Re: Solve for parameters of a truncated normal

*To*: mathgroup at smc.vnet.net*Subject*: [mg122950] Re: Solve for parameters of a truncated normal*From*: DrMajorBob <btreat1 at austin.rr.com>*Date*: Fri, 18 Nov 2011 06:22:42 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201111151050.FAA23783@smc.vnet.net>*Reply-to*: drmajorbob at yahoo.com

Try this: X = TruncatedDistribution[{0, \[Infinity]}, NormalDistribution[\[Mu], \[Sigma]]]; {mean, var} = FullSimplify@Through[{Mean, Variance}@X]; Manipulate[ p = {3.3532, .2242, .3637}; Column@{Style[ "mean can equal variance if contours intersect at height = " <> ToString[height], FontFamily -> "Arial", FontSize -> 11.5], Show[{ContourPlot[ mean == height, {\[Mu], 0.01, 3}, {\[Sigma], 0.01, 3}, ContourStyle -> {Red}, ImageSize -> 350], ContourPlot[var == height, {\[Mu], 0.01, 3}, {\[Sigma], 0.01, 3}, ContourStyle -> {Blue}]}, FrameLabel -> {"\[Mu]", "\[Sigma]"}], Plot3D[{mean, var, 1}, {\[Mu], 0.01, 3}, {\[Sigma], 0.01, 3}, PlotStyle -> {Blue, Green, Gray}, ImageSize -> 350, ViewPoint -> Dynamic[p]], ViewPoint -> Dynamic[p]}, {{height, 1}, 0.1, 3, 0.001}] I think the contour at 2 only adds confusion. Bobby On Thu, 17 Nov 2011 05:03:57 -0600, Barrie Stokes <Barrie.Stokes at newcastle.edu.au> wrote: > Andrzej , Bobby > > Speaking of nice graphics: > > If you combine Bobby's mods to my ContourPlot ... > > X = TruncatedDistribution[{0, \[Infinity]}, > NormalDistribution[\[Mu], \[Sigma]]]; > {mean, var} = FullSimplify@Through[{Mean, Variance}@X]; > Manipulate[ > Column@{Style[ "contour height is " <> ToString[ height], > FontFamily -> "Arial", FontSize -> 11.5 ], > Show[{ContourPlot[ > mean == height, {\[Mu], 0.01, 3}, {\[Sigma], 0.01, 3}, > ContourStyle -> {Red}, ImageSize -> 350], > ContourPlot[var == height, {\[Mu], 0.01, 3}, {\[Sigma], 0.01, 3}, > ContourStyle -> {Blue}]}, > FrameLabel -> {"\[Mu]", "\[Sigma]"}]}, {{height, 1}, 0.1, 3, > 0.001}] > > with these plots from Andrzej's code (rotate both plots to reveal the > "underside") ... > > X = TruncatedDistribution[{0, \[Infinity]}, > NormalDistribution[\[Mu], \[Sigma]]]; > m = Mean[X]; > v = Variance[X]; > > and then ... > > Plot3D[{m, v, 1}, {\[Mu], 0.01, 3}, {\[Sigma], 0.01, 3}, > PlotStyle -> {Blue, Green, Gray}] > > and ... > > Plot3D[{m, v, 2}, {\[Mu], 0.01, 3}, {\[Sigma], 0.01, 3}, > PlotStyle -> {Blue, Green, Gray}] > > the difference between height=1 and height=2 is clearly revealed. > > Cheers > > Barrie > > > > >>>> On 16/11/2011 at 8:46 pm, in message >>>> <201111160946.EAA06190 at smc.vnet.net>, > Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: > >> On 15 Nov 2011, at 11:50, paul wrote: >> >>> 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. >>> >>> >> >> Your first mistake is to use functions (Solve and NSolve) which are not >> intended for such purposes at all. NSolve can only solve (numerically) >> polynomial equations and systems of such. Your equations are certainly >> not of >> this kind. Solve (in version 8) can also solve certain univariate >> transcendental equations but not systems of such. So again, there is no >> point >> at all of trying either of these functions on your system. >> >> The only function that might work is FindRoot. However, before one even >> starts, one has to have some reason for believing such a solution >> exists. >> Now, looking at the graphs below, I see no such reason. So do you have >> one? >> >> X = TruncatedDistribution[{0, \[Infinity]}, >> NormalDistribution[\[Mu], \[Sigma]]]; >> >> m = Mean[X]; >> >> v = Variance[X]; >> >> Plot3D[{m, v, 1}, {\[Mu], 0.1, 2}, {\[Sigma], 0.1, 2}, >> PlotStyle -> {Blue, Green, Black}] >> >> Andrzej Kozlowski > > -- DrMajorBob at yahoo.com

**References**:**Solve for parameters of a truncated normal distribution***From:*paul <paulvonhippel@yahoo.com>