6 . I

s e s s i o n 6

PMFs a n d PDFSYo u s h o u l d k n o~6 . t l

P M F s - P D F s yBinomial / U n i f o r m -

Geome t r i c - Exponential -

Po i s s on v Gauss ian o r

-

' i t 6 . 3÷÷÷i÷÷i÷ia..ee.T h a t B h a s JacTred mayIneffinaussomethingaboutwhether

F E : Ior÷÷÷÷÷i÷÷i¥¥

I f B occured, A mu s t have.I f B occurred, A cannot h a v e .

I n general, knowing t h a t B h a s€ 1 µ3§ • ← a r e d may change your

belief

t h a t A h a s occurred.

Def-ni G i v e n ( S , F , P ) a n d 6 . 4

A , B EFTthe cond i t i ona l

prqgityofAF.tn}-

( " A given B " ) i s

PCAIBEFA.IT#=assamiTgBI.

hits. I f ① ( B ) = o , t h e n P l a n B ) = 0 , a n d t h i s

leaves u s w i t h P C A I B I = % ,which i s undefined.

6 . 5n i b ± . I f A C B , t h e n

P IA IB ) = PtpA%B#= ¥114

z PCA)

1 I f B C A , t h e n

H A I B ) = P¥I}I=Pp÷3) = L

Feet

I . I f A R B = ¢ , t h e n 6 . 6

P t - ftp.t#=Pff&=fFs,= I

I P f A I B ) = PlAp%B↳

z P l a n B )f o r

O <PCB) E ' t.

f a c t : I f 1.11 ( f r om ( S ,F , P ) ) 6 . 7

t h e n P C . I B ) i s a l s o a v a l i d"i÷÷÷÷÷÷÷÷÷"probabilitye a s u r e f o r any

Proofs: (exercise) verify t h ea x i o m s

probtability h o l d f o r P C I .

(S,F,P)%¥% ( S , F , P C . 113))n i b . ( S , F , PC .113) i s a v a l i d probability

s p a c e .( l o o k a t examples o f c o nd i t i o n a l probability i n Papoulis.)

tayormulathetotalprobability 6 . 8

%;:÷I"÷÷÷÷.pa.us#nL e t A , B E F .

T h e "p (Alps)=P¥% - - -

l ' t )

" " dPCBIA) = PlpA¥}l - - -

( 2 )

F r o m ( z ) w ehaves

¥ B ) = ①(BIA) PFA) - - -C z ' )

substituting Cz ' t i n t o 1 1 ) 6 . 9

P I A 1131 = PfApf}#=PCBlpttfgpcal-i.IE#s=PT3.?;:I?.I.u.

Ihe ton tap robab i l i t y : 6 . 1 0

L e t { A , , . . . , A n } b e a partition o fS

,and l e t B E F .

( n i b . A , , . . . , A , E F )

T h e n

P C B ) = P (BIA,)PCA,)tPlB/Az)P(Az)

t . I T # A nT PA n I

€ i ¥÷ .

6 . 11

Proofs: PCB) = P ( B h s )

= p ( B n (II.A i ) )= P (¥, ( B h a i ) )= ¥2, P f B h a i )

= ¥! PCBIA;)PlAi)L B

t a ye s t h e o re n 6 . 1 2

G i v e n ( S , F , P ) , a s s u m e t h a t

A [ ¥ # a partition o f f .

By Bayes formula

PCAi l t s ) = PlBlApifpµ

By t h e t o t a l prob. l a w

PlB)=¥¢PCBlAi)PCAµ

6 . 1 3suppose A m e { A , , . . . , A n }B y Bayes fo rmu l a

P lAmlB) = PlB¥A§lAm§

.÷÷÷::÷÷÷i . t . i s# . i i I iBaynestheoreumi L e t ( S , F , P ) be 6 . 1 4

a probability space T E A , , , , , , A n }b e a partition o f S . t e t h a t

¥ ¥ ¥ I , an d a s s u m e B e e f .

"±=÷÷÷i÷÷i÷.m= l ,c iµ

Proofs: Wejustdid

Statisticalindependency 6 . 1 5

Def f : Given ( S , F , P ) , l e t A . B E F .

T h e n theevents A-113a r e statistically independenti f a n d only i f ( i f f )

P\PlA)PCBµµ

f a c t : I f A a n d B a r e 6 . 1 6

s ta t i s t ica lpendent,t h r e a d ,

A a n d B , and A- a n d B - .

ProotffordB: w e w a n t t o s h o w t h a t

①( A n t s ) = ①( A ) . PCB) given PlAnB¥AlPlB).

P ( A ) = P lan ts ) + P l a n t s )

⇒ 1 ¥ 5 1 = P I A ) - P CAAB)= PCA) - PCA)PCB)= P ( A ) ( I B )

= p(A)PCB# •

Statisticadependence: 6 . 1 7

3 even t s A , B , C E F a r e

statistically independent i f f

P l a n B n c ) = PCA ) PCB ) P ( c ) ]P ( A R B ) = P C A ) P ( B )

P ( A n c ) = P I A ) P l c ) )① ( B n c ) = ① ( B ) P ( c ) y

6 . 1 8I n general, n e v e n t s a r e

statistically independent

i n t e r s e c t r o u s f a c t o r a s:÷i÷÷÷÷i:"PTA.si#Ajz)PAj)PAjz)a.iPlAjk)

where K = 2 , 3 , . . . , I .

T h e r e a r e 2 "combinat.in#IIII-lk.such

T h e r e a r e 2 " - ( h t t ) s u c h combinations.. 6 . 1 9

(8141+12)+13)t..- I ; )i n '

I t n

X t ( I t n ) = (no)I . I " t (Y) I ' . I n - ' t . . . t (2)I . t o

= £! (rd) Ikea-k=?'u%Yi:'

= Ts( I t 1 ) " = 2 4

1<=0

i . x = 2 " - ( h t t )•

Combineperiment 6 . 2 0

Suppose w e have t w o random experiments

( I , , F , , P, ) and ( h , Fa , P z ) .

F a n t t ocombine-themto

f o r m a "super experiment" w i t hprobability p a t e n t s ) , where

S - S , x s £E l em e n t s I f I a r e ordered pairs o f

o u t c ome s ⇒ E S , we-S, a n d BELL.

E t : Exp. 1 : f l i p a c o i n S, = { H i t } 6 . 2 1

Exp. 2 : R o l l a d i e S z = { 1 , 2 ,3 , 4 ,5,63

T h e comb i ned experiment h a s sample space

S = S , X S -= {(H, l ) , CH,2 ) , CH,3 ) , (H,4),CH,51, (H,b),

(T, l ) , CT,2 ) , CT,31,174), IT,5 ), CT,6)}

n d 1 1 1 = 1 1 , 1 . 1 1 1

6 . 2 2A n e v e n t i n o u r n e w experimentw i l l b e a s u b s e t o f t h e sample space

b = L , x S~I f A C S , a n d B C S >

t h e n C = A x B C S

i s a n e v e n t i n t h e n e w even t space#

(e.g . A - E H },B = { 3 , 6 3

A x ① = { ( H i s ) , CH,6 ) }B u t n o t a l l events c a n b e wri t ten

a s C a r t e s i a n products.

O u r e v e n t space I w i l l b e 6 . 2 3

t h e o - f i e l d generated by a l l

Cartesian products:

I = o f {AxB : V A E F , and tBEFz3)o r

T h i s w i l l b eo-euen-tp.ie?.nseIIN

combined experiment I S , F , P ) ,

T h e r e a r e e v e n t s t h a t c a n n o t b e writtena s t h e ca r te s i an product o f events f romJT a n d F z , b u t t h e c l o s u r e propertieso f t h e o - f i e l d produce t h e m :

Exampled S z

SIS,X t z

£1.• " £ \ [ µ

6 . 24

• 00.0 • • ooo -H • )• o o . . . . !

• .

C = { I t } x { 2 , 5 3 = {CH,2 ) , LH,5 ) 3

D = EI,-13=23 = { ( H R ) , CT,2 1 3

E = C U D = { I T ,2 ) , CH,2 ) , (H is ) }which c a n n o t b e wr i t t en a s a Cartesian

product.

How do w e assign probabilities 6 . 2 5

t o t h e combined experiment ( S F , P ) ?

F o r consistency w i t h P, a n d P z ,

P o f C S , F , P ) must-satisfyconsistencyPFAX.bz) = p , CA) ,

Y A e f ,conditions

{P(S , xB ) = Pz CB), V.B E Fa

-

How d o w e de f i ne P c c ) f o r

o t h e r e v e n t s C c - F

6 . 2 6

- w e k n o w t h a t P f c ) m u s t

satisfy t h e consistency condit ions

- ① ( c ) m u s t satisfy t h e a x i o m so f probability

- o t h e r t h a n t h a t w e can't saym u c h w i t h o u t making assumptions

§ I s t h e r e a l i n k o r mechan ism

between t h e t w o experimentsw e a r e combining?