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An Extension of Sub-Fractional Brownian Motion

499

S H,K exists. 2) It can be shown by the following theorem which gives
also a decomposition in law of the process. It will be useful in the sequel.
Theorem 1.2. 1) For any H ∈ (0, 1) and K ∈ (0, 1], the function S(., .)
is symmetric and positive definite.
2) We have the following decomposition in law
d

StHK = StH,K + C3 (K)XtKH ,

(3)
where C3 (K) =

q

K
Γ(1−K) ,

H ∈ (0, 1), and K ∈ (0, 1), and S H,K and

the Bm W are independent. S HK is sfBm with parameter HK ∈ (0, 1).
Proof: 1) The case K = 1 corresponds to sfBm with parameter H ∈ (0, 1).
First recall the following (easily verified) identity
Z +∞
K
K
λ =
(1 − e−λx )x−1−K dx,
Γ(1 − K) 0
valid for λ ≥ 0 and K ∈ (0, 1). Then, for any c1 , c2 , . . . , cn ∈ R,
n X
n
X

ci cj S(ti , tj )

i=1 j=1

=

=

K
Γ(1−K)

n X
n
+∞ X

Z
0

K
2Γ(1 − K)

+



H
H
H
H
1
1
ci cj −e−x(ti +tj ) + e−x(ti +tj ) + e−x|ti −tj | x−1−K dx
2
2
i=1 j=1

n X
n
+∞ X

Z
0

K
2Γ(1 − K)

H

ci cj e−x(ti

+tH
j )



H

ex(ti

H
+tH
j −(ti +tj ) )


− 1 x−1−K dx

i=1 j=1
n X
n
+∞ X

Z
0

H

ci cj e−x(ti

+tH
j )



H
H
H
ex(ti +tj −|ti −tj | ) − 1 x−1−K dx.

i=1 j=1

Since the functions tH + sH − (t + s)H and tH + sH − |t − s|H are positive
definite, so are




H
H
H
H
H
H
ex(t +s −(t+s) ) − 1
and
ex(t +s −|t−s| ) − 1 ,
for all x ≥ 0. Therefore the function S(., .) is positive definite.
2) Using the fact that S H,K is a Gaussian process, it suffices to see
that

E[StHK SsHK ] = tHK + sHK − (tH + sH )K + E[StH,K SsH,K ].
In the sequel C and Cp denote constants which will be different even
when they vary from one line to the next.