Solving a two equation system yields strange results

It sounds right, and, to the best of my knowledge, works right.

find_root(gamma_incomplete(k, k/0.2)/gamma(k) - 0.02, k, 0.1, 13);
(%o41) 0.58985858280192

find_root(gamma_incomplete(k, k/0.3)/gamma(k) - 0.02, k, 1, 10);
(%o42) 1.442704964541916

Thanks a mil,

Jason

On Thu, May 10, 2012 at 5:50 AM, Raymond Toy <toy.raymond at gmail.com> wrote:
> On 5/9/12 3:59 PM, Jason Filippou wrote:
>> Thanks for the help. Indeed, even after changing eq1 into eq1 : b (k -
>> 1) = a i ended up with a summation of parameterized terms in terms of
>> a, and proceeded to do a number of substitutions of a which all appear
>> to be minimized at 0.2.
>>
>> This concludes my Maxima - related questions for now, yet I am at a
>> loss as to what these results mean regarding the statistical nature of
>> my problem, as I mentioned it a couple of e-mails ago. I find it hard
>> to believe that one cannot choose the mean of a distribution such that
>> approximately 98% of its mass is bounded within [0,1]. Perhaps the
>> fact that the minimization does not fall below the 0.2 indiates that
>> I'm approaching my problem in the wrong way.
>
> Let's see if I understand your question.
>
> You have a gamma distribution with parameter k and theta. ?You want to
> find k and theta such that the cdf is 98% for x = 1. ?That means the
> tail of the cdf is 2% so with maxima's gamma_incomplete function we have
>
> gamma_incomplete(k, 1/theta)/gamma(k) = .02
>
> We also have theta*(k-1) = 0.1, or theta = 0.1/(k-1). ?Hence,
>
> gamma_incomplete(k, (k-1)/0.1) = 0.02.
>
> A quick plot shows that the root is somewhere between 1 and 2 and
>
> find_root(gamma_incomplete(k,(k-1)/0.1) - 0.02, k, 1, 2)
> -> 1.473593324038853
>
> Does that sound right? ?I suspect there's no analytical solution so this
> might be the best you can do.
>
> Ray
>
It does sound right, and it also appears to behave right.


-- 
Jason Filippou
Research Associate
NCSR Demokritos
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