       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*\[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*\[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:= InputForm[Integrate[(x+mu)*(1+Erf[(lambda*x)/(Sqrt*sigma)])/
(E^(x^2/(2*sigma^2))*(Sqrt[2*Pi]*sigma)),
{x,-Infinity,Infinity},
Assumptions->{Element[mu,Reals],sigma>0,lambda>0}]]

Out//InputForm= mu + (lambda*Sqrt[2/Pi]*sigma)/Sqrt[1 + lambda^2]

Daniel Lichtblau
Wolfram Research

```

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