# Fastest growing renewal process

There is a lemma in my lecture notes stating that the renewal function $$m(t)=\mathbb{E}[X_t]$$, of a renewal process $$X_t$$ with inter-arrival density $$f$$, satisfies the upper bound: $$m(t)\le Ce^t \text{ for some }C>0$$

Poisson processes of rate $$\lambda$$, as we know, satisfy $$m(t)=\lambda t$$ and have linear growth in $$m(t)$$. The elementary renewal theorem, $$\lim \limits_{t \to \infty}{\frac{m(t)}{t}}=\frac{1}{\mu}$$ (where $$\mu$$ is the inter-arrival distribution's expectation) seems to suggest that all renewal processes have $$m(t)$$ asymptotically linear in $$t$$.

Is this correct and the inequality in my notes a slack upper bound, or is it possible for a renewal process to grow exponentially? Which inter-arrival distribution leads to the fastest growing renewal process?

Suppose your renewal process has interarrival times $$T_1, T_2,\ldots$$. Fix $$c>0$$ and truncate to $$T_k^{(c)}:=T_k\wedge c$$. This results in a renewal process with more rapid renewal, hence a larger renewal function: $$m(t)\le m^{(c)}(t).$$ But by Wald's identity $$E\left(S^{(c)}_{X^{(c)}_t+1}\right)=(m^{(c)}(t)+1)\mu^{(c)},$$ where $$S^{(c)}_n=T^{(c)}_1+\cdots T^{(c)}_n$$ and $$\mu^{(c)}:=E(T^{(c)}_1)$$. A picture makes it clear that $$S^{(c)}_{X^{(c)}_t+1}\le t+c$$. Thus $$m(t)\le m^{(c)}(t)=E\left(S^{(c)}_{X^{(c)}_t+1}\right)/\mu^{(c)}-1\le {t+c\over\mu^{(c)}}-1.$$