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520 lines
22 KiB
C++
520 lines
22 KiB
C++
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// negative_binomial_example1.cpp
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// Copyright Paul A. Bristow 2007, 2010.
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// Use, modification and distribution are subject to the
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// Boost Software License, Version 1.0.
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// (See accompanying file LICENSE_1_0.txt
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// or copy at http://www.boost.org/LICENSE_1_0.txt)
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// Example 1 of using negative_binomial distribution.
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//[negative_binomial_eg1_1
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/*`
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Based on [@http://en.wikipedia.org/wiki/Negative_binomial_distribution
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a problem by Dr. Diane Evans,
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Professor of Mathematics at Rose-Hulman Institute of Technology].
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Pat is required to sell candy bars to raise money for the 6th grade field trip.
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There are thirty houses in the neighborhood,
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and Pat is not supposed to return home until five candy bars have been sold.
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So the child goes door to door, selling candy bars.
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At each house, there is a 0.4 probability (40%) of selling one candy bar
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and a 0.6 probability (60%) of selling nothing.
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What is the probability mass (density) function (pdf) for selling the last (fifth)
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candy bar at the nth house?
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The Negative Binomial(r, p) distribution describes the probability of k failures
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and r successes in k+r Bernoulli(p) trials with success on the last trial.
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(A [@http://en.wikipedia.org/wiki/Bernoulli_distribution Bernoulli trial]
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is one with only two possible outcomes, success of failure,
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and p is the probability of success).
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See also [@ http://en.wikipedia.org/wiki/Bernoulli_distribution Bernoulli distribution]
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and [@http://www.math.uah.edu/stat/bernoulli/Introduction.xhtml Bernoulli applications].
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In this example, we will deliberately produce a variety of calculations
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and outputs to demonstrate the ways that the negative binomial distribution
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can be implemented with this library: it is also deliberately over-commented.
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First we need to #define macros to control the error and discrete handling policies.
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For this simple example, we want to avoid throwing
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an exception (the default policy) and just return infinity.
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We want to treat the distribution as if it was continuous,
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so we choose a discrete_quantile policy of real,
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rather than the default policy integer_round_outwards.
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*/
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#define BOOST_MATH_OVERFLOW_ERROR_POLICY ignore_error
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#define BOOST_MATH_DISCRETE_QUANTILE_POLICY real
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/*`
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After that we need some includes to provide easy access to the negative binomial distribution,
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[caution It is vital to #include distributions etc *after* the above #defines]
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and we need some std library iostream, of course.
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*/
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#include <boost/math/distributions/negative_binomial.hpp>
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// for negative_binomial_distribution
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using boost::math::negative_binomial; // typedef provides default type is double.
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using ::boost::math::pdf; // Probability mass function.
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using ::boost::math::cdf; // Cumulative density function.
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using ::boost::math::quantile;
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#include <iostream>
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using std::cout; using std::endl;
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using std::noshowpoint; using std::fixed; using std::right; using std::left;
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#include <iomanip>
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using std::setprecision; using std::setw;
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#include <limits>
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using std::numeric_limits;
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//] [negative_binomial_eg1_1]
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int main()
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{
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cout <<"Selling candy bars - using the negative binomial distribution."
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<< "\nby Dr. Diane Evans,"
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"\nProfessor of Mathematics at Rose-Hulman Institute of Technology,"
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<< "\nsee http://en.wikipedia.org/wiki/Negative_binomial_distribution\n"
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<< endl;
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cout << endl;
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cout.precision(5);
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// None of the values calculated have a useful accuracy as great this, but
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// INF shows wrongly with < 5 !
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// https://connect.microsoft.com/VisualStudio/feedback/ViewFeedback.aspx?FeedbackID=240227
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//[negative_binomial_eg1_2
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/*`
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It is always sensible to use try and catch blocks because defaults policies are to
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throw an exception if anything goes wrong.
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A simple catch block (see below) will ensure that you get a
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helpful error message instead of an abrupt program abort.
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*/
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try
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{
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/*`
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Selling five candy bars means getting five successes, so successes r = 5.
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The total number of trials (n, in this case, houses visited) this takes is therefore
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= sucesses + failures or k + r = k + 5.
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*/
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double sales_quota = 5; // Pat's sales quota - successes (r).
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/*`
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At each house, there is a 0.4 probability (40%) of selling one candy bar
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and a 0.6 probability (60%) of selling nothing.
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*/
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double success_fraction = 0.4; // success_fraction (p) - so failure_fraction is 0.6.
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/*`
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The Negative Binomial(r, p) distribution describes the probability of k failures
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and r successes in k+r Bernoulli(p) trials with success on the last trial.
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(A [@http://en.wikipedia.org/wiki/Bernoulli_distribution Bernoulli trial]
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is one with only two possible outcomes, success of failure,
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and p is the probability of success).
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We therefore start by constructing a negative binomial distribution
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with parameters sales_quota (required successes) and probability of success.
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*/
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negative_binomial nb(sales_quota, success_fraction); // type double by default.
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/*`
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To confirm, display the success_fraction & successes parameters of the distribution.
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*/
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cout << "Pat has a sales per house success rate of " << success_fraction
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<< ".\nTherefore he would, on average, sell " << nb.success_fraction() * 100
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<< " bars after trying 100 houses." << endl;
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int all_houses = 30; // The number of houses on the estate.
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cout << "With a success rate of " << nb.success_fraction()
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<< ", he might expect, on average,\n"
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"to need to visit about " << success_fraction * all_houses
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<< " houses in order to sell all " << nb.successes() << " bars. " << endl;
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/*`
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[pre
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Pat has a sales per house success rate of 0.4.
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Therefore he would, on average, sell 40 bars after trying 100 houses.
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With a success rate of 0.4, he might expect, on average,
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to need to visit about 12 houses in order to sell all 5 bars.
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]
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The random variable of interest is the number of houses
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that must be visited to sell five candy bars,
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so we substitute k = n - 5 into a negative_binomial(5, 0.4)
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and obtain the __pdf of the distribution of houses visited.
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Obviously, the best possible case is that Pat makes sales on all the first five houses.
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We calculate this using the pdf function:
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*/
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cout << "Probability that Pat finishes on the " << sales_quota << "th house is "
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<< pdf(nb, 5 - sales_quota) << endl; // == pdf(nb, 0)
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/*`
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Of course, he could not finish on fewer than 5 houses because he must sell 5 candy bars.
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So the 5th house is the first that he could possibly finish on.
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To finish on or before the 8th house, Pat must finish at the 5th, 6th, 7th or 8th house.
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The probability that he will finish on *exactly* ( == ) on any house
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is the Probability Density Function (pdf).
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*/
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cout << "Probability that Pat finishes on the 6th house is "
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<< pdf(nb, 6 - sales_quota) << endl;
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cout << "Probability that Pat finishes on the 7th house is "
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<< pdf(nb, 7 - sales_quota) << endl;
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cout << "Probability that Pat finishes on the 8th house is "
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<< pdf(nb, 8 - sales_quota) << endl;
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/*`
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[pre
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Probability that Pat finishes on the 6th house is 0.03072
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Probability that Pat finishes on the 7th house is 0.055296
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Probability that Pat finishes on the 8th house is 0.077414
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]
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The sum of the probabilities for these houses is the Cumulative Distribution Function (cdf).
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We can calculate it by adding the individual probabilities.
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*/
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cout << "Probability that Pat finishes on or before the 8th house is sum "
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"\n" << "pdf(sales_quota) + pdf(6) + pdf(7) + pdf(8) = "
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// Sum each of the mass/density probabilities for houses sales_quota = 5, 6, 7, & 8.
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<< pdf(nb, 5 - sales_quota) // 0 failures.
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+ pdf(nb, 6 - sales_quota) // 1 failure.
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+ pdf(nb, 7 - sales_quota) // 2 failures.
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+ pdf(nb, 8 - sales_quota) // 3 failures.
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<< endl;
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/*`[pre
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pdf(sales_quota) + pdf(6) + pdf(7) + pdf(8) = 0.17367
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]
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Or, usually better, by using the negative binomial *cumulative* distribution function.
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*/
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cout << "\nProbability of selling his quota of " << sales_quota
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<< " bars\non or before the " << 8 << "th house is "
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<< cdf(nb, 8 - sales_quota) << endl;
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/*`[pre
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Probability of selling his quota of 5 bars on or before the 8th house is 0.17367
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]*/
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cout << "\nProbability that Pat finishes exactly on the 10th house is "
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<< pdf(nb, 10 - sales_quota) << endl;
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cout << "\nProbability of selling his quota of " << sales_quota
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<< " bars\non or before the " << 10 << "th house is "
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<< cdf(nb, 10 - sales_quota) << endl;
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/*`
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[pre
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Probability that Pat finishes exactly on the 10th house is 0.10033
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Probability of selling his quota of 5 bars on or before the 10th house is 0.3669
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]*/
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cout << "Probability that Pat finishes exactly on the 11th house is "
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<< pdf(nb, 11 - sales_quota) << endl;
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cout << "\nProbability of selling his quota of " << sales_quota
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<< " bars\non or before the " << 11 << "th house is "
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<< cdf(nb, 11 - sales_quota) << endl;
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/*`[pre
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Probability that Pat finishes on the 11th house is 0.10033
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Probability of selling his quota of 5 candy bars
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on or before the 11th house is 0.46723
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]*/
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cout << "Probability that Pat finishes exactly on the 12th house is "
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<< pdf(nb, 12 - sales_quota) << endl;
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cout << "\nProbability of selling his quota of " << sales_quota
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<< " bars\non or before the " << 12 << "th house is "
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<< cdf(nb, 12 - sales_quota) << endl;
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/*`[pre
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Probability that Pat finishes on the 12th house is 0.094596
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Probability of selling his quota of 5 candy bars
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on or before the 12th house is 0.56182
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]
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Finally consider the risk of Pat not selling his quota of 5 bars
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even after visiting all the houses.
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Calculate the probability that he /will/ sell on
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or before the last house:
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Calculate the probability that he would sell all his quota on the very last house.
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*/
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cout << "Probability that Pat finishes on the " << all_houses
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<< " house is " << pdf(nb, all_houses - sales_quota) << endl;
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/*`
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Probability of selling his quota of 5 bars on the 30th house is
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[pre
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Probability that Pat finishes on the 30 house is 0.00069145
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]
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when he'd be very unlucky indeed!
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What is the probability that Pat exhausts all 30 houses in the neighborhood,
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and *still* doesn't sell the required 5 candy bars?
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*/
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cout << "\nProbability of selling his quota of " << sales_quota
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<< " bars\non or before the " << all_houses << "th house is "
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<< cdf(nb, all_houses - sales_quota) << endl;
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/*`
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[pre
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Probability of selling his quota of 5 bars
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on or before the 30th house is 0.99849
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]
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/*`So the risk of failing even after visiting all the houses is 1 - this probability,
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``1 - cdf(nb, all_houses - sales_quota``
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But using this expression may cause serious inaccuracy,
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so it would be much better to use the complement of the cdf:
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So the risk of failing even at, or after, the 31th (non-existent) houses is 1 - this probability,
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``1 - cdf(nb, all_houses - sales_quota)``
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But using this expression may cause serious inaccuracy.
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So it would be much better to use the __complement of the cdf (see __why_complements).
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*/
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cout << "\nProbability of failing to sell his quota of " << sales_quota
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<< " bars\neven after visiting all " << all_houses << " houses is "
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<< cdf(complement(nb, all_houses - sales_quota)) << endl;
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/*`
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[pre
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Probability of failing to sell his quota of 5 bars
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even after visiting all 30 houses is 0.0015101
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]
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We can also use the quantile (percentile), the inverse of the cdf, to
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predict which house Pat will finish on. So for the 8th house:
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*/
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double p = cdf(nb, (8 - sales_quota));
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cout << "Probability of meeting sales quota on or before 8th house is "<< p << endl;
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/*`
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[pre
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Probability of meeting sales quota on or before 8th house is 0.174
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]
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*/
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cout << "If the confidence of meeting sales quota is " << p
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<< ", then the finishing house is " << quantile(nb, p) + sales_quota << endl;
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cout<< " quantile(nb, p) = " << quantile(nb, p) << endl;
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/*`
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[pre
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If the confidence of meeting sales quota is 0.17367, then the finishing house is 8
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]
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Demanding absolute certainty that all 5 will be sold,
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implies an infinite number of trials.
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(Of course, there are only 30 houses on the estate,
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so he can't ever be *certain* of selling his quota).
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*/
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cout << "If the confidence of meeting sales quota is " << 1.
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<< ", then the finishing house is " << quantile(nb, 1) + sales_quota << endl;
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// 1.#INF == infinity.
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/*`[pre
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If the confidence of meeting sales quota is 1, then the finishing house is 1.#INF
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]
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And similarly for a few other probabilities:
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*/
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cout << "If the confidence of meeting sales quota is " << 0.
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<< ", then the finishing house is " << quantile(nb, 0.) + sales_quota << endl;
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cout << "If the confidence of meeting sales quota is " << 0.5
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<< ", then the finishing house is " << quantile(nb, 0.5) + sales_quota << endl;
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cout << "If the confidence of meeting sales quota is " << 1 - 0.00151 // 30 th
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<< ", then the finishing house is " << quantile(nb, 1 - 0.00151) + sales_quota << endl;
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/*`
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[pre
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If the confidence of meeting sales quota is 0, then the finishing house is 5
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If the confidence of meeting sales quota is 0.5, then the finishing house is 11.337
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If the confidence of meeting sales quota is 0.99849, then the finishing house is 30
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]
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Notice that because we chose a discrete quantile policy of real,
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the result can be an 'unreal' fractional house.
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If the opposite is true, we don't want to assume any confidence, then this is tantamount
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to assuming that all the first sales_quota trials will be successful sales.
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*/
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cout << "If confidence of meeting quota is zero\n(we assume all houses are successful sales)"
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", then finishing house is " << sales_quota << endl;
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/*`
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[pre
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If confidence of meeting quota is zero (we assume all houses are successful sales), then finishing house is 5
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If confidence of meeting quota is 0, then finishing house is 5
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]
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We can list quantiles for a few probabilities:
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*/
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double ps[] = {0., 0.001, 0.01, 0.05, 0.1, 0.5, 0.9, 0.95, 0.99, 0.999, 1.};
|
||
|
|
// Confidence as fraction = 1-alpha, as percent = 100 * (1-alpha[i]) %
|
||
|
|
cout.precision(3);
|
||
|
|
for (int i = 0; i < sizeof(ps)/sizeof(ps[0]); i++)
|
||
|
|
{
|
||
|
|
cout << "If confidence of meeting quota is " << ps[i]
|
||
|
|
<< ", then finishing house is " << quantile(nb, ps[i]) + sales_quota
|
||
|
|
<< endl;
|
||
|
|
}
|
||
|
|
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
If confidence of meeting quota is 0, then finishing house is 5
|
||
|
|
If confidence of meeting quota is 0.001, then finishing house is 5
|
||
|
|
If confidence of meeting quota is 0.01, then finishing house is 5
|
||
|
|
If confidence of meeting quota is 0.05, then finishing house is 6.2
|
||
|
|
If confidence of meeting quota is 0.1, then finishing house is 7.06
|
||
|
|
If confidence of meeting quota is 0.5, then finishing house is 11.3
|
||
|
|
If confidence of meeting quota is 0.9, then finishing house is 17.8
|
||
|
|
If confidence of meeting quota is 0.95, then finishing house is 20.1
|
||
|
|
If confidence of meeting quota is 0.99, then finishing house is 24.8
|
||
|
|
If confidence of meeting quota is 0.999, then finishing house is 31.1
|
||
|
|
If confidence of meeting quota is 1, then finishing house is 1.#INF
|
||
|
|
]
|
||
|
|
|
||
|
|
We could have applied a ceil function to obtain a 'worst case' integer value for house.
|
||
|
|
``ceil(quantile(nb, ps[i]))``
|
||
|
|
|
||
|
|
Or, if we had used the default discrete quantile policy, integer_outside, by omitting
|
||
|
|
``#define BOOST_MATH_DISCRETE_QUANTILE_POLICY real``
|
||
|
|
we would have achieved the same effect.
|
||
|
|
|
||
|
|
The real result gives some suggestion which house is most likely.
|
||
|
|
For example, compare the real and integer_outside for 95% confidence.
|
||
|
|
|
||
|
|
[pre
|
||
|
|
If confidence of meeting quota is 0.95, then finishing house is 20.1
|
||
|
|
If confidence of meeting quota is 0.95, then finishing house is 21
|
||
|
|
]
|
||
|
|
The real value 20.1 is much closer to 20 than 21, so integer_outside is pessimistic.
|
||
|
|
We could also use integer_round_nearest policy to suggest that 20 is more likely.
|
||
|
|
|
||
|
|
Finally, we can tabulate the probability for the last sale being exactly on each house.
|
||
|
|
*/
|
||
|
|
cout << "\nHouse for " << sales_quota << "th (last) sale. Probability (%)" << endl;
|
||
|
|
cout.precision(5);
|
||
|
|
for (int i = (int)sales_quota; i < all_houses+1; i++)
|
||
|
|
{
|
||
|
|
cout << left << setw(3) << i << " " << setw(8) << cdf(nb, i - sales_quota) << endl;
|
||
|
|
}
|
||
|
|
cout << endl;
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
House for 5 th (last) sale. Probability (%)
|
||
|
|
5 0.01024
|
||
|
|
6 0.04096
|
||
|
|
7 0.096256
|
||
|
|
8 0.17367
|
||
|
|
9 0.26657
|
||
|
|
10 0.3669
|
||
|
|
11 0.46723
|
||
|
|
12 0.56182
|
||
|
|
13 0.64696
|
||
|
|
14 0.72074
|
||
|
|
15 0.78272
|
||
|
|
16 0.83343
|
||
|
|
17 0.874
|
||
|
|
18 0.90583
|
||
|
|
19 0.93039
|
||
|
|
20 0.94905
|
||
|
|
21 0.96304
|
||
|
|
22 0.97342
|
||
|
|
23 0.98103
|
||
|
|
24 0.98655
|
||
|
|
25 0.99053
|
||
|
|
26 0.99337
|
||
|
|
27 0.99539
|
||
|
|
28 0.99681
|
||
|
|
29 0.9978
|
||
|
|
30 0.99849
|
||
|
|
]
|
||
|
|
|
||
|
|
As noted above, using a catch block is always a good idea, even if you do not expect to use it.
|
||
|
|
*/
|
||
|
|
}
|
||
|
|
catch(const std::exception& e)
|
||
|
|
{ // Since we have set an overflow policy of ignore_error,
|
||
|
|
// an overflow exception should never be thrown.
|
||
|
|
std::cout << "\nMessage from thrown exception was:\n " << e.what() << std::endl;
|
||
|
|
/*`
|
||
|
|
For example, without a ignore domain error policy, if we asked for ``pdf(nb, -1)`` for example, we would get:
|
||
|
|
[pre
|
||
|
|
Message from thrown exception was:
|
||
|
|
Error in function boost::math::pdf(const negative_binomial_distribution<double>&, double):
|
||
|
|
Number of failures argument is -1, but must be >= 0 !
|
||
|
|
]
|
||
|
|
*/
|
||
|
|
//] [/ negative_binomial_eg1_2]
|
||
|
|
}
|
||
|
|
return 0;
|
||
|
|
} // int main()
|
||
|
|
|
||
|
|
|
||
|
|
/*
|
||
|
|
|
||
|
|
Output is:
|
||
|
|
|
||
|
|
Selling candy bars - using the negative binomial distribution.
|
||
|
|
by Dr. Diane Evans,
|
||
|
|
Professor of Mathematics at Rose-Hulman Institute of Technology,
|
||
|
|
see http://en.wikipedia.org/wiki/Negative_binomial_distribution
|
||
|
|
Pat has a sales per house success rate of 0.4.
|
||
|
|
Therefore he would, on average, sell 40 bars after trying 100 houses.
|
||
|
|
With a success rate of 0.4, he might expect, on average,
|
||
|
|
to need to visit about 12 houses in order to sell all 5 bars.
|
||
|
|
Probability that Pat finishes on the 5th house is 0.01024
|
||
|
|
Probability that Pat finishes on the 6th house is 0.03072
|
||
|
|
Probability that Pat finishes on the 7th house is 0.055296
|
||
|
|
Probability that Pat finishes on the 8th house is 0.077414
|
||
|
|
Probability that Pat finishes on or before the 8th house is sum
|
||
|
|
pdf(sales_quota) + pdf(6) + pdf(7) + pdf(8) = 0.17367
|
||
|
|
Probability of selling his quota of 5 bars
|
||
|
|
on or before the 8th house is 0.17367
|
||
|
|
Probability that Pat finishes exactly on the 10th house is 0.10033
|
||
|
|
Probability of selling his quota of 5 bars
|
||
|
|
on or before the 10th house is 0.3669
|
||
|
|
Probability that Pat finishes exactly on the 11th house is 0.10033
|
||
|
|
Probability of selling his quota of 5 bars
|
||
|
|
on or before the 11th house is 0.46723
|
||
|
|
Probability that Pat finishes exactly on the 12th house is 0.094596
|
||
|
|
Probability of selling his quota of 5 bars
|
||
|
|
on or before the 12th house is 0.56182
|
||
|
|
Probability that Pat finishes on the 30 house is 0.00069145
|
||
|
|
Probability of selling his quota of 5 bars
|
||
|
|
on or before the 30th house is 0.99849
|
||
|
|
Probability of failing to sell his quota of 5 bars
|
||
|
|
even after visiting all 30 houses is 0.0015101
|
||
|
|
Probability of meeting sales quota on or before 8th house is 0.17367
|
||
|
|
If the confidence of meeting sales quota is 0.17367, then the finishing house is 8
|
||
|
|
quantile(nb, p) = 3
|
||
|
|
If the confidence of meeting sales quota is 1, then the finishing house is 1.#INF
|
||
|
|
If the confidence of meeting sales quota is 0, then the finishing house is 5
|
||
|
|
If the confidence of meeting sales quota is 0.5, then the finishing house is 11.337
|
||
|
|
If the confidence of meeting sales quota is 0.99849, then the finishing house is 30
|
||
|
|
If confidence of meeting quota is zero
|
||
|
|
(we assume all houses are successful sales), then finishing house is 5
|
||
|
|
If confidence of meeting quota is 0, then finishing house is 5
|
||
|
|
If confidence of meeting quota is 0.001, then finishing house is 5
|
||
|
|
If confidence of meeting quota is 0.01, then finishing house is 5
|
||
|
|
If confidence of meeting quota is 0.05, then finishing house is 6.2
|
||
|
|
If confidence of meeting quota is 0.1, then finishing house is 7.06
|
||
|
|
If confidence of meeting quota is 0.5, then finishing house is 11.3
|
||
|
|
If confidence of meeting quota is 0.9, then finishing house is 17.8
|
||
|
|
If confidence of meeting quota is 0.95, then finishing house is 20.1
|
||
|
|
If confidence of meeting quota is 0.99, then finishing house is 24.8
|
||
|
|
If confidence of meeting quota is 0.999, then finishing house is 31.1
|
||
|
|
If confidence of meeting quota is 1, then finishing house is 1.#J
|
||
|
|
House for 5th (last) sale. Probability (%)
|
||
|
|
5 0.01024
|
||
|
|
6 0.04096
|
||
|
|
7 0.096256
|
||
|
|
8 0.17367
|
||
|
|
9 0.26657
|
||
|
|
10 0.3669
|
||
|
|
11 0.46723
|
||
|
|
12 0.56182
|
||
|
|
13 0.64696
|
||
|
|
14 0.72074
|
||
|
|
15 0.78272
|
||
|
|
16 0.83343
|
||
|
|
17 0.874
|
||
|
|
18 0.90583
|
||
|
|
19 0.93039
|
||
|
|
20 0.94905
|
||
|
|
21 0.96304
|
||
|
|
22 0.97342
|
||
|
|
23 0.98103
|
||
|
|
24 0.98655
|
||
|
|
25 0.99053
|
||
|
|
26 0.99337
|
||
|
|
27 0.99539
|
||
|
|
28 0.99681
|
||
|
|
29 0.9978
|
||
|
|
30 0.99849
|
||
|
|
|
||
|
|
*/
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|