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 $R \ge 1/4$, accept X = R, goto Step 3

  Step 2b.

  If $R < 1/4$, reject R, return to Step 1

  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 $1/4 \le a < b \le 1$,
    
    $$P(a < R \le b \vert 1/4 \le R \le 1) = \frac{P(a < R \le b)} {P(1/4 \le R \le 1)} = \frac{b-a}{3/4}$$
    
    
    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
    
    $$p(n) = P(N = n) = \frac{e^{-\alpha} \alpha^n} {n!}, n = 0, 1, 2, ...$$
    
    
    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 $A_1, A_2, ...$ are exponentially distributed with a mean of $\alpha$, i.e. $\alpha$ arrivals in one unit time.

  • Relation between the two distribution:
    
    $$N = n$$
    
    
    if and only if
    
    $$A_1 + A_2 + ... + A_n \le 1 < A_1 + A_2 + ... + A_n + A_{n+1}$$
    
    
    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 $A_i$s in the relation can be generated from uniformly distributed random number $A_i = (-1/\alpha) \ln R_i$, thus
    
    $$\sum_{i=1}^n \frac{-1}{\alpha} \ln R_i \le 1 < \sum_{i=1}^{n+1} \frac{-1}{\alpha} \ln R_i$$
    
    
    both sides are multiplied by $-\alpha$
    
    $$\ln \prod_{1}^n R_i = \sum_{i=1}^n \ln R_i \ge -\alpha > \sum_{i=1}^{n+1} \ln R_i = \ln \prod_{1}^{n+1} R_i$$
    
    
    that is
    
    $$\prod_{1}^n R_i \ge e^{-\alpha} > \prod_{1}^{n+1} R_i$$
    
    

  • 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 $R_{n+1}$ and replace P by $P * R_{n+1}$.

    Step 3.

    If $P < e^{-\alpha}$, 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).
    
    $$E(N+1) = \alpha + 1$$
    
    

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

  • When $\alpha$ is large, say $\alpha \le 15$, the acceptance-rejection technique described here becomes too expensive. Use normal distribution to approximate Poisson distribution. When $\alpha$ is large
    
    $$Z = \frac{N - \alpha}{\sqrt{\alpha}}$$
    
    
    is approximately normally distributed with mean 0 and variance 1, thus
    
    $$N = \lceil \alpha + \sqrt{\alpha} Z - 0.5 \rceil$$
    
    
    can be used to generate Poisson random variate.