Testing Random-Number Generators

Andy WangCIS 5930-03

Computer SystemsPerformance Analysis

Testing Random-Number Generators

• How do you know if your random number streams are good?– Plot histogram, CDF, QQ-plot– Run tests

• Necessary but not sufficient to prove the quality of a generator

– If it fails a test, the generator is bad– Passing a test is not a guarantee that a generator is

good

– Tests can also be used to verify distribution matching

2

3

Chi-Square Test

• Most commonly used test– Tests if a data set matches a distribution

• Steps– Prepare a histogram with k bins– Compare the observed and expected

frequencies via a formula– If the computed value < 2

[1-α; k-1] (Table A.5)• We have a match

Chi-Square Example

• N = 1000• k = 10• 2

[1-0.1; 10-1]

= 14.68– A match

4

i observed expected (observed – expected)2 / expected

1 100 100 0

2 96 100 .16

3 98 100 .04

4 85 100 2.25

5 105 100 .25

6 93 100 .49

7 97 100 .09

8 125 100 6.25

9 107 100 .49

10 94 100 .36

Total 1000 1000 10.38

Fine Prints

• When applied to a nonuniform distribution– Watch out for small expected values, which

can affect the outcome• Remedy

– Use variable bin sizes, so that the expected outcomes are equal

• Designed for discrete distribution with large sample sizes

5

6

Kolmogorov-Smirnov Test

• Tests if a sample of n observations is from a continuous distribution

• Observation– If two distributions match,

• | observed CDF Fo(x) - expected CDF Fe(x) | should be small

K-S Test

• Differences are measured by the maximum observed deviations above and below the expected CDFK+ = n maxx[Fo(xi) - Fe(xi)]

K- = n maxx[Fe(xi + 1) - Fo(xi)]• If both K+ and K- < K[1-α; n] (Table A.9)

– We have a match

7

K-S Example

• xn = 3xn-1 % 31n = 30

x0 = 15• K[0.9, 30]

= 1.042• A match

8

j xj xj/31 sorted xj/31 j/n – xj xj – (j-1)/n

1 14 .451 .032 .001 .032

2 11 .354 .064 .002 .031

3 2 .064 .096 .003 .030

4 6 .193 .129 .004 .029

5 18 .580 .161 .005 .027

…

30 15 .483 .967 .032 .001

maximum .032 .032

K-S Test vs. Chi-square Test

• K-S test– Designed for small

samples and continuous distributions

– Uses differences between CDFs

– Tests each sample without grouping

– Exact

• Chi-square test– Designed for large

samples and discrete distributions

– Uses differences between pdfs/pmfs

– Requires grouping– Approximate

9

10

Serial-Correlation Test

• Covariance: tests dependence of two random variables– If the covariance is nonzero, variables are

dependent– If the covariance is zero, variables can still

be dependent

Autocovariance

• Autocovariance at lag k (>1) Rk: covariance between numbers that are k values apart, 0 < random number U < 1

• For large n– Rk is normally distributed, with a mean of 0

• Variance of 1/[144(n – k)]

11

2

1

2

11

1ki

kn

iik UU

knR

Autocovariance

• Confidence interval• If confidence interval does not include

zero, the sequence has a significant correlation

• Example from slide 8– xn = 3xn-1 % 31– x0 = 15– Not so good

12

knzRk 12/2/1

0 2 4 6 8 10 12

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

lag k

Rk

Autocovariance

• Another example xn = 75xn-1 % (231 – 1)x0 = 1

• A better random-number generator

13

0 2 4 6 8 10 12

-0.0015

-0.001

-0.0005

0

0.0005

0.001

0.0015

lag k

Rk

14

Two-Level Tests

• If the sample test is too small, test results only apply locally

• If the sample test is large, test results only apply globally

15

• Solution– Apply a chi-square test on n bins of size k– Then apply a chi-square test on a set of n-

bin statistics– Can locate a nonrandom segment of a

random sequence

Two-Level Tests

16

k-Dimensional Uniformity or k-

Distributivity• So far, tests ensure that numbers are uniformly distributed in one dimensionFor 0 < a < u < b < 1

P(a < un < b) = b – a• It is known as the 1-distributivity of un

• For 2-distributivityP(a1 < un-1 < b1 and a2 < un < b2)

= (b1 – a1)(b2 – a2)

Visual Check

• Plot successive overlapping pairs of numbers in a 2-D space– E.g., (u0, u1), (u1, u2), …

• Example of 15-bit random numbers: – Tausworthe generator– x15 + x + 1– Period: 215 - 1– bn+15 bn+1 bn = 0– bn = bn-14bn-15

17

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.10.20.30.40.50.60.70.80.9

1

ui

ui + 1

Visual Check

• Another example of 15-bit random numbers: – Tausworthe generator– x15 + x4 + 1– bn+15 bn+4 bn = 0– bn = bn-11bn-15

18

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.10.20.30.40.50.60.70.80.9

1

ui

ui + 1

19

Serial Test

• Test for uniformity > 2D– Plot nonoverlapping pairs

• E.g., (u0, u1), (u2, u3), …– For 2D, divide the space into K2 cells

• Each cell is expected to have n/2K2 points• Use the chi-square test• With K2 – 1 degrees of freedom

– Can generalize it into k-dimension– Dependency likely to fail high-dimension

chi-square tests

Serial Test vs. Visual Check

• Serial Test– Uses nonoverlapping

points• Assumed by the chi-

square test– Has n/2 pairs

• Visual Check– Uses overlapping

points– Has n – 1 pairs

20

21

Spectral Test

• Plot successive overlapping pairs of numbers

• K-dimension tuples from an LCG fall on a finite number of parallel hyperplanes

22

• xn = 3xn-1 % 31– Can see three lines with positive slopes– xn = 3xn-1

– xn = 3xn-1 - 31– xn = 3xn-1 - 62

Spectral Test Example

0 5 10 15 20 25 30 350

5

10

15

20

25

30

35

nth number

random number

Observations

• k-tuples from an LCG fall on at most (k!m)1/k parallel hyperplanes, where m is the modulus– M = 232– 3-tuples will have < 2953 hyperplanes– 4-tuples, < 566 hyperplanes– 10-tuples, < 41

23

Another Spectral Test Example

• xn = 3xn-1 % 31– Period: 30– Pass chi-square test– 3 lines

• xn = 13xn-1 % 31– Period: 30– Pass chi-square test– 10 closer lines– Better 2-distributivity

24

0 5 10 15 20 25 30 350

5

10

15

20

25

30

35

nth number

random number

0 5 10 15 20 25 30 350

5

10

15

20

25

30

35

nth number

random number

25

White Slide