Editing Multimodal distribution (section)

===Tests for unimodality===

*When the components of the mixture have equal variances the mixture is unimodal [[if and only if]]<ref name=Holzmann2008>{{cite journal | last1 = Holzmann | first1 = Hajo | last2 = Vollmer | first2 = Sebastian | year = 2008 | title = A likelihood ratio test for bimodality in two-component mixtures with application to regional income distribution in the EU | journal = AStA Advances in Statistical Analysis | volume = 2 | issue = 1| pages = 57–69 | doi=10.1007/s10182-008-0057-2 | s2cid = 14470055 | url = http://resolver.sub.uni-goettingen.de/purl?gs-1/8526 }}</ref> <math display="block"> d \le 1 </math> or <math display="block">\left\vert \log( 1 - p ) - \log( p ) \right\vert \ge 2 \log( d - \sqrt{ d^2 - 1 } ) + 2d \sqrt{ d^2 - 1 } ,</math> where ''p'' is the mixing parameter and <math display="block"> d = \frac{ \left\vert \mu_1 - \mu_2 \right\vert }{ 2 \sigma }, </math> and where ''μ''<sub>1</sub> and ''μ''<sub>2</sub> are the means of the two normal distributions and ''σ'' is their standard deviation.
*The following test for the case ''p'' = 1/2 was described by Schilling ''et al''.<ref name=Schilling2002/> Let <math display="block"> r = \frac{ \sigma_1^2 }{ \sigma_2^2 } .</math> The separation factor (''S'') is <math display="block"> S = \frac{ \sqrt{ -2 + 3r + 3r^2 - 2r^3 + 2 \left( 1 - r + r^2 \right)^{ 1.5 } } }{ \sqrt{ r } \left( 1 + \sqrt{ r } \right) } .</math> If the variances are equal then ''S'' = 1. The mixture density is unimodal if and only if <math display="block"> | \mu_1 - \mu_2 | < S | \sigma_1 + \sigma_2 | .</math>
*A sufficient condition for unimodality is<ref name=Behboodian1970>{{cite journal | last1 = Behboodian | first1 = J | year = 1970 | title = On the modes of a mixture of two normal distributions | journal = Technometrics | volume = 12 | issue = 1| pages = 131–139 | doi=10.2307/1267357| jstor = 1267357 }}</ref><math display="block">|\mu_1-\mu_2| \le2\min (\sigma_1,\sigma_2).</math>
*If the two normal distributions have equal standard deviations <math>\sigma,</math> a sufficient condition for unimodality is<ref name=Behboodian1970/><math display="block">|\mu _1-\mu_2|\le 2\sigma \sqrt{1+\frac{ \left|\ln p-\ln (1-p)\right|}{2}}.</math>