Chapter I
:
Central limit
Theorem
Chapter Theorem limit Central I : tasting characterizations of - - PowerPoint PPT Presentation
Chapter Theorem limit Central I : tasting characterizations of weak convergence of equivalent Proof of distribution distribution follows theme conveyance of Observation : weak expectation interact with limits : of how IE ( " 7 Xn
:
Central limit
Theorem
Proof
characterizations of weak convergence
distribution
Observation : weak conveyance of distribution follows theme
how
limits
interact with
expectation
:
MCT : if
{ Xn) TX
Then
"nm IE (Xn) =
IE ( "7 Xn)
weak
when
: it rn
⇒p
then
"
"
nm Emf)
"
for
nice f
Recall
to
end
semester :
Thou ( Central
limit
Theorem)
be
iid with
finite
mean
m and
finite
variance
v .
let
Sn
Then
Marion , ← stand:L ,
distribution
Good
news :
CLT
wants
us
to
prove
some
weak
convergence
distributions , and
we
knew
a- lot
ways to
measure
that kind
conveyance
.Bad
news
:
we
need
a
new
tool
to pave at
Good
news :
the
new
tool
is
easy
to
define
and
relates
"
nicely
" to
weak
conveyance
.Defy ( characteristic faction)
let X
be
a
random
variable
. thenits
characteristic function
is
40×4)
=
E- Leith )
= Efoosltx)) t i E-faultXD
.Nate :
, It)
determined
by the
distribution
X
,
so
if X
and Y
are
identically
d.striated
, Thu
01×14--4, It)
.What's
so "characteristic "
about the
characteristic fndien ?
Thun
( Continuity Theorem)
Tt
X, Xi
, Xz ,
. . .be
random variables
. ThenL ( Xn) ⇒ LIX)
iff
"nm Kult)
, It )
.Pf ( half
continuity
Theorem )
EM
in
'
Assume
L ( Xn ) ⇒ L( x) .
Since
cosine is
a
bounded
Coatney
function , def
'n of
weak conveyance gives
"
nm IE ( cos ltxn ))
=
"I
Eun ( cos It xD
= IEµ ( cos ItxD
=
IE ( cos ItX ))
Same
idea
follows
with
Kam
IE ( smltxn))
Huwe
we
have
"
nm Ox. It)
IE ( cos ItXa)) t
i IEC Scutt Xa))
"
IE ( eos HX)) t
i IE( Sia LtX))
= oh
, It)
.with
continuity theorem
in
hand,
strategy for
at
is
to
show ky4s÷H)=4ma
First
:lets
compute
duraotwsti Anakin
9*14 for Hyun,"
,
By
definition
:
OHH
it × )
it")
±
§
, it"
e-
"%
Hd x)
= f: ein tr e
dx
.We
need
to compute this .
let
's
compute It
:
, Hif
. It!e ""
'
"
dx]
= !! # Lei
"
e
""
dx )
= f? ixeitx # e
dx
.ieitx
du = -teitx
lets integrate
by
parts !
re:* e-
"
"
10×14
'
"× # e
k
d x
.= ÷
, cite
.
t t.ci
* e- "
"
dx
=
O
, It)
.So :
,
' H)
⇐
talk
, It))
=
So
we
get
(after
exponentially )
with this
function
in
hand,
strategy for
at
is
to show uy9s÷÷lH=
How
will
we
do this ?
We'll
need
Them
( Taylor
expansion
for 0×14)
x
be
a
random
variable
with
E- ( Xk)
is finite
.Then
01×4)
IEC Xi)
t
junk
where
ttlk (junk) → o
as
t -so
.Pf
( sketch)
serie, expansion for exponential
,
H)
unit Z
"
details
to address
① deal
with complex
numbers
(easy
to
do)
② not
all
terms ar
non -
neg ,
so
we
might
not get
countable
twenty
e't E
.So : truncate .
D
Pf
( CLT
, sketch form )
Theall :
X, Xy
. . .are
iid
with
mean
m
and variance v
.Want :
fr
Su
we
have
L ( sniff) ⇒
Mmm,
.Equivalent
to :{ %y÷H=et%
Deline
Then
Etxi)
and NAH
.Then
the
random
variable
In
should
have
hum &
It)
so,
same ravttforgn
,So :
WLOG
assume m:O
and
r
Want :
. It)
→
e
Observe
Olsyr
⇐ ( eitxtrneitx.la
. . - eitxnlrn)i#t=
⇐ ( eitxtra) # feit Mra)
. . . E feit 'T")
") )
"
"
So
:
"in 0% It)
=
"
nm ( Ox.lt/rnl) "
n
Tay
hnm ( ⇐ I
t litY Elk.]tli%EH:) trunk)
=
"
nm (
I + O
, tjunk)
"
Naw
I
'
Hospital ( after
tu kg
aatml
leg
and writing
a
you hid this
equals
e
BEBO