Mahmoud Abas, Full-Stack Software Engineer & AI

Modeling and Simulation (2026-2027) - Random Numbers: Generation and Testing

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Modeling and Simulation - Lesson 3

Random Numbers: Generation and Testing

Two generators that produce random numbers, and three tests that check them: Exercise 3 solved step by step.

Objectives

Compute a multiplicative generator and find its cycle and period
Compute random numbers in [0, 1] with a linear congruential generator
Check uniformity with the Kolmogorov-Smirnov test
Check uniformity with the chi-square test
Check independence with the autocorrelation test
Read the critical value from the correct table and state the decision

Part 1

What a random number must do

The two properties every test checks

Two properties

Property	Meaning	Test
Uniformity	Every part of `[0, 1]` gets its fair share	Kolmogorov-Smirnov, chi-square
Independence	A number tells nothing about the next ones	Autocorrelation

Part 2

Generating random numbers

The multiplicative generator and the LCG

The multiplicative generator

Z_i = (a Z_i-1) mod m

Z0 is the seed, the starting value
a is the multiplier, m is the modulus
x mod m is the remainder of x ÷ m, so every Zi is from 0 to m - 1

Question 1

Q1 (a): Z0 = 1, a = 11, m = 16

i	(11 × Z) mod 16	Zi
1	11 × 1 = 11 → 11	11
2	11 × 11 = 121 → 9	9
3	11 × 9 = 99 → 3	3
4	11 × 3 = 33 → 1	1
5	11 × 1 = 11 → 11	11 (cycling)

Question 1

Q1 (b): Z0 = 2, a = 11, m = 16

i	(11 × Z) mod 16	Zi
1	11 × 2 = 22 → 6	6
2	11 × 6 = 66 → 2	2
3	11 × 2 = 22 → 6	6 (cycling)

Question 1, part 1 of 3

Q1 (c): Z0 = 1, a = 2, m = 13

i	(2 × Z) mod 13	Zi
1	2 × 1 = 2	2
2	2 × 2 = 4	4
3	2 × 4 = 8	8
4	2 × 8 = 16 → 3	3
5	2 × 3 = 6	6

Question 1, part 2 of 3

Q1 (c): Z0 = 1, a = 2, m = 13

i	(2 × Z) mod 13	Zi
6	2 × 6 = 12	12
7	2 × 12 = 24 → 11	11
8	2 × 11 = 22 → 9	9
9	2 × 9 = 18 → 5	5
10	2 × 5 = 10	10

Question 1, part 3 of 3

Q1 (c): Z0 = 1, a = 2, m = 13

i	(2 × Z) mod 13	Zi
11	2 × 10 = 20 → 7	7
12	2 × 7 = 14 → 1	1 (cycling)

Q1 (c)

A slip on the exercise sheet

Question 1

Q1 (d): Z0 = 2, a = 3, m = 13

i	(3 × Z) mod 13	Zi
1	3 × 2 = 6	6
2	3 × 6 = 18 → 5	5
3	3 × 5 = 15 → 2	2 (cycling)

Q1 in one table

Case	Z0, a, m	Period
(a)	1, 11, 16	4 = m / 4
(b)	2, 11, 16	2
(c)	1, 2, 13	12 = m - 1
(d)	2, 3, 13	3

The linear congruential generator (LCG)

X_i = (a X_i-1 + c) mod m

R_i = X_im

c is the increment, added before the remainder
Every Xi is an integer from 0 to m - 1
Divide by m to get a random number in [0, 1]

Question 2

Q2: X0 = 27, a = 8, c = 47, m = 100

i	8 × X + 47	Xi	Ri
1	216 + 47 = 263	63	0.63
2	504 + 47 = 551	51	0.51
3	408 + 47 = 455	55	0.55

A slip on the exercise sheet

Part 3

Testing uniformity

Kolmogorov-Smirnov and chi-square

Two tests, one decision rule

Test	Statistic	Table
Kolmogorov-Smirnov	`D`	KS critical values, by `N` and `α`
Chi-square	`χ0²`	Chi-square values, by degrees of freedom and `α`

The Kolmogorov-Smirnov test

Sort the numbers in ascending order: R(1) ≤ R(2) ≤ ... ≤ R(N)
Compute D+, the largest distance above the ideal line
Compute D-, the largest distance below it
Take D = max(D+, D-)
Read Dα from the KS table: if D < Dα, H0 is not rejected

D+ and D-

D⁺ = max [ iN - R_(i) ]

D^- = max [ R_(i) - i - 1N ]

D = max (D⁺, D^-)

Question 3

Q3: 0.54, 0.73, 0.98, 0.11, 0.68

Kolmogorov-Smirnov test with α = 0.05, N = 5

Question 3

Q3: computing D+

i	R(i)	i / N	i / N - R(i)
1	0.11	0.20	0.09
2	0.54	0.40	-0.14
3	0.68	0.60	-0.08
4	0.73	0.80	0.07
5	0.98	1.00	0.02

Question 3

Q3: computing D-

i	R(i)	(i - 1) / N	R(i) - (i - 1) / N
1	0.11	0.00	0.11
2	0.54	0.20	0.34
3	0.68	0.40	0.28
4	0.73	0.60	0.13
5	0.98	0.80	0.18

Question 3

Q3: the decision

D = max (0.09, 0.34) = 0.34

D_0.05 = 0.565 (N = 5)

The chi-square test

Divide [0, 1] into n equal class intervals
Count Oi, the values that fall in interval i
Expected count in every interval: Ei = N / n
Compute χ0², the sum of (Oi - Ei)² / Ei
Read χ² at α and n - 1 degrees of freedom: if χ0² is smaller, H0 is not rejected

The chi-square statistic

χ₀² = ∑ (O_i - E_i)²E_i

E_i = Nn

Question 4

Q4: 100 numbers, α = 0.05

N = 100 values, listed in full on the lesson page
n = 10 intervals: [0, 0.1], (0.1, 0.2], ... , (0.9, 1.0]
Every interval expects Ei = 100 / 10 = 10 values

Question 4

Q4: intervals 1 to 5

Interval	Oi	Oi - Ei	(Oi - Ei)² / Ei
[0, 0.1]	8	-2	0.4
(0.1, 0.2]	8	-2	0.4
(0.2, 0.3]	10	0	0
(0.3, 0.4]	9	-1	0.1
(0.4, 0.5]	12	2	0.4

Question 4

Q4: intervals 6 to 10

Interval	Oi	Oi - Ei	(Oi - Ei)² / Ei
(0.5, 0.6]	8	-2	0.4
(0.6, 0.7]	10	0	0
(0.7, 0.8]	14	4	1.6
(0.8, 0.9]	10	0	0
(0.9, 1.0]	11	1	0.1

Question 4

Q4: the decision

χ₀² = 0.4 + 0.4 + 0 + 0.1 + 0.4 + 0.4 + 0 + 1.6 + 0 + 0.1 = 3.4

χ²_{0.05, 9} = 16.9

A slip on the exercise sheet

Part 4

Testing independence

The autocorrelation test

The numbers we test

i: the position of the first tested number
m: the lag, the distance between two tested numbers (not the modulus)
N: the number of values in the sequence
M: the largest integer that keeps the last tested number inside the sequence

i + (M + 1) m ≤ N

The autocorrelation statistic

ρ_im = 1M + 1 [ ∑ R_i+km R_i+(k+1)m ] - 0.25

σ = √(13M + 7)12 (M + 1)

Z₀ = ρ_imσ

The decision rule

α = 0.05, so α/2 = 0.025
Look for the cumulative probability 0.975 in the normal table
zα/2 = 1.96: the limits are -1.96 and 1.96

Question 5

Q5: the 3rd, 8th, 13th, ... numbers

30 numbers, α = 0.05

Start at the 3rd number: i = 3
From the 3rd to the 8th: m = 5
Number of values: N = 30

3 + (M + 1) × 5 ≤ 30 ⇒ 8 + 5M ≤ 30 ⇒ M ≤ 4.4 ⇒ M = 4

Question 5

Q5: the tested numbers

Position	Value	Position	Value
R3	0.23	R18	0.27
R8	0.28	R23	0.05
R13	0.33	R28	0.36

Question 5

Q5: the five products

k	Pair	Product
0	0.23 × 0.28	0.0644
1	0.28 × 0.33	0.0924
2	0.33 × 0.27	0.0891
3	0.27 × 0.05	0.0135
4	0.05 × 0.36	0.0180

Question 5

Q5: the decision

ρ₃₅ = 0.27745 - 0.25 = 0.0555 - 0.25 = -0.1945

σ = √5960 = 7.6860 = 0.128

Z₀ = -0.19450.128 = -1.5195

Part 5

Common mistakes

and practice

Common mistakes

An LCG gives integers: the random number is Xi / m
The KS test needs the numbers sorted first
D+ uses i/N - R(i), D- uses R(i) - (i - 1)/N
Chi-square degrees of freedom: n - 1 intervals, not N - 1 values
A statistic inside the limit means H0 is not rejected
M is rounded down: M ≤ 4.4 gives M = 4

Practice

Multiplicative generator Z0 = 3, a = 5, m = 16: values until it cycles, and the period
LCG X0 = 7, a = 21, c = 13, m = 100: the first three numbers in [0, 1]
KS test, α = 0.05: 0.44, 0.81, 0.14, 0.05, 0.93
Chi-square, α = 0.05: counts 12, 7, 9, 13, 9 in 5 equal intervals
Autocorrelation on the Q5 data: the 1st, 7th, 13th, ... numbers

Key Takeaways

Multiplicative generator: Zi = (a Zi-1) mod m; seed, multiplier and modulus set the period
LCG: Xi = (a Xi-1 + c) mod m, then Ri = Xi / m
Kolmogorov-Smirnov: sort, compute D+ and D-, compare D with Dα
Chi-square: sum (Oi - Ei)² / Ei, compare with the table at n - 1 degrees of freedom
Autocorrelation: find M, compute Z0, compare with ±zα/2
Inside the limit: the hypothesis is not rejected

Open this lesson

Mahmoud AbasRandom Numbers: Generation and Testing

Property

Meaning

Test

Uniformity

Every part of [0, 1] gets its fair share

Kolmogorov-Smirnov, chi-square

Independence

A number tells nothing about the next ones

Autocorrelation

(11 × Z) mod 16

11 × 1 = 11 → 11

11 × 11 = 121 → 9

11 × 9 = 99 → 3

11 × 3 = 33 → 1

11 × 1 = 11 → 11

11 (cycling)

(11 × Z) mod 16

11 × 2 = 22 → 6

11 × 6 = 66 → 2

11 × 2 = 22 → 6

6 (cycling)

(2 × Z) mod 13

2 × 1 = 2

2 × 2 = 4

2 × 4 = 8

2 × 8 = 16 → 3

2 × 3 = 6

(2 × Z) mod 13

2 × 6 = 12

2 × 12 = 24 → 11

2 × 11 = 22 → 9

2 × 9 = 18 → 5

2 × 5 = 10

(2 × Z) mod 13

2 × 10 = 20 → 7

2 × 7 = 14 → 1

1 (cycling)

(3 × Z) mod 13

3 × 2 = 6

3 × 6 = 18 → 5

3 × 5 = 15 → 2

2 (cycling)

Case

Z0, a, m

Period

(a)

1, 11, 16

4 = m / 4

(b)

2, 11, 16

(c)

1, 2, 13

12 = m - 1

(d)

2, 3, 13

8 × X + 47

216 + 47 = 263

0.63

504 + 47 = 551

0.51

408 + 47 = 455

0.55

Test

Statistic

Table

Kolmogorov-Smirnov

D

KS critical values, by N and α

Chi-square

χ0²

Chi-square values, by degrees of freedom and α

R(i)

i / N

i / N - R(i)

0.11

0.20

0.09

0.54

0.40

-0.14

0.68

0.60

-0.08

0.73

0.80

0.07

0.98

1.00

0.02

R(i)

(i - 1) / N

R(i) - (i - 1) / N

0.11

0.00

0.11

0.54

0.20

0.34

0.68

0.40

0.28

0.73

0.60

0.13

0.98

0.80

0.18

Interval

Oi - Ei

(Oi - Ei)² / Ei

[0, 0.1]

-2

0.4

(0.1, 0.2]

-2

0.4

(0.2, 0.3]

(0.3, 0.4]

-1

0.1

(0.4, 0.5]

0.4

Interval

Oi - Ei

(Oi - Ei)² / Ei

(0.5, 0.6]

-2

0.4

(0.6, 0.7]

(0.7, 0.8]

1.6

(0.8, 0.9]

(0.9, 1.0]

0.1

Position

Value

Position

Value

0.23

R18

0.27

0.28

R23

0.05

R13

0.33

R28

0.36

Pair

Product

0.23 × 0.28

0.0644

0.28 × 0.33

0.0924

0.33 × 0.27

0.0891

0.27 × 0.05

0.0135

0.05 × 0.36

0.0180