II of large Number Lattin in probability almost convergence - - PowerPoint PPT Presentation

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II of large Number Lattin in probability almost convergence - - PowerPoint PPT Presentation

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Chapters : Proofs of the laws II of large Number Lattin in probability almost convergence convergence sure - - " weak " strong " " cow corn Two tests for stray convergence - O almost sure conveyance . ) - X

slide-1
SLIDE 1

Chapters

II

: Proofs of the

laws

  • f large Number
slide-2
SLIDE 2

Lattin

almost

sure

convergence ⇒

convergence

in probability

  • " strong

corn

" " weak

cow

"

Two tests for stray

convergence

① Fe > 0

All Xu

  • X Ise
  • i. o
.)
  • O ⇒ almost sure conveyance

② VE>O E P( IXn

  • Xl >E) as

⇒ almost sure convergence

slide-3
SLIDE 3

Newsprint

① Prove

S LLN

slide-4
SLIDE 4

Then I WLLN)

If

X , ,Xn,

. -

is

a sequence of independent random

variables and

there

exist

m , VER so

That IECXi)

  • m

and

W ( Xi) EV

fer

all

i

, then the

Ruden variables

Sn

  • I ( X , t
  • - - t Xa) converge to

m

in probability

:

V-E > 0

,

"I

IP ( Isn

  • ml >E) =D .

PI (Proof

relies

  • n Chebyshev
.)
slide-5
SLIDE 5

Observe

:

IECSN)

  • lE( tallit
  • -it Xd)
  • THEN.lt.at#Xnl/
= M .

Non)

  • NC 's

t-etxnll-ntwlxit.tk)

need'T ta ( NIX ,)t

. - -TNCXNDEI

indef

n

Chebyshev

says

:

IP ( Isn

  • ml > e)

a- *l s

E

  • Ea
.

Take

"I

can

both

sides

.

Ed

.
slide-6
SLIDE 6

Then ( SUN)

If

X , ,Xn,

. -

is

a sequence of independent random

variables and

there

exist

m , a EIR so

That IECXI)

  • m

and

⇐(Ni

  • m)

") Ea

fer

all

i

, then the

Ruden variables

Sh

  • I ( X , t
  • - - tha) converge to

m

almost surely

:

PC

"ash

  • m )
  • 1

Pt

WLOG

we assume

m

  • O .
slide-7
SLIDE 7

Its

'

a

fact

that

for all

ZEIR

we

have

z2sz4tl

.

Hence

LEI Xi ' )

= N (Xi) E

IE ( Xi't 1) E

att

.

Strategy

: let

Ta

  • X , t
  • txn

, and

we'll study

⇐ ( Tn

")

.

We'll

show

⇐( Tn' ) isn't

"too

big

" ,

from

which

we'll

be

able to use

Markov

to

prove

The

result

.
slide-8
SLIDE 8

IE ( Tn 4)

= IE( Nit
  • - -tXn)(X,t
  • - tXn )(X, t - utXn)(X,t - -t Xu))
  • E ( Ih Xi

"

t ¥ ,.bg?,i2Wt..?..KfixiXk

Vakil

+ §

;

4 Xi' Xi

t

e

24 Xi Xi Xk Xe)

  • IE

IE (Xi

") t

6

ECxi.IE Hilt 12 ?g

,.IE#7leYxi%Efxd

+ 4.7

, let xi7Ex 24 EE

Exile

slide-9
SLIDE 9

⇐ ( Tn

")

  • IE
,

IECXI

") t 6¥ ECxi.IE (Xi)

← ?

a

t

6 ÷

,

Cath Catt )

=

na

t

6 (2) (att) '

=

nut Kalla

  • Math'

← n' a

t

3 n

  • ( at, )'

s

na k

← K

  • atlatl

'

⇐( Tn 4) En

  • K
slide-10
SLIDE 10

So

: IEC Tn 4) E n

  • k

Markov

says

:
  • 5ptsfrAn

lP(l9nl)=lP(HHt-tXal)=lPlIlTnl)hh

IP ( I SnIs E)

= IP (

'

a Itn Is E)

  • IP ( Itn I > ne)
=

( Tn

" s

n

" E

")

e ⇐n!I

E nn¥g

,

= Kant

.

So

: q

H ( Isak E) E E E

. In. c is

.
slide-11
SLIDE 11

By The

Borel

  • Cartelli

version

  • f

the test

fer

almost

sure

conveyance

,

we

get

p##t

.

Sn converge

to

almost

DUE

surely

. .

Bh