Survival integro-differential equation for a classical risk model (${\phi }^{\prime }(u)=(\lambda /c)(\phi (u)-(\phi \ast f)(u))$)

In the classical risk model with premium rate $c>0$ and claim Poisson process rate $\lambda$, the first-claim decomposition for survival probability implies the displayed equation for ultimate survival probability. Change the first-claim integral to an integral over available capital and differentiate its lower limit. The bounded survival probability function convolved with the integrable claim probability density function is continuous, so this also establishes the needed differentiability. In particular ${\phi }^{\prime }(0)=(\lambda /c)\phi (0)$.

For an Erlang distribution claim density $f(x)={\beta }^{2}x{e}^{-\beta x}$ and $r=\lambda /c$, write $H=\varphi \ast f$. The survival integro-differential equation gives ${\varphi }^{\prime }=r(\varphi -H)$. The convolution satisfies $(D+\beta {)}^{2}H={\beta }^2\varphi$, because $H={\beta }^2{e}^{-\beta u}{\int }_0^u(u-t)\varphi (t)e^{\beta t}dt$. Applying $(D+\beta {)}^2$ to the first equation eliminates $H$ and gives the displayed third-order equation. The original integral equation also supplies ${\varphi }^{\prime }(0)=r\varphi (0)$ and ${\varphi }^{\prime \prime }(0)=r^2\varphi (0)$; these conditions must not be lost when solving the higher-order equation.

For a mixture distribution of exponential distributions, introduce one convolution state $A_j(u)={\int }_0^u\phi (z)e^{-{\beta }_j(u-z)}dz$ for each rate ${\beta }_j$. It satisfies $A_j^{\prime }=\phi -{\beta }_j{A}_j$. Together with the survival integro-differential equation these form a constant-coefficient first-order system; applying the differential operators $D+{\beta }_j$ eliminates the convolution states. For equal mixing weights at rates one and one half, with $r=\lambda /c$, the result is ${\phi }^{\prime \prime \prime }+(3/2-r){\phi }^{\prime \prime }+(1/2-3r/4){\phi }^{\prime }=0$. The original integral equation supplies initial conditions lost in elimination.