncontract is triggered if the price of the underlying hits some barrier,S = B, say, during [0, T ].

To be more specific, consider a contract which pays a specifiedamount at maturity T provided during 0 ≤ t ≤ T the price St doesnot cross a specified barrier B > 0 either from above (‘down-and-out’ barrier option) or from below (‘up-and-out’ barrieroption).

If the barrier is crossed before T , the option expires worthless(‘out’ barrier option). We assume here for convenience that B isconstant.



The 4 standard ‘out’ barrier options on a risky asset are

1. down-and-out call with terminal payoff (ST −K)+ with valueV Cdo(K,B, T ;S, t) at spot price S and time t,

2. down-and-out put with terminal payoff (K − ST )+ and withvalue V P

do(K,B, T ;S, t),

3. up-and-out put with terminal payoff (K − ST )+ and withvalue V P

uo(K,B, T ;S, t),

4. up-and-out call with terminal payoff (ST −K)+ and withvalue V C

uo(K,B, T ;S, t).

An “in” barrier option becomes the corresponding European vanillawhen the barrier B is crossed at time t ∈ (0, T ). E.g., theup-and-in call becomes the European call when B is crossed frombelow; we denote its price at spot S and time t byV Cui (K,B, T ;S, t).



A European plain vanilla option pays the same at T as adown-and-out plus an up-and-in with the same barrier B and ofthe same type (call/put) as the plain vanilla. The standard noarbitrage consideration shows that pricing an “in” barrier reducesto pricing the corresponding “out” barrier contract and a plainEuropean vanilla, i.e.

V Cdi (K,B, T ;S, t) = V C(K,T ;S, t)− V C

do(K,B, T ;S, t)

V Cui (K,B, T ;S, t) = V C(K,T ;S, t)− V C

uo(K,B, T ;S, t).(5.1)

Analogous identities hold for values of the other barrier contracts,too.



Out-Barrier

Assume that the riskless rate r > 0 is independent of t. Inlog-price x = logS, g(x) = (ex −K)+ = f(ex) is the terminalpayoff of a down-and-out call with barrier B = 1 for convenience(general B will be dealt with below). Let

v(x, t) = V Cdo(K, 1, T ;S, t)|x=logS

be the price of the down-and-out barrier call at t < T and at spotS > B = 1. Then,

v(x, t) = E[er(t−T )g(XT )1τ0>T|Xt = x

]where τ0 is the hitting time of (−∞, 0] = (−∞, logB] by theprocess Xt.



Consequently, v(x, t) is the solution of the BS equation in log-price

∂tv +ABS−logv − rv = 0 in (0,∞)× (0, T )v(0, t) = 0 in (0, T )v(x, T ) = g(x) in (0,∞)

In real price S and for a general barrier B > 0,V (S, t) = V C

do(K,B, T ;S, t) satisfies

V Cdo(K,B, T ;S, t) = E

[er(t−T )f(ST )1τB>T|St = S

]where τB is the first hitting time of [0, B] by the process St.



Therefore, V (S, t) = V Cdo(K,B, T ;S, t) is solution of the

deterministic BS equation

∂tV +ABSV − rV = 0 in (B,∞)× (0, T )V (B, t) = 0 in (0, T )V (S, T ) = f(S) in (B,∞)

(5.2)

As St →∞, the probability that St ≤ B gets small, so we expectfor 0 < t < T that

V (S, t) ∼ S as S →∞ . (5.3)

The above equations specify V (S, t) completely, and can be solvedapproximately with the FD methods of the previous sections.



In-Barriers

“In”-barriers become worthless unless St reaches B before T . If Stcrosses B before T , then the barrier option becomes a plain vanillawith identical payoff.

To derive the BS equations for “in” barrier contracts, we considera down-and-in European call and denote by C(S, t) a Europeanvanilla call with same K,T as the ‘down-and-in’ barrier call.

Since the BS equations are linear, we apply the principle ofsuperposition and use (5.1) to derive the BS equations for the “in”barrier.



We get for V (S, t) = V Cdi (K,B, T ;S, t) the BS PDE

∂tV +ABSV − rV = 0 (B,∞)× (0, T ),

and, by (5.1) and (5.3)

V (S, t)→ 0 as S →∞ .

Finally, again by (5.1) and superposition, we get the terminalcondition

V (S, T ) = 0 in (B,∞)

If S(t) = B for some t < T , then V Cdo(B, t) = 0 yields in (5.1)

V (B, t) = C(B, t) .

Therefore, we must only solve BS (5.2) for S > B.



European Average Strikes

Asians are path-dependent: their payoffs depend on price history.Consider a European contract with payoff depending on S and on

T∫0

ϕ(S(τ), τ)dτ,

where ϕ is a given function of S, t. E.g. an average strike call haspayoff function

f(S, T ) =

(S − 1

T

∫ T

0S(τ)dτ

)+

.



To get the PDE for Asians, we introduce the new variable

I(t) :=

t∫0

ϕ(S(τ), τ)dτ . (5.4)

Since the history of the asset price is independent of the currentprice we may treat S(t), I(t) and t as independent variables. Thevalue of an Asian will then be searched in the form V (S, I, t). Toderive a PDE for V , we need a SDE for I(t). Formally, for smalldt > 0,

I(t+ dt) = I + dI =

t+dt∫0

ϕ(S(τ), τ)dτ

=

t∫0

ϕ(S(τ), τ)dτ + ϕ(S(t), t)dt+O(dt) ,



hence, as dt→ 0, we get the deterministic equation:

dI = ϕ(S, t)dt .

Note that the vector process (S(t), I(t)) is a diffusion. In fact, forϕ(S, t) = S

dS(t) = r S(t)dt+ σ S(t) dWt ,

dI(t) = S(t)dt .

The infinitesimal generator AAS of this diffusion is given by

AASV =σ2

2S2∂2

SS V + r S∂SV + ϕ(S, t)∂IV.



Applying Ito’s lemma to V (S, I, t) gives the relation

dV = σS∂SV dW +(σ2

2S2∂2

SSV +rS∂SV +∂tV +ϕ(S, t)∂IV)

dt .

No-arbitrage arguments lead to the following PDE for V (S, I, t):

LASV := ∂tV+ϕ(S, t)∂IV+σ2

2S2∂2

SSV+rS∂SV−rV = 0 . (5.5)

The terminal condition is, for the average strike,

V (S, I, T ) = (S − I/T )+ =: f(S, I) ;

the boundary conditions in S are as for the BS case.



American Average Strikes

Denoting by LAS the operator in (5.5), we find for the value ofAmerican-Asian style contracts the parabolic differential inequality:

LAS V ≤ 0

and the constraint

V (S, I, t) ≥ f(S, I, t) = (S − I/t)+ , 0 < t < T

for the American average strike call. Also, the complementaritycondition

(V − f) · LAS V = 0

must hold, and we have the final condition

V (S, I, T ) = f(S, I, T ) .

The above equations/inequalities specify the value of the Americanaverage strike call.



Similarity Reduction

In many cases, the bivariate PDE (5.5) can be reduced to aunivariate one the numerical solution of which is less costly. Forillustration, consider the average strike European call:

LAS V = 0, V (S, I, T ) = (S − I/T )+, I(t) =

t∫0

S(τ)dτ .

Writing

f(S, I, T ) = (S − I/T )+ = Smax(

1− 1ST

T∫0

S(τ)dτ, 0),



and changing the variables

R :=1S

t∫0

S(τ)dτ =I

S,

we get the pay-off Smax(1−R/T, 0).

Making the Ansatz V (S, I, t) = SH(R, t), we find from LASV = 0that H must satisfy the PDE

∂tH +σ2

2R2∂2

RRH + (1− rR)∂RH = 0, 0 < t < T (5.6)

with terminal condition

H(R, T ) = (1−R/T )+ .



For the boundary conditions at R = 0, R→∞, we get (exercise)

∂tH + ∂RH = 0 at R = 0, 0 < t < T ,

H(R, t) = 0 as R→∞ .



FD Discretization

We solve (5.6) by Finite-Difference time-stepping from T = M∆tto t = 0, with ∆t = T/M , and truncate 0 ≤ R <∞ to0 ≤ R ≤ Rmax = N∆R.

Let tm = m∆t, Rn = n∆R, m = M, . . . , 0, n = 0, . . . , N . We useFinite Differences with Hm

n ∼ H(Rn, tm). Recall (for f ∈ C4, withfn := f(xn)):

f ′(xn) =1h

(fn+1 − fn

)+O(h) := (δxf)n +O(h)

f ′(xn) =1

2h(− fn+2 + 4fn+1 − 3fn

)+O(h2) := (δ+x f)n +O(h2)

f ′′(xn) =1h2

(fn+1 − 2fn − fn−2

)+O(h2) := (δ2xxf)n +O(h2)



We get from (5.6), i.e.

∂tH +σ2

2R2∂2

RRH + (1− rR)∂RH = 0 ,

the FD-approximation

Emn := (δtH)mn + θ

[σ2

2(n∆R)2(δ2RRH)mn + (1− rn∆R)(δ+RH)mn

]+(1− θ)

[σ2

2(n∆R)2(δ2RRH)m+1

n + (1− rn∆R)(δ+RH)m+1n

]= 0 .



The boundary condition H(R, t) = 0 as R→∞ becomes

HmN = 0 0 ≤ m ≤M .

The boundary condition

∂tH + ∂RH = 0, at R = 0, 0 < t < T

is automatically satisfied by the scheme Emn = 0 at n = 0.

The terminal condition H(R, T ) = (1−R/T )+ becomes

HMn = (1− n∆R/T )+, 0 ≤ n ≤ N .



Summary

Consider the price V of an ‘average strike’ call option with pay-off

max(S − 1

T

∫ T

0S(τ)dτ , 0

).

The similarity reduction allows to express the value V of the optionas V (S, I, t) = SH(R, t), R = I/S, with H solving

∂tH +σ2

2R2∂2

RRH + (1− rR)∂RH = 0

H(R, T ) = (1−R/T )+∂tH + ∂RH = 0 at R = 0

H(R, t) = 0 as R→∞ .

(5.7)



We restrict (5.7) to (0, Rmax) with Rmax > 0 sufficiently large,take ∆t = T/M and ∆R = Rmax/N with M,N ∈ N, anddiscretize (5.7) by Finite Differences and obtain

Emn = 0HMn = (1− n∆R/T )+, 0 ≤ n ≤ N,

HmN = 0, 0 ≤ m ≤M .

(5.8)

Problem (5.8) reads in matrix form: Given HM = HMn Nn=0, for

m = M − 1, . . . , 0 find Hm = Hmn Nn=0 such that

AHm = BHm+1.



The matrices A and B having the following band structure

A =

B0 C0 D0 0 0 . . . 0A1 B1 C1 D1 0 0

0 A2 B2 C2 D2...

0 0. . .

. . .. . .

. . . 0... AN−2 BN−2 CN−2 DN−2

0 0 AN−1 BN−1 CN−1

0 . . . 0 0 0 AN BN



B =

b0 c0 d0 0 0 . . . 0a1 b1 c1 d1 0 0

0 a2 b2 c2 d2...

0 0. . .

. . .. . .

. . . 0... aN−2 bN−2 cN−2 dN−2

0 0 aN−1 bN−1 cN−1

0 . . . 0 0 0 aN bN

,

where



An = −σ2

2θn2∆t

Bn = 1 +(σ2n2 +

3(1− r∆Rn)2∆R

)θ∆t

Cn =(− σ2

2n2 − 2

1− r∆Rn∆R

)θ∆t

Dn =1− r∆Rn

2∆Rθ∆t



and

an =σ2

2(1− θ)n2∆t

bn = 1−(σ2n2 +

3(1− r∆Rn)2∆R

)(1− θ)∆t

cn =(σ2

2n2 − 2

1− r∆Rn∆R

)(1− θ)∆t

dn = −1− r∆Rn2∆R

(1− θ)∆t.



function H = asian(N,M,Rmax)

sigma = 0.8; r = 0.05; T = 0.5; % model parameters

dR = Rmax/N; dt = T/M; theta = 0.5; % discretization parameters

A_vec = -sigma^2*0.5*dt*theta*[1:N+1]’.^2;

B_vec = 1 + (sigma^2*[0:N].^2 + 1.5*(1-r*dR*[0:N])/dR)’*theta*dt;

C_vec = (-sigma^2*0.5*[-1:N-1].^2 - 2.0*(1-r*dR*[-1:N-1])/dR)’*theta*dt;

D_vec = (1-r*dR*[-2:N-2])’/dR*0.5*theta*dt;

a_vec = sigma^2*0.5*dt*(1-theta)*[1:N+1]’.^2;

b_vec = 1 - (sigma^2*[0:N].^2 + 1.5*(1-r*dR*[0:N])/dR)’*(1-theta)*dt;

c_vec = (sigma^2*0.5*[-1:N-1].^2 + 2.0*(1-r*dR*[-1:N-1])/dR)’*(1-theta)*dt;

d_vec = -(1-r*dR*[-2:N-2])’/dR*0.5*(1-theta)*dt;

A = spdiags([A_vec B_vec C_vec D_vec], -1:2, N+1, N+1);

B = spdiags([a_vec b_vec c_vec d_vec], -1:2, N+1, N+1);

H = max(1-[0:N]’*dR/T, 0); % terminal condition, payoff

while T > 0

T = T - dt; H = A\(B*H); % solve linear system AH^m = BH^m+1

end

plot([0:dR:Rmax], H);


The idea of finite element methodsVariational formulation of parabolic problems

Computational Methods for Quantitative FinancePricing by Finite Element Methods

Prof. Ch. SchwabN. Hilber, Ch. Winter

Lecture 6



Variational formulationGalerkin discretization

Outline

The idea of finite element methodsVariational formulationGalerkin discretization

Variational formulation of parabolic problemsGeneral frameworkExamples




Problem to solve: (here, A is a linear differential operator)

∂u

∂t−Au = f in (0, T )× Iu(x, 0) = u0(x) in I

Multiply by a test function v(x) and integrate on I: Denote(u, v) :=

∫I uv and a(u, v) := −(Au, v) (bilinear form).

Variational formulation: Find u ∈ L2(0, T ;V ) ∩H1(0, T ;V ∗) suchthat

ddt

(u(t), v) + a(u(t), v) = (f(t), v) in (0, T ), ∀ v ∈ V ,u(0) = u0,

where V is some space of test functions.




Example: heat equation

∂u

∂t− ∂2u

∂x2= f in (0, T )× I

u(x, 0) = u0(x) in I

Let v be a differentiable function on I such that v = 0 on ∂I.

We have Au = ∂2u∂x2 , hence

a(u, v) = −∫I

∂2u

∂x2(x)v(x)dx =

∫I

∂u

∂x(x)

∂v

∂x(x)dx.

Note: here, u may be less regular than in the original equation.The variational formulation is a weak formulation.




Let V N be a finite (N) dimensional subspace of V .

Example: V N is the space of piecewise linear, continuous functionson a mesh T on I.

The idea is to approximate u(t, ·) by an element uN (t) of V N .The variational formulation is approximated by

ddt

(uN (t), vN ) + a(uN (t), vN ) = (f(t), vN ) in (0, T ), ∀ vN ∈ V N .




Let (bj)j=1,...,N be a basis of V N . For example:

. . .




We have uN (t, x) =∑N

j=1 uNj (t)bj(x) where

uN (t) = (uN1 (t), uN2 (t), . . . , uNN (t))>

is a vector of unknown functions. Similarly,

∀vN ∈ V N , vN (x) =N∑j=1

vNj bj(x).

As vN is arbitrary, the variational equation gives

ddt

MuN (t) + AuN (t) = fN (t) in (0, T ),

where M and A are N ×N matrices with Mij = (bi, bj),Aij = a(bj , bi), and fN is a vector with fNj = (f, bj).

This is a system of ordinary differential equations that can besolved using some discretization in time (e.g., explicit, implicit, orCrank-Nicolson scheme).



General frameworkExamples

Outline

The idea of finite element methodsVariational formulationGalerkin discretization

Variational formulation of parabolic problemsGeneral frameworkExamples




Recall that we consider the following problem:

∂u

∂t−Au = f in (0, T )× I

u(x, 0) = u0(x) in I

and its variational formulation:

ddt

(u(t), v) + a(u(t), v) = (f(t), v) in (0, T ), ∀ v ∈ V ,u(0) = u0.

I What are the conditions on the operator A, initial condition u0,right-hand side f , and the space of test functions V for thisproblem to have a unique solution?

I What is the regularity of this solution?

The answers are given in a general framework of Hilbert spaces.




Recall on Hilbert spaces

• Hilbert space: vector space H with scalar (= inner) product (u, v).Example: H = L2(I), (u, v) =

∫I u(x)v(x)dx.

• If X is a vector space, X∗ denotes the space of continuous linearfunctionals on X (dual space):

f ∈ X∗ is a continuous linear mapping from X to R.

If x ∈ X and f ∈ X∗, we denote f(x) = 〈f, x〉 (dual pairing).

• If H is a Hilbert space, H∗ may be identified with H:

f ∈ H∗ ⇐⇒ ∃ l ∈ H : ∀x ∈ H, 〈f, x〉 = (l, x).

We identify f with l and write H ∼= H∗.




Weak derivatives

Note: if u is smooth, then ∀φ ∈ C∞0 (I),∫Iuφ′ = −

∫Iu′φ (integration by parts).

Definition: g is called the weak derivative of u on I if, ∀φ ∈ C∞0 (I),∫Iuφ′ = −

∫Igφ (we write g = u′).

Remark: this expression is valid if u and g are (locally) integrable. Wecan generalize the definition to linear functionals on C∞0 (I):

〈u, φ′〉 = −〈g, φ〉.Examples: 1) u(x) = x1x≥0 =⇒ u′(x) = 1x≥0 (u′ is (locally)integrable).

2) u(x) = 1x≥x0 =⇒ u′ = δx0 (delta of Dirac).




Recall on Sobolev spaces

The weak derivatives always exist but are not always integrable.

Definition: W 1,2(I) ≡ H1(I) is the space of square integrablefunctions with square integrable first derivative:

H1(I) =u ∈ L2(I) such that u′ ∈ L2(I)

Note: more generally, we can define spaces Wn,p(I) of functions withn weak derivatives, all in Lp(I) (integrable with power p).

Remark: H1(I) is a Hilbert space with the following scalar product:

(u, v) =∫Iuv +

∫Iu′v′.




Some useful facts

• Functions u ∈ H1(I) are “essentially” continuous (for I ⊂ R):u ∈ H1(I) coincides almost everywhere with a continuous function u.

• In consequence, discontinuous functions like 1x≥x0 are not in H1(I)(their derivatives are delta-functions).

• By definition, H1 ⊂ L2 but L2 is strictly larger:for example, 1x≥x0 is in L2 but not in H1.

• The injection of H1 in L2 is continuous (denoted by H1 → L2):

‖u‖L2 ≤ C‖u‖H1

Recall: ‖u‖2L2(I) =∫I u

2dx, ‖u‖2H1(I) = ‖u‖2L2(I) + ‖u′‖2L2(I) .

• The injection of H1 in L2 is dense (denoted by H1 d→ L2):

∀u ∈ L2(I), ∃vn ∈ H1(I) : ‖u− vn‖L2(I) → 0.




General variational setting for parabolic equations

Given Hilbert spaces Vd→ H ∼= H∗

d→ V ∗

and u0 ∈ H, f ∈ L2(0, T ;V ∗),

find u ∈ L2(0, T ;V ) ∩H1(0, T ;V ∗) such that

u(0) = u0 in H ,

ddt

(u(t), w) + a(u(t), w) = (f(t), w) ∀t ∈ (0, T ), ∀w ∈ V .

Here, a(u, v) denotes the bilinear form associated with the operatorA ∈ L(V, V ∗) via

∀u, v ∈ V : a(u, v) = −〈Au, v〉

where 〈·, ·〉 denotes the V ∗ × V duality pairing.




Proposition

Assume that the bilinear form a(·, ·) is continuous: ∃C0 such that

∀u, v ∈ V : |a(u, v)| ≤ C0 ‖u‖V ‖v‖Vand satisfies the “Garding inequality”: ∃C1, C2 > 0 such that

∀u ∈ V : a(u, u) ≥ C1 ‖u‖2V − C2 ‖u‖2H .

Then, for every 0 < T <∞, the variational problem admits aunique solution u ∈ L2(0, T ;V ) ∩H1(0, T ;V ∗).

Moreover, u ∈ C0([0, T ];H), and there holds the a-priori estimate

‖u‖C0([0,T ];H)+‖u‖L2(0,T ;V )+‖u‖H1(0,T ;V ∗) ≤ C(‖u0‖H+‖f‖L2(0,T ;V ∗)

).

(norm of the solution is bounded by norms of the data)




BS-equation in log-price on R

The Black-Scholes equation in logarithmic price for v(t, x) := V (t, ex)reads

∂v

∂t+σ2

2∂2v

∂x2+(r − σ2

2

)∂v

∂x− rv = 0 in (0, T )× R

v(T, x) = h(x) := (ex −K)+.

Changing to time to maturity τ = T − t, this equation becomes:

∂u

∂τ− σ2

2∂2u

∂x2+(σ2

2− r)∂u

∂x+ ru = 0 in (0, T )× R

u(0, x) = h(x).




If h ∈ L2(R) then this problem fits into the general framework with

I H = L2(R)I V = H1(R)I u0 = h

I f = 0I and the bilinear form:

a(u, v) =σ2

2

∫R

u′(x)v′(x)dx+(σ2

2− r)∫

R

u′(x)v(x)dx+∫R

ru(x)v(x)dx.

It can be shown that a(·, ·) is continuous and satisfies the Gardinginequality. In consequence, the general existence result applies.




For most payoffs, h is not in L2(R). In the case of call options,h grows exponentially at infinity: h(x) = (ex −K)+.

To deal with such payoffs, we introduce weighted Sobolev spaces.

We define H1ν (R) (ν ∈ R) by

H1ν (R) := v ∈ L1

loc(R) | veν|x|, v′eν|x| ∈ L2(R).

Similarly,L2ν(R) := v ∈ L1

loc(R) | veν|x| ∈ L2(R).

Note: with this notation, the payoff function of a call belongs toH1−µ(R) for any µ > 1.

• L2ν(R) is a Hilbert space with scalar product

(u, v) =∫

Ru(x)v(x)e2ν|x|dx.




For the variational formulation of this problem, we choose:

I H = L2−µ(R), V = H1−µ(R), u0 = h, f = 0,

I and the bilinear form:

aBS−log−µ (u, v) =

σ2

2

∫R

u′(x)v′(x)e−2µ|x| dx

+∫R

ru(x)v(x)e−2µ|x| dx

+∫R

(− µσ2sign(x) +

(σ2

2− r))

u′(x)v(x)e−2µ|x| dx.

Again, it can be shown that aBS−log−µ (·, ·) is continuous and satisfies the

Garding inequality. So, there exists a unique solution

u ∈ L2((0, T );H1−µ(R)) ∩H1((0, T );H1

−µ(R)∗).




Localization. Excess to payoff

We localize the equation to a bounded computational domainΩR = (−R,R). This requires artificial boundary conditions at ±R.

To choose good conditions, we analyze the solution at infinity.

More precisely, we prove that the excess to payoff

U := u− e−rth(x+ rt)

decays exponentially at infinity.




Assume that h ∈ H1−ν(R) with ν > 0 (in particular, h is continuous).Remark: satisfied for puts and calls (but not for binary options).

Substituting u(τ, x) = U(τ, x) + e−rτh(x+ rτ) into the equation,we observe that U solves the following variational problem:

find U ∈ L2((0, T );H1ν (R)) ∩H1((0, T );H1

ν (R)∗) s.t. ∀ v ∈ H1ν (R)(

dUdτ

(τ, ·), v)L2ν(R)

+ aBS−logν (U(τ, ·), v) = e−rτ

σ2

2K〈f(τ, ·), ve2ν|x|〉

U(0, ·) = 0 in L2ν(R)

where we denote by 〈·, ·〉 the(H1ν (R)

)∗ ×H1ν (R) duality pairing.

Note: H = L2ν(R), V = H1

ν (R), u0 = 0, and, in the case of a call,

f = e−rτ σ2

2 Kδln(K)−rτ .

Corollary: excess to payoff U decays exponentially at infinity.In consequence, zero boundary conditions are a good choice for U .




Space H10(I)H10(I)H10(I)

For localized problems and barrier contracts, we needfunctional spaces with boundary conditions.

Definition

W 1,20 (I) ≡ H1

0 (I) = C10 (I) ‖·‖H1(I)

(closure of C10 (I) with respect to the norm of H1(I)).

Proposition

u ∈ H1(I) belongs to H10 (I) if and only if u = 0 on ∂I.




Localized problem

The localized problem for the excess to payoff reads:

∂UR∂τ−ABS−logUR + rUR = e−rτ

σ2

2Kδln(K)−rτ in (0, T )× ΩR

UR(τ, ·)|∂ΩR = 0 on ∂ΩR ∀ 0 < τ ≤ TUR|τ=0 = 0 in ΩR,

where

ABS−log =σ2

2∂2

∂x2+(r − σ2

2

)∂

∂x.

This equation cannot be discretized by Finite Differences,due to the presence of Dirac functional on the right hand side.

The variational formulation allows to handle it in a natural way.




Variational formulation

The variational form of this problem is as follows: ∀ v ∈ H10 (ΩR)(

dURdτ

(τ, ·), v)L2(ΩR)

+ aBS−log(UR(τ, ·), v)

= e−rtσ2

2K〈δln(K)−rτ , v〉H−1(ΩR)×H1

0 (ΩR)

UR(0, ·) = 0.

Note: 〈δln(K)−rτ , v〉 = v(ln(K)− rτ) is a computable quantity.




Localization error

The restriction of U from R to ΩR introduces a localization erroreR := UR − U , which goes to zero exponentially with increasing R.

TheoremLet ΩR/2 := |x| < R/2. Denote by UR the extension ofUR ∈ L2

(0, T ; H1

0 (ΩR))

with respect to x by zero to all of R.

Then there exist C = C(T ), α > 0 independent of R such thateR = UR − U satisfies the following estimate:

‖eR(τ, ·)‖2L2(ΩR/2) +∫ τ

0‖eR(s, ·)‖2H1(ΩR/2)ds ≤ Ce−αR.




BS-equation in normal form

Recall that the BS equation may be reduced to the heat equation.

If we localize it to ΩR = (−R,R) with zero Dirichlet conditions,its variational form reads:

find y ∈ L2(0, T ;H10 (ΩR)) ∩H1(0, T ;H−1(ΩR)) such that

ddτ

(y(τ), v)L2(ΩR) + (y′(τ), v′)L2(ΩR)︸︷︷︸a(y,v)

= (f(τ), v)L2(ΩR), ∀v ∈ H10 (ΩR) ,

y(0) = ψ in L2(ΩR).

Note: H = L2(ΩR), V = H10 (ΩR), H−1(ΩR) ≡ (H1

0 (ΩR))∗.




BS-equation in real price

Consider the BS-equation in real price and in time to maturity τ ,on the truncated price domain I = (0, R):

∂v

∂τ−ABSv + rv = 0 in (0, T )× (0, R) ,

v(S, 0) = g(S) in (0, R) ,

v(R, τ) = 0 in (0, T )

with artificial zero Dirichlet boundary conditions and where

ABS =σ2

2S2 ∂2

∂S2+ rS

∂

∂S.




Variational formulation

Multiply by a test function w(S) and integrate by parts. This gives thevariational formulation with f ≡ 0, u0 = g and

a(v, w) = aBS(v, w) :=σ2

2

R∫0

S2∂Sv ∂Sw dS

+(σ2 − r)R∫

0

Sw ∂Sv dS + r

R∫0

vw dS .

In which spaces V and H the bilinear form aBS(·, ·) is continuous andsatisfies the Garding inequality?




Weighted Sobolev space

The right functional spaces in this case are the following:

• H = L2(I), where I = (0, R);• V = W , where the weighted space W is defined by

W = C∞0 (I) ‖‖W with ‖u‖2W :=∫I

S2∣∣ dudS

∣∣2 + |u|2dS .

Properties

I Wd→ L2(I) (the embedding is continuous and dense).

I H10 (I) ⊂W ⊂ L2(I).

Functions in W have zero Dirichlet conditions at x = Rbut they may have nonzero boundary values at S = 0.


computational methods for quantitative finance

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