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GENERATING

RANDOM NUMBERSAND RANDOM

VARIABLESPresented By:

Dionisio, Adrian

Delos Reyes, Ma. Victorina

Dizon, Xaraenielle

Santos, Jeamaica

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GENERATING

RANDOM

NUMBERS

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WHAT ARE RANDOM NUMBERS

Random numbers are a necessary basic

ingredient in the simulation of almost all discrete

systems. Most computer languages have a subroutine,

object, or function that will generate a random

number. Similarly simulation languages generate

random numbers that are used to generate event

limes and other random variables.

It is impossible to produce an arbitrarily long

string of random digits and prove it is random.

Strangely, it is also very difficult for humans toproduce a string of random digits, and computer

programs can be written which, on average, actually

predict some of the digits humans will write down

based on previous ones.

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Intuitively, we can list a number of criteria that a

sequence of numbers must fulfill to pass as a random

number sequence:

unpredictability,

independence,

without pattern.

These criteria appear to be the minimum request foran algorithm to produce random numbers. More precisely

we can formulate:

uniform distribution,

uncorrelated,

passes every test of randomness,

large period before the sequence repeats ,

sequence repeatable and possibility to vary starting

values,

fast algorithm.

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Two important statistical properties:

y Uniformity

y Independence

PROPERTIES OF RANDOM NUMBERS

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Pseudo means false, so false random numbers

are being generated. The goal of any generation

scheme is to produce a sequence of numbers between

zero(0) and one(1) which simulates, or initiates, the

ideal properties of uniform distribution and

independence as closely as possible.

When generating pseudo-random numbers,

certain problems or errors can occur. These errors, or

departures from ideal randomness, are all related tothe properties stated previously. Some examples

include the following:

PSEUDORANDOM NUMBERS

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1. The generated numbers may not be uniformly

distributed.

2. The generated numbers may be discrete - valued

instead continuous valued.

3. The mean of the generated numbers may be too high

or too low.

4. The variance of the generated numbers may be toohigh or low

5. There may be dependence. The following are examples:

(a) Autocorrelation between numbers.(b) Numbers successively higher or lower than

adjacent numbers.

(c) Several numbers above the mean followed by

several numbers below the mean.

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Usually, random numbers are generated by a

digital computer as part of the simulation.

Numerous methods can be used to generate thevalues. In selecting among these methods, or

routines, there are a number of important

considerations.

Fast

Portable to different computers

Have sufficiently long cycle

Replicable

Closely approximate the ideal statistical

properties of uniformity and independences.

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Linear Congruential Method (LCM)

Combined Linear Congruential Generators(CLCG)

Random Number Streams

TECHNIQUES FOR GENERATING RANDOM

NUMBERS

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The linear congruential method, initially proposed by

Lehmer [1951], produces a sequence of integers, X1, X2,...between zero and m 1 according to the following recursive

relationship:

Xi+1 = (a Xi + c) mod m, i = 0,1, 2,...

The initial valueX0

is called the seed, a is called the

constant multiplier, c is the increment, and m is the modulus.

The selection of the values for a, c, m, andX0 drastically

affects the statistical properties and the cycle length.

LINEAR CONGRUENTIAL METHOD (LCM)

multiplier modulusincrement

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If c 0, the form is called

the mixed congruential method. When c =0, the

form is known as the multiplicativecongruential method. The selection of the values

for a, c, m and X 0 drastically affects the

statistical properties and the cycle length.

The random integers are being generated

[0, m-1], and to convert the integers to random

numbers:

X

Ri = ---- , i = 1,2,3

m

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Example:

Use the linear congruential method to

generate a sequence of random numbers with

X0 = 27, a = 17, c = 43, and m = 100. Here, the

integer values generated will all be betweenzero and 99 because of the value of the

modulus. These random integers should

appear to be uniformly distributed the

integers zero to 99. Random numbers betweenzero and 1 can be generated by:

Ri

= Xi

/ m, i=1,2,3.

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The sequence of Xi and subsequent Ri values is

computed as follows:

X0 = 27

X1 = (17 27 + 43) mod 100 = 502 mod 100 = 2

R1 = 2 100 = 0.02

X2= (17 2 + 43) mod 100 = 77 mod 100 = 77

R2 = 77 100 = 0.77

X3 = (1777+ 43) mod 100 = 1352 mod 100 = 52R3 = 52 100 = 0.52

X4 = (17 52 + 43) mod 100 = 927 mod 100 = 27

R4 = 27 100 = 0.27

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Example 2:

Find the sequence of random numbers given a = 13, c = 0,

m = 31, X0 = 1.

X0 = 1

X1 = (13 * 1 + 0) mod 31 = 13 mod 31 = 13

X2 = (13 * 13 + 0) mod 31 = 169 mod 31 = 14X3 = (13 * 14 + 0) mod 31 = 182 mod 31 = 27

X4 = (13 * 27 + 0) mod 31 = 351 mod 31 = 10

X5 = (13 * 10 + 0) mod 31 = 130 mod 31 = 6

X6 = (13 * 6 + 0) mod 31 = 78 mod 31 = 16

X7 = (13 * 16 + 0) mod 31 = 208 mod 31 = 22

X8 = (13 * 22 + 0) mod 31 = 286 mod 31 = 7

X9 = (13 * 7 + 0) mod 31 = 91 mod 31 = 29

X10 = (13 * 29 + 0) mod 31 = 377 mod 31 = 5

X11 = (13 * 5 + 0) mod 31 = 65 mod 31 = 3

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The sequence begins with:

1, 13, 14, 27, 10, 6, 16, 22, 7, 29, 5, 3,

What's the next value? Well, it looks

pretty unpredictable, but you've been

initiated. So you can compute. The first 30

terms in the sequence are a permutation of theintegers from 1 to 30 and then the sequence

repeats itself. It has a period equal to m - 1.

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If a pseudorandom integer sequence

with values between 0 and m is scaled bydividing by m, the result is floating-point

numbers uniformly distributed in the interval

[0, 1]. Our simple example begins with:

0.0323, 0.4194, 0.4516, 0.8710, 0.3226, 0.1935,

0.5161, .

There are only a finite number of values,

30 in this case. The smallest value is 1/31 and

the largest is 30/31.

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Seatwork:

Use the linear congruential

method to generate a sequence

of random numbers with X0 = 9,a = 5, c = 11, and m = 16. Find

the list of random numbers that

will be generated by the givenvalues before an occurrence of

any random number in the list

will occur. Show your solution.

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The desirable properties of random numbersare uniformity and independence. To insure that

these desirable properties are achieved, a number oftests can be performed. The first entry in the listbelow concerns testing for uniformity. The secondthrough fifth entries concern testing forindependence.

The five types of tests are:

1. Frequency Test

2. Runs Test3. Autocorrelation Test

4. Gap Test

5. Poker Test

TESTING RANDOM NUMBER RANDOMNESS

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When to Use These Tests:

If a well-known simulation languagesor random-number generators isused, it is probably unnecessary totest

If the generator is not explicitlyknown or documented, e.g.,spreadsheet programs,symbolic/numerical calculators, testsshould be applied to many samplenumbers.

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In testing for uniformity, thehypotheses are as follows:

H0: Ri ~ U[0,1]

H1: Ri ~ U[0,1]

The null hypothesis, H0 reads that thenumbers are distributed uniformly on theinterval [0, 1]. Failure to reject the nullhypothesis means that no evidence of non-

uniformity has been detected on the basisof this test. This does not imply thatfurther testing of the generator foruniformity is unnecessary.

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In testing for independence, the

hypotheses are as follows:

H0: Ri ~ independently

H1: Ri ~ independently

The null hypothesis, H0 reads that the

numbers are independent. Failure to reject

the null hypothesis means that no evidence

of dependence has been detected on thebasis of this test. This does not imply that

further testing of the generator for

independence is unnecessary.

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For each test, a level of

significance must be stated. The

level is the probability of rejecting

the null hypothesis given that the

null hypothesis is true, or

= P (reject H0|H0 true)

The decision maker sets the

value of for any test. Frequently,

is set to 0.01 or 0.05.

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If several tests are conducted on

the same set of numbers, the

probability of rejecting the nullhypothesis on at least one test, by

chance alone [i.e., making a Type I

(a) error], increases. Say that =0.05 and those five different tests

are conducted on a sequence of

numbers. The probability ofrejecting the null hypothesis on at

least one test, by chance alone, may

be as large as 0.25.

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A basic test that should always be

performed to validate a new generator isthe test of uniformity. Two different

methods of testing are available. They are

the Kolmogorov-Smirnov and the chi-

square test. Both of these tests measure thedegree of agreement between the

distribution of a sample of generated

random numbers and the theoretical

uniform distribution. Both tests are based

on the null hypothesis of no significant

difference between the sample distribution

and the theoretical distribution.

FREQUENCY TEST

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This test compares the continuous

cdf, F(x), of the uniform distribution

with the empirical cdf, SN(x), of the N

sample observations. By definition:

F(x) = x, 0 x 1

KOLMOGOROV-SMIRNOV TEST

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If the sample from the RN generator

is R1, R2, , RN, then the empirical cdf,SN(x) is:

As N becomes larger, SN(x) should

become a better approximation toF(x), provided that the null hypothesis is

true.

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The Kolmogorov-Smirnov test is

based on the largest absolute deviationbetween F(x) and SN(x) over the range of

the random variable. That is, it is based

on the statistic:

D = max| F(x) - SN(x)|

that is the absolute value of the

differences.

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Here D is a random variable.

If the calculated value is greaterthan the ones listed in the table,

the hypothesis (no disagreement

between the samples and the

theoretical value) should be

rejected; otherwise, we don'thave enough information to

reject it.

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Following steps are taken to perform this

test:

1. Rank the data from smallest to largest

R(1)R(2)... R(N)2. Compute

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3. Compute D = max(D+, D-)

4. Get D

for the significance

level

5. If D D accept, otherwisereject H0

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Example:

Suppose 5generated numbers are 0.44, 0.81,

0.14, 0.05, 0.93.

Step 1: Arrange R(i) from smallest to largest.

Step 2: Compute D+ and D-.

D+ = 0.26

D- = 0.21

R(i) 0.05 0.14 0.44 0.81 0.93

i/ N 0.20 0.40 0.60 0.80 1.00

D+ = max{i/N R(i)} 0.15 0.26 0.16 - 0.07

i/ N 0.05 - 0.04 0.21 0.13

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Step 3: D = max(D+

, D-

).D = 0.26

Step 4: Get D for the significance level .

For = 0.05,

D = 0.565 > D

Step 5: H0 is not rejected.

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The chi-square test looks at

the issue from the same anglebut uses different method.

Instead of measure the

difference of each point between

the samples and the true

distribution, chi-square checksthe ``deviation'' from the

``expected'' value.

CHI-SQUARE TEST

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where n is the number ofclasses (e.g. intervals), O

iis the

number of samples observed inthe interval or the observed # inthe i-th class, E

iis expected

number of samples in theinterval or expected # in the i-thclass.

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For the uniform distribution,Ei, the expected number in each

class is:

where N is the total # of observation.

This test is valid only on largesamples, e.g. N>=50.

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Example:

100 numbers from [0,1]

= 0.05

10 intervals

X20.05,9 = 16.9

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Interval Upper Limit Oi Ei Oi Ei (Oi Ei)2 (Oi Ei)

2/Ei

1 0.1 10 10 0 0 0

2 0.2 9 10 -1 1 0.1

3 0.3 5 10 -5 25 2.5

4 0.4 6 10 -4 16 1.6

5 0.5 16 10 6 36 3.6

6 0.6 13 10 3 9 0.9

7 0.7 10 10 0 0 0

8 0.8 7 10 -3 9 0.9

9 0.9 10 10 0 0 0

10 1.0 14 10 4 16 1.6

S 100 100 0 0 11.2

Accept, since

X

2

0 = 11.2 < X

2

0.05,9

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Seatwork:

Suppose there are 10

generated numbers which

are, 0.95, 0.06, 0.09, 0.92, 0.76,0.13, 0.43, 0.65, 0.63, 0.59. Test

if these numbers are

uniformly distributed form

the interval [0,1].