       Re: Re: solving a system of two equations

• To: mathgroup at smc.vnet.net
• Subject: [mg102025] Re: [mg102018] Re: solving a system of two equations
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Mon, 27 Jul 2009 05:53:29 -0400 (EDT)
• References: <200907260757.DAA19100@smc.vnet.net>

```Correct. My "solution" was OK algebraically, but spurious, since

Here are the extreme values of Mean and Variance (not all of them
achievable, hence the messages I've turned off):

Clear[a, b]
{m, v} = Through[{Mean, Variance}@BetaDistribution[a, b]];

f = #1[{#2, a > 0, b > 0}, {a, b}] &;
Off[Minimize::"natt", Maximize::"wksol"]
Transpose@Outer[f, {Minimize, Maximize}, {m, v}]

{{{0, {a -> Indeterminate,
b -> ComplexInfinity}}, {1, {a -> ComplexInfinity,
b -> Indeterminate}}}, {{0, {a -> 0, b -> Indeterminate}}, {1/
4, {a -> 0, b -> 0}}}}

That is, Mean varies from 0 to 1, and Variance varies from 0 to 1/4. Of
the four limits, only one (variance == 1/4) is achievable. The other
interval end-points are open.

Bobby

On Sun, 26 Jul 2009 02:57:46 -0500, Bill Rowe <readnews at sbcglobal.net>
wrote:

> On 7/25/09 at 4:17 AM, perfreem at gmail.com (per) wrote:
>
>> i am trying to find two parameters a, b of the Beta distribution
>> that make its mean equal to some given constant m and its variance
>> equal to some given constant v. this reduces to solving a system of
>> two equations based on the mean/variance definitions of the beta
>> distribution:
>
>> a/(a+b) = m a*b/((a + b)^2 (a + b + 1)) = v
>
>> i want to solve this equation for a and b. i tried to solve this in
>> mathematica, as follows (for m = .5, v = 1):
>
>> Solve[{a/(a + b) == .5, a*b/((a + b)^2 (a + b + 1)) == 2}, a] But it
>> returns: {}
>
> You are aware you are solving for v = 2 rather than v = 1 as you
> wrote right?
>
> But be that as it may, there is no solution for either value.
> The variance for any distribution is the expected value of (x -
> Mean[x])^2. For the standard beta distribution, x ranges from 0
> to 1. So, for all acceptable values of x, x - Mean[x] will be
> between -1 and 1. That is the variance of a standard beta
> distribution will be less than 1.
>
> Note, for the particular case where the mean of a beta
> distribution is 0.5, the distribution is symmetrical and the
> parameters are equal. In that case the variance reduces to
> 1/(4(2a+1)) which clearly is less than 1 for all a>0.
>
>

--
DrMajorBob at bigfoot.com

```

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