1. Random-Variate Generation

2. 2 Purpose & Overview Develop understanding of generating samples from a specified distribution as input to a simulation model. Illustrate some widely-used techniques for generating random variates.  Inverse-transform technique  Convolution technique  Acceptance-rejection technique  A special technique for normal distribution

3. 3 Inverse-transform Technique The concept: For cdf function: r = F(x)  Generate R sample from uniform (0,1)  Find X sample: X = F -1 (R) r1r1 x1x1 r = F(x)

4. 4 Exponential Distribution [Inverse-transform] Exponential Distribution:  Exponential cdf:  To generate X 1, X 2, X 3,… generate R 1, R 2, R 3,… r = F(x) = 1 – e - x for x  0 X i = F -1 (R i ) = -(1/  ln(1-R i ) Figure: Inverse-transform technique for exp( = 1)

5. 5 Example: Generate 200 variates X i with distribution exp( = 1) Matlab Code for i=1:200, expnum(i)=-log(rand(1)); end Exponential Distribution [Inverse-transform] R and (1 – R) have U(0,1) distribution

6. 6 Uniform Distribution:  Uniform cdf:  To generate X 1, X 2, X 3,…, generate R 1, R 2, R 3,… Uniform Distribution [Inverse-transform]

7. 7 Example: Generate 500 variates X i with distribution Uniform (3,8) Matlab Code for i=1:500, uninum(i)=3+5*rand(1); end Uniform Distribution [Inverse-transform]

8. 8 Discrete Distribution [Inverse-transform] All discrete distributions can be generated by the Inverse-transform technique. p1p1 p 1 + p 2 p 1 + p 2 + p 3 a b c F(x)F(x) R1R1 General Form

9. 9 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:  Method - Given R, the generation scheme becomes: Consider R 1 = 0.73: F(x i-1 ) < R <= F(x i ) F(x 0 ) < 0.73 <= F(x 1 ) Hence, x 1 = 1

10. 10 Empirical Continuous Dist’n [Inverse-transform] When theoretical distribution is not applicable To collect empirical data:  Resample the observed data  Interpolate between observed data points to fill in the gaps For a small sample set (size n):  Arrange the data from smallest to largest  Assign the probability 1/n to each interval where

11. 11 Empirical Continuous Dist’n [Inverse-transform] Example: Suppose the data collected for 100 broken- widget repair times are: Consider R 1 = 0.83: c 3 = 0.66 < R 1 < c 4 = 1.00 X 1 = x (4-1) + a 4 (R 1 – c (4-1) ) = 1.5 + 1.47(0.83-0.66) = 1.75

12. 12 Convolution Technique Use for X = Y 1 + Y 2 + … + Y n Example of application  Erlang distribution Generate samples for Y 1, Y 2, …, Y n and then add these samples to get a sample of X.

13. 13 Example: Generate 500 variates X i 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 Erlang Distribution [Convolution]

14. 14 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  Step 1: Generate Y with pdf h(x)  Step 2: Generate U ~ Uniform(0,1) independent of Y  Step 3: Acceptance-Rejection technique Generate Y,U Condition Output X=Y yes no Efficiency parameter is c

15. 15 Triangular Distribution [Acceptance-Rejection] Generate Y~U(0,0.5) and U~U(0,1) Output X = Y yes no

16. 16 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<=2-4*Y i=i+1; X(i)=Y; end

17. 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  Poisson Distribution [Acceptance-Rejection]

18. Step 1. Set n = 0, and P = 1 Step 2. Generate a random number R n+1 and let P = P. R n+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? Poisson Distribution [Acceptance-Rejection]

19. 19 Approach for normal(0,1):  Consider two standard normal random variables, Z 1 and Z 2, plotted as a point in the plane:  B 2 = Z 2 1 + Z 2 2 ~ chi-square distribution with 2 degrees of freedom = Exp( = 1/2). Hence,  The radius B and angle  are mutually independent. Normal Distribution [Special Technique] In polar coordinates: Z 1 = B cos  Z 2 = B sin   Uniform(0,2 

20. 20 Normal Distribution [Special Technique] Approach for normal( ,   ):  Generate Z i ~ N(0,1) Approach for lognormal( ,   ):  Generate X ~ N( ,   ) Y i = e X i X i =  +  Z i

21. 21 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