Asymptotic expansion of the renewal function for an exponential growing population

Question

I am looking for a formula giving the asymptotic expansion of the renewal equation when there is exponential growth (for lack of better terms).

Consider the renewal equation, for an unknown $M(t)$: \begin{equation} M(t)=a\int_0^\infty M(t-\tau)f(\tau)\mathrm{d}\tau \end{equation} where $f(\tau)$ is a probability density and $a>0$.

When $a=1$, $M(t)$ represents the average value of a renewal process, and it is possible to obtain an asymptotic expansion of it as $t\to \infty$: \begin{equation} \lim_{t\to\infty}\left [ M(t) - \frac{t}{\mu}\right]=\frac{\sigma^2-\mu^2}{2\mu^2} \end{equation}

See for example Chap 6. A first course in Stochastic Process by S. Karlin and HM. Taylor.

However, when $a>1$ the arguments used to derive this expansion can no longer be used (conditioning on the first renewal event), because the quantity is growing exponentially.

What would be the asymptotic expansion of $M(t)$ when $a>1$?

Answer 1

For any $\theta>0$, if you substitute $M(t)=e^{\theta t}N(t),$ then the equation becomes $$ N(t)=\int_0^\infty N(t-\tau)ae^{-\theta\tau}f(\tau)d\tau. $$ If $a>1$, then there is, by continuity, exactly one $\theta$ such that $ae^{-\theta \tau}f(\tau)$ is a probability density. For this $\theta$, the problem is reduced to the $a=1$ case.