1. CPSC 531:Random-Variate Generation
Instructor: Anirban Mahanti
Office: ICT 745
Class Location: TRB 101
Lectures: TR 15:30 – 16:45 hours
Class web page:
Notes from “Simulation” by Sheldon Ross, Academic Press, 2002.
CPSC 531: RV Generation

2. Outline
Talk about generating both discrete and continuous random variates
Inverse Transform Method (focus on this one)
The Rejection Method
The Composition Method
First consider the Inverse Transform Method
Discrete Random Variates
Continuous Random Variate
CPSC 531: RV Generation

3. Discrete Random Variates
Suppose we want to generate the value of a discrete random variable X that has the following probability mass function
The Inverse Transform Method is as follows: Generate u = U(0,1) and set X as follows:
CPSC 531: RV Generation

4. Discrete Random Variate Algorithm
Generate u=U(0,1)
If u<p0, set X=x0, return;
If u<p0+p1, set X=x1, return; …
After generating a random number u, we find the value of X by determining the interval [F(xj-1),F(xj)] in which u lies. That is, the inverse of F(u) is determined.
Time complexity of generating a random variable is proportional to the number of intervals to be searched.
CPSC 531: RV Generation

5. Example
Assume p1= 0.2, p2 = 0.1, p3 = 0.25, p4 = Determine RNG such that P{X=j} = pj?
Version 1
Generate u=U(0,1)
If u<0.2, set X=1, return;
If u<0.3, set X=2, return;
If u<0.55, set X=3, return;
Set X=4, return;
Version 2 (efficient)
Generate u=U(0,1)
If u<0.45, set X=4, return;
If u<0.7, set X=3, return;
If u<0.9, set X=1, return;
Set X=2, return;
CPSC 531: RV Generation

6. Geometric Random Variables
X is a geometric random variable with parameter p if it has the following probability mass function:
Generate value of X by generating random number u and setting X to j such that:
1 - qj-1 ≤ u < 1 – qj
CPSC 531: RV Generation

7. Geometric Random Variables
Note that 1 - qj-1 ≤ u < 1 – qj is equivalent to
qj < 1 - u ≤ qj-1
Therefore, X = min{j: qj < 1 – u}
CPSC 531: RV Generation

8. Generating Poisson Random Variables
Poisson Distribution with rate λ.
Poisson RV Algorithm (λ)
1. Generate u=U(0,1);
2. i=0, p = e-λ, F=p;
3. if u < F, set X = i and return;
4. p= λp/(i+1), F = F+p, i=i+1;
5. Go to step 3;
CPSC 531: RV Generation

9. Generating Binomial Random Variables
X is Binomial random variable with parameters n and p
Binomial RV Algorithm (n, p)
1. Generate u=U(0,1);
2. c=p/(1-p), i=0, d=(1-p)n, F=d
3. if u < F, set X = i and return;
4. d=[c(n-i)/(i+1)]d, F=F+d, i=i+1;
5. Go to step 3;
CPSC 531: RV Generation

10. Inverse Transform Method
Let r be a uniform (0, 1) random variable. For any continuous distribution F the random variable X defined by X=F-1(r) has distribution F. r1 x1
r = F(x)
CPSC 531: RV Generation

11. Exponential Distribution [BCNN05]
Exponential cdf:
To generate X1, X2, X3 …
r = F(x)
= 1 – e-lx for x ³ 0
Xi = F-1(Ri)
= -(1/l) ln(1-Ri) [Eq’n 8.3]
where r is a variate generated from Uniform (0,1)
and F-1(r) is the solution to the equation r = F(X)
Figure: Inverse-transform technique for exp(l = 1)
CPSC 531: RV Generation

12. Other Examples We will work out the following on the board:
Uniform in the interval [a,b]
Weibull Distribution
CPSC 531: RV Generation

13. Rejection technique [BCNN05]
Useful particularly when inverse cdf does not exist in closed form, a.k.a. thinning
Illustration: To generate random variates, X ~ U(1/4, 1)
R does not have the desired distribution, but R conditioned (R’) on the event {R ³ ¼} does.
Efficiency: Depends heavily on the ability to minimize the number of rejections.
Generate R
Condition
Output R’
yes no
Procedures:
Step 1. Generate R ~ U[0,1]
Step 2a. If R >= ¼, accept X=R.
Step 2b. If R < ¼, reject R, return to Step 1
CPSC 531: RV Generation

14. NSPP [BCNN05]
Non-stationary Poisson Process (NSPP): a Possion arrival process with an arrival rate that varies with time
Idea behind thinning:
Generate a stationary Poisson arrival process at the fastest rate, l* = max l(t)
But “accept” only a portion of arrivals, thinning out just enough to get the desired time-varying rate
Generate E ~ Exp(l*)
t = t + E no
Condition
R <= l(t)
yes
Output E ’~ t
CPSC 531: RV Generation

15. NSPP [BCNN05] Example: Generate a random variate for a NSPP
Data: Arrival Rates
Procedures:
Step 1. l* = max l(t) = 1/5, t = 0 and i = 1.
Step 2. For random number R = ,
E = -5ln(0.213) = 13.13
t = 13.13
Step 3. Generate R =
l(13.13)/l*=(1/15)/(1/5)=1/3
Since R>1/3, do not generate the arrival
Step 2. For random number R = ,
E = -5ln(0.553) = 2.96
t = = 16.09
Step 3. Generate R =
l(16.09)/l*=(1/15)/(1/5)=1/3
Since R<1/3, T1 = t = 16.09,
and i = i + 1 = 2
CPSC 531: RV Generation

16. Special Properties
Based on features of particular family of probability distributions
For example:
Direct Transformation for normal and lognormal distributions
Convolution
Beta distribution (from gamma distribution)
CPSC 531: RV Generation

17. Direct Transformation [BCNN05]
Approach for normal(0,1):
Consider two standard normal random variables, Z1 and Z2, plotted as a point in the plane:
B2 = Z21 + Z22 ~ chi-square distribution with 2 degrees of freedom = Exp(l = 2). Hence,
The radius B and angle f are mutually independent.
In polar coordinates:
Z1 = B cos f
Z2 = B sin f
CPSC 531: RV Generation

18. Direct Transformation [BCNN05]
Approach for normal(m,s2):
Generate Zi ~ N(0,1)
Approach for lognormal(m,s2):
Generate X ~ N((m,s2)
Xi = m + s Zi
Yi = eXi
CPSC 531: RV Generation

19. Summary Principles of random-variate generate via
Inverse-transform technique
Acceptance-rejection technique
Special properties
Important for generating continuous and discrete distributions
CPSC 531: RV Generation