1. Properties of Random Numbers
Lecture 5
Properties of Random Numbers

2. Random Number Generation
A simulation of any system or process in which there are inherently random components requires a method of generating or obtaining numbers that are random, in some sense.
The earliest methods were carried out by hands such as throwing dice.
As computers (and simulation) became more widely used, increasing attention was paid to methods of random number generation compatible with the computers work.

3. Random Number Generation (cont’)
Therefore, the research in the 1940’s and 1950’s turned to numerical or arithmetic ways to generate “random” numbers.
These method are sequential, with each new number being determined by one or several of its predecessors according to a fixed mathematical formula.
The first such arithmetic generate generator, proposed by von Neumann and Metropolic in the 1940’s is the famous midsquare method.

4. Midsquare Method
An example of midsquare method for generating a uniform [0, 1] random numbers: i Zi Ui
Zi2
7182 - 1
5811
0.5811 2
7677
0.7677 3
9363
0.9363 4
6657
0.6657 5
3156
0.3156 .

5. Midsquare Method (cont’)
Drawbacks for midsquare method
If Zi = 0 for some i, then Uj = Zj = 0 for all j > i.
Ui+1 is determined by Ui, i.e. Ui+1 is a function of Ui. Therefore, Ui and Ui+1 are not independent.
Nowadays, the random numbers generated from computers are more complicated and appear to be independent, in that they pass a series of statistical test.
But after all, the random numbers generated from computers are still not purely random
(pseudorandom, but too awkward to use this term).

6. Good Random Number Generation
A “good” arithmetic random number generator should posses several properties:
Above all, the numbers produced should appear to be distributed uniformly on [0, 1] and should not exhibit any correlation with each other; otherwise, the simulation’s results may be completely invalid.
From a practical standpoint, we would naturally like the generator to be fast and avoid the need for a lot of storage.
We would like to be able to reproduce a given stream of random numbers exactly, for at least two reasons. First, this can sometimes make debugging or verification f the computer program easier. More important, we might want to use identical random numbers in simulating different systems in order to obtain a more precise comparison.

7. Good Random Number Generation (cont’)
There should be provision in the generator for producing several separate “stream” of random numbers. As we shall see, a stream is simply a subsegment of the numbers produced by the generator, with one stream beginning where the previous stream ends.

8. Methods for Generation
Inverse Transform Technique (ITT)
Exponential distribution
Uniform distribution
Triangular distribution
Empirical discrete distribution
Empirical continuous distribution
Acceptance-Rejection Technique

9. Inverse Transform Technique
The inverse transform technique can be used to sample from the exponential, the Weibull and the uniform distributions, and empirical distribution.
Additionally, it is the underlying principle for sampling from a wide variety of discrete distributions.
A step by step procedure for the inverse transform techniques, illustrated by the exponential distribution, is as follows:
Step 1 Compute the cdf of the desired random variable.
For the exponential distribution, the cdf is

10. Inverse Transform Technique (cont’)
Step 2 Set F(x) = R on the range of X.
For the exponential distribution, it becomes
on the range
Since X is a random variable (with the exponential distribution in this case), it follows that
is also a random variable, here called R. As will be shown later, R has a uniform distribution over the interval (0, 1).

11. Inverse Transform Technique (cont’)
Step 3 Solve the equation F(X) = R for X in terms of R. For the exponential distribution, the solution proceeds as follows:

12. Inverse Transform Technique (cont’)
Step 4 Generate (as needed) uniform random number R1, R2, R3,…, and compute the desired random variable by
where
One simplification that is usually employed is to replace
by to yield
which is justified since both and are uniformly distributed on (0, 1).

13. Uniform Distribution
Consider a random variable X that is uniformly distributed on the interval [a, b].
The pdf of X is given by
Step 1 The cdf is given by
Step 2 Set F(X) = (X – a) / (b – a) = R
Step 3 Solving for X in terms of R yields X = a + (b – a) R b a

14. Triangular Distribution
Consider a random variable X which has pdf
This distribution is called a triangular distribution with endpoints (0, 2) and mode at 1.
Step 1 The cdf is given by
f(x) 1 1 2

15. Empirical Discrete Distributions
Service Time
Probability
Cumulative Probability
Random Number 1
0.10
0.01 – 0.10 2
0.20
0.30
0.11 – 0.30 3
0.60
0.31 – 0.60 4
0.25
0.85
0.61 – 0.85 5
0.95 6
0.05
1.00
0.96 – 000 1
p(x) 1 2 3 4
0.3
0.2
0.1 5 6 1
Draw a sample
from a 0-1 uniform
distribution
Convert the 0-1 sample to an equivalent sample from the target population 2

16. Empirical Continuous Distribution
If the modeler has been unable to find a theoretical distribution that provides a good model for the input data, then it may be necessary to use the empirical distribution of the data.
Suppose that 100 broken-widget repair times have been collected. The data are summarized in the following table in terms of the number of observations in various interval. For example, there were 31 observations between 0 and 0.5 hour, 10 between 0.5 and 1 hour, and so on.
Interval (Hours)
Frequency
Relative Frequency
Cumulative Frequency
0.0  x  0.5 31
0.31
0.5 < x  1.0 10
0.10
0.41
1.0 < x  1.5 25
0.25
0.66
1.5 < x  2.0 24
0.24
1.00

17. Empirical Continuous Distribution (cont’)
(1.5, 0.66)
(2.0, 1.0)
(1.0, 0.41)
(0.5, 0.31)
0.5
1.0
1.5
2.0
X1=1.75
0.8
0.6
0.4
0.2
R1=0.83
F(x) x
Repair times
Cumulative probability

18. Acceptance-Rejection Technique
Suppose that we need to devise a method for generating random variates, X, uniformly distributed between ¼ and 1.
Step 1: Generate a random number u ~ U (0, 1)
Step 2a: If , accept X = u, then go to step 3.
Step 2b: If , reject u, and return to step 1.
Step 3: If another uniform random variate on [1/4, 1] is needed, repeat the procedure beginning at step 1. If not, stop.

19. Tests for Random Numbers
The desirable properties of random numbers
uniformity and independence
To insure that these desirable properties are achieved, a number of tests can be performed.
The tests can be placed in two categories according to the properties of interest.
Test for uniformity
Test for independence

20. Tests for Random Numbers (cont’)
Frequency test: Uses the chi-square test to compare the distribution of the set of numbers generated to a uniform distribution.
Runs test: Tests the runs up and down or the runs above and below the mean by comparing the actual values to expected values. The statistic for comparison is the chi-square.

21. Frequency Test FREQUENCY TEST
Random numbers about from the uniform distribution and several tests have been developed to test for this condition. We will consider the 2 goodness-of-fit test.
The goodness-of-fit test requires that:
50 observations in total
Expected frequency of at least five in each class
The following table shows the results of placing a total of 100 observations in 10 evenly spaced classes

22. Frequency Test (cont’)
Classes
Observed Frequency
Expected Frequency
(fo – fe)2 / fe
0.00 – 0.10 9 10
0.10
0.10 – 0.20 12
0.40
0.20 – 0.30
0.00
0.30 – 0.40 11
0.40 – 0.50 8
0.50 – 0.60
0.60 – 0.70
0.70 – 0.80 7
0.90
0.80 – 0.90
0.90 – 1.00
100
2.40

23. Frequency Test (cont’)
The question is,
Do these numbers come from the uniform distribution?
Calculating the 2 statistic from the data using the equation
Gives a value of 2 = In testing the null hypothesis that the random numbers come from the uniform distribution,
H0 : Ri ~ U [0, 1]
one compares the calculated 2 to the value obtained from the table based on (10-1) = 9 degree of freedom and a  = 0.05.
This 2 value is found to be , which is larger than the calculated 2 value.
Therefore, we accepted the null hypothesis, and find our random number generation acceptable.

24. Run Up and Down Test RUNS UP AND DOWN TEST
Numbers can pass a uniformity test and still not be random.
For example, the numbers 0.00, 0.10, 0.20, 0.30, 0.40, …obviously are not random.
The numbers also must be sequentially random to be judged truly random.
A variety of runs test can be used for this purpose.
We will consider a run up and down test.

25. Run Up and Down Test (cont’)
In a sequence of numbers, if a number is followed by a larger number, this is an upward run.
Likewise, a number followed by a smaller number is a downstream run.
If the numbers are truly random, one would expect to find a certain numbers of runs up and down.
In a sequence of N numbers, one should expect to find runs equal to the following equation:

26. Run Up and Down Test (cont’)
As an example, assume that the following 40 numbers have been generated.
0.43, 0.32, 0.48, 0.23, 0.90, 0.72, 0.94, 0.11, 0.14, 0.67,
0.61, 0.25, 0.45, 0.56, 0.87, 0.54, 0.01, 0.64, 0.65, 0.32, 0.03,
0.93, 0.08, 0.58, 0.41, 0.32, 0.03, 0.18, 0.90, 0.74, 0.32,
0.75, 0.42, 0.71, 0.66, 0.03, 0.44, 0.99, 0.40, 0.51 – + – + – + – + – + – + – + – + – + – + – + – + – +

27. Run Up and Down Test (cont’)
One should expect to find runs
There were 26 runs in the sequence of numbers.
We consider to accept the generated numbers are random.

28. Table of Random Digits
10097
32533
76520
13586
34673
54876
80959
09117
39292
74945
37542
04805
64894
74296
24805
24037
20636
10402
00822
91655
08422
68953
19645
09303
23209
02560
15953
34764
35080
33606
99019
02529
09376
70715
38311
31165
88676
74397
04436
27659
12807
99970
80157
36147
64032
36653
98951
16877
12171
76833
66065
74717
34072
76850
36697
36170
65813
39885
11190
29170
31060
10805
45571
82406
35303
42614
86799
07439
23403
09732
85269
77602
02051
65692
68665
74818
73053
85247
18623
88579
63573
32135
05325
47048
90553
57548
28468
28709
83491
25624
73796
45753
03529
64778
35808
34282
60935
20344
35273
88435
98520
17767
14905
68607
22109
40558
60970
93433
50500
73998
11805
05431
39808
27732
50725
68248
29405
24201
52775
67851 …