On Mon, May 5, 2014 at 9:43 AM, Niloofar.Javanrouh
<javanrou...@yahoo.com> 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 this. First we solve for 'mu' in terms of the other variables
using the link equation:
> library(Ryacas)
> k <- Sym("k")
> mu <- Sym("mu")
> y <- Sym("y")
> L <- Sym("L")
> a <- Sym("a")
> b <- Sym("b")
> x <- Sym("x")
> sumy <- Sym("sumy")
> Solve(log(mu/(mu+k)) == a+b*x, "mu")
expression(list(mu == k * exp(a + b * x)/(1 - exp(a + b * x))))
>
Now in 'k*log(k/(k+mu)) + sumy * log(1-(k/mu+k))' substitute 'k *
exp(a + b * x)/(1 - exp(a + b * x)))' for 'mu' using 'Subst' and take
the derivative with respect to 'a' using 'deriv':
> s <- Subst(k*log(k/(k+mu)) + sumy * log(1-(k/mu+k)), mu,
+ k * exp(a + b * x)/(1 - exp(a + b * x)))
> deriv(s, a)
expression(sumy * (k * ((1 - exp(a + b * x)) * (k * exp(a + b *
x)) + k * exp(a + b * x)^2))/((1 - exp(a + b * x))^2 * (k *
exp(a + b * x)/(1 - exp(a + b * x)))^2 * (1 - (k * (1 - exp(a +
b * x))/(k * exp(a + b * x)) + k))) - k * ((k + k * exp(a +
b * x)/(1 - exp(a + b * x))) * (k * ((1 - exp(a + b * x)) *
(k * exp(a + b * x)) + k * exp(a + b * x)^2)))/((1 - exp(a +
b * x))^2 * ((k + k * exp(a + b * x)/(1 - exp(a + b * x)))^2 *
k)))
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