The (1,2)-skew Brownian motion: an explicit representation...

The generalized RS scheme The (β1, β2)-skew BM Sketch of the proofs References

The (β1, β2)-skew Brownian motion: an explicitrepresentation of its transition densities and its

exact simulation

Sara Mazzonetto(Universitat Potsdam and Universite Lille 1)

joint work with Sylvie Roelly and David Dereudre

Les Houches, April 21st 2016


Index

1 The generalized RS scheme

2 The (β1, β2)-skew BM

3 Sketch of the proofs


Generalized rejection sampling method

Rejection sampling methodAssume to know how to sample the r.v. Y with density g(x).Then one can sample the r.v. X with density h(x) if:

(i) ∃M > 0 such that h(x) ≤ Mg(x) for all x ∈ R;

(ii) f (x) := 1M

h(x)g(x) can be evaluated.

X(d)= (Y |U < f (Y )) i.e. an exact simulation is possible.

Theorem

Replacing (ii) by

(ii’) there exists a sequence of functions (fn)n converging pointwise to fat a decreasing rate (δn)n.Moreover for each x ∈ R fn(x), δn(x) can be evaluated,

then X(d)= (Y |∃n;U < fn(Y )− δn(Y )).


Toy example

Suppose p is an unknown parameter such that

∃(pn)n, (δn)n s.t. δnn→∞→ 0 decreasing, and |p − pn| < δn.

Then it is possible to simulate exactly a Bernoulli of parameter p sinceX := 1{∃n; |U−pn|>δn, U<pn} ∼ Bp .

0 1pu pn

pn − δn pn + δn

0 1p u′pn

pn − δn pn + δn


Heuristics on the (β1, β2)-skew BM

Let β1, β2 ∈ [−1, 1]

1+β1

21−β1

2

0

1+β2

21−β2

2

1

Example: y 7→ p(0.3,0.7)(1, 0.5, y) is the density of X1 where (Xt)t is the(0.3, 0.7)−SBM starting at 0.5.


The (β1, β2)-SBM with drift

The process is defined for β1, β2 ∈ [−1, 1], µ ∈ R, z1, z2 ∈ R,

through the SDE, by{dXt = dWt + µdt + β1dL

z1t (X ) + β2dL

z2t (X );

Lzit (X ) =∫ t

0I{Xs=zi}dL

zis (X )

the infinitesimal generator is Lf = 12 ∆f +µ∇f +

∑j=1,2 βj〈δzj ,∇f 〉.

Lemma (Divergence form operator)L = 1

2h(x)∇ (h(x)∇)

Dom(L) ={f ∈ H1

0 (h(x)dx); h∇f ∈ H1(h−1(x)dx)}

h(x) = e2µxk(x)

where k(x) =

14 (1− β1)(1− β2) x < z1,14 (1 + β1)(1− β2) z1 ≤ x < z2,14 (1 + β1)(1 + β2) x ≥ z2.


The transition densities of the (β1, β2)-SBM

Proposition

p(β1,β2)(t, x , y) = p(0,0)(t, x , y) · v (β1,β2)(t, x , y)

where, if z = z2 − z1,

v (β1,β2)(t, x , y) =∞∑k=0

(−β1β2)k4∑

j=1

cj(y)e−(aj (x,y)+2zk)2

2t e−|x−y |aj (x,y)+2zk

t

c1(y) ≡ 1

c2(y) =(21[z1,+∞)(y)− 1

)β1

c3(y) =(21[z2,+∞)(y)− 1

)β2

c4(y) =(1− 21[z1,z2)(y)

)β1β2

a1(x , y) ≡ 0

a2(x , y) = |y − z1|+ |x − z1| − |y − x |a3(x , y) = |y − z2|+ |y − z2| − |y − x |a4(x , y) = 2 (z2 −max(x , y , z1))+ + 2 (min(x , y , z2)− z1)+


Application of the GRS method

Lemma

There exists an upper bound for v (β1,β2)(t, x , y) uniform in x and y :

supx,y∈R

∣∣∣v (β1,β2)(t, x , y)∣∣∣ ≤ v t :=

(1 + |β1|)(1 + |β2|)1− |β1β2|e−

2z2

t

Lemma

The remainder of the truncated series is bounded uniformly in x , y :

|RNv (β1,β2)(t, x , y)| ≤ v t

(|β1β2|e−

2z2

t

)N+1

.


Approximation Vs generalized rejection sampling method

In the simulation we fix t = 1 and barriers z1 = 0 and z2 = 1, moreoverwe will sample 50000 times with GRSM y 7→ p(−0.7,0.3)(1, 0.5, y)

The 50000 simulations through GRS method are EXACT and the densityis approximated with its truncation at the 10th term.


Asymmetric cases with β1β2 > 0 and β1β2 < 0

y 7→ p(0.3,−0.7)(1, 0.5, y) y 7→ p(−0.8,−0.6)(1, 0.5, y)


Reflected skew Brownian motion and only one barrier

y 7→ p(1,−0.5)(1, 0.5, y) y 7→ p(0,−0.5)(1, 0.5, y)


Simulation of the process (0.7,−0.2)-SBM

Assume the initial condition X0 = 0.5

t

Xt


The transition density for the (β1, β2)-SBM with drift

Main result:Assume β1, β2 ∈ (−1, 1) and µ ∈ R.

p(β1,β2)µ (t, x , y) = p(0,0)

µ (t, x , y)v (β1,β2)µ (t, x , y)

where the function v(β1,β2)µ is given by a series of Fourier transforms. It is

a series involving two functions{J0(ω, βiµ

√t) := e−

ω2

2

√2πe

(βiµ√

t+ω)2

2 Φc(ω + βiµ√t) with i ∈ {1, 2}

J1(ω) := −e−ω2

2 ,

evaluated in ω ∈{ωj,k :=

aj (x,y)+2zk+|y−x|√t

, j = 1, 2, 3, 4, k ∈ N}

.


Representation of the transition density

Proposition

The infinitesimal generator L is self adjoint in L2(h(x)dx), and thefollowing equality holds:

p(β1,β2)µ (t, x , y) =

1

2πi

∫Γ

eλtG (β1,β2)µ (x , y ;λ)dλ,

where Γ is a complex contour of σ(L) ⊆ (−∞, 0] and

for y ∈ R and λ ∈ C \ (−∞, 0], x 7→ G(β1,β2)µ (x , y ;λ), the Green’s

function for the resolvent, solves{(λ− L)u(x) = δ{y}(x), u ∈ C2(R \ {z1, z2}) ∩ C(R)

h(z+i )u′(z+

i ) = h(z−i )u′(z−i ), i = 1, 2.

This is known as the Titchmarsh-Kodaira-Yoshida method (µ = 0).


The Green function for the (β1, β2)-SBM with drift

Lemma

The Green functions, if w :=√

2λ+ µ2 ∈ {v ∈ C;<(v) > 0}, are

G (x , y ;w) =1

weµ(y−x)

∑4j=1 cj(µ, y ;w)e−w(aj (x,y)+|x−y |)

β1β2e−2wz(w2 − µ2) + (w + β1µ)(w + β2µ).

cj(µ, y ;w) = w2cj,0(y) + wµcj,1(y) + µ2cj,2(y), wherec1,0 = 1,

c2,0 =(21[z1,+∞)(y)− 1

)β1

c3,0 =(21[z2,+∞)(y)− 1

)β2

c4,0 =(1− 21[z1,z2)(y)

)β1β2

c1,1 = β1 + β2

c2,1 = −β1 − c4,0

c3,1 = −β2 + c4,0

c4,1 = 0

c1,2 = β1β2

c2,2 = β1c3,0

c3,2 = −β2c2,0

c4,2 = −c4,0.


Ideas of the proof

If φ(λ) =√

2λ+ µ2, we recall the contour integral representation

p(β1,β2)µ (t, x , y) =

1

2πi

∫Γ

eλtG (β1,β2)µ (x , y ;φ(λ))dλ,

and we will use techniques of complex analysis:

R

iR

Γ

φ(Γ)

R

No poles

φ(Γ)

ρM

|µ|−βµ


Towards a series of Fourier transforms

For simplicity β1µ > 0, β2µ > 0, hence there is no pole.The formula for vµ

(β1,β2)(t, x , y) is

√t√

2πe

(x−y)2

2t

∫Re−

u2

2 t

∑4j=1−cj(µ, y ; iu)e−iu(aj (x,y)+|x−y |)

β1β2e−i2zu(u2 + µ2) + (u − iβ1µ)(u − iβ2µ)du.

Remark

For all u, µ ∈ R and β1, β2 ∈ (−1, 1)

|β1β2| (u2 + µ2) ≤ |(u − iβ1µ)(u − iβ2µ)|

Therefore

1

1 + β1β2(u2+µ2)e−i2zu

(u−iβ1µ)(u−iβ2µ)

=∞∑k=0

(−β1β2(u2 + µ2)

(u − iβ1µ)(u − iβ2µ)

)k

e−i2zuk


Density’s uniform bound

Under the assumption β1µ > 0, β2µ > 0,v(β1,β2)µ (t, x , y) = e

|x−y|22t∑∞

k=0

∑4j=1 Fj,k(ωj,k), ωj,k :=

aj (x,y)+2zk+|y−x|√t

Fj,k := (−β1β2)k F(w 7→ e−

w2

2 cj(y , µ√t; iw) · −(w2+µ2t)k

(w−iβ1µ√t)k+1(w−iβ2µ

√t)k+1

).

Lemma

There exists an upper bound for v(β1,β2)µ (t, x , y) uniform in x and y :

supx,y∈R

∣∣∣v (β1,β2)µ (t, x , y)

∣∣∣ ≤ C (β1, β2)

1− e−2z2

t

|RNv (β1,β2)µ (t, x , y)| ≤ C (β1, β2)

1− e−2z2

t

e−2z2

t (N+1).


Remerciements

Merci pour votre attention!


Bibliography

[1] David Dereudre, Sara Mazzonetto, and Sylvie Roelly. Exactsimulation of one-dimensional brownian diffusions with driftadmitting several jumps. Preprint.

[2] David Dereudre, Sara Mazzonetto, and Sylvie Roelly. An explicitrepresentation of the transition densities of the skew brownian motionwith drift and two semipermeable barriers. Monte Carlo MethodsAppl., 22(1):1–23, 2016.

[3] Pierre Etore and Miguel Martinez. Exact simulation for solutions ofone-dimensional stochastic differential equations with discontinuousdrift. ESAIM Probab. Stat., 18:686–702, 2014.

[4] Bernard Gaveau, Masami Okada, and Tatsuya Okada. Second orderdifferential operators and Dirichlet integrals with singular coefficients.I. Functional calculus of one-dimensional operators. Tohoku Math. J.(2), 39(4):465–504, 1987.

[5] Antoine Lejay, Lionel Lenotre, and Geraldine Pichot. One-dimensionalskew diffusions: explicit expressions of densities and resolvent kernel,2015, hal-01194187.

The (1,2)-skew Brownian motion: an explicit representation...

Documents

Transcript of The (1,2)-skew Brownian motion: an explicit representation...