mirror of
https://github.com/saitohirga/WSJT-X.git
synced 2025-10-27 11:00:32 -04:00
520 lines
22 KiB
C++
520 lines
22 KiB
C++
|
// negative_binomial_example1.cpp
|
||
|
|
|
||
|
|
// Copyright Paul A. Bristow 2007, 2010.
|
||
|
|
|
||
|
|
// 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)
|
||
|
|
|
||
|
|
// Example 1 of using negative_binomial distribution.
|
||
|
|
|
||
|
|
//[negative_binomial_eg1_1
|
||
|
|
|
||
|
|
/*`
|
||
|
|
Based on [@http://en.wikipedia.org/wiki/Negative_binomial_distribution
|
||
|
|
a problem by Dr. Diane Evans,
|
||
|
|
Professor of Mathematics at Rose-Hulman Institute of Technology].
|
||
|
|
|
||
|
|
Pat is required to sell candy bars to raise money for the 6th grade field trip.
|
||
|
|
There are thirty houses in the neighborhood,
|
||
|
|
and Pat is not supposed to return home until five candy bars have been sold.
|
||
|
|
So the child goes door to door, selling candy bars.
|
||
|
|
At each house, there is a 0.4 probability (40%) of selling one candy bar
|
||
|
|
and a 0.6 probability (60%) of selling nothing.
|
||
|
|
|
||
|
|
What is the probability mass (density) function (pdf) for selling the last (fifth)
|
||
|
|
candy bar at the nth house?
|
||
|
|
|
||
|
|
The Negative Binomial(r, p) distribution describes the probability of k failures
|
||
|
|
and r successes in k+r Bernoulli(p) trials with success on the last trial.
|
||
|
|
(A [@http://en.wikipedia.org/wiki/Bernoulli_distribution Bernoulli trial]
|
||
|
|
is one with only two possible outcomes, success of failure,
|
||
|
|
and p is the probability of success).
|
||
|
|
See also [@ http://en.wikipedia.org/wiki/Bernoulli_distribution Bernoulli distribution]
|
||
|
|
and [@http://www.math.uah.edu/stat/bernoulli/Introduction.xhtml Bernoulli applications].
|
||
|
|
|
||
|
|
In this example, we will deliberately produce a variety of calculations
|
||
|
|
and outputs to demonstrate the ways that the negative binomial distribution
|
||
|
|
can be implemented with this library: it is also deliberately over-commented.
|
||
|
|
|
||
|
|
First we need to #define macros to control the error and discrete handling policies.
|
||
|
|
For this simple example, we want to avoid throwing
|
||
|
|
an exception (the default policy) and just return infinity.
|
||
|
|
We want to treat the distribution as if it was continuous,
|
||
|
|
so we choose a discrete_quantile policy of real,
|
||
|
|
rather than the default policy integer_round_outwards.
|
||
|
|
*/
|
||
|
|
#define BOOST_MATH_OVERFLOW_ERROR_POLICY ignore_error
|
||
|
|
#define BOOST_MATH_DISCRETE_QUANTILE_POLICY real
|
||
|
|
/*`
|
||
|
|
After that we need some includes to provide easy access to the negative binomial distribution,
|
||
|
|
[caution It is vital to #include distributions etc *after* the above #defines]
|
||
|
|
and we need some std library iostream, of course.
|
||
|
|
*/
|
||
|
|
#include <boost/math/distributions/negative_binomial.hpp>
|
||
|
|
// for negative_binomial_distribution
|
||
|
|
using boost::math::negative_binomial; // typedef provides default type is double.
|
||
|
|
using ::boost::math::pdf; // Probability mass function.
|
||
|
|
using ::boost::math::cdf; // Cumulative density function.
|
||
|
|
using ::boost::math::quantile;
|
||
|
|
|
||
|
|
#include <iostream>
|
||
|
|
using std::cout; using std::endl;
|
||
|
|
using std::noshowpoint; using std::fixed; using std::right; using std::left;
|
||
|
|
#include <iomanip>
|
||
|
|
using std::setprecision; using std::setw;
|
||
|
|
|
||
|
|
#include <limits>
|
||
|
|
using std::numeric_limits;
|
||
|
|
//] [negative_binomial_eg1_1]
|
||
|
|
|
||
|
|
int main()
|
||
|
|
{
|
||
|
|
cout <<"Selling candy bars - using the negative binomial distribution."
|
||
|
|
<< "\nby Dr. Diane Evans,"
|
||
|
|
"\nProfessor of Mathematics at Rose-Hulman Institute of Technology,"
|
||
|
|
<< "\nsee http://en.wikipedia.org/wiki/Negative_binomial_distribution\n"
|
||
|
|
<< endl;
|
||
|
|
cout << endl;
|
||
|
|
cout.precision(5);
|
||
|
|
// None of the values calculated have a useful accuracy as great this, but
|
||
|
|
// INF shows wrongly with < 5 !
|
||
|
|
// https://connect.microsoft.com/VisualStudio/feedback/ViewFeedback.aspx?FeedbackID=240227
|
||
|
|
//[negative_binomial_eg1_2
|
||
|
|
/*`
|
||
|
|
It is always sensible to use try and catch blocks because defaults policies are to
|
||
|
|
throw an exception if anything goes wrong.
|
||
|
|
|
||
|
|
A simple catch block (see below) will ensure that you get a
|
||
|
|
helpful error message instead of an abrupt program abort.
|
||
|
|
*/
|
||
|
|
try
|
||
|
|
{
|
||
|
|
/*`
|
||
|
|
Selling five candy bars means getting five successes, so successes r = 5.
|
||
|
|
The total number of trials (n, in this case, houses visited) this takes is therefore
|
||
|
|
= sucesses + failures or k + r = k + 5.
|
||
|
|
*/
|
||
|
|
double sales_quota = 5; // Pat's sales quota - successes (r).
|
||
|
|
/*`
|
||
|
|
At each house, there is a 0.4 probability (40%) of selling one candy bar
|
||
|
|
and a 0.6 probability (60%) of selling nothing.
|
||
|
|
*/
|
||
|
|
double success_fraction = 0.4; // success_fraction (p) - so failure_fraction is 0.6.
|
||
|
|
/*`
|
||
|
|
The Negative Binomial(r, p) distribution describes the probability of k failures
|
||
|
|
and r successes in k+r Bernoulli(p) trials with success on the last trial.
|
||
|
|
(A [@http://en.wikipedia.org/wiki/Bernoulli_distribution Bernoulli trial]
|
||
|
|
is one with only two possible outcomes, success of failure,
|
||
|
|
and p is the probability of success).
|
||
|
|
|
||
|
|
We therefore start by constructing a negative binomial distribution
|
||
|
|
with parameters sales_quota (required successes) and probability of success.
|
||
|
|
*/
|
||
|
|
negative_binomial nb(sales_quota, success_fraction); // type double by default.
|
||
|
|
/*`
|
||
|
|
To confirm, display the success_fraction & successes parameters of the distribution.
|
||
|
|
*/
|
||
|
|
cout << "Pat has a sales per house success rate of " << success_fraction
|
||
|
|
<< ".\nTherefore he would, on average, sell " << nb.success_fraction() * 100
|
||
|
|
<< " bars after trying 100 houses." << endl;
|
||
|
|
|
||
|
|
int all_houses = 30; // The number of houses on the estate.
|
||
|
|
|
||
|
|
cout << "With a success rate of " << nb.success_fraction()
|
||
|
|
<< ", he might expect, on average,\n"
|
||
|
|
"to need to visit about " << success_fraction * all_houses
|
||
|
|
<< " houses in order to sell all " << nb.successes() << " bars. " << endl;
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
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.
|
||
|
|
]
|
||
|
|
|
||
|
|
The random variable of interest is the number of houses
|
||
|
|
that must be visited to sell five candy bars,
|
||
|
|
so we substitute k = n - 5 into a negative_binomial(5, 0.4)
|
||
|
|
and obtain the __pdf of the distribution of houses visited.
|
||
|
|
Obviously, the best possible case is that Pat makes sales on all the first five houses.
|
||
|
|
|
||
|
|
We calculate this using the pdf function:
|
||
|
|
*/
|
||
|
|
cout << "Probability that Pat finishes on the " << sales_quota << "th house is "
|
||
|
|
<< pdf(nb, 5 - sales_quota) << endl; // == pdf(nb, 0)
|
||
|
|
/*`
|
||
|
|
Of course, he could not finish on fewer than 5 houses because he must sell 5 candy bars.
|
||
|
|
So the 5th house is the first that he could possibly finish on.
|
||
|
|
|
||
|
|
To finish on or before the 8th house, Pat must finish at the 5th, 6th, 7th or 8th house.
|
||
|
|
The probability that he will finish on *exactly* ( == ) on any house
|
||
|
|
is the Probability Density Function (pdf).
|
||
|
|
*/
|
||
|
|
cout << "Probability that Pat finishes on the 6th house is "
|
||
|
|
<< pdf(nb, 6 - sales_quota) << endl;
|
||
|
|
cout << "Probability that Pat finishes on the 7th house is "
|
||
|
|
<< pdf(nb, 7 - sales_quota) << endl;
|
||
|
|
cout << "Probability that Pat finishes on the 8th house is "
|
||
|
|
<< pdf(nb, 8 - sales_quota) << endl;
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
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
|
||
|
|
]
|
||
|
|
|
||
|
|
The sum of the probabilities for these houses is the Cumulative Distribution Function (cdf).
|
||
|
|
We can calculate it by adding the individual probabilities.
|
||
|
|
*/
|
||
|
|
cout << "Probability that Pat finishes on or before the 8th house is sum "
|
||
|
|
"\n" << "pdf(sales_quota) + pdf(6) + pdf(7) + pdf(8) = "
|
||
|
|
// Sum each of the mass/density probabilities for houses sales_quota = 5, 6, 7, & 8.
|
||
|
|
<< pdf(nb, 5 - sales_quota) // 0 failures.
|
||
|
|
+ pdf(nb, 6 - sales_quota) // 1 failure.
|
||
|
|
+ pdf(nb, 7 - sales_quota) // 2 failures.
|
||
|
|
+ pdf(nb, 8 - sales_quota) // 3 failures.
|
||
|
|
<< endl;
|
||
|
|
/*`[pre
|
||
|
|
pdf(sales_quota) + pdf(6) + pdf(7) + pdf(8) = 0.17367
|
||
|
|
]
|
||
|
|
|
||
|
|
Or, usually better, by using the negative binomial *cumulative* distribution function.
|
||
|
|
*/
|
||
|
|
cout << "\nProbability of selling his quota of " << sales_quota
|
||
|
|
<< " bars\non or before the " << 8 << "th house is "
|
||
|
|
<< cdf(nb, 8 - sales_quota) << endl;
|
||
|
|
/*`[pre
|
||
|
|
Probability of selling his quota of 5 bars on or before the 8th house is 0.17367
|
||
|
|
]*/
|
||
|
|
cout << "\nProbability that Pat finishes exactly on the 10th house is "
|
||
|
|
<< pdf(nb, 10 - sales_quota) << endl;
|
||
|
|
cout << "\nProbability of selling his quota of " << sales_quota
|
||
|
|
<< " bars\non or before the " << 10 << "th house is "
|
||
|
|
<< cdf(nb, 10 - sales_quota) << endl;
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
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
|
||
|
|
]*/
|
||
|
|
cout << "Probability that Pat finishes exactly on the 11th house is "
|
||
|
|
<< pdf(nb, 11 - sales_quota) << endl;
|
||
|
|
cout << "\nProbability of selling his quota of " << sales_quota
|
||
|
|
<< " bars\non or before the " << 11 << "th house is "
|
||
|
|
<< cdf(nb, 11 - sales_quota) << endl;
|
||
|
|
/*`[pre
|
||
|
|
Probability that Pat finishes on the 11th house is 0.10033
|
||
|
|
Probability of selling his quota of 5 candy bars
|
||
|
|
on or before the 11th house is 0.46723
|
||
|
|
]*/
|
||
|
|
cout << "Probability that Pat finishes exactly on the 12th house is "
|
||
|
|
<< pdf(nb, 12 - sales_quota) << endl;
|
||
|
|
|
||
|
|
cout << "\nProbability of selling his quota of " << sales_quota
|
||
|
|
<< " bars\non or before the " << 12 << "th house is "
|
||
|
|
<< cdf(nb, 12 - sales_quota) << endl;
|
||
|
|
/*`[pre
|
||
|
|
Probability that Pat finishes on the 12th house is 0.094596
|
||
|
|
Probability of selling his quota of 5 candy bars
|
||
|
|
on or before the 12th house is 0.56182
|
||
|
|
]
|
||
|
|
Finally consider the risk of Pat not selling his quota of 5 bars
|
||
|
|
even after visiting all the houses.
|
||
|
|
Calculate the probability that he /will/ sell on
|
||
|
|
or before the last house:
|
||
|
|
Calculate the probability that he would sell all his quota on the very last house.
|
||
|
|
*/
|
||
|
|
cout << "Probability that Pat finishes on the " << all_houses
|
||
|
|
<< " house is " << pdf(nb, all_houses - sales_quota) << endl;
|
||
|
|
/*`
|
||
|
|
Probability of selling his quota of 5 bars on the 30th house is
|
||
|
|
[pre
|
||
|
|
Probability that Pat finishes on the 30 house is 0.00069145
|
||
|
|
]
|
||
|
|
when he'd be very unlucky indeed!
|
||
|
|
|
||
|
|
What is the probability that Pat exhausts all 30 houses in the neighborhood,
|
||
|
|
and *still* doesn't sell the required 5 candy bars?
|
||
|
|
*/
|
||
|
|
cout << "\nProbability of selling his quota of " << sales_quota
|
||
|
|
<< " bars\non or before the " << all_houses << "th house is "
|
||
|
|
<< cdf(nb, all_houses - sales_quota) << endl;
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
Probability of selling his quota of 5 bars
|
||
|
|
on or before the 30th house is 0.99849
|
||
|
|
]
|
||
|
|
|
||
|
|
/*`So the risk of failing even after visiting all the houses is 1 - this probability,
|
||
|
|
``1 - cdf(nb, all_houses - sales_quota``
|
||
|
|
But using this expression may cause serious inaccuracy,
|
||
|
|
so it would be much better to use the complement of the cdf:
|
||
|
|
So the risk of failing even at, or after, the 31th (non-existent) houses is 1 - this probability,
|
||
|
|
``1 - cdf(nb, all_houses - sales_quota)``
|
||
|
|
But using this expression may cause serious inaccuracy.
|
||
|
|
So it would be much better to use the __complement of the cdf (see __why_complements).
|
||
|
|
*/
|
||
|
|
cout << "\nProbability of failing to sell his quota of " << sales_quota
|
||
|
|
<< " bars\neven after visiting all " << all_houses << " houses is "
|
||
|
|
<< cdf(complement(nb, all_houses - sales_quota)) << endl;
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
Probability of failing to sell his quota of 5 bars
|
||
|
|
even after visiting all 30 houses is 0.0015101
|
||
|
|
]
|
||
|
|
We can also use the quantile (percentile), the inverse of the cdf, to
|
||
|
|
predict which house Pat will finish on. So for the 8th house:
|
||
|
|
*/
|
||
|
|
double p = cdf(nb, (8 - sales_quota));
|
||
|
|
cout << "Probability of meeting sales quota on or before 8th house is "<< p << endl;
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
Probability of meeting sales quota on or before 8th house is 0.174
|
||
|
|
]
|
||
|
|
*/
|
||
|
|
cout << "If the confidence of meeting sales quota is " << p
|
||
|
|
<< ", then the finishing house is " << quantile(nb, p) + sales_quota << endl;
|
||
|
|
|
||
|
|
cout<< " quantile(nb, p) = " << quantile(nb, p) << endl;
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
If the confidence of meeting sales quota is 0.17367, then the finishing house is 8
|
||
|
|
]
|
||
|
|
Demanding absolute certainty that all 5 will be sold,
|
||
|
|
implies an infinite number of trials.
|
||
|
|
(Of course, there are only 30 houses on the estate,
|
||
|
|
so he can't ever be *certain* of selling his quota).
|
||
|
|
*/
|
||
|
|
cout << "If the confidence of meeting sales quota is " << 1.
|
||
|
|
<< ", then the finishing house is " << quantile(nb, 1) + sales_quota << endl;
|
||
|
|
// 1.#INF == infinity.
|
||
|
|
/*`[pre
|
||
|
|
If the confidence of meeting sales quota is 1, then the finishing house is 1.#INF
|
||
|
|
]
|
||
|
|
And similarly for a few other probabilities:
|
||
|
|
*/
|
||
|
|
cout << "If the confidence of meeting sales quota is " << 0.
|
||
|
|
<< ", then the finishing house is " << quantile(nb, 0.) + sales_quota << endl;
|
||
|
|
|
||
|
|
cout << "If the confidence of meeting sales quota is " << 0.5
|
||
|
|
<< ", then the finishing house is " << quantile(nb, 0.5) + sales_quota << endl;
|
||
|
|
|
||
|
|
cout << "If the confidence of meeting sales quota is " << 1 - 0.00151 // 30 th
|
||
|
|
<< ", then the finishing house is " << quantile(nb, 1 - 0.00151) + sales_quota << endl;
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
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
|
||
|
|
]
|
||
|
|
|
||
|
|
Notice that because we chose a discrete quantile policy of real,
|
||
|
|
the result can be an 'unreal' fractional house.
|
||
|
|
|
||
|
|
If the opposite is true, we don't want to assume any confidence, then this is tantamount
|
||
|
|
to assuming that all the first sales_quota trials will be successful sales.
|
||
|
|
*/
|
||
|
|
cout << "If confidence of meeting quota is zero\n(we assume all houses are successful sales)"
|
||
|
|
", then finishing house is " << sales_quota << endl;
|
||
|
|
/*`
|
||
|
|
[pre
|
||
|
|
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
|
||
|
|
]
|
||
|
|
We can list quantiles for a few probabilities:
|
||
|
|
*/
|
||
|
|
|
||
|
|
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
|
||
|
|
|
||
|
|
*/
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|
||
|
|
|