fix: correct three distribution bugs and add CITATION.cff#25
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scipy.stats.triang expects its first argument (c) to be the normalized peak position in [0, 1], not the raw peak value. The raw peak c must be mapped to (c - a) / (b - a) before passing to triang. Closes #4
The old implementation used scipy.stats.exponweib (exponentiated Weibull), not the standard two-parameter Weibull distribution that the docstring and table describe. Switch to weibull_min(k, loc=0, scale=lamda) which matches the documented Weibull(lamda, k) parameterization. Closes #5
The previous call used loc=mu which shifts the distribution support in the original scale, not the log scale. The standard parameterization treats mu as E[ln(X)] (log-mean), so the correct scipy call is lognorm(sigma, loc=0, scale=exp(mu)) — the distribution's median is then exp(mu) as expected. Closes #3
GitHub surfaces CITATION.cff files automatically in the repository sidebar, making it easy for users to cite the software in academic work. Closes #8
Co-Authored-By: Claude Sonnet 4.6 <noreply@anthropic.com>
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Summary
Fixes three long-standing bugs in the distribution constructors and adds a citation file.
Tri()triangular distribution —scipy.stats.triangexpects its first argument to be the normalized peak(c - a) / (b - a)in[0, 1], not the rawcvalue. Closes Bug in Tri() constructor for triangular distribution #4Weib()Weibull distribution — replacedexponweib(exponentiated Weibull) withweibull_min(k, loc=0, scale=lamda)which matches the documented two-parameter Weibull. Closes Weib distribution #5LogN()log-normal parameterization —loc=muwas wrong (shifts support in original scale); changed toloc=0, scale=exp(mu)somucorrectly means E[ln(X)]. Closes Add shape param for LogN distribution #3CITATION.cff— GitHub surfaces this in the sidebar for easy software citation. Closes Add CITATION.cff file #8Test plan
Tri(1, 5, 3).mean≈ 3.0 (peak at midpoint → symmetric → mean at midpoint)Weib(4, 6).mean≈scipy.stats.weibull_min.mean(c=6, scale=4)≈ 3.78LogN(0, 1).mean≈scipy.stats.lognorm.mean(1)≈ 1.649 (median = exp(0) = 1)🤖 Generated with Claude Code