negative_binomial_example1.cpp

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      << 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 

*/