Proof of Lemma getdoc792e. 320KB Jun 04 2011 12:04:32 AM

4 Proof of the Proposition 3.2 As pointed out in Remark 3.4, Proposition 3.2 relies on the following key lemma: Lemma 4.1. For every ǫ and δ 0, there exists L 0 such that E Y ” g 1 Y Z a,b — 6 δPb − a ∈ τ 42 for every a 6 b in B 1 such that b − a ≥ ǫL. Given this lemma, the proof of Proposition 3.2 is very similar to the proof of [10, Proposition 2.3], so we will sketch only a few steps. The inequality 23 gives us E Y h g I Y Z I z,Y i 6 c |I | X a 1 ,b 1 ∈B i1 a 1 6 b 1 X a 2 ,b 2 ∈B i2 a 2 6 b 2 . . . X a l ∈B il Ka 1 E Y ” g i 1 Y Z a 1 ,b 1 — Ka 2 − b 1 E Y ” g i 2 Y Z a 2 ,b 2 — . . . . . . Ka l − b l −1 E Y ” g i l Y Z a l ,N — = c |I | X a 1 ,b 1 ∈B i1 a 1 6 b 1 X a 2 ,b 2 ∈B i2 a 2 6 b 2 . . . X a l ∈B il Ka 1 E Y ” g 1 Y Z a 1 −Li 1 −1,b 1 −Li 1 −1 — Ka 2 − b 1 . . . 43 . . . Ka l − b l −1 E Y ” g 1 Y Z a l −Lm−1,N−Lm−1 — . The terms with b i − a i ≥ ǫL are dealt with via Lemma 4.1, while for the remaining ones we just observe that E Y [g 1 Y Z a,b ] ≤ Pb − a ∈ τ since g 1 Y ≤ 1. One has then E Y h g I Y Z I z,Y i 6 c |I | X a 1 ,b 1 ∈B i1 a 1 6 b 1 X a 2 ,b 2 ∈B i2 a 2 6 b 2 . . . X a l ∈B il Ka 1 € δ + 1 {b 1 −a 1 6 ǫ L} Š Pb 1 − a 1 ∈ τ . . . Ka l − b l −1 € δ + 1 {N−a l 6 ǫ L} Š PN − a l ∈ τ. 44 From this point on, the proof of Theorem 3.2 is identical to the proof of Proposition 2.3 in [10] one needs of course to choose ǫ = ǫη and δ = δη sufficiently small. 4.1 Proof of Lemma 4.1 Let us fix a, b in B 1 , such that b − a ≥ ǫL. The small constants δ and ǫ are also fixed. We recall that for a fixed configuration of τ such that a, b ∈ τ, we have E Y W τ ∩ {a, . . . , b}, Y = 1 because z = 1. We can therefore introduce the probability measure always for fixed τ dP τ Y = W τ ∩ {a, . . . , b}, Y dP Y Y 45 where we do not indicate the dependence on a and b. Let us note for later convenience that, in the particular case a = 0, the definition 10 of W implies that for any function f Y E τ [ f Y ] = E X E Y f Y |X i = Y i ∀i ∈ τ ∩ {1, . . . , b} . 46 665 With the definition 20 of Z a,b := Z z=1 a,b , we get E Y ” g 1 Y Z a,b — = E Y E g 1 Y W τ ∩ {a, . . . , b}, Y 1 b ∈τ |a ∈ τ = b EE τ [g 1 Y ]Pb − a ∈ τ, 47 where b P · := P·|a, b ∈ τ, and therefore we have to show that bEE τ [g 1 Y ] 6 δ. With the definition 27 of g 1 Y , we get that for any K b EE τ [g 1 Y ] 6 ǫ K + b EP τ F 1 K . 48 If we choose K big enough, ǫ K is smaller than δ3 thanks to the Lemma 3.1. We now use two lemmas to deal with the second term. The idea is to first prove that E τ [F 1 ] is big with a b P −probability close to 1, and then that its variance is not too large. Lemma 4.2. For every ζ 0 and ǫ 0, one can find two constants u = uǫ, ζ 0 and L = L ǫ, ζ 0, such that for every a, b ∈ B 1 such that b − a ≥ ǫL, b P E τ [F 1 ] ≤ u p log L ≤ ζ, 49 for every L ≥ L . Choose ζ = δ3 and fix u 0 such that 49 holds for every L sufficiently large. If 2K = u p log L and therefore we can make ǫ K small enough by choosing L large, we get that b EP τ F 1 K 6 b EP τ F 1 − E τ [F 1 ] 6 − K + b P E τ [F 1 ] 6 2K 50 6 1 K 2 b EE τ ” F 1 − E τ [F 1 ] 2 — + δ3. 51 Putting this together with 48 and with our choice of K, we have b EE τ [g 1 Y ] 6 2δ3 + 4 u 2 log L b EE τ ” F 1 − E τ [F 1 ] 2 — 52 for L ≥ L . Then we just have to prove that b EE τ ” F 1 − E τ [F 1 ] 2 — = olog L. Indeed, Lemma 4.3. For every ǫ 0 there exists some constant c = cǫ 0 such that b EE τ ” F 1 − E τ [F 1 ] 2 — 6 c log L 3 4 53 for every L 1 and a, b ∈ B 1 such that b − a ≥ ǫL. We finally get that b EE τ [g 1 Y ] 6 2δ3 + clog L −14 , 54 and there exists a constant L 1 0 such that for L L 1 b EE τ [g 1 Y ] 6 δ. 55 666 4.2 Proof of Lemma 4.2 Up to now, the proof of Theorem 2.8 is quite similar to the proof of the main result in [10]. Starting from the present section, instead, new ideas and technical results are needed. Let us fix a realization of τ such that a, b ∈ τ so that it has a non-zero probability under bP and let us note τ ∩ {a, . . . b} = {τ R a = a, τ R a +1 , . . . , τ R b = b} recall that R n = |τ ∩ {1, . . . , n}|. We observe just go back to the definition of P τ that, if f is a function of the increments of Y in {τ n −1 + 1, . . . , τ n }, g of the increments in {τ m −1 + 1, . . . , τ m } with R a n 6= m ≤ R b , and if h is a function of the increments of Y not in {a + 1, . . . , b} then E τ f {∆ i } i ∈{τ n −1 +1,...,τ n } g {∆ i } i ∈{τ m −1 +1,...,τ m } h {∆ i } i ∈{a+1,...,b} 56 = E τ f {∆ i } i ∈{τ n −1 +1,...,τ n } E τ g {∆ i } i ∈{τ m −1 +1,...,τ m } E Y h {∆ i } i ∈{a+1,...,b} , and that E τ f {∆ i } i ∈{τ n −1 +1,...,τ n } = E X E Y f {∆ i } i ∈{τ n −1 +1,...,τ n } |X τ n −1 = Y τ n −1 , X τ n = Y τ n = E X E Y f {∆ i −τ n −1 } i ∈{τ n −1 +1,...,τ n } |X τ n −τ n −1 = Y τ n −τ n −1 . 57 We want to estimate E τ [F 1 ]: since the increments ∆ i for i ∈ B 1 \{a+1, . . . , b} are i.i.d. and centered like under P Y , we have E τ [F 1 ] := b X i, j=a+1 M i j E τ [−∆ i · ∆ j ]. 58 Via a time translation, one can always assume that a = 0 and we do so from now on. The key point is the following Lemma 4.4. 1. If there exists 1 ≤ n ≤ R b such that i, j ∈ {τ n −1 + 1, . . . , τ n }, then E τ [−∆ i · ∆ j ] = Ar r →∞ ∼ C X ,Y r 59 where r = τ n − τ n −1 in particular, note that the expectation depends only on r and C X ,Y is a positive constant which depends on P X , P Y ; 2. otherwise, E τ [−∆ i · ∆ j ] = 0. Proof of Lemma 4.4 Case 2. Assume that τ n −1 i ≤ τ n and τ m −1 j ≤ τ m with n 6= m. Thanks to 56-57 we have that E τ [∆ i · ∆ j ] = E X E Y [∆ i |X τ n −1 = Y τ n −1 , X τ n = Y τ n ] · E X E Y [∆ j |X τ m −1 = Y τ m −1 , X τ m = Y τ m ] 60 and both factors are immediately seen to be zero, since the laws of X and Y are assumed to be symmetric. Case 1. Without loss of generality, assume that n = 1, so we only have to compute E Y E X ” ∆ i · ∆ j X r = Y r — . 61 667 where r = τ 1 . Let us fix x ∈ Z 3 , and denote E Y r,x [·] = E Y [· Y r = x ]. E Y [∆ i · ∆ j Y r = x ] = E Y r,x h ∆ i · E Y r,x ” ∆ j ∆ i —i = E Y r,x ∆ i · x − ∆ i r − 1 = x r − 1 · E Y r,x ∆ i − 1 r − 1 E Y r,x h ∆ i 2 i = 1 r − 1 ‚ kxk 2 r − E Y r,x h ∆ 1 2 iŒ , where we used the fact that under P Y r,x the law of the increments {∆ i } i ≤r is exchangeable. Then, we get E τ [∆ i · ∆ j ] = E X E Y ” ∆ i · ∆ j 1 {Y r =X r } — P X −Y Y r = X r −1 = E X h E Y ” ∆ i · ∆ j Y r = X r — P Y Y r = X r i P X −Y Y r = X r −1 = 1 r − 1  E X   X r 2 r P Y Y r = X r   P X −Y Y r = X r −1 −E X E Y h ∆ 1 2 1 {Y r =X r } i P X −Y Y r = X r −1 = 1 r − 1  E X   X r 2 r P Y Y r = X r   P X −Y Y r = X r −1 − E X E Y h ∆ 1 2 Y r = X r i   . Next, we study the asymptotic behavior of Ar and we prove 59 with C X ,Y = t rΣ Y − t r € Σ −1 X + Σ −1 Y −1 Š . Note that t rΣ Y = E Y ||Y 1 || 2 = σ 2 Y . The fact that C X ,Y 0 is just a conse- quence of the fact that, if A and B are two positive-definite matrices, one has that A − B is positive definite if and only if B −1 − A −1 is [11, Cor. 7.7.4a]. To prove 59, it is enough to show that E X E Y h ∆ 1 2 Y r = X r i r →∞ → E X E Y h ∆ 1 2 i = σ 2 Y , 62 and that Br := E X k X r k 2 r P Y Y r = X r P X −Y X r = Y r r →∞ → t r € Σ −1 X + Σ −1 Y −1 Š . 63 To prove 62, write E X E Y h ∆ 1 2 Y r = X r i = E Y h ∆ 1 2 P X X r = Y r i P X −Y X r = Y r −1 = X y,z ∈Z d k yk 2 P Y Y 1 = y P Y Y r −1 = zP X X r = y + z P X −Y X r − Y r = 0 . 64 We know from the Local Limit Theorem Proposition 2.10 that the term P X X r = y+z P X −Y X r −Y r =0 is uniformly bounded from above, and so there exists a constant c 0 such that for all y ∈ Z d X z ∈Z d P Y Y r −1 = zP X X r = y + z P X −Y X r − Y r = 0 6 c. 65 668 If we can show that for every y fixed Z d the left-hand side of 65 goes to 1 as r goes to infinity, then from 64 and a dominated convergence argument we get that E X E Y h ∆ 1 2 Y r = X r i r →∞ −→ X y ∈Z d k yk 2 P Y Y 1 = y = σ 2 Y . 66 We use the Local Limit Theorem to get X z ∈Z d P Y Y r −1 = zP X X r = y + z = X z ∈Z d c X c Y r d e − 1 2r −1 z · Σ −1 Y z e − 1 2r y+z· Σ −1 X y+z + or −d2 = 1 + o1 X z ∈Z d c X c Y r d e − 1 2r z · Σ −1 Y z e − 1 2r z · Σ −1 X z + or −d2 67 where c X = 2π −d2 det Σ X −12 and similarly for c Y the constants are different in the case of simple random walks: see Remark 2.11, and where we used that y is fixed to neglect y p r. Using the same reasoning, we also have with the same constants c X and c Y P X −Y X r = Y r = X z ∈Z d P Y Y r = zP X X r = z = X z ∈Z d c X c Y r d e − 1 2r z · Σ −1 Y z e − 1 2r z · Σ −1 X z + or −d2 . 68 Putting this together with 67 and considering that P X −Y X r = Y r ∼ c X ,Y r −d2 , we have, for every y ∈ Z d X z ∈Z d P Y Y r −1 = zP X X r = y + z P X −Y X r − Y r = 0 r →∞ −→ 1. 69 To deal with the term Br in 63, we apply the Local Limit Theorem as in 68 to get E X   X r 2 r P Y Y r = X r   = c Y c X r d X z ∈Z d kzk 2 r e − 1 2r z · Σ −1 Y z e − 1 2r z · Σ −1 X z + or −d2 . 70 Together with 68, we finally get Br = c Y c X r d P z ∈Z d kzk 2 r e − 1 2r z · Σ −1 Y +Σ −1 X z + or −d2 c Y c X r d P z ∈Z d e − 1 2r z · Σ −1 Y +Σ −1 X z + or −d2 = 1 + o1E ” kN k 2 — , 71 where N ∼ N € 0, Σ −1 Y + Σ −1 X −1 Š is a centered Gaussian vector of covariance matrix Σ −1 Y + Σ −1 X −1 . Therefore, E ” kN k 2 — = t r € Σ −1 Y + Σ −1 X −1 Š and 63 is proven. Remark 4.5. For later purposes, we remark that with the same method one can prove that, for any given k ≥ 0 and polynomials U and V of order four so that E Y [|U € { ∆ k } k 6 k Š |] ∞ and E X [V ||X r || p r] ∞, we have E X E Y   U € { ∆ k } k 6 k Š V X r p r Y r = X r   r →∞ → E Y U € {k∆ k k} k 6 k Š E V kN k , 72 669 where N is as in 71. Let us quickly sketch the proof: as in 64, we can write E X E Y – U € {k∆ k k} k 6 k Š V kX r k p r Y r = X r ™ = 73 X y 1 ,..., y k0 ∈Z d U € {k y k k} k 6 k Š X z ∈Z d V kzk p r P X X r = z P Y Y r −k = z − y 1 − . . . − y k P X −Y X r − Y r = 0 ×P Y ∆ i = y i , i ≤ k . Using the Local Limit Theorem the same way as in 68 and 71, one can show that for any y 1 , . . . , y k X z ∈Z d V kzk p r P X X r = z P Y Y r −k = z − y 1 − . . . − y k P X −Y X r − Y r = 0 r →∞ → E V kN k . 74 The proof of 72 is concluded via a domination argument as for 62, which is provided by uniform bounds on P Y Y r −k = z − y 1 − . . . − y k and P X −Y X r − Y r = 0 and by the fact that the increments of X and Y have finite fourth moments. Given Lemma 4.4, we can resume the proof of Lemma 4.2, and lower bound the average E τ [F 1 ]. Recalling 58 and the fact that we reduced to the case a = 0, we get E τ [F 1 ] = R b X n=1    X τ n −1 i, j≤τ n M i j    A∆τ n , 75 where ∆ τ n := τ n − τ n −1 . Using the definition 29 of M , we see that there exists a constant c such that for 1 m ≤ L m X i, j=1 M i j ≥ c p L log L m 3 2 . 76 On the other hand, thanks to Lemma 4.4, there exists some r 0 and two constants c and c ′ such that Ar ≥ c r for r ≥ r , and Ar ≥ −c ′ for every r. Plugging this into 75, one gets p L log L E τ [F 1 ] ≥ c R b X n=1 p ∆τ n 1 {∆τ n ≥r } − c ′ R b X n=1 ∆τ n 3 2 1 {∆τ n 6 r } ≥ c R b X n=1 p ∆τ n − c ′ R b . 77 Therefore, we get for any positive B 0 independent of L b P E τ [F 1 ] 6 g p log L 6 b P   1 p L log L c R b X n=1 p ∆τ n − c ′ R b 6 u p log L   6 b P   1 p L log L c R b X n=1 p ∆τ n − c ′ p LB 6 u p log L   + bP € R b B p L Š 6 b P    R b 2 X n=1 p ∆τ n ≤ 1 + o1 u c p L log L    + b PR b B p L. 78 670 Now we show that for B large enough, and L ≥ L B, b PR b B p L 6 ζ2, 79 where ζ is the constant which appears in the statement of Lemma 4.2. We start with getting rid of the conditioning in b P recall b P · = P·|b ∈ τ since we reduced to the case a = 0. If R b B p L, then either |τ ∩ {1, . . . , b2}| or |τ ∩ {b2 + 1, . . . , b}| exceeds B 2 p L. Since both random variables have the same law under b P, we have b PR b B p L 6 2b P R b 2 B 2 p L ≤ 2cP R b 2 B 2 p L , 80 where in the second inequality we applied Lemma A.1. Now, we can use the Lemma A.3 in the Appendix, to get that recall b ≤ L P R b 2 B 2 p L ≤ P R L 2 B 2 p L L →∞ → P |Z | p 2 π ≥ B c K p 2 , 81 with Z a standard Gaussian random variable and c K the constant such that Kn ∼ c K n −32 . The inequality 79 then follows for B sufficiently large, and L ≥ L B. We are left to prove that for L large enough and u small enough b P    R b 2 X n=1 p ∆τ n 6 u c p L log L    6 ζ2. 82 The conditioning in b P can be eliminated again via Lemma A.1. Next, one notes that for any given A 0 independent of L P    R b 2 X n=1 p ∆τ n 6 u c p L log L    6 P    A p L X n=1 p ∆τ n 6 u c p L log L    + P € R b 2 A p L Š . 83 Thanks to the Lemma A.3 in Appendix and to b ≥ ǫL, we have lim sup L →∞ P R b 2 p L A ≤ P |Z | p 2 π Ac K r 2 ǫ , which can be arbitrarily small if A = A ǫ is small enough, for L large. We now deal with the other term in 83, using the exponential Bienaymé-Chebyshev inequality and the fact that the ∆ τ n are i.i.d.: P    1 p L log L A p L X n=1 p ∆τ n u c p log L    6 e uc p log L E – exp ‚ − r τ 1 L log L Œ™ A p L . 84 To estimate this expression, we remark that, for L large enough, E – 1 − exp ‚ − r τ 1 L log L Œ™ = ∞ X n=1 Kn  1 − e − Æ n L log L ‹ ≥ c ′ ∞ X n=1 1 − e − Æ n L log L n 3 2 ≥ c ′′ r log L L , 85 671 where the last inequality follows from keeping only the terms with n ≤ L in the sum, and noting that in this range 1 − e − Æ n L log L ≥ c p n L log L. Therefore, E – exp ‚ − r τ 1 L log L Œ™ A p L 6 1 − c ′′ r log L L A p L ≤ e −c ′′ A p log L , 86 and, plugging this bound in the inequality 84, we get P    1 p L log L A p L X n=1 p ∆τ n 6 u c p log L    6 e [uc−c ′′ A] p log L , 87 that goes to 0 if L → ∞, provided that u is small enough. This concludes the proof of Lemma 4.2. 4.3 Proof of Lemma 4.3