Section 23

Stochastic Processes. .

Given a set T and a probability space (Π, , P) a 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 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 X for t [0, ) such that t ⊂ ↓ EXt = 0 and 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 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 , 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 . 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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