Re: Mean of skew-normal distribution

*To*: mathgroup at smc.vnet.net*Subject*: [mg61905] Re: [mg61877] Mean of skew-normal distribution*From*: Daniel Lichtblau <danl at wolfram.com>*Date*: Fri, 4 Nov 2005 05:11:25 -0500 (EST)*References*: <200511030958.EAA26531@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Valeri Astanoff wrote: > Dear group, > > I want to prove with help of mathematica [5.1] that this integral : > > > Integrate[x*(1+Erf[(\[Lambda]*(x-\[Mu]))/ > (Sqrt[2]*\[Sigma])])/(E^((x-\[Mu])^2/(2*\[Sigma]^2))* > (Sqrt[2*Pi]*\[Sigma])),{x,-Infinity,Infinity}] > > - which is the mean of a so-called skew-normal distribution - > is equal to : > > \[Mu] + (Sqrt[2]*\[Lambda]*\[Sigma])/Sqrt[Pi(1 + \[Lambda]^2)] > > > Series expanding with lambda near 0 shows > that it is true at any order, but of course > that's not a proof, and I shall be grateful > to the good Samaritan that will help me. > > > Valeri Astanoff Not exactly a proff, but... translate x-->x+mu to make the Erf and exponential simpler, and Integrate gives a result, provided some assumptions are passed to it. In[1]:= InputForm[Integrate[(x+mu)*(1+Erf[(lambda*x)/(Sqrt[2]*sigma)])/ (E^(x^2/(2*sigma^2))*(Sqrt[2*Pi]*sigma)), {x,-Infinity,Infinity}, Assumptions->{Element[mu,Reals],sigma>0,lambda>0}]] Out[1]//InputForm= mu + (lambda*Sqrt[2/Pi]*sigma)/Sqrt[1 + lambda^2] Daniel Lichtblau Wolfram Research

**References**:**Mean of skew-normal distribution***From:*"Valeri Astanoff" <astanoff@yahoo.fr>