Stability analysis of approximation schemes for … Stability and convergence Numerical Stability Stability analysis of approximation schemes for BSDEs J-F Chassagneux (Imperial College

IntroductionStability and convergence

Numerical Stability

Stability analysis of approximation schemes for BSDEs

J-F Chassagneux (Imperial College London)joint work with .

A. Richou (Universite Bordeaux 1)

Advances in Financial Mathematics, Paris, 08-01-2014

J-F Chassagneux Stability of approx. for BSDEs


Numerical Stability

IntroductionFrameworkBTZ scheme - (implicit Euler)

Stability and convergenceLinear Multi-step schemesL2-StabilityNumerical illustration

Numerical StabilityDefinitionSemi-explicit EulerImplicit Euler



Numerical Stability

FrameworkBTZ scheme - (implicit Euler)

Outline






Numerical Stability


Backward Sto. Diff. Eq. - Markovian Setting

I Decoupled Forward Backward SDE on [0,T ]:

Xt = X0 +

∫ t

0b(Xs)ds +

∫ t

0σ(Xs)dWs

Yt = g(XT ) +

∫ T

tf (Ys ,Zs)ds −

∫ T

tZsdWs

↪→ b, σ, f and g are Lipschitz functions.

I PDE representation: Yt = u(t,Xt), u solution to

−L(0)u = f (u, L(1)u) and u(T , .) = g

where L(0)u := ∂tu + LXu and L(1)u = ∂xu.σ.If smoothness,

Zt = L(1)[u](t,Xt) and f (Yt ,Zt) = −L(0)[u](t,Xt).



Numerical Stability


Backward Sto. Diff. Eq. - Markovian Setting

I Decoupled Forward Backward SDE on [0,T ]:

Xt = X0 +

∫ t

0b(Xs)ds +

∫ t

0σ(Xs)dWs

Yt = g(XT ) +

∫ T

tf (Ys ,Zs)ds −

∫ T

tZsdWs

↪→ b, σ, f and g are Lipschitz functions.I PDE representation: Yt = u(t,Xt), u solution to

−L(0)u = f (u, L(1)u) and u(T , .) = g

where L(0)u := ∂tu + LXu and L(1)u = ∂xu.σ.If smoothness,

Zt = L(1)[u](t,Xt) and f (Yt ,Zt) = −L(0)[u](t,Xt).



Numerical Stability


Non-linearity and finance

Example

The stock price is given by

Xt = X0 +

∫ t

0µXsds +

∫ t

0σXsdWs

The price of an european option with payoff g , assuming different ratefor borrowing (R) and lending (r) is given by

Yt = g(XT ) +

∫ T

t(−rYs +

µ− r

σZs + (R − r)[Ys −

Zs

σ]−)ds −

∫ T

tZsdWs .

↪→ These are the dynamics of the value of the optimal hedging portfolio.

I Non-linearity coming from f (y , z) = −ry + µ−rσ z + (R − r)[y − z

σ ]−



Numerical Stability


Deriving the scheme

We are given an equidistant grid π = {0 = t0 < ... < ti < ... < tn = T},define h = T/n.

I Start with: Yti +∫ ti+1

tiZsdWs = Yti+1 +

∫ ti+1

tif (Ys ,Zs)ds (1)

I For the Y -part:

Take conditional expectation, Yti ' Eti

[Yti+1 + hf (Yti ,Zti )

]↪→ Yi := Eti[Yi+1 + hf (Yi ,Zi )]

I For the Z -part:

Multiply (1) by ∆Wi := Wti+1 −Wti , take conditional expectation:

Eti

[∫ ti+1

tiZsds

]' Eti

[∆WiYti+1

]Say Eti

[∫ ti+1

tiZsds

]' hZti , =⇒ hZti ' Eti[∆WiYi+1]

↪→ Zi := Eti[HiYi+1] with Hi := h−1∆Wi .



Numerical Stability


Deriving the scheme



tiZsdWs ' Yti+1 + hf (Yti ,Zti ) (1)

I For the Y -part:



]↪→ Yi := Eti[Yi+1 + hf (Yi ,Zi )]

I For the Z -part:


Eti

[∫ ti+1

tiZsds

]' Eti

[∆WiYti+1

]Say Eti

[∫ ti+1

tiZsds





Numerical Stability


Deriving the scheme




I For the Y -part:



]↪→ Yi := Eti[Yi+1 + hf (Yi ,Zi )]

I For the Z -part:


Eti

[∫ ti+1

tiZsds

]' Eti

[∆WiYti+1

]Say Eti

[∫ ti+1

tiZsds





Numerical Stability


Deriving the scheme




I For the Y -part:Take conditional expectation, Yti ' Eti


]

↪→ Yi := Eti[Yi+1 + hf (Yi ,Zi )]

I For the Z -part:


Eti

[∫ ti+1

tiZsds

]' Eti

[∆WiYti+1

]Say Eti

[∫ ti+1

tiZsds





Numerical Stability


Deriving the scheme






]↪→ Yi := Eti[Yi+1 + hf (Yi ,Zi )]

I For the Z -part:


Eti

[∫ ti+1

tiZsds

]' Eti

[∆WiYti+1

]Say Eti

[∫ ti+1

tiZsds





Numerical Stability


Deriving the scheme






]↪→ Yi := Eti[Yi+1 + hf (Yi ,Zi )]

I For the Z -part:


Eti

[∫ ti+1

tiZsds

]' Eti

[∆WiYti+1

]Say Eti

[∫ ti+1

tiZsds





Numerical Stability


Deriving the scheme






]↪→ Yi := Eti[Yi+1 + hf (Yi ,Zi )]

I For the Z -part:Multiply (1) by ∆Wi := Wti+1 −Wti , take conditional expectation:

Eti

[∫ ti+1

tiZsds

]' Eti

[∆WiYti+1

]

Say Eti

[∫ ti+1

tiZsds





Numerical Stability


Deriving the scheme






]↪→ Yi := Eti[Yi+1 + hf (Yi ,Zi )]


Eti

[∫ ti+1

tiZsds

]' Eti

[∆WiYti+1

]Say Eti

[∫ ti+1

tiZsds





Numerical Stability


Deriving the scheme






]↪→ Yi := Eti[Yi+1 + hf (Yi ,Zi )]


Eti

[∫ ti+1

tiZsds

]' Eti

[∆WiYti+1

]Say Eti

[∫ ti+1

tiZsds





Numerical Stability


Objectives

I The Scheme: given the terminal condition Yn = g(Xn), thetransition from step i + 1 to i is

Yi := Eti[Yi+1 + hf (Yi ,Zi )]

Zi := Eti[HiYi+1]

↪→ The curse of computing the conditional expectations.I Goals:

(i) Prove the convergence of the scheme (using a first notion of stability)(ii) Understand the behaviour of the scheme in practice (using a second

notion of stability)

I Remark: for this scheme, convergence has been proved by Zhangand Bouchard - Touzi (2004) and Gobet-Labart (2007).



Numerical Stability


Objectives



Zi := Eti[HiYi+1]

↪→ The curse of computing the conditional expectations.

I Goals:






Numerical Stability


Objectives



Zi := Eti[HiYi+1]







Numerical Stability


Objectives



Zi := Eti[HiYi+1]


(i) Prove the convergence of the scheme (using a first notion of stability)

(ii) Understand the behaviour of the scheme in practice (using a secondnotion of stability)




Numerical Stability


Objectives



Zi := Eti[HiYi+1]







Numerical Stability


Objectives



Zi := Eti[HiYi+1]







Numerical Stability

Linear Multi-step schemesL2-StabilityNumerical illustration

Outline






Numerical Stability


Designing more acurate scheme, Y -part

I We want to find coefficients (aj), (bj):

Yti = Eti

r∑j=1

ajYti+j + hr∑

j=0

bj f (Yti+j ,Zti+j )

+ εYi

with the truncation error εYi = Oi (hm).

I Observing that:

Eti

[Yti+j

]=

m∑k=0

cYk (ti )|jh|k + O(hm+1)

Eti

[f (Yti+j ,Zti+j )

]= −

m∑k=1

cYk (ti )|jh|k + O(hm)

we end up matching coefficients to cancel terms...



Numerical Stability


Designing more acurate scheme, Y -part

I We want to find coefficients (aj), (bj):

Yti = Eti

r∑j=1

ajYti+j + hr∑

j=0

bj f (Yti+j ,Zti+j )

+ εYi

with the truncation error εYi = Oi (hm).I Observing that:

Eti

[Yti+j

]=

m∑k=0

cYk (ti )|jh|k + O(hm+1)

Eti

[f (Yti+j ,Zti+j )

]= −

m∑k=1

cYk (ti )|jh|k + O(hm)

we end up matching coefficients to cancel terms...



Numerical Stability


Designing more acurate scheme, Z -part

I We want to find coefficients (αj), (βj):

Zti = Eti

r∑j=1

αjHYi ,jYti+j + h

r∑j=1

βjHfi ,j f (Yti+j ,Zti+j )

+ εZi .

such that εZi = Oi (hm)

I The following expansions hold true:

Eti

[HYi ,jYti+j

]=

m−1∑k=0

cZk (ti )|jh|k + O(hm)

Eti

[H fi ,j f (Yti+j ,Zti+j )

]= −

m−1∑k=1

cZk (ti )|jh|k + O(hm−2)

where Hi ,j = 1jh

∫ ti+j

tiϕ( s−tihj )dWs for some ’special’ functions ϕ.



Numerical Stability


Designing more acurate scheme, Z -part

I We want to find coefficients (αj), (βj):

Zti = Eti

r∑j=1

αjHYi ,jYti+j + h

r∑j=1

βjHfi ,j f (Yti+j ,Zti+j )

+ εZi .

such that εZi = Oi (hm)I The following expansions hold true:

Eti

[HYi ,jYti+j

]=

m−1∑k=0

cZk (ti )|jh|k + O(hm)

Eti

[H fi ,j f (Yti+j ,Zti+j )

]= −

m−1∑k=1

cZk (ti )|jh|k + O(hm−2)

where Hi ,j = 1jh

∫ ti+j

tiϕ( s−tihj )dWs for some ’special’ functions ϕ.



Numerical Stability


Linear multi-step scheme

Definition(i) To initialise the scheme with r steps, r ≥ 1, we are given r terminalconditions (Yn−j ,Zn−j), Ftn−j -measurable square integrable randomvariables, 0 ≤ j ≤ r − 1.(ii) For i ≤ n − r , the computation of (Yi ,Zi ) involves r steps and isgiven by

Yi = Eti

[∑rj=1 ajYi+j + h

∑rj=0 bj f (Yi+j ,Zi+j)

]Zi = Eti

[∑rj=1 αjH

Yi ,jYi+j + h

∑rj=1 βjH

fi ,j f (Yi+j ,Zi+j)

]where aj , bj , αj , βj are real numbers.



Numerical Stability


A pertubation approach

We consider a pertubed scheme: Yi = Eti

[∑rj=1 aj Yi+j + h

∑rj=0 bj f (Yi+j , Zi+j)

]+ ζYi

Zi = Eti

[∑rj=1 αjH

yi ,j Yi+j + h

∑rj=1 βjH

fi ,j fi ,j f (Yi+j , Zi+j)

]+ ζZi

set δYk := Yk − Yk , δZk := Zk − Zk .



Numerical Stability


Definition: L2-Stability

DefinitionThe multi-step scheme is said to be L2-stable if

max0≤i≤n−r

E[|δYi |2

]+

n−r∑i=0

hE[|δZi |2

]≤ C

(max

0≤j≤r−1E[|δYn−j |2 + h|δZn−j |2

]+ h

n−r∑i=0

E[

1

h2|ζYi |2 + |ζZi |2

])for all sequences ζYi ,ζZi of L2(Fti )-random variable and terminal values(Yn−j ,Zn−j), (Yn−j , Zn−j) belonging to L2(Ftn−j ), 0 ≤ j ≤ r − 1.

i.e. the overall error is the sum of the error done at each step +initialization error. C will depend ’dramatically’ on T and Lip(f ).



Numerical Stability


Main result

Recall the scheme:Yi = Eti

[∑rj=1 ajYi+j + h

∑rj=0 bj f (Yi+j ,Zi+j)

]Zi = Eti

[∑rj=1 αjH

Yi ,jYi+j + h

∑rj=1 βjH

fi ,j f (Yi+j ,Zi+j)

]TheoremAssume that f is Lipschitz continuous and the following holds

(H) The coefficients (aj) are non-negative,∑r

j=1 aj = 1 and for1 ≤ j ≤ r , aj = 0 =⇒ αj = 0,

then, the multi-step scheme is L2-stable, for h small enough.



Numerical Stability


Conclusion

I Stability + control of truncation error =⇒ convergence

I Possibly high order rate of convergence, if smoothness

I Scheme initialisation: use one-step method: Runge-Kutta method!



Numerical Stability


Example 1

Example

In this example, we work with d = 1. The decoupled FBSDE is given by

Xt = 1 +

∫ t

0µXsds +

∫ t

0σXsdWs

Yt =X1

1 + X1+

∫ 1

tZs(σYs −

µ

σ)ds −

∫ 1

tZsdWs , with (µ, σ) = (0.05, 0.4) .

A direct application of Ito formula yields that (Yt ,Zt) = ( Xt1+Xt

, σXt(1+Xt)2 ).



Numerical Stability


Graph Example 1



Numerical Stability


Example 2

Example

In this example, we work with d = 3 and X = W . We consider thefollowing BSDE

Yt =ω(1,W1)

1 + ω(1,W1)+

∫ 1

t(Z 1

s + Z 2s + Z 3

s )(5

6− Ys)ds −

∫ 1

tZsdWs .

where ω(t, x) = exp (x1 + x2 + x3 + t). Applying Ito formula, we verifythat the solution is given by

Yt =ω(t,Wt)

1 + ω(t,Wt)and Z l

t =ω(t,Wt)

(1 + ω(t,Wt))2, l ∈ {1, 2, 3} , t ∈ [0, 1] .



Numerical Stability


Graph Example 2



Numerical Stability

DefinitionSemi-explicit EulerImplicit Euler

Outline






Numerical Stability


Some questions to answer

I Many possible schemes, some better than other? (with same orderof convergence)

I L2-stability, convergence: asymptotic result when h is small... Whatif f has a big Lipschitz constant? Stiff problems.

I What if T is big?

I Will the scheme always return a reasonable value? Are there someconstraints on h?



Numerical Stability


Framework

We will discuss the two following schemes:

I BTZ-scheme ’Implicit Euler’:


Zi := Eti[HiYi+1]

I ’Semi explicit’ Euler:

Yi := Eti[Yi+1 + hf (Yi+1,Zi )]

Zi := Eti[HiYi+1]

I Eti[Hi ] = 0, Eti

[|Hi |2

]≤ Λ, for some given Λ > 0.



Numerical Stability


Assumptions

I f (0, 0) = 0 =⇒ (0, 0) is solution with terminal condition 0.

I f is Lipschitz continuous in Y (uniformly), i.e.

|f (y , z)− f (y ′, z)| ≤ LY |y − y ′| .

and

yf (y , 0) ≤ −lY |y |2

where LY , lY are non-negative real numbers.

I We also assume that f is Lipschitz continuous in z (uniformly) i.e.

|f (y , z)− f (y , z ′)| ≤ LZ |z − z ′| .



Numerical Stability


Assumptions



|f (y , z)− f (y ′, z)| ≤ LY |y − y ′| .

and

yf (y , 0) ≤ −lY |y |2



|f (y , z)− f (y , z ′)| ≤ LZ |z − z ′| .



Numerical Stability


Assumptions



|f (y , z)− f (y ′, z)| ≤ LY |y − y ′| .

and

yf (y , 0) ≤ −lY |y |2



|f (y , z)− f (y , z ′)| ≤ LZ |z − z ′| .



Numerical Stability


Remarks

I Things can go wrong already for ODEs: y ′ = f (y) withf (y) = −ay , a > 0.

Explicit Euler scheme satisfies: yn = (1− ah)ny0

if h > 2a and n is big, we get NaN!

h = 2a is a critical value.

I For this talk, numerical illustration done using binomial tree (or

trinomial tree) for brownian motion (W ).

↪→ The terminal condition is given by cos(WT ).

↪→ T will be big to get a sharp characterisation of thestability/unstability.



Numerical Stability


Remarks

I Things can go wrong already for ODEs: y ′ = f (y) withf (y) = −ay , a > 0.

Explicit Euler scheme satisfies: yn = (1− ah)ny0

if h > 2a and n is big, we get NaN!

h = 2a is a critical value.

I For this talk, numerical illustration done using binomial tree (or

trinomial tree) for brownian motion (W ).

↪→ The terminal condition is given by cos(WT ).

↪→ T will be big to get a sharp characterisation of thestability/unstability.



Numerical Stability


Some Definitions

Let ξ be a bounded terminal condition (random).

I Numerical stability: We say that the scheme is asymptoticallynumerically stable if there exists h∗ > 0, such that for all h ≤ h∗

|Y0| ≤ C ||ξ||∞,

where C is a constant that does not depend on T .

I A-stability: if, moreover, h∗ =∞, then we say that the scheme isA-stable.

↪→ In practice, we could expect 2 regimes for the scheme:

I h < h: scheme returns a ’reasonable’ value

I h > h: scheme is unstable



Numerical Stability


Some Definitions

Let ξ be a bounded terminal condition (random).

I Numerical stability: We say that the scheme is asymptoticallynumerically stable if there exists h∗ > 0, such that for all h ≤ h∗

|Y0| ≤ C ||ξ||∞,

where C is a constant that does not depend on T .

I A-stability: if, moreover, h∗ =∞, then we say that the scheme isA-stable.

↪→ In practice, we could expect 2 regimes for the scheme:

I h < h: scheme returns a ’reasonable’ value

I h > h: scheme is unstable



Numerical Stability


Strict monotony (lY > 0) and one dimensional Y

TheoremAssume that

1− h∗|LY |2

2lY− LZh∗max

i|Hi | ≥ 0 (1)

then the scheme is numerically asymptoticaly stable for the Y part.

↪→ Scheme based on binomial tree, Linear case: f (y , z) = −ay + bz .the condition reads

√h ≤

√(b

a

)2

+2

a− b

a> 0 .



Numerical Stability


Strict monotony (lY > 0) and one dimensional Y

TheoremAssume that

1− h∗|LY |2

2lY− LZh∗max

i|Hi | ≥ 0 (1)



√h ≤

√(b

a

)2

+2

a− b

a> 0 .



Numerical Stability


Strict monotony (lY > 0) and multidimensional Y

TheoremAssume that

1−

(LY√

h∗ + LZ√

Λ)2

2lY≥ 0 (2)



√h∗ =

√2

a− b

a.

More restrictive condition.



Numerical Stability


Strict monotony (lY > 0) and multidimensional Y

TheoremAssume that

1−

(LY√

h∗ + LZ√

Λ)2

2lY≥ 0 (2)



√h∗ =

√2

a− b

a.

More restrictive condition.



Numerical Stability


Numerical illustration f (y , z) = −ay + bz , b = 5

Yellow = unstable, correct Y0 value ' 0.



Numerical Stability


Strict monotony (lY > 0) and multidim Y

TheoremAssume that

1

Λ− |L

Z |2

2ly≥ 0 ,

then the scheme is A-stable.


1− b2

2a≥ 0 .



Numerical Stability


Strict monotony (lY > 0) and multidim Y

TheoremAssume that

1

Λ− |L

Z |2

2ly≥ 0 ,

then the scheme is A-stable.


1− b2

2a≥ 0 .



Numerical Stability


One dimensional Y , possibly lY = 0

TheoremAssume that that

1− LZh∗maxi|Hi | ≥ 0 (3)

then the scheme is asymptotically numerically stable.


h∗ ≤ 1

b2.



Numerical Stability


One dimensional Y , possibly lY = 0

TheoremAssume that that

1− LZh∗maxi|Hi | ≥ 0 (3)

then the scheme is asymptotically numerically stable.


h∗ ≤ 1

b2.



Numerical Stability


Financial application

I Consider the model with two different interest rates and thefollowing ’plausible’ parameters:

r = 0.03, R = 0.06, σ = 0.4.

I then (for binomial tree), we get

- Implicit scheme: A-stable!- Semi explicit scheme: h ≥ 12.

No worries, at least for d = 1...



Numerical Stability


Financial application

I Consider the model with two different interest rates and thefollowing ’plausible’ parameters:

r = 0.03, R = 0.06, σ = 0.4.

I then (for binomial tree), we get

- Implicit scheme: A-stable!- Semi explicit scheme: h ≥ 12.

No worries, at least for d = 1...



Numerical Stability


Numerical illustration −ay + bz , b = 5



Numerical Stability


Von Neumann Stability Analysis, f (y , z) = bz

I For k ∈ R, ξ = e ikWtn (binomial tree),

Yi = yieikWti

with

yi = λyi+1 with λ := E[(1 + b∆Wi )e ik∆Wi

]

I The scheme is stable iff |λ| ≤ 1 i.e. (after some computations)

h|b|2 ≤ 1 !

I Remark: observe that the dimension of b (so W ) impacts thestability of the scheme.



Numerical Stability




Yi = yieikWti

with


]I The scheme is stable iff |λ| ≤ 1 i.e. (after some computations)

h|b|2 ≤ 1 !




Numerical Stability




Yi = yieikWti

with


]I The scheme is stable iff |λ| ≤ 1 i.e. (after some computations)

h|b|2 ≤ 1 !




Numerical Stability


Numerical illustration f (z) = bz



Numerical Stability


Numerical illustration f (z) = b|z |



Numerical Stability


Numerical illustration f (z) = sin(bz)



Numerical Stability


Conclusion

I Above results hold true for US options.

I Convergent method for BSDEs with unbounded terminal time?

I Numerical stability condition for other schemes?

I Numerical experiments to run with other methods to computeconditional expectations, in greater dimension.



Numerical Stability


Non-linearity is beautiful!

Thank you!


Documents

Stability analysis of approximation schemes for … Stability and convergence Numerical Stability Stability analysis of approximation schemes for BSDEs J-F Chassagneux (Imperial College