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anderson_darling_cdf

PURPOSE ^

%

SYNOPSIS ^

function y = anderson_darling_cdf(z,n)

DESCRIPTION ^

% -*- texinfo -*-
% @deftypefn {Function File} @var{p} = anderson_darling_cdf (@var{A}, @var{n})
%
% Return the CDF for the given Anderson-Darling coefficient @var{A}
% computed from @var{n} values sampled from a distribution. For a 
% vector of random variables @var{x} of length @var{n}, compute the CDF
% of the values from the distribution from which they are drawn.
% You can uses these values to compute @var{A} as follows:
%
% @example
% @var{A} = -@var{n} - sum( (2*i-1) .* (log(@var{x}) + log(1 - @var{x}(@var{n}:-1:1,:))) )/@var{n};
% @end example
% 
% From the value @var{A}, @code{anderson_darling_cdf} returns the probability
% that @var{A} could be returned from a set of samples.
%
% The algorithm given in [1] claims to be an approximation for the
% Anderson-Darling CDF accurate to 6 decimal points.
%
% Demonstrate using:
%
% @example
% n = 300; reps = 10000;
% z = randn(n, reps);
% x = sort ((1 + erf (z/sqrt (2)))/2);
% i = [1:n]' * ones (1, size (x, 2));
% A = -n - sum ((2*i-1) .* (log (x) + log (1 - x (n:-1:1, :))))/n;
% p = anderson_darling_cdf (A, n);
% hist (100 * p, [1:100] - 0.5);
% @end example
%
% You will see that the histogram is basically flat, which is to
% say that the probabilities returned by the Anderson-Darling CDF 
% are distributed uniformly.
%
% You can easily determine the extreme values of @var{p}:
%
% @example
% [junk, idx] = sort (p);
% @end example
%
% The histograms of various @var{p} aren't  very informative:
%
% @example
% histfit (z (:, idx (1)), linspace (-3, 3, 15));
% histfit (z (:, idx (end/2)), linspace (-3, 3, 15));
% histfit (z (:, idx (end)), linspace (-3, 3, 15));
% @end example
%
% More telling is the qqplot:
%
% @example
% qqplot (z (:, idx (1))); hold on; plot ([-3, 3], [-3, 3], ';;'); hold off;
% qqplot (z (:, idx (end/2))); hold on; plot ([-3, 3], [-3, 3], ';;'); hold off;
% qqplot (z (:, idx (end))); hold on; plot ([-3, 3], [-3, 3], ';;'); hold off;
% @end example
%
% Try a similarly analysis for @var{z} uniform:
%
% @example
% z = rand (n, reps); x = sort(z);
% @end example
%
% and for @var{z} exponential:
%
% @example
% z = rande (n, reps); x = sort (1 - exp (-z));
% @end example
%
% [1] Marsaglia, G; Marsaglia JCW; (2004) 'Evaluating the Anderson Darling
% distribution', Journal of Statistical Software, 9(2).
%
% @seealso{anderson_darling_test}
% @end deftypefn

CROSS-REFERENCE INFORMATION ^

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