Niloofar.Javanrouh yahoo.com> writes:
>
>
> hello,
> i want to differentiate of L with respect to b
> when:
>
> L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k))
>#(negative binomial ln likelihood)
> and
> ln(mu/(mu+k)) = a+bx #link function
>
> how can i do it in R?
> thank you.
>
On Mon, May 5, 2014 at 10:16 PM, David Winsemius wrote:
>
> On May 5, 2014, at 6:05 PM, Gabor Grothendieck wrote:
>
>> On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh
>> wrote:
>>>
>>>
>>> hello,
>>> i want to differentiate of L with respect to b
>>> when:
>>>
>>> L= k*ln (k/(k+mu)) + sum(y) *
On May 5, 2014, at 6:05 PM, Gabor Grothendieck wrote:
> On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh
> wrote:
>>
>>
>> hello,
>> i want to differentiate of L with respect to b
>> when:
>>
>> L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k)) #(negative binomial ln
>> likelihood)
>> and
>>
On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh
wrote:
>
>
> hello,
> i want to differentiate of L with respect to b
> when:
>
> L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k)) #(negative binomial ln
> likelihood)
> and
> ln(mu/(mu+k)) = a+bx #link function
>
> how can i do it in R?
Try th
hello,
i want to differentiate of L with respect to b
when:
L= k*ln (k/(k+mu)) + sum(y) * ln (1-(k/mu+k)) #(negative binomial ln
likelihood)
and
ln(mu/(mu+k)) = a+bx #link function
how can i do it in R?
thank you.
_
Best Regards
Niloofar.Javanrouh
Ph.D Student of Bi
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