Re: fitting

• To: mathgroup at smc.vnet.net
• Subject: [mg104784] Re: fitting
• From: Ray Koopman <koopman at sfu.ca>
• Date: Tue, 10 Nov 2009 05:57:31 -0500 (EST)
• References: <hd3mt8\$9sd\$1@smc.vnet.net> <hd6ddt\$q2h\$1@smc.vnet.net>

```On Nov 8, 4:26 am, Ray Koopman <koop... at sfu.ca> wrote:
> On Nov 7, 3:50 am, Vadim Zaliva <kroko... at gmail.com> wrote:
>> Hi!
>>
>> I am very new to Mathematica and I am failing in trying to do
>> something very simple.
>> I am have the following data:
>>
>> x = {8, 36, 74, 96, 123, 152, 201, 269, 415, 460, 444, 579, 711,
>>      731, 602, 364, 151};
>>
>> Which is a curve, similar to PDF function of Beta distribution:
>>
>> ListLinePlot[x]
>>
>> And I am trying to fit it to:
>>
>>
>> finding Alpha and Beta values.
>>
>> I will appreciate it somebody can give me a hint how to do this
>> properly using FindFit or any other means. I am too embarrassed
>> to post my modest attempts here, but trust me, I've spent few
>> hours trying before posting :)
>>
>
> You haven't said what your 'x' represents or how it is supposed to
> relate to the Beta distribution. If we take the values as the counts
> in equal-width bins that cover [0,1], then we can get the sample mean
> and variance using the midpoints of the bins as the x-values, and use
> the method of moments to estimate 'a' and 'b', the parameters of the
> distribution.
>
> k = Length[f = {8, 36, 74, 96, 123, 152, 201, 269, 415, 460, 444,
>                 579, 711, 731, 602, 364, 151}]
> x = Range[1/2,k]/k; {n = Tr@f, N[m = f.x/n], N[v = f.x^2/n - m^2]}
> Solve[{m == a/(a+b), v == m(1-m)/(1+a+b)}, {a,b}]//Flatten//N
>
> 17
>
> {5416, 0.653065, 0.0391941}
>
> {a -> 3.12214, b -> 1.65861}

17 bins? That's a strange number! Were there 3 empty bins at the
start? In any case, this will get maximum likelihood estimates of
{a,b}. To keep them positive, we estimate their logs.

k = Length[f = { 0,0,0, 8, 36, 74, 96, 123, 152, 201, 269, 415, 460,
444, 579, 711, 731, 602, 364, 151}]
xx = N@Partition[Range[0,k]/k,2,1];
NMaximize[f.Log[BetaRegularized[Sequence@@#,E^ta,E^tb]&/@xx],{ta,tb}]
{a->E^ta, b->E^tb}/.%[[2]]

20

{-13831.2, {ta -> 1.53959, tb -> 0.673633}}

{a -> 4.66267, b -> 1.96135}

```

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