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Require a shorter test for the (optional) consistent probability sampler (#2319)
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‎specification/trace/tracestate-probability-sampling.md

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@@ -72,9 +72,9 @@
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- [Recommendation: Recognize inconsistent r-values](#recommendation-recognize-inconsistent-r-values)
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* [Appendix: Statistical test requirements](#appendix-statistical-test-requirements)
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+ [Test procedure: non-powers of two](#test-procedure-non-powers-of-two)
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- [Requirement: Pass 15 non-power-of-two statistical tests](#requirement-pass-15-non-power-of-two-statistical-tests)
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- [Requirement: Pass 12 non-power-of-two statistical tests](#requirement-pass-12-non-power-of-two-statistical-tests)
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+ [Test procedure: exact powers of two](#test-procedure-exact-powers-of-two)
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- [Requirement: Pass 5 power-of-two statistical tests](#requirement-pass-5-power-of-two-statistical-tests)
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- [Requirement: Pass 3 power-of-two statistical tests](#requirement-pass-3-power-of-two-statistical-tests)
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+ [Test implementation](#test-implementation)
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- [Appendix](#appendix)
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* [Methods for generating R-values](#methods-for-generating-r-values)
@@ -870,7 +870,7 @@ this a strict test for random behavior, we take the following approach:
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- Generate a pre-determined list of 20 random seeds
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- Use fixed values for significance level (5%) and trials (20)
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- Use a population size of one million spans
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- Use a population size of 100,000 spans
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- For each trial, simulate the population and compute ChiSquared
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test statistic
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- Locate the first seed value in the ordered list such that the
@@ -895,29 +895,26 @@ In this case there are two degrees of freedom for the Chi-Squared test.
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The following table summarizes the test parameters.
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| Test case | Sampling probability | Lower, Upper p-value when sampled | Expect<sub>lower</sub> | Expect<sub>upper</sub> | Expect<sub>unsampled</sub> |
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| --- | --- | --- | --- | --- | --- |
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| 1 | 0.900000 | 0, 1 | 100000 | 800000 | 100000 |
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| 2 | 0.600000 | 0, 1 | 400000 | 200000 | 400000 |
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| 3 | 0.330000 | 1, 2 | 170000 | 160000 | 670000 |
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| 4 | 0.130000 | 2, 3 | 120000 | 10000 | 870000 |
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| 5 | 0.100000 | 3, 4 | 25000 | 75000 | 900000 |
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| 6 | 0.050000 | 4, 5 | 12500 | 37500 | 950000 |
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| 7 | 0.017000 | 5, 6 | 14250 | 2750 | 983000 |
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| 8 | 0.010000 | 6, 7 | 5625 | 4375 | 990000 |
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| 9 | 0.005000 | 7, 8 | 2812.5 | 2187.5 | 995000 |
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| 10 | 0.002900 | 8, 9 | 1006.25 | 1893.75 | 997100 |
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| 11 | 0.001000 | 9, 10 | 953.125 | 46.875 | 999000 |
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| 12 | 0.000500 | 10, 11 | 476.5625 | 23.4375 | 999500 |
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| 13 | 0.000260 | 11, 12 | 228.28125 | 31.71875 | 999740 |
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| 14 | 0.000230 | 12, 13 | 14.140625 | 215.859375 | 999770 |
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| 15 | 0.000100 | 13, 14 | 22.0703125 | 77.9296875 | 999900 |
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|-----------|----------------------|-----------------------------------|------------------------|------------------------|----------------------------|
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| 1 | 0.900000 | 0, 1 | 10000 | 80000 | 10000 |
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| 2 | 0.600000 | 0, 1 | 40000 | 20000 | 40000 |
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| 3 | 0.330000 | 1, 2 | 17000 | 16000 | 67000 |
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| 4 | 0.130000 | 2, 3 | 12000 | 1000 | 87000 |
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| 5 | 0.100000 | 3, 4 | 2500 | 7500 | 90000 |
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| 6 | 0.050000 | 4, 5 | 1250 | 3750 | 95000 |
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| 7 | 0.017000 | 5, 6 | 1425 | 275 | 98300 |
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| 8 | 0.010000 | 6, 7 | 562.5 | 437.5 | 99000 |
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| 9 | 0.005000 | 7, 8 | 281.25 | 218.75 | 99500 |
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| 10 | 0.002900 | 8, 9 | 100.625 | 189.375 | 99710 |
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| 11 | 0.001000 | 9, 10 | 95.3125 | 4.6875 | 99900 |
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| 12 | 0.000500 | 10, 11 | 47.65625 | 2.34375 | 99950 |
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The formula for computing Chi-Squared in this case is:
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```
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ChiSquared = math.Pow(sampled_lowerP - expect_lowerP, 2) / expect_lowerP +
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math.Pow(sampled_upperP - expect_upperP, 2) / expect_upperP +
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math.Pow(1000000 - sampled_lowerP - sampled_upperP - expect_unsampled, 2) / expect_unsampled
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math.Pow(100000 - sampled_lowerP - sampled_upperP - expect_unsampled, 2) / expect_unsampled
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```
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This should be compared with 0.102587, the value of the Chi-Squared
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demonstrate a seed that produces exactly one ChiSquared value less
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than 0.102587.
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##### Requirement: Pass 15 non-power-of-two statistical tests
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##### Requirement: Pass 12 non-power-of-two statistical tests
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For the test with 20 trials and 1 million spans each, the test MUST
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For the test with 20 trials and 100,000 spans each, the test MUST
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demonstrate a random number generator seed such that the ChiSquared
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test statistic is below 0.102587 exactly 1 out of 20 times.
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In this case there is one degree of freedom for the Chi-Squared test.
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The following table summarizes the test parameters.
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| Test case | Sampling probability | P-value when sampled | Expect<sub>sampled</sub> | Expect<sub>unsampled</sub> | |
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| --- | --- | --- | --- | --- | |
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| 16 | 0x1p-01 (0.500000) | 1 | 500000 | n/a | 500000 |
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| 17 | 0x1p-04 (0.062500) | 4 | 62500 | n/a | 937500 |
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| 18 | 0x1p-07 (0.007812) | 7 | 7812.5 | n/a | 992187.5 |
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| 19 | 0x1p-10 (0.000977) | 10 | 976.5625 | n/a | 999023.4375 |
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| 20 | 0x1p-13 (0.000122) | 13 | 122.0703125 | n/a | 999877.9297 |
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| Test case | Sampling probability | P-value when sampled | Expect<sub>sampled</sub> | Expect<sub>unsampled</sub> |
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|-----------|----------------------|----------------------|--------------------------|----------------------------|
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| 13 | 0x1p-01 (0.500000) | 1 | 50000 | 50000 |
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| 14 | 0x1p-04 (0.062500) | 4 | 6250 | 93750 |
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| 15 | 0x1p-07 (0.007812) | 7 | 781.25 | 99218.75 |
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The formula for computing Chi-Squared in this case is:
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```
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ChiSquared = math.Pow(sampled - expect_sampled, 2) / expect_sampled +
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math.Pow(1000000 - sampled - expect_unsampled, 2) / expect_unsampled
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math.Pow(100000 - sampled - expect_unsampled, 2) / expect_unsampled
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```
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This should be compared with 0.003932, the value of the Chi-Squared
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demonstrate a seed that produces exactly one ChiSquared value less
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than 0.003932.
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##### Requirement: Pass 5 power-of-two statistical tests
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##### Requirement: Pass 3 power-of-two statistical tests
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For the teset with 20 trials and 1 million spans each, the test MUST
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For the test with 20 trials and 100,000 spans each, the test MUST
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demonstrate a random number generator seed such that the ChiSquared
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test statistic is below 0.003932 exactly 1 out of 20 times.
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#### Test implementation
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The recommended structure for this test uses a table listing the 20
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The recommended structure for this test uses a table listing the 15
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probability values, the expected p-values, whether the ChiSquared
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statistic has one or two degrees of freedom, and the index into the
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predetermined list of seeds.
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```
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for _, test := range []testCase{
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// Non-powers of two
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{0.90000, 1, twoDegrees, 5},
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{0.60000, 1, twoDegrees, 14},
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{0.33000, 2, twoDegrees, 3},
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{0.13000, 3, twoDegrees, 2},
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{0.90000, 1, twoDegrees, 3},
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{0.60000, 1, twoDegrees, 2},
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{0.33000, 2, twoDegrees, 2},
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{0.13000, 3, twoDegrees, 1},
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{0.10000, 4, twoDegrees, 0},
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{0.05000, 5, twoDegrees, 0},
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{0.01700, 6, twoDegrees, 2},
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{0.01000, 7, twoDegrees, 3},
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{0.00500, 8, twoDegrees, 1},
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{0.00290, 9, twoDegrees, 1},
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{0.00100, 10, twoDegrees, 5},
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{0.00050, 11, twoDegrees, 1},
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{0.00026, 12, twoDegrees, 3},
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{0.00023, 13, twoDegrees, 0},
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{0.00010, 14, twoDegrees, 2},
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{0.01000, 7, twoDegrees, 2},
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{0.00500, 8, twoDegrees, 2},
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{0.00290, 9, twoDegrees, 4},
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{0.00100, 10, twoDegrees, 6},
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{0.00050, 11, twoDegrees, 0},
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// Powers of two
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{0x1p-1, 1, oneDegree, 0},
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{0x1p-4, 4, oneDegree, 2},
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{0x1p-7, 7, oneDegree, 3},
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{0x1p-10, 10, oneDegree, 0},
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{0x1p-13, 13, oneDegree, 1},
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{0x1p-4, 4, oneDegree, 0},
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{0x1p-7, 7, oneDegree, 1},
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} {
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```
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Note that seed indexes in the example above have what appears to be
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the correct distribution. The five 0s, four 1s, four 2s, four 3s, and
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two 5s demonstrate that it is relatively easy to find examples where
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there is exactly one failure. Seed index 14, for probability 0.6 in
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the correct distribution. The five 0s, two 1s, five 2s, one 3s, and
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one 4 demonstrate that it is relatively easy to find examples where
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there is exactly one failure. Probability 0.001, with seed index 6 in
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this case, is a reminder that outliers exist. Further significance
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testing of this distribution is not recommended.
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