Many quantities of interest in the study of renewal processes can be described by a special type of integral equation known as a renewal equation. Renewal equations almost always arise by conditioning on the time of the first arrival and by using the defining property of a renewal process--the fact that the process restarts at each arrival time, independently of the past. However, before we can study renewal equations, we need to develop some additional concepts and tools involving measures, distribution functions, integrals, convolutions, and transforms. In many ways, these parallel our previous study of probability measures, probability distribution functions, expected value, convolution of probability density functions, and generating functions. Hopefully, these parallels will make the study easier.
Suppose that \(G\) is a positive measure (or distribution) on \([0, \infty)\) with the property that \( G[0, t] \lt \infty \) for each \( t \in [0, \infty) \) and \( G[0, \infty) \gt 0 \) (quite possible infinite). We will also use \( G \) to denote the corresponding distribution function:
\[ G(t) = G[0, t], \quad t \in [0, \infty) \]
Hopefully, the notation will not cause confusion and it will be clear from context whether \( G \) refers to the measure (a set function) or the distribution function (a point function). The basic structure of a positive measure and its associated distribution function occurred several times in our preliminary discussion of renewal processes:
Suppose that \( a \gt 0 \) and that \( G(a) \gt 0 \). Then
The function \( G \) satisfies many of the basic properties of a cumulative distribution function. Moreover, the proofs are essentially the same; in fact, the proofs follow from Exercise 1. As usual, we will let \( G(t^-) = \lim_{s \uparrow t} G(s) \) denote the limit of \( G \) from the left at \( t \). By convention, we will let \( G(0^-) = 0 \).
\( G \) satisfies the following properties, where \( s, \; t \in [0, \infty) \) and \( s \le t \):
A measure on \( [0, \infty) \) is completely determined by its values on intervals, so it follows that the measure \( G \) is completely determined by the distribution function \( G \). Equivalently, a function \( G \) that satisfies properties (a), (b), (c), and (d) of Exercise 2 defines a measure through (e), (f), (g), and (h). In most cases, the measure \( G \) will be discrete, continuous, or mixed, in analogy to the discrete, continuous, and mixed probability distributions that we have studied.
First, \( G \) is discrete if there exists a countable set \( S \subseteq [0, \infty) \) such that \(G\{t\} \gt 0\) for \(t \in S\) and \(G([0, \infty) \setminus S) = 0\). If we define \( g : S \to [0, \infty) \) by \( g(t) = G\{t\} \) then
\[ G(A) = \sum_{t \in A \cap S} g(t), \quad A \subseteq [0, \infty) \]Thus, the measure is concentrated at the discrete set of points \( S \) and \( g(t) \) is the measure at \( t \in S \). The function \( g \) is the density function of the measure \( G \) with respect to counting measure.
Next, \( G \) is continuous if \( G\{t\} = 0 \) for \( t \in [0, \infty) \), so that \( G \) has no points of positive measure. In most cases of interest, there exists a function \( g: [0, \infty) \to [0, \infty) \) such that
\[ G(A) = \int_A g(t) \, dt, \quad A \subseteq [0, \infty) \] The function \( g \) is the density function of the measure \( G \) with respect to standard Lebesgue measure (length measure) on \( [0, \infty) \). For our purposes, \( g \) will be a nice
function that is integrable in the ordinary calculus sense.
Finally, \( G \) is mixed if it is the mixture of a discrete and continuous measure. That is, there exists countable set \( S \) such that \( G\{t\} \gt 0 \) for \( t \in S \), \(G([0, \infty) \setminus S) \gt 0\), and \( G\{t\} = 0 \) for \( t \in [0, \infty) \setminus S \). Thus, part of the measure \( G \) is concentrated at the discrete set of points \( S \) and the rest of the measure is continuously spread out over \( [0, \infty) \). As before, we let \( g(t) = G\{t\} \) and we assume that there exists \( h: [0, \infty) \to [0, \infty) \) such that
\[ G(A) = \sum_{A \cap S} g(t) + \int_A h(t) \, dt, \quad A \subseteq [0, \infty) \]
Suppose now that \( u: [0, \infty) \to \R \), and that \( A \) is a (measurable) subset of \( [0, \infty) \). We will denote the integral of \( u \) on the set \( A \) with respect to the measure \( G \) by
\[ \int_A u(t) \, dG(t) \]We will not go into the technical details of the general definition of this integral. However, in the discrete case, and in the continuous and mixed cases with densities, the integral is very similar to the definitions that we have given for expected value. First, suppose that \( G \) is a discrete measure concentrated on \( S \) with discrete density \( g \) as defined above. Then
\[ \int_A u(t) \, dG(t) = \sum_{t \in A \cap S} u(t) \, g(t) \]Next, suppose that \( G \) is a continuous measure with density function \( g \) as defined above. Then
\[ \int_A u(t) \, dG(t) = \int_A u(t) \, g(t) \, dt \]Finally, suppose that \( G \) is a mixed measure with discrete density function \( g \) on \( S \) and continuous density \( h \), as defined above. Then
\[ \int_A u(t) \, dG(t) = \sum_{t \in A \cap S} u(t) \, g(t) + \int_A u(t) \, h(t) \, dt \]This general integral satisfies the essential properties of any integral, which are given in the following exercises. We assume that \( u \) and \( v \) are (measurable) functions on \( [0, \infty) \), \( c \) is a constant, and \( A \) is a (measurable) subset of \( [0, \infty) \). We assume also that the indicated integrals exist.
The integral is a linear operation:
The integral is a monotone operator:
In the remainder of this section, unless otherwise noted, we will denote the integral over the closed interval \( [0, t] \) by
\[ \int_0^t u(s) \, dG(s) \]Of course, if the measure \( G \) is continuous, the integrals over \( [0, t] \), \( (0, t] \), \( [0, t) \) and \( (0, t) \) are all the same.
As above, suppose that \( G \) is a positive measure on \( [0, \infty) \) and that \( u: [0, \infty) \to \R \). The convolution of the function \( u \) with the distribution \( G \) is the function \( u * G \) defined by
\[ (u * G)(t) = \int_0^t u(t - s) \, dG(s) \]This is a different use of the word than in our previous study of the convolution of probability density functions, but there is a close connection.
Suppose that \( G \) is a probability distribution with density function \( g \) and that \( u \) is a probability density function (either both discrete or both continuous). Then \( u * G = u * g \) where the convolution on the right is the ordinary convolution of probability density functions. Recall that this function is the probability density function of the sum of two independent random variables, one with probability density function \( g \) and one with probability density function \( u \).
Convolution is associative. Suppose that \( G \) and \( H \) are measures on \( [0, \infty) \) and that \( u \) is a function on \( [0, \infty) \). Then
\[ (u * G) * H = u * (G * H) \]Convolution is linear. Suppose that \( G \) is a measure on \( [0, \infty) \), that \( u \) and \( v \) are functions on \( [0, \infty) \), and that \( c \) is a constant. Then
In general, the commutative property does not make sense since the function \( u \) and the measure \( G \) are different types of objects. In the special case that the function is the cumulative function of a measure, the commutative property does hold.
Suppose that \( G \) and \( H \) are measures on \( [0, \infty) \). Then \( G * H = H * G \)
Armed with our new analytic machinery, we can return to the study of renewal processes. Thus, suppose that we have a renewal process with interarrival distribution function \( F \) and renewal function \( m \). Recall that each of these functions defines a positive measure on \( [0, \infty) \), as discussed above. Of course, the measure associated with \( F \) is actually a probability measure.
The distributions of the arrival times are the convolution powers of \( F \). That is, \( F_n = F^{\,*n} = F * F * \cdots * F \).
Suppose now that \( a \) and \( u \) are functions on \( [0, \infty) \), with \( a \) known and \( u \) unknown. An integral equation of the form
\[ u = a + u * F \]is called a renewal equation for \( u \). Often, \( u(t) = \E(U_t) \) where \( (U_t: t \ge 0) \) is a random process of interest associated with the renewal process. The renewal equation comes from conditioning on the first arrival time \( T_1 = X_1 \), and then using the defining property of the renewal process--the fact that the process starts over, interdependently of the past, at the arrival time.
The renewal function \( m \) satisfies \( m = F + m * F \).
This follows from conditioning on the frst arrival time.
Thus, the renewal function itself satisfies a renewal equation. Of course, we already have a formula
for \( m \), namely \( m = \sum_{n=1}^\infty F_n \). However, sometimes \( m \) can be computed more easily from the renewal equation directly.
The following exercises give the fundamental results on the solution of the renewal equation.
\( u = a + a * m \) is a solution of the renewal equation \( u = a + u * F \). Moreover, if \( a \) is locally bounded, then \( u \) is locally bounded and is the unique such solution.
First note that \( u * F = a * F + a * m * F = a * F + a * (m - F) = a * m = u - a \) and thus \( u \) is a solution. Next suppose that \( |a(s)| \le C_t \) for \( 0 \le s \le t \). Then \( |u(s)| \le [1 + m(t)]C_t \) for \( 0 \le s \le t \). Suppose that \( v \) is another locally bounded solution of the integral equation, and let \( w = u - v \). Then \( w \) is locally bounded and \( w = w * F \). Hence \( w = w * F_n \) for \( n \in \N_+ \). Suppose that \( |w(s)| \le D_t \) for \( 0 \le s \le t \). Then \( |w(t)| \le D_t \, F_n(t) \) for \( n \in \N_+ \). Since \( m(t) \lt \infty \) it follows that \( F_n(t) \to 0 \) as \( n \to \infty \). Hence \( w(t) = 0 \).
Let \( R_t \) denote the remaining life at time \( t \) and for \( y \ge 0 \), let
\[ u_y(t) = \P(R_t \gt y) = \P(N(t, t + y] = 0) \]We will derive and then solve a renewal equation for \( u_y \) by conditioning on the time of the first arrival. We can then find integral equations that describe the distribution of the current age and the joint distribution of the current and remaining ages. We need some additional notation. Let \( \bar{F}(t) = 1 - F(t) = \P(X \gt t) \) for \( t \ge 0 \) (the right-tail distribution function of an interarrival time), and for \( y \ge 0 \), let \( \bar{F}_y(t) = \bar{F}(t + y) \).
\( u_y \) satisfies the renewal equation \( u_y = \bar{F}_y + u_y * F \)
Note first that \( \P(R_t \gt y \mid X_1 = s) = \P(R_{t-s} \gt y) \) for \( s \in [0, t] \)
Next note that \( \P(R_t \gt y \mid X_1 = s) = 0 \) for \( s \in (t, t + y] \)
Finally note that \( \P(R_t \gt y \mid X_1 = s) = 1 \) for \( s \in (t + y, \infty) \)
The renewal equation now follows from conditioning on the time of the first arrival.
The remaining life at time \( t \ge 0 \) satisfies
\[ \P(R_t \gt y) = \bar{F}(t + y) + \int_0^t \bar{F}(t + y - s) \, dm(s), \quad y \ge 0 \]This follows from Theorems 11 and 12.
Now let \( C_t \) denote the current age at time \( t \ge 0 \).
The current age at time \( t \ge 0 \) satisfies
\[ \P(C_t \ge x) = \bar{F}(t) + \int_0^{t-x} \bar{F}(t - s) \, dm(s), \quad 0 \le x \le t \]This follows from Theorem 13 and the fact that \( \P(C_t \ge x) = \P(R_{t-x} \gt x) \) for \( 0 \le x \le t \).
The joint distribution of \( (C_t, R_t) \) is given by
\[ \P(C_t \ge x, R_t \gt y) = \bar{F}(t + y) + \int_0^{t-x} \bar{F}(t + y - s) \, dm(s), \quad 0 \le x \le t, \; 0 \le y \lt \infty \]Recall that \( \P(C_t \ge x, R_t \gt y) = \P(R_{t-x} \gt x + y) \). The result now follows from Theorem 13.
Consider the renewal process with interarrival times uniformly distributed on \( [0, 1] \). Thus the distribution function of an interarrival time is \( F(x) = x \) for \( 0 \le x \le 1 \). The renewal function \( m \) can be computed from the renewal equation in Exercise 10 by successively solving differential equations. The following two exercises give the first two cases.
\( m(t) = e^t - 1 \) for \( 0 \le t \le 1 \)
In the integral in the renewal equation, we first use the substitution \( y = t - s \). Next we differentiate the integral equation with respect to \( t \) to obtain the differential equation \( m^\prime(t) = 1 + m(t) \) for \( 0 \lt t \lt 1 \). Solving the differential equation subject to the initial condition \( m(0) = 0 \) gives the result.
\( m(t) = (e^t - 1) - (t - 1)e^{t-1} \) for \( 1 \le t \le 2 \)
In the integral in the renewal equation, we again use the substitution \( y = t - s \) and then differentiate the equation with respect to \( t \). This leads to the differential equation \( m^\prime(t) = 1 - e^{t-1} + m(t) \) for \( 1 \lt t \lt 2 \). Solving the differential equation subject to initial condition \( m(1) = e - 1 \) gives the result
Recall that the Poisson process has interarrival times that are exponentially distributed with rate parameter \( r \gt 0 \). Thus, the interarrival distribution function is \( F(x) = 1 - e^{-r\,x} \) for \( x \ge 0 \). The following exercises give alternate proofs of fundamental results obtained in the Introduction.
The renewal function is \( m(t) = r\,t \) for \( t \ge 0 \).
In the integral in the renewal equation, we use the substitution \( y = t - s \) and then differentiate the result with respect to \( t \). This leads to the differential equation \( m^\prime(t) = r \) for \( t \ge 0 \). Solving the differential equation subject to the initial condition \( m(0) = 0 \) gives the result.
The current and remaining life at time \( t \ge 0 \) satisfy the followin properties: .
These results follow from Theorem 15.
Consider the renewal process for which the interarrival times have the geometric distribution with parameter \( p \). Recall that the probability density function is
\[ f(n) = (1 - p)^{n-1}p, \quad n \in \N_+ \]The arrivals are the successes in a sequence of Bernoulli trials. The number of successes \( Y_n \) in the first \( n \) trials is the counting variable for \( n \in \N \). The renewal equations in this section can be used to give alternate proofs of some of the fundamental results in the Introduction.
The renewal function is \( m(n) = n\,p \) for \( n \in \N \).
This follows from the renewal equation in Exercise 10.
The current and remaining life at time \( n \in \N \) satisfy the following properties:.
These results follow from Exercise 15.