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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 Simple 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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