       Re: solving a system of two equations

• To: mathgroup at smc.vnet.net
• Subject: [mg101998] Re: [mg101984] solving a system of two equations
• From: DrMajorBob <btreat1 at austin.rr.com>
• Date: Sun, 26 Jul 2009 03:54:07 -0400 (EDT)
• References: <200907250817.EAA29655@smc.vnet.net>

```Solve for both parameters (since you need them both!) and don't use 0.5
when you mean 1/2:

Solve[{a/(a + b) == 1/2, a*b/((a + b)^2 (a + b + 1)) == 2}, {a, b}]
{a/(a + b) == 1/2, a*b/((a + b)^2 (a + b + 1)) == 2} /. %

{{a -> -(7/16), b -> -(7/16)}}

{{True, True}}

Symbolic solve doesn't work well with approximate numbers, so 0.5 was a
bad idea, although, in this case, it still works:

Solve[{a/(a + b) == 0.5, a*b/((a + b)^2 (a + b + 1)) == 2}, {a, b}]
{a/(a + b) == 1/2, a*b/((a + b)^2 (a + b + 1)) == 2} /. %

{{a -> -0.4375, b -> -0.4375}}

{{True, True}}

You certainly can't get a AND b without solving for them, however.

Bobby

On Sat, 25 Jul 2009 03:17:20 -0500, per <perfreem at gmail.com> wrote:

> hi all,
>
> 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: {}
>
> i want to get back values for a and b. does anyone know how i can do
> this? also, this is subject to the constraint that a and b are
> positive real numbers but i am not sure how to express that.
>
> thank you.
>

--
DrMajorBob at bigfoot.com

```

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