Clearly I is the union of all D
t ,u
, u ∈ {0, 1}
d
, so we can rewrite the left hand side of 23 under the form
P
i∈I
a
i
X
i
. For i ∈ I, put Ki :=
k ∈ {1, . . . , d}; i
k
= u
k
t
k
+ 1 .
25
Then observe that for i ∈ I, the u’s such that i ∈ D
t ,u
are exactly those which satisfy u
k
= 1 for
every k ∈ Ki. Using 21, this gives ∀i ∈ I,
a
i
= X
u∈{ 0,1}
d
,
∀k∈Ki, u
k
=1
wu =
Y
k∈Ki
s
k
. 26
On the other hand we have for every i ∈ I,
|R
n ,i
∩ [0, t ]| =
Y
k∈Ki
t
k
− b
k
u
k
t
k
Y
k ∈Ki
∆b
k
u
k
t
k
+ 1 =
π ∆BUt + 1
Y
k∈Ki
s
k
= a
i
π ∆BUt + 1
27 As |R
n ,i
|
−1
= π∆BUt + 1, 23 follows and the proof is complete.
For proving tightness of the process Ξ
n
it is convenient to get yet another representation of it. For this introduce the notations similar to [11]:
∆
j k
S
n
i = S
n
i
1
, . . . , i
j−1
, k, i
j+1
. . . , i
d
− S
n
i
1
, . . . , i
j−1
, k − 1, i
j+1
. . . , i
d
28 Clearly the operators ∆
j k
’s commute for different j’s. Note that when applied to S
n
i, ∆
j k
is really a difference operator acting on the j-th argument of a function with d arguments. Also since k
defines the differencing, ∆
j k
S
n
i does not depend on i
j
, and the following useful representation holds for
1 ≤ i ≤ k
n
, X
n ,i
= ∆
1 i
1
. . . ∆
d i
d
S
n
i. 29
Proposition 16. The process Ξ
n
admits the representation Ξ
n
t = S
n
Ut +
d
X
l=1
X
1≤i
1
i
2
···i
l
≤d l
Y
k=1
t
i
k
− b
i
k
u
i
k
∆b
i
k
u
i
k
t
i
k
+ 1
l
Y
k=1
∆
i
k
u
ik
t
ik
+1
S
n
Ut . 30
Proof. Recalling the notations 24, 25 and formula 27, we have Ξ
n
t = S
n
Ut +
X
i∈I
|R
n ,i
|
−1
|R
n ,i
∩ [0, t ]|X
n ,i
= S
n
Ut +
X
i∈I
Y
k∈Ki
s
k
X
n ,i
. This can be recast as
Ξ
n
t = S
n
Ut +
d
X
l=1
T
l
t 31
2269
with T
l
t :=
X
i∈I cardKi=l
Y
k∈Ki
s
k
X
n ,i
, 32
here by cardK we denote the cardinality of the set K. The expression 32 can be further recast as T
l
t =
X
K⊂{1,...,d} cardK=l
X
i∈I Ki=K
Y
k∈K
s
k
X
n ,i
= X
K⊂{1,...,d} cardK=l
Y
k∈K
s
k
X
i∈I Ki=K
X
n ,i
.
It should be clear that X
i∈I Ki=K
X
n ,i
= Y
k∈K
∆
k u
k
t
k
+1
S
n
Ut ,
where the symbol Π is intended as the composition product of differences operators. Recalling that s
k
= t
k
− b
k
u
k
t
k
∆b
k
u
k
t
k
+ 1, this leads to T
l
t =
X
K⊂{1,...,d} cardK=l
Y
k∈K
t
k
− b
k
u
k
∆b
k
u
k
t
k
+ 1 Y
k∈K
∆
k u
k
t
k
+1
S
n
Ut .
33
To complete the proof report this expression to the equation 31.
3.3 Rosenthal and Doob inequalities
When applied to our triangular array, Rosenthal inequality for independent non-identically dis- tributed random variables reads
E ¯
¯ ¯
¯ X
1≤ j ≤n
X
n , j
¯ ¯
¯ ¯
q
≤ c X
1≤ j ≤n
σ
2 n
, j
q2
+ X
1≤ j ≤n
E |X
n , j
|
q
, 34
for every q ≥ 2, with a constant c depending on q only. As in [11] we can also extend Doob inequality for independent non-identicaly distributed variables
E max
1≤k≤k
n
|S
n
k|
q
≤ p
p − 1
dq
E |S
n
k
n
|
q
, 35
for q 1.
4 Finite-dimensional distributions
4.1 Proof of the proposition 6
