I am trying to generate exponentially distributed random number with mean equal to 1. I know how to get random number for normal distribution with mean and standard deviation. We can get it by normal(mean, standard_deviation), but I don't know how to get random number for exponential distribution.
Can anyone help me with this?
With C++11 the standard actually guarantees that there is a RNG following the requirements of exponential-distribution available in the STL, and fittingly the object-type has a very descriptive name.
The mean in an exponentially distributed random generator is calculated by the formula E[X] = 1 / lambda1.
std::exponential_distribution has a constructor taking lambda as an argument, so we can easily create an object following your rules by calculating the value of lambda and passing this to our generator.
std::exponential_distribution rng (1/1); // lambda = 1 / E[X] Footnotes
1. according to en.wikipedia.org - Exponential distribution > Mean, variance, moments and median
#include <iomanip> #include <random> #include <map> #include <iostream> int main (int argc, char *argv[]) { double const exp_dist_mean = 1; double const exp_dist_lambda = 1 / exp_dist_mean; std::random_device rd; std::exponential_distribution<> rng (exp_dist_lambda); std::mt19937 rnd_gen (rd ()); /* ... */ std::map<int, int> result_set; for (int i =0; i < 100000; ++i) ++result_set[rng (rnd_gen) * 4]; for (auto& v : result_set) { std::cout << std::setprecision (2) << std::fixed; std::cout << v.first/4.f << " - " << (v.first+1)/4.f << " -> "; std::cout << std::string (v.second/400, '.') << std::endl; if (v.second/400 == 0) break; } }
0.00 - 0.25 -> ........................................................ 0.25 - 0.50 -> ........................................... 0.50 - 0.75 -> ................................. 0.75 - 1.00 -> ......................... 1.00 - 1.25 -> .................... 1.25 - 1.50 -> ............... 1.50 - 1.75 -> ............ 1.75 - 2.00 -> ......... 2.00 - 2.25 -> ....... 2.25 - 2.50 -> ..... 2.50 - 2.75 -> .... 2.75 - 3.00 -> ... 3.00 - 3.25 -> .. 3.25 - 3.50 -> .. 3.50 - 3.75 -> . 3.75 - 4.00 -> . 4.00 - 4.25 -> . 4.25 - 4.50 -> Generating an exponential distribution random variable can be done by:
-ln(U)/lambda (where U~Uniform(0,1)). More information can be found in this wikipedia article
In exponential distribution: lamda = 1/mean, so it gets you:
myVar = -ln(U) * mean (where U~Uniform(0,1)).