Re: Mean of skew-normal distribution

*To*: mathgroup at smc.vnet.net*Subject*: [mg61967] Re: Mean of skew-normal distribution*From*: Bill Rowe <readnewsciv at earthlink.net>*Date*: Sat, 5 Nov 2005 22:41:52 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Valeri Astanoff wrote: >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)] For solving problems like this you might want to consider getting mathStatica <http://www.mathstatica.com/> In[1]:=<<mathstatica` In[2]:= f = (1/(E^(x^2/2)*Sqrt[2*Pi]))*(1 + Erf[(\[Lambda]*x)/Sqrt[2]]); domain[f] = {x, -Infinity, Infinity} && {-Infinity < \[Lambda] < Infinity}; In[3]:=Expect[x, f] Out[3]= (Sqrt[2]*\[Lambda])/Sqrt[Pi*\[Lambda]^2 + Pi] In[4]:=$Version Out[4]="5.2 for Mac OS X (June 20, 2005)" -- To reply via email subtract one hundred and four