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1/*2 * Copyright Nick Thompson, 20193 * Use, modification and distribution are subject to the4 * Boost Software License, Version 1.0. (See accompanying file5 * LICENSE_1_0.txt or copy at http://www.boost.org/LICENSE_1_0.txt)6 */7 8#ifndef BOOST_MATH_STATISTICS_ANDERSON_DARLING_HPP9#define BOOST_MATH_STATISTICS_ANDERSON_DARLING_HPP10 11#include <cmath>12#include <algorithm>13#include <boost/math/statistics/univariate_statistics.hpp>14#include <boost/math/special_functions/erf.hpp>15 16namespace boost { namespace math { namespace statistics {17 18template<class RandomAccessContainer>19auto anderson_darling_normality_statistic(RandomAccessContainer const & v,20                                          typename RandomAccessContainer::value_type mu = std::numeric_limits<typename RandomAccessContainer::value_type>::quiet_NaN(),21                                          typename RandomAccessContainer::value_type sd = std::numeric_limits<typename RandomAccessContainer::value_type>::quiet_NaN())22{23    using Real = typename RandomAccessContainer::value_type;24    using std::log;25    using std::sqrt;26    using boost::math::erfc;27 28    if (std::isnan(mu)) {29        mu = boost::math::statistics::mean(v);30    }31    if (std::isnan(sd)) {32        sd = sqrt(boost::math::statistics::sample_variance(v));33    }34 35    typedef boost::math::policies::policy<36          boost::math::policies::promote_float<false>,37          boost::math::policies::promote_double<false> >38          no_promote_policy;39 40    // This is where Knuth's literate programming could really come in handy!41    // I need some LaTeX. The idea is that before any observation, the ecdf is identically zero.42    // So we need to compute:43    // \int_{-\infty}^{v_0} \frac{F(x)F'(x)}{1- F(x)} \, \mathrm{d}x, where F(x) := \frac{1}{2}[1+\erf(\frac{x-\mu}{\sigma \sqrt{2}})]44    // Astonishingly, there is an analytic evaluation to this integral, as you can validate with the following Mathematica command:45    // Integrate[(1/2 (1 + Erf[(x - mu)/Sqrt[2*sigma^2]])*Exp[-(x - mu)^2/(2*sigma^2)]*1/Sqrt[2*\[Pi]*sigma^2])/(1 - 1/2 (1 + Erf[(x - mu)/Sqrt[2*sigma^2]])),46    // {x, -Infinity, x0}, Assumptions -> {x0 \[Element] Reals && mu \[Element] Reals && sigma > 0}]47    // This gives (for s = x-mu/sqrt(2sigma^2))48    // -1/2 + erf(s) + log(2/(1+erf(s)))49 50 51    Real inv_var_scale = 1/(sd*sqrt(Real(2)));52    Real s0 = (v[0] - mu)*inv_var_scale;53    Real erfcs0 = erfc(s0, no_promote_policy());54    // Note that if erfcs0 == 0, then left_tail = inf (numerically), and hence the entire integral is numerically infinite:55    if (erfcs0 <= 0) {56        return std::numeric_limits<Real>::infinity();57    }58 59    // Note that we're going to add erfcs0/2 when we compute the integral over [x_0, x_1], so drop it here:60    Real left_tail = -1 + log(Real(2));61 62 63    // For the right tail, the ecdf is identically 1.64    // Hence we need the integral:65    // \int_{v_{n-1}}^{\infty} \frac{(1-F(x))F'(x)}{F(x)} \, \mathrm{d}x66    // This also has an analytic evaluation! It can be found via the following Mathematica command:67    // Integrate[(E^(-(z^2/2)) *(1 - 1/2 (1 + Erf[z/Sqrt[2]])))/(Sqrt[2 \[Pi]] (1/2 (1 + Erf[z/Sqrt[2]]))),68    // {z, zn, \[Infinity]}, Assumptions -> {zn \[Element] Reals && mu \[Element] Reals}]69    // This gives (for sf = xf-mu/sqrt(2sigma^2))70    // -1/2 + erf(sf)/2 + 2log(2/(1+erf(sf)))71 72    Real sf = (v[v.size()-1] - mu)*inv_var_scale;73    //Real erfcsf = erfc<Real>(sf, no_promote_policy());74    // This is the actual value of the tail integral. However, the -erfcsf/2 cancels from the integral over [v_{n-2}, v_{n-1}]:75    //Real right_tail = -erfcsf/2 + log(Real(2)) - log(2-erfcsf);76 77    // Use erfc(-x) = 2 - erfc(x)78    Real erfcmsf = erfc<Real>(-sf, no_promote_policy());79    // Again if this is precisely zero then the integral is numerically infinite:80    if (erfcmsf == 0) {81        return std::numeric_limits<Real>::infinity();82    }83    Real right_tail = log(2/erfcmsf);84 85    // Now we need each integral:86    // \int_{v_i}^{v_{i+1}} \frac{(i+1/n - F(x))^2F'(x)}{F(x)(1-F(x))}  \, \mathrm{d}x87    // Again we get an analytical evaluation via the following Mathematica command:88    // Integrate[((E^(-(z^2/2))/Sqrt[2 \[Pi]])*(k1 - F[z])^2)/(F[z]*(1 - F[z])),89    // {z, z1, z2}, Assumptions -> {z1 \[Element] Reals && z2 \[Element] Reals &&k1 \[Element] Reals}] // FullSimplify90 91    Real integrals = 0;92    int64_t N = v.size();93    for (int64_t i = 0; i < N - 1; ++i) {94        if (v[i] > v[i+1]) {95            throw std::domain_error("Input data must be sorted in increasing order v[0] <= v[1] <= . . .  <= v[n-1]");96        }97 98        Real k = (i+1)/Real(N);99        Real s1 = (v[i+1]-mu)*inv_var_scale;100        Real erfcs1 = erfc<Real>(s1, no_promote_policy());101        Real term = k*(k*log(erfcs0*(-2 + erfcs1)/(erfcs1*(-2 + erfcs0))) + 2*log(erfcs1/erfcs0));102 103        integrals += term;104        s0 = s1;105        erfcs0 = erfcs1;106    }107    integrals -= log(erfcs0);108    return v.size()*(left_tail + right_tail + integrals);109}110 111}}}112#endif113