Random Number Generation Concern with generating random numbers

Random Number Generation Concern with generating random numbers that have the following conditions: – Uniformity – Independence – Efficiency – Replicability – Long Cycle Length 1

Random Number Generation (cont. ) Each random number Rt is an independent sample drawn from a continuous uniform distribution between 0 and 1 ì 1 , 0 £ x £ 1 pdf: f(x) = í î 0 , otherwise 2

Random Number Generation (cont. ) PDF: f(x) 1 0 x 3

Techniques for Generating Random Number Mid. Square Example: X 0 = 7182 (seed) = 51581124 ==> R 1 = 0. 5811 = (5811) 2 = 33767721 ==> R 2 = 0. 7677 etc. 4

Techniques for Generating Random Number (cont. ) Note: Do not choose a seed that guarantees that the sequence will not degenerate and will have a long period. Also, zeros, once they appear, are carried in subsequent numbers. Ex 1: X 0 = 5197 (seed) = 27008809 ==> R 1 = 0. 0088 = 00007744 ==> R 2 = 0. 0077 Ex 2: X 0 = 4500 (seed) = 20250000 ==> R 1 = 0. 2500 = 06250000 ==> R 2 = 0. 2500 5

Techniques for Generating Random Number (cont. ) l Multiplicative Congruential Method: Basic Relationship Xi+1 = a Xi (mod m), where a ³ 0 and m ³ 0 – Most natural choice for m is one that equals to the capacity of a computer word. – m = 2 b (binary machine), where b is the number of bits in the computer word. – m = 10 d (decimal machine), where d is the number of digits in the computer word. 6

Techniques for Generating Random Number (cont. ) The max period (P) is: l. For m a power of 2, say m = 2 b, the longest possible period is P = m / 4 = 2 b-2 l. This is achieved provided that lthe seed X 0 is odd and lthe multiplier, a, is given by a = 3 + 8 k or a = 5 + 8 k, for some k = 0, 1, . . . 7

Techniques for Generating Random Number (cont. ) l For m a prime number, the longest possible period is P = m - 1, l This is achieved provided that the multiplier, a, has the property that lthe smallest integer k such that ak 1 is divisible by m is k = m - 1, 8

Techniques for Generating Random Number (cont. ) (Example) Using the multiplicative congruential method, find the period of the generator for a = 13, m = 26, and X 0 = 1, 2, 3, and 4. The solution is given in next slide. When the seed is 1 and 3, the sequence has period 16. However, a period of length eight is achieved when the seed is 2 and a period of length four occurs when the seed is 4. 9

Techniques for Generating Random Number (cont. ) Period Determination Using Various seeds i Xi Xi 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 13 41 21 17 29 57 37 33 45 9 53 49 61 25 5 1 2 26 18 42 34 58 50 10 2 3 39 59 63 51 23 43 47 35 7 27 31 19 55 11 15 3 4 52 36 20 4 10

Techniques for Generating Random Number (cont. ) SUBROUTINE RAN(IX, IY, RN) IY = IX * 1220703125 IF (IY) 3, 4, 4 3: IY = IY + 214783647 + 1 4: RN = IY RN = RN * 0. 4656613 E-9 IX = IY RETURN END 11

Techniques for Generating Random Number (cont. ) l Linear Congruential Method: Xi+1 = (a. Xi + c) mod m, i = 0, 1, 2. . (Example) let X 0 = 27, a = 17, c = 43, and m = 100, then X 1 = (17*27 + 43) mod 100 = 2 R 1 = 2 / 100 = 0. 02 X 2 = (17*2 + 43) mod 100 = 77 R 2 = 77 / 100 = 0. 77. . 12

Test for Random Numbers 1. Frequency test. Uses the Kolmogorov. Smirnov or the chi-square test to compare the distribution of the set of numbers generated to a uniform distribution. 2. Runs test. Tests the runs up and down or the runs above and below the mean by comparing the actual values to expected values. The statistic for comparison is the chisquare. 3. Autocorrelation test. Tests the correlation between numbers and compares the sample correlation to the expected correlation of zero. 13

Test for Random Numbers (cont. ) 4. Gap test. Counts the number of digits that appear between repetitions of a particular digit and then uses the Kolmogorov-Smirnov test to compare with the expected number of gaps. 5. Poker test. Treats numbers grouped together as a poker hand. Then the hands obtained are compared to what is expected using the chi-square test. 14

Test for Random Numbers (cont. ) In testing for uniformity, the hypotheses are as follows: H 0: Ri ~ U[0, 1] H 1: Ri ¹ U[0, 1] The null hypothesis, H 0, reads that the numbers are distributed uniformly on the interval [0, 1]. 15

Test for Random Numbers (cont. ) In testing for independence, the hypotheses are as follows; H 0: Ri ~ independently H 1: Ri ¹ independently This null hypothesis, H 0, reads that the numbers are independent. Failure to reject the null hypothesis means that no evidence of dependence has been detected on the basis of this test. This does not imply that further testing of the generator for independence is unnecessary. 16

The Chi squared test • Divide the interval [0, 1] into k equal subintervals. – The expected number of true random numbers in each interval is n/k. Let fi denote the number of pseudo random numbers that fall in the ith interval – Let – If we accept the hypothesis that these random umbers are truly random numbers with significance level a 17

Generating Bernoulli Random Variable • To generate B(p), p: success probability • Generate a uniform random number u between [0, 1]. • If u≤ p then let X = 1 • Else let X = 0 18

Generating general discrete Random Variats • We want to generate random variables given the following probability mass function: X Pmf p(X) CDF F(X) x 1 x 2 p 1 p 1+ p 2 xn pn p 1+ p 2+ …+ pn 19

Generating Discrete RV Cont. • Generate a uniform random number u between [0, 1]. • If u≤ p 1 then let X = x 1 • Else if p 1+ p 2+ …+ pj < u ≤ p 1+ p 2+ …+ p(j+1) Then let X = xj 0 p 1 p 2 p 3 p 4 p 5. . . 1 20

Generating Continuous RV. • Uniform [a, b] – Generate a uniform random number u on [0, 1] – Return X = a + (b-a) u 21

Inverse transformation Method • X has a pdf f(x) • And has a continuous CDF F(x) • Then – Generate a uniform RN u e [0, 1] – Return X = F-1 (u) • Example: Exponential RV with rate l : – Generate u e [0, 1] uniformly – Return X = -1/l ln(1 -u) (ln is the natural Logarithm) – or simply X = -1/l ln(u) 22