Exact linearization of one-dimensional jump-diffusion ...

Nonlinear Dyn DOI 10.1007/s11071-006-9165-2

ORIGINAL ARTICLE

Exact linearization of one-dimensional jump-diffusion stochastic differential equations Gazanfer Unal · Abdullah Sanver · Ismail Iyigunler

Received: 5 August 2006 / Accepted: 2 November 2006 C Springer Science + Business Media B.V. 2006 

Abstract Necessary and sufficient conditions for the linearization of the one-dimensional Itˆo jump-diffusion stochastic differential equations (JDSDE) are given. Stochastic integrating factor has been introduced to solve the linear JDSDEs. Exact solutions to some linearizable JDSDEs have been provided. Keywords Exact linearization . Jump-diffusion stochastic differential equations . Stochastic integrating factors

1 Introduction One-dimensional, nonlinear Itˆo SODEs (Stochastic Ordinary Differential Equation) arise in science, finance, and engineering [1, 3, 7, 8]. However, SODEs with jump terms (or driven by Poisson processes) appear to be more realistic in cases where rare events

G. Unal () Faculty of Commerce and Faculty of Science and Letters, Yeditepe University, Istanbul, Turkey e-mail: [email protected] A. Sanver Department of Mathematics, Koc University, Istanbul, Turkey I. Iyigunler Department of Computational Sciences and Engineering, Koc University, Istanbul, Turkey

play prominent role [4, 5]. Analytical (explicit) solutions of Itˆo SDEs not only allow us to study the underlying stochastic processes but also provide means to test the numerical schemes [2]. Linearizable onedimensional Itˆo SODEs may also be useful in the study of two-dimensional problems [9]. Therefore, analytical methods are of paramount importance. Here, we propose an analytical method that is based on exact linearization. We consider one-dimensional, nonlinear SODE with jump terms of the form d x = f (x) dt + g(x) d W (t) + r (x) d P(t), x(0) = x0 ,

(1)

where d W (t) is the infinitesimal increment of the Wiener process [8] and d P(t) is the infinitesimal increment of the Poisson process [5]. We assume that f (x) and g(x) satisfy the Ikeda–Watanabe conditions for the existence and the uniquenes of the solutions. We provide neccessary and sufficient conditions for Itˆo SDE given in (1) to be linearizable via an invertible transformation; hence, extending the Gard’s theorem to SODEs with jump terms. Furthermore, we introduce the stochastic integrating factors to solve the linear SODE with jump terms. It has been shown that mean-reverting Ornstein–Uhlenbeck, Cox–Ingersoll– Ross population growth models in noisy environment are linearizable. Springer

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2 Exact linearization

Substituting Equation (9) in Equation (5) we obtain 

We seek transformations of the form dh = 0 (at least locally) dx

y = h(x),

b2

+ (c1 y + c2 ) d P(t),

 (3)

where a1 , a2 , b1 , b2 , c1 , c2 are constants. Itˆo Lemma [4, 5] for h(x) yields,



(4) From Equations (3) and (4), we obtain (5) (6)

(7)

Ordinary differential Equation (6) has two distinct solutions for (i) b1 = 0 and (ii) b1 = 0. We now consider each case separately. (i) When b1 = 0, solution to Equation (6) reads 

dx . g(x)

(8)

Differentiating Equation (8), h  (x) = Springer

b2 , g(x)

h  (x) = −b2

g  (x) . g 2 (x)

= α 1 b2 .

(11)

  

f (x) 1  − g (x) g(x) 2

= 0.

(12)

h(x + r (x)) = h(x) (c1 + 1) + c2 .

(13)

From Equation (7), we have

Differentiating Equation (13) yields,

and h(x + r (x)) − h(x) = c1 h(x) + c2 .



f (x) 1  − g (x) g(x) 2

 g(x)

+ g(x)h  (x) d W + [h(x + r (x)) − h(x)] d P.

1 f (x)h  (x) + g 2 (x)h  (x) = a1 h(x) + a2 , 2 g(x)h  (x) = b1 h(x) + b2 ,

g(x)

Differentiating Equation (11) leads to,

 1 f (x)h  (x) + g 2 (x)h  (x) dt 2

h(x) = b2

= a1 h(x) + a2 ,

where α1 and α2 are constants. Differentiating Equation (10) and arranging yields,

dy = (a1 y + a2 ) dt + (b1 y + b2 ) d W (t)

dy =



f (x) 1  − g (x) = α1 h(x) + α2 , (10) g(x) 2

(2)

to transform Equation (1) into



f (x) 1 2 g  (x) − g (x) 2 g(x) 2 g (x)

(9)

h  (x + r (x))(1 + r  (x)) = h  (x) (c1 + 1) .

(14)

Substituting the results in Equation (9) into Equation (14) gives, b2 b2 (c1 + 1) , (1 + r  (x)) = g(x + r (x)) g(x) g(x) (1 + r  (x)) = c1 + 1. g(x + r (x))

(15)

Differentiating Equation (15), 



g(x) (1 + r  (x)) g(x + r (x))

= 0.

(16)

Hence, the linearization criterions (12) and (16) must be satisfied for Equation (1) to be linearizable via transformation (8). (ii) When b1 = 0 in Equation (6) without loss of generality the solution reads, h(x) = eb1



dx g(x)

.

(17)

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Differentiating Equation (17) yields, 

dx b1 b1  g(x) , h  (x) = e g(x)   dx b1 2 b1  g(x) + e . g(x)

h  (x) =

b1 g(x)

Substituting Equation (18) into Equation (24) gives, 

eb1



dx g(x)

(18)

b2 f (x) b1  b1 − g (x) + 1 g(x) 2 2

 e b1



b1 b1  e g(x)

dx g(x)

(1 + r  (x))

(c1 + 1) , dx

eb1 g(x+r (x)) g(x)  dx (1 + r  (x)) = c1 + 1. (25) g(x + r (x)) eb1 g(x) Differentiating Equation (25) and arranging leads to,

dx g(x)

= a1 h(x) + a2 = γ1 h(x) + γ2 .

(19)

Differentiating Equation (19) and arranging leads to,   f (x) 1  g(x) − g (x) g(x) 2   f (x) 1  1 + b1 (20) − g (x) + b1 = γ1 . g(x) 2 2 Differentiating Equation (20) yields, 

=

dx g(x+r (x))



Substituting Equation (18) into Equation (5) gives, 

 b1 e b1 g(x + r (x))



g(x)  (1 + r  (x)) g(x+r (x))  g(x)  1 (1 + r  (x)) g(x+r − (x)) g(x+r (x))

1 g(x)

(26) Differentiating Equation (26) yields, ⎞  g(x)   (1 + r  (x)) g(x+r (x)) g(x + r (x))g(x) ⎟ ⎜  ⎠ = 0. ⎝ g(x) g(x + r (x)) − g(x)  (1 + r (x)) g(x+r (x)) ⎛



  f (x) 1  − g (x) g(x) 2   f (x) 1  + b1 − g (x) = 0. g(x) 2

= −b1 .

(27)

g(x)

Hence, the linearization criteria (22) and (27) must be satisfied for Equation (1) to be linearizable via transformation (17). Hence, we have just proven the following proposition.

We now determine the coefficient b1 ;  f (x) 1    − 2 g (x) g(x) g(x) b1 = − .   f (x) 1 − 2 g  (x) g(x) 

(21)

Differentiating Equation (21) leads to,



 f (x) 1     g(x) g(x) − 2 g (x) = 0.   f (x) 1 − 2 g  (x) g(x)

(22)

(23)

Differentiating Equation (23) yields, h  (x + r (x))(1 + r  (x)) = h  (x) (c1 + 1) .

Using the linearization criterions, we can determine the drift function f (x) and the jump function r (x). We first consider the drift function f (x). From Equation (5), we have f (x) =

From Equation (7), we have h(x + r (x)) = h(x) (c1 + 1) + c2 .

Proposition 1. Stochastic ordinary differential Equation (1) is linearizable iff conditions (12, 16) or (22, 27) are satisfied.

(24)

a1 h(x) + a2 − 12 g 2 (x)h  (x) . h  (x)

(28)

We first consider the case b1 = 0. Suppose that diffusion and jump coefficient functions are given and they satisfy the equation given in Equation (16). We can determine the drift function such that Equation (1) can be linearized. Substituting the linearizing transformation Springer

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(8) and its derivatives into Equation (28) leads to   f (x) = g(x) a1

Differentiating Equation (33) yields,

 dx a2 1  + + g (x) . g(x) b2 2



a1 f (x) = g(x) b1 + a2 +



 b2 − e−b1 b1

dx g(x)



r (x) + (1 + r (x))  + b1  

We next consider the case b1 = 0. Suppose that diffusion and jump coefficient functions are given and they satisfy the equation given in Equation (27). We can determine the drift function such that Equation (1) can be linearized. Substituting the linearizing transformation (17) and its derivatives into Equation (28) yields 



g(x) − g(x + r (x)) × g(x)g(x + r (x))

 = 0,

where =

g(x) . g(x + r (x))

Letting r  (x) = p(x),

 +1

p  (x) + p(x) = −.



1  g (x) − b1 . 2

(34)

Solution to the first-order linear ODE in Equation (34) gives,

Suppose drift and diffusion functions are given and they satisfy the linearizing conditions (12, 16). The jump function can now be determined such that (1) is linearizable. From Equation (7), we have h(x + r (x)) − h(x) = c1 h(x) + c2 .

(29)

p(x) = e

 − d x

    d x C − e dx

where,  =

 2 g(x)  (x)) +b1 g (x)−g(x)g(x+r g(x+r (x)) g(x)g 2 (x+r (x)) .  g(x) g(x+r (x))

Differentiating Equation (29), Hence, the jump function r (x) is, 





h (x + r (x))(1 + r (x)) = h (x) (c1 + 1) .

(30)

Substituting the linearizing transformation (8) and its derivatives into the Equation (30) leads to, r  (x) =

g(x + r (x)) (c1 + 1) − 1, g(x)

r (x) =

Suppose drift and jump functions are given and they satisfy the linearizing conditions (12, 16). Diffusion function g(x) can now be determined by solving the

(31) g  (x)g 2 (x) + g  (x) f (x) + 2g(x)(a1 − f  (x)) = 0

while its solution reads  

       − d x d x r (x) = e C − e d x d x.

and Equation (16). The case b1 = 0 involves more complicated equations that will not be considered here.



g(x + r (x)) (c1 + 1) − 1 d x. g(x)

(32) 3 Stochastic integrating factors

Substituting the linearizing transformation (17) and its derivatives into Equation (30) gives, 

dx

g(x) eb1 g(x+r (x))  dx (1 + r  (x)) = c1 + 1. g(x + r (x)) eb1 g(x) Springer

(33)

Although there are other methods such as variation of parameters available (see, for instance, [8]) for the integration of one-dimensional Itˆo SDEs, here we will make use of the stochastic integrating factors.

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Definition 1. The function μ = μ(t, W, P) with property

The right-hand side of equation should not involve the variable y to comply with the definition of the integrating factor. This leads to

d(μy) = D1 (t)dt + D2 (t)d W + D3 (t)d P is an integrating factor for the one-dimensional linear JDSODE (3).

a1 (t)μ +

We now consider the chain rule [5] d(μy) = μdy + ydμ + dydμ.

(35)

∂μ 1 ∂ 2 μ ∂μ + + b1 (t) = 0, 2 ∂t 2 ∂W ∂W ∂μ b1 (t)μ + = 0 and ∂W (1 + c1 (t))μ(P + 1) = μ(P).

(39) (40) (41)

We now solve Equations (39) and (41). To achieve this goal, we let

Here dμ is [4]:  dμ =

∂μ 1 ∂ 2 μ + ∂t 2 ∂W2



μ(t, W, P) = M1 (t, W )M2 (t, P).

∂μ dt + dW ∂W

+ (μ(t, W, P + 1) − μ(t, W, P))d P

Equation (41) now reads (36) (1 + c1 (t))M2 (P + 1) = M2 (P).

(42)

and dμdy is dμdy =

∂μ (b1 y + b2 )dt + (μ(P + 1) ∂W − μ(P))(c1 y + c2 ) d P,

We now let M2 = α P in Equation (42) to obtain a solution of the form (37) α=

where μ(t, W, P) = μ(P) and μ(t, W, P + 1) = μ(P + 1). Following multiplication rules [4] have been made use of: dt d P(t) = 0, d P(t) d W (t) = 0, d P m (t) = d P, dt d W (t) = 0, d W (t) d W (t) = dt and dW m (t) = 0. Equation (35) can now be rewritten as   ∂μ 1 ∂ 2 μ ∂μ d(μy) = a1 (t)μ + + + b1 (t) y ∂t 2 ∂W2 ∂W

 ∂μ + a2 (t)μ + b2 (t) dt ∂W    ∂μ + b1 (t)μ + y + b2 (t)μ d W ∂W

1 , 1 + c1 (t)

which leads to M2 (t, P) = (1 + c1 (t))−Pt . Solution to Equation (40) becomes μ(t, W, P) = (1 + c1 (t))−Pt e−b1 (t)Wt +q(t) .

+ [c2 (t)μ + c2 (t)(μ(P + 1) − μ(P))]}d P. (38)

(44)

Substitution of Equation (44) into Equation (39) leads to dq Pt c˙1 (t) b2 (t) − 1 + a1 (t) − = 0. dt 2 (1 + c1 (t)) Its solution reads

+ {[c1 (t)μ + (μ(P + 1) − μ(P)) +c 1 (t)(μ(P + 1) − μ(P))]y

(43)

1 q(t) = 2





t

b12 (s)ds

0



+ 0

t

t

−

a1 (s)ds 0

Ps c˙1 (s) ds. 1 + c1 (s)

(45) Springer

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Rendering Equation (45) back into Equation (44) yields;   t  t 1 2 μ = (1 + c1 (t)) exp b (s)ds − a1 (s)ds 2 0 1 0   t Ps c˙1 (s) − b1 (t)Wt + ds . (46) 0 1 + c1 (s) −P

Invoking Equation (46) in Equation (38) leads to d(μy) = (a2 (t) − b1 (t)b2 (t))μ(P)dt + b2 (t)μ(P)d W + c2 (t)μ(P + 1)d P. (47) Integrating Equation (47), we obtain the solution 1 y= μ

 

+

t



t

b2 (s)μ(P) d W +

0

   f (x) 1  − g (x) g(x) 2

3  

1 ax 4 + 38 b2 x 2 3 −1 3 = bx 4 − bx 4 = 0, 3 8 bx 4   g(x)  (1 + r (x)) g(x + r (x)) ⎞ ⎛ 3  1 1 ⎠ = 0. = ⎝  1 4 3 x 4 + b x4 +b 4





g(x)

Hence, the transformation,

(a2 (s) − b1 (s)b2 (s))μ(P) ds

0 t

As it is seen later, Equation (49) passes the the linearization criterions (12, 16).

 c2 (s)μ(P +1) d P .

 y = h(x) =

x0

0

(48)

  3 2 1 3 3 d x = ax 4 + b x 2 dt + bx 4 dW 8  3 2 1 + 4x 4 b + 6x 4 b2 + 4x 4 b3 + b4 d P, x(0) = x0 . Table 1 Linearizable equations and solutions

dx bx

=

1 4 1 x 4 − x04 , b

dy =

a dt + d Wt + 4d Pt . b

Integration gives a t + Wt + 4Pt . b

y=

Hence, the solution xt is 

(49) Equation Criteria (12, 16) Equation Criteria (12, 16) Equation Criteria (22, 27) Equation Criteria (12, 16)

4

1 xt = x0 + (at + bWt + 4b Pt ) 4 1 4

1

1

.

d x = α(β − x)dt + σ x 2 dW + (σ x 2 + αβ)d P Solution 1 t α √ −αt  t α xt = (x02 + αβe 2 [ 0 e 2 s dWs + 0 e 2 s d Ps ])2 1 d x = μx(θ − ln x)dt + ρx 2 dW + ζ xd P Solution t t t ) 0 eμs ds + 0 eμs dWs + 0 eμs d Ps )] xt = x0 exp[ρe−μt (( ρ2 + μθ ρ d x = ξ x(η − x)dt + δxdW + λxd P Solution  t 1 (P −P ) [( δ2 +ξ η)(t−s)−δ(W −W )] −1 t s xt = x0 [ξ 0 ( λ+1 ) t se 2 ds] 1 2 1 3 3 23 3 13 1 3 d x = 3 x dt + x dW + ( 2 x + 4 x + 8 )d P Solution 1

xt = (x03 + 13 Wt + 12 Pt )3

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3 4

linearizes Equation (49)

4 Applications We now consider some linearizable nonlinear jump diffusion equations. We will give calculations in detail for the first example. The other examples will be partially discussed. The solutions will be given in Table 1. Our first example appeared in [4] and it reads

x

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The second example appeared in [1] with an additional jump term and it reads

(50)

Using the transformation 1  1  1 y=√ x 2 − x02 , (α = 2 αβ) αβ

(51)

x x0

−1

,

(57)

linearizes Equation (56) into dy = [(δ 2 − ξ η)y + ξ ]dt   λ + δyd W + − yd P. λ+1

(58)

Integrating Equation (58) via integrating factors and using Equation (57) leads to the solution given in Table 1. The final example appeared in [6] and it reads

linearizes Equation (50) into α dy = − ydt + dW + d P. 2

 y=

1

d x = α (β − x) dt + σ x 2 d W  1  + σ x 2 + αβ d P, x(0) = x0 .

Transformation

(52)

Integrating Equation (52) and using Equation (51) leads to the solution given in Table 1. The third example appeared in [8] as mean-reverting Ornstein–Uhlenbeck equation with an additional jump term and it reads

dx =

1 1 2 x 3 dt + x 3 d W 3   3 2 3 1 1 + x3 + x3 + d P, 2 4 8

x(0) = x0 . (59)

Transformation 1  1 y = 3 x 3 − x03 ,

(60)

1

d x = μx (θ − ln x) dt + ρx 2 d W + ζ xd P,

x(0) = x0 .

linearizes Equation (59) into (53)

Transformation 1 x y = ln , ρ x0

(54)

  ρ μθ dy = −μy + + dt 2 ρ 1 ln (1 + ζ ) d P. ρ

(55)

Integrating Equation (55) via integrating factors and using Equation (54) leads to the solution given in Table 1. The fourth example is a population growth model in a noisy environment with a jump term, and it reads d x = ξ x (η − x) dt + δxdW + λxd P,

x(0) = x0 .

Integrating Equation (52) and using Equation (51) leads to the solution given in Table 1. 5 Conclusions

linearizes Equation (53) into

+ dW +

3 dy = d W + d P. 2

(56)

Linearization conditions of the one-dimensional Itˆo jump-diffusion stochastic differential equations (JDSDE) are given. Integrating factor has been introduced to integrate (solve) the linear JDSDE. Several examples that appeared in the literature have been shown to be linearizable. References 1. Cox, J.C., Ingersoll, J.E., Ross, S.A.: A theory of the term structure of interest rates. Econometrica 53(2), 385–408 (1985) 2. Gard, T.C.: Introduction to Stochastic Differential Equations. Marcel Dekker, New York (1988)

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Nonlinear Dyn 3. Gardiner, C.W.: Handbook of Stochastic Methods for Physics, Chemistry, and the Natural Sciences. SpringerVerlag, Berlin, New York (1985) 4. Hanson, F.B.: Applied Stochastic Processes and Control for Jump-Diffusions: Modeling, Analysis and Computation. http://www.math.uic.edu/∼hanson/pub/SIAMbook (completed draft of book), SIAM Books: Advances in Design and Control Series (under contract) (2006) 5. Ikeda, N., Watanabe, S.: Stochastic Differential Equations and Diffusion Processes. North Holland, Amsterdam (1988)

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6. Maghsoodi, Y.: Mean square efficient numerical solution of jump-diffusion stochastic differential equations. Indian J. Stat. Ser. A 58(Pt. 1), 25–47 (1996) 7. Mikosch, T.: Elementary Stochastic Calculus with Finance in View. World Scientific, Singapore (1998) 8. Oksendal, B.: Stochastic Differential Equations: An Introduction with Applications. Springer-Verlag, Berlin (2003) 9. Wafo Soh, C., Mahomed, F.M.: Integration of stochastic ordinary differential equations from a symmetry standpoint. J. Phys. A: Math. Gen. 34(2), 177–192 (2001)