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Re: Mean of skew-normal distribution


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


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