Martingale Difference Central Limit Theorem Yichen Zhou May 9, 2016
Intuition Why martingale difference CLT. ◮ Statisticians rely on CLTs to make inference. ◮ CLTs we have seen before all require independence. ◮ Martingale difference CLT extends the scope by taking into account the dependence.
Setup ◮ Martingale array: { S ni , F ni , 1 ≤ i ≤ k n } zero-mean, square-integrable martingales for each n ≥ 1 . ◮ It can be derived from an ordinary martingale { S n , F n , 1 ≤ n } by setting k n = n, F ni = F n , S ni = s − 1 n S i 1 2 . s n = ( var ( S n ))
◮ Martingale difference: X ni = S ni − S n,i − 1 . ◮ Notice E ( X ni | F n,i − 1 ) = 0 . ◮ Conditional variance and squared variation of S ni : i i V 2 � E ( X 2 U 2 � X 2 ni = nj | F n,j − 1 ) , ni = nj . j =1 j =1
Martingale CLT Theorem (Martingale CLT I) Follow the notations above. Suppose η 2 is an a.s. finite r.v., and p p � X 2 → η 2 , max | X ni | − → 0 , − ni i i � � X 2 E max < M < ∞ , ni i F ni ⊆ F n +1 ,i . Therefore � d S nk n = X ni − → Z, i where Z has the characteristic function � � exp( − 1 E ( e itZ ) = E 2 η 2 t 2 ) .
Martingale CLT Theorem (Martingale CLT II) Follow the notations above. Suppse � � E max | X ni | → 0 , i p � X 2 → σ 2 , − ni i then d → N (0 , σ 2 ) . S nk n −
Proof of Martingale CLT Main idea of the proof: ◮ Truncate the martingale array by a stopping time. ◮ Show that the truncation results in a negligible difference. ◮ Show that the truncated array converges in distribution to normal.
Truncation ◮ WLOG, σ 2 = 1 . ◮ Define the stopping time t � X 2 T n = inf { t : ni > 2 } ∧ k n . i =1 ◮ Consider the new martingale array { ˜ S ni = S n,i ∧ T n , F ni , 1 ≤ i ≤ k n } with its differences Z ni = S n,i ∧ T n − S n, ( i − 1) ∧ T n = X ni ✶ { � i − 1 j =1 X 2 nj ≤ 2 } = X ni ✶ { T n >i − 1 } .
Truncation is Negligible ◮ Recall k n → σ 2 = 1 . p � X 2 − ni i =1 ◮ After truncation, P ( ˜ S nk n � = S nk n ) = P ( X n,i � = Z n,i for some i ≤ k n ) = P ( T n ≤ k n − 1) � k n � � X 2 ≤ P ni > 2 → 0 . i =1 ◮ Therefore k n p p ˜ � Z 2 S nk n − S nk n → 0 , − − → 1 . ni i =1
Convergence ◮ Suffices to show − t 2 � � E exp( it ˜ S nk n ) → exp . 2 ◮ Taylor expansion − x 2 � � exp( ix ) = (1 + ix ) exp 2 + r ( x ) , where | r ( x ) | < | x | 3 . Thus k n exp( it ˜ � S nk n ) = exp( itZ nj ) j =1 k n k n k n − t 2 � � � exp Z 2 . = (1 + itZ nj ) nj + r ( tZ nj ) 2 j =1 j =1 j =1
Martingale CLT Lemma For n ≥ 1 , let { U n } , { V n } be random variables satisfying the following conditions: p 1. U n − → a , 2. { V n } and { V n U n } are uniformly integrable sequences, 3. EV n → 1 . Then EV n U n → a. ◮ Based the lemma, let k n k n k n − t 2 � � Z 2 � . V n = (1+ itZ nj ) , U n = exp nj + r ( tZ nj ) 2 j =1 j =1 j =1
Martingale CLT ◮ Proof of the lemma: p V n ( U n − a ) is u.i.. Suffices to show V n ( U n − a ) − → 0 . P ( | V n ( U n − a ) | > ǫ ) ≤ P ( | U n − a | > ǫ/K ) + P ( | V n | > K ) → 0 . ◮ | V n U n | ≤ 1 , so { V n U n } is u.i.
Martingale CLT ◮ To show k n k n − t 2 − t 2 � � p � Z 2 � U n = exp nj + r ( tZ nj ) − → exp , 2 2 j =1 j =1 p notice � k n i =1 Z 2 − → 1 and ni � � k n k n � � � � � � ≤ | t | 3 | Z nj | 3 r ( tZ nj ) � � � � j =1 j =1 � � k n � ≤ | t | 3 | X nj | 3 j =1 k n ≤ | t | 3 max p � X 2 | X nj | − → 0 . nj j j =1
Martingale CLT ◮ To show k n T n � � V n = (1 + itZ nj ) = (1 + itX nj ) j =1 j =1 is u.i., by | 1 + iz | 2 ≤ exp( z 2 ) , � | 1 + itX nT n | | V n | = | 1 + itX nj | j<T n t 2 � X 2 (1 + | t || X nT n | ) ≤ exp nj 2 j<T n ≤ exp( t 2 )(1 + | t | max | X nj | ) . j Recall E (max j | X nj | ) → 0 , so max j | X nj | is u.i. So is V n .
Martingale CLT ◮ To show EV n → 1 , notice k n � EV n = E (1 + itZ nj ) j =1 � k n � � � = E E (1 + itZ nj ) F n,k n − 1 � � j =1 � k n − 1 � = E E (1 + itZ nk n | F n,k n − 1 ) (1 + itZ nj ) j =1 k n − 1 � = 1 . = E (1 + itZ nj ) j =1
Reference Peter Hall Martingale Limit Theory and Its Application . Academic Press, 1980. Sethuramas Sunder A Martingale Central Limit Theorem. http://math.arizona.edu/˜sethuram/notes/wi mart1.pdf
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