WSJT-X/boost/libs/math/doc/distributions/binomial_example.qbk

[section:binom_eg Binomial Distribution Examples]

See also the reference documentation for the __binomial_distrib.

[section:binomial_coinflip_example Binomial Coin-Flipping Example]

[import ../../example/binomial_coinflip_example.cpp]
[binomial_coinflip_example1]

See [@../../example/binomial_coinflip_example.cpp binomial_coinflip_example.cpp]
for full source code, the program output looks like this:

[binomial_coinflip_example_output]

[endsect] [/section:binomial_coinflip_example Binomial coinflip example]

[section:binomial_quiz_example Binomial Quiz Example]

[import ../../example/binomial_quiz_example.cpp]
[binomial_quiz_example1]
[binomial_quiz_example2]
[discrete_quantile_real]

See [@../../example/binomial_quiz_example.cpp binomial_quiz_example.cpp]
for full source code and output.

[endsect] [/section:binomial_coinflip_quiz Binomial Coin-Flipping example]

[section:binom_conf Calculating Confidence Limits on the Frequency of Occurrence for a Binomial Distribution]

Imagine you have a process that follows a binomial distribution: for each
trial conducted, an event either occurs or does it does not, referred
to as "successes" and "failures".  If, by experiment, you want to measure the
frequency with which successes occur, the best estimate is given simply
by /k/ \/ /N/, for /k/ successes out of /N/ trials.  However our confidence in that
estimate will be shaped by how many trials were conducted, and how many successes
were observed.  The static member functions 
`binomial_distribution<>::find_lower_bound_on_p` and
`binomial_distribution<>::find_upper_bound_on_p` allow you to calculate
the confidence intervals for your estimate of the occurrence frequency.

The sample program [@../../example/binomial_confidence_limits.cpp 
binomial_confidence_limits.cpp] illustrates their use.  It begins by defining
a procedure that will print a table of confidence limits for various degrees
of certainty:

   #include <iostream>
   #include <iomanip>
   #include <boost/math/distributions/binomial.hpp>

   void confidence_limits_on_frequency(unsigned trials, unsigned successes)
   {
      //
      // trials = Total number of trials.
      // successes = Total number of observed successes.
      //
      // Calculate confidence limits for an observed
      // frequency of occurrence that follows a binomial
      // distribution.
      //
      using namespace std;
      using namespace boost::math;

      // Print out general info:
      cout <<
         "___________________________________________\n"
         "2-Sided Confidence Limits For Success Ratio\n"
         "___________________________________________\n\n";
      cout << setprecision(7);
      cout << setw(40) << left << "Number of Observations" << "=  " << trials << "\n";
      cout << setw(40) << left << "Number of successes" << "=  " << successes << "\n";
      cout << setw(40) << left << "Sample frequency of occurrence" << "=  " << double(successes) / trials << "\n";

The procedure now defines a table of significance levels: these are the 
probabilities that the true occurrence frequency lies outside the calculated
interval:

      double alpha[] = { 0.5, 0.25, 0.1, 0.05, 0.01, 0.001, 0.0001, 0.00001 };

Some pretty printing of the table header follows:

   cout << "\n\n"
           "_______________________________________________________________________\n"
           "Confidence        Lower CP       Upper CP       Lower JP       Upper JP\n"
           " Value (%)        Limit          Limit          Limit          Limit\n"
           "_______________________________________________________________________\n";


And now for the important part - the intervals themselves - for each
value of /alpha/, we call `find_lower_bound_on_p` and 
`find_lower_upper_on_p` to obtain lower and upper bounds
respectively.  Note that since we are calculating a two-sided interval,
we must divide the value of alpha in two.

Please note that calculating two separate /single sided bounds/, each with risk
level [alpha][space]is not the same thing as calculating a two sided interval.
Had we calculate two single-sided intervals each with a risk
that the true value is outside the interval of [alpha], then:

* The risk that it is less than the lower bound is [alpha].

and

* The risk that it is greater than the upper bound is also [alpha].

So the risk it is outside *upper or lower bound*, is *twice* alpha, and the 
probability that it is inside the bounds is therefore not nearly as high as 
one might have thought.  This is why [alpha]/2 must be used in 
the calculations below.

In contrast, had we been calculating a 
single-sided interval, for example: ['"Calculate a lower bound so that we are P%
sure that the true occurrence frequency is greater than some value"]
then we would *not* have divided by two.

Finally note that `binomial_distribution` provides a choice of two
methods for the calculation, we print out the results from both 
methods in this example:

      for(unsigned i = 0; i < sizeof(alpha)/sizeof(alpha[0]); ++i)
      {
         // Confidence value:
         cout << fixed << setprecision(3) << setw(10) << right << 100 * (1-alpha[i]);
         // Calculate Clopper Pearson bounds:
         double l = binomial_distribution<>::find_lower_bound_on_p(
                        trials, successes, alpha[i]/2);
         double u = binomial_distribution<>::find_upper_bound_on_p(
                        trials, successes, alpha[i]/2);
         // Print Clopper Pearson Limits:
         cout << fixed << setprecision(5) << setw(15) << right << l;
         cout << fixed << setprecision(5) << setw(15) << right << u;
         // Calculate Jeffreys Prior Bounds:
         l = binomial_distribution<>::find_lower_bound_on_p(
               trials, successes, alpha[i]/2, 
               binomial_distribution<>::jeffreys_prior_interval);
         u = binomial_distribution<>::find_upper_bound_on_p(
               trials, successes, alpha[i]/2, 
               binomial_distribution<>::jeffreys_prior_interval);
         // Print Jeffreys Prior Limits:
         cout << fixed << setprecision(5) << setw(15) << right << l;
         cout << fixed << setprecision(5) << setw(15) << right << u << std::endl;
      }
      cout << endl;
   }

And that's all there is to it.  Let's see some sample output for a 2 in 10
success ratio, first for 20 trials:

[pre'''___________________________________________
2-Sided Confidence Limits For Success Ratio
___________________________________________

Number of Observations                  =  20
Number of successes                     =  4
Sample frequency of occurrence          =  0.2


_______________________________________________________________________
Confidence        Lower CP       Upper CP       Lower JP       Upper JP
 Value (%)        Limit          Limit          Limit          Limit
_______________________________________________________________________
    50.000        0.12840        0.29588        0.14974        0.26916
    75.000        0.09775        0.34633        0.11653        0.31861
    90.000        0.07135        0.40103        0.08734        0.37274
    95.000        0.05733        0.43661        0.07152        0.40823
    99.000        0.03576        0.50661        0.04655        0.47859
    99.900        0.01905        0.58632        0.02634        0.55960
    99.990        0.01042        0.64997        0.01530        0.62495
    99.999        0.00577        0.70216        0.00901        0.67897
''']

As you can see, even at the 95% confidence level the bounds are
really quite wide (this example is chosen to be easily compared to the one
in the __handbook
[@http://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm
here]).  Note also that the Clopper-Pearson calculation method (CP above)
produces quite noticeably more pessimistic estimates than the Jeffreys Prior
method (JP above).


Compare that with the program output for
2000 trials:

[pre'''___________________________________________
2-Sided Confidence Limits For Success Ratio
___________________________________________

Number of Observations                  =  2000
Number of successes                     =  400
Sample frequency of occurrence          =  0.2000000


_______________________________________________________________________
Confidence        Lower CP       Upper CP       Lower JP       Upper JP
 Value (%)        Limit          Limit          Limit          Limit
_______________________________________________________________________
    50.000        0.19382        0.20638        0.19406        0.20613
    75.000        0.18965        0.21072        0.18990        0.21047
    90.000        0.18537        0.21528        0.18561        0.21503
    95.000        0.18267        0.21821        0.18291        0.21796
    99.000        0.17745        0.22400        0.17769        0.22374
    99.900        0.17150        0.23079        0.17173        0.23053
    99.990        0.16658        0.23657        0.16681        0.23631
    99.999        0.16233        0.24169        0.16256        0.24143
''']

Now even when the confidence level is very high, the limits are really
quite close to the experimentally calculated value of 0.2.  Furthermore
the difference between the two calculation methods is now really quite small.

[endsect]

[section:binom_size_eg Estimating Sample Sizes for a Binomial Distribution.]

Imagine you have a critical component that you know will fail in 1 in
N "uses" (for some suitable definition of "use").  You may want to schedule
routine replacement of the component so that its chance of failure between
routine replacements is less than P%.  If the failures follow a binomial
distribution (each time the component is "used" it either fails or does not)
then the static member function `binomial_distibution<>::find_maximum_number_of_trials`
can be used to estimate the maximum number of "uses" of that component for some
acceptable risk level /alpha/.

The example program 
[@../../example/binomial_sample_sizes.cpp binomial_sample_sizes.cpp]
demonstrates its usage.  It centres on a routine that prints out
a table of maximum sample sizes for various probability thresholds:

   void find_max_sample_size(
      double p,              // success ratio.
      unsigned successes)    // Total number of observed successes permitted.
   {

The routine then declares a table of probability thresholds: these are the
maximum acceptable probability that /successes/ or fewer events will be
observed.  In our example, /successes/ will be always zero, since we want
no component failures, but in other situations non-zero values may well
make sense.

   double alpha[] = { 0.5, 0.25, 0.1, 0.05, 0.01, 0.001, 0.0001, 0.00001 };

Much of the rest of the program is pretty-printing, the important part
is in the calculation of maximum number of permitted trials for each
value of alpha:

   for(unsigned i = 0; i < sizeof(alpha)/sizeof(alpha[0]); ++i)
   {
      // Confidence value:
      cout << fixed << setprecision(3) << setw(10) << right << 100 * (1-alpha[i]);
      // calculate trials:
      double t = binomial::find_maximum_number_of_trials(
                     successes, p, alpha[i]);
      t = floor(t);
      // Print Trials:
      cout << fixed << setprecision(5) << setw(15) << right << t << endl;
   }

Note that since we're
calculating the maximum number of trials permitted, we'll err on the safe
side and take the floor of the result.  Had we been calculating the
/minimum/ number of trials required to observe a certain number of /successes/
using `find_minimum_number_of_trials` we would have taken the ceiling instead.

We'll finish off by looking at some sample output, firstly for
a 1 in 1000 chance of component failure with each use:

[pre
'''________________________
Maximum Number of Trials
________________________

Success ratio                           =  0.001
Maximum Number of "successes" permitted =  0


____________________________
Confidence        Max Number
 Value (%)        Of Trials
____________________________
    50.000            692
    75.000            287
    90.000            105
    95.000             51
    99.000             10
    99.900              0
    99.990              0
    99.999              0'''
]

So 51 "uses" of the component would yield a 95% chance that no
component failures would be observed.

Compare that with a 1 in 1 million chance of component failure:

[pre'''
________________________
Maximum Number of Trials
________________________

Success ratio                           =  0.0000010
Maximum Number of "successes" permitted =  0


____________________________
Confidence        Max Number
 Value (%)        Of Trials
____________________________
    50.000         693146
    75.000         287681
    90.000         105360
    95.000          51293
    99.000          10050
    99.900           1000
    99.990            100
    99.999             10'''
]

In this case, even 1000 uses of the component would still yield a 
less than 1 in 1000 chance of observing a component failure 
(i.e. a 99.9% chance of no failure).

[endsect] [/section:binom_size_eg Estimating Sample Sizes for a Binomial Distribution.]

[endsect][/section:binom_eg Binomial Distribution]

[/ 
  Copyright 2006 John Maddock and Paul A. Bristow.
  Distributed under 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).
]