CPE 619 Testing RandomNumber Generators Aleksandar Milenkovi The

CPE 619 Testing Random-Number Generators Aleksandar Milenković The La. CASA Laboratory Electrical and Computer Engineering Department The University of Alabama in Huntsville http: //www. ece. uah. edu/~milenka http: //www. ece. uah. edu/~lacasa

Overview n n n n Chi-square test Kolmogorov-Smirnov Test Serial-correlation Test Two-level tests K-dimensional uniformity or k-distributivity Serial Test Spectral Test 2

Testing Random-Number Generators Goal: To ensure that the random number generator produces a random stream n Plot histograms n Plot quantile-quantile plot n Use other tests n Passing a test is necessary but not sufficient n Pass ¹ Good Fail Bad n New tests Old generators fail the test n Tests can be adapted for other distributions 3

Chi-Square Test n n Most commonly used test Can be used for any distribution Prepare a histogram of the observed data Compare observed frequencies with theoretical n n n k = Number of cells oi = Observed frequency for ith cell ei = Expected frequency D=0 Exact fit n D has a chi-square distribution with k-1 degrees of freedom. Compare D with c 2[1 -a; k-1] Pass with confidence a if D is less n 4

Example 27. 1 n 1000 random numbers with x 0 = 1 n n n Observed difference = 10. 380 Observed is Less Accept IID U(0, 1) 5

Chi-Square for Other Distributions n n Errors in cells with a small ei affect the chi-square statistic more Best when ei's are equal n n n Use an equi-probable histogram with variable cell sizes Combine adjoining cells so that the new cell probabilities are approximately equal The number of degrees of freedom should be reduced to k-r-1 (in place of k-1), where r is the number of parameters estimated from the sample Designed for discrete distributions and for large sample sizes only Lower significance for finite sample sizes and continuous distributions If less than 5 observations, combine neighboring cells 6

Kolmogorov-Smirnov Test n n n Developed by A. N. Kolmogorov and N. V. Smirnov Designed for continuous distributions Difference between the observed CDF (cumulative distribution function) Fo(x) and the expected cdf Fe(x) should be small 7

Kolmogorov-Smirnov Test n n n n K+ = maximum observed deviation below the expected cdf K- = minimum observed deviation below the expected cdf K+ < K[1 -a; n] and K- < K[1 -a; n] Pass at a level of significance Don't use max/min of Fe(xi)-Fo(xi) Use Fe(xi+1)-Fo(xi) for KFor U(0, 1): Fe(x)=x Fo(x) = j/n, where x > x 1, x 2, . . . , xj-1 8

Example 27. 2 n n 30 Random numbers using a seed of x 0=15: The numbers are: 14, 11, 2, 6, 18, 23, 7, 21, 1, 3, 9, 27, 19, 26, 17, 20, 29, 25, 13, 8, 24, 10, 30, 28, 22, 4, 12, 5, 15. 9

Example 27. 2 (cont’d) n The normalized numbers obtained by dividing the sequence by 31 are: 0. 45161, 0. 35484, 0. 06452, 0. 19355, 0. 58065, 0. 22581, 0. 67742, 0. 03226, 0. 09677, 0. 29032, 0. 61290, 0. 83871, 0. 51613, 0. 54839, 0. 64516, 0. 80645, 0. 41935, 0. 25806, 0. 77419, 0. 32258, 0. 90323, 0. 70968, 0. 12903, 0. 38710, 0. 16129, 0. 74194, 0. 87097, 0. 93548, 0. 96774, 0. 48387. 10

Example 27. 2 (cont’d) n n K[0. 9; n] value for n = 30 and a = 0. 1 is 1. 0424 Observed<Table Pass 11

Chi-square vs. K-S Test 12

Serial-Correlation Test n n Nonzero covariance Dependence. The inverse is not true Rk = Autocovariance at lag k = Cov[xn, xn+k] For large n, Rk is normally distributed with a mean of zero and a variance of 1/[144(n-k)] 100(1 -a)% confidence interval for the autocovariance is: For k 1 Check if CI includes zero n For k = 0, R 0= variance of the sequence Expected to be 1/12 for IID U(0, 1) 13

Example 27. 3: Serial Correlation Test 10, 000 random numbers with x 0=1: 14

Example 27. 3 (cont’d) n All confidence intervals include zero All covariances are statistically insignificant at 90% confidence. 15

Two-Level Tests If the sample size is too small, the test results may apply locally, but not globally to the complete cycle. n Similarly, global test may not apply locally n Use two-level tests Use Chi-square test on n samples of size k each and then use a Chi-square test on the set of n Chi-square statistics so obtained Chi-square on Chi-square test. n Similarly, K-S on K-S n Can also use this to find a ``nonrandom'' segment of an otherwise random sequence. n 16

k-Distributivity n n k-Dimensional Uniformity Chi-square uniformity in one dimension Given two real numbers a 1 and b 1 between 0 and 1 such that b 1 > a 1 This is known as 1 -distributivity property of un. The 2 -distributivity is a generalization of this property in two dimensions: For all choices of a 1, b 1, a 2, b 2 in [0, 1], b 1>a 1 and b 2>a 2 17

k-Distributivity (cont’d) n k-distributed if: For all choices of ai, bi in [0, 1], with bi>ai, i=1, 2, . . . , k. n k-distributed sequence is always (k-1)-distributed. The inverse is not true. n Two tests: 1. Serial test 2. Spectral test 3. Visual test for 2 -dimensions: Plot successive overlapping pairs of numbers n 18

Example 27. 4 n n n Tausworthe sequence generated by: The sequence is kdistributed for k up to d /l e, that is, k=1. In two dimensions: Successive overlapping pairs (xn, xn+1) 19

Example 27. 5 n n Consider the polynomial: Better 2 -distributivity than Example 27. 4 20

Serial Test n n Goal: To test for uniformity in two dimensions or higher. In two dimensions, divide the space between 0 and 1 into K 2 cells of equal area xn +1 xn 21

Serial Test (cont’d) n n n n n Given {x 1, x 2, …, xn}, use n/2 non-overlapping pairs (x 1, x 2), (x 3, x 4), … and count the points in each of the K 2 cells Expected= n/(2 K 2) points in each cell Use chi-square test to find the deviation of the actual counts from the expected counts The degrees of freedom in this case are K 2 -1 For k-dimensions: use k-tuples of non-overlapping values k-tuples must be non-overlapping Overlapping number of points in the cells are not independent chi-square test cannot be used In visual check one can use overlapping or non-overlapping In the spectral test overlapping tuples are used Given n numbers, there are n-1 overlapping pairs, n/2 nonoverlapping pairs 22

Spectral Test n n Goal: To determine how densely the k-tuples {x 1, x 2, …, xk} can fill up the k-dimensional hyperspace The k-tuples from an LCG fall on a finite number of parallel hyper-planes Successive pairs would lie on a finite number of lines In three dimensions, successive triplets lie on a finite number of planes 23

Example 27. 6: Spectral Test n n Plot of overlapping pairs n All points lie on three straight lines. Or: 24

Example 27. 6 (cont’d) n In three dimensions, the points (xn, xn-1, xn-2) for the above generator would lie on five planes given by: Obtained by adding the following to equation Note that k+k 1 will be an integer between 0 and 4. 25

Spectral Test (More) n n n Marsaglia (1968): Successive k-tuples obtained from an LCG fall on, at most, (k!m)1/k parallel hyper-planes, where m is the modulus used in the LCG. Example: m = 232, fewer than 2, 953 hyper-planes will contain all 3 -tuples, fewer than 566 hyper-planes will contain all 4 -tuples, and fewer than 41 hyper-planes will contain all 10 -tuples. Thus, this is a weakness of LCGs. Spectral Test: Determine the max distance between adjacent hyper-planes. Larger distance worse generator In some cases, it can be done by complete enumeration 26

Example 27. 7 n Compare the following two generators: n Using a seed of x 0=15, first generator: n Using the same seed in the second generator: 27

Example 27. 7 (cont’d) Every number between 1 and 30 occurs once and only once Both sequences will pass the chi-square test for uniformity n 28

Example 27. 7 (cont’d) n First Generator: 29

Example 27. 7 (cont’d) n n Three straight lines of positive slope or ten lines of negative slope Since the distance between the lines of positive slope is more, consider only the lines with positive slope Distance between two parallel lines y=ax+c 1 and y=ax+c 2 is given by The distance between the above lines is or 9. 80 30

Example 27. 7 (cont’d) n Second Generator: 31

Example 27. 7 (cont’d) n n n All points fall on seven straight lines of positive slope or six straight lines of negative slope. Considering lines with negative slopes: The distance between lines is: or 5. 76. The second generator has a smaller maximum distance and, hence, the second generator has a better 2 -distributivity The set with a larger distance may not always be the set with fewer lines 32

Example 27. 7 (cont’d) n Either overlapping or non-overlapping k-tuples can be used n n n With overlapping k-tuples, we have k times as many points, which makes the graph visually more complete. The number of hyperplanes and the distance between them are the same with either choice. With serial test, only non-overlapping k-tuples should be used. For generators with a large m and for higher dimensions, finding the maximum distance becomes quite complex. See Knuth (1981) 33

Summary n n n Chi-square test is a one-dimensional test Designed for discrete distributions and large sample sizes K-S test is designed for continuous variables Serial correlation test for independence Two level tests find local non-uniformity k-dimensional uniformity = k-distributivity tested by spectral test or serial test 34

Random Variate Generation

Overview n n n Inverse transformation Rejection Composition Convolution Characterization 36

Random-Variate Generation n General Techniques Only a few techniques may apply to a particular distribution Look up the distribution in Chapter 29 37

Inverse Transformation n Used when F-1 can be determined either analytically or empirically 1. 0 CDF F(x) u 0. 5 0 x 38

Proof 39

Example 28. 1 n n n For exponential variates: If u is U(0, 1), 1 -u is also U(0, 1) Thus, exponential variables can be generated by: 40

Example 28. 2 n The packet sizes (trimodal) probabilities: n The CDF for this distribution is: 41

Example 28. 2 (cont’d) n n The inverse function is: Note: CDF is continuous from the right the value on the right of the discontinuity is used The inverse function is continuous from the left u=0. 7 x=64 42

Applications of the Inverse. Transformation Technique 43

Rejection n n Can be used if a pdf g(x) exists such that c g(x) majorizes the pdf f(x) c g(x) > f(x) 8 x Steps: n n n 1. Generate x with pdf g(x) 2. Generate y uniform on [0, cg(x)] 3. If y < f(x), then output x and return Otherwise, repeat from step 1 Continue rejecting the random variates x and y until y > f(x) Efficiency = how closely c g(x) envelopes f(x) Large area between c g(x) and f(x) Large percentage of (x, y) generated in steps 1 and 2 are rejected If generation of g(x) is complex, this method may not be efficient 44

Example 28. 2 n n Beta(2, 4) density function: Bounded inside a rectangle of height 2. 11 Steps: n n n Generate x uniform on [0, 1] Generate y uniform on [0, 2. 11] If y < 20 x(1 -x)3, then output x and return Otherwise repeat from step 1 45

Composition n n Can be used if CDF F(x) = Weighted sum of n other CDFs. Here, , and Fi's are distribution functions. n CDFs are composed together to form the desired CDF Hence, the name of the technique. The desired CDF is decomposed into several other CDFs Also called decomposition Can also be used if the pdf f(x) is a weighted sum of n other pdfs: 46

Steps: n Generate a random integer I such that: n n This can easily be done using the inversetransformation method. Generate x with the ith pdf fi(x) and return. 47

Example 28. 4 n n n pdf: Composition of two exponential pdf's Generate If u 1<0. 5, return; otherwise return x=a ln u 2. Inverse transformation better for Laplace 48

Convolution n n n Sum of n variables: Generate n random variate yi's and sum For sums of two variables, pdf of x = convolution of pdfs of y 1 and y 2. Hence the name Although no convolution in generation If pdf or CDF = Sum Composition Variable x = Sum Convolution 49

Convolution: Examples n n n n Erlang-k = åi=1 k Exponentiali Binomial(n, p) = åi=1 n Bernoulli(p) Generated n U(0, 1), return the number of RNs less than p c 2(n) = åi=1 n N(0, 1)2 G(a, b 1)+G(a, b 2)=G(a, b 1+b 2) Non-integer value of b = integer + fraction åi=1 n Any = Normal å U(0, 1) = Normal åi=1 m Geometric = Pascal åi=12 Uniform = Triangular 50

Characterization n n n Use special characteristics of distributions characterization Exponential inter-arrival times Poisson number of arrivals Continuously generate exponential variates until their sum exceeds T and return the number of variates generated as the Poisson variate. The ath smallest number in a sequence of a+b+1 U(0, 1) uniform variates has a b(a, b) distribution. The ratio of two unit normal variates is a Cauchy(0, 1) variate. A chi-square variate with even degrees of freedom c 2(n) is the same as a gamma variate g(2, n/2). If x 1 and x 2 are two gamma variates g(a, b) and g(a, c), respectively, the ratio x 1/(x 1+x 2) is a beta variate b(b, c). If x is a unit normal variate, em+s x is a lognormal(m, s) variate. 51

Summary Is CDF invertible? Yes Is CDF a sum of other CDFs? Yes Is pdf a sum of other pdfs? Yes Use inversion Use composition Use Composition 52

Summary (cont’d) Is the variate a sum of other variates Is the variate related to other variates? Does a majorizing function exist? No Yes Yes Use convolution Use characterization Use rejection Use empirical inversion 53

Homework #6 n n Submit answers to exercise 27. 1 Submit answers to exercise 27. 4 Due: Monday, April 7, 2008, 12: 45 PM Submit a hard copy to instructor 54