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Two main cases Systems break down because of cumulative effect of - PDF document

Shock models Allan Gut G oteborg August 16, 2005 Much of this is joint work with J urg H usler Two main cases Systems break down because of cumulative effect of shocks; extreme individual shock. Notation { X k }


  1. Shock models ∗ Allan Gut G¨ oteborg August 16, 2005 ∗ Much of this is joint work with J¨ urg H¨ usler

  2. Two main cases Systems break down because of • cumulative effect of shocks; • extreme individual shock. Notation • { X k } magnitude of shocks; time between shocks; • { Y k } • { ( X k , Y k ) } i.i.d.; • S n = � n T n = � n k =1 X k , k =1 Y k ; µ x , µ y , σ 2 x , σ 2 • Means, variances: y , . . . 1

  3. Models The cumulative case ν ( t ) = min { n : S n > t } , t ≥ 0 . Lifetime/failure time: T ν ( t ) . The extreme case τ ( t ) = min { n : X n > t } , t ≥ 0 . Lifetime/failure time: T τ ( t ) . Stopping times behave differently — however... Failure times are Stopped Random Walks. 2

  4. But first ... a general problem Suppose ⋄ { Y n , n ≥ 1 } arbitrary; as n → ∞ ; ⋄ Y n → Y ⋄ { N ( t ) , t ≥ 0 } positive integer valued; ⋄ N ( t ) → ∞ as t → ∞ ; p a.s. r d ⋄ in some sense, → → → → . What about p a.s. r d Y N ( t ) → ? → ? → ? → ? ? ? 3

  5. Almost sure convergence Proposition 1 Suppose ⋄ { Y n , n ≥ 1 } arbitrary; a.s. as n → ∞ ; ⋄ Y n → Y ⋄ { N ( t ) , t ≥ 0 } positive integer valued; ⋄ N ( t ) a.s. → ∞ as t → ∞ . Then a.s. as Y N ( t ) → Y t → ∞ . Proof Union of two nullsets. 4

  6. The central limit theorem Proposition 2 — Anscombe (R´ enyi) Suppose ⋄ { X k , k ≥ 1 } i.i.d.; Var X = σ 2 < ∞ ; ⋄ E X = 0, ⋄ S n = � n k =1 X k , n ≥ 1; p ⋄ N ( t ) → θ as t → ∞ (0 < θ < ∞ ). t Then S N ( t ) d √ → N (0 , 1) as t → ∞ . σ tθ Proof CLT + Kolmogorov’s inequality. 5

  7. Two dimensions (with Svante Janson) ♣ { ( U ( x ) , U ( y ) ) , n ≥ 1 } r.w., n n ♣ i.i.d. increments { ( X k , Y k ) , k ≥ 1 } , ♣ µ y = E Y 1 > 0, µ x = E X 1 exists. ♣ First passage time process: τ ( t ) = min { n : U ( y ) t ≥ 0 . > t } , n Problem: What about { U ( x ) τ ( t ) , t ≥ 0 } ? 6

  8. LLN for U ( x ) τ ( t ) U ( x ) → µ x τ ( t ) a.s. as t → ∞ t µ y CLT for U ( x ) τ ( t ) γ 2 = Var ( µ y X 1 − µ x Y 1 ) > 0, If then U ( x ) τ ( t ) − µ x µ y t d → N (0 , 1) as t → ∞ . y γ 2 t � µ − 3 Many applications Typically: • { Y k } times, marks / rewards ..... • { X k } Stopped random walks . Springer (1988). Back to shocks ... 7

  9. Cumulative shocks ♯ { X k } magnitude of shocks, ♯ { Y k } time between shocks, ♯ S n = � n T n = � n k =1 X k , k =1 Y k , ♯ ν ( t ) = min { n : S n > t } . Theorem 1 (i) If µ x > 0, and | µ y | < ∞ , then T ν ( t ) → µ y a.s. as t → ∞ . t µ x γ 2 > 0, (ii) If, in addition, then T ν ( t ) − µ y µ x t d → N (0 , 1) as t → ∞ . � µ − 3 x γ 2 t 9

  10. Extreme shocks ♭ x F := sup { x : F ( x ) < 1 } , ♭ p t = P ( X 1 > t ). ♭ Stopping times: τ ( t ) = min { n : X n > t } , t ≥ 0 . Then, τ ( t ) geometric, mean 1 /p t . Theorem 2 If p t → 0 as t → x F , then (i) p t τ ( t ) d → Exp(1) as t → x F . (ii) Suppose | µ y | < ∞ . Then d p t T τ ( t ) → µ y Exp(1) as t → x F . 10

  11. Cont’d Proof T τ ( t ) τ ( t ) · p t τ ( t ) d p t T τ ( t ) = → µ y Exp(1) (= Exp( µ y ) if µ y > 0) . Note: No LLN for τ ( t ); no Anscombe. Also Weak convergence in D [0 , ∞ ). 11

  12. Mixed shock models The system breaks down when • the cumulative shocks reach “some high” level or • a single “large” shock appears whichever comes first, viz., the system breaks down at min { ν ( t ) , τ ( t ) } . 12

  13. However, ν ( t ) = O ( t ) τ ( t ) = O (1 /p t ) , so that, necessary for nontrivial results: • ν ( t ) ∼ τ ( t ), • x F = ∞ . Define λ t , the θ/t -quantile: P ( X 1 > λ t ) = θ/t, and set τ λ ( t ) = min { n : X n > λ t } , t ≥ 0 . The system breaks down at time κ ( t ) = min { ν ( t ) , τ λ ( t ) } . 13

  14. Applications/examples ...... • Boxing. A knock-out may be caused by many small punches or a real big one. • Rain in Uppsala. On August 17, 1997, Uppsala had extreme rain during one hour; the basement at home was flooded. A ye- ar later again, but due to rain, on and off, for some days. More generally: Flooding in rivers or dams. • Fatigue, tenacity. A rope, a wire. Less generally: A coat hanger. • Environmental damage. A factory may on and off leak poisonous waste products into a river killing the vegetation and the fish. Or: some catastrophy. • Radioactivity. A variation on the pre- vious example; many minor emissions or a sudden melt-down. 14

  15. Theorem If µ x > 0, | µ y | < ∞ , then d (a) κ ( t ) as t → ∞ , where → Z t  f Z ( y ) = θe − θy , 0 < y < 1 /µ x ,  P ( Z = 1 /µ x ) = e − θ/µ x ,  or, equivalently,  1 − e − θy , for 0 < y < 1 /µ x ,  F Z ( y ) = 1 , for y ≥ 1 /µ x ,  T κ ( t ) d (b) → µ y Z as t → ∞ , t S κ ( t ) d (c) → µ x Z as t → ∞ , t X κ ( t ) p (d) → 0 as t → ∞ . t Results for moments also exist. 15

  16. Basic tool Proposition 3 { U t , t ≥ 0 } and { V t , t ≥ 0 } . Suppose that p d U t → a ∈ R and V t → V as t → ∞ . Then, as t → ∞ ,  P ( V > y ) , for y < a,  P (min { U t , V t } > y ) → 0 , for y > a,  and  0 , for y < a,  P (max { U t , V t } ≤ y ) → P ( V ≤ y ) , for y > a.  Note: Point masses at y = a . 16

  17. Comparing stopping times For example, as t → ∞ : E ν ( t ) 1 → , t µ x E τ λ ( t ) 1 → θ, t 1  µ x , E κ ( t ) 1 � 1 − e − θ/µ x �  → ≤ 1 t θ θ .  Note lim t →∞ E κ ( t ) /t smallest (of course). In particular θ = µ x : E ν ( t ) E τ λ ( t ) ∼ 1 ∼ , t t µ x E κ ( t ) 1 ∼ 0 . 632 1 � 1 − e − 1 � ∼ . t µ x µ x 17

  18. Comparing failure times E T ν ( t ) µ y as → t → ∞ , t µ x E T τ λ ( t ) µ y → as t → ∞ , t θ E T κ ( t ) µ y θ (1 − e − µ y /θ ) → as t → ∞ . t In particular θ = µ x : E T τ λ ( t ) E T ν ( t ) ∼ µ y ∼ , t t µ x E T κ ( t ) (1 − e − 1 ) µ y ≈ 0 . 632 µ y ∼ . t µ x µ x 18

  19. More realistically ♥ “Minor shocks” have no long time effect; ♥ “Discount” of earlier shocks; ♥ Level varies as t ր . Which necessitates limit theorems for ♥ Delayed sums; ♥ Windows; ♥ with/without random size. 19

  20. More general setup • Delayed sums, lag sums, windows n � S k,n = X j , 1 ≤ k ≤ n. j = n − k +1 • Let k n ∼ cn γ , 0 < γ < 1. Consider n � S k n ,n = X j , n ≥ 1 . j = n − k n +1 • ν ( t ) = min { n : S k n ,n > t } , t ≥ 0. Note T ν ( t ) = time until failure, T k ν ( t ) ,ν ( t ) = duration of the fatal window. 20

  21. Strong laws as t → ∞ k ν ( t ) 1 a.s. → , t µ x ν ( t ) 1 a.s. → ( cµ x ) 1 /γ , t 1 /γ S k ν ( t ) ,ν ( t ) a.s. → 1 , t µ 1 − (1 /γ ) S ν ( t ) a.s. x → , t 1 /γ c 1 /γ T k ν ( t ) ,ν ( t ) µ y a.s. → , t µ x T ν ( t ) µ y a.s. → t 1 /γ ( cµ x ) 1 /γ 21

  22. Interpretation • Size of the fatal window = O ( t ); • Total number of shocks at failure = O ( t 1 /γ ); • Shock load of fatal window is = O ( t ); • Complete shock load at failure = O ( t 1 /γ ); • Duration of fatal window = O ( t ); • Total lifetime = O ( t 1 /γ ). Proofs follow the same technique Asymptotic normality also provable. 22

  23. A further extension Recall the Extreme shock model : One “very” large shock is fatal. What about “some” rather large shocks ? In addition to fatal/nonfatal shocks, introduce harmful shocks that weaken the system. Let t = α t (0) ≥ α t (1) ≥ α t (2) ≥ · · · ≥ β t . A shock X is  fatal , if X > t,    harmful, nonfatal , if β t < X < t,   innocent , if X < β t .  If X i is harmful, nonfatal, then X j , j > i is  fatal , if X j > α t (1) ,    harmful, nonfatal , if β t < X j < α t (1) ,  innocent , if  X < β t .  And so on. 23

  24. Results With L t ( n ) = # { i ≤ n : X i ≥ β t } ( L t (0) = 0) , and τ ( t ) = min { n : X n ≥ α t ( L t ( n − 1)) } , one obtains m � m � F m − j ( β t ) � P ( τ ( t ) > m ) = j j =0 j − 1 � � � F ( α t ( k )) − F ( β t ) × . k =0 Special cases α t ( k ) = t nonfatal=harmless, ← → β t = t ← → extreme model 24

  25. Stopping asymptotics Theorem 3 If 1 − F ( β t ) → 0, and 1 − F ( α t ( k )) for k = 1 , 2 , . . . , → c k 1 − F ( β t ) then j − 1 e − z z j � � � � (1 − F ( β t )) τ ( t ) > z (1 − c k ) . P → j ! j ≥ 0 k =0 Theorem 4 If 1 − F ( α t ( ∞ )) → 0, and 1 − F ( α t ( k )) → ( t →∞ ) a k ր ( k →∞ ) 1 , 1 − F ( α t ( ∞ )) 1 − F ( α t ( ∞ )) → 0 , 1 − F ( β t ) then � � → e − z . (1 − F ( α t ( ∞ )) τ ( t ) > z P 25

  26. Lifetime asymptotics Theorem 5 Suppose that | µ y | < ∞ . Under conditions of Theorems 1 or 2, d p t T τ ( t ) → µ y Z as t → ∞ , where Z is as in Theorem 3 or 4. 26

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