Section 7
Stopping times, Wald’s identity.
Another proof of SLLN.
Consider a sequence (Xi )i∗1 of independent r.v.s and an integer valued random variable V ⊂ {1, 2, . . .}.
We say that V is independent of the future if {V → n} is independent of ε((Xi )i∗n+1 ). We say that V is a
stopping time (Markov time) if {V → n} ⊂ ε(X1 , . . . , Xn ) for all n. Clearly, stopping time is independent of
the future. An example of stopping time is V = min{k ∗ 1, Sk ∗ 1}.
Suppose that V is independent of the future. We can write
�
�
ESV =
ESV I(V = k) =
ESk I(V = k)
k∗1
=
k∗1
��
(�)
EXn I(V = k) =
��
EXn I(V = k) =
n∗1 k∗n
k∗1 n→k
�
n∗1
EXn I(V ∗ n).
In (*) we can interchange the order of summation if, for example, the double sequence is absolutely summable,
by Tonelli’s theorem. The event
{V ∗ n} = {V → n − 1}c ⊂ ε(X1 , . . . , Xn−1 )
is independent of ε(Xn ) and we get
ESV =
�
n∗1
EXn × P(V ∗ n).
(7.0.1)
This implies the following.
Theorem 13 (Wald’s identity.) If (Xi )i∗1 are i.i.d., E|X1 | < ↓ and EV < ↓, then ESV = EX1 EV.
Proof. By (7.0.1) we have,
ESV =
�
n∗1
EXn P(V ∗ n) = EX1
�
n∗1
P(V ∗ n) = EX1 EV.
The reason we can interchange the order of summation in (*) is because the same holds for |X i | and we can
apply Tonelli’s theorem.
Suppose that random variables (Xi )i∗1 , V are defined on a probability space (Π, F, P). Given a stopping
time V, let us define by εV a ε-algebra of events expressed in terms of a sequence (X1 , . . . , XV ) up to time
V, i.e.
�
�
εV = B ⊂ F : �n ∗ 1, {V → n} ⇐ B ⊂ ε(X1 , . . . , Xn ) .
In other words, if we know that V = n then event B is expressed in terms of X1 , . . . , Xn .
23
Theorem 14 (Markov property) Suppose that (Xi )i∗1 are i.i.d. and V is a stopping time. Then (V, X1 , . . . , XV )
(i.e. εV ) is independent of (XV +1 , XV +2 , . . .) and
dist
(XV +1 , XV +2 , . . .) = (X1 , X2 , . . .).
Proof. Given a set B ⊂ εV , denote
An = {V = n} ⇐ B ⊂ ε(X1 , . . . , Xn ).
Let D ⊂ B ↓ be a set in the cylindrical Borel ε-algebra. We have,
�
�
P(B ⇐{(XV +i )i∗1 ⊂ D}) =
P ({V = n} ⇐ B ⇐ {(XV +i )i∗1 ⊂ D}) =
P (An ⇐ {(Xn+i )i∗1 ⊂ D}) = (↑).
n∗1
n∗1
Since An ⊂ ε(X1 , . . . , Xn ) and {(Xn+i )i∗1 ⊂ D} ⊂ ε(Xn+1 , . . .), these events are independent and
�
�
(↑) =
P(An )P ({(Xn+i )i∗1 ⊂ D}) =
P(An )P ({(Xi )i∗1 ⊂ D}) = P(B)P ({(Xi )i∗1 ⊂ D}) .
n∗1
n∗1
Theorem 15 Suppose that (Xi )i∗1 are i.i.d. such that EX1 > 0. If Z = inf n∗1 Sn then P(Z > −↓) = 1.
(Partial sums can not drift down to −↓ if EX1 > 0.)
Proof. Let us define (see figure 7.1),
(2)
φ1 = min{k ∗ 1, Sk ∗ 1}, Z1 = min Sk , Sk = Sρ1 +k − Sρ1 ,
k→ρ1
�
�
(2)
(2)
(3)
(2)
φ2 = min k ∗ 1, Sk ∗ 1 , Z2 = min Sk , Sk = Sρ2 +k − Sρ(2)
.
2
k→ρ2
By induction,
�
�
(n)
(n)
(n+1)
(n)
φn = min k ∗ 1, Sk ∗ 1 , Zn = min Sk , Sk
= Sρn +k − Sρ(n)
.
n
k→ρn
Z1 , ..., Zn are i.i.d. by Markov property. By construction, Sρ1 +···+ρn−1 ∗ n − 1. Let
1
0
1
0
z1
z2
τ2
τ1
Figure 7.1: A sequence of stopping times.
Z = inf Sk = inf{Z1 , Sρ1 + Z2 , Sρ1 +ρ2 + Z3 , ...}.
k∗1
We have,
{Z → −N } =
�
�
{Sρ1 +...+ρk−1 + Zk → −N } ∧
{k − 1 + Zk → −N }.
k∗1
k∗1
24
Therefore,
P(Z → −N ) →
=
�
k∗1
�
k∗1
P(k − 1 + Zk → −N ) =
�
k∗1
P(Zk → −N − k + 1)
P(Z1 → −N − k + 1) = {j = N + k − 1}
if
�
j∗1
�
j∗N
P(Z1 → −j) →
�
j∗N
N �↓
P(|Z1 | ∗ j) −∃ 0
P(|Z1 | ∗ j) → E|Z1 | < ↓.
We have,
E|Z1 | → E
�
i→ρ1
|Xi |
Wald
= E|X1 |Eφ1 < ↓
if we can show that Eφ1 < ↓, which is left as an exercise (hint: truncate Xi ’s and φ1 and use Wald’s identity).
N �↓
We proved that P(Z → −N ) −∃ 0 which, of course, implies that P(Z > −↓) = 1.
This result gives another proof of the SLLN.
Theorem 16 If (Xi )i∗1 are i.i.d. and EX1 = 0 then
Sn
n
∃ 0 a.s.
Proof. Given π > 0 we define Xiπ = Xi + π so that EX1π = π > 0. By the above result,
�
Z π := inf Skπ where Skπ =
Xiπ = Sk + kπ
n∗1
i→k
satisfies P(Z π > −↓) = 1. For any π > 0,
�
�
�
�
�
�
Sk
P inf
< −2π
= P ≈k ∗ n, Sk + kπ < −kπ = P ≈k ∗ n, Skπ < −kπ
k∗n k
�
�
�
�
�
�
n�↓
→ P ≈k ∗ n, Skπ < −nπ → P ≈k ∗ 1, Skπ < −nπ = P Z π = inf Skπ < −nπ −∃ 0.
k∗1
Take π = 1/2m. For large enough n(m)
P
By Borel-Cantelli,
P
which implies that
�
��
inf
k∗n(m)
Sk
1
<−
k
m
Sk
1
inf
<−
k∗n(m) k
m
�
25
→
1
.
m2
⎪
i.o.
�
= 1.
Sk
P lim inf
∗0
k�↓ k
Similarly, lim sup Skk → 0 with probability 1.
�
�
=0