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Re: Solve for parameters of a truncated

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122954] Re: Solve for parameters of a truncated
  • From: Barrie Stokes <Barrie.Stokes at newcastle.edu.au>
  • Date: Fri, 18 Nov 2011 06:23:25 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201111151050.FAA23783@smc.vnet.net>

Hi Bobby

Very nice.  My final mod of your mod of ...

Manipulate[
   Column@{Style[
         "mean equals variance equals " <>
           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, height}, {\[Mu], 0.01, 3}, {\[Sigma], 
     0.01, 3},
         PlotStyle -> {Blue, Green, {Opacity[0.5], Gray}}, 
    ImageSize -> 350]}, {{height, 1},
     0.1, 3, 0.001}]


Best,

Barrie

>>> On 18/11/2011 at 6:19 am, in message <op.v43um9zhtgfoz2 at bobbys-imac.local>,
DrMajorBob <btreat1 at austin.rr.com> wrote:
> 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
>>
>>
> 



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