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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 #include #include #include namespace boost { namespace math { namespace statistics { template auto anderson_darling_normality_statistic(RandomAccessContainer const & v, typename RandomAccessContainer::value_type mu = std::numeric_limits::quiet_NaN(), typename RandomAccessContainer::value_type sd = std::numeric_limits::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, boost::math::policies::promote_double > 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::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(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(-sf, no_promote_policy()); // Again if this is precisely zero then the integral is numerically infinite: if (erfcmsf == 0) { return std::numeric_limits::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(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