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- // Copyright Paul A. 2007, 2010
- // Copyright John Maddock 2007
- // Use, modification and distribution are subject to the
- // Boost Software License, Version 1.0.
- // (See accompanying file LICENSE_1_0.txt
- // or copy at http://www.boost.org/LICENSE_1_0.txt)
- // Simple example of computing probabilities for a binomial random variable.
- // Replication of source nag_binomial_dist (g01bjc).
- // Shows how to replace NAG C library calls by Boost Math Toolkit C++ calls.
- // Note that the default policy does not replicate the way that NAG
- // library calls handle 'bad' arguments, but you can define policies that do,
- // as well as other policies that may suit your application even better.
- // See the examples of changing default policies for details.
- #include <boost/math/distributions/binomial.hpp>
- #include <iostream>
- using std::cout; using std::endl; using std::ios; using std::showpoint;
- #include <iomanip>
- using std::fixed; using std::setw;
- int main()
- {
- cout << "Using the binomial distribution to replicate a NAG library call." << endl;
- using boost::math::binomial_distribution;
- // This replicates the computation of the examples of using nag-binomial_dist
- // using g01bjc in section g01 Somple Calculations on Statistical Data.
- // http://www.nag.co.uk/numeric/cl/manual/pdf/G01/g01bjc.pdf
- // Program results section 8.3 page 3.g01bjc.3
- //8.2. Program Data
- //g01bjc Example Program Data
- //4 0.50 2 : n, p, k
- //19 0.44 13
- //100 0.75 67
- //2000 0.33 700
- //8.3. Program Results
- //g01bjc Example Program Results
- //n p k plek pgtk peqk
- //4 0.500 2 0.68750 0.31250 0.37500
- //19 0.440 13 0.99138 0.00862 0.01939
- //100 0.750 67 0.04460 0.95540 0.01700
- //2000 0.330 700 0.97251 0.02749 0.00312
- cout.setf(ios::showpoint); // Trailing zeros to show significant decimal digits.
- cout.precision(5); // Might calculate this from trials in distribution?
- cout << fixed;
- // Binomial distribution.
- // Note that cdf(dist, k) is equivalent to NAG library plek probability of <= k
- // cdf(complement(dist, k)) is equivalent to NAG library pgtk probability of > k
- // pdf(dist, k) is equivalent to NAG library peqk probability of == k
- cout << " n p k plek pgtk peqk " << endl;
- binomial_distribution<>my_dist(4, 0.5);
- cout << setw(4) << (int)my_dist.trials() << " " << my_dist.success_fraction()
- << " " << 2 << " " << cdf(my_dist, 2) << " "
- << cdf(complement(my_dist, 2)) << " " << pdf(my_dist, 2) << endl;
- binomial_distribution<>two(19, 0.440);
- cout << setw(4) << (int)two.trials() << " " << two.success_fraction()
- << " " << 13 << " " << cdf(two, 13) << " "
- << cdf(complement(two, 13)) << " " << pdf(two, 13) << endl;
- binomial_distribution<>three(100, 0.750);
- cout << setw(4) << (int)three.trials() << " " << three.success_fraction()
- << " " << 67 << " " << cdf(three, 67) << " " << cdf(complement(three, 67))
- << " " << pdf(three, 67) << endl;
- binomial_distribution<>four(2000, 0.330);
- cout << setw(4) << (int)four.trials() << " " << four.success_fraction()
- << " " << 700 << " "
- << cdf(four, 700) << " " << cdf(complement(four, 700))
- << " " << pdf(four, 700) << endl;
- return 0;
- } // int main()
- /*
- Example of using the binomial distribution to replicate a NAG library call.
- n p k plek pgtk peqk
- 4 0.50000 2 0.68750 0.31250 0.37500
- 19 0.44000 13 0.99138 0.00862 0.01939
- 100 0.75000 67 0.04460 0.95540 0.01700
- 2000 0.33000 700 0.97251 0.02749 0.00312
- */
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