Renewal Theory. Chapter 7 1

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1 Renewal Theory Definiions, Limi Theorems, Renewal Reward Processes, Alernaing Renewal Processes, Age and Excess Life Disribuions, Inspecion Paradox Chaper 7 1

2 Poisson Process: Couning process iid exponenial imes beween arrivals Relax couning process Relax exponenial inerarrival imes Coninuous Time Markov Chain: Exponenial imes beween ransiions Renewal Process: Couning process iid imes beween arrivals Chaper 7 2

3 Couning Process A sochasic process {N(), } is a couning process if N() represens he oal number of evens ha have occurred in [, ] Then {N(), } mus saisfy: N() N() is an ineger for all If s <, hen N(s) N() For s <, N() - N(s) is he number of evens ha occur in he inerval (s, ]. Chaper 7 3

4 Renewal Process A couning process {N(), } is a renewal process if for each n, X n is he ime beween he (n-1)s and nh arrivals and {X n, n 1} are independen wih he same disribuion F. n The ime of he nh arrival is Sn = X 1 i, n 1, i= wih S =. Can wrie N( ) = max { n: Sn } and if µ = E[X n ], n 1, hen he srong law of large numbers says ha P S n n µ as n = 1 Noe: µ is now a ime inerval, no a rae; 1/ µ will be called he rae of he r. p. Chaper 7 4

5 Fundamenal Relaionship ( ) n N n S I follows ha P N = n = P N n P N n+ 1 = { () } { ( ) } () { } { } ( ) ( ) { } P Sn P Sn+ 1 = Fn Fn+ 1 where F n () is he n-fold convoluion of F wih iself. The mean value of N() is () () () { } { n } n( ) m = E N = PN n = PS = F n= 1 n= 1 n= 1 Condiion on he ime of he firs renewal o ge he renewal equaion: m () = F () + m ( x) f ( xdx ) Chaper 7 5

6 Exercise 1 Is i rue ha: ( ) < >? N n S ( )? N n S ( ) > <? N n S n n n Chaper 7 6

7 Exercise 3 If he mean-value funcion of he renewal process {N(), } is given by m ( ) = 2, Then wha is P{N(5) = }? Chaper 7 7

8 Exercise 6 Consider a renewal process {N(), } having a gamma λx (r,λ) inerarrival disribuion wih densiy λe λx ( ) λ e λ (a) Show ha P{ N() n} = i= nr i! (b) Show ha i λ i e ( λ) m () =, where x is he larges ineger x i= r r i! Hin: use he relaionship beween he gamma (r,λ) disribuion and he sum of r independen exponenials wih rae λ o define N() in erms of a Poisson process wih rae λ. ( ) ( ) ( r 1! ) r 1 f x =, x> i Chaper 7 8

9 Limi Theorems ( ) 1 as Wih probabiliy 1, N µ m Elemenary renewal heorem: µ ( ) 1 as Cenral limi heorem: For large, N() is approximaely 2 3 normally disribued wih mean /µ and variance σ µ where σ 2 is he variance of he ime beween arrivals; in paricular, Var 2 N( ) σ as 3 µ Chaper 7 9

10 Exercise 8 A machine in use is replaced by a new machine eiher when i fails or when i reaches he age of T years. If he lifeimes of successive machines are independen wih a common disribuion F wih densiy f, show ha (a) he long-run rae a which machines are replaced is 1 T ( ) ( ( )) xf x + T 1 F T (b) he long-run rae a which machines in use fail equals F( T) T xf ( x) + T ( 1 F ( T )) Hin: condiion on he lifeime of he firs machine Chaper 7 1

11 Renewal Reward Processes Suppose ha each ime a renewal occurs we receive a reward. Assume R n is he reward earned a he nh renewal and {R n, n 1} are independen and idenically disribued (R n may N( ) depend on X n ). The oal reward up o ime is R () = n = R 1 If E[ R] < and X] < hen R( ) E[ R] P E[ X] as = 1 and E R( ) E[ R ] E X [ ] as n Chaper 7 11

12 Age & Excess Life of a Renewal Process The age a ime is A() = he amoun of ime elapsed since he las renewal. The excess life Y() is he ime unil he nex renewal: S N() A() A( ) = S N () Y() Wha is he average value of he age lim s () A d s Chaper 7 12

13 Average Age of a Renewal Process Imagine we receive paymen a a rae equal o he curren age of he renewal process. Our oal reward up o ime s is s A () d and he average reward up o ime s is s Ad () E[ reward during a renewal cycle ] s E[ lengh of a renewal cycle] If X is he lengh of a renewal cycle, hen he oal reward 2 during he cycle is X X d = So, he average age is 2 2 E X 2E X [ ] Chaper 7 13

14 Average Excess or Residual Now imagine we receive paymen a a rae equal o he curren excess of he renewal process. Our oal reward up o ime s is s Y () d and he average reward up o ime s is s Y () d E [ reward during a renewal cycle ] s E[ lengh of a renewal cycle] If X is he lengh of a renewal cycle, hen he oal reward 2 X during he cycle is X X d = ( ) 2 So, he average excess is (also) 2 E X 2E X [ ] Chaper 7 14

15 Inspecion Paradox Suppose ha he disribuion of he ime beween renewals, F, is unknown. One way o esimae i is o choose some sampling imes 1, 2, ec., and for each i, record he oal amoun of ime beween he renewals jus before and jus afer i. This scheme will overesimae he iner-renewal imes Why? For each sampling ime,, we will record X = S S Find is disribuion by condiioning on he ime of he las renewal prior o ime () () () N N N Chaper 7 15

16 Inspecion Paradox (con.) S N() S N()+1 -s { } N() + 1 N() + 1 N() { } P X > x = E P X > x S = s { } N() + 1 N() If s > x hen P X > x S = s = 1 { } N() N() N() N() { } If s x hen P X > x S = s = P X > x X > s { } N() + 1 { > s} N + 1 () ( ) ( ) P X > x 1 F x = = > 1 P X 1 F s ( ) F x Chaper 7 16

17 Inspecion Paradox (con.) S N() S N()+1 -s { } ( ) { } For any s, P X () > x S 1 () = s 1 F x N + N so P{ X } () > x = E P X N 1 N() x S 1 N() s + > = + 1 F( x) = P{ X > x} where X is an ordinary iner-renewal ime. Inuiively, by choosing random imes, i is more likely we will choose a ime ha falls in a long ime inerval. Chaper 7 17

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