# Find $r>0$ such that $\mathrm E[\mathrm e^{r(X-cW)}] = 1$ where $c>0$, $X\sim$ Exp$(\lambda)$ and $W\sim$ Gamma$(2,\alpha)$

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?

• 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. – JKL Feb 1 at 11:39
• Well, there are analytical solutions to this, but if I understand the theory correctly, it should have a unique solution. – Hölderlin Feb 1 at 12:38
• 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 $r = \frac{2\alpha c - \lambda + \sqrt{\lambda} \sqrt{4 \alpha c + \lambda}}{2c}$. – JKL Feb 1 at 23:23
• Oh boy. Of course I can just divide by $r$ and solve the resulting quadratic equation. Thanks. :) – Hölderlin Feb 2 at 11:21