Editing Multimodal distribution (section)

==Mixture of two normal distributions==

It is not uncommon to encounter situations where an investigator believes that the data comes from a mixture of two normal distributions. Because of this, this mixture has been studied in some detail.<ref name=Robertson1969>{{cite journal | last1 = Robertson | first1 = CA | last2 = Fryer | first2 = JG | year = 1969 | title = Some descriptive properties of normal mixtures | journal = Skandinavisk Aktuarietidskrift | volume = 69 | issue = 3–4| pages = 137–146 |doi=10.1080/03461238.1969.10404590}}</ref>

A mixture of two normal distributions has five parameters to estimate: the two means, the two variances and the mixing parameter. A mixture of two [[normal distribution]]s with equal [[standard deviation]]s is bimodal only if their means differ by at least twice the common standard deviation.<ref name="Schilling2002"/> Estimates of the parameters is simplified if the variances can be assumed to be equal (the [[homoscedastic]] case).

If the means of the two normal distributions are equal, then the combined distribution is unimodal. Conditions for [[unimodality]] of the combined distribution were derived by Eisenberger.<ref name=Eisenberger1964>{{cite journal | last1 = Eisenberger | first1 = I | year = 1964 | title = Genesis of bimodal distributions | journal = Technometrics | volume = 6 | issue = 4| pages = 357–363 | doi=10.1080/00401706.1964.10490199}}</ref> Necessary and sufficient conditions for a mixture of normal distributions to be bimodal have been identified by Ray and Lindsay.<ref name=Ray2005>{{cite journal | last1 = Ray | first1 = S | last2 = Lindsay | first2 = BG | year = 2005 | title = The topography of multivariate normal mixtures | journal = Annals of Statistics | volume = 33 | issue = 5| pages = 2042–2065 | doi=10.1214/009053605000000417| arxiv = math/0602238 | s2cid = 36234163 }}</ref>

A mixture of two approximately equal mass normal distributions has a negative kurtosis since the two modes on either side of the center of mass effectively reduces the tails of the distribution.

A mixture of two normal distributions with highly unequal mass has a positive kurtosis since the smaller distribution lengthens the tail of the more dominant normal distribution.

Mixtures of other distributions require additional parameters to be estimated.

===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>