processus martingales

8/18/2019 processus martingales

1/46

STAT-F-407

Martingales

Thomas Verdebout

Université Libre de Bruxelles

2015


2/46

Outline of the course

1. A short introduction.

2. Basic probability review.

3. Martingales.

4. Markov chains.

5. Markov processes, Poisson processes.

6. Brownian motions.


3/46


1. A short introduction.2. Basic probability review.

3. Martingales.

3.1. Definitions and examples.

3.2. Sub- and super- martingales, gambling strategies.

3.3. Stopping times and the optional stopping theorem.

3.4. Limiting results.

4. Markov chains.




4/46

Definitions and basic comments

Let (Ω,A,P ) be a measure space.

Definition: a filtration is a sequence {An |n ∈ N} of σ-algebrassuch that A0 ⊂ A1 ⊂ . . . ⊂ A.

Definition: The SP {Y n |n ∈ N} is adapted to the filtration {An |n ∈ N}⇔ Y n is An -measurable for all n .

Intuition: growing information An ... And the value of Y n isknown as soon as the information An is available.

The SP {Y n |n ∈N} is a martingale w.r.t. the filtration{An |n ∈ N} ⇔

(i) {Y n |n ∈ N} is adapted to the filtration {An |n ∈ N}.

(ii) E[|Y n |] < ∞ for all n .

(iii) E

[Y n +1|An ] = Y n a.s. for all n .


5/46


Remarks:

(iii) shows that a martingale can be thought of as the

fortune of a gambler betting on a fair game.

(iii) ⇒ E[Y n ] = E[Y 0] for all n (mean-stationarity). Using (iii),we also have (for k = 2, 3, . . .)

E[Y n +k |An ] = EE[Y n +k |An +k −1]An = E[Y n +k −1|An ] == . . . =

E[Y n +1|An ] = Y n

a.s. for all n .


6/46


Let σ(X 1, . . . ,X n ) be the smallest σ-algebra containing

{X −1i (B ) |B ∈ B , i = 1, . . . , n }.

The SP {Y n |n ∈ N} is a martingale w.r.t. the SP {X n |n ∈ N} ⇔

{Y n |n ∈N

}is a martingale w.r.t. the filtration

{σ(X

1, . . . ,X n )|

n ∈N

}

(i) Y n is σ(X 1, . . . ,X n )-measurable for all n .

(ii) E[|Y n |] < ∞ for all n .

(iii) E[Y n +1

|X 1

, . . . ,X n

] = Y n

a.s. for all n .

Remark:

(i) just states that “Y n is a function of X 1, . . . ,X n only".


7/46


A lot of examples...


8/46

Example 1

Let X 1,X 2, . . . be ⊥⊥ integrable r.v.’s, with common mean 0.

Let Y n :=n

i =1 X i .Then {Y n } is a martingale w.r.t. {X n }.

Indeed,

(i) is trivial.

(ii) is trivial. (iii): with An := σ(X 1, . . . ,X n ), we have

E[Y n +1|An ] = E[Y n + X n +1|An ] = E[Y n |An ] + E[X n +1|An ]

= Y n + E[X n +1] = Y n + 0 = Y n a.s. for all n ,

where we used that Y n is An -measurable and that X n +1 ⊥⊥ An .

Similarly, if X 1,X 2, . . . are ⊥⊥ and integrable with meansµ1, µ2, . . ., respectively, {

n i =1(X i − µi )} is a martingale w.r.t.

{X n }.


9/46

Example 2

Let X 1,X 2, . . . be ⊥⊥ integrable r.v.’s, with common mean 1.

Let Y n :=n

i =1 X i .Then {Y n } is a martingale w.r.t. {X n }.

Indeed,

(i) is trivial.

(ii) is trivial. (iii): with An := σ(X 1, . . . ,X n ), we have

E[Y n +1|An ] = E[Y n X n +1|An ] = Y n E[X n +1|An ] = Y n E[X n +1] = Y n

a.s. for all n , where we used that Y n is An -measurable(and hence behaves as a constant in E[ . |An ]).

Similarly, if X 1,X 2, . . . are ⊥⊥ and integrable with meansµ1, µ2, . . ., respectively, {

n i =1(X i /µi )} is a martingale w.r.t.

{X n }.


10/46

Example 2

The previous example is related to basic models for stock

prices.

Assume X 1,X 2, . . . are positive, ⊥⊥ , and integrable r.v.’s.Let Y n := c

n i =1 X i , where c is the initial price.

The quantity X i − 1 is the change in the value of the stock overa fixed time interval (one day, say) as a fraction of its current

value. This multiplicative model (i) ensures nonnegativeness ofthe price and (ii) is compatible with the fact fluctuations in the

value of a stock are roughly proportional to its price.

Various models are obtained for various distributions of the X i ’s:

Discrete Black-Scholes model: X i = e ηi , where

ηi ∼ N (µ, σ2) for all i .

Binomial model: X i = e −r (1 + a )2ηi −1, where ηi ∼ Bin(1,p )

for all i (r = interest rate, by which one discounts future rewards)


11/46

Example 3: random walks

Let X 1,X 2, . . . be i.i.d., with P [X i = 1] = p and P [X i = −1] = q = 1−Let Y n :=

n i =1 X i .

The SP {Y n } is called a random walk.

Remarks:

If p = q , the RW is said to be symmetric.

Of course, from Example 1, we know that

{(n i =1 X i )− n (p − q )} is a martingale w.r.t. {X n }.But other martingales exist for RWs...


12/46


Consider the non-symmetric case (p = q ) and let S n :=q p Y n

.

Then {S n } is a martingale w.r.t. {X n }.

Indeed,

(i) is trivial. (ii): |S n | ≤ max((q /p )n , (q /p )−n ). Hence, E[|S n |] < ∞.

(iii): with An := σ(X 1, . . . ,X n ), we have

E[S n +1|An ] = E[(q /p )Y n (q /p )X n +1 |An ] = (q /p )

Y n E[(q /p )X n +1 |An ]

= S n E[(q /p )X n +1 ] = S n

(q /p )1×p +(q /p )−1×q

= S n , a.s. for all

where we used that Y n is An -measurable and that X n +1 ⊥⊥ An .


13/46


Consider the symmetric case (p = q ) and let S n := Y 2n − n .

Then {S n } is a martingale w.r.t. {X n }.Indeed,

(i) is trivial.

(ii): |S n | ≤ n 2 − n . Hence, E[|S n |] < ∞.


E[S n +1|An ] = E[(Y n + X n +1)2 − (n + 1)|An ]

= E[(Y 2n + X 2n +1 + 2Y n X n +1)− (n + 1)|An ]

= Y 2n +E[X 2n +1|An ]+2Y n E[X n +1|An ]−(n +1)

= Y 2n + E[X 2n +1] + 2Y n E[X n +1]− (n + 1)

= S n a.s. for all n ,

where we used that Y n

is An

-measurable and that X n +1

⊥⊥ An

.

D


14/46

Example 4: De Moivre’s martingales

Let X 1,X 2, . . . be i.i.d., with P [X i = 1] = p and P [X i = −1] = q = 1−Let Y 0 := k ∈ {1, 2, . . . ,m − 1} be the initial state.Let Y n +1 := (Y n + X n +1) I [Y n /∈ {0,m }] + Y n I [Y n ∈ {0,m }].

The SP {Y n } is called a random walk with absorbingbarriers.

Remarks:

Before being caught either in 0 or m , this is just a RW. As soon as you get in 0 or m , you stay there forever.

E l 4 D M i ’ ti l


15/46

Example 4: De Moivre’s martingales

Let X 1,X 2, . . . be i.i.d., with P [X i = 1] = p and P [X i = −1] = q = 1−Let Y 0 := k ∈ {1, 2, . . . ,m − 1} be the initial state.Let Y n +1 := (Y n + X n +1) I [Y n /∈ {0,m }] + Y n I [Y n ∈ {0,m }].

In this new setup and with this new definition of Y n ,

In the non-symmetric case,

{S n := q

p Y n } is still a martingale w.r.t. {X n }.

(exercise).

E l 5 b hi


16/46

Example 5: branching processes

Consider some population, in which each individual i of the Z n

individuals in the n th generation gives birth toX n +1,i children (theX n ,i ’s are i.i.d., take values in N, and have common mean µ < ∞).

Assume that Z 0 = 1.

Then {Z n /µn } is a martingale w.r.t.

{An := σ(the X m ,i ’s,m ≤ n )}.

Indeed,

(i), (ii): exercise...

(iii): EZ n +1µn +1 |An

=

1µn +1 E

Z n i =1 X n +1,i |An

= 1µn +1

Z n i =1 E[X n +1,i |An ] =

1µn +1

Z n i =1 E[X n +1,i ] =

Z n µn a.s. for all n .

In particular, EZ n µn = E

Z 0µ0 = 1. Hence, E[Z n ] = µ

n for all n .

E ample 6 Pol a’s rn


17/46

Example 6: Polya’s urn

Consider an urn containing b blue balls and r red ones.Pick randomly some ball in the urn and put it back in the urn

with an extra ball of the same color. Repeat this procedure.

This is a so-called contamination process.

Let X n be the number of red balls in the urn after n steps.Let

R n = X n

b + r + n

be the proportion of red balls in the urn after n steps.

Then {R n } is a martingale w.r.t. {X n }.



18/46


Indeed,

(i) is trivial.

(ii): 0 ≤ |R n | ≤ 1. Hence, E[|R n |] < ∞.


E[X n +1|An ] = (X n + 1) X n

r + b + n + (X n + 0)

1− X n

r + b + n

= (X n + 1)X n + X n ((r + b + n )− X n )

r + b + n

= (r + b + n + 1)X n

r + b + n = (r + b + n + 1)R n a.s. for all n ,

so that E[R n +1|An ] = R n a.s. for all n .



19/46




3. Martingales.

3.1. Definitions and examples.

3.2. Stopping times and the optional stopping theorem.

3.3. Sub- and super- martingales, Limiting results.

4. Markov chains.



Stopping times


20/46

Stopping times

Let (Y n ) be a martingale w.r.t. (An )Let T : (Ω,A,P) → N := N ∪ {∞} be a r.v.

Definition: T is a stopping time w.r.t. (An ) ⇔(i) T is a.s. finite (i.e., P[T < ∞] = 1).

(ii) [T = n ] ∈ An for all n .Remarks:

(ii) is the crucial assumption:

it says that one knows, at time n , on the basis of the

“information" An , whether T = n or not, that is, whetherone should stop at n or not.

(i) just makes (almost) sure that one will stop at some point.

Stopping times (examples)


21/46


(Kind of) examples...

Let (Y n ) be a martingale w.r.t. (An ). Let B ∈ B .

(A) Let T := inf{n ∈ N|Y n ∈ B } be the time of 1st entry of (Y n )into B . Then,

[T = n ] = [Y 0 /∈ B ,Y 1 /∈ B , . . . ,Y n −1 /∈ B ,Y n ∈ B ] ∈ An .

Hence, provided that T is a.s. finite, T is a ST.

(B) Let T := sup{n ∈ N|Y n ∈ B } be the time of last escape of(Y n ) out of B . Then,

[T = n ] = [Y n ∈ B ,Y n +1 /∈ B ,Y n +2 /∈ B , . . .] /∈ An .

Hence, T is not a ST.



22/46


(C) Let T := k a.s. (for some fixed integer k ). Then, of course,(i) T < ∞ a.s. and (ii)

[T = n ] = ∅ if n = k

Ω if n = k ,

which is in An for all n . Hence, T is a ST.

Stopping times (properties)


23/46

Stopping times (properties)

Properties:

[T = n ] ∈ An ∀n (1)

⇔ [T ≤ n ] ∈ An ∀n (2)

⇔ [T > n ] ∈ An ∀n .

Indeed,(1)⇒ follows from [T ≤ n ] =

n k =1[T = k ].

(1)

⇐ follows from [T = n ] = [T ≤ n ]\[T ≤ n − 1].(2)⇔ follows from [T ≤ n ] = Ω\[T > n ].

T 1,T 2 are ST ⇒ T 1 + T 2, max(T 1,T 2), and min(T 1,T 2) areST (exercise).

Let (Y n ) be a martingale w.r.t. (An ).Let T be a ST w.r.t. (An ).Then Y T :=

∞n =0 Y n I[T =n ] is a r.v.

Indeed, [Y T ∈ B ] = ∞n =0{[T = n ] ∩ [Y n ∈ B ]} ∈ A.

Stopped martingale


24/46

Stopped martingale

A key lemma:

Lemma: Let (Y n ) be a martingale w.r.t. (An ). Let T be a STw.r.t. (An ).Then {Z n := Y min(n ,T )} is a martingale w.r.t. (An ).

Proof: note that

Z n = Y min(n ,T ) =n −1k =0

Y k I[T =k ] + Y n I[T ≥n ].

So

(i): Z n is An -measurable for all n . (ii): |Z n | ≤

n −1k =0 |Y k |I[T =k ] + |Y n |I[T ≥n ] ≤

n k =0 |Y k |.

Hence, E[|Z n |] < ∞.

Stopped martingale


25/46

Stopped martingale

(iii): we have

E[Z n +1|An ] − Z n = E[Z n +1 − Z n |An ]

= E[(Y n +1 − Y n )I[T ≥n +1]|An ]

= E[(Y n +1 − Y n )I[T >n ]|An ]= I[T >n ] E[Y n +1 − Y n |An ]

= I[T >n ]E[Y n +1|An ]− Y n

= 0 a.s. for all n ,

where we used that I[T >n ] is An -measurable.

Stopped martingale


26/46

pp g

Corollary: Let (Y n ) be a martingale w.r.t. (An ). Let T be a STw.r.t. (An ). Then E[Y min(n ,T )] = E[Y 0] for all n .

Proof: the lemma and the mean-stationarity of martingales yield

E[Y min(n ,T )] = E[Z n ] = E[Z 0] = E[Y min(0,T )] = E[Y 0] for all n .

In particular, if the ST is such that T ≤ k a.s. for some k , wehave that, for n ≥ k ,

Y min(n ,T ) = Y T a.s.,

so that

E[Y T ] = E[Y 0].

Stopped martingale


27/46

pp g

E[Y T ] = E[Y 0] does not always hold.

Example: the doubling strategy, for which the winnings are

Y n =n

i =1C i X i ,

where the X i ’s are i.i.d. P[X i = 1] = P[X i = −1] = 12 and C i = 2

i −1b .

The SP (Y n ) is a martingale w.r.t. (X n ) (exercise).Let T = inf{n ∈ N|X n = 1} (exercise: T is a ST).

As we have seen, Y T = b a.s., so that E[Y T ] = b = 0 = E[Y 0].

However, as shown by the following result, E[Y T ] = E[Y 0] holdsunder much broader conditions than "T ≤ k a.s."

Optional stopping theorem (examples)


28/46

p pp g ( p )

Let p k := P[Y T = 0].

Then

E[Y T ] = 0× p k + m × (1− p k )

and

E[Y 0] = E[k ] = k ,

so that E[Y T ] = E[Y 0] yields

m (1− p k ) = k ,

that is, solving for p k ,

p k = m − k

m .



29/46

p pp g p

Let X 1,X 2, . . . be i.i.d., with P[X i = 1] = p andP[X i = −1] = q = 1− p .Let Y 0 := k ∈ {1, 2, . . . ,m − 1} be the initial state.Let Y n +1 := (Y n + X n +1) I[Y n /∈{0,m }] + Y n I[Y n ∈{0,m }].

The SP (Y n ) is called a random walk with absorbing barriers.

In the non-symmetric case,

S n :=

q p

Y n is a martingale

w.r.t. (X n ). Let T := inf{n ∈ N|S n ∈ {0,m }} (exercise: T is astopping time, and the assumptions of the optional stopping

thm are satisfied). E[S T ] = E[S 0].



30/46

Let again p k := P[Y T = 0]. Then, since

E[S T ] = E

q p

Y T

=q p

0 × p k +

q p

m (1− p k ),

andE[S 0] = E

q p

Y 0= E

q p

k =q p

k ,

we deduce thatq p

0× p k +

q p

m (1− p k ) =

q p

k ,

that is, solving for p k ,

p k =

q p

k −q p

m 1−

q p

m .



31/46

Is there a way to get E[T ] here as well?

In the non-symmetric case, (R n := Y n −min(n ,T )(p − q )) is amartingale w.r.t. (X n ) (exercise: check this, and check that theoptional stopping thm applies with (R n ) and T ).

E[R T ] = E[R 0], where

E[R T ] = E[Y T −T (p −q )] = E[Y T ]−(p −q )E[T ] =

0×p k +m ×(1−p k )

and

E[R 0] = E[Y 0 − min(0,T )(p − q )] = E[Y 0] = E[k ] = k .

Hence,

E[T ] = m (1− p k )− k

p − q =

m

1−q p

k − k

1−

q p

m (p − q )

1−

q p

m .



32/46



3. Martingales.

3.1. Definitions and examples.3.2. Stopping times and the optional stopping theorem.

3.3. Sub- and super- martingales, Limiting results.

4. Markov chains.



Sub- and super-martingales


33/46

Not every game is fair...

There are also favourable and defavourable games.

Therefore, we introduce the following concepts:

The SP (Y n )n ∈N is a submartingale w.r.t. the filtration (An )n ∈N ⇔

(i) (Y n )n ∈N is adapted to the filtration (An )n ∈N. (ii)’ E[Y +n ] < ∞ for all n .

(iii)’ E[Y n +1|An ] ≥ Y n a.s. for all n .

The SP (Y n )n ∈N is a supermartingale w.r.t. the filtration (An )n ∈N ⇔

(i) (Y n )n ∈N is adapted to the filtration (An )n ∈N.

(ii)” E[Y −n ] < ∞ for all n .

(iii)” E

[Y n +1|An ] ≤ Y n a.s. for all n .



34/46

Remarks:

(iii)’ shows that a submartingale can be thought of as the

fortune of a gambler betting on a favourable game.

(iii)’ ⇒ E[Y n ] ≥ E[Y 0] for all n .

(iii)” shows that a supermartingale can be thought of as thefortune of a gambler betting on a defavourable game.

(iii)” ⇒ E[Y n ] ≤ E[Y 0] for all n .

(Y n ) is a submartingale w.r.t. (An )

⇔ (−Y n ) is a supermartingale w.r.t. (An ).

(Y n ) is a martingale w.r.t. (An )⇔ (Y n ) is both a sub- and a supermartingale w.r.t. (An ).



35/46

Consider the following strategy for the fair version of roulette

(without the "0" slot):

Bet b euros on an even result. If you win, stop.

If you lose, bet 2b euros on an even result. If you win, stop.

If you lose, bet 4b euros on an even result. If you win, stop...

And so on...

How good is this strategy?

(a) If you first win in the n th game, your total winning is

−

n −2i =0

2i b + 2n −1b = b

⇒ Whatever the value of n is, you win b euros with thisstrategy.



36/46

(b) You will a.s. win. Indeed, let T be the time index of first

success. Then

P[T < ∞] =∞n =1

P[n − 1 first results are "odd", then "even"]

=∞

n =11

2n −1 1

2 =

∞

n =11

2n

= 1.

But

(c) The expected amount you lose just before you win is

0×

1

2 +b ×1

22

+(b +2b )×1

23

+. . .+n −2

i =02i

b 1

2n

+. . . = ∞

⇒ Your expected loss is infinite!

(d) You need an unbounded wallet...



37/46

Let us try to formalize strategies...

Consider the SP (X n )n ∈N, where X n is your winning per unitstake in game n . Denote by (An )n ∈N the corresponding filtration(An = σ(X 1, . . . ,X n )).

Definition: A gambling strategy (w.r.t. (X n )) is a SP (C n )n ∈Nsuch that C n is An −1-measurable for all n .

Remarks:

C n = C n (X 1, . . . ,X n −1) is what you will bet in game n . A0 = {∅, Ω).



38/46

Using some strategy (C n ), your total winning after n games is

Y (C )n =

n i =1

C i X i .

A natural question:

Is there any way to choose (C n ) so that (Y (C )n ) is "nice"?.

Consider the "blind" strategy C n = 1 (for all n ), that consists inbetting 1 euro in each game, and denote by (Y n =

n i =1 X i ) the

corresponding process of winnings.

Then, here is the answer:

Theorem: Let (C n ) be a gambling strategy with nonnegative

and bounded r.v.’s. Then if (Y n ) is a martingale, so is (Y (C )n ). If

(Y n ) is a submart., so is (Y (C )n ). And if (Y n ) is a supermart., so

is (Y (C )n ).



39/46

Proof:

(i) is trivial.

(ii): |Y (C )n | ≤

n i =1 a i |X i |. Hence, E[|Y (C )n |] < ∞. (iii),(iii)’,(iii)”: with An := σ(X 1, . . . ,X n ), we have

E[Y (C )n +1|An ] = E[Y

(C )n + C n +1X n +1|An ]

= Y

(C )

n + C n +1E

[X n +1|An ]= Y

(C )n + C n +1 E[Y n +1 − Y n |An ]

= Y (C )n + C n +1

E[Y n +1|An ]− Y n

,

where we used that C n +1 is An -measurable. SinceC n +1 ≥ 0, the result follows.

Remark: The second part was checked for martingales only.

Exercise: check (ii)’ and (ii)”...

Convergence of martingales


40/46

Theorem: let (Y n ) be a submartingale w.r.t. (An ). Assume that,for some M , E[Y +n ] ≤ M for all n . Then

(i) ∃Y ∞ such that Y n a .s .→ Y ∞ as n →∞.(ii) If E[|Y 0|] < ∞, E[|Y ∞|] < ∞.

The following results directly follow:

Corollary 1: let (Y n ) be a submartingale or a supermartingalew.r.t. (An ). Assume that, for some M , E[|Y n |] ≤ M for all n .

Then ∃Y ∞ (satisfying E[|Y ∞|] < ∞) such that Y n a .s .→ Y ∞ as

n →∞

.

Corollary 2: let (Y n ) be a negative submartingale or a positive

supermartingale w.r.t. (An ). Then ∃Y ∞ such that Y n a .s .→ Y ∞ as

n →∞.

Example: products of r.v.’s


41/46

Let X 1,X 2, . . . be i.i.d. r.v.’s, with common distribution

distribution of X i values 0 2

probabilities 12

12

The X i ’s are integrable r.v.’s with common mean 1,

so that (Y n = n i =1 X i ) is a (positive) martingale w.r.t. (X n )(example 2 in the previous lecture).

Consequently, ∃Y ∞ such that Y n a .s .→ Y ∞ as n →∞.

We showed, in Lecture #4, that Y n P → 0 as n →∞ so that

Y ∞ = 0 a.s.

But we also showed there that convergence in L1 does not

hold. To ensure L1-convergence, one has to require uniform

integrability of (Y n ).

Example: Polya’s urn


42/46

Consider an urn containing b blue balls and r red ones.

Pick randomly some ball in the urn and put it back in the urn

with an extra ball of the same color. Repeat this procedure.

Let X n be the number of red balls in the urn after n steps.

Let R n = X n b +r +n be the proportion of red balls after n steps.

We know that (R n ) is a martingale w.r.t. (X n ).

Now, |R n | ≤ 1 (⇒ E[|R n |] ≤ 1), so that ∃R ∞ (satisfying E[|R ∞|] < ∞)

such that R n a .s .→ R ∞ as n →∞.

Clearly, uniform integrability holds. Hence, R n L1→ R ∞ as

n →∞.

Remark: it can be shown that R ∞ has a beta distribution:

P[R ∞ ≤ u ] = b + r

r u

0

x r −1(1− x )b −1 dx , u ∈ (0, 1).

Example: branching processes


43/46

Consider some population, in which each individual i of the Z n

individuals in the n th generation gives birth to X n ,i children (theX n ,i ’s are i.i.d., take values in N, and have common mean µ < ∞).

Assume that Z 0 = 1.

Then (Y n := Z n /µn ) is a martingale w.r.t.

(An := σ(X n ,i , n fixed)).

What can be said about the long-run state?

E

[|Y n |] = E

[Y 0] = 1 for all n ⇒ ∃Y ∞ (satisfying E

[|Y ∞|] < ∞)such that Y n a .s .→ Y ∞ as n →∞.

Z n ≈ Y ∞µn for large n .



44/46

Case 1: µ ≤ 1.

If p 0 := P[X n ,i = 0] > 0, Z n a .s .→ Z ∞ := 0 as n →∞.

Proof:

Let k ∈N

0. Then

P[Z ∞ = k ] = P[Z n = k ∀n ≥ n 0] = P[Z n 0 = k ] (P[Z n 0+1 = k |Z n 0 = k ])∞

where

P[Z n 0+1 = k |Z n 0 = k ] ≤ P[Z n 0+1 = 0|Z n 0 = k ] = 1 − (p 0)k


45/46

Case 2: µ > 1.

Then Z n a .s .

→ Z ∞ as n →∞, where the distribution of Z ∞ is givenby

distribution of Z ∞values 0 ∞

probabilities π 1 − π

where π ∈ (0, 1) is the unique solution of g (s ) = s , where,defining p k := P[X n ,i = k ], we let g (s ) :=

∞k =0 p k s

k .

Proof:

Clearly, since Z n ≈ Y ∞µn for large n , Z n a .s .→ Z ∞ as n →∞,where Z ∞ only assumes the values 0 and ∞, with proba a and1− a , respectively, say.

We may assume that p 0 > 0 (clearly, if p 0 = 0, Z ∞ = ∞ a.s.)



46/46

There is indeed a unique π ∈ (0, 1) such that g (π) = π, since g

is monotone increasing, g (0) = p 0 > 0, and g

(1) = µ > 1. (πZ n ) is a martingale w.r.t. (An ), since

E[πZ n +1 |An ] = E[πZ n

i =1 X n +1,i |An ] =

E[πX n +1,1 |An ]

Z n

=E[πX n +1

,1 ]Z n =

∞k =0

πk p k Z n

=g (π)

Z n = πZ n

a.s. for all n .

(πZ n ) is also uniformly integrable, so that πZ n L1→ πZ ∞ as n →∞.

Therefore, E[πZ ∞] = π0 × a + π∞ × (1− a ) = a is equal toE[πZ 0 ] = E[π1] = π.

Hence, P[Z ∞ = 0] = π and P[Z ∞ = ∞] = 1− π.

Documents

processus martingales