probability and statistics for engineering

PROBABILITY AND STATISTICS FOR ENGINEERING

Hossein Sameti

Department of Computer EngineeringSharif University of Technology

The Weak Law and the Strong Law of Large Numbers

Timeline…

Bernoulli, weak law of large numbers (WLLN)

Poisson, generalized Bernoulli’s theorem

Tchebychev, discovered his method

Markov, used Tchebychev’s reasoning to extend Bernoulli’s theorem to dependent random variables as well.

Borel, the strong law of large numbers that further generalizes Bernoulli’s theorem.

Kolmogorov, necessary and sufficient conditions for a set of mutually independent r.vs.

1700

1800

1866

1909

1926

Weak Law of Large Numbers

s: i.i.d Bernoulli r.vs such that

: the number of “successes”in n trials.

Then the weak law due to Bernoulli states that

i.e., the ratio “total number of successes to the total number of trials”

tends to p in probability as n increases.

iX

,1)0( ,)( qpXPpXP ii

nXXXk 21

.2

npq

pP nk

Strong Law of Large Numbers

Borel and Cantelli:

this ratio k/n tends to p not only in probability, but with probability 1.

This is the strong law of large numbers (SLLN).

What is the difference?

SLLN states that if is a sequence of positive numbers converging to zero, then

From Borel-Cantelli lemma, when this formula is satisfied the events

can occur only for a finite number of indices n in an infinite sequence,

or equivalently, the events occur infinitely often, i.e., the

event k/n converges to p almost-surely.

}{ n

. 1

nnpP n

k

n

pnk

= n nknA p

Proof

Since

we have

and hence

where

pPpPnnkpnpk nk

nk

n

n

k

)()( 4444

0

4

440

4 )()(

n

kpnpkpP

n

kn

nk

knkn

iin qpkXPkp

k

n

1

)(

444 nnpkpn

k

Proof – continued

since

So we obtain

,3

)]1(3[))(1(3)(

)()()1(3)()()1(4)(

)(}){(

2

233

2

1 1

2

1 1

3

1

4

1 1 1 1

4

1

pqn

pqnnnpqnnpqqpn

YEYEnnYEYEnnYE

YYYYEYE

j

n

i

n

jij

n

i

n

ji

n

ii

n

i

n

k

n

j

n

lljki

n

ii

can coincide withj, k or l, and the second variabletakes (n-1) values

ni 1

12/1 ,133)( 22333 pqpqqpqpqp

2 4

3kn

pqP p

n

0

}{}{1

44

10

4 )( )()()()(

n

ii

n

ii

n

kn pXEnpXEkpnpk

Proof – continued

Let so that the above integral reads

and hence

thus proving the strong law by exhibiting a sequence of positive numbers

that converges to zero and satisfies

3/ 21/8 3/ 2 1

1 1

1 13 3 1

3 (1 2) 9 ,

( )n n

knP p pq pq x dx

n n

pq pq

1/81

n

. 1

nnpP n

k 1/81/n n

What is the difference?

The weak law states that for every n that is large enough, the

ratio is likely to be near p with certain probability that

tends to 1 as n increases.

It does not say that k/n is bound to stay near p if the number of trials is increased.

Suppose is satisfied for a given in a certain number of trials

If additional trials are conducted beyond the weak law does not

guarantee that the new k/n is bound to stay near p for such trials.

There can be events for which for in some regular manner.

.2

npq

pP nk

1( ) / /

n

iX n k ni

.0n

,0n

,/ pnk 0nn

What is the difference?

The probability for such an event is the sum of a large number of very small probabilities

The weak law is unable to say anything specific about the convergence of that sum.

However, the strong law states that not only all such sums converge, but the total number of all such events:

where is in fact finite!

. 1

nnpP n

k

pnk /

Bernstein’s inequality

This implies that the probability of the events as n increases becomes and remains small,

since with probability 1 only finite violations to the above inequality takes place as

It is possible to arrive at the same conclusion using a powerful bound known as Bernstein’s inequality that is based on the WLLN.

}{ pnk

.n

Bernstein’s inequality

Note that

and for any this gives

Thus

)( pnkpnk

,0 .1))(( pnke

n

k

knkknpnk

knkn

pnkknpnk

n

pnk

knkkn

qpe

qpe

qppP nk

0

))((

)(

))((

)(

}{

Bernstein’s inequality

Since for any real x,

We can obtain

2xx exe

.

)()(22222

2222

eqepe

epqeqpqepepq

pqpq

.}{2 nnepP n

k

.)(

)()(}{0

npqn

knpkqn

k

n

qepee

qepeepP kn

nk

Bernstein’s inequality

But is minimum for and hence

and hence we obtain Bernstein’s inequality

This is more powerful than Tchebyshev’s inequality as it states that the

chances for the relative frequency k /n exceeding its probability p tends to zero exponentially fast as

nn 2 2/

0. ,}{ 4/2

nepP nk

4/2

}{ nepP nk Similarly,

2 / 4 { } 2 .nk

nP p e

.n

Chebyshev’s inequality gives the probability of k /n to lie between

and for a specific n.

We can use Bernstein’s inequality to estimate the probability for k /n to

lie between and for all large n.

Towards this, let

so that

To compute the probability of note that its complement is given by

p p

pp

}{ ppy nk

n

2 / 4 ( ) { } 2c n

nnkP y P p e

,n

mny

cn

mn

cn

mnyy

)(

We have

This gives

or,

Thus k /n is bound to stay near p for all large enough n, in probability, as stated by the SLLN.

2

2

/ 4

/ 4

2( ) {1 ( )} 1 1 as

1

m

n nn m n m

eP y P y m

e

. as 1} ,{ mmnall forppP nk

2

2

2

/ 4/ 4

/ 4

2( ) ( ) 2 .

1

mc c nn n

n m n m n m

eP y P y e

e

Let Thus if we toss a fair coin 1,000 times, from the weak law

Thus on the average 39 out of 40 such events each with 1000 or more trials will satisfy the inequality

It is quite possible that one out of 40 such events may not satisfy it.

Continuing the experiment for 1000 more trials, with k successes out of

n, for it is quite possible that for few such n the above inequality may be violated.

}{ 1.021 n

k

,20001000n

Discussion

.40

11.0

21

nkP

.1.0

.2

npq

pP nk

This is still consistent with the weak law

but according to the strong law such violations can occur only a finite number of times each with a finite probability in an infinite sequence of trials,

hence almost always the above inequality will be satisfied, i.e., the

sample space of k /n coincides with that of p as

Discussion - continued

.n

2n red cards and 2n black cards (all distinct) are shuffled together to form a single deck, and then split into half.

What is the probability that each half will contain n red and n black cards?

From a deck of 4n cards, 2n cards can be chosen in different ways.

Consider the unique draw consisting of 2n red cards and 2n black cards in each half.

Example

Solution

n

n

2

4

Among those 2n red cards, n of them can be chosen in different ways;

Similarly for each such draw there are ways of choosing n black cards.

Thus the total number of favorable draws containing n red and n black cards in each half are among a total of draws.

This gives the desired probability to be

Example – continued

n

n2

n

n2

n

n

2

4

n

n2

n

n2

np

.)! ()!4(

)! 2(4

4

2

4

22

nnn

p

n

n

n

n

n

n

n

For large n, using Stirgling’s formula we get

For a full deck of 52 cards, we have which gives

For a partial deck of 20 cards, we have and

nennennenn

pnnnn

nn

n 2

] 2[)4( )4(2])2( )2(2[

444

422

,13n 221.0np

5n .3568.0n

p

Example – continued

An Experiment

20 cards were given to a 5 year old child to split them into two equal halves

the outcome was declared a success if each half contained exactly 5 red and 5 black cards.

With adult supervision (in terms of shuffling) the experiment was repeated 100 times that very same afternoon. The results are tabulated below in next slides.

Expt Number of successes

Expt Number of successes

Expt Number of successes

Expt Number of successes

Expt Number of successes

1 0 21 8 41 14 61 23 81 29

2 0 22 8 42 14 62 23 82 29

3 1 23 8 43 14 63 23 83 30

4 1 24 8 44 14 64 24 84 30

5 2 25 8 45 15 65 25 85 30

6 2 26 8 46 16 66 25 86 31

7 3 27 9 47 17 67 25 87 31

8 4 28 10 48 17 68 25 88 32

9 5 29 10 49 17 69 26 89 32

10 5 30 10 50 18 70 26 90 32

11 5 31 10 51 19 71 26 91 33

12 5 32 10 52 20 72 26 92 33

13 5 33 10 53 20 73 26 93 33

14 5 34 10 54 21 74 26 94 34

15 6 35 11 55 21 75 27 95 34

16 6 36 12 56 22 76 27 96 34

17 6 37 12 57 22 77 28 97 34

18 7 38 13 58 22 78 29 98 34

19 7 39 14 59 22 79 29 99 34

20 8 40 14 60 22 80 29 100 35

0.3437182

Results of an experiment of 100 trialsThis figure shows the convergence of k/n to p.

n

np