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Random Number
Generation

© neo
INDEPENDENCE TEST
• Autocorrelation Test
• Gap Test
• Poker Test

© neo
AUTOCORRELATION TEST

© neo
AUTOCORRELATION TEST
• The tests for Autocorrelation are
concerned with the dependence between
numbers in a sequence. For eg.
0.12
0.99
0.68

0.01
0.15
0.49

0.23
0.33
0.05

0.28
0.35
0.43

0.89
0.91
0.95

0.31
0.41
0.58

0.64
0.60
0.19

0.28
0.27
0.36

0.83
0.75
0.69

0.93
0.88
0.87

© neo
AUTOCORRELATION TEST
• From a visual inspection, these numbers
appear to be random, and they would
probably pass all tests presented to this
point.
0.12
0.99
0.68

0.01
0.15
0.49

0.23
0.33
0.05

0.28
0.35
0.43

0.89
0.91
0.95

0.31
0.41
0.58

0.64
0.60
0.19

0.28
0.27
0.36

0.83
0.75
0.69

0.93
0.88
0.87

© neo
AUTOCORRELATION TEST
• From a visual inspection, these numbers
appear to be random, and they would
probably pass all tests presented to this
point.
0.12
0.99
0.68

0.01
0.15
0.49

0.23
0.33
0.05

0.28
0.35
0.43

0.89
0.91
0.95

0.31
0.41
0.58

0.64
0.60
0.19

0.28
0.27
0.36

0.83
0.75
0.69

0.93
0.88
0.87

© neo
AUTOCORRELATION TEST
• The test requires the computation of
autocorrelation between every m numbers,
starting with the ith number .
• Thus the autocorrelation ρim between the
following numbers would be of interest:
Ri , Ri+m, Ri+2m, .. .. .. .., Ri+(M+1)m.
AUTOCORRELATION TEST
• Where M is the largest integer such that
i+(M+1)m<=N , where N is total number of
values in sequence.
• A nonzero autocorrelation implies a lack
of independence, so following two tailed test
is appropriate: H0 : ρim = 0
H1 : ρim ×= 0
AUTOCORRELATION TEST
• For large values of M, the distribution of
the estimator of ρim , denoted ρim is
̂
approximately normal if the values Ri , Ri+m,
Ri+2m, .. .. .. .., Ri+(M+1)m are uncorrelated, then
the statistics can be as follows:
Z0 = ρ̂ im

σρ

im
GAP TEST

© neo
GAP TEST
• For each Uj in certain range, this test
examines the length of ‘Gap’ between this
element and the next element to fall in that
range.
•So if ä and ß are two real numbers such that
0 <= ä < ß <= 1 we are looking for the length
of consecutive subsequences Uj, Uj+1,.. , Uj+(r+1)

© neo
GAP TEST
• Such that Uj and Uj+(r+1) are between ä and ß
but the other elements in the subsequence are
not (this is a gap of length r).
•We would then perform chi-squared test on
the results using the different lengths of
gaps as the categories, and the probabilities
are as follows:

© neo
GAP TEST
• Such that Uj and Uj+(r+1) are between ä and ß
but the other elements in the subsequence are
not (this is a gap of length r).
•We would then perform chi-squared test on
the results using the different lengths of
gaps as the categories, and the probabilities
are as follows:

© neo
GAP TEST
2

k

p0 = p, p1= p(1-p), p2= p(1-p) , .. .., pk= p(1-p)
Here p= ß - ä which is probability
that any element Uj is between ä
and ß

© neo
POKER TEST

© neo
POKER TEST
• As with the gap test, the name of the poker
test suggests its description. We examine n
groups of five consecutive integers, and put
each of these groups into one of the
following categories:

© neo
POKER TEST
•All different: ABCDE
•One pair:
AABCD
•Two pairs:
AABBC
•Three of a kind: AAABC
•Full house:
AAABB
•Four of a kind: AAAAB
•Five of a kind: AAAAA

© neo
POKER TEST
• In a more intuitive way, let us consider a
hand of k cards from k dierent cards.
• The probability to have exactly c different
cards is
P(C=c) = 1

k!
k

k

(k-c)!

c

2Sk

© neo
Thank you

© neo

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Random Number Generation

  • 2. INDEPENDENCE TEST • Autocorrelation Test • Gap Test • Poker Test © neo
  • 4. AUTOCORRELATION TEST • The tests for Autocorrelation are concerned with the dependence between numbers in a sequence. For eg. 0.12 0.99 0.68 0.01 0.15 0.49 0.23 0.33 0.05 0.28 0.35 0.43 0.89 0.91 0.95 0.31 0.41 0.58 0.64 0.60 0.19 0.28 0.27 0.36 0.83 0.75 0.69 0.93 0.88 0.87 © neo
  • 5. AUTOCORRELATION TEST • From a visual inspection, these numbers appear to be random, and they would probably pass all tests presented to this point. 0.12 0.99 0.68 0.01 0.15 0.49 0.23 0.33 0.05 0.28 0.35 0.43 0.89 0.91 0.95 0.31 0.41 0.58 0.64 0.60 0.19 0.28 0.27 0.36 0.83 0.75 0.69 0.93 0.88 0.87 © neo
  • 6. AUTOCORRELATION TEST • From a visual inspection, these numbers appear to be random, and they would probably pass all tests presented to this point. 0.12 0.99 0.68 0.01 0.15 0.49 0.23 0.33 0.05 0.28 0.35 0.43 0.89 0.91 0.95 0.31 0.41 0.58 0.64 0.60 0.19 0.28 0.27 0.36 0.83 0.75 0.69 0.93 0.88 0.87 © neo
  • 7. AUTOCORRELATION TEST • The test requires the computation of autocorrelation between every m numbers, starting with the ith number . • Thus the autocorrelation ρim between the following numbers would be of interest: Ri , Ri+m, Ri+2m, .. .. .. .., Ri+(M+1)m.
  • 8. AUTOCORRELATION TEST • Where M is the largest integer such that i+(M+1)m<=N , where N is total number of values in sequence. • A nonzero autocorrelation implies a lack of independence, so following two tailed test is appropriate: H0 : ρim = 0 H1 : ρim ×= 0
  • 9. AUTOCORRELATION TEST • For large values of M, the distribution of the estimator of ρim , denoted ρim is ̂ approximately normal if the values Ri , Ri+m, Ri+2m, .. .. .. .., Ri+(M+1)m are uncorrelated, then the statistics can be as follows: Z0 = ρ̂ im σρ im
  • 11. GAP TEST • For each Uj in certain range, this test examines the length of ‘Gap’ between this element and the next element to fall in that range. •So if ä and ß are two real numbers such that 0 <= ä < ß <= 1 we are looking for the length of consecutive subsequences Uj, Uj+1,.. , Uj+(r+1) © neo
  • 12. GAP TEST • Such that Uj and Uj+(r+1) are between ä and ß but the other elements in the subsequence are not (this is a gap of length r). •We would then perform chi-squared test on the results using the different lengths of gaps as the categories, and the probabilities are as follows: © neo
  • 13. GAP TEST • Such that Uj and Uj+(r+1) are between ä and ß but the other elements in the subsequence are not (this is a gap of length r). •We would then perform chi-squared test on the results using the different lengths of gaps as the categories, and the probabilities are as follows: © neo
  • 14. GAP TEST 2 k p0 = p, p1= p(1-p), p2= p(1-p) , .. .., pk= p(1-p) Here p= ß - ä which is probability that any element Uj is between ä and ß © neo
  • 16. POKER TEST • As with the gap test, the name of the poker test suggests its description. We examine n groups of five consecutive integers, and put each of these groups into one of the following categories: © neo
  • 17. POKER TEST •All different: ABCDE •One pair: AABCD •Two pairs: AABBC •Three of a kind: AAABC •Full house: AAABB •Four of a kind: AAAAB •Five of a kind: AAAAA © neo
  • 18. POKER TEST • In a more intuitive way, let us consider a hand of k cards from k dierent cards. • The probability to have exactly c different cards is P(C=c) = 1 k! k k (k-c)! c 2Sk © neo