ASSOCIATED WITH A PIECEWISE CONTINUOUS PROCESS
B. IFTIMIE, I. MOLNAR and C. VÂRSAN
We provide a quasimartingale representation and a stochastic rule of dierentia- tion for test functions belonging to some class of polynomials, which depend on the solution of a stochastic dynamic system dened via a piecewise continuous func- tionλ(t). This is achieved by applying the classical Itô's dierentiation rule on each continuity interval. The quasimartingale representation on each continuity interval is obtained by solving some possibly degenerate elliptic equations.
AMS 2000 Subject Classication: 35B40, 35J70, 60G44, 91B28.
Key words: stochastic rule of dierentiation, piecewise continuous process, dege- nerate elliptic equation.
1. INTRODUCTION
The evolution of a multi-dimensional piecewise continuous process {(z(t), µ(t)); t>0}is analyzed in connection with the stochastic rule of deriva- tion by using polynomial solutions for some (possibly degenerate) elliptic equa- tions. In our setting, the process{z(t)∈Rn;t>0}is generated as a solution of a stochastic dierential system depending on a piecewise constant process {µ(t) ∈ S ⊆ Rd;t > 0}. The set of test functions ϕ ∈ Pp(z;µ) consists of all polynomial functions of degree p in the state variables z = (z1, . . . , zn), whose coecients are continuous bounded functions ofµ∈Rd. It agrees with the elliptic operators Lµdepending on the parameterµ∈Rd which determine the drift part when a pondered functional {exp(γt)ϕ(z(t); µ(t)); t > 0} is considered and the decomposition formula
exp(γt)ϕ(z(t);µ(t)) =ψt(z(·), µ(·)) + Z t
0
exp(γs)[γϕ+Lµ(ϕ)](z(s);µ(s))ds+
+Mt(z(·), µ(·)), t>0
is valid. Here, Lµ:Pp(z;µ)→ Pp(z;µ) is an elliptic operator, {Mt(z(·), µ(·));
t > 0} is a continuous martingale and {ψt(z(·);µ(·)); t > 0} is a piecewise constant process.
REV. ROUMAINE MATH. PURES APPL., 53 (2008), 4, 323338
The above decomposition formula is meaningfull when describing the asymptotic behaviour of the augmented process {exp(γt)kz(t)k2 | t > 0}, provided the nontrivial solution ϕγ∈ P2(z;µ) of the elliptic equation
γϕγ(z;µ) +kyk2+Lµ(ϕγ)(z, µ) = 0, ∀z∈Rn, µ∈Rd, satises the nonsingularity condition
δ(γ)kzk2 6ϕγ(z, µ)6 1 C(γ)kzk2 for some positive constants δ(γ),C(γ).
2. DEFINITIONS, SETTING OF THE PROBLEM AND AUXILIARY RESULTS
LetW(t) be a standard m-dimensional Wiener process over a complete probability space {Ω,F,{Ft}t>0,P}. On the same probability space we have piecewise constant adapted processes{µ(t);t>0}and{y(t);t>0}of dimen- sion d, respectivelyn. Set λ(t) = (y(t), µ(t)),t>0, and
λ(t, ω) =λk(ω) = (yk(ω), µk(ω)), t∈[tk(ω), tk+1(ω)),
where {tk;k >0} is an increasing sequence of positive adapted random vari- ables with t0 = 0, tk → ∞, a.s., as k → ∞ and (yk, µk), k > 0, are mul- tidimensional Ftk-measurable random variables. We assume that the process W(t) and the sequence{(tk, λk);k>1} are independent.
We are next given a nite set of vector elds gi(z;µ) ∈ Rn, linear in z, dened as
(1) gi(z;µ) =ai(µ) +Ai(µ)z, i= 0,1, . . . , m, µ∈Rd, z ∈Rn,
where the (n×n)-matrices Ai(µ) and the n-dimensional vectors ai(µ) are assumed to be continuous and bounded.
For eachx∈Rn dene a piecewise continuous process {z(t, x);t>0} as the unique solution of the dynamical system
(2)
dz(t) =g0(z(t);µ(t)) +
m
X
j=1
gj(z(t);µ(t))dWj(t), t∈[tk, tk+1), z(tk) =z_(tk) +y(tk), for any k>1,
z(0) =x, wherez_(tk) = lim
t↑tkz(t).
By denition, the unique solution of system (2) is a piecewise continuous and {Ft}-adapted process{z(t, x);t>0}such that, at each jump timetk, the jumpz(tk, x)−z_(tk, x) =y(tk) occurs.
Denition 1. Denote by Pp(z;µ) the set consisting of the polynomial functions of degree p in the variables (z1, z2, . . . , zn) = z, whose coecients are continuous bounded functions of µ ∈ Rd; Pp(z) ⊆ Pp(z;µ) consists of polynomials with constant coecients.
Consider the piecewise continuous scalar process (3) Vzγ(t, x) = exp(γt)ϕ(z(t, x);µ(t)) +
Z t 0
exp(γs)f(z(s, x))ds, t>0, whereγ is a constant,f ∈ Pp(z) and ϕ∈ Pp(z;µ).
We provide a stochastic rule of dierentiation involving the piecewise con- tinuous process{Vzγ(t, x);t>0}which contains both a continuous component and a piecewise constant process dened as
(4) Dγz(t, x) =
P
0<tk6t
exp(γtk)ψk(z(·, x)), t>t1,
0, t06t < t1,
whereψk(z(·, x)) =ϕ(z(tk, x);µ(tk))−ϕ(z_(tk, x);µ(tk−1)), for any k>1. Lemma 1. Let f ∈ Pp(z), ϕ ∈ Pp(z;µ) dene the piecewise continu- ous process {Vzγ(t, x);t > 0} as in (3), corresponding to the unique solution {z(t, x);t>0} of system (2). Then Vzγ(t, x) satises the integral equation (5) Vzγ(t, x) =ϕ(z(0, x);µ(0)) +
Z t 0
exp(γs) [γϕ+f+L(ϕ)](z(s, x);µ(s))ds
+
m
X
j=1
Z t 0
exp(γs)h∇zϕ(z(s, x);µ(s)), gj(z(s, x);µ(s))idWj(s) +Dγz(t, x), for t> 0 and x ∈ Rn, where the piecewise constant process {Dzγ(t, x);t >0}
is dened by (4) and the elliptic operator L:Pp(z;µ)→ Pp(z;µ) is given by (6) L(ϕ)(z;µ) =h∇zϕ(z;µ), g0(z;µ)i+ 1
2
m
X
j=1
h∇z∇zϕ(z;µ)gj(z;µ), gj(z;µ)i.
Proof. The stochastic rule of dierentiation from (5) can be obtained by applying the standard rule of stochastic dierentiation associated with the continuous process{Vzγ(t, x);t∈[tk, tk+1)}for eachk. In this respect, rewrite (7) Vzγ(t, x) =Vzγ(tk, x) +Uk(t, z(t, x))−Uk(tk, z(tk, x))
+ Z t
tk
exp(γs)f(z(s, x))ds,
for t ∈ [tk, tk+1), where Uk(t, z) = exp(γt)ϕ(z;µ(tk)). According to the sto- chastic rule of dierentiation, we get
Uk(t, z(t, x))−Uk(tk, z(tk, x)) = Z t
tk
exp(γs) [γϕ+L(ϕ)](z(s, x);µ(tk))ds +
m
X
j=1
Z t tk
exp(γs)h∇zϕ(z(s, x);µ(tk)), gj(z(s, x);µ(tk))idWj(s), for every t∈[tk, tk+1). Using also (7), it follows that
(8) Vzγ(t, x) =Vzγ(tk, x) + Z t
tk
exp(γs) [γϕ+f+L(ϕ)](z(s, x);µk)ds
+
m
X
j=1
Z t tk
exp(γs)h∇zϕ(z(s, x);µk), gj(z(s, x);µk)idWj(s), for t∈[tk, tk+1) where, by denition,µk=µ(tk) and
(9) Vzγ(tk, x) =Uk(tk, z(tk, x)) + Z tk
0
exp(γs)f(z(s, x))ds.
On the other hand, we notice that
Uk(tk, z(tk, x)) =Uk−1(tk, z_(tk, x)) + exp(γtk)ψk(z(·, x)) for k>1. Uk−1(tk, z_(tk, x))is computed as
Uk−1(tk, z_(tk, x)) =Uk−1(tk−1, z(tk−1, x)) +
Z tk
tk−1
exp(γs)[γϕ+L(ϕ)](z(s, x);µk−1)ds
+
m
X
j=1
Z tk
tk−1
exp(γs)h∇zϕ(z(s, x);µk−1), gj(z(s, x);µk−1)idWj(s).
Inserting now the last two formulas into (9), we get (10) Vzγ(tk, x) =Vzγ(tk−1, x) +
Z tk
tk−1
exp(γs)[γϕ+f +L(ϕ)](z(s, x);µk−1)ds
+
m
X
j=1
Z tk
tk−1
exp(γs)h∇zϕ(z(s, x);µk−1), gj(z(s, x);µk−1)idWj(s) + exp(γtk)ψk(z(·, x)).
Now, iterating the last formula, we can rewrite equation (8) as it appears in (5).
Problem A. For a xedf∈ Pp(z), nd a constantγ 6= 0andϕγ ∈ Pp(z;µ), such that the piecewise continuous process {Vzγ(t, x); t>0} dened by (3) is a quasi-martingale, i.e.,
(11) Vzγ(t, x) =ϕγ(z(0, x);µ(0)) +Dγz(t, x) +
m
X
j=1
Z t 0
exp(γs)h∇zϕγ(z(s, x);µ(s)), gj(z(s, x);µ(s))idWj(s).
Problem B. For a xedf∈ Pp(z), nd a constantγ 6= 0andϕγ∈ Pp(z;µ) solution of the elliptic equation
(12) γϕγ(z;µ) +f(z) +L(ϕγ)(z;µ) = 0 for any z∈Rn, µ∈Rd, where the linear operator L:Pp(z;µ)→ Pp(z;µ)is dened by (6).
Remark 1. Notice that each solution(γ, ϕγ) of Problem B is a solution of Problem A. In this respect, the scalar process {Vzγ(t, x); t > 0} dened by (3) is a quasi-martingale expressed as in (11), provided the stochastic rule of dierentiation from Lemma 1 is used for the given solution of Problem B.
Lemma 2. Let f ∈ P2(z) and the vector elds gi(z;µ), i= 0,1, . . . , m, be given as in (1). Let γ <0 be such that
(13) |γ|>4
kg0k+1 2
m
X
j=1
kgjk2
, where kgjk= sup
µ∈Rd
(kaj(µ)k+kAj(µ)k). Then there existsϕγ ∈ P2(z;µ) satis- fying the elliptic equation (12).
In addition,ϕγ is the sum of the convergent series (14) ϕγ(z;µ) = 1
|γ|
∞ X
k=0
Lk|γ|(f)(z;µ)
, (z, µ)∈Rn×Rd, where L|γ|= |γ|1 L.
Proof. Estimates of the polynomialsLk|γ|(f)(z;µ)∈ P2(z;µ),k>1, can be obtained by an induction argument. Write each ϕ∈ P2(z;µ) as
ϕ(z;µ) =C(µ) +hh0(µ), zi+hH(µ)z, zi,
whereH(µ) = (h1(µ), . . . , hn(µ) andC(µ)∈R,hi(µ)∈Rn,i= 0,1, . . . , n are continuous bounded functions with respect toµ. Denotekϕk=kCk+Pn
i=0
khik,
wherekCk= sup
µ∈Rd
|C(µ)|andkhik= sup
µ∈Rd
khi(µ)k, usingkhi(µ)k= Pn j=1
|hji(µ)|, i= 0,1, . . . , n. It is easily seen that
|ϕ(z;µ)|6kϕk(1 +kzk2), ∀(z, µ)∈Rn×Rd, and an elementary computation shows that
(15)
(a) kL|γ|(ϕ)k6 4
|γ|
kg0k2+1 2
n
X
j=1
kgjk2
kϕk,
(b) |L|γ|(ϕ)(z;µ)|6kϕk 4
|γ|
kg0k2+1 2
n
X
j=1
kgjk2
(1 +kzk2) for any (z, µ).
Using an induction argument, we can easily prove that
(16)
Lk|γ|(f)(z;µ) 6
4
|γ|
kg0k2+1 2
n
X
j=1
kgjk2
k
kfk(1 +kzk2) for any (z, µ)∈Rn×Rd and k>0.
In this respect, recall thatLk+1|γ| :P2(z;µ)→ P2(z;µ) can be written as Lk+1|γ| (f)(z, µ) =L|γ| Lk|γ|(f)
(z, µ), k>0, and assume that Lk|γ|(f)∈ P2(z;µ)and
Lk|γ|(f) 6
4
|γ|
kg0k2+ 1 2
n
X
j=1
kgjk2
k
kfk
for some xedk>1. Then, using estimates (15) for ϕ=Lk|γ|(f), we get
Lk+1|γ| (f) =
L|γ|(ϕ) 6 4
|γ|
kg0k2+1 2
n
X
j=1
kgjk2
kϕk
6
4
|γ|
kg0k2+1 2
n
X
j=1
kgjk2
k+1
kfk=ρk+1kfk and
Lk+1|γ| (f)(z, µ)
6ρk+1kfk(1 +kzk2) for any (z, µ), whereρ:= |γ|4
h
kg0k2+12
n
P
j=1
kgjk2i .
This shows that estimate (16) is satised for any k > 0 and assuming that ρ < 1 (see (13)), we easily get that the series in (14) is convergent. In addition, using (16) we obtain that
|ϕγ(z, µ)|6 1
|γ|
∞ X
k=0
ρk
kfk(1 +kzk2), where
1
|γ|
∞
X
k=0
ρk= 1
|γ|
1
1−ρ = 1
|γ| −4
"
kg0k2+1 2
n
P
j=1
kgjk2
#>0.
By denition, ϕγ∈ P2(z;µ)and using (14) we get that ϕγ satises the elliptic equation (12), as the following simple computation shows
L(ϕγ)(z;µ) =L|γ|
"∞ X
k=0
Lk|γ|(f)(z, µ)
#
=
∞
X
k=1
Lk|γ|(f)(z, µ) =−γϕγ(z;µ)−f(z).
The proof is complete.
Remark 2. The functional ϕγ ∈ P2(z;µ) constructed in Lemma 2 is meaningful for the asymptotic analysis of the piecewise continuous process {Vzγ(t, x);t>0} providedϕγ(z;µ) is a quadratic positive form satisfying
δ(γ)kzk2 6ϕγ(z;µ)6 1
C(γ)kzk2, for any µ∈Rd. Here, δ(γ) and C(γ) are positive constants.
This can be achieved by imposing the additional conditions on the vector elds gi(z;µ),i= 0,1, . . . , mgiven in (1)
(17)
(a) gi(z;µ) =Ai(µ)z, i= 0,1, . . . , m,
(b) Aj(µ)Ak(µ) =Ak(µ)Aj(µ) for anyj, k∈ {0,1, . . . , m}, µ∈Rd, where recall that Ai(µ)are continuous and bounded symmetric matrices.
Lemma 3. Assume that the vector eldsgi(z;µ) given by (1) satisfy as- sumptions (17). Let f(z) =kzk2 and let the constant γ full
(18) γ <0, |γ|>4
kg0k+1 2
m
X
j=1
kgjk2
. Then there exists ϕγ ∈ P2(z;µ) satisfying the equations (19)
(a) γϕγ(z;µ) +kzk2+L(ϕγ)(z;µ) = 0, ∀(z, µ)∈Rn×Rd, (b) ϕγ(z;µ) =hRγ(µ)z, zi, δ(γ)kzk2 6ϕγ(z;µ)6 1
C(γ)kzk2,
where
L(ϕγ)(z;µ) =h∇zϕ(z, µ), A0(µ)zi+1 2
m
X
j=1
h∇z∇zϕ(z;µ)Aj(µ)z, Aj(µ)zi,
C(γ), δ(γ) are positive constants, Rγ(µ) =
|γ|In−B(µ)−1
and B(µ) = 2A0(µ) +
m
P
j=1
A2j(µ).
Proof. The assumptions of Lemma 2 are satised. Letϕγ∈ P2(z;µ)be the solution of the elliptic equation (12) given by the series
ϕγ(z;µ) = 1
|γ|
"∞ X
k=0
Lk|γ|(f)(z, µ)
#
, (z, µ)∈Rn×Rd, wheref(z) =kzk2 and the linear operator L|γ|= |γ|1 Lsatises L|γ|(ϕ)(z, µ) = 1
|γ|
h∇zϕ(z, µ), A0(µ)zi+1 2
m
X
j=1
h∇z∇zϕ(z;µ)Aj(µ)z, Aj(µ)zi
. By denition,
L|γ|(f)(z, µ) = 1
|γ|
*
2A0(µ) +
m
X
j=1
A2j(µ)
z, z +
= 1
|γ|hB(µ)z, zi.
It is obvious that the symmetric matrix B(µ) commutes with each Ai(µ), i= 0,1, . . . , m. Therefore,
Lk|γ|(f)(z, µ) =
* B(µ)
|γ|
k
z, z +
. Here, the matrix B(µ) satises
kB(µ)k62kA0(µ)k+
m
X
j=1
kAj(µ)k262kg0k+
m
X
j=1
kgjk2 =ν, for any µ∈Rd. DenekBk= sup
µ∈Rd
kB(µ)k. Using (18), we get
(20) ρ= kBk
|γ| 6 ν
|γ| < 1 2 and the matrix series
(21) Rγ(µ) = 1
|γ|
∞
X
k=0
B(µ)
|γ| k
, µ∈Rd,
is convergent if |γ|>2ν. Hence ϕγ(z;µ) = 1
|γ|
* ∞ X
k=0
B(µ)
|γ|
k
z, z +
=hRγ(µ)z, zi, provided |γ|>2ν. Recall that the symmetric matrixRγ(µ) fulls
(22) Rγ(µ) = 1
|γ|
In−B(µ)
|γ|
−1
.
Conclusion (19) is a direct application of the following argument. Let z = T(µ)y be an orthogonal mapping, i.e. T∗(µ) = [T(µ)]−1, such that
T−1(µ)B(µ)T(µ) =D(µ) =diag(γ1(µ), . . . , γn(µ)) and write ϕγ(T(µ)y;µ) =hT−1(µ)Rγ(µ)T(µ)y, yi as
ϕγ(T(µ)y;µ) = 1
|γ|
* ∞ X
k=0
D(µ)
|γ| k
y, y +
=hQγ(µ)y, yi, y∈Rn, whereQγ(µ) =diag
(|γ| −γ1(µ))−1, . . . ,(|γ| −γn(µ))−1
. UsingB(µ)ek(µ) = γk(µ)ek(µ), k= 1, . . . , n, where T(µ) =col[e1(µ), . . . , en(µ)], it is easily seen that kB(µ)k > max
16k6n|γk(µ)|=|bγ(µ)|and |bγ(µ)|6kBk for any µ∈Rd. The hypothesis|γ|>2ν(see (18)) implies that the matrixQγ(µ)is positive dened, uniformly with respect to µ, and
δ(γ)kyk26ϕγ(T(µ)y;µ)6 1
C(γ)kyk2,
with some positive constants δ(γ), C(γ). Take now y = T−1(µ)z. Using kyk2 =kzk2, we obtain
1
C(γ)kzk2>ϕγ(z;µ) =hRγz, zi>δ(γ)kzk2, ∀z∈Rn.
Remark 3. The stochastic rule of dierentiation from Lemma 1 and the result stated in Lemma 3 are relevant for the piecewise continuous process {z(t, x);t>0}, dened as the unique solution of system (2). In this respect, we use the decomposition
(23) z(t, x) =zc(t, x) +y(t), t>0,
where the continuous component {zc(t, x);t>0}is the unique solution of the linear stochastic system
(24)
dzc(t) =A0(µ(t))zc(t)dt+
m
X
j=1
A0(µ(t))zc(t)dWj(t), t>0, zc(0) =x.
The stochastic rule of dierentiation written for zc(t, x)reads as (25) Vzγc(t, x) =ϕγ(x;µ(0)) +Dγzc(t, x)
+ Z t
0
exp(γs)[γϕγ+f +L(ϕγ)](zc(s, x);µ(s))ds +
m
X
j=1
Z t 0
exp(γs)h∇zϕγ(zc(s, x);µ(s)), Aj(s, x);µ(s))zc(s, x)idWj(s), where
Dzγc(t, x) = X
0<tk6t
exp(γtk)ψk(zc(·, x)) and
ψk(zc(·, x)) =ϕγ(zc(tk, x);µ(tk))−ϕγ(zc(tk, x);µ(tk−1)), k>1.
Lemma 4. Let the hypotheses of Lemma 3 hold and denote Uγ(t, x) = exp(γt)ϕγ(zc(t, x);µ(t)), t>0, x∈Rn,
forf(z) =kzk2 andϕγ(z;µ) =hRγz, zi. Then the piecewise continuous process Uγ(t, x) satises the integral equation
(26) Uγ(t, x) =ϕγ(zc(0, x);µ(0)) +Dγzc(t, x) +
Z t 0
h−C(γ)Uγ(s, x) + exp(γs)βγ(zc(s, x);µ(s))i ds +
m
X
j=1
Z t 0
exp(γs)h∇zϕγ(zc(s, x);µ(s)), Aj(µ(s))zc(s, x)idWj(s), where
βγ(z;µ) =C(γ)ϕγ(z;µ)− kzk2 60, ∀(z, µ)∈Rn×Rd, Dγzc(t, x) = X
0<tk6t
exp(γtk)
ϕγ(zc(tk, x);µ(tk))−ϕγ(zc(tk, x);µ(tk−1)) . Proof. By hypothesis, the stochastic rule of dierentiation from Lemma 1 is fullled for the continuous process {zc(t, x);t > 0}. If we apply it for {Vzγc(t, x)}, we get equation (25). Using the elliptic equation
γϕγ(z;µ) +kzk2+L(ϕγ)(z;µ) = 0, ∀(z, µ)∈Rn×Rd
and the integral equation (25), we obtain that the piecewise continuous process {Uγ(t, x);t>0} satises
(27) Uγ(t, x) =ϕγ(x;µ(0)) +Dγzc(t, x)− Z t
0
exp(γs)kzc(s, x)k2ds
+
m
X
j=1
Z t 0
exp(γs)h∇zϕγ(zc(s, x);µ(s)), Aj(µ(s))zc(s, x)idWj(s).
Conclusion (19) of Lemma 3 shows that the equation
−kzk2=−C(γ)ϕγ(z;µ) +βγ(z;µ), is satised, where βγ(z;µ)60 for any(z, µ)∈Rn×Rd.
Rewrite now (27) using the last formula. We thus get conclusion (26).
Remark 4. Assuming that the setS ⊆Rdis chosen such that Rγ(µ(t)) =Rγ(µ(0)), for anyt=tk, k>0 (seeµ(tk)∈S),
the piecewise constant process {Dzγc(t, x); t > 0} vanishes and the integral equation (26) can be written accordingly. In addition, the same equation reduces to
(28) Uγ(t, x) =ϕγ(x;µ(0))− Z t
0
C(γ)Uγ(s, x)ds
+ 2
m
X
j=1
Z t 0
Uγ(s, x)θj(s)dWj(s) + Z t
0
exp(γs)βγ(zc(s, x);µ(s))ds, where θj(µ) : S → R,j = 1, . . . , m, are given such that Aj(µ) = θj(µ)In and θbj(t) =µj(µ(t)).
If it is the case, then dene an exponential martingale (29) ξ(t) = exp
2
m
X
j=1
Z t 0
θbj(s)dWj(s)− Z t
0
(θbj(s))2ds
, t>0, which satises the equation
(30)
dξ(t) = 2ξ(t)
m
X
j=1
bθj(t)dWj(t), t>0;
ξ(0) = 1.
Lemma 5. Let hypotheses (17), (18) and (19) hold. In addition, assume that Aj(µ) = θj(µ)In, j = 1, . . . , m, and that the matrix B(µ) = 2A0(µ) +
m
P
j=1
A2j(µ) is constant. Denote hγ(t, x) = exp(γt)kzc(t, x)k2, t>0. Then (31) 06Uγ(t, x)6ϕγ(x;µ(0)) exp(−C(γ)t)ξ(t),
(32) 06hγ(t, x)6 ϕγ(x;µ(0))
δ(γ) exp(−C(γ)t)ξ(t), for any t>0, x∈Rn.
Proof. By hypothesis, the conclusions of Lemma 4 hold. Using Remark 2 we rewrite the integral equation (26) as in (28). Set Φγ(t) = exp(−C(γ)t)ξ(t), t>0. Represent the solution Uγ(t, x)of equation (26) as
(33) Uγ(t, x) = Φγ(t)
ϕγ(x;µ(0))+
Z t 0
Φ−1γ (s) exp(γs)βγ(zc(s, x);µ(s))ds
. Recall that βγ(z;µ) 6 0 for any (z, µ) ∈ Rn×S. Thus, (31) follows. Notice that hγ(t, x)6 δ(γ)1 Uγ(t, x). We easily get (32).
Remark 5. A relaxation of the assumptions made consists of admitting a common eigen-subspace generated by the constant vectors {q1, . . . , qd} ⊆Rn, d6n. In this respect, real symmetric(d×d)-matrices Bi(µ)such that
Q∗Ai(µ)z=Bi(µ)y, i= 0,1, . . . , m, µ∈S, z∈Rn,
will exist, where Q = [q1, . . . , qd], y = col(q∗1z, . . . , qd∗z). The original vector elds gi(z, µ) =ai(µ) +Ai(µ)z will be replaced by
hi(y, µ) =bi(µ) +Bi(µ)y, bi(µ) =Q∗ai(µ),
wherebi(µ)∈Rdand the(d×d)-matricesBi(µ) are continuous and bounded.
For each x ∈ Rn, dene a piecewise continuous process {z(t, x);b t > 0}
as the unique solution of the dynamic system (34)
dbz(t) =h0(z(t);b µ(t))dt+
m
X
j=1
hj(z(t);b µ(t))dWj(t), t∈[tk, tk+1), bz(tk) =zb_(tk) +Q∗µ(tk) for anyk>1, bz(0) =Q∗[x+µ(0)].
Remark 6. The analysis contained in Lemmas 15 has an adequate ver- sion for the piecewise continuous process {bz(t, x); t > 0} and, in particular, the decomposition
z(t, x) =b bzc(t, x) +Q∗µ(t)
can be used. Here, the continuous argument bzc(t, x) stands for the unique solution of the linear stochastic system
(35)
dzbc(t) =h0(zbc(t);µ(t))dt+
m
X
j=1
hj(bzc(t);µ(t))dWj(t), t>0, zbc(0) =Q∗x.
Problem C. Prove that under suitable conditions imposed on the matrices Bi(µ), we get the asymptotic behaviour of the continuous process {˜z(t, x); t>0}, satisfying
(a)exp(γt)Ek˜z(t, x)k2→0ast→ ∞for eachx∈Rnprovidedγ <0and
|γ|>kB0k+12
m
P
j=1
kBjk2, wherehi(y;µ) =Bi(µ)y and kBik= sup
µ∈S
kBi(µ)k.
(b) For anyγ as above, lim
t↑∞exp(γt)k˜z(t, x)k2 = 0 P-a.s.
3. SOLUTIONS OF PROBLEM C
Assume that there existq1, q2, . . . , qd∈Rn,d6n, and symmetricd×d- matrices Bi(µ) such that
Q∗gi(z;µ) =Bi(µ)y, i∈ {0,1, . . . , m}, µ∈S, z ∈Rn,
where gi(z;µ) =ai(µ) +Ai(µ)z, Q= [q1, q2, . . . , qd] and y=col(q1∗z, q∗2z, . . . , q∗dz). Associate the linear stochastic system
(36)
˜
z(t) =B0(µ(t))˜z(t)t+
m
X
j=1
Bj(µ(t))˜z(t)dWj(t), t>0,
˜
z(0) =Q∗x, x∈Rn.
Moreover, assume the matrices Bi(µ) are bounded, continuous and mutually commuting, i.e.,
(a)Bi(µ)Bj(µ) =Bj(µ)Bi(µ)for any µ∈S; (b)B(µ) = 2B0(µ) +
m
P
j=1
Bj2(µ) is a constant matrix.
Theorem 1. Let z(t)˜ be the unique solution of system (36) and γ < 0 such that |γ|>4
h
kB0k+12Pm
j=1kBjk2i . Then
(37) lim
t↑∞exp(γt)E
k˜z(t, x)k2
= 0, for anyx∈Rn.
Proof. Let ϕγ ∈ P2(z;µ) be the positive dened quadratic functional ϕγ(y;µ) = hRγ(µ)y, yi such that the conclusions in Lemma 3 hold. Here, z∈Rnis replaced by y∈Rd and the equations mentioned are replaced by (38)
(a) γϕγ(y;µ) +kyk2+L(ϕγ)(y;µ) = 0, ∀(y, µ)∈Rn×S, (b) δ(γ)kyk2 6ϕγ(y;µ)6 1
C(γ)kyk2, where
L(ϕγ)(y;µ) =h∇yϕ(y, µ), B0(µ)yi+1 2
m
X
j=1
h∇y∇yϕ(y;µ)Bj(µ)y, Bj(µ)yi, C(γ), δ(γ) are positive constants, Rγ(µ) = [|γ|Id−B(µ)]−1 and B(µ) = 2B0(µ) +Pm
j=1Bj2(µ).
SetUbγ(t, x) = exp(γt)ϕγ(˜z(t, x);µ(t)),t>0,x∈Rn. Then the piecewise continuous process {Ubγ(t, x);t>0} fulls all the necessary conditions needed
to apply Lemma 4. As a consequence, Ubγ(t, x) is a solution of the integral equation
(39) Ubγ(t, x) =ϕγ(˜z(0, x);µ(0)) +Dbyγ(t, x) +
Z t 0
h
−C(γ)Ubγ(s, x) + exp(γs)βγ(˜z(s, x);µ(s)) i
ds +
m
X
j=1
Z t 0
exp(γs)h∇yϕγ(˜z(s, x);µ(s)), Bj(µ(s))˜z(s, x)idWj(s), where
βγ(y;µ) =C(γ)ϕγ(y;µ)− kyk260, ∀(y, µ)∈Rn×S, Dbγy(t, x) = X
0<tk6t
exp(γtk) [ϕγ(˜z(tk, x);µ(tk))−ϕγ(˜z(tk, x);µ(tk−1))]. Notice that the last term in (39) is a martingale on the ltered probability space {Ω,F,{Ft}t>0,P}. Taking now the expectation in equation (39) and setting v(t, x) =EUbγ(t, x), we get
(40) v(t, x) =v(0, x) +D(t, x) + Z t
0
[−C(γ)v(s, x) + exp(γs)β(s, x)] ds, where
D(t, x) = X
0<tk6t
exp(γtk)E{h[Rγ(µ(tk))−Rγ(µ(tk−1))] ˜z(tk, x),z(t˜ k, x)i}
and
β(s, x) =E[βγ(˜z(s, x);µ(s))]b 60 for anys>0, x∈Rn. Notice that
Rγ(µ) = 1
|γ|
∞
X
k=0
B(µ)
|γ|
k
, whereB(µ) = 2B0(µ)+
m
P
j=1
B2j(µ). In addition, sinceB(µ)is a constant matrix, we easily get that Rγ(µ) is also constant for any µ∈S and γ <0. Therefore, the piecewise constant component{D(t, x); t>0}vanishes and by a standard integral representation formula we have
(41) 06v(t, x) = exp[−C(γ)t]
v(0, x) + Z t
0
exp(C(γ)s) exp(γs)β(s, x)ds
6exp(−C(γ)t)v(0, x) for anyt>0, x∈Rn. Notice that
exp(γt)Ekz(t, x)k˜ 2 6 1
δ(γ)v(t, x).
Using (41) we get (37).
Theorem 2. Let the assumptions of Theorem1hold. Moreover, letBj(µ) = θj(µ)Id, j= 1, . . . , m, and assume µj(µ) continuous and bounded.
Dene the exponential martingale (42) ξ(t) = exp
2
m
X
j=1
Z t 0
θbj(s)dWj(s)− Z t
0
(θbj(s))2ds
, t>0, where θbj(t) =θj(µ(t)). Then
exp(γt)k˜z(t, x)k2 6exp[−C(γ)t]ξ(t)
k˜z(0, x)k2 δ(γ)
.
Proof. By hypothesis,Ubγ(t, x)dened in Theorem 1, satises the integral equation (27). Since the matrix Rγ(µ) is constant, we get that Dbγy(t, x) = 0 for any t>0,x∈Rn. Hence, (27) becomes
(43) Ubγ(t, x) =ϕγ(˜z(0, x);µ(0)) +
Z t 0
h
−C(γ)Ubγ(s, x) + exp(γs)βγ(˜z(s, x);µ(s)) i
ds +2
m
X
j=1
Z t 0
Ubγ(s, x)bθj(s)dWj(s).
A standard integral representation formula allows us to write (44) 06Ubγ(t)6exp[−C(γ)t]ξ(t) [ϕγ(˜z(0, x), µ(0))], ∀t>0.
Notice that
exp(γt)k˜z(t, x)k26 1
δ(γ)Ubγ(t, x), t>0, x∈Rn and estimate (44) leads us to the conclusion.
Acknowledgement. The authors were supported by Contract 2-CEx06-11-18/2006.
REFERENCES
[1] R. Cont and P. Tankov, Financial Modelling with Jump Processes. Chapman and Hall, 2004.
[2] I. Karatzas and S. Shreve, Brownian Motion and Stochastic Calculus, 2nd Edition.
Springer Verlag, 1991.
[3] D. Lamberton and B. Lapeyre, Introduction to Stochastic Calculus Applied to Finance.
Chapman and Hall, 1996.
[4] M. Musiela and M. Rutkovski, Martingale Methods in Financial Modelling. Springer Verlag, 1998.
[5] B. Oksendal, Stochastic Dierential Equations. An Introduction with Applications, 5th Edition. Springer, 2000.
Received 12 November 2007 B. Iftimie
Academy of Economic Studies Department of Mathematics
Piaµa Roman , 6 010374 Bucharest, Romania
and
I. Molnar, C. Vârsan Romanian Academy
S. Stoilow Institute of Mathematics P.O. Box 1-764
014700 Bucharest, Romania