Re: solving a system of two equations

• To: mathgroup at smc.vnet.net
• Subject: [mg102018] Re: solving a system of two equations
• From: Bill Rowe <readnews at sbcglobal.net>
• Date: Sun, 26 Jul 2009 03:57:46 -0400 (EDT)

```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.

```

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