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Random-Number
Generation
Properties of Random Numbers

Random Number, Ri, must be independently drawn from a
uniform distribution with pdf:
U(0,1)
1, 0  x  1
f ( x)  
0, otherwise
1

2
x
E ( R)  xdx 
0
2

1
0
1

2
Figure: pdf for
random numbers
Two important statistical properties:


Uniformity
Independence.
2
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.
3
Linear Congruential Method

[Techniques]
To produce a sequence of integers, X1, X2, … between 0
and m-1 by following a recursive relationship:
X i 1  (aX i  c) mod m, i  0,1,2,...
The
multiplier


The
increment
The
modulus
The selection of the values for a, c, m, and X0 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:
Ri 
Xi
, i  1,2,...
m
4
Examples


[LCM]
Use X0 = 27, a = 17, c = 43, and m = 100.
The Xi and Ri values are:
X1 = (17*27+43) mod 100 = 502 mod 100 = 2,
X2 = (17*2+43) mod 100 = 77,
X3 = (17*77+43) mod 100 = 52,
X4 = (17* 52 +43) mod 100 = 27,
R1 = 0.02;
R2 = 0.77;
R3 = 0.52;
R3 = 0.27
“Numerical Recipes in C” advocates the generator
a = 1664525, c = 1013904223, and m = 232
 Classical LCG’s can be found on

http://random.mat.sbg.ac.at
5
Characteristics of a Good Generator
[LCM]

Maximum Density



Such that the values assumed by Ri, i = 1,2,…, leave no large
gaps on [0,1]
Problem: Instead of continuous, each Ri is discrete
Solution: a very large integer for modulus m


Maximum Period



Approximation appears to be of little consequence
To achieve maximum density and avoid cycling.
Achieve by: proper choice of a, c, m, and X0.
Most digital computers use a binary representation of
numbers

Speed and efficiency are aided by a modulus, m, to be (or close
to) a power of 2.
6
A Good LCG Example
X=2456356; %seed value
for i=1:10000,
X=mod(1664525*X+1013904223,2^32);
U(i)=X/2^32;
end
edges=0:0.05:1;
M=histc(U,edges);
bar(M);
hold;
figure;
hold;
for i=1:5000,
plot(U(2*i-1),U(2*i));
end
7
Randu (from IBM, early 1960’s)
X=1; %seed value
for i=1:10000,
X=mod(65539*X+57,2^31);
U(i)=X/2^31;
end
edges=0:0.05:1;
M=histc(U,edges);
bar(M);
hold;
figure;
hold;
for i=1:3333,
plot3(U(3*i-2),U(3*i-1),U(3*i));
end
Marsaglia
Effect (1968)
8
Random-Numbers Streams

[Techniques]
The seed for a linear congruential random-number generator:

Is the integer value X0 that initializes the random-number sequence.
 Any value in the sequence can be used to “seed” the generator.

A random-number stream:
Refers to a starting seed taken from the sequence X0, X1, …, XP.
 If the streams are b values apart, then stream i could defined by starting
seed:
Si  X b ( i 1)




Older generators: b = 105; Newer generators: b = 1037.
A single random-number generator with k streams can act like k
distinct virtual random-number generators
To compare two or more alternative systems.

Advantageous to dedicate portions of the pseudo-random number
sequence to the same purpose in each of the simulated systems.
9
Random-Variate
Generation
Purpose & Overview

Develop understanding of generating samples
from a specified distribution as input to a
simulation model.

Illustrate some widely-used
generating random variates.
techniques
for
 Inverse-transform
technique
 Convolution technique
 Acceptance-rejection technique
 A special technique for normal distribution
11
Inverse-transform Technique

The concept:
For cdf function: r = F(x)
 Generate R sample from uniform (0,1)
 Find X sample:
r = F(x)
r1
X = F-1(R)
x1

1
P
r
(
X

x
)

P
r
(
F
(
R
)

x
)

P
r
(
R

F
(
x
)
)(
F
x
)
12
Exponential Distribution

[Inverse-transform]
Exponential Distribution:

Exponential cdf:
r = F(x) = 1 – e-x

for x 0
To generate X1, X2, X3 ,…
generate R1, R2, R3 ,…
Xi = F-1(Ri) = -(1/ ln(1-Ri)
Figure: Inverse-transform
technique for exp( = 1)
13
Exponential Distribution


[Inverse-transform]
Example: Generate 200 variates Xi with distribution
exp(= 1)
Matlab Code
for i=1:200,
expnum(i)=-log(rand(1));
end
R and (1 – R) have
U(0,1) distribution
14
Uniform Distribution

[Inverse-transform]
Uniform Distribution:

Uniform cdf:
xa
0
xa

rFx
( )
axb
ba
bx

1

To generate X1, X2, X3 ,…, generate R1, R2, R3 ,…
X
a
(b
aR
)i
i
15
Uniform Distribution


[Inverse-transform]
Example: Generate 500 variates Xi with distribution
Uniform (3,8)
Matlab Code
for i=1:500,
uninum(i)=3+5*rand(1);
end
16
Discrete Distribution
[Inverse-transform]
All discrete distributions can be generated by the
Inverse-transform technique.
F(x)
x
p(x)
a
b
c
p1
p2
p3
p1 + p 2 + p 3
p1 + p 2
R1
p1
General Form
X

m
in
{
xF
: ()
x
r
}
a
b
c
17
Discrete Distribution

[Inverse-transform]
Example: Suppose the number of shipments, x, on the
loading dock of IHW company is either 0, 1, or 2


Data - Probability distribution:
x
p(x)
F(x)
0
1
2
0.50
0.30
0.20
0.50
0.80
1.00
Method - Given R, the generation
scheme becomes:
,
R0
.5
0

x 
1
, 0
.5R0
.8
2
.8R1
.0
, 0
Consider R1 = 0.73:
F(xi-1) < R <= F(xi)
F(x0) < 0.73 <= F(x1)
Hence, x1 = 1
18
Convolution Technique

Use for X = Y1 + Y2 + … + Yn

Example of application
 Erlang

distribution
Generate samples for Y1 , Y2 , … , Yn and then
add these samples to get a sample of X.
19
Erlang Distribution


[Convolution]
Example: Generate 500 variates Xi with distribution
Erlang-3 (mean: k/
Matlab Code
for i=1:500,
erlnum(i)=-1/6*(log(rand(1))+log(rand(1))+log(rand(1)));
end
20
Acceptance-Rejection technique



Useful particularly when inverse cdf does not exist in closed form
a.k.a. thinning
Steps to generate X with pdf f(x)

Step 0: Identify a majorizing function g(x) and a pdf h(x) satisfying
x, g(x)  f (x)

Efficiency parameter is c
c   g(x)dx  

g(x)
h(x) 
c

Generate Y,U
no
Step 1: Generate Y with pdf h(x)
 Step 2: Generate U ~ Uniform(0,1) independent of Y
f(Y
)
 Step 3:
U


se
tX
Y
Condition
yes
Output X=Y
g
(Y
)
e
lsere
p
e
a
tfro
m
S
te
p1
.
21
Triangular Distribution
[Acceptance-Rejection]
Generate Y~U(0,0.5)
and U~U(0,1)
no
yes
Output X = Y
22
Triangular Distribution
[Acceptance-Rejection]
Matlab Code: (for exactly 1000 samples)
i=0;
while i<1000,
Y=0.5*rand(1);
U=rand(1);
if Y<=0.25 & U<=4*Y | Y>0.25 & U<=2-4*Y
i=i+1;
X(i)=Y;
end
end
180
160
140
120
100
80
60
40
20
0
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
23
Poisson Distribution
[Acceptance-Rejection]
e  n
p ( n)  P ( N  n) 
, n  0,1, 2, ...
n!
– N can be interpreted as number of arrivals from a
Poisson arrival process during one unit of time
– Then, the time between the arrivals in the process are
exponentially distributed with rate 
N n 
n
A
i 1
i
n 1
 1   Ai
i 1
Poisson Distribution
n
A
i 1
i
n 1
n
 1   Ai   
i 1
i 1
[Acceptance-Rejection]
1

n 1
ln Ri  1   
i 1
n
n 1
i 1
i 1
1

ln Ri
  Ri  e    Ri
• Step 1. Set n = 0, and P = 1
• Step 2. Generate a random number Rn+1 and let P =
P. Rn+1
• Step 3. If P < e-, then accept N = n. Otherwise,
reject current n, increase n by one, and return to step 2
• How many random numbers will be used on the
average to generate one Poisson variate?
Normal Distribution

[Special Technique]
Approach for normal(0,1):

Consider two standard normal random variables, Z1 and Z2,
plotted as a point in the plane:
 Uniform(0,2
In polar coordinates:
Z1 = B cos 
Z2 = B sin 


B2 = Z21 + Z22 ~ chi-square distribution with 2 degrees of freedom
2ln
R
)1/2
= Exp( = 1/2). Hence, B(
The radius B and angle  are mutually independent.
1
/
2
Z

(

2
l
n)
R
o
s
(
2

R
)
1
1 c
2
1
/
2
Z

(

2
l
n)
R
i
n
(
2

R
)
2
1 s
2
26
Normal Distribution

[Special Technique]
Approach for normal(,):
 Generate
Zi ~ N(0,1)
Xi =  +  Zi

Approach for lognormal(,):
 Generate
X ~ N(,)
Yi = eXi
27
Normal Distribution
[Special Technique]
Generate 1000 samples of Normal(7,4)
 Matlab Code
for i=1:500,
R1=rand(1);
R2=rand(1);
Z(2*i-1)=sqrt(-2*log(R1))*cos(2*pi*R2);
Z(2*i)=sqrt(-2*log(R1))*sin(2*pi*R2);
end

for i=1:1000,
Z(i)=7+2*Z(i);
end
28