     • date post

23-Jun-2020
• Category

## Documents

• view

0

0

Embed Size (px)

### Transcript of Lecture 3: Stochastic Differential Equations â€؛ ~song â€؛ Gene Golub Summer...آ  Linear...

• Lecture 3: Stochastic Differential Equations

David Nualart

Department of Mathematics Kansas University

Gene Golub SIAM Summer School 2016 Drexel University

David Nualart (Kansas University) July 2016 1 / 54

• Strong solutions

Let B = {Bjt , t ≥ 0, j = 1, . . . ,d} be a d-dimensional Brownian motion and ξ an m-dimensional random vector independent of B.

Let Ft be the σ-field generated by {Bs,0 ≤ s ≤ t} , ξ and the null sets.

Consider measurable coefficients bi (t , x) and σij (t , x), 1 ≤ i ≤ m, 1 ≤ j ≤ d from [0,∞)× Rm to R.

Our aim to give a meaning to the stochastic differential equation on Rm :

dX it = bi (t ,Xt )dt + d∑

j=1

σij (t ,Xt )dB j t , 1 ≤ i ≤ m (1)

with initial condition X0 = ξ.

David Nualart (Kansas University) July 2016 2 / 54

• Definition We say that an adapted and continuous process X = {Xt , t ≥ 0} is a solution to equation (1) if for all t ≥ 0,

Xt = ξ + ∫ t

0 b(s,Xs)ds +

∫ t 0 σ(s,Xs)dBs, a.s.

or

X it = ξi + ∫ t

0 bi (s,Xs)ds +

d∑ j=1

∫ t 0 σij (s,Xs)dB

j s, 1 ≤ i ≤ m, a.s.

b is called the drift and σ is called the diffusion coefficient.

David Nualart (Kansas University) July 2016 3 / 54

• Theorem Suppose that the coefficients are locally Lipschitz in the space variable, that is, for each N ≥ 1, there exists KN > 0 such that for each ‖x‖, ‖y‖ ≤ N and t ≥ 0

‖b(t , x)− b(t , y)‖+ ‖σ(t , x)− σ(t , y)‖ ≤ KN‖x − y‖.

Then, two solutions with the same initial condition coincide almost surely (strong uniqueness holds).

In the absence of the locally Lipschitz condition equation (1) might fail to be solvable or have multiple solutions.

Example : Xt = ∫ t

0 |Xs| αds, where α ∈ (0,1). Then Xt = 0 is a solution

and also, for any s ≥ 0,

Xt = (

t − s β

)β 1[s,∞)(t), β = 1/(1− α),

is also a solution.

David Nualart (Kansas University) July 2016 4 / 54

• Lemma (Gronwall lemma) Let u be a nonnegative continuous function on [0,∞) such that

u(t) ≤ α(t) + ∫ t

0 β(s)u(s)ds, t ≥ 0

with β ≥ 0 and α non-decreasing. Then,

u(t) ≤ α(t) exp

(∫ t 0 β(s)ds

) , t ≥ 0.

In particular, if α and β are constant, we get

u(t) ≤ αeβt .

David Nualart (Kansas University) July 2016 5 / 54

• Proof :

Let X and X̃ two solutions. Define

Sn = inf{t ≥ 0 : ‖Xt‖ ≥ n or ‖X̃t‖ ≥ n}.

Clearly Sn are stopping times such that Sn ↑ ∞. Then,

E‖Xt∧Sn − X̃t∧Sn‖ 2 ≤ 2E

[∫ t∧Sn 0

‖b(u,Xu)− b(u, X̃u‖du

]2

+2E m∑

i=1

∣∣∣∣∣∣ d∑

j=1

∫ t∧Sn 0

(σij (u,Xu)− σij (u, X̃u))dBju

∣∣∣∣∣∣ 2

≤ 2tE ∫ t∧Sn

0 ‖b(u,Xu)− b(u, X̃u‖2du

+2E ∫ t∧Sn

0 ‖σ(u,Xu)− σ(u, X̃u‖2du.

David Nualart (Kansas University) July 2016 6 / 54

• Using the local Lipschitz property, we obtain for any t ≥ 0,

E‖Xt∧Sn − X̃t∧Sn‖ 2 ≤ 2(t + 1)K 2n

∫ t 0

E‖Xu∧Sn − X̃u∧Sn‖ 2du.

Then g(t) = E‖Xt∧Sn − X̃t∧Sn‖2 satisfies

g(t) ≤ 2(t + 1)K 2n ∫ t

0 g(u)du,

which, by Gronwall’s lemma, implies that g = 0.

Letting n→∞ we conclude that Xt = X̃t .

David Nualart (Kansas University) July 2016 7 / 54

• A local Lipschitz condition is not sufficient to guarantee global existence of a solution. Example :

Xt = 1 + ∫ t

0 X 2s ds.

the solution is Xt = 11−t , which explodes as t ↑ 1.

David Nualart (Kansas University) July 2016 8 / 54

• Exercise : Given x ∈ Rm, we can find a strictly positive stopping time τ and a stochastic process {Xt , t < τ} such that

Xt = x + ∫ t

0 b(s,Xs)ds +

∫ t 0 σ(s,Xs)dBs, t < τ.

The process {Xt , t < τ} is unique in the sense that if ρ is another strictly positive stopping time and {Yt , t < ρ} satisfies

Yt = x + ∫ t

0 b(s,Ys)ds +

∫ t 0 σ(s,Ys)dBs, t < ρ,

then ρ ≤ τ and for every t ≥ 0, Yt1{t

• Theorem Suppose that the coefficients b and σ satisfy the global Lipschitz and linear growth conditions :

‖b(t , x)− b(t , y)‖+ ‖σ(t , x)− σ(t , y)‖ ≤ K (‖x − y‖, ‖b(t , x)‖2 + ‖σ(t , x)‖2 ≤ K 2(1 + ‖x‖2),

for every x , y ∈ Rm, t ≥ 0. Suppose also that

E‖ξ‖2 0

E

( sup

0≤t≤T ‖Xt‖2

) ≤ CT ,K (1 + E‖ξ‖2),

where CT ,K depends on T and K .

David Nualart (Kansas University) July 2016 10 / 54

• Proof :

(i) Define the Picard iterations by putting X (0)t = ξ and for k ≥ 0,

X (k+1)t = ξ + ∫ t

0 b(s,X (k)s )ds +

∫ t 0 σ(s,X (k)s )dBs.

It is easy to check that

E

( sup

0≤t≤T ‖X (1)t ‖

2

) ≤ CT ,K (1 + E‖ξ‖2).

Then X (k+1)t − X (k) t = Vt + Mt , where

Vt = ∫ t

0 [b(s,X (k)s )− b(s,X

(k−1) s )]ds

and

Mt = ∫ t

0 [σ(s,X (k)s )− σ(s,X

(k−1) s )]dBs.

David Nualart (Kansas University) July 2016 11 / 54

• (ii) By the maximal inequality for square integrable martingales,

E

[ sup

0≤t≤T ‖Mt‖2

] ≤ 4E

∫ T 0 ‖σ(s,X (k)s )− σ(s,X

(k−1) s )‖2ds

≤ 4K 2 ∫ T

0 E‖X (k)s − X

(k−1) s ‖2ds.

On the other hand,

E

[ sup

0≤t≤T ‖Vt‖2

] ≤ K 2T

∫ T 0

E‖X (k)s − X (k−1) s ‖2ds,

E

[ sup

0≤t≤T ‖X (k+1)t − X

(k) t ‖

2

] ≤ L

∫ T 0

E‖X (k)s − X (k−1) s ‖2ds,

where L = 2K 2(4 + T ).

David Nualart (Kansas University) July 2016 12 / 54

• (iii) By iteration,

E

[ sup

0≤t≤T ‖X (k+1)t − X

(k) t ‖

2

] ≤ C∗ (LT )

k

k ! ,

where C∗ = E [ max0≤t≤T ‖X (1) − ξ‖2

]

• Under the assumptions of the theorem, if E‖ξ‖p

• Linear stochastic differential equations

The geometric Brownian motion

Xt = ξe ( µ−σ22

) t+σBt

solves the linear SDE

dXt = µXtdt + σXtdBt .

More generally, the solution of the homogeneous linear SDE

dXt = b(t)Xtdt + σ(t)XtdBt ,

where b(t) and σ(t) are continuous functions, is

Xt = ξ exp [∫ t

0

( b(s)− 12σ

2(s) )

ds + ∫ t

0 σ(s)dBs ] .

David Nualart (Kansas University) July 2016 15 / 54

• Ornstein-Uhlenbeck process

Consider the SDE (Langevin equation)

dXt = a (µ− Xt ) dt + σdBt

with initial condition X0 = x , where a, σ > 0 and µ is a real number.

The process Yt = Xteat , satisfies

dYt = aXteatdt + eatdXt = aµeatdt + σeatdBt .

Thus,

Yt = x + µ(eat − 1) + σ ∫ t

0 easdBs,

which implies

Xt = µ+ (x − µ)e−at + σe−at ∫ t

0 e asdBs.

David Nualart (Kansas University) July 2016 16 / 54

• The process Xt is Gaussian with mean and covariance given by :

E(Xt ) = µ+ (x − µ)e−at ,

Cov (Xt ,Xs) = σ2e−a(t+s)E

[(∫ t 0

ear dBr

)(∫ s 0

ear dBr

)]

= σ2e−a(t+s) ∫ t∧s

0 e2ar dr

= σ2

2a

( e−a|t−s| − e−a(t+s)

) .

The law of Xt is the normal distribution

N ( µ+ (x − µ)e−at , σ

2

2a ( 1− e−2at

)) and it converges, as t tends to infinity to the normal law

ν = N(µ, σ2

2a ).

This distribution is called invariant or stationary.

Exercise : Show that if L(ξ) = ν, then Xt has law ν, ∀t ≥ 0. David Nualart (Kansas University) July 2016 17 / 54

• General linear SDEs

Consider the equation

dXt = (a(t) + b(t)Xt ) dt + (c(t) + d(t)Xt ) dBt ,

with initial condition ξ = x , where a, b, c and d are continuous functions. The solution to this equation is given by

Xt = Ut

( x +

∫ t 0

[a(s)− c(s)d(s)] U−1s ds + ∫ t

0 c(s) U−1s dBs

) ,

where

Ut = exp

(∫ t 0

b(s)ds + ∫ t

0 d(s)dBs −

1 2

∫ t 0

d2(s)ds

) .

Proof : Write Xt = UtVt , where dVt = α(t)dt + β(t)dBt , and find α and β. �

David Nualart (Kansas University) July 2016 18 / 54

• Stochastic flows

Suppose that the coefficients are globally Lipschitz with linear growth. Denote by X xt the solution with initial condition x ∈ Rm.

Proposition Let T > 0. For every p ≥ 2, there exists a constant Cp,T such that for each 0 ≤ s ≤ t ≤ T and x , y ∈ Rm,

E ( ‖X xt − X

y s ‖p ) ≤ Cp,T (‖x − y‖p + |t − s|p/2).

As a consequence, there exists a version {X̃ xt , t ≥ 0, x ∈ Rm} of the process {X xt , t ≥ 0,