Tirthankar is pointing in the right direction here.
Note also the existing FAQ:
How do I calculate values of the beta function?
http://www.stata.com/support/faqs/stat/beta.html
which -search beta- would have revealed.
However, caution is advised when using -lngamma()-. In essence the reason for -lngamma()- is that the gamma function grows explosively as x becomes large and positive. So, we work with its logarithm instead. Thus, use -lngamma()- as much as possible and use -exp()- as late as possible.
Thus the FAQ advises calculating the beta function this way
exp(lngamma(a) + lngamma(b) - lngamma(a + b))
and the same advice holds for related functions and integrals. Thus, Tirthankar's example
(exp(lngamma(2))*exp(lngamma(1))/exp(lngamma(3)))
would be better as
exp(lngamma(2) + lngamma(1) - lngamma(3))
which needs fewer function evaluations as well as being better practice to avoid overflows.
Nick
[email protected]
Tirthankar Chakravarty
Here's an example:
*** begin
clear*
sysuse auto
g x = log(price)/10
g new = (exp(lngamma(2))*exp(lngamma(1))/exp(lngamma(3)))*ibeta(2,1,x)
// alpha = 2, beta = 1
su new
*** end
On Fri, Jul 10, 2009 at 4:17 PM, Tirthankar
Chakravarty<[email protected]> wrote:
> help ibeta
>
>
> What you want can be written as:
> (gamma(a)*gamma(b)/gamma(a+b))*ibeta(a,b,x)
> On Fri, Jul 10, 2009 at 3:58 PM, Juanita Riano<[email protected]> wrote:
>> Thanks for the suggestion Roy, maybe I was not clear enough with my
>> posting. What I need is to transform my data solving the following
>> integral:
>>
>> Integral of [X^(alpha-1)*(1-X)^(Beta-1)] where alpha and beta are known
>> parameters
>>
>>> I am wondering if anyone knows of any routine in Stata that I could
>> use
>>> for standardizing data using a beta transformation technique for which
>> I
>>> have already the parameters.
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