Stochastic

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Unformatted text preview: t; y . It is easy to see that Z Ti Ic (s) ds = (ti (x + y ))+ Ti 1 To check this we consider two cases. Ignoring the contribution from the last incomplete cycle [TN (t) , t], we have Z 0 t N (t) Ix,y (s) ds ⇡ X (ti (x + y ))+ i=1 The right-hand side is a renewal reward process with so it follows from Theorem 3.3 that the limit is E (t1 (x + y ))+ /Et1 . Applying (A.22) to X = (t1 (x + y ))+ now gives the desired result. 112 CHAPTER 3. RENEWAL PROCESSES Setting x = 0 and then y = 0 we see that the limiting distribution for the age and residual life have the density function given by g (z ) = P (ti > z ) Eti (3.9) which looks the same as the result in discrete time. Multiplying by z , integrating from 0 to 1 and using (A.21) the limit has expected value Et2 /2Eti i (3.10) Di↵erentiating twice we see that if ti has density function fT then the limiting joint density of (At , Zt ) is fT (a + z )/Et1 (3.11) Example 3.10. Exponential. In this case the limiting density given in (3.11) is e (x+y)...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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