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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
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