Section 23
Stochastic Processes. Brownian Motion.
Given a set T and a probability space (Π, , P) a stochastic process is a function F X (�) = X(t, �) : T Π R t × ∃ such that for each t T , X : Π R is a random variable, i.e. a measurable function. A stochastic process ⊂ t ∃ is often defined by specifying finite dimensional (f.d.) distributions PF = ( Xt t F ) for all finite subsets F T. Kolmogorov’s consistency theorem ensures the existence of a probabilitL {y space} ∩ on which the process is defined∧ if f.d. distributions are consistent, i.e. if for
F1 F2, PF1 = PF2 . √ RF1 � One can also think of a process as a random variable from Π �with values in RT = f : T R , because for T � { ∃ } a fixed � Π, Xt(�) R is a (random) function of t T. In Kolmogorov’s theorem, given consistent f.d. ⊂ ⊂ ⊂ T distributions a process was defined on a probability space (R , BT ) where BT was a cylindrical ε-algebra T F F generated by cylindrical sets B R \ for Borel sets B in R . In general, some very natural sets such as ×
sup Xy > 1 = Xt > 1 t T { } � ∩ ⎪ t T �∩ may be not measurable and some additional assumptions on a process might be helpful.
Definition. If (T, d) is a metric space then a process X is called sample continuous if for all � Π, t ⊂ Xt(�) C(T ) - a space of continuous function on (T, d). Xt is called continuous in probability if Xt Xt0 in probabilit⊂ y whenever t t . ∃ ∃ 0 Example. Let (Π, P) = ([0, 1], ϕ) where ϕ is the Lebesgue measure and let T = [0, 1]. Define
Xt(�) = I(t = �) and Xt�(�) = 0. F.d. distributions of both processes are the same because for a fixed t,
P(Xt = 0) = P(Xt� = 0) = 1.
However, P( X is continuous ) = 0 but for X � this probability is 1. { t } t
Definition. Let (T, d) be a metric space. Process Xt is called measurable if X (�) : T Π R t × ∃ is jointly measurable on (T, ) (Π, ) where is the Borel ε-algebra on T. B × F B
95 Lemma 44 If (T, d) is a separable metric space and Xt is sample continuous then Xt is measurable.
1 Proof. Let (Sj )j 1 be a measurable partition of T such that diam(Sj ) . For each non empty Sj take a ∗ → n point tj Sj and define ⊂ n X (�) = X (�) for t S . t tj ⊂ j Xn(�) is obviously measurable on T Π because for any Borel set A on R, t ×
n (�, t) : X (�) A = � : X (�) A S . t ⊂ tj ⊂ × j j 1 � � �∗ � � X (w) is continuous and X n(�) X (�). Therefore, X is also measurable. t t ∃ t Consider a metric space (C(T ), ) of continuous functions C(T ) on a metric space (T, d). Let be a ε-algebra of Borel sets on C(T ) generated|| · ||↓ by open balls B
Bg(π) = f C(T ) : f g < π { ⊂ || − ||↓ } and let S = B C(T ) : B - cylindrical ε-algebra T ⇐ ⊂ BT � � be an intersection of cylindrical ε-algebra on RT with the space of continuous functions.
Lemma 45 If (T, d) is separable space then = S . B T
Proof. Let us first show that ST . An element of cylindrical algebra that generates cylindrical ε-algebra is given by ∧ B T F F B R \ for a finite F T and for some Borel set B R . × ∧ ∧ Then T F B R \ C(T ) = x C(T ) : (xt)t F B = α(x) B × ⊂ ∩ ⊂ ⊂ F ⎩ � � where α : C(T ) R � is a finite �dimensional projection such that α(x) = (x�t)t F . Pro�jection α is, obviously, ∃ ∩ continuous in norm and, therefore, measurable on Borel ε-algebra generated by open sets in || · ||↓ B � · �↓ norm. It implies that the above set α(x) B which proves that S . ⊂ ∧ B T ∧ B
Let us show that ST . Let�T � T be�a countable dense set. Then by continuity any closed π-ball in C(T ) can be writtenB as∧ ∧
f C(T ) : f g π = f C(T ) : f(t) g(t) π ST . ⊂ || − ||↓ → ⊂ | − | → ⊂ t T � � ⎩∩ �� � This finished the proof.
Remark. The meaning of this lemma is that, since Borel ε-algebra on C(T ) is equal to restriction of cylindrical ε algebra to C(T ), any law on C(T ) is completely determined by its finite dimensional distribu tions.
Brownian motion. Brownian motion is a sample continuous process Xt on T = R+ such that for each 2 t 0 the distribution of Xt is centered Gaussian, X0 = 0, EX1 = 1, Xt and Xs Xt are independent for ∗ 2 − t < s and Xt and (Xs Xt) = (Xs t). If we denote by ε (t) = Var(Xt) the variance of Xt then from the above properties L − L − t 1 ε2(nt) = nε2(t), ε2 = ε2(t) and ε2(qt) = qε2(t) m m � � for all rational q. Since ε2(1) = 1, for all rational q, ε2(q) = q and by sample continuity, ε2(t) = t for all t 0. For all s < t, ∗ EX X = EX (X + (X X )) = s = min(t, s). s t s s t − s
96 As a result one can give an equivalent definition.
Definition. Brownian motion is a sample continuous Gaussian process X for t [0, ) such that t ⊂ ↓ EXt = 0 and covariance Cov(Xt, Xs) = min(t, s). Without the requirement of sample continuity the existence of such process would follow from Kol mogorov’s consistency theorem since finite dimensional distributions are well defined. However, we need to prove that one can construct a sample continuous version of the process. Let us start with a simple estimate.
Lemma 46 If f(c) = (0, 1)(c, ) is the tail probability of the standard normal distribution then N ↓ 2 c f(c) e− 2 for all c > 0. → Proof. We have 2 2 2 1 ↓ x 1 ↓ x x 1 1 c f(c) = e− 2 dx e− 2 dx = e− 2 . ∩2α → ∩2α c ∩2α c �c �c If c > 1/∩2α then f(c) exp( c2/2). If c 1/∩2α then a simpler estimate gives the result → − → 2 2 1 1 1 c f(c) f(0) = exp e− 2 . → 2 → −2 ∩2α → � � � �
Theorem 51 There exists a continuous version of Brownian motion.
Proof. It is enough to consider Xt on the interval t [0, 1]. Given Xt that has finite dimensional distributions of a Brownian process but is not necessarily contin⊂uous, let us define for n 1 ∗ n Vk = X k+1 X k for k = 0, . . . , 2 1. 2n − 2n − n Clearly, Var(Vk ) = 1/2 and using the above lemma,
n 1 1 n 1 n+1 2 − P max Vk 2 P V1 2 exp . k | | ∗ n2 → | | ∗ n2 → − n4 � � � � � � The right hand side is summable for n 1 and by Borel-Cantelli lemma ∗ 1 P max Vk i.o. = 0. (23.0.1) k | | ∗ n2 �� � � t n t ↓ j j Given t [0, 1] and its dyadic decomposition t = j=1 2j for tj 0, 1 let us define t(n) = j=1 2j so that ⊂ ⊂ { } � n � Xt(n) Xt(n 1) 0 Vk : k = 0, . . . , 2 1 . − − ⊂ { } ⇒ { − } Then a sequence Xt(n) = 0 + (Xt(j) Xt(j 1)) − − 1 j n →�→ converges almost surely to some limit Zt because by (23.0.1), with probability one
2 Xt(n) Xt(n 1) n− | − − | →
for large enough (random) n n (�). By construction, Z = X for dyadic t. Therefore, Z agrees with ∗ 0 t t t Xt on a dense set in [0, 1]. If we can show that Zt is continuous then finite dimensional distribution of Zt will be the same as of X so Z is a continuous version of Brownian motion. Take any t, s [0, 1] such that t t ⊂
97 n k m t s 2− . If t(n) = 2n and s(n) = 2n , then k m 0, 1 . As a result, Xt(n) Xs(n) is either equal | − | → 2 | − | ⊂ { } | − | to 0 or one of V and X X n− . Finally, | k| | t(n) − s(n)| → Z Z Z X + X X + X Z | t − s| → | t − t(n)| | t(n) − s(n)| | s(n) − s| 1 1 1 c + + → l2 n2 l2 → n l n l n �∗ �∗ which proves continuity of Zt. On the set of probability zero where (23.0.1) does not hold we set Zt = 0.
Definition. A sample continuous Gaussian process B for t [0, 1] is called a Brownian bridge if t ⊂ EB = 0 and EB B = s(1 t) for s < t. t t s − Such process exists because if Xt is Brownian motion then Bt = Xt tX1 is a Brownian bridge since for s < t − EB B = E(X tX )(X sX ) = s st ts + st = s(1 t). s t t − 1 s − 1 − − − Notice that B0 = B1 = 0. Empirical c.d.f. and Kolmogorov-Smirnov test. Let us give some motivation for the upcoming lectures. Suppose that u1, . . . , un are i.i.d. uniform random variables on [0, 1]. By law of large numbers, for any t [0, 1], empirical c.d.f. ⊂ n 1 I(u t) P(u t) = t n i → ∃ 1 → i=1 � converges to true c.d.f. and by CLT
1 n X = ∩n I(u t) t (0, t(1 t)). n,t n i → − ∃ N − i=1 � � � Stochastic process Xn,t is called an empirical process. The covariance for any t, s
EX X = E(I(u t) t)(I(u s) s) = s ts ts + ts = s(1 t) n,t n,s 1 → − 1 → − − − − is the same as covariance of a Brownian bridge and by multivariate CLT finite dimensional distributions of empirical process (Xn,t)t F (Bt)t F (23.0.2) L ∩ ∃ L ∩ converge to f.d. distributions of a Brownian� bridge. Ho� wever,� we would� like to show the convergence in some stronger sense that would imply convergence of some general functionals of the process.
Kolmogorov-Smirnov test could be one motivation for this. Suppose that i.i.d. (Xi)i 1 have continuous 1 n ∗ distribution with c.d.f. F (t) = P(x t). Let Fn(t) = n i=1 I(Xi t) be an empirical c.d.f. Then it is easy to see that → → � sup ∩n Fn(t) F (t) =L sup Xn,t t | − | t | |
because F (Xi) have uniform distribution. In order to test whether the data come from distribution F we would like to know the distribution of the above supremum or, as an approximation, a distribution of its limit. (23.0.2) suggests that, possibly,
(sup Xn,t ) (sup Bt )? L t | | ∃ L t | |
Since we proved that Bt is sample continuous, it defines a law on a metric space (C[0, 1], ). Even though 1/2 �·�↓ Xn,t is not continuous, its jumps are of order n− so it has a ”close” continuous version Xn,t� . Since is a continuous functional on C[0, 1], the answer to the above question would follow if we can prove conv�ergence·�↓
Xn,t� L Bt in the setting of a metric space (C[0, 1], ). Remark 2 at the end of Lecture 19 shows that ∃ � · �↓ we only need to prove uniform tightness of Xn,t� because (23.0.2) already identifies a Brownian motion as a unique possible limit. In the following lectures we will address the question of uniform tightness on this metric space.
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