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# Acceptance-Rejection Technique to Generate Random Variate

• Example: use following steps to generate uniformly distributed random numbers between 1/4 and 1.
Step 1.
Generate a random number R
Step 2a.
If , accept X = R, goto Step 3
Step 2b.
Step 3.
If another uniform random variate on [1/4, 1] is needed, repeat the procedure begining at Step 1. Otherwise stop.
• Do we know if the random variate generated using above methods is indeed uniformly distributed over [1/4, 1]? The answer is Yes. To prove this, use the definition. Take any ,

which is the correct probability for a uniform distribution on [1/4,1].
• The efficiency: use this method in this particular example, the rejection probability is 1/4 on the average for each number generated. The number of rejections is a geometrically distributed random variable with probability of success'' being p = 3/4, mean number of rejections is (1/p - 1) = 4/3 - 1 = 1/3 (i.e. 1/3 waste).
• For this reason, the inverse transform (X = 1/4 + (3/4) R) is more efficient method.

• Poisson Distribution
• pmf

where N can be interpreted as the number of arrivals in one unit time.

• From the original Poisson process definition, we know the interarrival time are exponentially distributed with a mean of , i.e. arrivals in one unit time.

• Relation between the two distribution:

if and only if

essentially this means if there are n arrivals in one unit time, the sum of interarrival time of the past n observations has to be less than or equal to one, but if one more interarrival time is added, it is greater then one (unit time).

• The s in the relation can be generated from uniformly distributed random number , thus

both sides are multiplied by

that is

• Now we can use the Acceptance-Reject method to generate Poisson distribution.
Step 1.
Set n = 0, P = 1.
Step 2.
Generate a random number and replace P by .
Step 3.
If , then accept N = n, meaning at this time unit, there are n arrivals. Otherwise, reject the current n, increase n by one, return to Step 2.
• Efficiency: How many random numbers will be required, on the average, to generate one Poisson variate, N? If N = n, then n+1 random numbers are required (because of the (n+1) random numbers product).

• Example 9.10 on page 346, Example 9.11 on page 347

• When is large, say , the acceptance-rejection technique described here becomes too expensive. Use normal distribution to approximate Poisson distribution. When is large

is approximately normally distributed with mean 0 and variance 1, thus

can be used to generate Poisson random variate.

Next: Input Modeling Up: Random Variate Generation Previous: Convolution Method
Meng Xiannong 2002-10-18