Re: Solve for parameters of a truncated normal distribution

*To*: mathgroup at smc.vnet.net*Subject*: [mg122924] Re: Solve for parameters of a truncated normal distribution*From*: "Dr. Wolfgang Hintze" <weh at snafu.de>*Date*: Wed, 16 Nov 2011 04:46:25 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <j9tga2$n91$1@smc.vnet.net>

"paul" <paulvonhippel at yahoo.com> schrieb im Newsbeitrag news:j9tga2$n91$1 at smc.vnet.net... > I'm trying to solve the following problem: > X = TruncatedDistribution[{0, \[Infinity]}, > NormalDistribution[\[Mu], \[Sigma]]] > Solve[Mean[X] == 1 && Variance[X] == 1, {\[Mu], \[Sigma]}, Reals] > > I get an error message: "This system cannot be solved with the > methods > available to Solve." It doesn't help if I replace Solve with NSolve. > > In case I've made a mistake in defining the problem, I should say > that > I'm looking for the parameters of a normal distribution so that, if > the normal is truncated on the left at zero, the result will be a > truncated distribution whose mean and variance are both 1. It seems > to > me Mathematica should be able to solve this, at least numerically. > > Many thanks for any suggestions. > > Writing down the problem in explicit form I find a solution mu = 0, sigma = 2 Sqrt[2/Pi] But I don't know if there are more. Here we dive into the details: << "Statistics`ContinuousDistributions`" X = NormalDistribution[\[Mu], \[Sigma]] NormalDistribution[\[Mu], \[Sigma]] Mean[X] \[Mu] Variance[X] \[Sigma]^2 PDF[X, x] 1/(E^((x - \[Mu])^2/(2*\[Sigma]^2))*(Sqrt[2*Pi]*\[Sigma])) (* Y is the truncated distribution *) Y = (1/2)*PDF[X, x] 1/(E^((x - \[Mu])^2/(2*\[Sigma]^2))*(2*Sqrt[2*Pi]*\[Sigma])) (* the first few moments of Y *) m0 = Integrate[Y, {x, 0, Infinity}, Assumptions -> {\[Sigma] > 0, \[Mu] > 0}] (1/4)*(1 + Erf[\[Mu]/(Sqrt[2]*\[Sigma])]) m1 = Integrate[x*Y, {x, 0, Infinity}, Assumptions -> {\[Sigma] > 0, \[Mu] > 0}]/m0 (\[Mu] + (Sqrt[2/Pi]*\[Sigma])/E^(\[Mu]^2/(2*\[Sigma]^2)) + \[Mu]*Erf[\[Mu]/(Sqrt[2]*\[Sigma])])/ (1 + Erf[\[Mu]/(Sqrt[2]*\[Sigma])]) m2 = Integrate[x^2*Y, {x, 0, Infinity}, Assumptions -> {\[Sigma] > 0, \[Mu] > 0}]/m0 (4*((\[Mu]*\[Sigma])/(E^(\[Mu]^2/(2*\[Sigma]^2))*(2*Sqrt[2*Pi])) + (1/4)*(\[Mu]^2 + \[Sigma]^2)* (1 + Erf[\[Mu]/(Sqrt[2]*\[Sigma])])))/ (1 + Erf[\[Mu]/(Sqrt[2]*\[Sigma])]) s2 = Simplify[m2 - m1^2] (\[Sigma]*((-E^(-(\[Mu]^2/(2*\[Sigma]^2))))*Sqrt[2/Pi]*\[Mu] + \[Sigma] - (2*\[Sigma])/(E^(\[Mu]^2/\[Sigma]^2)*Pi) + ((-E^(-(\[Mu]^2/(2*\[Sigma]^2))))*Sqrt[2/Pi]*\[Mu] + 2*\[Sigma])* Erf[\[Mu]/(Sqrt[2]*\[Sigma])] + \[Sigma]*Erf[\[Mu]/(Sqrt[2]*\[Sigma])]^2))/ (1 + Erf[\[Mu]/(Sqrt[2]*\[Sigma])])^2 (* solving with Solve or NSolve does not lead to a result *) NSolve[m1 == 1 && s2 == 1, {\[Mu], \[Sigma]}] $Aborted (* now setting mu -> 0 *) (m1 /. \[Mu] -> 0) == 1 Sqrt[2/Pi]*\[Sigma] == 1 (s2 /. \[Mu] -> 0) == 1 \[Sigma]*(\[Sigma] - (2*\[Sigma])/Pi) == 1 (* ... and solving for sigma gives the result *) Solve[Sqrt[2/Pi]*\[Sigma] == 1 && \[Sigma]*(\[Sigma] - (2*\[Sigma])/Pi) == 1] {{\[Sigma] -> 2*Sqrt[2/Pi]}} (* The case mu != 0 can be studied graphically plotting the two quantities m1 and s2 versus sigma *) (* I could not find a point of intersection at m1 == s2 == 1 *) Regards, Wolfgang