I am reading this paper of Lin and Willmot. I dont understand how they come up with formula $2.7$ and why $\tilde{\phi}(s)=\frac{\tilde{H}(s)}{1+\beta-\tilde{g}(s)}$. Can someone help me?
So i want to show that $$\phi(u)=\frac{1}{1+\beta}\int_0^u\phi(u-x)dG(x)+\frac{1}{1+\beta}H(u)$$ is the same as $$\phi(u)=\frac{1}{\beta}\int_0^uH(u-x)dK(x)+\frac{1}{1+\beta}H(u)$$ where $\beta>0, G(x)=1-\overline{G}(x)$ is a distribution function with $G(0)=0$ and $H(u)$ continuous for $u\geq 0$ and $K(u)=1-\overline{K}(u)$ with $\overline{K}(u)=\sum_{n=1}^\infty\frac{\beta}{1+\beta}\left(\frac{1}{1+\beta}\right)^n\overline{G}^{*n}(u)$. $f^{*n}(x)$ is the $n$ fold convolution of $f$. \ Can someone explain me why they are the same?
The Laplace transform of a convolution is the product of its individual Laplace transforms. Therefore $$\mathcal{L}f^{\ast n}=(\mathcal{L}f)^n$$
Especially if $F$ is differentiable with $F'=f$, you find
$$\mathcal{L}f(s)=\int_0^\infty e^{-su}f(u)du=\int_0^\infty e^{-su} dF(u).$$
Recall, that $\bar G^{\ast n}=1-G^{\ast n}$ for $n\in\mathbb{N}$, since it is defined as the tail of the $n$-convolution of $G$. Moreover, it holds $G^{\ast n}(0)=0$ for $n\in\mathbb{N}$. (The special case $n=1$ is covered by $G(0)=0$.)
It holds
$$ \begin{aligned} \frac{\beta}{1+\beta}+\bar K(0) &=\frac{\beta}{1+\beta}+\frac{\beta}{1+\beta}\sum_{n=1}^\infty\big(\frac{1}{1+\beta}\big)^n(1-G^{\ast n}(0))\\ &=\frac{\beta}{1+\beta}+\frac{\beta}{1+\beta}\sum_{n=1}^\infty\big(\frac{1}{1+\beta}\big)^n\\ &=\frac{\beta}{1+\beta}\sum_{n=0}^\infty\big(\frac{1}{1+\beta}\big)^n\\ &=\frac{\beta}{1+\beta}\frac{1}{1-\frac{1}{1+\beta}}\\ &=1, \end{aligned} $$ such that
$$K(0)=1-\bar K(0)=\frac{\beta}{1+\beta}.$$
Moreover, it holds
$$dK=-d\bar K = -\frac{\beta}{1+\beta}\sum_{n=1}^\infty\big(\frac{1}{1+\beta}\big)^nd\bar G^{\ast n}=\frac{\beta}{1+\beta}\sum_{n=1}^\infty\big(\frac{1}{1+\beta}\big)^nd G^{\ast n},$$
such that
$$ \begin{aligned} K(0)+\mathcal{L}k(s) &=K(0)+\int_0^\infty e^{-su}dK(u)\\ &=K(0)+\frac{\beta}{1+\beta}\sum_{n=1}^\infty\big(\frac{1}{1+\beta}\big)^n\int_0^\infty e^{-su}dG^{\ast n}(u)\\ &=K(0)+\frac{\beta}{1+\beta}\sum_{n=1}^\infty\big(\frac{1}{1+\beta}\big)^n\mathcal{L}g^{\ast n}(s)\\ &=K(0)+\frac{\beta}{1+\beta}\sum_{n=1}^\infty\big(\frac{1}{1+\beta}\big)^n(\mathcal{L}g)^n(s)\\ &=\frac{\beta}{1+\beta}\sum_{n=0}^\infty\big(\frac{\mathcal{L}g(s)}{1+\beta}\big)^n\\ &=\frac{\beta}{1+\beta}\frac{1}{1-\frac{\mathcal{L}g(s)}{1+\beta}}\\ &=\frac{\beta}{1+\beta-\mathcal{L}g(s)}. \end{aligned} $$
Lastly, your initial equation can be written as
$$\phi(s)=\frac{1}{1+\beta} (\phi\ast g)(s)+\frac{1}{1+\beta}H(s).$$
Taking the Laplace transform results in
$$ \begin{aligned} \mathcal{L}\phi(s) &=\frac{1}{1+\beta} \mathcal{L}(\phi \ast g)(s)+\frac{1}{1+\beta}\mathcal{L}H(s)\\ &=\frac{1}{1+\beta} \mathcal{L}\phi(s) \mathcal{L}g(s)+\frac{1}{1+\beta}\mathcal{L}H(s), \end{aligned}$$
such that
$$\big(1-\frac{\mathcal{L}g(s)}{1+\beta}\big) \mathcal{L}\phi(s) = \frac{1}{1+\beta}\mathcal{L}H(s).$$
This means,
$$ \begin{aligned} \mathcal{L}\phi(s) = \frac{\mathcal{L}H(s)}{1+\beta-\mathcal{L}g(s)}&=\frac{1}{\beta}\mathcal{L}H(s)(\mathcal{L}k(s)+K(0))\\ &=\frac{1}{\beta}\mathcal{L}H(s)\mathcal{L}k(s)+\frac{K(0)}{\beta}\mathcal{L}H(s)\\ &=\frac{1}{\beta}\mathcal{L}(H\ast k)(s)+\frac{1}{1+\beta}\mathcal{L}H(s). \end{aligned}$$
Taking the Inverse Laplacian transform yields the result.