Computation of the limit

3.5 Computation of the limit

Recall notation of Section 3.2. Let A ǫ s be the Lebesgue measure of [0, s] T €S k ≥0 [T ǫ k , S ǫ k+ 1 ] Š . The process t 7→ sup{i ∈ N; P i j= 1 e ǫ j ≤ A ǫ t } is a Poisson process with intensity 1ǫ and the process s 7→ N ǫ,t , where N ǫ,t = sup{k ∈ N; A ǫ S ǫ k ≤ t} = sup{k ∈ N; M ǫ k X j= 1 e ǫ j ≤ A ǫ t }, is a marked Poisson process with intensity Pm τ 6= 0ǫ, where τ is an exponential random variable with mean ǫ independent of S . We first study the process t 7→ N ǫ,t . Lemma 3.7. The process t 7→ N ǫ,t is a Poisson process with intensity φ 1 γ ǫψ γ , where γ = ψ −1 1ǫ. Proof. We have, by the similar computations as in the proof of Lemma 3.5, Pm τ = 0 = 1 ǫ E –Z ∞ d t e −tǫ 1 {m t =0} ™ = 1 ǫ Z ∞ ds e −γs N –Z σ d t e −tǫ 1 {m t =0} ™ = 1 ǫγ N –Z σ d t e −tǫ 1 {m t =0} ™ . By time reversibility and using optional projection and 15, we have N –Z σ d t e −tǫ 1 {m t =0} ™ = N –Z σ d t e −σ−tǫ 1 {m t =0} ™ = N –Z σ d t e −γ〈ρ t ,1 〉 1 {m t =0} ™ . The proof of Lemma 2.1, see 29 and 31, gives that Pm τ = 0 = 1 ǫψ γ . Since ǫ −1 = ψγ = ψ γ − φ 1 γ, we get 1 ǫ Pm τ 6= 0 = φ 1 γ ǫψ γ . We then get the following Corollary. Corollary 3.8. There exists a sub-sequence ǫ j , j ∈ N decreasing to 0, s.t. P-a.s. for any t ≥ 0 and any continuous function h defined on R + × M f R + such that hu, µ = 0 for u ≥ t , we have, with γ j = ψ −1 1ǫ j , lim j →∞ φ 1 γ j −1 ∞ X k= 1 hA S ǫ j k , ρ − S ǫ j k = Z ∞ hu , ˜ ρ u du. 1456 Proof. Notice that as a direct consequence of 9 and 20, we get lim ǫ→0 ǫψ γ = 1. Recall that A ǫ S ǫ k , k ≥ 1 are the jumping time of the Poisson process t 7→ N ǫ,t with parameter φ 1 γǫψ γ. Standard results on Poisson process implies the vague convergence in distribution see also Lemma XI.11.1 in [18] of φ 1 γ −1 P ∞ k= 1 δ A ǫ Sǫ k d r towards the Lebesgue measure on R + as ǫ goes down to 0. Since the limit is deterministic, the convergence holds in probability and a.s. along a decreasing sub-sequence ǫ j , j ∈ N. In particular, if g is a continuous function on R + with compact support hence bounded, we have that a.s. lim j →∞ φ 1 γ j −1 ∞ X k= 1 gA ǫ j S ǫ j k = Z ∞ gu du . Notice that A ǫ s ≥ A s and that a.s. A ǫ s → A s as ǫ goes down to 0. This implies that a.s. A ǫ s , s ≥ 0 converges uniformly on compacts to A s , s ≥ 0. Therefore, if g is continuous with compact support, we have a.s. lim j →+∞ φ 1 γ j −1 ∞ X k= 1 gA ǫ j S ǫ j k − gA S ǫ j k = 0. So we have that lim j →∞ φ 1 γ j −1 ∞ X k= 1 gA S ǫ j k = Z ∞ gu du 51 and this convergence also holds for a càd-làg function g with compact support as the Lebesgue measure does not charge the point of discontinuity of g. Let h be a continuous function defined on R + × M f R + such that hu, µ = 0 for u ≥ t . First let us remark that ρ − S ǫ k = ρ T ǫ k and that m T ǫ k = 0. Using the strong Markov property at time T ǫ k and the second part of Corollary 2.2, we deduce that P-a.s. for all k ∈ N ∗ , C A T ǫ k = T ǫ k . 52 Therefore, as A S ǫ k = A T ǫ k , we have P-a.s. ˜ ρ A Sǫ k = ˜ ρ A T ǫ k = ρ T ǫ k = ρ − S ǫ k . This gives φ 1 γ j −1 ∞ X k= 1 hA S ǫ j k , ρ − S ǫ j k = φ 1 γ j −1 ∞ X k= 1 hA S ǫ j k , ˜ ρ A S ǫ j k and applying the convergence 51 to the càd-làg function gu = hu , ˜ ρ u gives the result of the lemma. We now study K ǫ given by 43. We keep the same notation as in Lemma 3.5. 1457 Lemma 3.9. There exists a deterministic function R s.t. lim ǫ→0 Rǫ = 0 and for all ǫ 0 and µ ∈ M f R + , we have: sup r ≥0 φ 1 γ logK ǫ r, µ − α 1 vr , µ − Z 0,∞ π 1 dℓ 1 1 − wℓ 1 , r, µ ≤ Rǫ. Proof. We have K ǫ r, µ = ψγ ψγ − ψvr, µ γ − vr, µ γ 1 φ 1 γ α 1 γ + γ Z 1 du Z 0,∞ ℓ 1 π 1 dℓ 1 wuℓ 1 , r, µ e −γ1−uℓ 1 = ψγ ψγ − ψvr, µ γ − vr, µ γ 1 φ 1 γ α 1 γ + Z 0,∞ π 1 dℓ 1 Z γℓ 1 e −s ds w ℓ 1 − s γ , r, µ = ψγ ψγ − ψvr, µ γ − vr, µ γ 1 φ 1 γ φ 1 γ − Z 0,∞ π 1 dℓ 1 Z γℓ 1 e −s ds 1 − wℓ 1 − s γ , r, µ . In particular, we have φ 1 γ logK ǫ r, µ = −A 1 + A 2 + A 3 , where A 1 r = φ 1 γ log 1 − ψvr, µψγ , A 2 r = φ 1 γ log1 − vr, µγ, A 3 r = φ 1 γ log 1 − Z 0,∞ π 1 dℓ 1 Z γℓ 1 e −s ds 1 − wℓ 1 − s γ , r, µ φ 1 γ . Thanks to h 3 , there exists a finite constant a 0 s.t. P-a.s. vr, µ a for all r ≥ 0. We deduce there exists ǫ 0 and a finite constant c 0 s.t. P-a.s for all ǫ ∈ 0, ǫ ], sup r ≥0 |A 1 r| ≤ c φ 1 γ ψγ and sup r ≥0 |A 2 r − α 1 vr , µ| ≤ c γ + c| φ 1 γ γ − α 1 |. 53 We have Z 0,∞ π 1 dℓ 1 Z γℓ 1 e −s ds 1 − wℓ 1 − s γ , r, µ − Z 0,∞ π 1 dℓ 1 1 − wℓ 1 , r, µ = Z 0,∞ π 1 dℓ 1 e −γℓ 1 wℓ 1 , r, µ − 1 + Z 0,∞ π 1 dℓ 1 Z ∞ e −s ds w ℓ 1 , r, µ − wℓ 1 − s γ , r, µ 1 {s≤γℓ 1 } . 1458 It is then easy to get, using h 3 and 35, that P-a.s φ 2 γ = sup r ≥0 Z 0,∞ π 1 dℓ 1 Z γℓ 1 e −s ds 1 − wℓ 1 − s γ , r, µ − Z 0,∞ π 1 dℓ 1 1 − wℓ 1 , r, µ converges to 0 as γ goes to infinity. Recall that we assumed that lim γ→∞ φ 1 γ = +∞. Thus, there exist ǫ 0 and a finite constant c 0 s.t. P-a.s for all ǫ ∈ 0, ǫ ], sup r ≥0 A 3 r − Z 0,∞ π 1 dℓ 1 1 − wℓ 1 , r, µ ≤ c φ 1 γ + φ 2 γ. 54 Using 53 and 54, we get that there exists a deterministic function R s.t. P-a.s sup r ≥0 φ 1 γ logK ǫ r, µ − α 1 vr , µ − Z 0,∞ π 1 dℓ 1 1 − wℓ 1 , r, µ ≤ Rǫ, where lim ǫ→0 Rǫ = 0, thanks to 9 and 20. The previous results allow us to compute the following limit. We keep the same notation as in Lemma 3.5. Lemma 3.10. Let ϕ satisfying condition h 1 –h 3 . There exists a sub-sequence ǫ j , j ∈ N decreasing to 0, s.t. P-a.s. lim j →∞ ∞ Y k= 1 K ǫ j A S ǫ j k , ρ − S ǫ k = exp − Z ∞ du α 1 vu + Z 0,∞ π 1 dℓ 1 − wℓ, u, µ . Proof. Notice that thanks to h 1 , the functions v and u, µ 7→ wℓ, u, µ are continuous and that for r ≥ t, vr, µ = 0 and wℓ, r, µ = 1. The result is then a direct consequence of Corollary 3.8 and Lemma 3.9.

3.6 Proof of Theorem 3.2