       Re: Mean of skew-normal distribution

• To: mathgroup at smc.vnet.net
• Subject: [mg61967] Re: Mean of skew-normal distribution
• 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*\[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)]

For solving problems like this you might want to consider getting mathStatica <http://www.mathstatica.com/>

In:=<<mathstatica`
In:=
f = (1/(E^(x^2/2)*Sqrt[2*Pi]))*(1 + Erf[(\[Lambda]*x)/Sqrt]);
domain[f] = {x, -Infinity, Infinity} &&
{-Infinity < \[Lambda] < Infinity};

In:=Expect[x, f]

Out=
(Sqrt*\[Lambda])/Sqrt[Pi*\[Lambda]^2 + Pi]

In:=\$Version
Out="5.2 for Mac OS X (June 20, 2005)"
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```

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