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.

**Follow-Ups**:**Re: Re: solving a system of two equations***From:*DrMajorBob <btreat1@austin.rr.com>

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