2%2brandom%2bnumbers

7/23/2019 2%2BRandom%2Bnumbers

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Kleber Barcia V.

Chapter 2

Random Numbers

1. Random-number generation

2. Random-variable generation


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Kleber Barcia V.

Chapter 2

2

1. Objectives. Random numbers

Discuss how the random numbers are

generated.

Introduce the subsequent testing for

randomness:

Frequency test

Autocorrelation test.


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Kleber Barcia V.

Chapter 2

3

Properties of random numbers

Two important statistical properties:

Uniformity

Independence.

Random Number, Ri, must be independently drawn from auniform distribution with pdf:

Figure: pdf for random numbers

2

1

2)(

1

0

21

0

xxdxrE

12

1

4

1

3

1

2

1

3)(

21

0

31

0

22

xrEdxxrV

12

22

abrV

barE

otherwise,0

10,1)(

xxf


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Chapter 2

4

Generation of pseudo-random numbers

Pseudo, because generating numbers using a known

method removes the potential for true randomness.

Goal: To produce a sequence of numbers in [0,1] that

simulates, or imitates, the ideal properties of random numbers

(RN).

Important considerations in RN routines:

Fast

Portable to different computers

Have sufficiently long cycle

Replicable

Closely approximate the ideal statistical properties of uniformity

and independence.


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Kleber Barcia V.

Chapter 2

5

Techniques to generate random numbers

Middle-square algorithm

Middle-product algorithm

Constant multiplier algorithm

Linear algorithm

Multiplicative congruential algorithm

Additive congruential algorithm

Congruential non-linear algorithms

Quadratic

Blumy Shub


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Chapter 2

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1. Select a seed x0 with D digits (D > 3)

2. Rises to the square: (x0)2

3. Takes the four digits of the Center = x1

4. Rises to the square: (x1)2

5. Takes the four digits of the Center = x2

6. And so on. Then convert these values to r1between

0 and 1

If it is not possible to obtain the D digits in the Center,

add zeros to the left.


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Chapter 2

7


If x0 = 5735

Then,

Y0 = (5735)2 = 32890225 x1= 8902 r1= 0.8902

Y1 = (8902)2

= 79245604 x2= 2456 r2= 0.2456Y2 = (2456)

2 = 06031936 x3= 0319 r3= 0.0319

Y3 = (0319)2 = 101760 x4= 0176 r4= 0.0176

Y4 = (0176)2 = 030976 x5= 3097 r5= 0.3097

1. Small cycles

2. May lose the series (e.g.: X0 = 0012)



Chapter 2

8


It requires 2 seeds of D digits

1. Select 2 seeds x0and x1with D > 3

2. Y0= x0* x1 xi = digits of the center3. Yi= xi*xi+1 xi+1 = digits of the center

If it is not possible to obtain the D digits in the center, add

zeros to the left.



Chapter 2

9


Example:

Seeds x0= 5015 y x1= 5734

Y0= (5015)(5734) = 28756010 x2= 7560 r1= 0.756

Y1= (5734)(7560) = 43349040 x3= 3490 r2= 0.349

Y2= (7560)(3490) = 26384400 x4= 3844 r3= 0.3844

Y3= (3490)(3844) = 13415560 x5= 4155 r4= 0.4155

Y4= (3844)(4155) = 15971820 x6= 9718 r5= 0.9718



Chapter 2

10


Similar to Middle-product algorithm

1. Select a seed x0(D > 3)

2. Select a constant "a

3. Y0= a*x0 xi = D digits of the center

4. Yi = a*xi xi+1= D digits of the center

5. Repeat until you get the desired random numbers



Chapter 2

11


Example:

If: seed x0= 9803 and constant a = 6965

Y0= (6965)(9803) = 68277895 x1= 2778 r1= 0.2778

Y1= (6965)(2778) = 19348770 x2= 3487 r2= 0.3487

Y2= (6965)(3487) = 24286955 x3= 2869 r3= 0.2869

Y3= (6965)(2869) = 19982585 x4= 9825 r4= 0.9825

Y4= (6965)(9825) = 68431125 x5= 4311 r5= 0.4311



Chapter 2

12

Linear algorithm

To produce a sequence of integers,X1, X2, between 0andm-1by following a recursive relationship:

The selection of the values for a, c, m, andX0drastically affects

the statistical properties and the cycle length, must be > 0

"mod (m)" is the residue of the term (axi+ c) /m

The random integers are being generated [0,m-1], and toconvert the integers to random numbers:

,...2,1,0,mod)(1 imcaXX ii

The

multiplier

The

increment

Themodulus

,...2,1, im

XR i

i



Chapter 2

13

Linear algorithm

Example:

UseX0= 27, a = 17, c = 43y m = 100

The values ofxiand riare:

X1= (17*27+43) mod 100 = 502 mod 100 = 2 r1= 2/99 = 0.02

X2= (17*2+43) mod 100 = 77 mod 100 = 77 r2= 77/99 = 0.77

X3= (17*77+43) mod 100 = 1352 mod,100 = 52 r3= 52/99 = 0.52

The maximum life period Nis equal to m



Chapter 2

14


It arises from the linear algorithm when c = 0. The recursiveequation is:

The values of a, m, andx0must be > 0 and integers

The conditions that must be fulfilled to achieve the

maximum period are:

m = 2g

a = 3 + 8 k or 5 + 8 k where k = 0, 1, 2, 3,...

X0= should be an odd number

g must be integer

With these conditions we achieve N = m/4 = 2g-2

,...

,...2,1,0,mod1 imaXX ii

,...2,1,1

im

Xr ii



Chapter 2

15


Example: Usex0= 17, k = 2, g = 5

Then: a = 5+8(2) = 21 and m=25= 32

x1= (21*17) mod 32 = 5 r1= 5/31 = 0.1612

x2= (21*5) mod 32 = 9 r2= 9/31 = 0.2903

x3= (21*9) mod 32 = 29 r3= 29/31 = 0.9354

x4

= (21*29) mod 32 = 1 r4

= 1/31 = 0.3225

x5= (21*1) mod 32 = 21 r5= 21/31 = 0.6774

x6= (21*21) mod 32 = 25 r6= 25/31 = 0.8064

x7= (21*25) mod 32 = 13 r7= 13/31 = 0.4193

x8

= (21*13) mod 32 = 17 r8

= 17/31 = 0.5483


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Chapter 2

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Requires a sequence of n integer numbers x1, x2, x3,xn

to generate new random numbers starting at xn+1, xn+2,

The recursive equation is:

1m

Xr ii

Nnnimxxx niii ,...,2,1,mod1


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Chapter 2

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Example:

x1=65, x2= 89, x3= 98, x4= 03, x5= 69, m = 100

x6=(x5+x1)mod100 = (69+65)mod100 = 34 r1= 34/99 = 0.3434

x7=(x6+x2)mod100 = (34+89)mod100 = 23 r1= 23/99 = 0.2323

x8=(x7+x3)mod100 = (23+98)mod100 = 21 r1= 21/99 = 0.2121

x9=(x8+x4)mod100 = (21+03)mod100 = 24 r1= 24/99 = 0.2424

x10=(x9+x5)mod100 = (24+69)mod100 = 93 r1= 93/99 = 0.9393

x11=(x10+x6)mod100 = (93+34)mod100 = 27 r1= 27/99 = 0.2727

x12=(x11+x7)mod100 = (27+23)mod100 = 50 r1= 50/99 = 0.5050


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Chapter 2

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Quadratic:

Conditions to achieve a maximum period N = m

m= 2g

amust be an even number

cmust be an odd number

gmust be integer

(b-1)mod4 = 1

1m

Xr ii

Nimcbxaxxi ,...,3,2,1,0mod21


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Chapter 2

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Congruential non-linear algorithms (quadratic)

Example:

x0 = 13, m = 8, a = 26, b = 27, c = 27

x1 = (26*132+27*13+27)mod 8 = 4

x2 = (26*42+27*4+27)mod 8 = 7

x3 = (26*72+27*7+27)mod 8 = 2

x4 = (26*22+27*2+27)mod 8 = 1

x5 = (26*12+27*1+27)mod 8 = 0

x6 = (26*02+27*0+27)mod 8 = 3

x7 = (26*32+27*3+27)mod 8 =6

x8 = (26*62+27*6+27)mod 8 = 5

x9 = (26*52+27*5+27)mod 8 = 4

Nimcbxaxxi ,...,3,2,1,0mod21


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Chapter 2

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Algorithm Blum, Blumy Shub

It occurs when Quadraticis:

a = 1

b = 0

c = 0

nimxx ii ,...,3,2,1,0mod2

1

h


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Chapter 2

21

Characteristics of a good generator

Maximum Density Such that he values assumed by Ri, i = 1,2,, leave no large

gaps on [0,1]

Problem: Instead of continuous, each Riis discrete

Solution: a very large integer for modulus m Approximation appears to be of little consequence

Maximum Period

To achieve maximum density and avoid cycling.

Achieve by: proper choice of a, c, m, andX0.

Most digital computers use a binary representation ofnumbers

Speed and efficiency depend on the modulus, m, to be (or close

to) a (2b

)

Ch 2


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Chapter 2

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Tests for Random Numbers

Two categories:

Testing for uniformity:

H0: Ri~ U[0,1]

H1: Ri~ U[0,1]

Test for independence:H0: Ri~ independently

H1: Ri~ independently

Mean test

H0: ri= 0.5H1: ri= 0.5

We can use the normal distribution with 2= 1/12

or = 0.2887 with a significance level = 0.05

/

/

/

Ch t 2


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Chapter 2

23

Tests for Random Numbers

When to use these tests:

If a well-known simulation languages or random-number

generators is used, it is probably unnecessary to test

If the generator is not explicitly known or documented, e.g.,

spreadsheet programs, symbolic/numerical calculators, testsshould be applied to many sample numbers.

Types of tests:

Theoretical tests: evaluate the choices of m, a, and c without

actually generating any numbers

Empirical tests: applied to actual sequences of numbers

produced. Our emphasis.

Ch t 2


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Chapter 2

24

Kolmogorov-Smirnov test [Uniformity Test]

Very useful for little data.

For discrete and continuous variables.

Procedure:

1. Arrange the ri from lowest to highest

2. Determine D+, D-

3. Determine D4. Determine the critical value of D, n

5. If D> D, nthen riare not uniform

Chapter 2


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Chapter 2

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Kolmogorov-Smirnov test [Uniformity Test]

Example: Suppose that 5 generated numbers are 0.44,0.81, 0.14, 0.05, 0.93.

Step 1:

Step 2:

Step 3: D = max(D+, D-) = 0.26

Step 4: For a= 0.05,

Da

n= 0.565 > D

So not reject H0.

Sort R (i) Ascending

D+= max {i/NR(i)}

D-= max {R(i) - (i-1)/N}

R(i) 0.05 0.14 0.44 0.81 0.93

i/N 0.20 0.40 0.60 0.80 1.00

i/NR(i) 0.15 0.26 0.16 - 0.07R(i)(i-1)/N 0.05 - 0.04 0.21 0.13

Chapter 2


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Chapter 2

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Chi-square test [Uniformity Test]

The Chi-square test uses the statistical sample:

Approximately the chi-square distribution with n-1 degrees of

freedom is tabulated in Table A.6

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

is:

Valid only for large samples, example: N> = 50

If X02>X2(; n-1) reject the hypothesis of uniformity

n

i i

ii

E

EO

1

22

0

)(

nobservatioof#totaltheisN where,nNEi

nis the # of classes

Oiis the observed # of

the class ith

Eiis the expected # of

the class ith

Chapter 2


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Chapter 2

27


Example: Use Chi-square with a= 0.05 to test that the data shownare uniformly distributed

The test uses n = 10 intervals of equal length:

[0, 0.1), [0.1, 0.2), , [0.9, 1.0)

0.34 0.90 0.25 0.89 0.87 0.44 0.12 0.21 0.46 0.67

0.83 0.76 0.79 0.64 0.70 0.81 0.94 0.74 0.22 0.74

0.96 0.99 0.77 0.67 0.56 0.41 0.52 0.73 0.99 0.02

0.47 0.30 0.17 0.82 0.56 0.05 0.45 0.31 0.78 0.05

0.79 0.71 0.23 0.19 0.82 0.93 0.65 0.37 0.39 0.42

0.99 0.17 0.99 0.46 0.05 0.66 0.10 0.42 0.18 0.49

0.37 0.51 0.54 0.01 0.81 0.28 0.69 0.34 0.75 0.49

0.72 0.43 0.56 0.97 0.30 0.94 0.96 0.58 0.73 0.05

0.06 0.39 0.84 0.24 0.40 0.64 0.40 0.19 0.79 0.62

0.18 0.26 0.97 0.88 0.64 0.47 0.60 0.11 0.29 0.78

Chapter 2


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Chapter 2

28


The X02value = 3.4

Compared to the value of Table A.6, X20.05, 9= 16.9

Then H0is not rejected.

Intervalo Oi Ei Oi-Ei (Oi-Ei)2

(Oi-Ei)2

Ei

1 8 10 -2 4 0.4

2 8 10 -2 4 0.4

3 10 10 0 0 0.0

4 9 10 -1 1 0.1

5 12 10 2 4 0.4

6 8 10 -2 4 0.4

7 10 10 0 0 0.0

8 14 10 4 16 1.6

9 10 10 0 0 0.0

10 11

100

10

100

1

0

1 0.1

3.4

Chapter 2


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Kleber Barcia V.

Chapter 2

29

Tests for Autocorrelation [Test of Independence]

It is a test between each mnumbers, starting with thenumber i The auto-correlationrimbetween the numbers Ri, Ri+m, Ri+2m,

Ri+(M+1)m

M is the largest integer such that, , where N is

the total number of values in the sequence Hypothesis:

If the values are not correlated: For large values of M, the distribution estimationrim, is

approximately a normal distribution.

N)m(Mi 1

tindependenNOTarenumbers,0:

,0:

1

0 tindependenarenumbers

im

im

H

H

r

r

Chapter 2


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Kleber Barcia V.

Chapter 2

30

Tests for Autocorrelation [Test of Independence]

The statistical test is:

Z0is distributed normally with mean = 0and variance = 1, and:

Ifrim> 0, the subsequence has positive autocorrelation

High random numbers tend to be followed by high ones, and vice versa.

Ifrim< 0, the subsequence has negative autocorrelation

Low random numbers tend to be followed by high ones, and vice versa.

im

imZr

r

0

)(M

M

.RRM

im

M

k

)m(kikmiim

112

713

2501

1

0

1

Chapter 2


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Chapter 2

31

Example [Test for autocorrelation]

Test if the 3rd, 8th, 13thand so on, are auto-correlated.Use a= 0.05

Then i = 3, m= 5, N= 30

Then M= 4

0.12 0.01 0.23 0.28 0.89 0.31 0.64 0.28 0.83 0.93

0.99 0.15 0.33 0.35 0.91 0.41 0.60 0.27 0.75 0.88

0.68 0.49 0.05 0.43 0.95 0.58 0.19 0.36 0.69 0.87

3028

305143

1

N)m(Mi

Chapter 2


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Kleber Barcia V.

p

32

Example [Test for autocorrelation]

From Table A.3, z0.025= 1.96. then the hypothesis is not

rejected.

516.11280.0

1945.0

128.01412

7)4(13

1945.0

250)36.0)(05.0()05.0)(27.0(

)27.0)(33.0()33.0)(28.0()28.0)(23.0(

14

1

0

35

35

Z

)(

.

Chapter 2


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Kleber Barcia V.

p

33

Other tests of Independence

Test runs up and down Test runs up and below the mean

Test poker

Test series

Test holes

-----HOMEWORK 3 at SIDWeb-----

Bonus: (3 points for the partial exam)

Write a computer program that generates random numbers (4 digits)using the Multiplicative Congruential Algorithm. Allow user to entervalues x0, k, g

Give an example printed and the program file

Chapter 2


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Kleber Barcia V.

p

34

Understand how to generate random variables

from a specific distribution as inputs of a

simulation model.

Illustrate some widely used techniques for

generating random variables.

Inverse-transform technique

Acceptance- rejection technique

2. Objectives. Random Variables

Chapter 2


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Kleber Barcia V.35

Inverse-transform technique

It can be used to generate random variables from anexponential distribution, uniform, Weibull, triangular and

empirical distribution.

Concept:

For cdf function: r = F(x)

Generate r (0,1)

Find x:

x = F-1(r)

r1

x1

r = F(x)

Chapter 2


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Kleber Barcia V.36

Exponential Distribution [inverse-transform]

The exponential cdf:

To generateX1, X2, X3

r = F(x)

r = 1e-lx forx 0

Xi= F-1(ri)

= -(1/lln(1-ri)

= -(1/l) ln(ri)

Figure: Inverse-transform

technique for exp (l

= 1)

)1ln(1

)1ln(

1

rX

rX

re X

l

l

l

Chapter 2


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Kleber Barcia V.37

Table of Distributions [inverse-transform]

Distribution Cumulative Inverse-Transform

Uniform

Weibull

Triangular

Normal

bxaabaxxF

01

xexFx a

21221

102

2

2

xxxF

xxxF

ii rabax

a1

1ln rx

121122

2102

rrx

rrx

zx

Chapter 2


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Kleber Barcia V.38

Exponential Distribution [inverse-transform]

Example: Generate 200 variables Xi with exponential distribution(l= 1)

The histograms of riand xiare:

Uniform distribution Exponential distribution

i 1 2 3 4 5

Ri 0.1306 0.0422 0.6597 0.7965 0.7696

Xi 0.1400 0.0431 1.078 1.592 1.468

Chapter 2


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Kleber Barcia V.39

Uniform Distribution [Example]

The temperature of an ovenbehaves uniformly within the

range of 95 to 100 C.

Modeling the behavior of the

temperature.

ii rabax

Measurements ri TemperatureoC

1 0.48 95+5*0.48=97.40

2 0.82 99.103 0.69 98.45

4 0.67 98.35

5 0.00 95.00

Chapter 2


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Kleber Barcia V.40

Exponential Distribution [Example]

Historical data service time a Bankbehave exponentially with an

average of 3 min / customer.

Simulate the behavior of this

random variable

Customer ri Service Time(min)

1 0.36 3.06

2 0.17 5.313 0.97 0.09

4 0.50 2.07

5 0.21 0.70

ii rx 1ln3

Chapter 2


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Kleber Barcia V.41

Poisson Distribution [Example]

The number of pieces entering a production systemfollows a Poisson distribution with mean 2 pcs/hour.

Simulate the arrival behavior of the pieces to the system.

x P(x) F(x) Random digits

0 0.1353 0.1353 0-0.1353

1 0.2706 0.4060 0.1353-0.4060

2 0.2706 0.6766 0.4060-0.6766

3 0.1804 0.8571 0.6766-0.8571

4 0.0902 0.9473 0.8571-0.9473

5 0.0369 0.9834 0.9473-0.9834

6 0.0120 0.9954 0.9834-0.9954

7 0.0034 0.9989 0.9954-0.9989

8 0.0008 0.9997 0.9989-0.9997

9 0.0001 0.9999 0.9997-0.9999

10 0.00003 0.9999 0.9999-0.9999

Arrival

Time

ri Pieces/hour

1 0.6754 2

2 0.0234 0

3 0.7892 3

4 0.5134 2

5 0.3331 1

x

i

i

x

i

exF

x

x

exp

0 !)(

,...1,0,

!a

a

a

a

Chapter 2


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Kleber Barcia V.42

Convolution Method

In some probability distributions, the random variable

can be generated by adding other random variables

Erlang Distribution

The sum of exponential variables of equal means

(1/)

Y = x1+x2++xk

Y = (-1/k)ln(1-r1)+ (-1/k)ln(1-r2)++ (-1/k)ln(1-rk)

Where:

Y = Eri= (-1/k)ln(1-ri) The factor productoria

Chapter 2


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Kleber Barcia V.43

Convolution Method

The processing time of certain machine follows a 3-erlang distribution with mean 1/ = 8 min/piece

Y = -(8/3)[ln(1-ri)] = -(8/3)ln[(1-r1) (1-r2) (1-r3)]

Piece 1-r1 1-r2 1-r3 Process Time

1 0.28 0.52 0.64 6.328

2 0.96 0.37 0.83 3.257

3 0. 04 0.12 0.03 23.588

4 0.35 0.44 0.50 6.837

5 0.77 0.09 0.21 11.279

Chapter 2


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Kleber Barcia V.44

Empirical continuous distribution [inverse-transform]

When theoretical distribution is not applicable To collect empirical data:

Resample the observed data

Interpolate between observed data points to fill in the gaps

For a small sample set (size n):

Arrange the data from smallest to largest

Assign the probability 1/n to each interval

Where 1 / n is the probability

(n)(2)(1) xxx

(i)1)-(i xxx

niRaxRFX ii )1()(

)1(

1

n

xx

nin

xxa

iiii

i

/1/)1(/1

)1()()1()(

Chapter 2


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Kleber Barcia V.45

Empirical continuous distribution [inverse-transform]

Example: Suppose that the observed data of 100 repair times of a machineare:

i

Interval

(Hours) Frecuency

Relative

Frecuency

Cumulative

Frecuencia, ci Slot, ai

1 0.25 x 0.5 31 0.31 0.31 0.81

2 0.5 x 1.0 10 0.10 0.41 5.0

3 1.0 x 1.5 25 0.25 0.66 2.0

4 1.5 x 2.0 34 0.34 1.00 1.47

Consider r1= 0.83:

c3= 0.66 < r1< c4= 1.00

X1= x(4-1)+ a4(r1c(4-1))

= 1.5 + 1.47(0.83-0.66)

= 1.75

Chapter 2


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Kleber Barcia V.46

Empirical Discrete Distribution

Example: Suppose the number of shipments, x, on theloading dock of IHW company is either 0, 1, or 2

Data - Probability distribution:

Method - Given R, the generation

scheme becomes:

0.18.0

8.05.0

5.0

,2

,1

,0

R

R

R

x

Consider R1= 0.73:

F(xi-1) < R


47/47

Acceptance-Rejection technique

Useful particularly when inverse cdf does not exist in closedform, a.k.a. thinning

To generate random variables X ~ U (1/4, 1) we must:

This procedure is only for uniform distributions

Acceptance-rejection technique is also applied to Poisson,nonstationary Poisson, and gamma distributions

-----HOMEWORK 4 at SIDWeb-----

procedure :

Step 1. Generate R ~ U[0,1]Step 2a. If R >= , accept X=R.

Step 2b. If R < , reject R,back to step 1

Generate R

Condition

output R

si

no

2%2brandom%2bnumbers

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