Use the renewal equation to show that the renewal function of a Poisson process

Question

Use the renewal equation to show that the renewal function of a Poisson process with rate $\lambda > 0$ is $m(t) = \lambda t$.

Answer 1

The interrenewal distribution is exponential with rate $\lambda$, so $F(t)=1-e^{-\lambda t}$ and so the renewal equation gives us \begin{align} m(t) &= F(t) + F\star m(t)\\ &= 1-e^{-\lambda t} + \int_0^t m(t-s)\lambda e^{-\lambda s}\ \mathsf ds\\ &= 1-e^{-\lambda t} + \lambda e^{-\lambda t}\int_0^t m(x)e^{\lambda x}\ \mathsf dx \end{align} (with the substitution $x=t-s$). Multiplying by $e^{\lambda t}$ yields $$e^{\lambda t}m(t) = e^{\lambda t} -1 +\lambda\int_0^t m(x)e^{\lambda x}\ \mathsf dx, $$ and after differentiating with respect to $t$ we have $$\lambda e^{\lambda t}m(t) + e^{\lambda t}m'(t) = \lambda e^{\lambda t}+\lambda e^{\lambda t}m(t). $$ Therefore $m'(t)=\lambda$, and as $m(0)=0$, it follows that $m(t)=\lambda t$.