THE RENEWAL FUNCTION

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Theorem 811 Assume that the probability distribution function F (x) of the interoccurrence times has a probability density f (x) Then the renewal function M(t) satis es the integral equation M(t) = F (t) +

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M(t x)f (x) dx,

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t 0

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(813)

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This integral equation has a unique solution that is bounded on nite intervals Proof The proof of (813) is instructive Fix t > 0 To compute E[N (t)], we condition on the time of the rst renewal and use that the process probabilistically starts over after each renewal Under the condition that X1 = x, the random variable N (t) is distributed as 1+N (t x) when 0 x t and N (t) is 0 otherwise Hence, by conditioning upon X1 , we nd E[N (t)] =

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E[N (t) | X1 = x]f (x) dx =

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E[1 + N (t x)]f (x) dx,

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which gives (813) To prove that the equation (813) has a unique solution, t suppose that H (t) = F (t) + 0 H (t x)f (x) dx, t 0 for a function H (t) that is bounded on nite intervals We substitute this equation repeatedly into itself and use the convolution formula Fn (t) =

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F (t x)fn 1 (x) dx,

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where fk (x) denotes the probability density of Fk (x) This gives

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H (t) =

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Fk (t) +

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H (t x)fn (x) dx,

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n = 1, 2,

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(814)

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Fix now t > 0 Since H (x) is bounded on [0, t], the second term on the right-hand side of (814) is bounded by cFn (t) for some c > 0 Since M(t) < , we have Fn (t) 0 as n By letting n in (814), we nd H (t) = Fk (t) k=1 showing that H (t) = M(t) Theorem 811 allows for the following important generalization Theorem 812 Assume that F (x) has a probability density f (x) Let a(x) be a given, integrable function that is bounded on nite intervals Suppose the function Z(t), t 0, is de ned by the integral equation Z(t) = a(t) +

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Z(t x)f (x) dx,

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t 0

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(815)

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Then this equation has a unique solution that is bounded on nite intervals The solution is given by

ADVANCED RENEWAL THEORY

Z(t) = a(t) +

a(t x)m(x) dx,

t 0,

(816)

where the renewal density m(x) denotes the derivative of M(x) Proof We give only a sketch of the proof The proof is similar to the proof of the second part of Theorem 811 Substituting the equation (815) repeatedly into itself yields

Z(t) = a(t) +

k=1 0

a(t x)fk (x) dx +

Z(t x)fn+1 (x) dx

Next, by letting n , the desired result readily follows It is left to the reader to verify that the various mathematical operations are allowed The integral equation (815) is called the renewal equation This important equation arises in many applied probability problems As an application of Theorem 812, we derive an expression for the second moment of the excess life at time t

2 Lemma 813 Assuming that 2 = E(X1 ) is nite,

E( t2 ) = 2 [1 + M(t)] 2 1 t +

M(x) dx + t 2 ,

t 0

(817)

Proof Fix t 0 Given that the epoch of the rst renewal is x, the random variable t is distributed as t x when x t and t equals x t otherwise Thus E( t2 ) = =

0 0 t

E( t2 | X1 = x)f (x) dx

2 E( t x )f (x) dx +

(x t)2 f (x) dx

and a(t) = t (x t)2 f (x) dx, we obtain a Hence, by letting Z(t) = renewal equation of the form (815) Next it is a question of tedious algebra to derive (817) from (816) The details of the derivation are omitted 812 Computation of the Renewal Function The following tools are available for the numerical computation of the renewal function: (a) the series representation, (b) numerical Laplace inversion, (c) discretization of the renewal equation

E( t2 )

THE RENEWAL FUNCTION

In Section 211 we have already seen that the renewal function can be directly computed from the series representation (811) when the interoccurrence times have a gamma distribution If the interoccurrence times have a Coxian-2 distribution an explicit expression can be given for the renewal function; see Exercise 81 In general the renewal function M(x) can be computed by numerical inversion of its Laplace transform The Laplace transform M (s) = 0 e sx M(x) dx is given by M (s) =

f (s) , s[1 f (s)]

where f (s) = 0 e sx f (x) dx denotes the Laplace transform of the probability density of the interoccurrence times; see Appendix E How to proceed with numerical Laplace inversion is discussed in Appendix F In this appendix it is also discussed how to proceed when the Laplace transform f (s) is analytically intractable Next we discuss a simple but useful discretization method The renewal equation (813) for M(t) is a special case of an integral equation which is known in numerical analysis as a Volterra integral equation of the second kind Many numerical methods have been proposed to solve such equations Unfortunately, these methods typically suffer from the accumulation of round-off errors when t gets larger However, using basic concepts from the theory of Riemann Stieltjes integration, a simple and direct solution method with good convergence properties can be given for the renewal equation (813) This method discretizes the time and computes recursively the renewal function on a grid of points For xed t > 0, let [0, t] be partitioned according to 0 = t0 < t1 , < < tn = t, where ti = ih for a given grid size h > 0 Put for abbreviation Mi = M(ih), Fi = F ((i 05)h) and Ai = F (ih), 1 i n

The recursion scheme for computing the Mi is as follows: i 1 1 Ai + (Mj Mj 1 )Fi j +1 Mi 1 F1 , Mi = 1 F1

j =1

1 i n,

starting with M0 = 0 This recursion scheme is a minor modi cation of the Riemann Stieltjes method proposed in Xie (1989) (the original method uses Fi instead of Ai ) The recursion scheme is easy to program and gives surprisingly accurate results It is remarkable how well the recursion scheme is able to resist the accumulation of round-off errors as t gets larger How to choose the grid size h > 0 depends not only on the desired accuracy in the answers, but also on the shape of the distribution function F (x) and the length of the interval [0, t] The usual way to nd out whether the answers are accurate enough is to do the computations for both a grid size h and a grid size h/2 In many cases of practical interest a fourdigit accuracy is obtained for a grid size h in the range 005 001 In Table 811 some results are given for the renewal function of the Weibull distribution, where