Processes 80-646-08 Calcul stochastique Geneviève Gauthier HEC Montréal
Let (Ω, F) be a measurable space. A stochastic process X = fx t : t 2 T g is a family of random variables, all built on the same measurable space (Ω, F) where T represents a set of indices.
s I s A family F = ff t : t 2 T g of σ algebras on Ω is a ltration on the measurable space (Ω, F) if (F 1) 8t 2 T, F t F, (F 2) 8t 1, t 2 2 T such that t 1 t 2, F t1 F t2. A stochastic process X = fx t : t 2 T g is said to be adapted to the ltration F = ff t : t 2 T g if 8t 2 T, X t is F t measurable.
II s s The ltration F = ff t : t 2 T g is said to be generated by the stochastic process X = fx t : t 2 T g if 8t 2 T, F t = σ fx s : s 2 T, s tg.
s I 1. Let s assume that the sample space is Ω = fω 1, ω 2, ω 3, ω 4 g and that T = f0, 1, 2, 3g. The stochastic process X = fx t : t 2 f0, 1, 2, 3gg represents the evolution of a stock price, X t = the stock price at close of market on the t th day, while time t = 0 represents today. ω X 0 (ω) X 1 (ω) X 2 (ω) X 3 (ω) ω 1 1 0, 50 1 0, 50 ω 2 1 0, 50 1 0, 50 ω 3 1 2 1 1 ω 4 1 2 2 2
II s Question. What is the ltration generated by this stochastic process?
s III ω X 0 (ω) X 1 (ω) X 2 (ω) X 3 (ω) ω 1 1 0, 50 1 0, 50 ω 2 1 0, 50 1 0, 50 ω 3 1 2 1 1 ω 4 1 2 2 2 Answer. F 0 = σ fx 0 g = f?, Ωg, F 1 = σ fx 0, X 1 g = σ ffω 1, ω 2 g, fω 3, ω 4 gg, F 2 = σ fx 0, X 1, X 2 g = σ ffω 1, ω 2 g, fω 3 g, fω 4 gg, F 3 = σ fx 0, X 1, X 2, X 3 g = σ ffω 1, ω 2 g, fω 3 g, fω 4 gg. Note that any σ algebra F containing the sub-σ algebra F 3 make X 0, X 1, X 2 and X 3 F measurable.
s Recall that F 0 = σ fx 0 g = f?, Ωg, IV F 1 = σ fx 0, X 1 g = σ ffω 1, ω 2 g, fω 3, ω 4 gg, F 2 = σ fx 0, X 1, X 2 g = σ ffω 1, ω 2 g, fω 3 g, fω 4 gg, F 3 = σ fx 0, X 1, X 2, X 3 g = σ ffω 1, ω 2 g, fω 3 g, fω 4 gg. Interpretation. Ω represents states of nature. X t (ω i ) represents the stock price at time t if it is the i i th state of nature that has occurred. At time 0 (today), we know with certitude the stock price and we cannot identify which of the states of nature has occurred. That s why the sub-σ algebra F 0 is the trivial σ algebra, since it doesn t contain any information.
s Recall that V F 1 = σ fx 0, X 1 g = σ ffω 1, ω 2 g, fω 3, ω 4 gg. 1 At time t = 1, we know a bit more. Indeed, if we observe a stock price of 0.50, then we know that the state of nature that has occurred is ω 1 or ω 2 but certainly not ω 3 or ω 4. As a result, we can deduce that the stock price for the following two periods (t = 2 and t = 3) will be 1 and 0.50 dollar respectively. 2 On the contrary, if at time t = 1, we observe a stock price of 2 dollars, then we know that the state of nature that has occurred is either ω 3 or ω 4. We can deduce from there that the stock price won t fall back under the one-dollar level: because, after observing the process at time t = 1, we ll be able to determine whether event fω 1, ω 2 g or event fω 3, ω 4 g has happened, F 1 = σ ffω 1, ω 2 g, fω 3, ω 4 gg.
s Recall that VI F 2 = σ fx 0, X 1, X 2 g = σ ffω 1, ω 2 g, fω 3 g, fω 4 gg. Let s now assume that in time t = 2, we observe a price of one dollar. A frequently made mistake is to conclude that the sub-σ algebra associated with that time is σ ffω 1, ω 2, ω 3 g, fω 4 gg since by observing X 2 we are able to distinguish between the events fω 1, ω 2, ω 3 g and fω 4 g. That would be true if we were just beginning to observe the process, which is not the case. We must take into account the information obtained since time t = 0. But the paths (X 0 (ω), X 1 (ω), X 2 (ω)) enable us to distinguish between the three following events: fω 1, ω 2 g, fω 3 g and fω 4 g. Indeed, after observing the prices until time two, we will know with certitude which state of nature ω has occurred, unless we have observed path (1, 2 1, 1), in which case we ll be unable to distinguish between states of nature ω 1 and ω 2. 1 Throughout this chapter, we go further into an example initiated in Calculus, A Tool for Finance by Daniel Dufresne.
Introduction We will realize how very useful the concept of stopping time is when we will attempt to price American-style derivative products. The main role of stopping times is to help determine the time when the option holder will exercise his or her right.
Let (Ω, F) be a measurable space such that Card (Ω) < and equipped with the ltration F = ff t : t 2 f0, 1,...gg. A stopping time τ is a (Ω, F) random variable that takes its values in f0, 1,...g and is such that fω 2 Ω : τ (ω) tg 2 F t 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 F t for all t 2 f0, 1,...g.
I. Let s return to the example described earlier: X represents a stock price. ω X 0 (ω) X 1 (ω) X 2 (ω) X 3 (ω) ω 1 1 0, 50 1 0, 50 ω 2 1 0, 50 1 0, 50 ω 3 1 2 1 1 ω 4 1 2 2 2
II We had determined that the ltration containing the information revealed by the process at each time is F 0 = σ fx 0 g = f?, Ωg, F 1 = σ fx 0, X 1 g = σ ffω 1, ω 2 g, fω 3, ω 4 gg, F 2 = σ fx 0, X 1, X 2 g = σ ffω 1, ω 2 g, fω 3 g, fω 4 gg, F 3 = σ fx 0, X 1, X 2, X 3 g = σ ffω 1, ω 2 g, fω 3 g, fω 4 gg.
Recall that: III ω X 0 (ω) X 1 (ω) X 2 (ω) X 3 (ω) ω 1 1 0, 50 1 0, 50 ω 2 1 0, 50 1 0, 50 ω 3 1 2 1 1 ω 4 1 2 2 2 We won t sell our stocks today (t = 0) but we will sell them as soon as the price is greater than or equal to 1. The random time representing that situation is τ (ω 1 ) = 2, τ (ω 2 ) = 2, τ (ω 3 ) = 1 and τ (ω 4 ) = 1.
IV Such a random variable truly is a stopping time since fω 2 Ω : τ (ω) = 0g =? 2 F 0, fω 2 Ω : τ (ω) = 1g = fω 3, ω 4 g 2 F 1, fω 2 Ω : τ (ω) = 2g = fω 1, ω 2 g 2 F 2, fω 2 Ω : τ (ω) = 3g =? 2 F 3.
Recall that: V ω X 0 (ω) X 1 (ω) X 2 (ω) X 3 (ω) ω 1 1 0, 50 1 0, 50 ω 2 1 0, 50 1 0, 50 ω 3 1 2 1 1 ω 4 1 2 2 2 Let s now consider the random time τ modelling the following situation: we ll buy stock as soon as it enables us to make a pro t later. Such a random value takes values τ (ω 1 ) = 1, τ (ω 2 ) = 1, τ (ω 3 ) = 0 and τ (ω 4 ) = 0. τ is not a stopping time since fω 2 Ω : τ (ω) = 0g = fω 3, ω 4 g /2 F 0.
VI Intuitively, the time τ when one makes a decision is a stopping time if the decision is made based on the information available at that time. In the case of stopping times, using a crystal ball is prohibited.
I transformation Theorem Let (Ω, F), be a measurable space such that Card (Ω) < and equipped with the ltration F = ff t : t 2 f0, 1,...gg. If the random variables τ 1 and τ 2 are stopping times with respect to the ltration F, then τ 1 ^ τ 2 min fτ 1, τ 2 g and τ 1 _ τ 2 max fτ 1, τ 2 g are also stopping times.
II transformation Proof of the theorem. If τ is a stopping time, then 8t 2 f0, 1,...g 8k 2 f0, 1,...g, fω 2 Ω : τ (ω) tg 2 F t. fω 2 Ω : τ 1 (ω) ^ τ 2 (ω) kg = fω 2 Ω : τ 1 (ω) k or τ 2 (ω) kg = fω 2 Ω : τ 1 (ω) kg {z } 2F k 2 F k. Exercise. Prove the above result for τ 1 _ τ 2. [ fω 2 Ω : τ 2 (ω) kg {z } 2F k
I time Let (Ω, F) be a measurable space such that Card (Ω) < and equipped with the ltration F = ff t : t 2 f0, 1,...gg. X = fx t : t 2 f0, 1,...gg represents a stochastic process adapted to that ltration. Let B R a subset of the real numbers. We de ne the time until the stochastic process X rst enters the set B as τ B (ω) = min ft 2 f0, 1,...g : X t (ω) 2 Bg. If it happened that the path t! X t (ω) never hits the set B then we de ne τ B (ω) =.
II time Theorem The random variable τ B is a stopping time. Proof of the theorem. Since Card (Ω) <, then 8t 2 f0, 1,...g, X t can only take a nite number of values. Let s denote them by x (t) 1 <... < x (t) m t. 8t 2 f0, 1,...g,
fω 2 Ω : τ B (ω) = tg III time = fω 2 Ω : X 0 (ω) /2 B,..., X t 1 (ω) /2 B, X t (ω) 2 Bg! = = t 1 \ fω 2 Ω : X k (ω) /2 Bg k=0 0 t B \ 1 [ n @ k=0 x (k) i /2B 0 B \ @ [ n 2 F t x (t) i 2B ω 2 Ω : X k (ω) = x (k) i ω 2 Ω : X t (ω) = x (t) i \ fω 2 Ω : X t (ω) 2 Bg 1 o C A 1 o C A
IV time since t\ 1 [ n ω 2 Ω : X k (ω) = x (k) o i k=0 x (k) i /2B {z } 2F k since X is adapted. {z } 2F k F t since F k is a σ algebra {z } 2F t since F t is a σ algebra. \ [ n ω 2 Ω : X t (ω) = x (t) o i x (t) i 2B {z } 2F t since X t is F t measurable. {z } 2F t since F t is a σ algebra.
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