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Stochastic processes in discrete time
Stefan Geiss
June 1, 2009
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Contents
1 Introduction 51.1 The gamblers ruin problem . . . . . . . . . . . . . . . . . . . . 51.2 Branching processes . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Some notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2 Sums of independent random variables 112.1 Zero-One laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Convergence of sums . . . . . . . . . . . . . . . . . . . . . . . 20
2.2.1 The Two-Series-Theorem and its applications . . . . . 202.2.2 Proof of the Two-Series-Theorem . . . . . . . . . . . . 27
2.3 The law of iterated logarithm . . . . . . . . . . . . . . . . . . 33
3 Martingales in discrete time 37
3.1 Some prerequisites . . . . . . . . . . . . . . . . . . . . . . . . 373.2 Definition and examples of martingales . . . . . . . . . . . . . 463.3 Some elementary properties of martingales . . . . . . . . . . . 513.4 Stopping times . . . . . . . . . . . . . . . . . . . . . . . . . . 543.5 Doob-decomposition . . . . . . . . . . . . . . . . . . . . . . . 583.6 Optional stopping theorem . . . . . . . . . . . . . . . . . . . . 603.7 Doobs maximal inequalities . . . . . . . . . . . . . . . . . . . 643.8 Uniformly integrable martingales . . . . . . . . . . . . . . . . 673.9 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
3.9.1 A martingale proof forKolmogorovs zero-one law . 743.9.2 The branching process example . . . . . . . . . . . . . 75
3.9.3 The Theorem of Radon-Nikodym. . . . . . . . . . . . 773.9.4 On a theorem ofKakutani . . . . . . . . . . . . . . . 80
3.10 Backward martingales . . . . . . . . . . . . . . . . . . . . . . 85
4 Exercises 914.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914.2 Sums of independent random variables . . . . . . . . . . . . . 914.3 Martingales in discrete time . . . . . . . . . . . . . . . . . . . 93
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4 CONTENTS
Important remark for the reading: These notes are only lecturenotes which combines material from different sources, for example from thebeautiful books [6] and [7]. Consequently, these lecture notes cannot replacethe careful study of the literature as the mentioned books of Williams [7]and Shirjaev [6], and of Neveu [5]. Furthermore, there is a lecture notes ofHitczenko [3] which concerns the Central Limit Theorem for martingales nottouched in these notes.
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Chapter 1
Introduction
In the Introduction we want to motivate by two examples the main two partsof the lecture which deal with
sums of independent random variables and the martingale theory.
The examples are given at this stage in an intuitive way without being rig-orous. In the course we will come back to the examples and treat them in arigorous way.
Throughout this lecture we assume that the students are familiar withthe basic probability theory. In particular, we assume that the material ofthe lecture notes [1] is known and will refer from time to time to these notes.
1.1 The gamblers ruin problem
We assume that there is a gambler starting with x0 Euro as his startingcapital at time n= 0. The game is as follows: we fix some p(0, 1) as ourprobability of lossand at each time-step n = 1, 2,...the gambler
loses with probability pthe amount of 1 Euro,
wins with probability 1 pthe amount of 1 Euro.To realize this we can assume a coin which gives by flipping
tails with probability p, heads with probability 1 p,
so that the gambler loses or wins if tails or heads, respectively, occurs. Ofcourse, ifp= 1/2, then the coin is not fair.
Iffn is the amount of capital of the gambler at time n, then we obtain a(random) sequence (fn)n=1 with f0 = x0 such that
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6 CHAPTER 1. INTRODUCTION
f1 = x0 + 1
f1 = x0 1
f2 = x0 + 2
f2 = x0
f2 = x0 2
f3 = x0 + 3
f3 = x0 + 1
f3 = x0 1
f3 = x0 3
f0 =x0
To make our example more realistic, one would need to have at least alower bound forfnto bound the losses. Here we take two bounds < A 0, q0 (0, 1), andMn := fnn . Let(Fn)n=0 be the filtration generated by(fn)n=0. Then one has the following:
(i) (Mn)n=0 is a martingale with respect to the filtration (Fn)n=0, thatmeansIE(Mn+1|Fn) =Mn a.s.,
(ii) M:= limn Mn exists almost surely.
(iii) If1, thenM = 0 a.s. and limnIP(Mn= 0) = 1.(iv) If >1, thenIEM = 1 and limnIP(Mn= 0)(0, 1).
The interpretation of is the expected number of sons in each generationof a single family. Then, n can be seen as some kind of average number of
families with name Hakkinen in the n-th generation. So fn/
n
should bethe right quantity which stabilizes after some time.
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10 CHAPTER 1. INTRODUCTION
1.3 Some notation
The symbol IN stands for the natural numbers
{1, 2,...
}. For a sequence of
real numbers (n)n=1 we recall that
lim supn
n := limn
supkn
k,
lim infn
n := limn
infkn
k.
Ifa, b IR, then we let a b := min {a, b} and a+ := max {a, 0}. Besidesthis notation we will need
Lemma 1.3.1 [Lemma of Borel-Cantelli] 1 2 Let(, F, IP)be a prob-ability space andA1, A2,... F. Then one has the following:(1) If
n=1IP(An)
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Chapter 2
Sums of independent randomvariables
First let us recall the independence of random variables.
Definition 2.0.1 Let(, F, IP) be a probability space andf: IR with I be a family of random variables. The family is called independentprovided that for all pairwise different1,...,nIand allB1,...,Bn B(IR)one has that
IP(f1
B1,...,fn
Bn) = IP(f1
B1)
IP(fn
Bn).
2.1 Zero-One laws
It is known that
n=11n
= but thatn=1(1)n1n does converge. Whatis going to happen if we take a random sequence of signs?
Definition 2.1.1 For p (0, 1) we denote by (p)1 , (p)2 ,... : IR a se-quence of independent random variables such that
IP (p)n =1 =p and IP (p)n = 1 = 1 p.If p = 1/2 we write n =
(1/2)n and call this sequence a sequence of
Bernoulli 1 random variables.
Now we are interested in
IP
n=1
nn
converges
.
1
Jacob Bernoulli, 27/12/1654 (Basel, Switzerland)- 16/08/1705 (Basel, Switzerland),Swiss mathematician.
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12 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
Remark 2.1.2 Let 1, 2,... : IR be random variables over (, F, IP).Then
A:= : n=1
n() converges F.This is not difficult to verify because A if and only if
N=1,2,...
n0=1,2,...
m>nn0
:
m
k=n+1
k()
< 1N
.
What is a typical property of the set A={ : n=1 n() converges}?The condition does not depend on the first realizations 1(),...,N() sincethe convergence ofn=1 n() andn=N+1 n() are equivalent so that
A=
:
n=N+1
n() converges
.
We shall formulate this in a more abstract way. For this we need
Definition 2.1.3 Let (, F, IP) be a probability space and : IR bea family of random variables. Then( : I) is the smallest-algebrawhich contains all sets of form
{
:
()
B}
where
I and B B
(IR).
Exercise 2.1.4 Show that (:I) is the smallest -algebra on suchthat all random variables : IR are measurable.Example 2.1.5 For random variables 1, 2,...: IR and
A=
:
n=1
n() converges
one has that A
N=1 (N, N+1, N+2,...).
The above example leads us straight to the famous Zero-One law ofKol-mogorov 2:
Proposition 2.1.6 (Zero-One law ofKolmogorov) Assume inde-pendent random variables 1, 2,... : IR on some probability space(, F, IP) and let
FN :=(N, N+1,...) and F :=
Fn .
ThenIP(A) {0, 1} for allA F.2
Andrey Nikolaevich Kolmogorov 25/04/1903 (Tambov, Russia) - 20/10/1987 (Mos-cow, Russia), one of the founders of modern probability theory, Wolf prize 1980.
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2.1. ZERO-ONE LAWS 13
For the proof the following lemma is needed:
Lemma 2.1.7 Let
A Fbe an algebra and let(
A) =
F. Then, for all
>0 andB F, there is anA Asuch thatIP(AB)< .
Proof of Proposition 2.1.6. The idea of the proof is to show that IP(A) =IP(A)2. Define the algebra
A:=
n=1
(1,...,n).
We have thatF (A). Hence Lemma 2.1.7 implies that for A Fthere are An(1,...,Nn) such that
IP(AAn)n 0.We get also that
IP(An A)n IP(A) and IP(An)nIP(A).The first relation can be seen as follows: since
IP(An A) + IP(AnA) = IP(An A)IP(A)we get that
lim infn
IP(An A)IP(A)lim sup IP(An A).
The second relation can be also checked easily. But now we get, by indepen-dence, that
IP(A) = limn
IP(A An) = limn
IP(A)IP(An) = IP(A)2
so that IP(A) {0, 1}. Corollary 2.1.8 Let 1, 2,... : IR be independent random variablesover(, F, IP). Then
IP
n=1
nconverges
{0, 1} .
We also obtain an early version of the law of the iterated logarithm (LIL).
But for this we need the central limit theorem (CLT) for the Bernoullirandom variables, which we will not prove here.
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14 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
Proposition 2.1.9 (CLT) Let 1, 2,... : IR be independent Ber-noulli random variables andcIR. Then one has that
limn
IP1+ + nn
c = c
ex22 dx
2.
As a second application ofKolmogorovs Zero-One law we get
Corollary 2.1.10 Let 1, 2,... : IR be independentBernoulli ran-dom variables. Then one has that
IP
lim sup
n
fnn
=, lim infn
fnn
=
= 1
wherefn:=1+ + n.
Proof. The assertion is equivalent to
IP
lim sup
n
fnn
=
= 1 and IP
lim inf
n
fnn
=
= 1.
By symmetry it is sufficient to prove the first equality only. Letting
Ac:= lim sup
n
fnnc
for c0 we get that
IP
lim sup
n
fnn
=
= IP
m=1
Am
= lim
mIP
Am
because ofAmAm+1. Hence we have to prove that IP
Am
= 1. Since
lim supn
1() + + n()n
= lim s upn,nN
1() + + N1()n
+N() + + n()n
= lim s up
n,nN
N() + + n()n
because of
limn,nN
1() + + N1()n
= 0,
where we assume that N2, we get that
Am
N=1 (N, N+1,...) .
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2.1. ZERO-ONE LAWS 15
Consequently, in order to prove IP
Am
= 1 we are in a position to ap-plyKolmogorovs Zero-One law (Proposition 2.1.6) and the only thing toverify is
IP Am >0.To get this, we first observe that
Am=
lim sup
n
fnnm
n=1
k=n
fk
km
since n=1 k=n fkk m implies that for all n = 1, 2,... there is ak n such that fk()/
k m, which gives a subsequence (kl)l=1 with
fkl
()/
kl
m and finally lim supn
fn()/
n
m. Applying Fatou 3slemma we derive from that
IP(Am)IP
n=1
k=n
fk
km
lim sup
nIP
fn
nm
.
Since the central limit theorem (Proposition 2.1.9) gives
limn
IP
fn
nm
=
m
ex2
2dx
2>0
we are done. Now let us come back to question (Q6) of the ruin problem in Section 1.1.
Let p, q (0, 1), p+ q= 1, and fn :=n
i=1 (p)i where n = 1, 2,... and the
random variables(p)i were defined in Definition 2.1.1. Consider the event
A:={ : # {n: fn() = 0}=} .
In words, A is the event, that the path of the process (fn)n=1 reaches 0 in-
finitely many often. We would like to have a Zero-One law for this event.However, we are not able to apply Kolmogorovs Zero-One law (Proposi-
tion 2.1.6). But what is the typical property ofA? We can rearrangefinitelymany elements of the sum
ni=1
(p)i and we will get the same event since
from some Non, this rearrangement does not influence the sum anymore.To give a formal definition of this property we need
Definition 2.1.11 A map : ININ is calledfinite permutation if(i) the map is a bijection,
(ii) there is someN IN such that(n) =n for allnN.3
Pierre Joseph Louis Fatou, 28/02/1878 (Lorient, France) - 10/08/1929 (Pornichet,France), French mathematician, dynamical systems, Mandelbrot set.
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16 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
Moreover, we recall that B(IRIN) is the smallest -algebra on IRIN ={(1, 2,...) :nIR} which contains all cylinder sets Zof form
Z :={(1, 2,...) :an < n< bn, n= 1, 2,...}for some< an < bn ) = IP(l> )
for all k, lIN and IR.
Proposition 2.1.14 (Zero-One law ofHewitt and Savage) 4 As-
sume a sequence of independent and identically distributed random variables1, 2,...: IR over some probability space(, F, IP). If the eventA Fis symmetric, thenIP(A) {0, 1}.
Proof. (a) We are going to use the permutations
n(k) :=
n + k : 1knk n : n + 1k2n
k : k >2n.
4
Leonard Jimmie Savage 20/11/1917 (Detroit, USA) - 1/11/1971 (New Haven, USA),American mathematician and statistician.
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2.1. ZERO-ONE LAWS 17
Now we approximate the set A. By Lemma 2.1.7 we find Bn B(IRn) suchthat
IP (AnA)n0 for An:={: (1(), . . . , n())Bn} .(b) Our goal is to show that
IP (An)IP(A),IP (n(An))IP(A),
IP (An n(An))IP(A),as n , where
n(An) := : n(k)()n
k=1
Bn= {: (n+1(), . . . , 2n())Bn} .
(c) Assuming for a moment that (b) is proved we derive
IP (An n(An)) = IP(An)IP(n(An))sinceAnis a condition on1,...,n,n(An) is a condition on n+1,...,2n, and1,...,2n are independent. By n this equality turns into
IP(A) = IP(A)IP(A)
so that IP(A)
{0, 1
}.
(d) Now we prove (b). The convergence
IP(An)IP(A)is obvious since IP(AnA)0. This implies
IP(n(An))IP(A)since IP(n(An)) = IP(An) which follows from the fact that
(n+1,...,2n, 1,...,n) and (1,...,n, n+1,...,2n)
have the same law in IR2n as a consequence that we have an identicallydistributed sequence of random variables. Finally
IP(AAn) = IP (n(AAn))
= IP (n(A)n(An))
= IP (An(An))
where the first equality follows because the random variables (1, 2,...) havethe same distribution and the last equality is a consequence of the symmetryofA. Hence
IP (AAn)n0 and IP (An(An))n0
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18 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
which implies
IP (A (An
n(An)))
n0 and IP (An
n(An))
IP(A).
As an application we consider
Proposition 2.1.15 Let1, 2,...: IRbe independentBernoulliran-dom variables andfn =
ni=1 i, n= 1, 2,... Then
IP(# {n: fn= 0}=) = 1
Proof. Consider the sets
A+ := { : # {n: fn() = 0} 0} ,A := { : # {n: fn() = 0}=} ,A := { : # {n: fn() = 0}
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2.1. ZERO-ONE LAWS 19
(b) The set A:={ : n() = 0 for all n= 1, 2, . . .} does not belongtoF but is symmetric, because
A = { : (n())n=1B}=
: (n)()n=1B
for B={(0, 0, 0, . . . )} B(IRIN).(c) The set A := { : n=1|n()|exists and n=1n()< 1} does
not belong toF but is symmetric, becauseA = { : (n())n=1B}
=
: (n)()n=1BifB is the set of all summable sequences with sum strictly less than
one.
We finish by the non-symmetric random walk. As preparation we needStir-ling 5s formula
n! =
2nn
e
ne 12n
for n= 1, 2,...and some (0, 1) depending on n. This gives2n
n
=
(2n)!
(n!)2 =
2(2n)
2ne
2ne 1122n
(
2n)2
ne
2n
e 212n
2
4nn
.
Proposition 2.1.17 Letp= 1/2 andfn=n
i=1 (p)i , n= 1, 2,..., where the
random variables(p)1 ,
(p)2 ,...: IR are given by Definition 2.1.1. ThenIP (# {n: fn = 0}=) = 0.
Proof. Letting Bn:={f2n = 0} we obtain
IP(Bn) =
2n
n
(pq)n (4pq)
n
n
byStirlings formula. Since p=qgives 4pq
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20 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
2.2 Convergence of sums
If 1, 2, . . . are independent Bernoulli random variables, that means
IP (n = 1) = IP (i=1) = 12 , then we know from Section 2.1 that
IP
n=1
nn
converges
{0, 1} .
But, do we get probability zero or one? For this there is a beautiful completeanswer: It consists of the Three-Series-Theorem ofKolmogorovwhich wewill deduce from the Two-Series-Theorem ofKolmogorov. First we formu-late the Two-Series-Theorem, then we give some examples and consequences(including the Three-Series-Theorem), and finally we prove the Two-Series-
Theorem.
Besides the investigation of sums of independent random variables thissection also provides some basic concepts going into the direction of themartingale-theory.
2.2.1 The Two-Series-Theorem and its applications
Proposition 2.2.1 (Two-Series-Theorem ofKolmogorov) Assumea probability space(, F, IP) and random variables1, 2,...: IR whichare assumed to be independent.
(i) If
n=1IEn converges and
n=1IE (n IEn)2 0 is someconstant. Then the following assertions are equivalent:
(a) IP (n=1 n converges) = 1.
(b)n=1IEn converges andn=1IE (n IEn)2
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2.2. CONVERGENCE OF SUMS 21
Proof. Assuming the conditions on (n)n=1 and (n)
n=1 we get, for n :=
n+ nn, that
n=1
IEn=
n=1
[n+ nIEn] =
n=1
n
and n=1
IE[n IEn]2 =
n=1
IE[nn]2 =
n=1
2n
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22 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
converges, we get that supn |n|= supn |n||n(0)| cc : f()0 the following three conditions are satisfied:
(a)
n=1IEcn converges,
(b)
n=1IE (cn IEcn)2 c)
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2.2. CONVERGENCE OF SUMS 23
The lemma ofBorel-Cantelli implies that
IP (
{
: #
{n:
Bn
}=
}) = 0.
This implies that IP (n=1 n converges ) = 1 so thatIP
n=1
n converges
= IP
n=1
[n+ cn] converges
= 1.
(ii) =(iii) is trivial.(i) = (ii) The almost sure convergence ofn=1 n() implies that
limn |n()|= 0 a.s. so that
IPlim supn { :|n()| c} = 0.The Lemma ofBorel-Cantelli (note that the random variables n areindependent) gives that
n=1
IP (|n| c)
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24 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
and
n=1 IP(|n| 1)2
n=1pn 0 such that
(i) n ,
(ii) IEn= 0,
(iii)
n=1IE2n2n
0, n ,and(xn)
n=1IR with limn xn = xIR. Then one has that
1
n(1x1+ + nxn)nx.
Proof. Let >0 and find some n0 such that|xn x|< for nn0. Then,for n > n0,
1nn
i=1 ixi x = 1nn
i=1 ixi 1nn
i=1 ix 1
n
ni=1
i|xi x|
= 1
n
n0i=1
i|xi x| + 1n
ni=n0+1
i|xi x|
1n
n0i=1
i|xi x| + .
6
Otto Toplitz 01/08/1881 (Breslau, Germany) - 15/02/1940 (Jerusalem), worked oninfinite linear and quadratic forms.
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2.2. CONVERGENCE OF SUMS 25
Since n there is some n1 > n0 such that for all nn1 one has that1
n
n0
i=1
i|xi x| which implies that 1n
ni=1
ixi x 2
for nn1.
Lemma 2.2.7 (Kronecker) 7 Letn>0,n , and(xn)n=1IRsuchthat
n=1 xn exists. Then
1
n
ni=1
ixin 0.
Proof. Let 0 = S0 = 0 and Sn := x1+ + xn. Then1
n
ni=1
ixi = 1
n
ni=1
i(Si Si1)
= 1
n nSn 0S0
n
i=1Si1(i i1)
= Sn 0S0n
1n
ni=1
Si1(i i1).
Observing that limn Sn=
n=1 xnIR,0S0/n0 since n , and
limn
1
n
ni=1
Si1(i i1) = limn
Sn =
n=1
xn
by Lemma 2.2.6 we finally obtain that
limn
1n
ni=1
ixi = 0.
Proofof Proposition 2.2.5. From the Two-Series-Theorem we know that
IP
n=1
nn
converges
= 1.
7Leopold Kronecker, 7/12/1823 (Liegnitz, Prussia ; now Legnica, Poland)- 29/12/1891
(Berlin, Germany), major contributions in elliptic functions and the theory of algebraicnumbers.
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2.2. CONVERGENCE OF SUMS 27
dyadic expansion 0, 123 where we do not allow an infinite series ofdigits 1. Then considering 1, 2,...: IR we get random variables whichsatisfy the assumptions of Example 2.2.9. The assertion of this propositionsays that zeros and ones are equally likely for the dyadic representation of anumberx[0, 1).
Earlier we have shown for independent Bernoulli random variables1, 2,...: IR that
IP
lim sup
n
fnn
=, lim infn
fnn
=
= 1
wherefn:=1+ + n. As an application of Proposition 2.2.5 we show anopposite statement now.
Proposition 2.2.10 Let1, 2,...: IRbe independentBernoulliran-dom variables,fn:=1+ + n, and >0. Then
IP
limn
fnn(log(n + 1))
12+
= 0
= 1.
Proof. Let n :=
n(log(n + 1))12+ nso that
n=1
IE2nn(log(n + 1))1+2 =
n=1
1
n(log(n + 1))1+2
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28 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
for >0 so that
IP (|f| ) IE |f|2
2 .
Assuming independent random variables 1, 2,... : IR and fn := 1+ + n this gives that
IP (|fn| ) IE|fn|2
2 .
Now one can enlarge the left-hand side by replacing|fn|by supk=1,...,n |fk|sothat we get a maximal inequality.
Lemma 2.2.11 (Inequality ofKolmogorov) Let 1, 2,... : IR bea sequence of independent random variables such that
(i) IEn= 0 forn= 1, 2,...,(ii) IE2n0, andn= 1, 2,..., then one has that
IP
max
1kn|fk|
IEf
2n
2 .
Proof. Let n= 1, 2, . . . be fixed, A :={max1kn |fk| }, andBk:={|f1|< , . . . , |fk1|< , |fk| }
for k2 and B1 :={|f1| }. We get a disjoint union A= nk=1 Bk. Now2IP(A) = 2 IP
nk=1
Bk
=n
k=1
2 IP(Bk) =n
k=1
Bk
2dIP
n
k=1
Bk
f2kdIPn
k=1
Bk
f2ndIP
IEf2
n,
which proves our assertion, where the second to the last inequality can beverified as follows:
Bk
f2ndIP =
Bk
(fk + k+1 + + n)2 dIP
=
Bk
f2kdIP + 2
Bk
fk(k+1+ + n) dIP
+
Bk
(k+1+ + n)2 dIP
= Bk
f2kdIP + Bk
(k+1+ + n)2 dIP
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2.2. CONVERGENCE OF SUMS 29
since, by independence,
Bk fk(k+1+ + n) dIP = (fkBk) (k+1+ + n) dIP=
fkBkdIP
(k+1+ + n) dIP
= 0.
Remark 2.2.12 There are other inequalities, called Levy 10 -Octavianiinequalities: assuming independent random variables1, 2,...: IR and >0 one has
(i) IP (max1kn |fk| > )3 max1kn IP |fk|> 3,(ii) IP (max1kn |fk| > )2IP (|fn|> ) if the sequence (n)n=1 is addi-
tionally symmetric that means that for all signs1,...,n {1, 1}thedistributions of the vectors (1,...,n) and (11,...,nn) are the same.
Example 2.2.13 For Bernoulli variables (n)n=1 it follows from Lemma
2.2.11 that
IP
max
1kn|1+ + k|
IEf
2n
2 =
n
2.
Letting = n, >0, this givesIP
max
1kn|1+ + k|
n
1
2.
The left-hand side describes the probability that the random walk exceedsn or n upto step n (not onlyat step n).
Now we need the converse of the above inequality:
Lemma 2.2.14 (Converse inequality ofKolmogorov) Let 1, 2,... :
IR be a sequence of independent random variables such that
(i) IEn= 0 forn= 1, 2,...,
(ii) there exists a constantc > 0 such that|n()| c for all andn= 1, 2,....
Iffn:=1+ + n, >0, n {1, 2,...}, andIEf2n >0, then
IP
max
1kn|fk|
1 (c + )
2
IEf2n.
10Paul Pierre Levy, 15/09/1886 (Paris, France) - 15/12/1971 (Paris, France), influenced
greatly probability theory, also worked in functional analysis and partial differential equa-tions.
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30 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
The next lemma, we need, describes Cauchy 11 sequences with respect tothe convergence in probability.
Lemma 2.2.15 Letf1, f2,... : IR be a sequence of random variables.Then the following conditions are equivalent:
(i) IP ({ : (fn())n=1 is aCauchy sequence}) = 1,(ii) For all >0 one has that
limn
IP
supk,ln
|fk fl|
= 0.
(iii) For all >0 one has that
limn
IP
supkn
|fk fn|
= 0.
Proof. (ii)(iii) follows fromsupkn
|fk fn| supk,ln
|fk fl| sup
kn|fk fn| + sup
ln|fn fl|
= 2 supkn |fk fn|.(i)(ii) Let
A:={ : (fn())n=1 is a Cauchysequence} .Then we get that
A=
N=1,2,...
n=1,2,...
k>ln
:|fk() fl()| 1
N
.
Consequently, we have that IP(A) = 1 if and only if
IP
n=1,2,...
k>ln
:|fk() fl()| 1
N
= 1
for all N= 1, 2,..., if and only if
limn
IP
k>ln
:|fk() fl()| 1
N
= 1
11
Augustin Louis Cauchy, 21/08/1789 (Paris, France)- 23/05/1857 (Sceaux, France),study of real and complex analysis, theory of permutation groups.
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2.2. CONVERGENCE OF SUMS 31
for allN= 1, 2,..., if and only if
limn IPk>ln
:|fk() fl()|> 1N = 0for allN= 1, 2,.... We can finish the proof by remarking that
k>ln
:|fk() fl()|> 1
N
=
supk>ln
|fk() fl()|> 1N
.
Proofof Proposition 2.2.1. (i) We letn:= n IEnso that IEn = 0 and
n=1
IE2n=
n=1
IE(n IEn)2 0. But this follows from
IP
supkn
|gk gn|
= limN
IP
sup
nkn+N|gk gn|
lim
NIE(gn+N gn)2
2
= limN Nk=1IE2n+k2
l=n+1IE
2l
2
where we have used theKolmogorovinequality Lemma 2.2.11. Obviously,the last term converges to zero as n .
(ii) Because of step (i) we only have to prove that (a)(b). We use againa symmetrization argument and consider a new sequence 1,
2,...:
IRof independent random variables on (, F, IP) having the same distributionas the original sequence 1, 2,..., that means
IP(n) = IP(n)
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32 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
for all n = 1, 2,... and all IR. We also may assume that|n()| d for all and n = 1, 2, .... Taking the product space (M, , ) =(,
F, IP)
(,
F, IP) we may considern,
n
:M
IR with the conventionthat n(,
) =n() and n(, ) =n(
). Now we let
n(, ) =n(, ) n(, )
and get
IEn = IEn IEn= 0,|n(, )| |n(, )| + |n(, )| 2d,
and
(, )M :
n=1
n(, ) converges = IP IP
(, ) :
n=1
(n() n()) converges
= 1.
Letting gn:=1+ + n and >0, Lemma 2.2.15 implies that there is ann {1, 2,...} such that
supk
n|gk gn|
0
IP(g > )
2
e 2
22
= 1.
Proof. By the change of variables y= xwe get that
lim>0
IP(g > )
2
e 2
22
= lim>0
e y2
22 dy
2
2
e 2
22
= lim>0
ex2
2 dx
2
2
e 2
22
= lim
>0
ex2
2 dx
2
12e
2
2
= 1
where we apply the rule ofLHospital 13.
Next we need again a maximal inequality similar to one of that alreadymentioned in Remark 2.2.12.
Lemma 2.3.4 Let1,...,n: IRbe independent random variables whichare symmetric, that means
IP(k) = IP(k)for allk= 1,...,nandIR. Then one has that
IP
maxk=1,...,n
fk>
2IP(fn> )
wherefk :=1+ + k and >0.12Norbert Wiener, 26/11/1894 (Columbia, USA) - 18/03/1964 (Stockholm, Sweden),
worked on Brownian motion, from where he progressed to harmonic analysis, and won theBocher prize from his studies on Tauberian theorems.
13Guillaume Francois Antoine Marquis de LHopital, 1661 (Paris, France)- 2/2/1704
(Paris, France), French mathematician who wrote the first textbook on calculus, whichconsisted of the lectures of his teacher Johann Bernoulli.
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2.3. THE LAW OF ITERATED LOGARITHM 35
Proof. Let 1kn. Because of IP(fn fk >0) = IP(fn fk 0) + IP(fn
fk } and
Bk :={f1, ..., fk1, fk > }for k= 2,...,n. Then we get that, where 0/0 := 1,
IP
maxk=1,...,n
fk>
=
nk=1
IP(Bk)
=
n
k=1
IP(Bk)
IP(fn
fk)
IP(fnfk)
=n
k=1
IP(Bk {fnfk})IP(fnfk)
where we used the independence ofBk andfn fk=k+1 + + n ifk < n.Using that
1
IP(fnfk)2we end up with
IP maxk=1,...,n
fk > 2 n
k=1
IP(Bk {fnfk})
2IP(fn> ).
Proofof Proposition 2.3.1 for n = gn. By symmetry we only need toshow that
IP
lim sup
nn3
fn(n)
1
= 1 and IP
lim sup
nn3
fn(n)
1
= 1.
This is equivalent that for all (0, 1) one has that
IP
:n03 nn0 fn()(1 + )(n)
= 1 (2.2)
and
IP
: # {n3 fn()(1 )(n)}
=
= 1. (2.3)
First we turn to (2.2): let := 1 + and nk := k for k k0 such that
nk0
3. Define
Ak :={ :n(nk, nk+1] fn()> (n)}
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36 CHAPTER 2. SUMS OF INDEPENDENT RANDOM VARIABLES
for k k0. Then IP(lim supkAk) = 0 would imply (2.2). According to theLemma ofBorel-Cantelliit is sufficient to show that
k=k0
IP(Ak) (nk)) 2IP f[nk+1]> (nk) 2c (f[nk+1])
2((nk))e ((nk))
2
2(f[nk+1])2
where we have used the maximal inequality from Lemma 2.3.4 and the esti-mate from Lemma 2.3.3 and where (fn)
2 is the variance offn. Finally, by(fn)
2 =n we get (after some computation) that
k=k0
IP(Ak)
k=k0
c1k k0 and := 1 with(0, 1). Assume
that we can show thatYk > (nk) + 2(nk1) (2.5)
happens infinitely often with probability one, which means by definition that
f[nk] f[nk1]> (nk) + 2(nk1)
happens infinitely often with probability one. Together with (2.4) this wouldimply that
f[nk]> (nk)
happens infinitely often with probability one. Hence we have to show (2.5).For this one can prove that
IP (Yk > (nk) + 2(nk1)) c2k log k
so that k=k0+1
IP (Yk> (nk) + 2(nk1)) =.
An application of the Lemma ofBorel-Cantelli implies (2.5).
Now we have also answered questions (Q5) and (Q7) from Section 1.1.
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Chapter 3
Martingales in discrete time
Originally martingales were intended to model fair games. Having this inmind, it is not surprising that martingale theory plays an important role inareas connected to stochastic modeling such as Stochastic Finance, Biology,and Physics. On the other hand side, martingale theory enters many branchesof pure mathematics as well: for example it is a powerful tool in harmonicand functional analysis and gives new insight into certain phenomena.
In this lecture notes we want to develop some basic parts of the mar-tingale theory. One can distinguish between two cases: discrete time andcontinuous time. The first case deals with sequences of random variablesand is sometimes easier to treat. In the second case we have to work with
families of random variables having uncountably many elements, which istechnically partially more advanced, is however based on the discrete timecase in many cases.
3.1 Some prerequisites
First we summarize some notation and basic facts used from now on.
The Lebesgue spaces. The Lebesgue spaces are named after HenriLebesgue, 1875-1941.
Definition 3.1.1 (i) Let (, F, IP) be a probability space and 1p
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38 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Example 3.1.2
(a) Let = [0, 1),F =B([0, 1)) and IP = be the Lebesgue-measure,that means
([a, b)) =b afor 0 a < b 1. Assume a continuous f : [0, 1) IR. Thenf Lp([0, 1)) if and only if
fp= 1
0
|f(x)|pdx 1
p
0,
f(x) := 1
x Lp([0, 1))
if and only ifp
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3.1. SOME PREREQUISITES 39
(iv) Equivalence-classes: forf Lp(, F, IP) one has thatfp= 0 if andonly if IP(f= 0) = 0. The relationf g if IP(f =g) = 1 defines anequivalence class relation.
Remark 3.1.4
(i) Because of assertion (ii) of Proposition 3.1.3 the expression p is asemi-normonLp. Moreover, (ii) and (iv) imply that Lp is a normon Lp, that means [Lp, Lp ] is a normed space.
(ii) Items (iii) and (iv) say that everyCauchy-sequencein Lp converges toa limit inLp. Hence [Lp, p] is acompletenormed space. A completenormed space is called Banach3 space.
(iii) Ifp= 2, then [L2, 2] is a Hilbert 4 spacewhere the inner product isgiven by
f , g:=
f gdIP for f f and gg.
Conditional expectation. There are different ways to approach condi-tional expectations. Let us explain the approach of the best approximation.Assume = [0, 1),F =B([0, 1)) and IP = to be the Lebesgue measure.Define the sub--algebras
Fdyadn = 0, 12n , 12n , 22n , . . . , 2n12n , 1 , n= 0, 1, 2, ...,which is the system of all possible unions of sets of type
k12n
, k2n
for k =
1, 2, 3, . . . , 2n where n 1 is fixed. ThenFdyadn FdyadN F for 1 nN
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40 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Proposition 3.1.5 (i) The best approximationg is of form
g(x) = 2n
k2n
k12n
f(y)dy forx k12n , k2n .(ii) For allA Fdyadn one has that
A
f(x)dx=
A
g(x)dx.
Proof. (i) Take anotherFdyadn -measurable functionh: [0, 1)IR. Then
1
0
|f(x) h(x)|2dx=2n
k=1 k2n
k12n
|f(x) h(x)|2dx.
We will prove that k2n
k12n
|f(x) h(x)|2dx k
2n
k12n
|f(x) g(x)|2dx
=
k2n
k12n
f(x) 2n k
2n
k12n
f(y)dy
2
dx
Consider
(y) : = k2nk12n
|f(x) y|2dx
=
k2n
k12n
|f(x)|2dx 2y k
2n
k12n
f(x)dx + 1
2ny2.
Then the minimum is attained for y = 2n k
2nk12n
f(x)dxand we are done.
(ii) A Fdyadn means A =
kIk12n
, k2n
. Then
A
f(x)dx=kI
k2nk12n
f(x)dx
=kI
k2n
k12n
g(x)dx
=
A
g(x)dx.
Behind this example there is the general notion of the expected value weintroduce now.
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3.1. SOME PREREQUISITES 41
Proposition 3.1.6 (Conditional expectation) Let(, F, IP) be a proba-bility space andG Fbe a sub--algebra. Letf L1(, F, IP).
(i) There exists ag L1(, G, IP) such thatB
f dIP =
B
gdIP for allB G.
(ii) Ifg andg are like in(i), thenIP (g=g ) = 0.
The main point of the theorem above is that g isG-measurable. This leadsto the following definition:
Definition 3.1.7 (Conditional expectation) TheG
measurable and in-tegrable random variable g from Proposition 3.1.6 is called conditional ex-pectationoffwith respect toG and is denoted by
g= IE(f| G).
One has to keep in mind that the conditional expectation is only unique upto null sets fromG.
We will prove Proposition 3.1.6 later. First we continue with some basicproperties of conditional expectations.
Proposition 3.1.8 Letf , g , f 1, f2 L1(, F, IP) and letH G Fbe sub-algebras ofF. Then the following holds true:
(i) Linearity: if, IR, thenIE(f+ g| G) =IE(f| G) + IE(g| G) a.s.
(ii) Monotonicity: iff1 f2 a.s., thenIE(f1| G) IE(f2| G) a.s.(iii) Positivity: Iff 0 a.s., thenIE(f| G) 0 a.s.
(iv) Convexity: one has that|IE(f| G)| IE(|f| | G) a.s.(v) Projection property: iff isG-measurable, thenIE(f| G) =f a.s.
(vi) Advanced projection property:
IE(IE(f| G)| H) = IE(IE(f| H)| G) = IE(f| H)a.s.
(vii) Ifh: IR isG-measurable andf h L1(, F, IP), thenIE(hf| G) =hIE(f| G)a.s.
(viii) IfG ={, }, thenIE(f|G) = IEf a.s.
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42 CHAPTER 3. MARTINGALES IN DISCRETE TIME
(ix) The case that f is independent fromG : if for all B B(IR) and allA Gone has that
IP ({f B} A) = IP(f B)IP(A),
thenIE(f|G) = IEf a.s.(x) Monotone convergence: assumef 0 a.s. and random variables0
hnf a.s. Thenlimn
IE(hn|G) = IE(f|G)a.s..
Proof. (i) is an exercise.
(ii) Assume that (ii) does not hold. We find an A G
, IP(A) > 0 and < such that
IE(f2| G) < IE(f1| G) on A.
Hence A
f2dIPIP(A)< IP(A)A
f1dIP
which is a contradiction.
(iii) Apply (ii) to 0 =f1f2 = f.
(iv) The inequality f |f| gives IE(f|G) IE(|f||G) a.s. andf |f|givesIE(f|G) IE(|f||G) a.s., so that we are done.
(v) follows directly from the definition.
(vi) Since IE(f|H) isH-measurable and henceG-measurable, item (v)implies that IE(IE(f|H)|G) = IE(f|H) a.s. so that one equality is shown. Forthe other equality we have to show that
A
IE(IE(f|G)|H)dIP =A
f dIP
for A H. Letting h:= IE(f|G) this follows fromA
IE(IE(f|G)|H)dIP =A
IE(h|H)dIP
=
A
hdIP
=
A
IE(f|G)dIP
=
A
f dIP
since A H G.
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3.1. SOME PREREQUISITES 43
(vii) Assume first that h =N
n=1 nAn , whereN
n=1
An = is a disjoint
partition with An G
. For A
Gwe get, a.s., that
A
hf dIP =N
n=1
n
A
Anf dIP
=N
n=1
n
AAn
f dIP
=N
n=1
n
AAn
IE(f|G)dIP
= A N
n=1nAn IE(f|G)dIP
=
A
hIE(f|G)dIP.
Hence IE(hf|G) = hIE(f|G) a.s. For the general case we can assume thatf, h 0 since we can decompose f = f+ f and h = h+ h withf+ := max {f, 0} and f := max {f, 0} (and in the same way we proceedwith h). We find step-functions 0hn h such that hn()h(). Then,by our fist step, we get that
hnIE(f|G) = IE(hnf|G) a.s.By n the left-hand side follows. The right-hand side is a consequenceof the monotone convergence given in (x).
(viii) Clearly, we have that
f dIP = 0 =
(IEf)dIP and
f dIP = IEf=
(IEf)dIP
so that (viii) follows.
(ix) is an exercise.(x) is an exercise.
Proof of Proposition 3.1.6. For this purpose we need the theorem ofRadon 5 -Nikodym6 .
5Johann Radon, 16/12/1887 (Tetschen, Bohemia)- 25/05/1956 (Vienna, Austria),worked on the calculus of variations, differential geometry and measure theory.
6Otton Marcin Nikodym, 13/8/1887 (Zablotow, Austria-Hungary ; now Ukraine)-4/05/1974 (Utica, USA), worked in measure theory, functional analysis, projections onto
convex sets with applications to Dirichlet problem, generalized solutions of differentialequations, descriptive set theory and the foundations of quantum mechanics.
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44 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Definition 3.1.9 (Signed measures) Let (, F) be a measurable space.(i) A map :F IR is called (finite) signed measureif and only if
= + ,where , 0 and+ and are probability measures onF.
(ii) Assume that (, F, IP) is a probability space and that is a signedmeasure on (, F). Then IP ( is absolutely continuous withrespect to IP) if and only if
IP(A) = 0 implies (A) = 0.
Example 3.1.10 Let L L1(, F, IP) and
(A) := A
LdIP.
Then is a signed measure and IP.Proof. We letL+ := max {L, 0} andL:= max {L, 0} so thatL= L+L.Assume that
LdIP> 0 and define
(A) =
ALdIP
LdIP
.
Now we check that are probability measures. First we have that
() = LdIP
LdIP = 1.
Then assume An F to be disjoint sets such that A =
n=1
An. Set :=
L+dIP. Then
+ n=1
An
=
1
n=1
AnL+dIP
= 1
n=1 An()L+dIP
= 1
limN
Nn=1
An
L+dIP
= 1
limN
Nn=1
An
L+dIP
=
n=1
+(An)
where we have usedLebesgues dominated convergence theorem. The same
can be done for L.
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3.1. SOME PREREQUISITES 45
Theorem 3.1.11 (Radon-Nikodym) Let(, F, IP) be a probability spaceand a signed measure withIP. Then there exists anL L1(, F, IP)such that
(A) = A
L()dIP(), A F. (3.1)The random variable L is unique in the following sense. If L and L arerandom variables satisfying (3.1), then
IP(L=L) = 0.Definition 3.1.12 Lis called Radon-Nikodymderivative. We shall write
L= d
dIP.
We should keep in mind the rule
(A) =
Ad=
ALdIP,
so that d= LdIP.
Proof of Proposition 3.1.6. Define
(A) :=
A
f dIP for A G
so that is a signed measure onG. Applying the Theorem of Radon-Nikodym gives an g L1(, G, IP) such that
(A) =
A
gdIP
so that A
gdIP =(A) =
A
f dIP.
Assume now another g L1(, G, IP) withA
gdIP =
A
gdIP
for all A G and assume that IP(g= g) > 0. Hence we find a set A Gwith IP(A)> 0 and real numbers < such that
g() < g () for Aor
g() < g() for A.Consequently (for example in the first case)
A
gdIPIP(A)< IP(A)A
gdIP
which is a contradiction.
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46 CHAPTER 3. MARTINGALES IN DISCRETE TIME
3.2 Definition and examples of martingales
Looking up the word martingale from an encyclopedia (for example in
www.dict.org/bin/Dict) gives the following:
(i) A strap fastened to a horses girth, passing between his fore legs, andfastened to the bit, or now more commonly ending in two rings, throughwhich the reins pass. It is intended to hold down the head of the horse,and prevent him from rearing.
(ii) (Naut.) A lower stay of rope or chain for the jib boom or flying jibboom, fastened to, or reeved through, the dolphin striker. Also, thedolphin striker itself.
(iii) (Gambling) The act of doubling, at each stake, that which has beenlost on the preceding stake; also, the sum so risked; metaphoricallyderived from the bifurcation of the martingale of a harness. [Cant]Thackeray.
We start with the notation of a filtration which describes the information wehave at a certain time-point.
Definition 3.2.1 (Filtration and adapted process) (i) Let (, F) bea measurable space. An increasing sequence (
Fn)n=0 of -algebras
F0 F1 . . . Fn . . . F is called filtration. If (, F, IP) isa probability space, then (, F, IP, (Fn)n=0) is called filteredprobabil-ity space.
(ii) A stochastic process (Xn)n=0, whereXn: IR are random variables,
is called adapted with respect to the filtration (Fn)n=0 (or (Fn)n=0-adapted) provided that Xn isFn-measurable for all n= 0, 1,...
(iii) Given a stochastic processX= (Xn)n=0of random variablesXn :
IR, then the filtration (FXn )n=0 given by
FXn :=(X0,...,Xn)is called natural filtrationof the process X.
It is clear that a process Xis adapted with respect to (FXn )n=0.
Definition 3.2.2 (Martingale) LetM= (Mn)n=0withMn L1(, F, IP)
for n = 0, 1,...and let (Fn)n=0 be a filtration on (, F, IP).(i) The process M is called martingaleifM is (Fn)n=0-adapted and
IE(Mn+1| Fn) =Mn a.s. for n= 0, 1,...
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3.2. DEFINITION AND EXAMPLES OF MARTINGALES 47
(ii) The process M is called sub-martingale or super-martingale, if M is(Fn)n=0-adapted and
IE(Mn+1| Fn) Mn a.s. or IE(Mn+1| Fn) Mn a.s.,for n= 0, 1,..., respectively.
The definition of a (sub-, super-) martingale depends on the filtrationand the measure. To emphasize this, one sometimes uses the phrases(Fn)n=0-martingaleor IP-martingale(and the same for the sub- and super-martingales). Now we consider some examples.
Random walk. Let 0 < p, q < 1, p+q= 1 and (p)1 ,
(p)2 , . . .: IR be
independent random variables such that
IP((p)n =1) =p and IP((p)n = 1) =q(see Definition 2.1.1). As filtration we useF0 :={, } and
Fn := ((p)1 , . . . , (p)n )for n1 so thatFn consists of all possible unions of sets of type
{(p)1 =1, . . . , (p)n =n}with 1, . . . , n {1, 1}. In dependence on pwe get that our random walkconsidered in Chapter 2 is a (sub-, super-) martingale.
Proposition 3.2.3 LetM= (Mn)n=0 be given byMn :=
(p)1 + . . . +
(p)n for
n1 andM0:= 0. Then one has the following:(i) Ifq=p = 1
2, thenM is a martingale.
(ii) Ifq > 12
, thenMis a sub-martingale.
(iii) Ifq < 12
, thenM is a super-martingale.
Proof. Using the rules for conditional expectations we get that
IE(Mn+1| Fn) = IE(Mn+ n+1| Fn)= IE(Mn| Fn) + IE(n+1| Fn)= Mn+ IEn+1.
It remains to observe that IEn+1 = qp is negative, zero, or positive independence on p.
The random walk from Proposition 3.2.3 has an additivestructure since,for 0kn,
Mn Mk is independent fromFk, Mn Mk has the same distribution as Mnk.
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48 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Exponential random walk. In applications, for example in StochasticFinance, one needs instead of the additive structure a multiplicative struc-ture. For instance, changes in share price processes are always relative tothe current price. Hence we are looking for a process Mn >0 such that, for0kn,Mn/Mk is independent fromFk and has the same distribution asMnk. One process of this type is given by
Proposition 3.2.4 Letp(0, 1), >0, cIR, M0:= 1,
Mn:=ePn
i=1(p)i +cn
forn = 1, 2,...,F0 ={, }, andFn = ((p)1 , . . . , (p)n ). Then one has thefollowing:
(i) Ifp(e
e
) =ec
e
, thenM= (Mn)n=0 is a martingale.(ii) Ifp(e e)ec e, thenM= (Mn)n=0 is a sub-martingale.
(iii) Ifp(e e)ec e, thenM= (Mn)n=0 is a super-martingale.
Proof. We have that
IE(Mn+1| Fn) = IE(e(p)n+1+cMn| Fn) =MnIEe
(p)n+1+c =Mn(pe
c + qec+)
and obtain a martingale if and only ifpe + (1 p)e =ec or
p(e e) =ec e.If we have an inequality, then we get a sub- or super-martingale.
Dyadic martingales. Let = [0, 1),F =B([0, 1)), f L1([0, 1)), andIP = be the Lebesguemeasure. As filtration we define
Fdyadicn =
0, 12n
,
12n
, 22n
, . . . ,
2n1
2n , 1
for n = 0, 1, . . .Finally, we let
Mn(t) := IE(f|Fdyad
n )(t).The process (Mn)
n=0is calleddyadicmartingale. This process is a martingale
as shown by the more general
Proposition 3.2.5 For f L1(, F, IP) and a filtration (Fn)n=0 the se-quence(Mn)
n=0 withMn:= IE(f|Fn) is a martingale.
Proof. One has that, a.s.,
IE(Mn+1| Fn) = IE(IE(f| Fn+1)| Fn) = IE(f| Fn) =Mn.
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3.2. DEFINITION AND EXAMPLES OF MARTINGALES 49
Remark 3.2.6 The function fin Proposition 3.2.5 is called the closureofthe martingale (IE(f|Fn))n=0.
Remark 3.2.7 Assume a continuous f : [0, 1] IR and Mn(t) :=IE(f|Fdyadn ) as above. Then one can easily show that
limn
Mn(t) =f(t)
for all t [0, 1) and obtain our first limit theorem. To check this, fix t0[0, 1), find k0, k1,...such that
kn
1
2n t0 0, f0 := 1,
and
fn+1() :=X(n+1)0 () + + X(n+1)fn()() and Mn:=
fnn
.
Then M = (Mn)n=0 is a martingale with respect to the filtration (Fn)n=0
given byFn:=(M0, . . . , M n) forn0.
Proof. Clearly, Mn isFn-measurable by definition ofFn. Next we showMn L1(, F, IP). We proceed by induction and assume that IEMn1
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50 CHAPTER 3. MARTINGALES IN DISCRETE TIME
or equivalently, that IEfn1
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3.3. SOME ELEMENTARY PROPERTIES OF MARTINGALES 51
3.3 Some elementary properties of martin-gales
Here we collect some very basic properties of martingales. It will be use-ful to introduce for a martingale M = (Mn)
n=0 the sequence ofmartingale
differencesdM0:= 0, and
dMn:=Mn Mn1 for n= 1, 2,...
We start with
Proposition 3.3.1 Let M = (Mn)n=0 be a martingale on a filtered proba-bility space(, F, IP, (Fn)n=0). Then one has
(i) IE(Mm| Fn) =Mn a.s. for0n < m
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52 CHAPTER 3. MARTINGALES IN DISCRETE TIME
(iii) Assume that 0m < n. Then
dMndMmdIP=
(Mn Mn1)(Mm Mm1)dIP
=
MnMmdIP
MnMm1dIP
Mn1MmdIP +
Mn1Mm1dIP
=
IE(MnMm|Fm)dIP
IE(MnMm1| Fm1)dIP
IE(Mn1Mm|Fm)dIP +
IE(Mn1Mm1|Fm1)dIP
=
MmIE(Mn| Fm)dIP
Mm1IE(Mn| Fm1)dIP
MmIE(Mn1|Fm)dIP +
Mm1IE(Mn1|Fm1)dIP
=
M2mdIP
M2m1dIP
M2mdIP +
M2m1dIP
= 0
where we have used Proposition 3.1.8.
Lemma 3.3.2 (Jensens inequality) 7
Let : IRIR be a convex func-tion, f L1(, F, IP) such that(f) L1(, F, IP), andG F be a sub-algebra. Then
(IE(f|G))(IE((f)|G)) a.s.
Proof. Any convex function : IRIR is continuous, hence measurable. Tocheck the continuity we fixxIR and observe first that for xnx we havethat
lim supn
(xn)(x)
(if this is not true we would get, by drawing a picture, a contradiction tothe convexity of). Assume now that an x and bn x and that eitherlim infn (an) or lim infn (bn) (or both) is (are) strictly less than (x). Pick-ing n(0, 1) such that x= (1 n)an+ nbn we would get that
(1 n)(an) + n(bn)< (x)for nn0 which is a contradiction to the convexity of. Next we observethat there exists a countable set D of linear functions L : IRIR such that,
7Johan Ludwig William Valdemar Jensen, 08/05/1859 (Nakskov, Denmark)- 05/03/1925 (Copenhagen, Denmark), studied infinite series, the gamma function and in-
equalities for convex functions. Only did mathematics in his spare time, his actual jobwas in a telephone company.
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3.3. SOME ELEMENTARY PROPERTIES OF MARTINGALES 53
for allxIR, one has that(x) = sup
LDL(x).
To get that D we proceed as follows. For q Q we let
aq := limxq
(q) (x)q x and bq := limxq
(q) (x)q x .
By convexity of we have that < aq bq
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54 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Definition 3.3.5 Assume a filtered probability space (, F, IP, (Fn)n=0).(i) A sequence of random variables (vn)
n=1, vn :
IR, is called pre-
dictableifvn isFn1-measurable forn = 1, 2,...(ii) Given a martingale M = (Mn)
n=0 we define its martingale transform
Mv = (Mvn)n=0 byM
v0 :=M0 and
Mvn :=M0+n
k=1
vk(Mk Mk1) for n= 1, 2,...
Proposition 3.3.6 If M = (Mn)n=0 is a martingale and v = (vn)
n=1 is
predictable such that IE|vndMn|
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3.4. STOPPING TIMES 55
Let us start with some examples. The most easiest one is
Example 3.4.2 Consider :
{0, 1,...
}with ()
n0. Then is a
stopping time, since
{=n}=
: n= n0
: n=n0.
One typical and often used class of stopping times is the class ofhitting times.
Example 3.4.3 (Hitting time) Let B B(IR), the stochastic processX= (Xn)
n=0 be adapted, and
B() := inf
{n
0 :Xn()
B
}with inf:=. Then B is a stopping time and called hitting timeofB .
Proof. That B is a stopping time follows from
{B =n}={X0B} . . . {Xn1B} {XnB} Fnfor n1 and{B = 0}={X0B} F0.
Example 3.4.4 Let X = (Xn)n=0 and Y = (Yn)
n=0 be adapted processes
and
() := inf{n0 : Xn() =Yn()}with inf=:. Then is a stopping time. In fact, if we let Zn:=Xn Ynand B :={0}, then is the hitting time ofB with respect to the adaptedprocessZ.
Example 3.4.5 Let X= (Xn)n=0 be adapted. Then
() := inf{n0 : Xn+1()> 1}with inf:=is, in general, nota stopping time.
Now we continue with some general properties of stopping times.
Proposition 3.4.6 Let , : {0, 1, 2, . . .} {} be stopping times.Then one has the following:
(i) The map : IR{}is a random variable, that means1()F and1(B) F for allB B(IR).
(ii) The timemax{, } is a stopping time.(iii) The timemin{, } is a stopping time.
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56 CHAPTER 3. MARTINGALES IN DISCRETE TIME
(iv) The time+ is a stopping time.
Proof. Item (i) follows from
1(B) =
nB,n01({n})
n0Fn F
and
1() = \
n=0
1({n}) F.
Items (ii) and (iii) are consequences of
{max{, }= n}={= n, n} {n, =n}= (
{=n
} {
n
})
(
{
n
} {=n
})
Fnand
{min{, }= n}= ({=n} { n}) ({n} {=n})= [{=n} ( \ { < n})] [{=n} ( \ { < n})] Fn.
(iv) is an exercise.
Now we introduce the -algebraFfor a stopping time . The-algebrawill contain all events one can decide until the stopping time occurs. For
example one can decideA={=n0}
until the stopping time: We take and wait untilis going to happen,that means on{ = 0} the stopping time occurs at time 0, on{ = 1} attime 1, and so on. If() is going to happen, and ()=n0, then A, if() =n0, then A. The abstract definition is as follows:Definition 3.4.7 Let : {0, 1, 2, . . .}{}be a stopping time, thena set A Fbelongs toFif and only if
A {
=n} F
n
for all n= 0, 1, 2, . . ..
Proposition 3.4.8F is a-algebra.Proof. We have that { = n} = Fn and { = n} ={ =n} Fn for all n {0, 1, 2,...} so that, F. Now let A F. ThenAc {=n}={ =n} \ (A {=n}) Fn so that Ac F. Finally, letA1, A2, . . . F. ThenAk {=n} Fn and (
k=1
Ak) { = n} Fn and
k=1Ak F.
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3.4. STOPPING TIMES 57
Example 3.4.9 If n0, thenF =Fn0. In fact, by definitionA F ifand only ifA { = n} Fn for n = 0, 1, 2,.... Sincen= n0 immediatelygives that A
{ = n
}=
Fn the condition reduces to n = n0 which
means that A F if and only ifA { = n0} Fn0 which simply meansthat A Fn0.
Proposition 3.4.10 Let be a stopping time. Then A F if and onlyif A = A
n=0
An with An Fn and An {=n}, and A F andA {=}.
Let us illustrate this by an example.
Example 3.4.11 Let Xn=1+ . . . + n, X0= 0, and
() = inf{n0 : Xn()2}.
Then one has
A = A k{,0,1,...}
{=k}=k{,0,1,...}
[A {=k}]
=:
k{,0,1,...}Ak
with{= 0}={= 1}=(after 0 or 1 steps one cannot have the value 2)and
{= 2} = {1=2 = 1},{= 3} = ,{= 4} = {1=1, 2=3 =4 = 1} {2 =1, 1=3 = 4= 1}
{3=1, 1=2 =4 = 1} {4 =1, 1=2 = 3= 1},
We continue with some general properties.
Proposition 3.4.12 LetX = (Xn)n=0 be an adapted process and :
{0, 1, . . .} a stopping time. Then
Z() :=X()() : IR
isF-measurable.
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58 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Proof. We have to show that
{
: Z()
B
} F for all B
B(IR).
By definition this is equivalent to
{ : Z()B} { : () =n} Fn,or
{ : X()()B, () =n} Fn,or
{ : Xn()B, () =n} Fn,or
{XnB} {=n} Fn,which is true.
Proposition 3.4.13 Let0S T be stopping times. ThenFS FT.
Proof. We have thatA FSif and only ifA{S=n} Fnfor n= 0, 1, 2, . . .Hence
A {
T n}
= A {
S n} {
T n} F
n
since A {S n} Fn and{T n} Fn. But this implies that A {T =n} Fn.
3.5 Doob-decomposition
TheDoob8 -decomposition of a sub-martingale consists in a martingale andin a predictable increasing process. In this way, we get two components whichcan be treated separately.
Proposition 3.5.1 LetY = (Yn)n=0 be a sub-martingale. Then there exists
a processA= (An)n=0 such that
(i) 0 =A0()A1(). . . a.s.,(ii) An isFn1-measurable forn= 1, 2, . . .,
(iii) Mn := Yn An is a martingale.8Joseph Leo Doob, 27/02/1910 (Cincinnati, USA)- 7/06/2004 (Clark-Lindsey Village,
USA),Doobs work was in probability and measure theory. Made major contributions instochastic processes, martingales, optimal stopping, potential theory.
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3.5. DOOB-DECOMPOSITION 59
The processA = (An)n=0 is unique in the following sense: if there are two
such processesA andA, thenIP(An=An) = 1 forn= 0, 1, 2, . . ..
Proof. In order to deduce the formula for An let us assume that we alreadyhave the decomposition Yn = Mn+ An. We get that, a.s.,
IE(Yn+1 An+1| Fn) = IE(Mn+1| Fn) =Mn=Yn Anand
An+1= IE(Yn+1| Fn) Yn+ An.Now let us go the other way round. Define
A0:= 0 and An+1:= IE(Yn+1| Fn) Yn+ An.The properties ofA = (An)
n=0 are that
An+1 isFn-measurable forn0, that An+1 An = IE(Yn+1| Fn) Yn 0 a.s. since Y = (Yn)n=0 is a
sub-martingale, and that
the process M is a martingale because, a.s.,IE(Mn+1
| Fn) = IE(Yn+1
An+1
| Fn)
= IE(Yn+1 IE(Yn+1| Fn) + Yn An| Fn)=Yn An=Mn.
The uniqueness can be seen as follows: if (An)n=0 and (A
n)n=0 are such
processes, then, a.s.,
An+1= IE(Yn+1| Fn) Yn+ AnAn+1= IE(Yn+1| Fn) Yn+ An
with A0=A00, which implies by induction that An = An.
Remark 3.5.2 In fact, we have an equivalence: ifYn=Mn+ An, where Mis a martingale and A an increasing, predictable, and integrable process asabove, then Y is a sub-martingale.
Definition 3.5.3 The decomposition
Yn = Mn+ An
from Proposition 3.5.1 is called Doob-decomposition.
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60 CHAPTER 3. MARTINGALES IN DISCRETE TIME
One might think that the process (An)n=0measures how far a sub-martingale
is from a martingale. Let us consider now some examples.
Example 3.5.4 Let 0 = Y0 Y1 Y2 . . . be real numbers. For anyfiltration we get a sub-martingale since IE(Yn+1| Fn) =Yn+1Yn a.s. TheDoob-decomposition is then Mn= 0 a.s. and An= Yn a.s.
Example 3.5.5 IfY = (Yn)n=0 is a martingale, then An = 0 a.s.
Example 3.5.6 We consider independent random variables (p)1 ,
(p)2 ,... :
IR such that IP((p)n =1) =p and IP((p)n = 1) =q, where 0< p, q 1
2, then the process Y = (Yn)
n=0 with Y0= 0 and
Yn=(p)1 + . . . + (p)n
is a sub-martingale with respect toFn := ((p)1 ,...,(p)n ) for n 1 andF0 :={, }, and
An+1 = IE(Yn+1| Fn)Yn + An = IEn+1 + An= (n + 1)IE1 = (n +1)(qp).
Example 3.5.7 Using the same filtration and random variables as inthe previous example we consider the exponential random walk Sn =
ePn
i=1(p)i +cn withS0:= 1. Ifp(e
e)ec e we get a sub-martingaleand
An+1= IE(Sn+1| Fn) Sn+ An=SnIE
e
(p)n+1+c
Sn+ An
=Sn[ 1] + An=Sn[ 1] + Sn1[ 1] + An1= ( 1)(Sn+ Sn1+ + S0).
Moreover, we have that 1 if and only ifp(e e)ec e.
3.6 Optional stopping theorem
IfM= (Mn)n=0 is a martingale and 0ST 1, we would like to win by the following game: if you have at
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3.6. OPTIONAL STOPPING THEOREM 61
time n an amount ofNn, we toss a fair coin for 2n + |Nn| Euro, that means
thatN0
1 and Nn+1 := Nn+ n+1(2
n +|Nn
|)
where 1, 2,... are independent symmetric Bernoulli random variables.Letting also 0 = 1, we takeFn:=(0,...,n). We obtain
Nn+1 2n ifn+1= 1,
this is what we want, and that the process (Nn)n=0 is a martingale transform
of (0+ + n)n=0 sinceNn+1 Nn= (2n + |Nn|)n+1.
Now we introduce our stopping strategy : {0, 1, 2, . . .}{} by() := inf{n0 : Nn()c}.
The properties ofare summarized in
Proposition 3.6.2 The random time is a stopping time with
(i) IP({ :() T})> 0.Proof. The process (Nn)
n=0 is adapted and can be considered as the
hitting time of the set B= [c, ). Hence we have a stopping time.(i) We observe that one has, for n0,
IP(Nn+1 2n) IP(n+1 = 1) =
1
2.
Assuming now an n00 with 2n0 c gives that
IP( supnn0 Nn+1 c) IP(n0+1 = 1) + IP(n0+1=1, n0+2= 1)
+IP(n0+1 =1, n0+2=1, n0+3 = 1) + = 1
2+ 1
2 1
2+ 1
2 1
2 1
2+
= 1.
(ii) follows by the very definition of.
(iii) We can take 1=. . .= T =1 and have for this path thatN0()< c, . . . , N T()< c
so that IP( > T)2T.
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62 CHAPTER 3. MARTINGALES IN DISCRETE TIME
From that we get
N()()
c a.s. and IE(N{ T0 sothat
IE(YT YS) =m
n=1
IE(Hn(Yn Yn1))
=m
n=1
IE(IE(Hn(Yn Yn1)|Fn1))
=m
n=1IE(HnIE((Yn Yn1)|Fn1))
= 0
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3.6. OPTIONAL STOPPING THEOREM 63
because IE((Yn Yn1)|Fn1) = 0 a.s. so that IEYT = IEYS. Let nowB FSand define the new random times
SB :=S B+ mBc,
TB :=T B+ mBc.
The random times SB and TB are stopping times since, for example for SB,one has
{SB =n}= (B {S=n}) (Bc {m= n})where the first term belongs toFn by the definition of the -algebraFSandthe second term belongs toFn since it is equal the empty set ifm=n andBc FS Fm=Fn ifm= n. Hence we have stopping times with
0 SB TB m
so that
IE(YTB+ YmBc) = IEYTB = IEYSB = IE(YSB+ YmBc)
and B
YTdIP =
B
YSdIP.
Together with Proposition 3.4.12 this implies that IE(YT| FS) =YS a.s.(b) Now we assume that we have a sub-martingale: we first apply the
Doob-decomposition and get that
Yn=Mn+ An.
By step (a) we haveIE(MT| FS) =MSa.s.,
so that, a.s.,
IE(YT| FS) = IE(MT+ AT| FS)= MS+ IE(AT| FS) MS+ IE(AS
| FS)
= YS.
Example 3.6.1(continued) Assume now that we have to stop our game atsome time T0>0. Our new strategy is then
T() := min{(), T0}which is as the minimum of two stopping times again a stopping time. More-overT T0. Letting S
0, we have
0 S T T0
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64 CHAPTER 3. MARTINGALES IN DISCRETE TIME
and, by the optional stopping theorem, IE(MT|FS) =MSa.s. so thatIEMT= IE(IE(MT
|FS)) = IEMS= 1.
Hence our unfair doubling strategy does not work anymore as before, wecannot become rich, if we only have finite time at our disposal.
3.7 Doobs maximal inequalities
Doobs maximal inequality is a replacement ofKolmogorovs inequality inLemma 2.2.11. In the following we also work with (sub-, super-) martingalesX = (Xn)
Nn=0 over the finite time scale{0,...,N}. Moreover, given X =
(Xn)Nn=0 or X= (Xn)
n=0, we need the maximal functions
Xk() := sup0nk
|Xn()| and X() := supn=0,1,...
|Xn()|,
where the latter is used for X= (Xn)n=0 only.
Proposition 3.7.1 (Doobs maximal inequalities) LetX= (Xn)Nn=0 be
a martingale or non-negative sub-martingale. Then the following holds true:
(i) For0 one hasIP (X
N)
IE
{X
N}|X
N|.
(ii) Forp(1, ) one has
XNLp p
p 1XNLp .
Proof. If X = (Xn)Nn=0 is a martingale, then (|Xn|)Nn=0 is a non-negative
sub-martingale, so that it remains to consider this case only.
(i) For 0 we define the stopping times() := inf{n : Xn} N and () :=N
so that 0N. By the Optional Stopping Theorem we may deduceXIE(X| F) = IE(XN| F) a.s.
and
IEXN IEX = IE(X{XN}) + IE(X{XN
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3.7. DOOBS MAXIMAL INEQUALITIES 65
(ii) Let c >0 be fixed. Then
IE((XN c)p
) = IE (XNc)
0 pp
1
d
= IE
c0
pp1{XN}d
=
c0
pp1IP(XN)d.
By step (i) we continue with
IE((XN c)p) c
0
pp2IE(|XN|{XN})d
= pIE[|XN| XNc0
p2d]
= p
p 1 IE[|XN|(XN c)p1]
pp 1 (IE|XN|
p)1p (IE((XN c)p
(p1))1
p
with 1 = 1p
+ 1p
. Because ofp(p 1) =p this implies that
IE((XN c)p)11p p
p
1(IE|XN|p)
1p .
By c we conclude the proof.
Now we reprove Lemma 2.2.11.
Corollary 3.7.2 (Inequality ofKolmogorov) Let1, 2,...: IR bea sequence of independent mean zero and quadratic integrable random vari-ables. IfMn:=1+ + n forn1 andM0 := 0, then one has that
IP (Mn ) IEM2n
2
for all >0.
Proof. Applying Proposition 3.3.3 we get that (M2n)n=0 is a sub-martingale.
Hence we may apply Proposition 3.7.1 and get that
IP
(Mn)2 IEM2n
or2IP (Mn)IEM2n.
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66 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Corollary 3.7.3 (Exit probability of a random walk) Let 1, 2, . . . : IR be independent Bernoulli random variables, M0 : 0, andMn=1+ . . . + n forn= 1, 2,...Then, for
0 andn= 1, 2,...,
IP
1
nMn
2e
2
2 .
Proof. We can assume >0. We define the stochastic exponential
Zn:=eMn
2
2 n
where , > 0 are fixed later. The process Z = (Zn)n=0 is a positive
martingale if 12
(e +e) = e2
2 under the natural filtration (
Fn)n=0 given
byF0 ={, } andFn=(1, . . . , n) for n= 1, 2,...In fact, we haveIE(Zn+1| Fn) =Zn
if and only if
IE(en+12
2 ) = 1,
or12
(e + e) =e2
2 .
Hence we are in a position to apply Doobs maximal inequality and obtain
IP sup0kn
Mk = IPe sup0knMk22 n e22 n IP
e
sup0kn
Mk
2
2 k
e
2
2 n
= IP
Zne
2
2 n
e+
2
2 nIEZn
= e+2
2 n
and
IP
1n
sup0kn
Mk
en+
2
2 n =e
n
e + e
2
ne
ne
2
2 n =e
2
2 nn,
where the inequalitye+e 2e22 can be verified by a Taylor9 -expansionof both sides and the comparison of the expansions. Finally, we have to find
9Brook Taylor, 18/08/1685 (Edmonton, England) - 29/12/1731 (Somerset House, Eng-
land), calculus of finite differences, integration by parts, discovered the formula knownas Taylors expansion.
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3.8. UNIFORMLY INTEGRABLE MARTINGALES 67
the minimum of
f() := 2
2n n.
Since f() =n n, we get 0 = n and
IP
1n
sup0kn
Mk
e 2
22 =e2
2 .
We finish the proof by
IP
Mnn
IP
1n
sup0kn
Mk
+IP
1n
sup0kn
(Mk)
2e2
2 .
3.8 Uniformly integrable martingales
In this section we consider uniformly integrable martingales. Let us recallthe definition ofuniformly integrable.
Definition 3.8.1 A family (fi)iIof random variables fi : IR is calleduniformly integrableprovided that for all >0 there exists a c >0 such that
supiI
|fi|c
|fi|dIP.
An equivalent condition to the above one is
limc
supiI
|fi|c
|fi|dIP = 0.
Moreover, we need the following criteria for uniform integrability:
Lemma 3.8.2 LetG: [0, )[0, ) be non-negative and increasing suchthat
limt
G(t)
t =
and(fi)iIbe a family of random variablesfi: IR such that
supiI
IEG(|fi|)
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68 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Proof. We let >0 and M:= supiIIEG(|fi|) and find a c >0 such thatM
G(t)
t
for tc. Then |fi|c
|fi|dIP M
|fi|c
G(|fi|)dIP.
Corollary 3.8.3 Let p (1, ) and X = (Xn)n=0 be a martingale or anon-negative sub-martingale with supn=0,1,... XnLp
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3.8. UNIFORMLY INTEGRABLE MARTINGALES 69
Proposition 3.8.6 Let(, F, IP, (Fn)n=0) be a stochastic basis with
F= n=0
Fn .andM= (Mn)
n=0 be a martingale. Then the following assertions are equiv-
alent:
(i) The sequence(Mn)n=0 is a Cauchy-sequence inL1, that means for all
>0 there exists ann00 such that for allm, nn0 one has thatMm MnL1 < .
(ii) There exists an Z L1 such that limn Mn = Z inL1 that meanslimn Mn ZL1 = 0.(iii) There exists anM L1 such thatMn= IE(M| Fn) a.s. forn0.(iv) The familyM= (Mn)n0 is uniformly integrable.
If these equivalent conditions hold, then
(a) Z=M a.s.,
(b) M= limn Mn a.s.
(c) IfsupnIE|Mn|p
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70 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Since limm
|Z Mm|dIP = 0 we conclude that IE|Mn IE(Z| Fn)| = 0
and Mn = IE(Z| Fn) a.s.(iii)=
(iv) For c >0 we get that
{|Mn|>c}|Mn|dIP =
{|Mn|>c}
|IE(M| Fn)|dIP
{|Mn|>c}
IE(|M| | Fn)dIP
=
{|Mn|>c}
|M|dIP
{supm |Mm|>c}
|M|dIP.
Since for c >0 Doobs inequality gives
IP
supm0
|Mm|> c
= supN1
IP
sup
0mN|Mm|> c
sup
N1
1
cIE|MN|
= supN1
1
cIE|IE(M|FN)|
supN1
1
cIEIE(|M||FN)
= 1c
IE|M| 0as c , we are done.
Now we interrupt the proof to prepare the implication (iv) (ii). Weneed the notion ofdown-crossings of a function: let I IR, f : I IR {, }, < a < b b},s2 : = inf
{ti> s1 : f(ti)< a
},
...
s2n+1 : = inf{ti> s2n : f(ti)> b},s2n+2 : = inf{ti> s2n+1 : f(ti)< a},
where we use inf:= td, andD0(f,F, [a, b]) := #{n : s2n< td}.
Definition 3.8.7
D(f , I, [a, b]) := sup
{D0(f,F, [a, b]) : F
I finite
}is the number of down-crossingsoffover [a, b].
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3.8. UNIFORMLY INTEGRABLE MARTINGALES 71
Proposition 3.8.8 Let X = (Xn)n=0 be a sub-martingale and let I
{0, 1, 2,...} be non-empty. Then one has for all< a < b b and on A2n we have that Xs2n < a.Hence
(b a)IP(A2n)A2n1\A2n
(Xs2n b)dIP.
Since s2n=N on Ac2n this implies that
(b a)IP(A2n) A2n1\A2n
(XN b)dIP.
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72 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Summing over none gets
(b a)
n=1 IP(D0(X,F, [a, b]) n) = (b a)
n=1 IP(A2n) (XN b)+dIP.
Corollary 3.8.9 Let(Xn)n=0 be a sub-martingale such that
supn
IEX+n
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3.8. UNIFORMLY INTEGRABLE MARTINGALES 73
Proposition 3.8.10 Let(fn)n=0 L1(, F, IP) be uniformly integrable and
f : IR be a random variable such thatf = limn fn a.s. Then onehas that
(i) fL1(, F, IP),(ii) limnIE|f fn|= 0.
Proof. (i) Let >0 and find c >0 such that|fn|c
|fn|dIP
for n = 0, 1, 2,... Then IE|fn| c+ and supnIE|fn|
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74 CHAPTER 3. MARTINGALES IN DISCRETE TIME
ifA, B A with AB , then B\ A A,
ifA1
A2
and A1, A2,...
A, then
n=1 An
A.
From the implication (ii)(iii) we get thatIE(M|Fn) =Mn= IE(Z|Fn) a.s.
so that B:= n=1 Fn A. Since Bis a-system (closed under intersections)which generates F, the--Theorem implies(B) A. Hence A=FwhichgivesZ=M a.s.
(b) From Corollary 3.8.9 we know that limn Mn exists almost surely. Atsame time we know that there is a subsequence such that limkMnk =Za.s.Hence limn Mn = Za.s. Applying (a) gives the assertion of (b).
(c) ByDoobs maximal inequality we have
IE(M)p
p
p 1p
supn
IE|Mn|p
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3.9. APPLICATIONS 75
3.9.2 The branching process example
The key-point is that we use Corollary 3.8.9 to get a limit of our re-normalized
branching process.
Proof of Proposition 1.2.1: Corollary 3.8.9 says that there exists a limitM = limn Mn a.s. Let
n:= IP(Mn= 0)
for n= 0, 1,... Since by construction Mn() = 0 implies that Mn+1() = 0,we get a bounded increasing sequence
0 =0< q0=123 1with the limit := limn n. We shall investigate the value ofby a functionalmethod. For that purpose we let the generating function offn, n= 0, 1,...,be
n() :=k=0
kIP(fn=k) = IEfn, [0, 1],
and, in particular,
() := 1() =N
k=0
qkk
the generating function of the off-spring distribution. Then
n+1= n 1 = 1 1 1since
n+1() = IEfn+1
=k=0
IP(fn=k)IEX
(n+1)1 ++X(n+1)k
=
k=0 IP(fn=k)IEX(n+1)1
k
= n
IEX
(n+1)1
= n 1()
with the convention that 00 := 1 and empty sums are treated as zero. Weget that
(i) 0< q0=(0)< (1) = 1,
(ii) n+1 = (n),
(iii) = (1),
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76 CHAPTER 3. MARTINGALES IN DISCRETE TIME
(iv) () =,
Assertion (i) is clear by our assumptions. Part (ii) follows from n = n(0)
andn+1= n+1(0) = (n(0)) =(n).
To check (iii) we see that
() =N
k=1
qkkk1 and =
Nk=1
qkk= (1).
Finally, (iv) follows from (ii), limn n =, and the continuity of. Now weconsider the different cases for the expected number of sons .
(a) 0 <
1: The only fix-point of is = 1 so that = 1. Hence
limnIP(fn= 0) = 1 and
IP(M(, )) = limn
IP(Mn(, ))limn
IP(fn= 0) = 1
asMn= fnn
. Since this is true for all >0 we conclude that IP(M = 0) = 1.
(b) >1: Here we have exactly one fix-point 0 in (0, 1). Moreover, bydrawing a picture we check that
= limn
n = 0(0, 1).
The equality IEM= 1 we do not prove here. To understand the different cases better, here are two figures:
Figure 3.1: On the left 0 < 1, on the right >1 and 0= 0.25.
Remark 3.9.1 In dependence on the parameter one distinguishes the fol-lowing cases for our branching process:
0< 1 : supercritical case
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3.9. APPLICATIONS 77
3.9.3 The Theorem of Radon-Nikodym.
The following proof is according to P.-A. Meyer [4] (see also D. Williams
[7]). Assume a measurable space (, F) and two finite measures IP andon(, F). We recall that is absolutely continuous with respect to IP providedthat IP(A) = 0 implies that (A) = 0. The aim of this section is to provethe following fundamental theorem by martingale methods.
Proposition 3.9.2 (Theorem of Radon-Nikodym) Assume a probabil-ity space(, F, IP) and a finite measure on(, F). If is absolutely con-tinuous with respect to IP, then there exists a non-negative random variableL L1(, F, IP) such that
(A) = A LdIP for all A F.First we show that our assumption implies some quantitative uniformity.
Lemma 3.9.3 Assume a probability space(, F, IP) and a finite measureon(, F)such that is absolutely continuous with respect to IP. Then, given >0 there is some(0, 1) such thatIP(A)implies that(A).
Proof. Assume the contrary, that means that there is an 0 > 0 such that
for all(0, 1) there is a set A() such thatIP(A()) but (A())> 0.
In particular, we get a sequence Bn := A(2n) for n = 1, 2, 3,... From the
Lemma ofFatou we know that
0lim supn
(Bn)
n=1
k=n
Bk
.
On the other hand, for all n01,
IP
n=1
k=n
Bk
IP
k=n0
Bk
k=n0
IP(Bk) 22n0
so that
IP
n=1
k=n
Bk
= 0 but
n=1
k=n
Bk
0
which is a contradiction to the absolute continuity of with respect to IP.
Proofof Proposition 3.9.2. (a) First we assume a filtration (Fn)n=0 suchthat
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78 CHAPTER 3. MARTINGALES IN DISCRETE TIME
Fn=
A(n)1 ,...,A
(n)Ln
,
the A(n)
1 ,...,A
(n)
Ln are pair-wise disjoint andLnl=1 A(n)l = , every A(n)l is a union of elements from
A(n+1)1 ,...,A
(n+1)Ln+1
,
F=
A(n)l :n = 0, 1,...and l= 1,...,Ln
.
We define the random variables Mn : IR by
Mn() :=
(A(n)l
)
IP(A(n)l )
: IP(A(n)l )> 0
1 : IP(A(n)l ) = 0
whenever A(n)l . The process M= (Mn)n=0 is a martingale with respectto the filtration (Fn)n=0. Clearly IE|Mn| 0. By assumption we can express
A(n)l as disjoint union
A(n)l = mI
A(n+1)m
for some index set I {1,...,Ln+1}. HenceA(n)l
Mn+1dIP =mI
A(n+1)m
Mn+1dIP
=mI
IP(A(n+1)m )(A(n+1)m )
IP(A(n+1)m )
= mI(A(n+1)m )= (A
(n)l )
=
A(n)l
MndIP
with the convention that 0/0 := 1 (note that our assumption of absolute con-
tinuity is used for the equality IP(A(n+1)m )((A
(n+1)m )/IP(A
(n+1)m )) =(A
(n+1)m )).
In particular, the above computation shows that Mn is the density of withrespect to IP onFn, i.e.
(A) = A
MndIP for all A Fn.
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3.9. APPLICATIONS 79
Next we show that the martingaleM= (Mn)n=0 is uniformly integrable. We
know that, for c >0, one has that
IP(Mnc) IEMnc
= ()c
.
Applying Lemma 3.9.3, for any > 0 we find a c > 0 such that IP(A)()/cimplies that (A). Hence
MncMndIP =(Mnc)
and the uniform integrability is proved. Applying Proposition 3.8.6 we havea limit L
L1(,
F, IP) such that
Mn= IE (L|Fn) a.s.
The random variableL is the density we were looking for: defining a measureonF by
(A) :=
A
LdIP
we get = on
n=0 Fn which is a generating algebra ofF. Hence = onFaccording to Caratheodory10s extension theorem.
(b) Next we assume the general case. Let
Abe the collection of all sub-
-algebrasG ofFsuch thatG =(B1,...,BL)
for L = 1, 2,...andBl F. As we saw in step (a) for everyG Athere is anon-negative LG L1(, G, IP) such that
d
dIP|G =LG.
Fact 3.9.4 For all >0 there is an
G
Asuch that
IE|LG1 LG2|<
for allG1 G andG2 G.
Proof. Assuming the contrary gives an 0 > 0 and a sequenceG1 G2G3 fromAsuch that
IE|LGn+1 LGn| 010Constantin Caratheodory, 13/09/1873 (Berlin, Germany)- 2/02/1950 (Munich, Ger-
many), significant contributions to the calculus of variations, the theory of measure, andthe theory of functions of a real variable.
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80 CHAPTER 3. MARTINGALES IN DISCRETE TIME
forn = 1, 2,...However, using the argument from step (a) says that (LGn)n=1
should be a uniformly integrable martingale with respect to the filtration(Gn)n=1
which yields to a contradiction so that the fact is proved.
Now we define our candidate for the density L: We takeG(n) A suchthat
IE|LG1 LG2|< 1
2n+1
for allG1 G(n) andG2 G(n). LettingHn := (G(1), ...,G(n)) we obtain amartingale (LHn)
n=1 with respect to the filtration (Hn)n=1 which converges
inL1 and almost surely to a limit L. Now, for B Fwe get
(B) B LdIP = B L(Hn,B)dIP B LdIP IE L(Hn,B) L L(Hn,B) LHnL1+ LHn LL1
1
2n+1+ LHn LL1n0
so that (B) =B
LdIP.
3.9.4 On a theorem ofKakutani
Martingales that are generated by products of independent random vari-ables play an important role in stochastic modeling and in understanding
equivalent probabilities on a given measurable space. Let us start with theKakutanis theorem
Proposition 3.9.5 Let X0, X1, X2,... : IR be independent and non-negative random variables withX01 andIEXn = 1. Define
Mn:=X0X1X2 XnandFn:=(X0,...,Xn) forn= 0, 1, 2,...
(i) Then M = (Mn)n=0 is a non-negative martingale with respect to
(Fn)n=0 so thatM := lim Mn exists almost surely.(ii) The following assertions are equivalent:
(a)
n=0(1 IE
Xn) 0.
(c) IEM
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3.9. APPLICATIONS 81
Before we prove the proposition we consider a simple example.
Example 3.9.6 Let 1, 2,...:
IR be independent Bernoulli random
variables, i.e. IP(n =1) = IP(n = 1) = 1/2. Define M0:= 1 and
Mn:=e1++ncn such that IEe1c = 1 with c= log
e + 1
e
2
.
Then Xn = enc for n1 and
n=1
(1 IEe nc2 ) =
because we have that IEenc
2 = IEe1c2 0 (note thatn= 0 would imply thatXn = 0 a.s. so that IEXn= 1would not be possible).
(a)(b) is a simple fact from analysis.(b)= (c) Defining
Nn:=
X0
0
Xn
n=
Mn
0 nwe get again a martingale N= (Nn)
n=0. Since
IEN2n = 1
20
2n
k=01
2k
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82 CHAPTER 3. MARTINGALES IN DISCRETE TIME
almost surely, so that IP(M = 0) = 1. This proves (e)(b) and at sametime item (iii).
Now let = IRIN the space of all real valued sequences = (n)n=1IR
and
Yn: IR given by Yn() :=n.As filtration we useF0 :={, } and
Fn:=(Y1,...,Yn).
Finally, we letF :=n=0Fn. Assume Borel-measurable fn, gn : IR IRsuch that fn(x)> 0 and gn(x)> 0 for all xIR and such that
IR
fn(x)dx=
IR
gn(x)dx= 1
for all n = 1, 2,... which means that fn and gn are densities of probabilitymeasures n and n, respectively, on IR. Define the measures
IP :=n=1n and Q:=n=1non the measurable space (, F) and
Xn() := gn(Yn())fn(Yn())= gn(n)fn(n)
.
The random variablesXn form aLikelihood-Ratio Test. We get the following
Proposition 3.9.7 The following assertions are equivalent:
(i) The measureQ is absolutely continuous with respect to IP, that meansthatIP(A) = 0 implies thatQ(A) = 0.
(ii)
n=1IEIP
Xn=
n=1 IR
fn(x)gn(x)dx >0.
(iii)n=1 IR(fn(x) gn(x))2dx
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3.9. APPLICATIONS 83
for n1. Moreover,
IEIPXn = IR gn(x)fn(x) fn(x)dx= IR gn(x)fn(x)dxand
n=0
(1 IEIP
Xn) =
n=1
1
IR
gn(x)
fn(x)fn(x)dx
=
n=1
1
IR
fn(x)gn(x)dx
= 1
2
n=1 IR(fn(x) gn(x))2dx.Applying Kakutani11 s theorem Proposition 3.9.5 gives that conditions(ii) and (iii) are equivalent to the uniform integrability of the martingaleM= (Mn)
n=0 given by
Mn:= X0 Xn.Moreover, we have that
Mn= dQ
dIP|Fn
since
nk=1k(B) = B
g1(x1) gn(xn)dx1...dxn
=
B
g1(x1) gn(xn)f1(x1) fn(xn)d(
nk=1k)(x).
(ii) = (i) Here we use the proof of the Radon-Nikodym theorem wherewe exploit that Mis uniformly integrable.
(i)=(ii) Here we get a density L = dQdIP
and
IE(L|Fn) =Mn a.s.
But then Proposition 3.8.6 gives the uniform integrability ofM.Finally, ifQ is not absolutely continuous with respect to IP, then M isnot uniformly integrable and Kakutanis theorem says that in this caseMn 0 a.s. The other way round: ifMn 0 a.s. but, IEMn = 1 for alln = 1, 2,..., then (Mn)
n=0 cannot be uniformly integrable and Q cannot be
absolutely continuous with respect to IP.
Looking at the symmetry in conditions (ii) and (iii) of Proposition 3.9.7with respect to the densities fn and gn one gets immediately the following
11Shizuo Kakutani, 28/08/1911 (Osaka, Japan)- 17/08/2004 (New Haven, USA), con-tributed to several areas of mathematics: complex analysis, topological groups, fixed point
theorems, Banach spaces and Hilbert spaces, Markov processes, measure theory, flows,Brownian motion, and ergodic theory.
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84 CHAPTER 3. MARTINGALES IN DISCRETE TIME