Comprehensive Analysis on Donsker Theorem and Its Proof

Vadym Omelchenko

Definition ( ) = + + + 2 Given the random X variables , ,...., , i.i.d. 0, and .... S 1 2 n 1 2 n n define we follows as n 1 1 ( ) = + 1.1) ( , ) ( ) ( ) X t S n t n t + 1 n n t n t n n

Donsker Theorem Suppose random the variables { independen are } identicall and t y n 2 distribute with d mean and 0 finite, = positive = variance : 2 E var , 0 X n n Then the random functions satisfy .1) (1 by defined X n D W. n

Proof The proof of theorem this consists of two steps : finite all for it prove We 1. - dimensiona l distributi ons S n tightness the prove We 2. of mesures the {S } n Hence 3. it will proven be by the following theorem : Theorem 1. Let P measures prob be P , n on (C, - alg(C)) . If finitedime the nsional distributi ons of P converge , weakly to those of and P, if {P is } n n ti ght, then Pn P

Proof Consider t when the case he dimension one. is We must prove that D ( ) X s W n s Since 1 1 P ( ) 0 X s S + 1 n ns ns n n Theorem X 2. X ( ) If and then , 0 . X , Y Y X n n n n 1 D S W ns s n

Proof But this a is direct consequenc e of Lindeberg the - Levy CLT ns fact that the and / . n s Theorem 3. (Linderber g - Levy) ( ) 2 If , ,.... are 0 then , = iid , 1 2 1 n ( ) 1 0 , D N k n 1 k

Proof Consider now two time points and with prove to are We . s t s t ( ) ( ) D ( ), ( ) , , X s X t W W n n s t = D D If and , 0 then X X P{X D } h(X ) h(X) n h n ~ ( ) ( ) D ( ), ( ) ( ) , , X s X t X s W W W n n n s t s enough to is It prove that ( ) 1 1 ( ) D (2) ( ), ( ) ( ) , , S s S t S s W W W n n n s t s n shown n because it was as above 1 P ( ) 0 as n X S s n n n

Proof independen are left on the components the Since independen by the t, ce of the n following the and rem limit theo central by the follows (2) , theorem : Theorem 4. ' '' ' '' ' ' '' '' separable, is S If then if and only if and P P P P P P P P . n n n n A set of treated be can points time more or, three in the same way, and hence the finite - dimensiona distributi l converge ons properly .

Proof of the tightness following the use we tightness prove To lemma : LEMMA Let 1. ,..., independen be + = i random t variables with mean and 0 1 m S + 2 finite variance Put ; ... . 1 m Then S . (3) P max i S 2 P ( 2 ) m m i m m

Proof (Proof of the Lemma) Consider t he sets = max j E S m S i j i i Clearly, ( ) + (A) max j 2 P S m P S m j m m ( ( ) ). 1 m- 1 = i 2 P E S m i m

Proof (Proof of the Lemma) ( ) Since S m and S 2 m together imply i m independen S S 2 m Chebyshev' by follows it inequality s and the m i asumed ce of most at is (A) sum that the : ( ) E i m 2 1 1 ( ) m- m- m 2 i ( ) E = i = i = (B) S S 2 m P P P m i i 2 1 1 m 1 1 ( ) 1 1 m- m- m 2 i ( ) E ( ) E = i = i max P P P S m i i i 2 2 1 1

Proof (Proof of the Lemma) Hence both (A) and (B) imply (3) which is the affirmation of the theorem. QED

Proof our theore For m we have : S ( ) max i 2 2 P S n P n n i n ( ) if 2 2 then 2 / hence 2 max i 2 S P S n P n n i 2 n Chebyshev and CLT By inequality we have 8 3 S N P n P E N n 3 2 2

Proof Hence follows : lim sup max i P S n i 2 n n sufficient for large. ly Tightness now following the by the follows theorem : Theorem 5. is sequence The X n there with , integer an and 1, tight iffor n each n positive exist a such n 0 that, if then , 0 P max S S n + k i k 2 holds for all k.

Proof Having proved the assertion of this theorem for finite- dimensional distributions and having proved the tightness we have proved the theorem. QED

Application of Donsker Theorem Donsker' qualitativ this has theorem s e is D interpreta tion : says that, if small, then particle a subject to X W n . . . independen displaceme t nts , 1 successive at . . . , times 2 , will, 2 appear to afar, from viewed perform approximat Brownian a ely motion .

Unit Dimension {-1,+1} N=20 N=60 N=1000

Application of Donskers Theorem More important than this qualitative interpretation is the use of Donsker's theorem to prove limit theorems for various functions of the partial sums

Application of Donskers Theorem = ( ) sup ( continuous a is ) function on C, implies by the theorem 6 h x x t X W D t n that sup ( ) sup X t W D t n t t obvious The relation 1 = sup 0 t ( ) max i X t S n i n n 1 Now implies 1 (* * *) max i sup S W D i t t n n technique The we shall use to find this latter distributi compute to is on limiting the distributi on of max max i S i n = with the , 1 in an easy special case iid : probabilit same y 0.5 n

Random Walk and Reflection Principle max S 0 i n i = + = (*) max 2 P put S a P S a P S a 0 i n i n n = Assume a , 0 max M = S 0 i j i j = + , , P and M a P S , a P M a S a P M a S a n n n n n n : = P M follow a if S a prove P S a n n n = (*) will we , , P M a S a P M a S a n n n n

Hence we have: = a n n 1 = + = (*) max 2 P S P S a P S a 0 i n i n n n n n Central By Limit Theorem a S n we have N P : (**) , 0. P n

Combining the results (**) and (***) we have: 2 1 = sup exp , . 0 2 P W u du t t 2 2 0 1 2 1 max i exp , . 0 2 P S u du i 2 2 n n 0 assumption the drop can we principle invariance By the 1 = = = = 1 1 . P P n n 2 Replace by : , 0 ( ) 2 iid n

() is . If h continuous continuous or except fot points having Wiener ( ) limiting the find can then we 0 measure distributi on of h if we X n ( ) W know distributi the on of and vice versa. h Donsker' theorem s is often called 'invarianc of e principl e' 'functiona the l central limit theo rem'

Functions of Brownian M. Paths Maximum Minimum and = = Let inf and m n sup and = let m W M W t t t = min 0 x , max 0 x S M S i n i i n i n ( ) mapping The : : inf ( ), sup ( ), ) 1 ( x 3 f C t x t R t t continuous is everywhere that so 1 ( ) , , ( , , ) m M S m M W D 1 n n n n

Functions of Brownian M. Paths 2 ( P = k = ) 1 ) 1 ) 1 + sup ( 2 ( P W b k b N k b k t t

The Arc Sine Law

The Arc Sine Law

The Arc Sine Law = For in (x) , let supremum the be (x) of for which 1] [0, in those t ( ) ; 0 x C h x t 1 let Lebegue the be (x) h measure of for which 1] [0, in t those ( ) ; 0 h x t 2 And let Lebegue the be t measure of those t in [0, ] h (x) 3 1 for which ( ) ; 0 x ( ) is = Then U at which t time the is Wiener he W path last passes through h = T = h W W 1 ( ) W ( ) find 0, the total W amount of time spends above and 0, T h V W 2 3 amount the is of time spends (T, U, V, above W in the 0 ) interval [0, We . ] shall the joint distributi on of . 1

The Arc Sine Law ( because ( ) ( ) ( ) ) ( ) , , , ) 1 ( n , , , h X h X h X X T U V W D 1 2 3 1 n n n () , . () , . () mappings the measurable are . contunous and h h h 1 2 3 except on set a of Wiener measure 0.

The Arc Sine Law ( ) is = = maximum the = 1 , i for which n 0 T n h X i S 1 n n i ( ) ( ) = = , , / . U n h X V n h X X S n 2 3 n n n n n n 1 n 1 n 1 n 1 ( ) , , , , , , , T U V S T U V W D 1 n n n n n ( ) distributi the find will We on of iid , , , using basic random walk T U V W 1 ( , 0 ) generalize the and it to 2 2 , 1 = n ,.. n

The Arc Sine Law 1 2 ds u = = . 1 arcsin , 0 P U u u u 1 ( ds ) s s 0 1 2 t = = . 1 arcsin , 0 P T u t t 1 ( s ) s 0 Hence / where, n number the is such od which greater th are an 0, distributi sine arc has limit. in on T T S n n i

Example(1) Normal and Student-t = Normal Law, 200 n = Normal Law, 20 n = Student - Law, t 50 n = Student - Law, t 200 n

Example (2) = = ( ) X Random walk, 500 Random walk, 20 n n = = = sgn , ( ) 1 , 0 ( C ), 200 Y X L X n 5 = difference Exp , 50 n ( ) ( ) = i = 5 [ ], L X RayleyDist = i = 5 [ = ], , L Y RayleyDist X Y 2 , 1 ,.., , 50 n n

Brownian Bridge Random Gaussian requiremen with Process = ts : and E W 0 o t = E W W 1 ( s ) if s t. o o t t s directly be can Bridge Brownian The constructe means by d of the Wiener Process : = W = W W , 0 t 1. o t t t t = b Note that W sup P W with 0 = b probabilit 2 + k y ) 1 1. o o 1 0 P sup W = 2 2 1 ( , . 0 2 e b o k k b t t = 1 2 1 , . 0 2 W e b o b t t