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.)