The CLT without characteristic functions.
Easy version: assume third momemt is finite.
Theorem. Let X1, X 2 ,. be i.i.d. rvs. Assume
E( X ) 0, E( X 2 ) 1 and E( X 3 ) .
Step 1. It is enough to show that
Ecfw_ f (
Then
Sn
Z , where Z ~ N(0,1)
n
Sn
(
) E
Inequalities that are related to the proofs of CLT
1. Let zi , wi ,
i 1,., n be complex numbers and let i maxcfw_ zi ,  wi ,1 .
n
n
n
i 1
i 1
i 1
Then
n
k 1
 zi wi  [  zi wi ] k
Proof.
n
n
 zi wi 
i 1
i 1
n
ni
i 1
k 1
( z k ) z ni 1 (
n
ni
k n
STT881 FALL 2011_ Final examSOLUTION
Some facts:
()
1.
(
()
2.
)
cfw_
Name:
( ).
, where
(
where
)
3.  e ix 1  2  x 
()
4. 

5.
for each real number x
for each
,
complex number
6. Let cfw_ a n , m : 1 m n , n 1 be triangular array of complex numb
X n converges in distribution to X ( X n X ) if Eg( X n ) Eg( X )
g CB (R) (all bounded and continuous functions, g : R R )
Definition:
for all
1. First observation: It is enough to check the definition for all g CB (R) that are
uniformly continuous.
Proo
Shlomo Levental, Michigan State University
Remarks on Durretts Probability: Theory and examples, 3rd edition.
The following Lemma will replace 3 different calculations in the
text:
(1) In section 1.7 (Etmadi SLLN, Lemma (b) and (c), p. 5556)
(2) In secti
STT881 MIDTERM EXAM SOLUTION FALL 2012.
Problem 1.
Let be a random variable and assume ( )
.
(

)
1. Prove that
as
.
( 
2. Prove that
)
as
NAME:
.
3. The following is an integration by parts (IBP) formula that we have studied: Let
be numbers and let (
Let cfw_ X k be a sequence of independent and symmetric ( i.e. X k X k in
distribution) random variables. We denote S n
n
Xk .
k 1
Levy inequality.
t 0 we have
P( max S k t ) 2 P(S n t )
1 k n
Proof.
Observe that cfw_ max S k t
1 k n
n
cfw_
max S j t
Proof of the Inversion formula. Let
characteristic function ( c.f.). Let
numbers. The Inversion formula is
be a r.v. and let
where
(1)
Proof. We can consider
where
represents its
are real
and
as independent. Obviously we get
. Here are 3 easy observations
Assignment 1, due Friday 9/7/2012
Problem 1.3 page 438.
Assignment 2, due 9/21/2012
Problems 4.1, 4.2, p. 457
Hints for 4.2: (i) Prove first that for g
am1Bm
m0
( i.e.
Bm ,
m0
where cfw_Bm is a partition of
Bm Bk 0 for m k ) and am 0 we have
g d am
Let cfw_ X k be a sequence of independent random variables. We denote
n
Sn X k ,
k 1
*
Sn
max  S k ,
1k n
*
X n max  X k 
1k n
Hoffman Jorgensen( HJ) inequality.
n 1, t 0 we have
*
*
*
P( S n 4t ) P( X n t ) P 2 ( S n t )
Proof.
*
P( S n 4t )
*
*
*
A remark on Uniform Integrability.
Definition . cfw_ X n is Uniformly Integrable (UI) if
where ( M ) sup E ( X n  1cfw_  X n  M ) .
(M ) M 0 ,
n
The following is an equivalent definition:
Theorem. cfw_ X n is UI iff
(i) sup E ( X n ) , and
n
(ii