example martingales - hec montréalneumann.hec.ca/~p240/c80646en/4martingaleen.pdf · martingales...

Martingales

Denition

Example

Stoppedprocess

OptionalStoppingTheorem

Markovianprocess

Martingales80-646-08

Stochastic Calculus I

Geneviève Gauthier

HEC Montr éal

Martingales

DenitionLemma 1Examplelemma 2

Example

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DenitionOn the ltered probability space (Ω,F ,F,P), where F is theltration fFt : t 2 f0, 1, 2, . . .gg, the stochastic process

M = fMt : t 2 f0, 1, 2, . . .gg

is a discrete-time martingale if

(M1) 8t 2 f0, 1, 2, . . .g, EP [jMt j] < ∞;(M2) 8t 2 f0, 1, 2, . . .g, Mt is Ftmeasurable;(M3) 8s, t 2 f0, 1, 2, . . .g such that s < t, EP [Mt jFs ] = Ms .

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MartingaleConstant expectation process

Lemma. Let M = fMt : t 2 f0, 1, 2, . . .gg be a martingalebuilt on the ltered probability space (Ω,F ,F,P). Then

8t 2 f1, 2, . . .g , EP [Mt ] = EP [M0] .

Proof of the lemma. 8t 2 f1, 2, . . .g,

EP [Mt ] = EPhEP [Mt jF0 ]

iby (EC3)

= EP [M0] by (M3) .

Interpretation. A martingale is a stochastic process that,on average, is constant. This doesnt mean however thatsuch a process varies little, since the variance VarP [Mt ],at every time, can be innite.

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Example IExample. Let fξt : t 2 f1, 2, . . .gg be a sequence of(Ω,F )independent and identically distributed randomvariables with respect to the measure P and such that

EP [ξt ] = 0 and EPξ2t< ∞.

Lets dene

F0 = f?,Ωg ;8t 2 f1, 2, . . .g ,Ft = σ fξs : s 2 f1, . . . , tgg ;

and

M0 = 0, Mt =t

∑s=1

ξs .

The stochastic process M is a martingale on the space(Ω,F ,F,P).

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Example II

Indeed,

EP [jMt j] = EP

" t∑s=1 ξs

#

t

∑s=1

EP [jξs j] t

∑s=1

qEPξ2s< ∞

where the second inequality comes from the fact that, for anyrandom variable,

0 Var [jX j] = EhjX j2

i (E [jX j])2 ) E [jX j]

rEhjX j2

i.

Given the selected ltration, M is adapted (which is to say that8t 2 f0, 1, 2, . . .g, Mt is Ftmesurable).

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Example III

Lastly, 8s, t 2 f0, 1, 2, . . .g such that s < t,

EP [Mt jFs ]

= EP

"Ms +

t

∑u=s+1

ξu jFs

#

= EP [Ms jFs ] +t

∑u=s+1

EP [ξu jFs ]

= Ms +t

∑u=s+1

EP [ξu jFs ]| z =EP[ξu ]

from (EC1) since Ms is Fs measurable, and

from (EC7) since ξu is independent from ξ1, . . . , ξs .= Ms .

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Martingale I

LemmaIn the denition of a martingale, the condition (M3) isequivalent to

M3* 8t 2 f1, 2, . . .g, EP [Mt jFt1 ] = Mt1.

Proof of the lemma. Clearly, (M3)) (M3)since (M3) isonly a special case of (M3). Indeed, it is su¢ cient to denes = t 1.So we must show that (M3)) (M3). This can be proved byinduction. Intuitively, if s < t then

EP [Mt jFs ] = EPhEP [Mt jFt1 ] jFs

ifrom (EC3),

= EP [Mt1 jFs ] from (M3)

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Martingale II

But if s < t 1, then we can use the same logic again, and weget

EP [Mt1 jFs ] = EPhEP [Mt1 jFt2 ] jFs

ifrom (EC3),

= EP [Mt2 jFs ] from (M3).

Now just substitute this result into the rst equation:

EP [Mt jFs ] = EP [Mt2 jFs ] .

By iterating such an algorithm, we will eventually obtain

EP [Mt jFs ] = EP [Ms+1 jFs ]= Ms from (M3).

Martingales

Denition

Example

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Example I

ω ξ1 ξ2 ξ3 P Q

ω1 1 1 1 18

12

ω2 1 1 1 18

114

ω3 1 1 1 18

114

ω4 1 1 1 18

114

ω ξ1 ξ2 ξ3 P Q

ω5 1 1 1 18

114

ω6 1 1 1 18

114

ω7 1 1 1 18

114

ω8 1 1 1 18

114

On the sample space Ω = fω1, . . . ,ω8g, we will use theσ-algebra F = the set of all events in Ω. The ltration F ismade up of the σ-subalgebras

F0 = f?,Ωg ,F1 = σ fξ1g = σ ffω1,ω2,ω3,ω4g , fω5,ω6,ω7,ω8gg ,F2 = σ fξ1, ξ2g = σ ffω1,ω2g , fω3,ω4g , fω5,ω6g , fω7,ω8gg ,F3 = σ fξ1, ξ2, ξ3g = F .

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Example

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Example II

The stochastic process M, built on the ltered measurablespace (Ω,F ,F) is dened as follows :

M0 = 0,

M1 = ξ1M2 = ξ1 + ξ2 and

M3 = ξ1 + ξ2 + ξ3.

By construction, M is adapted to the ltration F.

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Example III

M = fMt : t 2 f0, 1, 2, 3gg is a martingale on (Ω,F ,F,P).

Indeed, condition (M2) is already veried since M isFadapted.Condition (M1) is also satised since 8t 2 f0, 1, 2, 3g,

EP [jMt j] EP [jξ1j] + EP [jξ2j] + EP [jξ3j] = 3.

Lets verify condition (M3).

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Example IV

EP [M1 jF0 ] = EP [M1 ] from (EC4)

= 0

= M0

8ω 2 fω1,ω2,ω3,ω4g ,

EP [M2 jF1 ] (ω) =112

2 1

8 2 1

8+ 0 1

8+ 0 1

8

= 1

M1 (ω) = 1;

8ω 2 fω5,ω6,ω7,ω8g ,

EP [M2 jF1 ] (ω) =112

0 1

8+ 0 1

8+ 2 1

8+ 2 1

8

= 1

M1 (ω) = 1

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Example V

8ω 2 fω1,ω2g ,

EP [M3 jF2 ] (ω) =114

3 1

8 1 1

8

= 2 and M2 (ω) = 2;

8ω 2 fω3,ω4g ,

EP [M3 jF2 ] (ω) =114

1 1

8+ 1 1

8

= 0 and M2 (ω) = 0;

8ω 2 fω5,ω6g ,

EP [M3 jF2 ] (ω) =114

1 1

8+ 1 1

8

= 0 and M2 (ω) = 0;

8ω 2 fω7,ω8g ,

EP [M3 jF2 ] (ω) =114

1 1

8+ 3 1

8

= 2 and M2 (ω) = 2.

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Example VI

By contrast M = fMt : t 2 f0, 1, 2, 3gg is not a martingale on(Ω,F ,F,Q). Indeed,

EQ [M1 jF0 ]= EQ [M1] from (EC4)

=12+114+114+114+114+114+114+114

= 614

6= 0

= M0.

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Example VII

Conclusion. For a stochastic process, the property ofbeing a martingale depends on the ltration and on themeasure. Thats why the notation (F,P)martingalemay sometimes be seen.

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Denition

Example

StoppedprocessDenitionExampleTheorem


Markovianprocess

Denition

DenitionThe stochastic process X and the stopping time τ are built onthe same ltered measurable space (Ω,F ,F). The stochasticprocess X τdened by

X τt (ω) = Xt^τ(ω) (ω) (1)

is called a stopped process with stopping time τ.

Martingales

Denition

Example



Markovianprocess

Example I

ω X0 X1 X2 X3 τ X τ0 X τ

1 X τ2 X τ

3

ω1 1 12 1 1

2 0 1 1 1 1

ω2 1 12 1 1

2 3 1 12 1 1

2

ω3 1 2 1 1 1 1 2 2 2

ω4 1 2 2 1 1 1 2 2 2

Martingales

Denition

Example



Markovianprocess

Stopped martingale ITheorem

TheoremIf the martingale M and the stopping time τ are built on thesame ltered probability space (Ω,F ,F,P) then the stoppedprocess Mτ is also a martingale on that space.

Martingales

Denition

Example



Markovianprocess

Stopped martingale IITheorem

Proof of the theorem. The key to the proof is to express Mτt

in terms of the components of the process M.

Mτ0 = M0

and 8t 2 f1, 2, . . .g , Mτt = Mτ

t

t1∑k=0

Ifτ=kg + Ifτtg

!

=t1∑k=0

Ifτ=kgMτt + IfτtgM

τt

=t1∑k=0

Ifτ=kgMk + IfτtgMt .

Martingales

Denition

Example



Markovianprocess

Stopped martingale IIITheorem

Verifying condition (M1) :

EP [jMτ0 j] = EP [jM0j] < ∞

and 8t 2 f1, 2, . . .g ,

EP [jMτt j] = EP

"t1∑k=0

Ifτ=kgMk + IfτtgMt

#

t1∑k=0

EPIfτ=kgMk

+ EPIfτtgMt

t1∑k=0

EP [jMk j] + EP [jMt j] < ∞

since, M being a martingale, we have that 8t 2 f0, 1, 2, . . .g,EP [jMt j] < ∞.

Martingales

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Example



Markovianprocess

Stopped martingale IVTheorem

Verifying condition (M2) :

Mτ0 = M0 is F0 measurable. (2)

Now, 8t 2 f1, 2, . . .g,

Mτt =

t1∑k=0

Ifτ=kg| z Fk measurablesince fτ=kg2Fk

Mk|zFk measurablesince M is adapted.| z

Ft measurable since k<t)FkFt

+ Ifτtg| z Ft1 measurable

sincefτtg=fτt1gc2Ft1

Mt|zFt measurablesince M is adapted.

is Ftmeasurable.

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Example



Markovianprocess

Stopped martingale VTheorem

Verifying condition (M3) : 8t 2 f1, 2, . . .g ,

Mτt Mτ

t1

=

t1∑k=0

Ifτ=kgMk + IfτtgMt

!

t2∑k=0

Ifτ=kgMk + Ifτt1gMt1

!= Ifτ=t1gMt1 + IfτtgMt Ifτt1gMt1

= IfτtgMt Ifτt1g Ifτ=t1g

Mt1

= IfτtgMt IfτtgMt1

since, fτ = t 1g and fτ tg being disjoint,Ifτ=t1g + Ifτtg = Ifτ=t1g[fτtg = Ifτt1g.

= Ifτtg (Mt Mt1) .

Martingales

Denition

Example



Markovianprocess

Stopped martingale VITheorem

As a consequence, since Ifτtg is Ft1measurable

EP [Mτt jFt1 ]Mτ

t1= EP [Mτ

t Mτt1 jFt1 ]

= EPIfτtg (Mt Mt1) jFt1

= IfτtgE

P [Mt Mt1 jFt1 ]

= Ifτtg

EP [Mt jFt1 ] EP [Mt1 jFt1 ]

= Ifτtg (Mt1 Mt1) = 0

henceEP [Mτ

t jFt1 ] = Mτt1 .

Martingales

Denition

Example

Stoppedprocess


Markovianprocess

Optional Stopping Theorem

Theorem(Optional Stopping Theorem). Let

X = fXt : t 2 f0, 1, 2, . . .gg

be a process built on the ltered probability space(Ω,F ,F,P), where F is the ltration fFt : t 2 f0, 1, 2, . . .gg.Lets assume that the stochastic process X is Fadapted andthat it is integrable, i.e. EP [jXt j] < ∞. Then X is amartingale if and only if

EP [Xτ] = EP [X0]

for any bounded stopping time τ, i.e for any given stoppingtime τ, there exists a constant b such that

8ω 2 Ω , 0 τ (ω) b.

Martingales

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Example

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Markovianprocess

Optional Stopping Theorem IProof of the theorem

First part. Lets assume that X is a martingale and letsshow that, in such a case, EP [Xτ] = EP [X0] for anybounded stopping time.

Let τ be any bounded stopping time. Then, there exists aconstant b such that 8ω 2 Ω, 0 τ (ω) b. As a

Martingales

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Optional Stopping Theorem IIProof of the theorem

consequence,

EP [Xτ ] = EP

"b

∑k=0

XkIfτ=kg

#

= EP

"b

∑k=0

Xk

Ifτkg Ifτk+1g#

=b

∑k=0

EPhXkIfτkg

i

b

∑k=0

EPhXkIfτk+1g

i= EP [X0 ] +

b

∑k=1

EPhXkIfτkg

ib1∑k=0

EPhXkIfτk+1g

isince Ifτ0g = IΩ = 1 and Ifτb+1g = I? = 0.

= EP [X0 ] +b

∑k=1

EPhXkIfτkg

i

b

∑k=1

EP[Xk1Ifτkg

Martingales

Denition

Example

Stoppedprocess


Markovianprocess

Optional Stopping Theorem IIIProof of the theorem

= EP [X0 ] +b

∑k=1

EPh(Xk Xk1) Ifτkg

i= EP [X0 ] +

b

∑k=1

EPhEPh(Xk Xk1) Ifτkg jFk1

iifrom (EC3),

= EP [X0 ] +b

∑k=1

EPhIfτkgEP [Xk Xk1 jFk1 ]

ifrom (EC6),

= EP [X0 ]

since, X being a martingale,

EP [Xk Xk1 jFk1 ] = EP [Xk jFk1 ] EP [Xk1 jFk1 ]= Xk1 Xk1 = 0.

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Example

Stoppedprocess


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Optional Stopping Theorem IVProof of the theorem

Second part. Lets now assume that, for any boundedstopping time τ, EP [Xτ] = EP [X0] and lets show that, insuch a case, the adapted and integrable stochastic processis a martingale.

By hypothesis, X already satises conditions (M1) and (M2).The only thing left to verify is that 8s, t 2 f0, 1, 2, . . .g suchthat s < t, EP [Xt jFs ] = Xs .

So, lets set s and t 2 f0, 1, 2, . . .g such that s < t.We denote by P s =

nA(s)1 , . . . ,A(s)ns

othe nite

partition generated by Fs .

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Optional Stopping Theorem VProof of the theorem

For any i 2 f1, . . . , nsg we build a random time :

Si (ω) =

8><>:s if ω 2 A(s)i

t if ω /2 A(s)i .

Si is a stopping time (obviously bounded) since8u 2 f0, 1, 2, . . .g ,

fω 2 Ω : Si (ω) = ug =

8>>>>><>>>>>:

A(s)i if u = s 2 FsA(s)i

cif u = t 2 Fs Ft

? otherwise 2 F0 Fu

.

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Optional Stopping Theorem VIProof of the theorem

So, by hypothesis, we have that

EP [XSi ] = EP [X0] .

Besides, since the random time τt dened as 8ω 2 Ω,τt (ω) = t is also a bounded stopping time, we have, again byhypothesis, that

EP [Xt ] = EP [Xτt ] = EP [X0]

henceEP [XSi ] = EP [Xt ] .

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Example

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Optional Stopping Theorem VIIProof of the theorem

As a consequence, 8i 2 f1, . . . , nsg,

0 = EP [Xt ] EP [XSi ]

= EP [Xt XSi ]

= EP

(Xt XSi ) I

A(s)i+ (Xt XSi ) I

A(s)ic

= EP

(Xt Xs ) I

A(s)i+ (Xt Xt ) I

A(s)ic

= EPh(Xt Xs ) I

A(s)i

i= ∑

ω2A(s)i

(Xt (ω) Xs (ω))P (ω)

hence

∑ω2A(s)i

Xt (ω)P (ω) = ∑ω2A(s)i

Xs (ω)P (ω) .

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Optional Stopping Theorem VIIIProof of the theorem

Now we can conclude the proof, since

EP [Xt jFs ] =ns

∑i=1

IA(s)i

PA(s)i

∑ω2A(s)i

Xt (ω)P (ω)

=ns

∑i=1

IA(s)i

PA(s)i

∑ω2A(s)i

Xs (ω)P (ω)

= EP [Xs jFs ]= Xs .

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Denition

Example

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Markovian process IDenition

DenitionA stochastic process X = fXt : t 2 T g, where T is a set ofindicesa, is said to be markovian if, for anyt1 < t2 < . . . < tn 2 T , the conditional distribution of Xtngiven Xt1 , . . . ,Xtn1 is equal to the conditional distribution ofXtn given Xtn1 , i.e. for any x1, . . . , xn 2 R,

P [Xtn xn jXt1 = x1, . . . ,Xtn1 = xn1 ]= P [Xtn xn jXtn1 = xn1 ] .

aExamples: T = f0, 1, 2, . . .g, T = f0, 1, 2, . . . ,T g where T is apositive integer, T = [0,T ] where T is a positive real number, T = [0,∞),etc.

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Markovian process IIDenition

Intuitively, if we assume that t are temporal indices, theprocess X is markovian if its distribution in the future, giventhe present and the past, only depends on the present.

A Markov chain is then a memoryless randomphenomenon: the distribution of an observation tocome, given our present knowledge of the system andits whole history, is the same when only its presentstate is known.1

The set of values that the process may take is called the statespace of X and we denote it by EX .

1Jean Vaillancourt.

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Markovian process IExample

Example. We throw a dice repeatedly.The random variable ξn represents the number of pointsobtained on the n th throw.The stochastic process X represents the total cumulativenumber of points obtained at any time, i.e. for any naturalinteger t,

Xt =t

∑n=1

ξn.

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Markovian process IIExample

X is a Markovian process. Indeed,

Xt =t

∑n=1

ξn =t1∑n=1

ξn + ξt = Xt1 + ξt .

But the outcome of the t th throw of dice, ξt , is independentfrom the results obtained on the rst t 1 th throws,σ fξn : n 2 f1, . . . , t 1gg. As a consequence, the distributionof X t depends on the past of the stochastic process,σ fξn : n 2 f1, . . . , t 1gg , through σ fXt1g only.

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Example

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Markovian processRemark

Question. Why do we need a probability space? Wouldnta measurable space have been su¢ cient?

Answer. A probability measure is required to ensure thatthe independence property is satised.

Martingales

Denition

Example

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Markovian processRandom walk

The stochastic process X , built on the probability space(Ω,F ,P), is a random walk if it admits the representation

X0 = 0 et 8t 2 f1, 2, . . .g , Xt =t

∑n=1

ξn

where the sequence fξt : t 2 f1, 2, . . .gg is made up ofindependent and identically distributed random variables.Random walks are Markovian processes.

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