       Re: Solve for parameters of a truncated normal distribution

• To: mathgroup at smc.vnet.net
• Subject: [mg122917] Re: Solve for parameters of a truncated normal distribution
• From: Paul von Hippel <paulvonhippel at yahoo.com>
• Date: Wed, 16 Nov 2011 04:45:08 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201111151050.FAA23783@smc.vnet.net> <4EC38F2B.813B.006A.0@newcastle.edu.au>
• Reply-to: Paul von Hippel <paulvonhippel at yahoo.com>

```Thanks, Barrie! You're the third person to suggest it can't be done. It surprises me, but I'm starting to believe it. Your demonstration has exceptionally spiffy graphics.

I think you can push the common mean and variance of the truncated variablebelow 1.75 -- if you allow mu to be negative. For example, the common mean and variance of the truncated variable is 1.15 if {\[Mu] -> -7.37968, \[Sigma] -> 3.39052}.

And if you don't require the mean and variance of the truncated variable to be equal, you can get even closer to mean=variance=1. E.g.,

FindRoot[{Mean[X] == 1.1, Variance[X] == 1}, {\[Mu], 1, -20,
6}, {\[Sigma], 2, 1, 10}]

________________________________
From: Barrie Stokes <Barrie.Stokes at newcastle.edu.au>
To: mathgroup at smc.vnet.net; paul <paulvonhippel at yahoo.com>
Sent: Tuesday, November 15, 2011 5:23 PM
Subject: [mg122917] Re: [mg122903] Solve for parameters of a truncated normal distribution

Hi Paul

I think this code:

Manipulate[
Show[ {ContourPlot[
mean == height, {\[Mu], 0.01, 3}, {\[Sigma], 0.01, 3},
ContourStyle -> {Red} ],
ContourPlot[ var == height, {\[Mu], 0.01, 3}, {\[Sigma], 0.01, 3},
ContourStyle -> {Blue} ]},
FrameLabel -> {"\[Mu]", "\[Sigma]"} ], {{height, 1}, 0.1, 3,
0.001} ]

shows that this can't be done for the common value for the mean and variance of 1.

The minimum value for a solution is around 1.757 (after 30 seconds playing with the above Manipulate).

Cheers

Barrie

>>> On 15/11/2011 at 9:50 pm, in message <201111151050.FAA23783 at smc.vnet.net>, paul
<paulvonhippel at yahoo.com> 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.

```

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