Result as to uniformly integrable sequence

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Result as to Uniformly Integrable Sequence
Result Suppose that there exists r >1 and M < 1 such that EjXtXt kj
r < M for all t , k and
that
P 1
j = 1 jhj j < 1 . Define Yt =
P 1
j = 1 hj Xt j . Then fYtYt kg is uniformaly integrable for
any integer k. This implies that a product of uniformly integrable sequence is also uniformly
integrable.
Proof
Firstly the productYtYt k takes the form of
YtYt k =
1X
i= 1
hiXt i
1X
j = 1
hj Xt k j =
1X
i= 1
1X
j = 1
hihj Xt iXt k j
According to the deﬁnition of uniformly integrability, we consider whether or notE[jYtYt kjfjYtYt k j cg]
converges to zero, wherefjYtYt k j cg equals to one if the argument is met and equals to zero otherwise.
Now we have
E
h
jYtYt kj fjYtYt k j cg
i
= E
h
1X
i= 1
hiXt i
1X
j = 1
hj Xt k j
fjYtYt k j cg
i
= E
h
1X
i= 1
1X
j = 1
hihj Xt iXt k j
fjYtYt k j cg
i
E
h 1X
i= 1
1X
j = 1
jhihj jjXt iXt k j j fjYtYt k j cg
i
(1)
Here we can change the operation of expectaion and summation. In order to conduct this changing,
we need to conﬁrm thatE[jXt iXt k j j fjYtYt k j cg] is bounded and fhihj g is absolutely summable,
however, both of which are proved lastly(). Then we can rewrite Eq.(1) as
E
h
jYtYt kj fjYtYt k j cg
i
E
h 1X
i= 1
1X
j = 1
jhihj jjXt iXt k j j fjYtYt k j cg
i
=
1X
i= 1
1X
j = 1
jhihj jE
jXt iXt k j j fjYtYt k j cg
(2)
As to the expectation term, using H¨older’s inequality, we can get
E
jXt iXt k j j fjYtYt k j cg
EjXt iXt k j j
r
1=r
E[fjYtYt k j cg]
r=(r 1)
(r 1)=r
=
EjXt iXt k j j
r
1=r
E[fjYtYt k j