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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*: Sun, 6 Nov 2005 05:11:55 -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)"
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