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Given a risk process $U_t = U_0 + \Pi_t + S_t = u + c t - \sum_{i=1}^{N_t} X_i$ with payment rate $c>0$ and starting capital $u>0$ where $N = \{N_t : t\geq0\}$ is a Poisson($\lambda$) process and the damages $X_i$ are Gamma$(2,\alpha)$ distributed. Let $X=X_1$ and $W=W_1$ where $W_i$ are the Exp$(\lambda)$ waiting times of $N$. I want to compute the adjustment coefficient, that is an $r>0$ that suffices $$ \mathrm E[\mathrm e^{r(X-cW)}] = 1 $$ under the net profit condition (NPC) $\mathrm E[X] < c \mathrm E[W_1]$.

My idea so far: By independence of $X$ and $W$ we have $$\mathrm E[\mathrm e^{rX}]=\mathrm E[\mathrm e^{-crW}]^{-1}.$$ With the pdfs $f(x) = \alpha^2 x \mathrm e^{-\alpha x}$ of $X$ and $g(x) = \lambda \mathrm e^{- \lambda x}$ of $W$ we can just compute the expected values as $$\mathrm E[\mathrm e^{rX}] = \int\limits_0^\infty \mathrm e^{r x} \alpha^2 x \mathrm e^{-\alpha x} \mathrm d x = \frac{\alpha^2}{(\alpha- r)^2}$$ and $$\mathrm E[\mathrm e^{-crW}] = \int\limits_0^\infty \mathrm e^{-crx} \lambda \mathrm e^{- \lambda x} \mathrm d x = \frac{\lambda}{\lambda + c r}.$$

Plugging that into the equality gives $$\frac{\alpha^2}{(\alpha- r)^2} = \frac{\lambda + c r}{\lambda}.$$ Arriving at that cubic equation and quite an ugly expression for $r$ I have to wonder whether I made some mistake or have I overlooked something to simplify this?

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    $\begingroup$ It is possible that the resulting equation to compute the adjustment coefficient is not easy/possible to solve analytically. The equation you have is possible to solve numerically, if that will suffice. $\endgroup$ – JKL Feb 1 at 11:39
  • $\begingroup$ Well, there are analytical solutions to this, but if I understand the theory correctly, it should have a unique solution. $\endgroup$ – Hölderlin Feb 1 at 12:38
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    $\begingroup$ Oh yep. If you try and solve the cubic for $r$, it is actually possible (albeit a bit messy) -- or pop it into Wolfram Alpha. One solution is $r = 0$, and the other two are given by the quadratic formula. Given your NPC, $2 \alpha c - \lambda > 0$ and so the largest positive solution is $r = \frac{2\alpha c - \lambda + \sqrt{\lambda} \sqrt{4 \alpha c + \lambda}}{2c}$. $\endgroup$ – JKL Feb 1 at 23:23
  • $\begingroup$ Oh boy. Of course I can just divide by $r$ and solve the resulting quadratic equation. Thanks. :) $\endgroup$ – Hölderlin Feb 2 at 11:21

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