BU_FCAI_SCC430_Modeling&Simulation_Ch05-P2.pptx

SCC430
Modeling and Simulation
Chapter 05 Random-Number Generation
Part 02
Dr. Ahmed Hagag
Faculty of Computers and Artificial Intelligence
Benha University
Fall 2020

• Properties of Random Numbers.
• Generation of Pseudo-Random Numbers.
• Techniques for Generating Pseudo-Random Numbers.
• Tests for Random Numbers.
©Ahmed Hagag SCC430 Modeling and Simulation 2
Chapter 5: Ran. Num. Gen.

Tests for PRNs (1/7)
Introduction (1/5)
The properties of random numbers: Uniformity and
Independence. Number of tests can be performed to check
these properties been achieved or not.
• Frequency Test: Uses the Kolmogorov-Smirnov or the
Chi-square test to compare the distribution of the set of
numbers generated to a uniform distribution.
• Autocorrelation Test: Tests the correlation between
numbers and compares the sample correlation to the
expected correlation, zero.

Introduction (2/5)

Introduction (3/5)
In testing for uniformity, the hypotheses are as follows:
The null hypothesis, 𝐻0, reads that the numbers are
distributed uniformly on the interval [0, 1]. Failure to reject
the null hypothesis means that evidence of nonuniformity
has not been detected by this test. This does not imply that
further testing of the generator for uniformity is
unnecessary.

Introduction (4/5)
In testing for independence, the hypotheses are as follows:
This null hypothesis, 𝐻0, reads that the numbers are
independent. Failure to reject the null hypothesis means
that evidence of dependence has not been detected by this
test. This does not imply that further testing of the
generator for independence is unnecessary.

Introduction (5/5)
For each test, a level of significance 𝛼 must be stated. The
level 𝛼 is the probability of rejecting the null hypothesis
when the null hypothesis is true:
The decision maker sets the value of 𝛼 for any test.

Kolmogorov-Smirnov (K-S) Test (1/5)
Compares the continuous cumulative distribution function
(cdf), F(x), of the uniform distribution with the empirical
distribution fun. 𝑆𝑁(𝑥) of the sample of N observations.
Uniformity
𝐹 𝑥 =
0, 𝑥 ≤ 𝑎
𝑥 − 𝑎
,
𝑏 − 𝑎
1,
𝑎 < 𝑥 < 𝑏
𝑥 ≥ 𝑏

Compares the continuous cumulative distribution function
(cdf), F(x), of the uniform distribution with the empirical
distribution fun. 𝑆𝑁(𝑥) of the sample of N observations.
𝐷 = max 𝐹 𝑥 − 𝑆𝑁 𝑥
The sampling distribution of 𝐷 is known; it is tabulated as
a function of 𝑁 in the following Table.
Uniformity

𝒊𝒇 𝑵 = 𝟖 and 𝑎 = 𝟎. 𝟎𝟓
𝑫𝟎.𝟎𝟓 = 𝟎. 𝟒𝟓𝟕

𝑁
𝟏. 𝟔𝟑
𝑵
𝟏. 𝟑𝟔
𝑵
𝟏. 𝟐𝟐
𝑵
𝟏. 𝟏𝟒
𝑵
𝟏. 𝟎𝟕
𝑵

For testing against a uniform cdf, the test procedure
follows these steps:
Step1: Rank the data from smallest to largest. Let 𝑅𝑖
denote 𝑖th smallest observation, so that 𝑅1 ≤ 𝑅2 ≤ ⋯ 𝑅𝑁
𝑖
Step2: Null hypothesis 𝐻0: 𝑅′s~U 0,1
Uniformity
Distributed as uniform distribution

Step3: Compute 𝐷+ and 𝐷−, where
𝐷+ = max
𝑁 𝑖 𝑖
𝑖 𝑖 − 1
− 𝑅 , 𝐷− = max 𝑅 −
𝑁
Step4: Compute 𝐷 = max 𝐷+, 𝐷−
Uniformity

Step3: Compute 𝐷+ and 𝐷−, where
𝑁 𝑖
𝐷+ = max
𝑖
− 𝑅 , 𝑖
𝐷− = max 𝑅 −
𝑖 − 1
𝑁
Step4: Compute 𝐷 = max 𝐷+, 𝐷−
Uniformity
𝒊 = 𝟏, 𝟐, … , 𝑵

Step5: Locate the critical value 𝐷𝛼 in the K-S Table for the
specified significance level 𝛼 and the sample size 𝑁.
Uniformity

Step6: If the sample statistic 𝐷 is greater than the critical
value 𝐷𝛼, the null hypothesis is rejected. If 𝐷 ≤ 𝐷𝛼,
conclude that we fail to reject the null hypothesis and these
PRNs can be accepted according to K-S test.
Uniformity

Example – (K-S) Test (1/5)
Suppose 5 generated numbers are 0.44, 0.81, 0.14, 0.05, 0.93.
For the specified significance level 𝛼 = 0.05.
Uniformity

For the specified significance level 𝛼 = 0.05.
Uniformity
𝑵 = 𝟓 and 𝑎 = 𝟎. 𝟎𝟓

𝑵
Uniformity
𝑵 = 𝟓 and 𝑎 = 𝟎. 𝟎𝟓
𝑫𝟎.𝟎𝟓 = 𝟎. 𝟓𝟔𝟓

The 5 generated numbers are 0.44, 0.81, 0.14, 0.05, 0.93.
First, the numbers must be sorted from smallest to largest.
𝑹𝒊 0.05 0.14 0.44 0.81 0.93
Uniformity

Example – (K-S) Test (3/5) Uniformity
𝑹𝒊 0.05 0.14 0.44 0.81 0.93
𝒊 = 𝟏, 𝟐, … , 𝑵
𝑁=5

Example – (K-S) Test (3/5) Uniformity
𝑹𝒊 0.05 0.14 0.44 0.81 0.93
𝒊 = 𝟏, 𝟐, … , 𝟓
𝑁=5

𝑹𝒊 0.05 0.14 0.44 0.81 0.93
𝒊/𝑵 0.20 0.40 0.60 0.80 1.00
Uniformity
𝑁=5

𝑹𝒊 0.05 0.14 0.44 0.81 0.93
𝒊/𝑵 0.20 0.40 0.60 0.80 1.00
𝒊/𝑵 – 𝑹𝒊 0.15 0.26 0.16 − 0.07 𝑫+ = 𝟎. 𝟐𝟔
Uniformity
𝑁=5

𝑹𝒊 0.05 0.14 0.44 0.81 0.93
𝒊/𝑵 0.20 0.40 0.60 0.80 1.00
𝒊/𝑵 – 𝑹𝒊 0.15 0.26 0.16 − 0.07
(𝒊 − 𝟏)/𝑵 0.00 0.20 0.40 0.60 0.80
𝑫+ = 𝟎. 𝟐𝟔
Uniformity
𝑁=5

𝑹𝒊 0.05 0.14 0.44 0.81 0.93
𝒊/𝑵 0.20 0.40 0.60 0.80 1.00
𝒊/𝑵 – 𝑹𝒊 0.15 0.26 0.16 − 0.07
(𝒊 − 𝟏)/𝑵 0.00 0.20 0.40 0.60 0.80
𝑹𝒊 – (𝒊 − 𝟏)/𝑵 0.05 − 0.04 0.21 0.13
𝑁=5
𝑫+ = 𝟎. 𝟐𝟔
𝑫− = 𝟎. 𝟐𝟏
Uniformity

𝐷 = max 𝐷+, 𝐷− = 0.26
Uniformity
𝑫+ = 𝟎. 𝟐𝟔
𝑫− = 𝟎. 𝟐𝟏

Since the computed value, 0.26, is less than the critical value,
0.565, the hypothesis that the distribution of the generated
numbers is the uniform distribution is “failed to reject".
𝐷 = max 𝐷+, 𝐷− = 0.26 < 𝐷𝛼 𝐷𝛼 = 0.565
Uniformity

Chi-Square ( 𝝌𝟐 ) Test (1/4)
We have 𝑁 generated PRNs are 𝑅1, 𝑅2, … , 𝑅𝑁.
Step1: We divide [0, 1) into 𝑘 subintervals of equal length
(i.e., Classes). (As a general rule, 𝑘 should be at least 100)
𝑖
Uniformity

Step3: Chi-square test uses the sample statistic:
𝑖
=1
𝑘
𝑂 − 𝐸
𝜒2 = ෍ 𝑖 𝑖
2
𝐸𝑖
Uniformity

𝑖
=1
𝑘
𝑂 − 𝐸
𝜒2 = ෍ 𝑖 𝑖
2
𝐸𝑖
Uniformity
𝒌 is the#
of classes

𝑘
𝜒2 = ෍
𝑖=1
𝑂𝑖 − 𝐸𝑖
2
𝐸𝑖
Uniformity
is theobserved #
in the 𝒊𝒕𝒉 class
How many 𝑹′𝒔
in the 𝒊𝒕𝒉 class 𝒌 is the#
of classes

𝑘
𝜒2 = ෍
𝑖=1
2
𝐸𝑖
Uniformity
is theobserved #
is theexpected #
𝑵
𝑬𝒊 =
𝒌
Uniform
Distribution
𝒌 is the#
of classes

Step4: 𝜒2 will have an approximate chi-square distribution
with 𝑘 − 1 degrees of freedom (df) at significance level 𝛼.
𝛼, 𝑘−1
𝜒2
= critical value, where the critical values are
tabulated in the following Table.
Uniformity

𝜒
2
𝛼, 𝑘−1

0.025, 9
𝜒2
= 19
𝛼 = 0.025,
𝑘 = 10

Step5:
If 𝜒2 > 𝜒2 , we reject 𝐻0, else we fail to reject 𝐻0.
𝛼, 𝑘−1
Uniformity

Example – ( 𝝌𝟐) Test (1/7)
We have 100 generated PRNs 𝑅′s are shown below. Use
𝑖
Chi-square test with 𝛼 = 0.05 and 𝑘 = 10.
Uniformity

𝑖
Step1: We divide [0, 1) into 𝟏𝟎 subintervals of equal length.
[0, 0.1), [0.1, 0.2), [0.2, 0.3), …, [0.9,1.0)
Or we divide (0, 1] into 𝟏𝟎 subintervals of equal length.
(0, 0.1], (0.1, 0.2], (0.2, 0.3], …, (0.9,1.0]
𝑖
Uniformity

𝑖
𝑘
𝜒2 = ෍
𝑖=1
2
𝐸𝑖
Uniformity
is theobserved #
is theexpected #
𝑵
𝑬𝒊 =
𝒌
𝒌 is the#
of classes

𝑖
𝑘
𝜒2 = ෍
𝑖=1
2
𝐸𝑖
Uniformity
is theobserved #
is theexpected #
𝟏𝟎𝟎
𝑬𝒊 =
𝟏𝟎
= 𝟏𝟎
𝒌 is the#
of classes
=10

𝑖
𝑖
=1
10
𝜒2 = ෍ 𝑖
𝑂 − 10 2
10
Uniformity
is theobserved #
is theexpected #
𝟏𝟎𝟎
𝑬𝒊 =
𝟏𝟎
= 𝟏𝟎

𝑖
Uniformity
(0.0, 0.1]
(0.1, 0.2]
(0.2, 0.3]
(0.9, 1.0]

𝑖
Uniformity
𝑘
𝜒2 = ෍
𝑖=1
𝑂𝑖 − 𝐸 2
𝐸𝑖
𝑖
= 3.4
(0.0, 0.1]
(0.1, 0.2]
(0.2, 0.3]
(0.9, 1.0]

𝑖
Step4: Calculate 𝜒2 = 𝜒2
𝛼, 𝑘−1 0.05, 9
Uniformity

0.05, 9
𝜒2
= 16.9

𝑖
Step5:
𝜒2 = 3.4
0.05, 9
𝜒2
= 16.9
0.05, 9
∵ 𝜒2 < 𝜒2 , then we fail to reject 𝐻0, and these PRNs are
accepted according to chi-square test.
Uniformity

Autocorrelation (1/3)
Correlation between a signal and itself signals for detecting
dependency. Efficient calculation of correlation coefficient
between two sequences (listed in time order of collection)
𝑥(𝑖), 𝑦(𝑖), where 𝑖 = 1, 2, … , 𝑛.
Independence

𝑥(𝑖), 𝑦(𝑖), where 𝑖 = 1, 2, … , 𝑛.
Independence
𝑖
=1
expected value (sample mean) = σ𝑛 𝑥𝑖 /𝑛

𝑥(𝑖), 𝑦(𝑖), where 𝑖 = 1, 2, … , 𝑛.
Independence
𝑟𝑥𝑦 =
1
𝑛 − 1
𝑖
=1
σ𝑛
𝑥𝑖 − 𝑥ҧ𝑦𝑖 − 𝑦
ത
𝑠𝑥𝑠𝑦
standard deviation 𝑠𝑥

Efficient calculation of correlation
sequence 𝑥(𝑖), where 𝑖 = 1, 2, … , 𝑛.
coefficient for a
𝑥
𝑥
𝑟 𝑘 =
1
𝑛 − 𝑘
𝑖
=1
σ𝑛−𝑘
𝑥𝑖 − 𝑥ҧ𝑥 𝑖+𝑘 − 𝑥
ҧ
𝑠𝑥𝑠𝑥
𝑟𝑥𝑥 𝑘 is calculated for 𝑘 = 1, 2, … , 𝑛 − 1 for 𝑛 PRNs.
Independence

𝑟𝑥𝑥 𝑘 is calculated for 𝑘 = 1,2, … , 𝑛 − 1 for 𝑛 PRNs.
If the samples are independent, then all the values should be
theoretically zero. But in practice we can say:
• Small values (close to zero) indicate independence.
• Large values (close to 1) indicate dependence.
Independence

Example (1/10)
Independence

Example (2/10)
𝒙𝒊 0.44 0.81 0.14 0.05 0.93
Independence
1 2 3 4 5
𝒏 = 𝟓

Example (3/10)
𝒙𝒊 0.44 0.81 0.14 0.05 0.93
Independence
1 2 3 4 5
𝒏 = 𝟓
𝑥ҧ
=
𝑖
=1
σ𝑛
𝑥𝑖
𝑛
=
0.44 + 0.81 + 0.14 + 0.05 + 0.93
5
= 0.474
𝑠𝑥

Example (4/10)
𝒙𝒊 0.44 0.81 0.14 0.05 0.93
𝒙𝒊 − 𝒙
ഥ -0.034 0.336 -0.334 -0.424 0.456
Independence
1 2 3 4 5
𝑥ҧ = 0.474
𝑠𝑥
𝑠𝑥 = 0.15333

Example (5/10)
𝒙𝒊 0.44 0.81 0.14 0.05 0.93
Independence
1 2 3 4 5
𝑥ҧ = 0.474
𝑠𝑥 = 0.15333
𝑟𝑥𝑥 𝑘 =
1
5 − 𝑘
𝑖
=1
σ5−𝑘
𝑥𝑖 − 0.474 𝑥 𝑖+𝑘 − 0.474
0.15333
𝑘 = 1,2,3,4

Example (6/10)
𝒙𝒊 0.44 0.81 0.14 0.05 0.93
Independence
1 2 3 4 5
𝑥ҧ = 0.474
𝑠𝑥 = 0.15333
𝑟𝑥𝑥 𝑘 =
1
5 − 𝑘
𝑖
=1
σ5−𝑘
𝑥𝑖 − 0.474 𝑥 𝑖+𝑘 − 0.474
0.15333
𝑘 = 1,2,3,4
𝑟𝑥𝑥
1 −0.17538
1 =
4 0.15333
= −0.28595

Example (7/10)
𝒙𝒊 0.44 0.81 0.14 0.05 0.93
Independence
1 2 3 4 5
𝑥ҧ = 0.474
𝑠𝑥 = 0.15333
𝑟𝑥𝑥 𝑘 =
1
5 − 𝑘
𝑖
=1
σ5−𝑘
𝑥𝑖 − 0.474 𝑥 𝑖+𝑘 − 0.474
0.15333
𝑘 = 1,2,3,4
𝑟𝑥𝑥
1 −0.28341
2 =
3 0.15333
= −0.61613

Example (8/10)
𝒙𝒊 0.44 0.81 0.14 0.05 0.93
Independence
1 2 3 4 5
𝑥ҧ = 0.474
𝑠𝑥 = 0.15333
𝑟𝑥𝑥 𝑘 =
1
5 − 𝑘
𝑖
=1
σ5−𝑘
𝑥𝑖 − 0.474 𝑥 𝑖+𝑘 − 0.474
0.15333
𝑘 = 1,2,3,4
𝑟𝑥𝑥
1 0.167632
3 =
2 0.15333
= 0.546638

Example (9/10)
𝒙𝒊 0.44 0.81 0.14 0.05 0.93
Independence
1 2 3 4 5
𝑥ҧ = 0.474
𝑠𝑥 = 0.15333
𝑟𝑥𝑥 𝑘 =
1
5 − 𝑘
𝑖
=1
σ5−𝑘
𝑥𝑖 − 0.474 𝑥 𝑖+𝑘 − 0.474
0.15333
𝑘 = 1,2,3,4
𝑟𝑥𝑥
1 −0.0155
4 =
1 0.15333
= −0.10112

Example (10/10)
𝒓𝒙𝒙(𝒌) −0.28595 −0.61613 0.546638 −0.10112
Independence
1 2 3 4
𝑘

Example (10/10)
Independence
You can
indicate independence
𝒓𝒙𝒙(𝒌) −0.28595 −0.61613 0.546638 −0.10112
1 2 3 4
𝑘
𝐴𝑣𝑔 𝐴𝑏𝑠 𝑟𝑥𝑥 𝑘 = 0.387456

Another forAutocorrelation (1/6)
As an example, consider the following sequence of numbers,
read from left to right:
Independence

As an example, consider the following sequence of numbers,
read from left to right:
From a visual inspection, these numbers appear random, and
they would probably pass all the tests presented to this point.
However, an examination of the 5th, 10th, 15th (every five
numbers beginning with the fifth), and so on, indicates a very
large number in that position.
Independence

The test to be described shortly requires the computation of the
autocorrelation between every 𝑃 numbers (𝑃 is also known as
the lag), starting with the 𝑖th number. Thus, the autocorrelation
𝜌𝑖𝑃 between the following numbers would be of interest:
𝑅𝑖 , 𝑅𝑖+𝑃, 𝑅𝑖+2𝑃, … , 𝑅𝑖+(𝑀+1)𝑃.
The value 𝑀 is the largest integer such that 𝑖 + (𝑀 + 1)𝑃 ≤ 𝑁,
where 𝑁 is the total number of values in the sequence.
Independence
𝑵 − 𝒊 − 𝒍
𝑴 =
𝒍

A nonzero autocorrelation implies a lack of independence, so
the following two-tailed test is appropriate:
Independence

For large values of 𝑀, the distribution of the estimator of 𝜌𝑖𝑃,
denoted 𝜌
ො 𝑖𝑃, is approximately normal if the values
𝑅𝑖 , 𝑅𝑖+𝑃, 𝑅𝑖+2𝑃, … , 𝑅𝑖+(𝑀+1)𝑃 are uncorrelated. Then the test
statistic can be formed as follows:
which is distributed normally with a mean of zero and a
variance of 1, under the assumption of independence, for large
𝑀.
Independence

Another forAutocorrelation (5/6) Independence

After computing 𝑍0, do not reject the null hypothesis of
−𝑧𝛼/2 ≤ 𝑍0 ≤ 𝑧𝛼/2
independence if where 𝛼 is the level of
significance and 𝑧𝛼/2 is obtained from the following Table.
Independence

−𝑧𝛼/2 ≤ 𝑍0 ≤ 𝑧𝛼/2
Independence

Another for Autocorrelation (6/3)
−𝑧𝛼/2 ≤ 𝑍0 ≤ 𝑧𝛼/2
Independence
Hagag S Modeling and tion 81

Example (1/7) Independence

𝑴 =
𝑵 − 𝒊 − 𝒍
𝒍
=
𝟑𝟎 − 𝟑 − 𝟓
𝟓
= 𝟒

Therefore, the hypothesis of independence
cannot be rejected on the basis of this test.

Discussion Question
Draw the bar chart for the following data:
Object Speed (km/hr.)
Ant 0.36
Car 200
Rocket 5,760
Light 1,079,252,848

Discussion Question
Draw the bar chart for the following data:

Dr. AhmedHagag
ahagag@
f
c
i
.bu.edu.eg

BU_FCAI_SCC430_Modeling&Simulation_Ch05-P2.pptx

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