Now assume that sup n n e x n then sup n n e x n

Now assume that sup
n ∈- N
E [ | X
n
| ] < ∞ ; then sup
n ∈- N
| E [ X
n
] | < ∞ . Since n 7→ E [ X
n
] is increasing,
lim
n →-∞
E [ X
n
] exists and is finite. Fix ε > 0; then there is an N
0
∈ - N such that E [ X
n
] > E [ X
N
] - ε if
n ≤ N . For any K > 0, we calculate that for any n ≤ N ,
E | X
n
| χ
{| X
n
|≥ K }
= E X
n
χ
{ X
n
≥ K }
- E X
n
χ
{ X
n
≤- K }
= E X
n
(
χ
{ X
n
≥ K }
+ χ
{ X
n
> - K }
)
- E [ X
n
]
≤ E X
N
(
χ
{ X
n
≥ K }
+ χ
{ X
n
> - K }
)
- E [ X
N
] + ε/ 2 = E | X
N
| χ
{| X
n
|≥ K }
+ ε/ 2
1
Taken from Revuz and Yor
30

In the last inequality, the first term comes from the submartingale inequality, and the last term comes from
the choice of N . By Markov’s inequality,
sup
n ∈- N
P {| X
n
| ≥ K } ≤ K
- 1
sup
n ∈- N
E [ | X
n
| ] .
Since X
N
is integrable, we have that
lim
K →∞
sup
n ≤ N
E | X
n
| χ
{| X
n
|≥ K }
≤ ε ;
since all of the X
n
’s are integrable, we also have that
lim
K →∞
sup
N<n ≤ 0
E | X
n
| χ
{| X
n
|≥ K }
≤ ε ;
putting these two together, we have that the X
n
’s are uniformly integrable. Hence, the P -a.s. convergence
of X
n
to X
-∞
also holds in L
1
. For any A ∈ G
-∞
and any n , we thus have
E [ X
-∞
χ
A
] = lim
m →-∞
E [ X
m
χ
A
] ≤ E [ X
n
χ
A
] .
The last inequality follows from the submartingale inequality. Thus we get the final claim.
We can use this result to easily prove the Strong Law of Large Numbers
Theorem 0.31 (Strong Law of Large Numbers) . Let { ξ
1
, ξ
2
. . . } be a collection of independent and
identically distributed integrable random variables with common law μ . Then
lim
n →∞
n
- 1
n X
k =1
ξ
k
=
Z
R
xμ ( dx ) ,
this limit being both almost-sure and in L 1 .
Proof. Set
S
n
def
=
n X
k =1
ξ
k
n ∈ N
For each n ∈ N , define
G
- n
def
= σ { S
k
; k ≥ n } .
Set
X
n
def
= E [ ξ
1
| G
n
] . n ∈ - N
Then X is a martingale. Clearly
sup
n ∈- N
E [ | X
n
| ] ≤ E [ | ξ
1
| ] ,
so X
-∞
def
= lim
n →-∞
X
n
exists both in L
1
and P -a.s. Note that for any n ≥ 1 and any 1 ≤ k ≤ n ,
E [ ξ
k
| G
- n
] = E [ ξ
1
| G
- n
];
thus
n X
k =1
X
- n
=
n X
k =1
E [ ξ
1
| G
- n
] =
n X
k =1
E [ ξ
k
| G
- n
] = E [ S
n
| G
- n
] = S
n
so in fact
X
- n
=
1
n
S
n
for all n ≥ 1. Thus S
n
= X
- n
for all n ∈ N . Hence
X
-∞
= lim
n →∞
S
n
n
this limit being both P -a.s. and in L
1
. We thus only need show that X
-∞
=
R
R
xμ ( dx ). Note that for every
k ,
X
-∞
= lim
n →∞
S
n
n
= lim
n →∞
∑
n
j =1
ξ
k + j
n
.
31

so by Kolmogorov’s zero-one law, X
-∞
is almost-surely constant. Hence
X
-∞
= E [ X
-∞
] = lim
n →∞
n
- 1
E [ S
n
] =
Z
R
xμ ( dx )
This completes the proof.
Exercises
(1) Show that for any stopping time τ , F
τ
is indeed a sigma-algebra.
(2) Show that for any stopping time τ , τ is itself F
τ
-measurable.
(3) Show that for any fixed t ∈ I , the mapping τ : Ω 7→ t is a stopping time. Show that F
τ
= F
t
.
(4) Let { τ
1
, τ
2
. . . } be a countable collection of stopping times. Show that sup
n ≥ 1
τ
n
is also a stopping time.
Show that if { τ
1
, τ
2
. . . , τ
n
} is a finite collection of stopping times, then min
1 ≤ k ≤ n
τ
k
is also a stopping
time. stopping times.