Stochasticprocesses

Stochasticprocesses

Filtration

Stopping time

References

Stochastic Processes80-646-08

Calcul stochastique

Geneviève Gauthier

HEC Montréal

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping time

References

Stochastic processesDenition

DenitionLet (Ω,F ) be a measurable space. A stochastic process

X = fXt : t 2 T g

is a family of random variables, all built on the samemeasurable space (Ω,F ) where T represents a set of indices.

Stochasticprocesses

Stochasticprocesses

FiltrationDenitionsExample

Stopping time

References

Filtration IDenitions

DenitionA family F = fFt : t 2 T g of σalgebras on Ω is a ltrationon the measurable space (Ω,F ) if

(F1) 8t 2 T , Ft F ,(F2) 8t1, t2 2 T such that t1 t2, Ft1 Ft2 .

DenitionA stochastic process X = fXt : t 2 T g is said to be adaptedto the ltration F = fFt : t 2 T g if

8t 2 T , Xt is Ft measurable.

Stochasticprocesses

Stochasticprocesses

FiltrationDenitionsExample

Stopping time

References

Filtration IIDenitions

DenitionThe ltration F = fFt : t 2 T g is said to be generated by thestochastic process X = fXt : t 2 T g if

8t 2 T , Ft = σ fXs : s 2 T , s tg .

Stochasticprocesses

Stochasticprocesses

FiltrationDenitionsExample

Stopping time

References

Filtration IExample

Example1. Lets assume that the sample space isΩ = fω1,ω2,ω3,ω4g and that T = f0, 1, 2, 3g. Thestochastic process X = fXt : t 2 f0, 1, 2, 3gg represents theevolution of a stock price, Xt = the stock price at close ofmarket on the t th day, while time t = 0 represents today.

ω X0 (ω) X1 (ω) X2 (ω) X3 (ω)

ω1 1 0, 50 1 0, 50ω2 1 0, 50 1 0, 50ω3 1 2 1 1ω4 1 2 2 2

Stochasticprocesses

Stochasticprocesses

FiltrationDenitionsExample

Stopping time

References

Filtration IIExample

Question. What is the ltration generated by this stochasticprocess?

Stochasticprocesses

Stochasticprocesses

FiltrationDenitionsExample

Stopping time

References

Filtration IIIExample

ω X0 (ω) X1 (ω) X2 (ω) X3 (ω)

ω1 1 0, 50 1 0, 50ω2 1 0, 50 1 0, 50ω3 1 2 1 1ω4 1 2 2 2

Answer.

F0 = σ fX0g = f?,Ωg ,F1 = σ fX0,X1g = σ ffω1,ω2g , fω3,ω4gg ,F2 = σ fX0,X1,X2g = σ ffω1,ω2g , fω3g , fω4gg ,F3 = σ fX0,X1,X2,X3g = σ ffω1,ω2g , fω3g , fω4gg .

Note that any σalgebra F containing the sub-σalgebra F3 make X0,X1, X2 and X3 Fmeasurable.

Stochasticprocesses

Stochasticprocesses

FiltrationDenitionsExample

Stopping time

References

Filtration IVExample

Recall that

F0 = σ fX0g = f?,Ωg ,F1 = σ fX0,X1g = σ ffω1,ω2g , fω3,ω4gg ,F2 = σ fX0,X1,X2g = σ ffω1,ω2g , fω3g , fω4gg ,F3 = σ fX0,X1,X2,X3g = σ ffω1,ω2g , fω3g , fω4gg .

Interpretation. Ω represents states of nature. Xt (ωi ) represents the stockprice at time t if it is the i i th state of nature that has occurred. At time0 (today), we know with certitude the stock price and we cannot identifywhich of the states of nature has occurred. Thats why the sub-σalgebraF0 is the trivial σalgebra, since it doesnt contain any information.

Stochasticprocesses

Stochasticprocesses

FiltrationDenitionsExample

Stopping time

References

Filtration VExample

Recall that

F1 = σ fX0,X1g = σ ffω1,ω2g , fω3,ω4gg .

1 At time t = 1, we know a bit more. Indeed, if we observe a stockprice of 0.50, then we know that the state of nature that hasoccurred is ω1 or ω2 but certainly not ω3 or ω4. As a result, we candeduce that the stock price for the following two periods (t = 2 andt = 3) will be 1 and 0.50 dollar respectively.

2 On the contrary, if at time t = 1, we observe a stock price of 2dollars, then we know that the state of nature that has occurred iseither ω3 or ω4. We can deduce from there that the stock pricewont fall back under the one-dollar level: because, after observingthe process at time t = 1, well be able to determine whether eventfω1,ω2g or event fω3,ω4g has happened,F1 = σ ffω1,ω2g , fω3,ω4gg.

Stochasticprocesses

Stochasticprocesses

FiltrationDenitionsExample

Stopping time

References

Filtration VIExample

Recall that

F2 = σ fX0,X1,X2g = σ ffω1,ω2g , fω3g , fω4gg .

Lets now assume that in time t = 2, we observe a price of one dollar. Afrequently made mistake is to conclude that the sub-σalgebra associatedwith that time is σ ffω1,ω2,ω3g , fω4gg since by observing X2 we areable to distinguish between the events fω1,ω2,ω3g and fω4g. Thatwould be true if we were just beginning to observe the process, which isnot the case. We must take into account the information obtained sincetime t = 0. But the paths (X0(ω),X1(ω),X2(ω)) enable us to distinguishbetween the three following events: fω1,ω2g, fω3g and fω4g. Indeed,after observing the prices until time two, we will know with certitude whichstate of nature ω has occurred, unless we have observed path (1, 12 , 1), inwhich case well be unable to distinguish between states of nature ω1 andω2.

1Throughout this chapter, we go further into an example initiated inStochastic Calculus, A Tool for Finance by Daniel Dufresne.

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping timeIntroduction

We will realize how very useful the concept of stopping time iswhen we will attempt to price American-style derivativeproducts. The main role of stopping times is to help determinethe time when the option holder will exercise his or her right.

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping timeDenition

DenitionLet (Ω,F ) be a measurable space such that Card (Ω) < ∞and equipped with the ltration F = fFt : t 2 f0, 1, ...gg. Astopping time τ is a (Ω,F )random variable that takes itsvalues in f0, 1, ...g and is such that

fω 2 Ω : τ (ω) tg 2 Ft for all t 2 f0, 1, ...g . (1)

Exercise. Show that the condition (1) above is equivalent to

fω 2 Ω : τ (ω) = tg 2 Ft for all t 2 f0, 1, ...g .

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time IExample

Example. Lets return to the example described earlier: Xrepresents a stock price.

ω X0 (ω) X1 (ω) X2 (ω) X3 (ω)

ω1 1 0, 50 1 0, 50ω2 1 0, 50 1 0, 50ω3 1 2 1 1ω4 1 2 2 2

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time IIExample

We had determined that the ltration containing theinformation revealed by the process at each time is

F0 = σ fX0g = f?,Ωg ,F1 = σ fX0,X1g = σ ffω1,ω2g , fω3,ω4gg ,F2 = σ fX0,X1,X2g = σ ffω1,ω2g , fω3g , fω4gg ,F3 = σ fX0,X1,X2,X3g = σ ffω1,ω2g , fω3g , fω4gg .

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time IIIExample

Recall that:

ω X0 (ω) X1 (ω) X2 (ω) X3 (ω)

ω1 1 0, 50 1 0, 50ω2 1 0, 50 1 0, 50ω3 1 2 1 1ω4 1 2 2 2

We wont sell our stocks today (t = 0) but we will sellthem as soon as the price is greater than or equal to 1.

The random time representing that situation isτ (ω1) = 2, τ (ω2) = 2, τ (ω3) = 1 and τ (ω4) = 1.

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time IVExample

Such a random variable truly is a stopping time since

fω 2 Ω : τ (ω) = 0g = ? 2 F0,fω 2 Ω : τ (ω) = 1g = fω3,ω4g 2 F1,fω 2 Ω : τ (ω) = 2g = fω1,ω2g 2 F2,fω 2 Ω : τ (ω) = 3g = ? 2 F3.

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time VExample

Recall that:

ω X0 (ω) X1 (ω) X2 (ω) X3 (ω)

ω1 1 0, 50 1 0, 50ω2 1 0, 50 1 0, 50ω3 1 2 1 1ω4 1 2 2 2

Lets now consider the random time τ modelling thefollowing situation: well buy stock as soon as it enablesus to make a prot later.

Such a random value takes values τ (ω1) = 1,τ (ω2) = 1, τ (ω3) = 0 and τ (ω4) = 0. τ is not astopping time since

fω 2 Ω : τ (ω) = 0g = fω3,ω4g /2 F0.

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time VIExample

Intuitively, the time τ when one makes a decision is astopping time if the decision is made based on theinformation available at that time. In the case of stoppingtimes, using a crystal ball is prohibited.

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time IStopping time transformation

TheoremLet (Ω,F ), be a measurable space such that Card (Ω) < ∞and equipped with the ltration F = fFt : t 2 f0, 1, ...gg. Ifthe random variables τ1 and τ2 are stopping times with respectto the ltration F, then τ1 ^ τ2 min fτ1, τ2g andτ1 _ τ2 max fτ1, τ2g are also stopping times.

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time IIStopping time transformation

Proof of the theorem. If τ is a stopping time, then8t 2 f0, 1, ...g

fω 2 Ω : τ (ω) tg 2 Ft .

8k 2 f0, 1, ...g ,

fω 2 Ω : τ1 (ω) ^ τ2 (ω) kg= fω 2 Ω : τ1 (ω) k or τ2 (ω) kg= fω 2 Ω : τ1 (ω) kg| z

2Fk

[ fω 2 Ω : τ2 (ω) kg| z 2Fk

2 Fk .

Exercise. Prove the above result for τ1 _ τ2.

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time IFirst passage time

DenitionLet (Ω,F ) be a measurable space such that Card (Ω) < ∞and equipped with the ltration F = fFt : t 2 f0, 1, ...gg.X = fXt : t 2 f0, 1, ...gg represents a stochastic processadapted to that ltration. Let B R a subset of the realnumbers. We dene the time until the stochastic process Xrst enters the set B as

τB (ω) = min ft 2 f0, 1, ...g : Xt (ω) 2 Bg .

If it happened that the path t ! Xt (ω) never hits the set Bthen we dene τB (ω) = ∞.

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

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Stopping time IIFirst passage time

TheoremThe random variable τB is a stopping time.

Proof of the theorem. Since Card (Ω) < ∞, then8t 2 f0, 1, ...g, Xt can only take a nite number of values.Lets denote them by

x (t)1 < ... < x (t)mt .

8t 2 f0, 1, ...g,

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time IIIFirst passage time

fω 2 Ω : τB (ω) = tg= fω 2 Ω : X0 (ω) /2 B , ...,Xt1 (ω) /2 B ,Xt (ω) 2 Bg

=

t1\k=0

fω 2 Ω : Xk (ω) /2 Bg!\ fω 2 Ω : Xt (ω) 2 Bg

=

0B@t1\k=0

[x (k )i /2B

nω 2 Ω : Xk (ω) = x

(k )i

o1CA\

0B@ [x (t)i 2B

nω 2 Ω : Xt (ω) = x

(t)i

o1CA2 Ft

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping timeDenitionExampleTransformationsFirst passage

References

Stopping time IVFirst passage time

since

t1\k=0

[x (k )i /2B

nω 2 Ω : Xk (ω) = x

(k )i

o| z

2Fk since X is adapted.| z 2FkFt since Fk is a σalgebra| z 2Ft since Ft is a σalgebra.

\[

x (t)i 2B

nω 2 Ω : Xt (ω) = x

(t)i

o| z 2Ft since Xt is Ftmeasurable.| z

2Ft since Ft is a σalgebra.

Stochasticprocesses

Stochasticprocesses

Filtration

Stopping time

References

References

BILLINGSLEY, Patrick (1986). Probability and Measure,Second Edition, Wiley, New York.

DUFRESNE, Daniel (1996). Stochastic Calculus, A Toolfor Finance, Department of Mathematics and Statistics,Université de Montréal.

KARLIN Samuel and TAYLOR Howard M. (1975). A FirstCourse in Stochastic Processes, Second Edition, AcademicPress, New York.