Return the CDF for the given Anderson-Darling coefficient A computed from n values sampled from a distribution. For a vector of random variables x of length n, compute the CDF of the values from the distribution from which they are drawn. You can uses these values to compute A as follows:
A = -n - sum( (2*i-1) .* (log(x) + log(1 - x(n:-1:1,:))) )/n;
From the value A, anderson_darling_cdf returns the probability
that 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:
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);
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 p:
[junk, idx] = sort (p);
The histograms of various p aren’t very informative:
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));
More telling is the qqplot:
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;
Try a similarly analysis for z uniform:
z = rand (n, reps); x = sort(z);
and for z exponential:
z = rande (n, reps); x = sort (1 - exp (-z));
[1] Marsaglia, G; Marsaglia JCW; (2004) "Evaluating the Anderson Darling distribution", Journal of Statistical Software, 9(2).
See also: anderson_darling_test.
Package: statistics