Renewal equation

Example

Let $F$ be the distribution function of a nonnegative random variable with $F(0) = 0$, and let $g$ be a locally bounded function on $[0, \infty)$. The renewal equation for an unknown locally bounded function $u$ on $[0, \infty)$ is $$u(t) = g(t) + \int_0^t u(t - s) \, dF(s), \quad t \ge 0.$$ When $F$ has a density $k = F'$, the integral is $\int_0^t k(t - s) \, u(s) \, ds$, and the equation is a Volterra equation of convolution type of the second kind.

In renewal theory, let $X_1, X_2, \dots$ be independent waiting times with distribution function $F$, and let $N(t)$ be the number of $n \ge 1$ with $X_1 + \cdots + X_n \le t$. Conditioning on the first waiting time $X_1$ shows that the renewal function $m(t) = \mathbb{E}[N(t)]$ solves the renewal equation with $g = F$: $$m(t) = F(t) + \int_0^t m(t - s) \, dF(s).$$

Check on the exponential case: let $F(t) = 1 - e^{-\lambda t}$ with $\lambda > 0$, so $dF(s) = \lambda e^{-\lambda s} \, ds$. Then $N$ is a Poisson process of rate $\lambda$, and $m(t) = \lambda t$. Indeed $\int_0^t \lambda (t - s) \lambda e^{-\lambda s} \, ds = \lambda t - 1 + e^{-\lambda t}$, and adding $F(t) = 1 - e^{-\lambda t}$ gives $\lambda t$.