Let $X$ be a random variable with density $f(x)=2x\cdot I_{0\le x\le 1}$, a cdf $F(x)=I_{\{x>1\}}+x^2I_{\{0\le x\le 1\}}$ and let $F_I(x):=1/\mu\int_0^x (1-F(y))dy$ where $\mu=E[X]$
(Note: $F_I$ is again a distribution function.)
I try to find a short argument that the moment generating function of $F_I$
which is $\hat{f_I}(t)=\int_0^\infty e^{tx}dF_I(x)= 1/\mu\int_0^x e^{tx}(1-F(y))dy$
hits a value $K>1$, i.e. $\exists t\in \mathbb R$ such that $\hat{f_I}(t)=K>1$ where $K$ is arbitrary.
This condition is called the small claims condition in the Cramér-Lundberg model. This question concerns the existence of the so called adjustment coefficient.
If we consider an arbitrary moment generating function $\hat{f}$ of a random variable. If $\hat{f}(t)<\infty \quad \forall t>0$ then the adjustment coefficient exists, i.e. $\exists t\in \mathbb R$ such that $\hat{f_I}(t)=K>1$ where $K>1$ is arbitrary. So what does the existence of a mgf imply for the existence of the adjustment coefficient?
I try to formulate a differential equation in the interval (0,1) for $\Phi(x):=\Phi(0)+\frac{\lambda}{c}\int_0^x \Phi(x-y)(1-F(y))dy$ by differentiating it once.
I used that the convolution is commutative and then by Leibniz rule $\Phi'(x)=\frac{\lambda}{c}\int_0^x \Phi(y)\frac{d}{dx}(1-F(x-y))dy + \frac{\lambda}{c}\Phi(u)(1-F(0))=\frac{\lambda}{c}\int_0^x \Phi(y)\frac{d}{dx}[1-(x-y)^2]dy + \frac{\lambda}{c}\Phi(u)=-2\frac{\lambda}{c}\int_0^x \Phi(y)(x-y)dy + \frac{\lambda}{c}\Phi(u)$
You see I can't made a differential equation of that.