The modied Euler scheme for a weak approximation of solutions of stochastic dierential equations driven by a Wiener process

The modied Euler scheme for a weak approximation of solutions of stochastic dierential equations driven by a Wiener process

Semen Bodnarchuk

Igor Sikorsky Kyiv Polytechnic Institute

Potsdam, 1 April 2019

joint result with prof. Alexey Kulik

Let X be a diusion process in R^dof the form

Xt= x +

t

Z

0

a(Xs)ds +

t

Z

0

σ(Xs)dWs, 0 ≤ t ≤ T, (1)

where W ∈ R^mis a Wiener process, a : R^d→ R^d, σ : R^d→ R^d×m.

Main goal

E x f (X _T ) =?

Monte-Carlo method

If XT is known then one can simulate N independent copies of it, e.g. XT⁽ⁱ⁾, and approximate

Exf (XT) ≈ 1 N

N

X

i=1

f (X_T⁽ⁱ⁾).

Unfortunately, in most cases the law of XT is not known. In this case one can use an approximation of it, e.g. ˆXT. Then

E^xf (XT) − 1 N

N

X

i=1

f ( ˆX_T⁽ⁱ⁾)     

≤ 

E^xf (XT) − E^xf ( ˆXT)   +

E^xf ( ˆXT) − 1 N

N

X

i=1

f ( ˆX_T⁽ⁱ⁾)      .

Time discrete approximation

Consider a time discretization of the interval [0; T ] with step h = T n: tk= kh, k = 0, . . . , n.

Let Ft= σ{Xs: s ≤ t}. Denition

We call cadlag process Y^h= Y = {Yt, 0 ≤ t ≤ T }a time discrete approximation if Yt_k is Ft_k-measurable and Yt_k+1 can be expressed as a function of Y0, . . . , Yt_k, t0, . . . , tk, tk+1

and a nite number of Ft_k+1-measurable variables.

Example.(Maruyama'55) The simplest time discrete approximation of solution to (1) is the Euler approximation (or the Euler-Maruyama approximation)

Yt_k+1= Yt_k+ a(Yt_k)h + σ(Yt_k)(Wt_k+1− Wt_k), k = 0, . . . , n − 1 (2) with Y = x.

Time discrete approximation

Denition

We say that a time discrete approximation Y^hconverges weakly with order β > 0 to X at time T as h → 0 if for all 'good enough' functions f there exist a positive constant C, which does not depend on h, such that

|Ef (XT) − Ef (Y_T^h)| ≤ Ch^β. (3) Example.(Milshtein'78) The Euler approximation converges to X with the weak order β = 1.

How to improve the order of approximation?

1) The classical approach: by using the Ito-Taylor expansion.

P. E. Kloeden, E. Platen, Numerical solution of stochastic dierential equations, Springer, Berlin, 1995.

2) Our approach.

One-step approximation

It can be shown (Milshtein'78) that for proving (3) it's sucient to prove

|Ef (Xh) − Ef (Yh^h)| ≤ Ch^β+1. (4)

Euler approximation

On the interval [0, h] the Euler scheme has the form

X˜t= x + a(x)t + σ(x)Wt, 0 ≤ t ≤ h.

Let us prove that Euler approximation has the weak order β = 1. We need to show that 

E^xf (Xh) − Exf ( ˜Xh)  ≤ Ch².

Ito formula

If X is a process of form (1) and g ∈ C²(R) then

g(Xt) = g(x) +

t

Z

0

Ag(Xs)ds +

t

Z

0

Lg(Xs)dWs, (5)

where

Ag(y) = a(y)g⁰(y) +1

2b(y)g⁰⁰(y), Lg(y) = σ(y)g⁰(y).

Here b(y) = σ²(y).

⇒ E^xg(Xt) = g(x) +

t

Z

0

E^xAg(Xs)ds. (6)

Ito formula

For ˜X the Ito formula gives

Exg( ˜Xt) = g(x) +

t

Z

0

ExAg( ˜˜ Xs)ds, where

Ag(y) = a(x)g˜ ⁰(y) +1

2b(x)g⁰⁰(y).

Euler approximation

If f ∈ Cb⁴(R), a, σ ∈ Cb²(R), then

Exf (Xh) = f (x) +

h

Z

0

ExAf (Xs)ds = f (x) +

h

Z

0



Af (x) +

s

Z

0

ExA(Af (Xr))dr



ds =

= f (x) + Af (x)h +

h

Z

0 s

Z

0

E^xA(Af (Xr))drds = f (x) + Af (x)h + O(h²).

Euler approximation

For ˜X we have

E^xf ( ˜Xh) = f (x) + a(x)

h

Z

0

E^xf⁰( ˜Xs)ds +1 2b(x)

h

Z

0

E^xf⁰⁰( ˜Xs)ds =

= f (x) + a(x)

h

Z

0



f⁰(x) +

s

Z

0

E^xAf˜ ⁰( ˜Xr)dr



ds+

+1 2b(x)

h

Z

0



f⁰⁰(x) +

s

Z

0

ExAf˜ ⁰⁰( ˜Xr)



ds =

= f (x) + Af (x)h + O(h²).

Euler approximation

E^xf (Xh) = f (x) + Af (x)h +1

2A(Af (x))h²+ O(h³) (7) and

E^xf ( ˜Xh)= f (x) + Af (x)h +^? 1

2A(Af (x))h²+ O(h³), (8) where

A(Af (x)) =



a(x)a⁰(x) +1

2a⁰⁰(x)b(x)

 f⁰(x)+

+



a²(x) +1

2a(x)b⁰(x) + a⁰(x)b(x) +1

4b(x)b⁰⁰(x)

 f⁰⁰(x)+

+



a(x)b(x) +1

2b(x)b⁰(x)



f⁰⁰⁰(x) + 1

4b²(x)f^{(IV )}(x).

Euler approximation

Exf ( ˜Xh) = f (x) + Af (x)h+

+1 2



a²(x)f⁰⁰(x) + a(x)b(x)f⁰⁰⁰(x) +1

4b²(x)f^{(IV )}(x)



h²+ O(h³).

Modied Euler scheme

Instead of the Euler scheme we consider its modication of the form Xˆt= ˜Xt+ ∆t, 0 ≤ t ≤ h,

where the corrector ∆ = ∆t( ˜Xt)has to be chosen in a such way that 

E^xf (Xh) − Exf ( ˆXh)  ≤ Ch³ for all 'good enough' functions f.

Hermite polynomials

For constructing the corrector ∆twe introduce the notion of Hermite polynomials. Let us remind their denition. Transition probability density pt(x, y)of the process ˜Xt has the form

pt(x, y) = 1

p2πtb(x)exp



−(y − x − a(x)t)² 2tb(x)

 .

Denition

The Hermite polinomials are the family of functions

H_t^(m)(x, y) : x, y ∈ R, t > 0, m ∈ N ∪ {0} , each of them satises an equality

∂^m

∂y^mpt(x, y) = (−1)^mH_t^(m)(x, y)pt(x, y).

Hermite polynomials

H_t⁽⁰⁾(x, y) = 1, H_t⁽¹⁾(x, y) =(y − x − a(x)t)

b(x)t , H_t⁽²⁾(x, y) = (y − x − a(x)t)²

b²(x)t² − 1 b(x)t and so on.

Hermite polynomials

In what follows we need the next relations



H_t⁽¹⁾(x, y)2

= H_t⁽²⁾(x, y) + 1

tb(x), (9)

H_t⁽¹⁾(x, y)H_t⁽²⁾(x, y) = H_t⁽³⁾(x, y) + 2

tb(x)H_t⁽¹⁾(x, y), (10)



H_t⁽²⁾(x, y)2

= H_t⁽⁴⁾(x, y) + 4

tb(x)H_t⁽²⁾(x, y) + 2

t²b²(x). (11)

Hermite polynomials

Lemma

Let ˜Xt= x + a(x)t + σ(x)Wt, t ≥ 0. Then 1) for f ∈ Cb^m(R) the following formula holds true

Exf ( ˜Xt)H_t^(m)(x, ˜Xt) = Exf^(m)( ˜Xt). (12) 2) for f, a, σ ∈ Cb(R) there exists constant C, which doesn't depend on t and x, such

that 

E^xf ( ˜Xt)H_t^(m)(x, ˜Xt) 

≤ Ct^−m2. (13)

Corrector

We dene

∆t:= t²

c0(x) + c1(x)H_t⁽¹⁾(x, ˜Xt) + c2(x)H_t⁽²⁾(x, ˜Xt)

, (14)

with

c0(x) = 1

2a(x)a⁰(x) +1

4a⁰⁰(x)b(x), c1(x) =1

4a(x)b⁰(x) +1

2a⁰(x)b(x) +1

8b(x)b⁰⁰(x) − 1

16(b⁰(x))², c2(x) = 1

4b(x)b⁰(x).

Main theorem (one-step approximation)

Theorem

If f ∈ Cb⁶(R), a, σ ∈ Cb⁴(R) then the following bound holds true 

E^xf (Xh) − E^xf ( ˆXh)   ≤ Ch³.

Exf (Xh) = f (x) + Af (x)h +1

2A(Af (x))h²+ O(h³) (15) and

E^xf ( ˆXh)= f (x) + Af (x)h +^? 1

2A(Af (x))h²+ O(h³). (16)

Consider the scheme ˆX with corrector ∆ dened as in (14) with arbitrary coecients c0(x), c1(x)òà c2(x). By the Taylor formula

Exf ( ˆXh) = Exf ( ˜Xh+ ∆h) = Exf ( ˜Xh) + Exf⁰( ˜Xh)∆h+1

2Exf⁰⁰( ˜Xh)∆²_h+ +1

6Exf⁰⁰⁰( ˜Xh+ θh∆h)∆³_h, θh∈ (0, 1).

Consider each term separately.

E

f ( ˜ X

)

Exf ( ˜Xh) = f (x) + Af (x)h+

+1 2



a²(x)f⁰⁰(x) + a(x)b(x)f⁰⁰⁰(x) +1

4b²(x)f^{(IV )}(x)



h²+ O(h³).

E

f

( ˜ X

)∆

E^xf⁰( ˜Xh)∆h=

= h²E^xf⁰( ˜Xh)

c0(x) + c1(x)H_h⁽¹⁾(x, ˜Xh) + c2(x)H_h⁽²⁾(x, ˜Xh) . By (12) we have

Exf⁰( ˜Xh)H_h⁽¹⁾(x, ˜Xh) = Exf⁰⁰( ˜Xh), Exf⁰( ˜Xh)H_h⁽²⁾(x, ˜Xh) = Exf⁰⁰⁰( ˜Xh).

Then by the Ito formula

E^xf⁰( ˜Xh)∆h=

= h² c0(x)f⁰(x) + c1(x)f⁰⁰(x) + c2(x)f⁰⁰⁰(x)
+ O(h³).

1

2 E

f

( ˜ X

)∆

1

2E^xf⁰⁰( ˜Xh)∆²h=

=h⁴

2E^xf⁰⁰( ˜Xh)

c0(x) + c1(x)H_h⁽¹⁾(x, ˜Xh) + c2(x)H_h⁽²⁾(x, ˜Xh)2

.

1

2 E

f

( ˜ X

)∆

E^xf⁰⁰( ˜Xh) = O(1),

Exf⁰⁰( ˜Xh)H_h⁽¹⁾(x, ˜Xh) = Exf⁰⁰⁰( ˜Xh) = O(1),

E^xf⁰⁰( ˜Xh)H_h⁽²⁾(x, ˜Xh) = E^xf^{(IV )}( ˜Xh) = O(1),

E^xf⁰⁰( ˜Xh)

H_h⁽¹⁾(x, ˜Xh)2

= E^xf⁰⁰( ˜Xh)



H_h⁽²⁾(x, ˜Xh) + 1 b(x)h



=

= Exf^{(IV )}( ˜Xh) + 1

Exf⁰⁰( ˜Xh) = O 1 ,

1

2 E

f

( ˜ X

)∆

E^xf⁰⁰( ˜Xh)H_h⁽¹⁾(x, ˜Xh)H_h⁽²⁾(x, ˜Xh) =

= Exf⁰⁰( ˜Xh)



H_h⁽³⁾(x, ˜Xh) + 2

b(x)hH_h⁽¹⁾(x, ˜Xh)



=

= Exf^{(V )}( ˜Xh) + 2

b(x)hE^xf⁰⁰⁰( ˜Xh) = O 1 h

 ,

E^xf⁰⁰( ˜Xh)



H_h⁽²⁾(x, ˜Xh)

2

=

= E^xf⁰⁰( ˜Xh)



H_h⁽⁴⁾(x, y) + 4

b(x)hH_h⁽²⁾(x, y) + 2 b²(x)h²



=

= Exf^{(V I)}( ˜Xh) + 4

b(x)hExf^{(IV )}( ˜Xh) + 2

b²(x)h²Exf⁰⁰( ˜Xh) =

 1 2

Last term

1

6Exf⁰⁰⁰( ˜Xh+ θh∆h)∆³_h= O(h³).

Exf ( ˆXh) = f (x) + Af (x)h +h² 2



2c0(x)f⁰(x) +



2c1(x) + 2c²2(x)

b²(x) + a²(x)

 f⁰⁰(x)

+(2c2(x) + a(x)b(x))f⁰⁰⁰(x) + 1

4b²(x)f^{(IV )}(x)



+ O(h³).

Recall that

A(Af (x)) =



a(x)a⁰(x) +1

2a⁰⁰(x)b(x)

 f⁰(x)+

+



a²(x) +1

2a(x)b⁰(x) + a⁰(x)b(x) +1

4b(x)b⁰⁰(x)

 f⁰⁰(x)+

+



a(x)b(x) +1

2b(x)b⁰(x)



f⁰⁰⁰(x) + 1

4b²(x)f^{(IV )}(x).

c0(x) = 1

2a(x)a⁰(x) +1

4a⁰⁰(x)b(x), c1(x) =1

4a(x)b⁰(x) +1

2a⁰(x)b(x) +1

8b(x)b⁰⁰(x) − 1

16(b⁰(x))², c2(x) = 1

4b(x)b⁰(x).

Ito-Taylor expansion

If a, σ ∈ C⁶ then

Xh= x +

h

Z

0

a(Xs)ds +

h

Z

0

σ(Xs)dWs=

= x +

h

Z

0



a(x) +

s

Z

0

Aa(Xr)dr +

s

Z

0

La(Xr)dWr



ds+

+

h

Z

0



σ(x) +

s

Z

0

Aσ(Xr)dr +

s

Z

0

Lσ(Xr)dWr



dWs=

= x + a(x)

h

Z

0

ds + σ(x)

h

Z

0

dWs+ R2=

Ito-Taylor expansion

= x + a(x)h + σ(x)Wh+

+Aa(x)

h

Z

0 s

Z

0

drds + La(x)

h

Z

0 s

Z

0

dWrds + Aσ(x)

h

Z

0 s

Z

0

drdWs+ Lσ(x)

h

Z

0 s

Z

0

dWrdWs+

+R3.

Ito-Taylor expansion

Let Z1 and Z2 are two independent N(0, 1) random variables. Then

Wh=√ hZ1,

h

Z

0 s

Z

0

dWrds = 1 2h³²

 Z1+ 1

√3Z2

 .

h

Z

0 s

Z

0

dWrdWs= 1

2 W_h²− h
.

h

Z

0 s

Z

0

dW_rⁱdW_s^k=?

Thank you for attention!