dx(t) = f(x(t), x(t − τ), t)dt + σ(t, x(t))dw(t).
dx(t) = f(x(t), x(t − τ), t)dt.
dx(t) = f(x(t), x(t−τ), t)dt+σ(t)dw(t). (1.1)
dx(t) = f(x(t), x(t−τ), t)dt+σ(t, x(t))dw(t). (1.2)
(Ω, , {
t
}
t0
, P )
{
t
}
t0
|x| x ∈ R
n
A
A = sup{|Ax| : |x| = 1} B
T
1
A = (a
ij
) Trace(A) =

a
ii
τ
C([−τ, 0]; R
d
) R
d
−

σ : R
d
×R
+
→ R
d×m
w ξ ∈ L
2

0
([−τ, 0]; R
d
)
x(t, ξ)
(2.1)
δ K
ξ ∈ L
2

0
([−τ, 0]; R
d
)
E|x(t, ξ)|  Kξ
2
E
e
−δt
, ∀t ≥ 0. (2.2)
δ K

1
− c
3
2x
T
f(x, y, t)  −c
1
|x|
2
+ c
2
|y|
2
,
T race(σ(t, x)σ
T
(t, x))  c
3
|x|
2
,
c
2
e
c
1
τ
+ c
3
< c

s
x
T
(s)σ(s, x(s))dw(s).
N(t) =

t
0
e
c
1
s
(c
1
|x(s)|
2
+ 2x(s)
T
f(x(s), x(s − τ), s) + trace(σ(s, x(s))σ
T
(s, x(s)))ds.
By (i), (ii) we have
e
c
1
t
|x(t)|
2
 |x(0)|
2

0
e
c
1
s
|x(s − τ)|
2
ds +

max {τ,t}
τ
e
c
1
s
|x(s − τ)|
2
ds


τ
0
e
c
1
s
|x(s−τ)|
2
ds+e
c

2
|x(s − τ)|
2
ds +

t
0
(c
3
+ c
2
e
c
1
τ
)e
c
1
s
|x(s)|
2
ds
because EM(t) = 0, moreover we have

τ
0
e
c
1
s

+

τ
0
e
c
1
s
c
2
E|x(s − τ )|
2
ds +

t
0
(c
3
+ c
2
e
c
1
τ
)e
c
1
s
E|x(s)|
2

2
ds. (4.6).
From (4.6) and applying lemma 2.2 with
u(t) = e
c
1
t
E|x(t)|
2
; v(t) = c
3
+ c
2
e
c
1
τ
; N
0
= (1 +
c
2
c
1
(e
c
1
τ
− 1))ξ
2

.
Hence we obtain
E|x(t)|
2
 (1 +
c
2
c
1
(e
c
1
τ
− 1))ξ
2
E
e
(c
3
+c
2
e
c
1
τ
−c
1
)t
By assumptions (iii) we can rewrite
E|x(t)|

dx(t) = f(x(t), x(t − τ), t)dt + σ(t, x(t))dw(t).
dx(t) = f(x(t), x(t − τ), t)dt.