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- // geometric_examples.cpp
- // Copyright Paul A. Bristow 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)
- // This file is written to be included from a Quickbook .qbk document.
- // It can still be compiled by the C++ compiler, and run.
- // Any output can also be added here as comment or included or pasted in elsewhere.
- // Caution: this file contains Quickbook markup as well as code
- // and comments: don't change any of the special comment markups!
- // Examples of using the geometric distribution.
- //[geometric_eg1_1
- /*`
- For this example, we will opt to #define two 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
- /*`
- [caution It is vital to #include distributions etc *after* the above #defines]
- After that we need some includes to provide easy access to the negative binomial distribution,
- and we need some std library iostream, of course.
- */
- #include <boost/math/distributions/geometric.hpp>
- // for geometric_distribution
- using ::boost::math::geometric_distribution; //
- using ::boost::math::geometric; // 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 <boost/math/distributions/negative_binomial.hpp>
- // for negative_binomial_distribution
- using boost::math::negative_binomial; // typedef provides default type is double.
- #include <boost/math/distributions/normal.hpp>
- // for negative_binomial_distribution
- using boost::math::normal; // typedef provides default type is double.
- #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;
- //] [geometric_eg1_1]
- int main()
- {
- cout <<"Geometric distribution example" << endl;
- cout << endl;
- cout.precision(4); // But only show a few for this example.
- try
- {
- //[geometric_eg1_2
- /*`
- It is always sensible to use try and catch blocks because defaults policies are to
- throw an exception if anything goes wrong.
- Simple try'n'catch blocks (see below) will ensure that you get a
- helpful error message instead of an abrupt (and silent) program abort.
- [h6 Throwing a dice]
- The Geometric distribution describes the probability (/p/) of a number of failures
- to get the first success in /k/ Bernoulli trials.
- (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).
- Suppose an 'fair' 6-face dice is thrown repeatedly:
- */
- double success_fraction = 1./6; // success_fraction (p) = 0.1666
- // (so failure_fraction is 1 - success_fraction = 5./6 = 1- 0.1666 = 0.8333)
- /*`If the dice is thrown repeatedly until the *first* time a /three/ appears.
- The probablility distribution of the number of times it is thrown *not* getting a /three/
- (/not-a-threes/ number of failures to get a /three/)
- is a geometric distribution with the success_fraction = 1/6 = 0.1666[recur].
- We therefore start by constructing a geometric distribution
- with the one parameter success_fraction, the probability of success.
- */
- geometric g6(success_fraction); // type double by default.
- /*`
- To confirm, we can echo the success_fraction parameter of the distribution.
- */
- cout << "success fraction of a six-sided dice is " << g6.success_fraction() << endl;
- /*`So the probability of getting a three at the first throw (zero failures) is
- */
- cout << pdf(g6, 0) << endl; // 0.1667
- cout << cdf(g6, 0) << endl; // 0.1667
- /*`Note that the cdf and pdf are identical because the is only one throw.
- If we want the probability of getting the first /three/ on the 2nd throw:
- */
- cout << pdf(g6, 1) << endl; // 0.1389
- /*`If we want the probability of getting the first /three/ on the 1st or 2nd throw
- (allowing one failure):
- */
- cout << "pdf(g6, 0) + pdf(g6, 1) = " << pdf(g6, 0) + pdf(g6, 1) << endl;
- /*`Or more conveniently, and more generally,
- we can use the Cumulative Distribution Function CDF.*/
- cout << "cdf(g6, 1) = " << cdf(g6, 1) << endl; // 0.3056
- /*`If we allow many more (12) throws, the probability of getting our /three/ gets very high:*/
- cout << "cdf(g6, 12) = " << cdf(g6, 12) << endl; // 0.9065 or 90% probability.
- /*`If we want to be much more confident, say 99%,
- we can estimate the number of throws to be this sure
- using the inverse or quantile.
- */
- cout << "quantile(g6, 0.99) = " << quantile(g6, 0.99) << endl; // 24.26
- /*`Note that the value returned is not an integer:
- if you want an integer result you should use either floor, round or ceil functions,
- or use the policies mechanism.
- See __understand_dis_quant.
- The geometric distribution is related to the negative binomial
- __spaces `negative_binomial_distribution(RealType r, RealType p);` with parameter /r/ = 1.
- So we could get the same result using the negative binomial,
- but using the geometric the results will be faster, and may be more accurate.
- */
- negative_binomial nb(1, success_fraction);
- cout << pdf(nb, 1) << endl; // 0.1389
- cout << cdf(nb, 1) << endl; // 0.3056
- /*`We could also the complement to express the required probability
- as 1 - 0.99 = 0.01 (and get the same result):
- */
- cout << "quantile(complement(g6, 1 - p)) " << quantile(complement(g6, 0.01)) << endl; // 24.26
- /*`
- Note too that Boost.Math geometric distribution is implemented as a continuous function.
- Unlike other implementations (for example R) it *uses* the number of failures as a *real* parameter,
- not as an integer. If you want this integer behaviour, you may need to enforce this by
- rounding the parameter you pass, probably rounding down, to the nearest integer.
- For example, R returns the success fraction probability for all values of failures
- from 0 to 0.999999 thus:
- [pre
- __spaces R> formatC(pgeom(0.0001,0.5, FALSE), digits=17) " 0.5"
- ] [/pre]
- So in Boost.Math the equivalent is
- */
- geometric g05(0.5); // Probability of success = 0.5 or 50%
- // Output all potentially significant digits for the type, here double.
- #ifdef BOOST_NO_CXX11_NUMERIC_LIMITS
- int max_digits10 = 2 + (boost::math::policies::digits<double, boost::math::policies::policy<> >() * 30103UL) / 100000UL;
- cout << "BOOST_NO_CXX11_NUMERIC_LIMITS is defined" << endl;
- #else
- int max_digits10 = std::numeric_limits<double>::max_digits10;
- #endif
- cout << "Show all potentially significant decimal digits std::numeric_limits<double>::max_digits10 = "
- << max_digits10 << endl;
- cout.precision(max_digits10); //
- cout << cdf(g05, 0.0001) << endl; // returns 0.5000346561579232, not exact 0.5.
- /*`To get the R discrete behaviour, you simply need to round with,
- for example, the `floor` function.
- */
- cout << cdf(g05, floor(0.0001)) << endl; // returns exactly 0.5
- /*`
- [pre
- `> formatC(pgeom(0.9999999,0.5, FALSE), digits=17) [1] " 0.25"`
- `> formatC(pgeom(1.999999,0.5, FALSE), digits=17)[1] " 0.25" k = 1`
- `> formatC(pgeom(1.9999999,0.5, FALSE), digits=17)[1] "0.12500000000000003" k = 2`
- ] [/pre]
- shows that R makes an arbitrary round-up decision at about 1e7 from the next integer above.
- This may be convenient in practice, and could be replicated in C++ if desired.
- [h6 Surveying customers to find one with a faulty product]
- A company knows from warranty claims that 2% of their products will be faulty,
- so the 'success_fraction' of finding a fault is 0.02.
- It wants to interview a purchaser of faulty products to assess their 'user experience'.
- To estimate how many customers they will probably need to contact
- in order to find one who has suffered from the fault,
- we first construct a geometric distribution with probability 0.02,
- and then chose a confidence, say 80%, 95%, or 99% to finding a customer with a fault.
- Finally, we probably want to round up the result to the integer above using the `ceil` function.
- (We could also use a policy, but that is hardly worthwhile for this simple application.)
- (This also assumes that each customer only buys one product:
- if customers bought more than one item,
- the probability of finding a customer with a fault obviously improves.)
- */
- cout.precision(5);
- geometric g(0.02); // On average, 2 in 100 products are faulty.
- double c = 0.95; // 95% confidence.
- cout << " quantile(g, " << c << ") = " << quantile(g, c) << endl;
- cout << "To be " << c * 100
- << "% confident of finding we customer with a fault, need to survey "
- << ceil(quantile(g, c)) << " customers." << endl; // 148
- c = 0.99; // Very confident.
- cout << "To be " << c * 100
- << "% confident of finding we customer with a fault, need to survey "
- << ceil(quantile(g, c)) << " customers." << endl; // 227
- c = 0.80; // Only reasonably confident.
- cout << "To be " << c * 100
- << "% confident of finding we customer with a fault, need to survey "
- << ceil(quantile(g, c)) << " customers." << endl; // 79
- /*`[h6 Basket Ball Shooters]
- According to Wikipedia, average pro basket ball players get
- [@http://en.wikipedia.org/wiki/Free_throw free throws]
- in the baskets 70 to 80 % of the time,
- but some get as high as 95%, and others as low as 50%.
- Suppose we want to compare the probabilities
- of failing to get a score only on the first or on the fifth shot?
- To start we will consider the average shooter, say 75%.
- So we construct a geometric distribution
- with success_fraction parameter 75/100 = 0.75.
- */
- cout.precision(2);
- geometric gav(0.75); // Shooter averages 7.5 out of 10 in the basket.
- /*`What is probability of getting 1st try in the basket, that is with no failures? */
- cout << "Probability of score on 1st try = " << pdf(gav, 0) << endl; // 0.75
- /*`This is, of course, the success_fraction probability 75%.
- What is the probability that the shooter only scores on the fifth shot?
- So there are 5-1 = 4 failures before the first success.*/
- cout << "Probability of score on 5th try = " << pdf(gav, 4) << endl; // 0.0029
- /*`Now compare this with the poor and the best players success fraction.
- We need to constructing new distributions with the different success fractions,
- and then get the corresponding probability density functions values:
- */
- geometric gbest(0.95);
- cout << "Probability of score on 5th try = " << pdf(gbest, 4) << endl; // 5.9e-6
- geometric gmediocre(0.50);
- cout << "Probability of score on 5th try = " << pdf(gmediocre, 4) << endl; // 0.031
- /*`So we can see the very much smaller chance (0.000006) of 4 failures by the best shooters,
- compared to the 0.03 of the mediocre.*/
- /*`[h6 Estimating failures]
- Of course one man's failure is an other man's success.
- So a fault can be defined as a 'success'.
- If a fault occurs once after 100 flights, then one might naively say
- that the risk of fault is obviously 1 in 100 = 1/100, a probability of 0.01.
- This is the best estimate we can make, but while it is the truth,
- it is not the whole truth,
- for it hides the big uncertainty when estimating from a single event.
- "One swallow doesn't make a summer."
- To show the magnitude of the uncertainty, the geometric
- (or the negative binomial) distribution can be used.
- If we chose the popular 95% confidence in the limits, corresponding to an alpha of 0.05,
- because we are calculating a two-sided interval, we must divide alpha by two.
- */
- double alpha = 0.05;
- double k = 100; // So frequency of occurrence is 1/100.
- cout << "Probability is failure is " << 1/k << endl;
- double t = geometric::find_lower_bound_on_p(k, alpha/2);
- cout << "geometric::find_lower_bound_on_p(" << int(k) << ", " << alpha/2 << ") = "
- << t << endl; // 0.00025
- t = geometric::find_upper_bound_on_p(k, alpha/2);
- cout << "geometric::find_upper_bound_on_p(" << int(k) << ", " << alpha/2 << ") = "
- << t << endl; // 0.037
- /*`So while we estimate the probability is 0.01, it might lie between 0.0003 and 0.04.
- Even if we relax our confidence to alpha = 90%, the bounds only contract to 0.0005 and 0.03.
- And if we require a high confidence, they widen to 0.00005 to 0.05.
- */
- alpha = 0.1; // 90% confidence.
- t = geometric::find_lower_bound_on_p(k, alpha/2);
- cout << "geometric::find_lower_bound_on_p(" << int(k) << ", " << alpha/2 << ") = "
- << t << endl; // 0.0005
- t = geometric::find_upper_bound_on_p(k, alpha/2);
- cout << "geometric::find_upper_bound_on_p(" << int(k) << ", " << alpha/2 << ") = "
- << t << endl; // 0.03
- alpha = 0.01; // 99% confidence.
- t = geometric::find_lower_bound_on_p(k, alpha/2);
- cout << "geometric::find_lower_bound_on_p(" << int(k) << ", " << alpha/2 << ") = "
- << t << endl; // 5e-005
- t = geometric::find_upper_bound_on_p(k, alpha/2);
- cout << "geometric::find_upper_bound_on_p(" << int(k) << ", " << alpha/2 << ") = "
- << t << endl; // 0.052
- /*`In real life, there will usually be more than one event (fault or success),
- when the negative binomial, which has the neccessary extra parameter, will be needed.
- */
- /*`As noted above, using a catch block is always a good idea,
- even if you hope not 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(g, -1)`` for example,
- we would get an unhelpful abort, but with a catch:
- [pre
- Message from thrown exception was:
- Error in function boost::math::pdf(const exponential_distribution<double>&, double):
- Number of failures argument is -1, but must be >= 0 !
- ] [/pre]
- */
- //] [/ geometric_eg1_2]
- }
- return 0;
- } // int main()
- /*
- Output is:
- Geometric distribution example
- success fraction of a six-sided dice is 0.1667
- 0.1667
- 0.1667
- 0.1389
- pdf(g6, 0) + pdf(g6, 1) = 0.3056
- cdf(g6, 1) = 0.3056
- cdf(g6, 12) = 0.9065
- quantile(g6, 0.99) = 24.26
- 0.1389
- 0.3056
- quantile(complement(g6, 1 - p)) 24.26
- 0.5000346561579232
- 0.5
- quantile(g, 0.95) = 147.28
- To be 95% confident of finding we customer with a fault, need to survey 148 customers.
- To be 99% confident of finding we customer with a fault, need to survey 227 customers.
- To be 80% confident of finding we customer with a fault, need to survey 79 customers.
- Probability of score on 1st try = 0.75
- Probability of score on 5th try = 0.0029
- Probability of score on 5th try = 5.9e-006
- Probability of score on 5th try = 0.031
- Probability is failure is 0.01
- geometric::find_lower_bound_on_p(100, 0.025) = 0.00025
- geometric::find_upper_bound_on_p(100, 0.025) = 0.037
- geometric::find_lower_bound_on_p(100, 0.05) = 0.00051
- geometric::find_upper_bound_on_p(100, 0.05) = 0.03
- geometric::find_lower_bound_on_p(100, 0.005) = 5e-005
- geometric::find_upper_bound_on_p(100, 0.005) = 0.052
- */
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