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 /*
  * Copyright Nick Thompson, 2019
  * 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)
  */
 
 #ifndef BOOST_MATH_STATISTICS_ANDERSON_DARLING_HPP
 #define BOOST_MATH_STATISTICS_ANDERSON_DARLING_HPP
 
 #include <cmath>
 #include <algorithm>
 #include <boost/math/statistics/univariate_statistics.hpp>
 #include <boost/math/special_functions/erf.hpp>
 
 namespace boost { namespace math { namespace statistics {
 
 template<class RandomAccessContainer>
 auto anderson_darling_normality_statistic(RandomAccessContainer const & v,
                                           typename RandomAccessContainer::value_type mu = std::numeric_limits<typename RandomAccessContainer::value_type>::quiet_NaN(),
                                           typename RandomAccessContainer::value_type sd = std::numeric_limits<typename RandomAccessContainer::value_type>::quiet_NaN())
 {
     using Real = typename RandomAccessContainer::value_type;
     using std::log;
     using std::sqrt;
     using boost::math::erfc;
 
     if (std::isnan(mu)) {
         mu = boost::math::statistics::mean(v);
     }
     if (std::isnan(sd)) {
         sd = sqrt(boost::math::statistics::sample_variance(v));
     }
 
     typedef boost::math::policies::policy<
           boost::math::policies::promote_float<false>,
           boost::math::policies::promote_double<false> >
           no_promote_policy;
 
     // This is where Knuth's literate programming could really come in handy!
     // I need some LaTeX. The idea is that before any observation, the ecdf is identically zero.
     // So we need to compute:
     // \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}})]
     // Astonishingly, there is an analytic evaluation to this integral, as you can validate with the following Mathematica command:
     // 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]])),
     // {x, -Infinity, x0}, Assumptions -> {x0 \[Element] Reals && mu \[Element] Reals && sigma > 0}]
     // This gives (for s = x-mu/sqrt(2sigma^2))
     // -1/2 + erf(s) + log(2/(1+erf(s)))
 
 
     Real inv_var_scale = 1/(sd*sqrt(Real(2)));
     Real s0 = (v[0] - mu)*inv_var_scale;
     Real erfcs0 = erfc(s0, no_promote_policy());
     // Note that if erfcs0 == 0, then left_tail = inf (numerically), and hence the entire integral is numerically infinite:
     if (erfcs0 <= 0) {
         return std::numeric_limits<Real>::infinity();
     }
 
     // Note that we're going to add erfcs0/2 when we compute the integral over [x_0, x_1], so drop it here:
     Real left_tail = -1 + log(Real(2));
 
 
     // For the right tail, the ecdf is identically 1.
     // Hence we need the integral:
     // \int_{v_{n-1}}^{\infty} \frac{(1-F(x))F'(x)}{F(x)} \, \mathrm{d}x
     // This also has an analytic evaluation! It can be found via the following Mathematica command:
     // Integrate[(E^(-(z^2/2)) *(1 - 1/2 (1 + Erf[z/Sqrt[2]])))/(Sqrt[2 \[Pi]] (1/2 (1 + Erf[z/Sqrt[2]]))),
     // {z, zn, \[Infinity]}, Assumptions -> {zn \[Element] Reals && mu \[Element] Reals}]
     // This gives (for sf = xf-mu/sqrt(2sigma^2))
     // -1/2 + erf(sf)/2 + 2log(2/(1+erf(sf)))
 
     Real sf = (v[v.size()-1] - mu)*inv_var_scale;
     //Real erfcsf = erfc<Real>(sf, no_promote_policy());
     // This is the actual value of the tail integral. However, the -erfcsf/2 cancels from the integral over [v_{n-2}, v_{n-1}]:
     //Real right_tail = -erfcsf/2 + log(Real(2)) - log(2-erfcsf);
 
     // Use erfc(-x) = 2 - erfc(x)
     Real erfcmsf = erfc<Real>(-sf, no_promote_policy());
     // Again if this is precisely zero then the integral is numerically infinite:
     if (erfcmsf == 0) {
         return std::numeric_limits<Real>::infinity();
     }
     Real right_tail = log(2/erfcmsf);
 
     // Now we need each integral:
     // \int_{v_i}^{v_{i+1}} \frac{(i+1/n - F(x))^2F'(x)}{F(x)(1-F(x))}  \, \mathrm{d}x
     // Again we get an analytical evaluation via the following Mathematica command:
     // Integrate[((E^(-(z^2/2))/Sqrt[2 \[Pi]])*(k1 - F[z])^2)/(F[z]*(1 - F[z])),
     // {z, z1, z2}, Assumptions -> {z1 \[Element] Reals && z2 \[Element] Reals &&k1 \[Element] Reals}] // FullSimplify
 
     Real integrals = 0;
     int64_t N = v.size();
     for (int64_t i = 0; i < N - 1; ++i) {
         if (v[i] > v[i+1]) {
             throw std::domain_error("Input data must be sorted in increasing order v[0] <= v[1] <= . . .  <= v[n-1]");
         }
 
         Real k = (i+1)/Real(N);
         Real s1 = (v[i+1]-mu)*inv_var_scale;
         Real erfcs1 = erfc<Real>(s1, no_promote_policy());
         Real term = k*(k*log(erfcs0*(-2 + erfcs1)/(erfcs1*(-2 + erfcs0))) + 2*log(erfcs1/erfcs0));
 
         integrals += term;
         s0 = s1;
         erfcs0 = erfcs1;
     }
     integrals -= log(erfcs0);
     return v.size()*(left_tail + right_tail + integrals);
 }
 
 }}}
 #endif

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