Integral equations Qualitative behavior of solutions Theorem

Asymptotic behavior of solutions of renewal equations

Builds on Renewal theory Cumulative distribution function Lebesgue–Stieltjes integration Volterra integral equation

Statement

Let $F$ be a cumulative distribution function with $F(0) = 0$ , so that $F$ is the law of a positive random variable. Assume that $F$ is not arithmetic, that is, $F$ is not concentrated on a set $\set{0, h, 2h, \dots}$ for any $h > 0$ . Let $\mu = \int_{[0, \infty)} t \, dF(t)$ be its mean, with $0 < \mu \le \infty$ . Let $g \colon [0, \infty) \to \R$ be directly Riemann integrable. This means that, for the partition of $[0, \infty)$ into intervals of length $h$ , the upper and lower Riemann sums of $g$ are finite. It also means that they tend to the same limit as $h \to 0$ . Consider the renewal equation $u(t) = g(t) + \int_{[0, t]} u(t - s) \, dF(s), \qquad t \ge 0,$ where the integral is a Lebesgue–Stieltjes integral. Then:

The equation has exactly one solution $u$ that is bounded on bounded intervals, namely $u(t) = \int_{[0, t]} g(t - s) \, dU(s)$ . Here $U = \sum_{n \ge 0} F^{*n}$ , where $F^{*n}$ is the $n$ -fold convolution power of $F$ and $F^{*0}$ is the unit mass at $0$ .
As $t \to \infty$ , $u(t) \to \frac{1}{\mu} \int_0^\infty g(s) \, ds,$ where $1/\mu$ is read as $0$ when $\mu = \infty$ .

Remarks

Part 1 follows by iterating the equation: the series of convolution powers converges because $F^{*n}(t) \to 0$ fast for each fixed $t$ . Part 2 is the key renewal theorem. Its proof first treats $g$ equal to the indicator function of $[0, h)$ , which is Blackwell's renewal theorem. It then approximates a directly Riemann integrable $g$ from above and below by step functions. A nonnegative, nonincreasing, Lebesgue integrable function on $[0, \infty)$ is directly Riemann integrable.

Example: for $g = 1 - F$ , the constant $u = 1$ solves the equation, since $1 - F(t) + F(t) = 1$ . The limit in part 2 is $\frac{1}{\mu} \int_0^\infty (1 - F(s)) \, ds = \mu / \mu = 1$ , as it must be.

The hypothesis that $F$ is not arithmetic is needed. If $F$ is the unit mass at $1$ , then $U$ puts mass $1$ on each integer $n \ge 0$ . For $g$ the indicator function of $[0, 1/2)$ , the solution is $u(t) = 1$ when the fractional part of $t$ is less than $1/2$ and $u(t) = 0$ otherwise, so $u(t)$ has no limit.

Going deeper

For a student

In renewal theory, $U(t)$ is the expected number of renewals in $[0, t]$ , counting the start. Many quantities of a renewal process, such as the law of the residual waiting time at time $t$ , satisfy a renewal equation with an explicit $g$ . The theorem then gives their limits as $t \to \infty$ . The same method describes the long-time behaviour of age-structured populations through the renewal equation and the Euler–Lotka equation, after a change of variable that makes $F$ a probability distribution. From the point of view of Volterra integral equations, it is a statement about the Volterra equation of convolution type whose kernel is a probability measure.

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Cumulative distribution function concept
Lebesgue–Stieltjes integration concept
Renewal theory concept
Volterra integral equation concept
Asymptotic behavior of solutions of renewal equations statement

Tag	`C82WDEQ3DC`
URI	`m2://C82WDEQ3DC`
Kind	statement, theorem
Writing	W0 stub
MSC	45M05, 60K05, 45E10
Status	This copy of the site has no build data. Run `m2 status C82WDEQ3DC`.
About	Renewal theory
Uses	Renewal theory, Cumulative distribution function, Lebesgue–Stieltjes integration, Volterra integral equation