Adjustment coefficient (Lundberg coefficient) for a Weibull distribution I am very glad that I have found this community and I wanted, beforehand, to thank anyone reading my post. I appreciate that very much!
I am estimating the ruin probability for Weibull (in the light tail case) distributed claims. For this purpose I need to calculate the adjustment coefficient.
The CDF is given by:
$$
F(x)=1-e^{-ax^b}\text{,  }a>0\text{, }b\ge1.
$$
The formula for the adjustment coefficient r is:
$$
\int_0^\infty e^{rx}(\overline{F}(x)) = \int_0^\infty e^{rx}e^{-ax^b}.
$$
Now this integral cannot be solved. However, I need some sort of an approximation for the general case. Meaning: I need a formula for the adjustment coefficient, without using specific values for a and b.
Is there a way to get this without differentiating and using the Newton-Raphson Formula? What would be the easiest way to calculate the adjustment coefficient?
Thank you so much for your time!
Regards,
Chris
 A: As far as I understand, the problem is to solve the equation $M(r) =
m$ where $m>1$ is given and $M(r)$ is the integral at the left hand
side in the question.  No usable closed form seems to exist for
$M(r)$, so you will indeed have to implement a numerical solution. A
few iterations of Newton-Raphson (NR) should be enough to reach a good
precision since $M(r)$ is well shaped, being an increasing and
log-convex function of $r$. However, the two integrals needed at each
NR iteration (one for $M(r)$ and one for its derivative) might be
difficult to evaluate by numerical quadrature when $b$ is small, say
$b \le 1.5$ because their integrand tends quite slowly to $0$. This
will become a problem if large values of $r$ are required, in which case 
an approximation based on Laplace's method could be useful.
The scale parameter of the Weibull distribution is easy to cope with,
and you can assume that $a=1$, i.e. that $ \overline{F}(x)= \exp(-x^b)$.
Using the exponential distribution with the same expectation, with
survival $\overline{G}(x)$, the function $N(r):=\int e^{rx}
\,\overline{G}(x) \,\mathrm{d}x$ has a closed form and can be used to
get an initial value.  The case with shape $b= 2$ can be used as a
test since $M(r)$ can then be expressed using the normal probability
function and requires no quadrature.
By restricting to $b > 1.5$ (say), you could implement 
the solution in a R function with a few lines of code.
