4 1

S e s s i o n 3

We w e r e l a s t looking a t s o m e

simple examples o f probability

spaces . . .

4 2Recall...

E x ¥ : S = { o , I }I = { d , {03 , { 1 3 , 1 3

A

P l a t e {to; I:{:3,O

,A = ¢

I, A = L ,

where o s d s t .

n.be P I A ) sa t i s f i e s t h e a x i oms o f probability.

E ¥ L e t S b e any sample4 3

space.

L e t F I S ) = { lo, s }

By axioms o f probability onlyo n e probability m e a s u r e w i l lw o r k w i t h t h i s even t space

PCA) = {1, A= L

O,A = ¢

I t fo l l ow s f r o m t h e ano ins 4 4

o f probability t h a t f o r any S a n d

F I L ) P (ol) = 0 .

Proofs: S = S UOl

i pdisjoint Shot = ¢

a x . 3

①( S ) = P (Sud) = PCS) + ① lo l)

¥2 I = I t ①lol) ± Pldt-0.

n.b.es PCI) = l e P l a t 4 5

Proofs: S = A V IA 9

disjoint

P C S ) = P I AV E ) ¥3 PCA) + PTA)↳I = PCA) t PTA)

⇒ PCA) = I - P IA ) ,

B r i e r y : 4 6

• Random Experiment ⇒ random outcome.

• S i s t h e s e t o f a l l possible outcomes.

• E v e n t s a r e described a s subsets o f L

• I f A C S , w e say event A o c c u r s

i f t h e r a n d om outcome W E S i s

i n A C i . e . , w e A ) . (n.b.wteA-i.E.h.gg)

• Events A o f interest a r e co l l ec tedi n t h e e v e n t space F C S ) ( o - f i e l d )F C S ) s a t i s f i e s c e r t a i n c losure properties.

4 7

• T h e probability t h a t a n event

A- E F C S ) o c c u r s i s given by PCA),

P f . ) : F C S ) → I R

satisfying t h e axioms o f probability.

Wenowtakeamoredetailedloot 4 8

a t S , F , a u d #

S a m p i e s :

Intuitively listing o f a l l

possible ou t c omes

Mathematically: A n abstractspaces

o r universal .

C x ± : S = { w k ; k =L , . . . , n } 4 9

specificallyS = { 0 , 1 3 ,

S = { H , T 3

& = { 1 , 2 , 3,4, 5 , 6 3

E x i t : A countable sample space

& = { w e ; K = 1 , 2 , 3 , . . . }specifically ,

t h e following a r e countable

se t s I N = { 1 , 2 , 3 , . . . }I t = { o , l , 2 , . . . }€ = { . . . , - I , O , l , . . . }

Z i s countable because i t 5 0

c a n b e put i n t o o n e - t o - o n e

correspondence w i t h 1 N :

Z :. . . - 3 , - 2 , - l , O , l , 2 , 3 , . . .

I n nI I I b 1 1

IN : 7 , 5 , 3 , 1 , 2 , 4,6, . . .- a - y

i . T h e integers a r e countable.

between the elements of the set and the set of natural numbers.Notice, the infinite case is the same as giving the elements of the seta waiting number in an infinite line :).

And here is how you can order rational numbers (fractions in otherwords) into such a "waiting line." It's just for positive fractions, butafter you have these ordered, you could just slip each negativefraction after the corresponding positive one in the line, and placethe zero leading the crowd. I like this proof because it is so simpleand intuitive, yet convincing.

The numbers in red/blue table cells are not part of the proof but justshow you how the fractions are formed. You start at 1/1 which is 1,and follow the arrows. You will encounter equivalent fractions, whichare skipped.

If you think about it, all possible fractions will be in the list. Forexample, 145/8793 will be in the table at the intersection of the145th row and 8793rd column, and will eventually get listed in the"waiting line."

See also

What are rational numbers?

What are irrational numbers?

T h e Rat ional Numbers a r e countab le 5 1#

3 1 I t s i t s i n t h i s array a t-•Consider 2 5 5 t h e in tersect ion o f r o w 3 1

a n d c o l u m n 2 5 5 .

¥ 3 S = ( L , B ) , o r, pc.IR 5 2

a - p

= { x e l R : d < x < p }T h i s i s a n uncountable s e t .

[ d i p ] = { X E R : L E X E R } #i s uncountable

1 R = C-os, t o o ) i s uncountable.

* L e t 2 = 0 a n d @=/. t h i s m e a n s

[ o , i ] i s uncoun tab l e .

E x - 4 T a k e S t o b e t h e 5 3-

s e t o f a l l K- d am vectors

who's coordinates (elements)

c o m e f r o m anyo f t h e examples

E x . 1 through E x . 3 .

( c a l l t h e s e t A )k

S = A x A x . - - × A = I T A✓ i s t

k - f o l dExample: - s t a t e o f a contro l system (AER)

- Length k binary codeword 1 A ={0113)- Length k d i s c r e t e - t o n e signal

( A = R o r G ) .

H ow c a n w e s e e t h a t t h e 5 4

s e t [ d i p ] i s uncountable?

←Prove t h i sLet's l o o k a t [ 0 , I ] . i s uncountable.

Suppose [ 0 , 1 ] i s countable. Then w e

c a n l i s t t h e elements o f [ o , , ]i n l - l correspondence with 1 N :

O . d'" di" d j". . . dj'#c- {0,13X . value o f t he(2) 1 2 (z ) a . .

O . d, d , d , K-th l is tedn u m b e r

O.§%3). . .

I , d,"'I l Z , -

',

'a j e t 2 J

wha t a b o u t t h e n u m b e r 5 5

T h i s argument i s0 . I [ I s . . . ca l l e d t h e

( k )"Canetordiagonalization"

whe re I , I dk , 6 = 1 , 2 , . . . .

b u t Jae {0 ,1 3,K e l , 2 , . . .

T h e numbe r O . I I I i s n o t

i n t h e coun tab le l i s t . Howe ve r

0 . F I . . . = II, II,-E [ o i l ]⇒ con t r a d i c t i o n ! [ o i l ] i s n o tcountable

[ o , ' ] i s uncoun t a b l e

EX . se A space o f countable 5 6

sequences d r a w n f r o m

examples 1 - 3 .

S = A x A x A x . . . + A x . . .• I N

= I T A = I T A = A

i e I N i c e

Exampley: I f w e t h i n k o f A - { H i t }t h e n S - I I , { H i t } ,

a n d a typical sequence i s o f

t h e f o r m* T H A T . ' '

E v e n i f A i s d i s c r e t e (even finite) 5 7

I w i l l b e uncountable.

why? B e c a u s e e a c h sequencec a n b e mapped t o a

n u m b e r i n [ o , I ] , which

L e ti s uncountable ( I - 1

correspondence)

I NS = A whe re A - {0 ,13

A typical e l em e n t i n S wou ld l o o k

l i ke < a , , a z , a > , . . . 7 , d i f { O i l }

• a , 9 2 9 3 . . - = 1¥, 9¥ c - [0,1 ]Abinary point

s o S - A " c a n be put i n t o 5 8

o n e - t o - o n e correspondence w i t h

[ o , I ],wh i c h i s uncountable

I N

i . S = A i s uncountable.