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Re: Mean of skew-normal distribution
*To*: mathgroup at smc.vnet.net
*Subject*: [mg61949] Re: Mean of skew-normal distribution
*From*: "Ray Koopman" <koopman at sfu.ca>
*Date*: Sat, 5 Nov 2005 01:52:56 -0500 (EST)
*References*: <dkcnof$q74$1@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[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
I've run into similar equations (in the context of sampling with
selection), and I too have been unable to get Mathematica to do
the integrals. For samples from a standard normal distribution,
whose pdf is f[z_] = Exp[-z^2/2]/Sqrt[2Pi]
and cdf is P[z_] = Erfc[-z/Sqrt[2]]/2,
using the selection function P[a+b*z],
the pdf is P[a+b*z]f[z]/I0[a,b],
the mean is m[a_,b_] = I1[a,b]/I0[a,b]
and the variance is I2[a,b]/I0[a,b] - m[a,b]^2,
where
I0[a_,b_] = Integrate[P[a+b*z]f[z],{z,-Infinity,Infinity}]
= P[alpha],
I1[a_,b_] = Integrate[z P[a+b*z]f[z],{z,-Infinity,Infinity}]
= beta f[alpha],
I2[a_,b_] = Integrate[z^2 P[a+b*z]f[z],{z,-Infinity,Infinity}]
= P[alpha] - alpha beta^2 f[alpha],
and {alpha,beta} = {a,b}/Sqrt[1+b^2].
Using Assumptions->Element[{a,b},Reals], version 5.2 gets I1 but gives
up on I0 and I2. (That's all in terms of a & b, of course, not alpha &
beta, which I introduced here only for economy of representation.)
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