{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecturenotes05

# lecturenotes05 - MS&E 223 Simulation Peter J Haas Lecture...

This preview shows pages 1–4. Sign up to view the full content.

MS&E 223 Lecture Notes #5 Simulation Generation Of Non-Uniform Random Numbers Peter J. Haas Spring Quarter 2009-10 Generation of Non-Uniform Random Numbers Refs : Law, Ch. 8, and Devroye, Non-Uniform Random Variate Generation (watch for typos!) Problem : Given a uniform random variable U, generate a random variable X having a prescribed cumulative distribution function F X ) ( . We previously discussed the inversion method. While inversion is a very general method, it may be computationally expensive. In particular, computing 1 X F ( ) may have to be implemented via a numerical root-finding method in many cases. Therefore, we will now describe other methods for non-uniform random number generation. 1. Acceptance-Rejection Method We now discuss a method that is well-suited to generating random variates with an easily-calculated density. This method is called acceptance-rejection. Goal : Generate a random variate X having given probability density f X (x), where f X (x) is positive only on the interval [a, b] (where < a < b < - ). Enclose the density in a rectangle R having base (b a) and height m, where m = sup f (x) a x b X . f X (x) a b x m Suppose we throw down points uniformly in the rectangle R (denote these points by the symbol x). Throw away (or reject) the points above the density f X (x). Claim : The x-coordinate of each accepted point (denoted by the symbol ) has density f X (x). Proof : Let be the (x, y) coordinates of a random point distributed uniformly in R. Then, for a x b, 1 2 (Z ,Z ) . 1 1 P(Z x, acceptance) P(Z x , Z f (Z )) = 2 X 1 1 But is just the probability that falls into the shaded region below: 1 2 X P(Z x , Z f (Z )) 1 2 (Z ,Z ) Page 1 of 9

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document