PREVIOUS SECTION: Introduction, Convergences

1.3 Uniform integrability

Let eq080 be a sequence of real-valued r.v.'s.

Definition 1

R.v.'s eq081 are uniformly integrable (UI), if eq082  eq083 and, moreover,

 eq084

We can assume the upper bound eq085 to be monotone and right-continuous.

Lemma 1 The following are equivalent:

(i) eq081 are UI;

(ii) eq049 a function eq087:

(a) eq088 ; eq089 ; eq090 ;

(b) eq091

Note: eq088 is not essential!

Proof.

eq094

eq095 as

For eq098, put

eq099

and, for eq100, put eq101. From eq102 get eq088.

Note: eq103 is an interval, and if eq104 is its left boundary point, then eq105. Therefore,

eq106

.

Remark 1 As a corollary, one can get the following:

if eq107, then eq049 eq028 from Lemma 1 such that eq108.

"If eq049 the first moment" eq109 " eq049something more".


Lemma 2 Assume eq030. Then holds

(1) eq081 are UI eq109 eq107 and eq110;

(2) eq111 ; eq112 ; eq110 eq109 eq081 are UI.

Remark 2. In (2), the condition eq114 may be weakened in a natural way.

Problem No.2. How?

But it cannot be eliminated.

Example.

eq115 eq116 eq117 0
 

eq118

eq118

eq119

eq109 , but eq081 are not UI!

Proof of Lemma 2.

First, note that both statements (1) and (2) are "marginal", i.e. only marginal distributions are involved. So, we can construct a coupling: eq019.

Prove (1).

(a) Assume that eq121: eq122 eq123 (this is a special case of UI).

Then eq124 and, eq022,

eq126

Therefore, eq110.

(b) Assume now that eq127 eq123 and for some eq093.

Since eq019, then eq128 eq129

Then, eq130 and

eq131

(c) Prove: eq107. Indeed,

eq132

(d) eq022, choose eq009: eq134 and eq135.

Then,

eq136

Since eq137 eq083 and eq139, then

eq140 and for any

QDE

Prove (2).

eq141

Use (b) from the proof of (1): for a given eq142,

eq143 eq109

Therefore, eq144:

eq145

Now, eq146 eq147

eq148

Set eq149. Then

eq150

Therefore,

eq151

QDE

NEXT SECTION: Properties of UI


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