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Some corrections, but final example 7.5 I don't understand and needs doing or deleting.

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[SVN r38670]
This commit is contained in:
Paul A. Bristow
2007-08-15 10:12:51 +00:00
parent 65b1f877df
commit e16e81bf10
1 changed files with 44 additions and 30 deletions
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74
example/negative_binomial_example3.cpp
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@@ -46,33 +46,39 @@ int main()
// K. Krishnamoorthy, Handbook of Statistical Distributions with Applications,
// ISBN 1 58488 635 8, page 100, example 7.3.1
// r successes = 20, k failures = 18, and success probability (fraction) = 0.6
negative_binomial nbk(20, 0.6); // successes, success fraction.
negative_binomial nbk(20, 0.6); // successes, success_fraction.
cout.precision(6); // default precision.
cout << "probability of < 18 = cdf(18) = " << cdf(nbk, 18) << endl; // = 0.862419
cout << "probability of >= 18 = 1 - cdf(17) = " << 1 - cdf(nbk, 17) << endl; // = 0.181983
// But this may suffer from inaccuracy, so mcuh better to use the complement.
cout << "probability of >= 18 = cdf(complement(nbk, 17)) = "
cout << "Probability of <= 18, P(X <= 18) == cdf(18) = " << cdf(nbk, 18) << endl; // = 0.862419
cout << "Probability of > 18, P(X < 18) == 1 - cdf(18) = " << 1 - cdf(nbk, 18) << endl; // = 0.137581
// But this may suffer from inaccuracy, so much better to use the complement.
cout << "Probability of > 18, P(X < 18) == 1 - cdf(18) = " << cdf(complement(nbk, 18)) << endl; // 0.137581
// And, of course, the sum of probability of X <= 18 and X > 18 really should be unity!
BOOST_ASSERT(cdf(nbk, 18) + cdf(complement(nbk, 18))); // = 0.862419 + 0.137581 == 1
cout << "Probability of < 18 == <= 17 == cdf(17) = " << cdf(nbk, 17) << endl; //
cout << "Probability of >= 18 == > 17 == 1 - cdf(17) = " << 1 - cdf(nbk, 17) << endl; // 0.181983
// But this may suffer from inaccuracy, so much better to use the complement.
cout << "Probability of >= 18 == > 17 == cdf(complement(nbk, 17)) = "
<< cdf(complement(nbk, 17)) << endl; // = 0.181983
cout << "probability of == 18 = pdf(18) = " << pdf(nbk, 18) << endl; // 0.044402
cout << "probability of exactly 18 = pdf(18) = " << pdf(nbk, 18) << endl; // 0.044402
cout << "Probability of exactly == 18, P(X =18) = pdf(18) = " << pdf(nbk, 18) << endl; // 0.044402
cout << endl;
//] [/negative_binomial_example3_1 Quickbook end]
}
{ // Example 7.3.2 to find the success probability when r = 4, k = 5, P(X <= k) = 0.56 and
// calculates success probability = 0.417137.
negative_binomial nb(4, 0.417137);
cout << cdf(nb, 5) << endl; // expect P(X <= k) = 0.56 got 0.560001
cout << "P(X <= k) = " << cdf(nb, 5) << endl; // expect P(X <= k) = 0.56 got 0.560001
// Now check the inverse by calculating the k failures.
cout << quantile(nb, 0.56) << endl; // expect to get k = 5. and got 5.000000
cout << " k = " << quantile(nb, 0.56) << endl; // expect to get k = 5. and got 5.000000
// P(X <= k) = 0.56001
// P(X >= k) = 0.55406
// P(X == k) = 0.114061
// Compute moments:
// mean 5.58918, sd 3.66045, mode = 4, cv = 0.654918, skew 1.03665, Mean dev 2.86877, kurtosis 4.57463
cout << "Mean = " << mean(nb) // 5.589176
<< ", sd = " << standard_deviation(nb)
<< ", Coefficient of variation (sd/mean) = " << standard_deviation(nb) / mean(nb)
<< ", mode = " << mode(nb) << ", skew " << skewness(nb)
<< ", mean deviation = " << "??" // perhaps todo?
// << ", mean deviation = " << "??" // perhaps todo?
<< ", kurt = " << kurtosis(nb) << endl;
}
{ // 7.3.3 Coin flipping. What are chances that 10th head will occur at the 12th flip.
@@ -130,45 +136,51 @@ int main()
cout << "The rounding style for a double type is: "
<< numeric_limits<double>::round_style << endl;
}
{ // K. Krishnamoorthy, ISBN 1 58488 635 8, page 102, section 7.4
{ // K. Krishnamoorthy, ISBN 1 58488 635 8, page 102, section 7.4 & 7.5
// Suppose required k + r trials to get the rth success.
double r = 5; double k = 25; // Example 7.5.1 values
double rm1 = r -1; // r-1 is the penultimate success.
double p = rm1/ (rm1 + k);
// A point estimate of the actual proportion of defective items.
// A point estimate phat of the actual proportion of defective items.
// 0.137931
// 'True' estimate, if this was all the items ever available is successes/failures = r/k = 5/25 = 0.2
// so the point estimate 'how we are doing from the info so far' is rather less at 0.14.
// 'True' estimate, if this was all the items ever available, is successes/failures = r/k = 5/25 = 0.2
// so the point estimate 'how we are doing from the info so far' is rather less at 0.138.
cout << "Uniformly minimum variance unbiased estimator of success probability is " << p << endl;
double v = 0.5 * p * p * (1 - p) * (k + k + 2 - p) / (k * (k - p + 2));
cout << "Variance of estimator is " << v << endl; // 0.000633
cout << "Standard deviation of estimator is " << sqrt(v) << endl; // 0.025 - seems small?
cout << "Standard deviation of estimator is " << sqrt(v) << endl; // 0.025 - seems smallish?
// Would expect point estimate 0.14 + a couple of sd = 0.25 * 2 = 0.05 = 1.9 ~= 0.2?
// So perhaps near enough?
// Section 7.5 testing the true success probability.
// 7.5.1 A lot is inspected randomly. 5th defective found at 30th inspection.
// so 25 are OK and 5 are duds.
// Using StatCalc r = 5, k = 25, so r + r = 30, value for p0 0.3
negative_binomial nb(5, 0.2); // 1/3rd are defective?
cout << cdf(nb, 25) << endl; // 0.812924
cout << quantile(nb, 0.3) << endl; // 7.344587
// TODO?
cout << "P(X <= 25) = " << cdf(nb, 25) << endl; // 0.744767
cout << "quantile(nb, 0.3) = " << quantile(nb, 0.3) << endl; // 13
// Don't understand what the p-value calculation does.
// TODO????
}
return 0;
} // int main()
/*
Example 3 Negative_binomial . ..\..\..\..\..\..\boost-sandbox\math_toolkit\libs\math\example\negative_binomial_example3.cpp Mon Aug 13 14:32:28 2007 140050727
probability of < 18 = cdf(18) = 0.862419
probability of >= 18 = 1 - cdf(17) = 0.181983
probability of >= 18 = cdf(complement(nbk, 17)) = 0.181983
probability of == 18 = pdf(18) = 0.0444024
probability of exactly 18 = pdf(18) = 0.0444024
0.560001
5
Mean = 5.58918, sd = 3.66045, Coefficient of variation (sd/mean) = 0.654918, mode = 4, skew 1.03665, mean deviation = ??, kurt = 4.57463
Output is:
Example 3 Negative_binomial, Krishnamoorthy applications. ..\..\..\..\..\..\boost-sandbox\math_toolkit\libs\math\example\negative_binomial_example3.cpp Wed Aug 15 11:08:14 2007
Probability of <= 18, P(X <= 18) == cdf(18) = 0.862419
Probability of > 18, P(X < 18) == 1 - cdf(18) = 0.137581
Probability of > 18, P(X < 18) == 1 - cdf(18) = 0.137581
Probability of < 18 == <= 17 == cdf(17) = 0.818017
Probability of >= 18 == > 17 == 1 - cdf(17) = 0.181983
Probability of >= 18 == > 17 == cdf(complement(nbk, 17)) = 0.181983
Probability of exactly == 18, P(X =18) = pdf(18) = 0.0444024
P(X <= k) = 0.560001
k = 5
Mean = 5.58918, sd = 3.66045, Coefficient of variation (sd/mean) = 0.654918, mode = 4, skew 1.03665, kurt = 4.57463
Probability that 10th head will occur at the 12th flip is 0.0134277
P(X == k) 0.0134277
Probability that 10th head will occur before the 12th flip is 0.00585938
@@ -186,8 +198,10 @@ The rounding style for a double type is: 1
Uniformly minimum variance unbiased estimator of success probability is 0.137931
Variance of estimator is 0.000633295
Standard deviation of estimator is 0.0251654
0.744767
P(X <= 25) = 0.744767
13
*/