Independence and Bernoulli Trials. Sharif University of Technology 2 Independence A, B independent...

Independence and Bernoulli Trials

Sharif University of Technology 2

Independence

A, B independent implies: are also independent.

Proof for independence of :

Events A and B are independent if

).()()( BPAPABP

BAABBAAB )(

, BAAB

),()()())(1()()()()( BPAPBPAPBPAPBPBAP

)()()()()()()( BAPBPAPBAPABPBAABPBP


Application

let :

Then

Also

Hence it follows that Ap and Aq are independent events!

" the prime divides the number "pA p N

" the prime divides the number ".qA q N

1 1{ } , { }p qP A P A

p q

1{ } {" divides "} { } { }p q p qP A A P pq N P A P A

pq

Example


Some Properties of Independence

Let P(A)=0, Then:

Thus: P(AB) = P(A) P(B) = 0

The event of zero probability is independent of every other event

Other independent events cannot be ME

Since: 0)( ,0)( BPAP

.0)( ABP

AAB

,0)(0)()( ABPAPABP


If and Ais are independent,

Application in solving number theory problems.

Application

A family of events {Ai} are said to be independent, if for every finite sub collection

Remark

Application

,,,,21 niii AAA

n

ki

n

ki kk

APAP11

).(

,321 nAAAAA

nAAAA 21

.))(1()()()(11

21

n

ii

n

i

in APAPAAAPAP

,))(1(1)(1)(1

n

iiAPAPAP


Example

Each switch remains closed with probability p.

(a) Find the probability of receiving an input signal at the output.

Three independent parallel switches

Let Ai = “Switch Si is closed” and R = “input signal is received at the output”. Then

,)( pAP i

).()()()( );()()( 321321 APAPAPAAAPAPAPAAP jiji .31 i

Solution

.321 AAAR .33)1(1)()( 323

321 ppppAAAPRP


Example – continued

Since any event and its complement form a trivial partition,

Thus:

Another Solution

).()|()()|()( 1111 APARPAPARPRP

,1)|( 1 ARP2

321 2)()|( ppAAPARP

,33)1)(2()( 322 pppppppRP

Note:The events A1, A2, A3 do not form a partition, since they are not ME.

Moreover, .1)()()( 321 APAPAP


Example – continued

From Bayes’ theorem

Also, because of the symmetry :

Solution

.33

32

33

)1)(2(

)(

)()|()|(

2

2

32

211

1pp

pp

ppp

ppp

RP

APARPRAP

).|()|()|( 321 RAPRAPRAP

(b) Find the probability that switch S1 is open given that an input signal is received at the output.


Repeated Trials

A joint performance of the two experiments with probability models (1, F1, P1) and (2, F2, P2).

Two models’ elementary events: 1, 2

How to define the combined trio (, F, P)?

= 1 2

Every elementary event in is of the form = (, ).

Events: Any subset A B of such that AF1 and B F2

F : all such subsets A B together with their unions and complements.


Repeated Trials

In this model, The events A 2 and 1 B are such that

Since

the events A 2 and 1 B are independent for any A F1 and B F2.

So, for all A F1 and B F2

Now, we’re done with definition of the combined model.

).()( ),()( 2112 BPBPAPAP

,)()( 12 BABA

)()()()()( 2112 BPAPBPAPBAP


Generalization

Experiments with

let

Elementary events

Events in are of the form

and their unions and intersections.

If s are independent, and is the probability of the event in then

,,,, 21 n ,1 , and niPF ii

,,,, 21 n .ii

nAAA 21

n 21

,ii FA

)( ii APiA iF

1 2 1 1 2 2( ) ( ) ( ) ( ).n n nP A A A P A P A P A

iA


Example

A has probability p of occurring in a single trial. Find the probability that A occurs exactly k times, k n in n

trials.

The probability model for a single trial : (, F, P)

Outcome of n experiments is :

where and

A occurs at trial # i , if Suppose A occurs exactly k times in .

,,,, 021 n i 0

Solution

.Ai


Example - continued

If belong to A,

Ignoring the order, this can happen in N disjoint equiprobable ways, so:

Where

1 2, , ,

ki i i

.)()()( )()()(

}) ({}) ({}) ({}) ({}) ,,,,, ({)(21210

knk

knk

iiiiiiii

qpAPAPAPAPAPAP

PPPPPPnknk

,)()(

) trials" in timesexactly occurs ("

01

0knk

N

ii qNpNPP

nkAP

k

n

kkn

n

k

knnnN

!)!(

!

!

)1()1(


Bernoulli Trials

Independent repeated experiments Outcome is either a “success” or a “failure”

The probability of k successes in n trials is given by the above formula,

where p represents the probability of “success” in any one trial.

,,,2,1,0 ,

) trials" in timesexactly occurs (")(

nkqpk

n

nkAPkP

knk

n


Example

Toss a coin n times. Obtain the probability of getting k heads in n trials.

“head” is “success” (A) and let Use the mentioned formula.

In rolling a fair dice for eight times, find the probability that either 3 or 4 shows up five times ?

Thus

).(HPp

. } 4or 3either { success"" 43 ffA

3

1

6

1

6

1)()()( 43 fPfPAP

Example

Example

5 38

8( ) (5) (1/ 3) (2 / 3) 0.068282

5k n k

n

nP k p q P

k


Bernoulli Trials

Consider a Bernoulli trial with success A, where Let:

As are mutually exclusive,

so,

.)( qAP ,)( pAP

. trials" in soccurrence exactly " nkX k

.1)( 10 nXXXP

ji XX ,

n

k

n

k

knkkn qp

k

nXPXXXP

0 010 .)()(

,)(0

knkn

k

n bak

nba

1)( nqp


Bernoulli Trials

For a given n and p what is the most likely value of k ?

Thus if or

Thus as a function of k increases until

The is

the most likely number of successes

in n trials..2/1 ,12 pn

Fig. 2.2

)(kPn

k

.1!

!)!(

)!1()!1(

!

)(

)1( 11

p

q

kn

k

qpn

kkn

kkn

qpn

kP

kPknk

knk

n

n

),1()( kPkP nn pknpk )1()1( .)1( pnk

)(kPn pnk )1(

],)1[(max pnk


Effects Of p On Binomial Distribution


Effects Of n On Binomial Distribution


Example

In a Bernoulli experiment with n trials, find the probability that the

number of occurrences of A is between k1 and k2.

.)()(

) " and between is of sOccurrence ("

2

1

2

1

211 1

21

k

kk

knkk

kkkkkk qp

k

nXPXXXP

kkAP

Solution


Example

Suppose 5,000 components are ordered. The probability that a part is defective equals 0.1. What is the probability that the total number of defective parts does not exceed 400 ?

".components 5,000 among defective are parts " kYk

.)9.0()1.0(5000

)()(

5000400

0

400

040010

kk

k

kk

k

YPYYYP

Solution


If kmax is the most likely number of successes in n trials,

or

So,

This connects the results of an actual experiment to the axiomatic definition of p.

pnkpn )1(1)1( max

,max

n

pp

n

k

n

qp

.lim pn

km

n


Bernoulli’s Theorem

A : an event whose probability of occurrence in a single trial is p.

k : the number of occurrences of A in n independent trials

Then ,

i.e. the frequency definition of probability of an event and its

axiomatic definition ( p) can be made compatible to any degree of accuracy.

.2

n

pqp

n

kP

n

k


Proof of Bernoulli’s Theorem

Direct computation gives:

Similarly,

.)(

!)!1(

)!1(

!)!1(

!

)!1()!(

!

!)!(

!)(

1

11

0

111

0

1

1

10

npqpnp

qpiin

nnpqp

iin

n

qpkkn

nqp

kkn

nkkPk

n

inin

i

inin

i

knkn

k

n

k

knkn

kn

.)!1()!(

!

)!2()!(

!

)!1()!(

!)(

22

1

210

2

npqpnqpkkn

n

qpkkn

nqp

kkn

nkkPk

knkn

k

knkn

k

n

k

knkn

kn


Proof of Bernoulli’s Theorem - continued

Equivalently,

Using equations derived in the last slide,

,)( toequivalent is 222 nnpkpn

k

.)()()( 22

0

22

0

2 nkPnkPnpk n

n

kn

n

k

.2

)(2)()()(

2222

22

00

2

0

2

npqpnnpnpnpqpn

pnkPknpkPkkPnpk n

n

kn

n

kn

n

k

.

)( )()(

)()()()()()(

22

222

22

0

2

nnpkPn

kPnkPnpk

kPnpkkPnpkkPnpk

nnnpk

nnnpk

nnnpk

nnnpk

n

n

k


Proof of Bernoulli’s Theorem - continued

Using last two equations,

For a given can be made arbitrarily small by letting n become large.

Relative frequency can be made arbitrarily close to the actual probability of the event A in a single trial by making the number of experiments large enough.

As the plots of tends to concentrate more and more around

.2

n

pqp

n

kP

2/ ,0 npq

maxk,n )(kPn

Conclusion


Example: Day-trading strategy

A box contains n randomly numbered balls (not necessarily 1 through n)

m = np (p<1) of them are drawn one by one by replacement The drawing continues until a ball is drawn with a number larger than the

first m numbers.

Determine the fraction p to be initially drawn, so as to maximize the

probability of drawing the largest among the n numbers using this strategy.


Day-trading strategy: Solution

Let drawn ball has the largest number among:

all n balls,

the first k balls is in the group of first m balls, k > m.

= where

A = “largest among the first k balls is in the group of first

m balls drawn”

B = (k+1)st ball has the largest number among all n balls”.

Since A and B are independent , So,

P (“selected ball has the largest number among all balls”)

stk kX )1(

kX ,A B

. 1

1

)()()(kp

knp

nkm

nBPAPXP k

1 1

1 1( ) ln

ln .

n n n n

k npnpk m k m

P X p p p kk k

p p


Day-trading strategy: Solution

To maximize with respect to p, let:

So, p = e-1 ≈ 0.3679

This strategy can be used to “play the stock market”. Suppose one gets into the market and decides to stay up to 100 days. When to get out?

According to the solution, when the stock value exceeds the maximum among the first 37 days.

In that case the probability of hitting the top value over 100 days for the stock is also about 37%.

0)ln1()ln( pppdpd


Example: Game of Craps

A pair of dice is rolled on every play,- Win: sum of the first throw is 7 or 11

- Lose: sum of the first throw is 2, 3, 12

- Any other throw is a “carry-over”

If the first throw is a carry-over, then the player throws the dice repeatedly until:- Win: throw the same carry-over again

- Lose: throw 7

Find the probability of winning the game.


Game of Craps - Solution

kkkkk

PkCBPkCPkCBPBPP

TPTPTPQ

TPTPP

10

7,4

10

7,42

1

1

)|()()|()(

9

1

36

1

36

2

36

1)12()3()2(

9

2

36

2

36

6)11()7(

B: winning the game by throwing the carry-overC: throwing the carry-over

P1: the probability of a win on the first throw

Q1: the probability of loss on the first throw



3/15/211/511/55/23/1

1098654

)|(

,...,3,2,1,

6/11

1

1

1

k

j

jkkk

jkk

kk

a

k

rpkCBPa

jrp

pr

The probability of win by throwing the number of Plays that do not count is:

The probability that the player throws the carry-over k on the j-th throw is:



492929.0495

134

9

2

495

134

36

3

3

1

36

4

5

2

36

5

11

5

36

5

11

5

36

4

5

2

36

3

3

1

21

10

7,42

PP

paPkk

kk

the probability of winning the game of craps is 0.492929 for the player.

Thus the game is slightly advantageous to the house.


Game Of Craps Using Biased Dice

Now assume that for each dice,

- Faces 1, 2 and 3 appear with probability

- Faces 4, 5 and 6 appear with probability

If T represents the combined total for the two dice, we get

16 0,

16

24

2 25

2 26

7

16

1 136 6

1 136 6

{ 4} {(1,3),(2,2),(1,3)} 3( )

{ 5} {(1,4),(2,3),(3,2),(4,1)} 2( ) 2( )

{ 6} {(1,5),(2,4),(3,3),(4,2),(5,1)} 4( ) ( )

{ 7} {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1

p P T P

p P T P

p P T P

p P T P

2

2 28

2 29

210

11

136

1 136 6

1 136 6

16

16

)} 6( )

{ 8} {(2,6),(3,5),(4,4),(5,3),(6,2)} 4( ) ( )

{ 9} {(3,6),(4,5),(5,4),(6,3)} 2( ) 2( )

{ 10} {(4,6),(5,5),(6,4)} 3( )

{ 11} {(5,6),(6,5)} 2(

p P T P

p P T P

p P T P

p P T P

2) .



This gives the probability of win on the first throw to be

and the probability of win by throwing a carry-over to be

Thus

Although perfect dice gives rise to an unfavorable game,

T = k 4 5 6 7 8 9 10 11

pk = P{T = k} 0.0706 0.1044 0.1353 0.1661 0.1419 0.1178 0.0936 0.0624

1 ( 7) ( 11) 0.2285P P T P T

210

24 77

0.2717k

k kk

p

p pP

1 2{winning the game} 0.5002P P P



Even if we let the two dice to have different loading factors and (for the situation described above), similar conclusions do follow.

For example,

gives:

1 2

1 20.01 and 0.005

{winning the game} 0.5015.P

Euler, Ramanujan and Bernoulli Numbers


Euler’s Identity

S. Ramanujan in (J. of Indian Math Soc; V, 1913): if are numbers less than unity where the subscripts are the series of prime numbers,

then

The terms above are arranged so that the product obtained by multiplying the subscripts are the series of all natural numbers

This follows by observing that the natural numbersare formed by multiplying

primes and their powers.

2 3 5 7 11, , , , ,a a a a a 2,3,5,7,11,

2,3,4,5,6,7,8,9, .

2 3 2 2 52 3 5 7

2 3 7 2 2 2 3 3

1 1 1 11

1 1 1 1

.

a a a a aa a a a

a a a a a a a a


Euler’s Identity

This is used to derive a variety of interesting identities including the Euler’s identity.

By letting This gives the Euler identity

The right side can be related to the Bernoulli numbers (for s even). Bernoulli numbers are positive rational numbers defined through the power

series expansion of the even function

Thus if we write Then

2 3 51/ 2 , 1/ 3 , 1/ 5 ,s s sa a a

1

prime 1

1(1 ) 1/ .ss

p np n

2 cot( / 2).x x

1 2 3 4 51 1 1 1 16 30 42 30 66, , , , , .B B B B B

2 4 6

1 2 3cot( / 2) 12 2! 4! 6!x x x x

x B B B


Euler’s Identity

Alse we can obtain :

This is another way to define Bernoulli numbers Further

which gives

Thus

62 431 21

1 2 2! 4! 6!x

B xx x B x B xe

2 1 2 1 2 420 0

2 2 2 2 2

14 ( )

2(2 )! 1 1 1 1(2 ) 1 2 3 4

n n x xxn

n n n n n

xe

B n dx x e e dx

n

2 42 4

1 1

1/ ; 1/ etc.6 90k k

k k

22

21

(2 )1/

2(2 )!

nn n

nk

BS k

n

Independence and Bernoulli Trials. Sharif University of Technology 2 Independence A, B independent...

Documents

Transcript of Independence and Bernoulli Trials. Sharif University of Technology 2 Independence A, B independent...