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Thesis Presentation
Limiting Distributions and Large Deviations forRandom Walks in Random Environments
Jonathon Peterson
School of MathematicsUniversity of Minnesota
July 24, 2008
Jonathon Peterson 7/24/2008 1 / 35
Thesis Presentation Model
RWRE in Zd with i.i.d. environment
An environment ω = ω(x , y)x ,y∈Zd , such that∑y∈Zd
ω(x , y) = 1, ∀x ∈ Zd .
ω(x , ·)x∈Zd i.i.d. with distribution P.
Quenched law Pω: fix an environment.Xn a random walk: X0 = 0, and
Pω(Xn+1 = x + y |Xn = x) := ω(x , y).
Annealed law P: average over environments.
P(G) :=
∫Ω
Pω(G)dP(ω)
Jonathon Peterson 7/24/2008 2 / 35
Thesis Presentation Model
RWRE in Zd with i.i.d. environment
An environment ω = ω(x , y)x ,y∈Zd , such that∑y∈Zd
ω(x , y) = 1, ∀x ∈ Zd .
ω(x , ·)x∈Zd i.i.d. with distribution P.
Quenched law Pω: fix an environment.Xn a random walk: X0 = 0, and
Pω(Xn+1 = x + y |Xn = x) := ω(x , y).
Annealed law P: average over environments.
P(G) :=
∫Ω
Pω(G)dP(ω)
Jonathon Peterson 7/24/2008 2 / 35
Thesis Presentation Model
RWRE in Zd with i.i.d. environment
An environment ω = ω(x , y)x ,y∈Zd , such that∑y∈Zd
ω(x , y) = 1, ∀x ∈ Zd .
ω(x , ·)x∈Zd i.i.d. with distribution P.
Quenched law Pω: fix an environment.Xn a random walk: X0 = 0, and
Pω(Xn+1 = x + y |Xn = x) := ω(x , y).
Annealed law P: average over environments.
P(G) :=
∫Ω
Pω(G)dP(ω)
Jonathon Peterson 7/24/2008 2 / 35
Thesis Presentation Model
Definitions
Nearest neighbor:
ω(x , y) > 0 ⇐⇒ |y | = 1.
Elliptic:P(ω(x , y) ∈ (0,1), ∀x ∈ Zd , ∀|y | = 1
)= 1.
Uniformly Elliptic: ∃κ > 0 such that
P(ω(x , y) ∈ [κ,1− κ], ∀x ∈ Zd , ∀|y | = 1
)= 1.
Jonathon Peterson 7/24/2008 3 / 35
Thesis Presentation Part I
Part I: Limit Distributions forTransient, One-Dimensional
RWRE
Jonathon Peterson 7/24/2008 4 / 35
Thesis Presentation Review of RWRE in Z
RWRE in Z: Recurrence / Transience
A crucial statistic is:ρx :=
ω(x ,−1)
ω(x ,1)
Theorem (Solomon ’75)
Transience or recurrence is determined by EP(log ρ0):
(a) EP(log ρ0) < 0⇒ limn→∞
Xn = +∞, P− a.s.
(b) EP(log ρ0) > 0⇒ limn→∞
Xn = −∞, P− a.s.
(c) EP(log ρ0) = 0⇒ Xn is recurrent, P− a.s.
Jonathon Peterson 7/24/2008 5 / 35
Thesis Presentation Review of RWRE in Z
RWRW in Z: Law of Large Numbers
Assume EP(log ρ) < 0 (transience to the right).Assume EPρ
s = 1 for some s > 0.
Theorem (LLN, Solomon ’75)P− a.s.:
(a) s > 1 (EPρ < 1) ⇒ limn→∞
Xn
n=
1− EP(ρ)
1 + EP(ρ)> 0
(b) s ≤ 1 (EPρ ≥ 1) ⇒ limn→∞
Xn
n= 0
Denote limn→∞Xnn =: vP .
Jonathon Peterson 7/24/2008 6 / 35
Thesis Presentation Review of RWRE in Z
RWRE in Z: Annealed Limit Laws
Theorem (Kesten, Kozlov, Spitzer ’75)There exists a constant b such that
(a) s ∈ (0,1)⇒ limn→∞
P(
Xn
ns ≤ x)
= 1− Ls,b(x−1/s)
(b) s ∈ (1,2)⇒ limn→∞
P(
Xn − nvP
n1/s ≤ x)
= 1− Ls,b(−x)
(c) s > 2⇒ limn→∞
P(
Xn − nvP
b√
n≤ x
)= Φ(x)
where Ls,b is an s-stable distribution function.
Characteristic Function of Ls,b:
exp−b|t |s
(1− i
t|t |
tan(πs/2)
)Jonathon Peterson 7/24/2008 7 / 35
Thesis Presentation Review of RWRE in Z
RWRE in Z: Annealed Limit Laws
Proof: First prove stable limit laws for hitting times
Tn := infk ≥ 0 : Xk = n
Theorem (Kesten, Kozlov, Spitzer ’75)There exists a constant b such that
(a) s ∈ (0,1)⇒ limn→∞
P(
Tn
n1/s ≤ x)
= Ls,b(x)
(b) s ∈ (1,2)⇒ limn→∞
P
(Tn − nv−1
Pn1/s ≤ x
)= Ls,b(x)
(c) s > 2⇒ limn→∞
P
(Tn − nv−1
P
b√
n≤ x
)= Φ(x)
where Ls,b is an s-stable distribution function.
Jonathon Peterson 7/24/2008 8 / 35
Thesis Presentation Review of RWRE in Z
RWRE in Z: Annealed Limit Laws
Proof: First prove stable limit laws for hitting times
Tn := infk ≥ 0 : Xk = n
Theorem (Kesten, Kozlov, Spitzer ’75)There exists a constant b such that
(a) s ∈ (0,1)⇒ limn→∞
P(
Tn
n1/s ≤ x)
= Ls,b(x)
(b) s ∈ (1,2)⇒ limn→∞
P
(Tn − nv−1
Pn1/s ≤ x
)= Ls,b(x)
(c) s > 2⇒ limn→∞
P
(Tn − nv−1
P
b√
n≤ x
)= Φ(x)
where Ls,b is an s-stable distribution function.
Jonathon Peterson 7/24/2008 8 / 35
Thesis Presentation Quenched Central Limit Theorem
Quenched Limit Laws (Gaussian Regime)
Theorem (Goldsheid ’06, P. ’06)If s > 2 then
limn→∞
Pω
(Tn − EωTn
σ√
n≤ x
)= Φ(x), P − a.s.
where σ2 = EP(VarωT1), and
limn→∞
Pω
(Xn − nvP + Zn(ω)
v3/2P σ√
n≤ x
)= Φ(x), P − a.s.
where Zn(ω) depends only on the environment.
Main results of thesis are for s < 2.Do we get quenched stable laws?
Jonathon Peterson 7/24/2008 9 / 35
Thesis Presentation Quenched Central Limit Theorem
Quenched Limit Laws (Gaussian Regime)
Theorem (Goldsheid ’06, P. ’06)If s > 2 then
limn→∞
Pω
(Tn − EωTn
σ√
n≤ x
)= Φ(x), P − a.s.
where σ2 = EP(VarωT1), and
limn→∞
Pω
(Xn − nvP + Zn(ω)
v3/2P σ√
n≤ x
)= Φ(x), P − a.s.
where Zn(ω) depends only on the environment.
Main results of thesis are for s < 2.Do we get quenched stable laws?
Jonathon Peterson 7/24/2008 9 / 35
Thesis Presentation Quenched Central Limit Theorem
Sketch of Proof
Ti − Ti−1∞i=1 are independent under Pω. Lindberg-Feller⇒
limn→∞
Pω
(Tn − EωTn
σ√
n≤ x
)= Φ(x), P − a.s.
Define X ∗t := maxXn : n ≤ t. Then,
limn→∞
Pω
(X ∗n − nvP + Rn(ω,X ∗n )
v3/2P σ√
n≤ x
)= Φ(x), P − a.s.
Difficulty is to replace Rn(ω,X ∗n ) by Zn(ω), which only depends on theenvironment.
Jonathon Peterson 7/24/2008 10 / 35
Thesis Presentation Quenched Central Limit Theorem
Sketch of Proof
Ti − Ti−1∞i=1 are independent under Pω. Lindberg-Feller⇒
limn→∞
Pω
(Tn − EωTn
σ√
n≤ x
)= Φ(x), P − a.s.
Define X ∗t := maxXn : n ≤ t. Then,
limn→∞
Pω
(X ∗n − nvP + Rn(ω,X ∗n )
v3/2P σ√
n≤ x
)= Φ(x), P − a.s.
Difficulty is to replace Rn(ω,X ∗n ) by Zn(ω), which only depends on theenvironment.
Jonathon Peterson 7/24/2008 10 / 35
Thesis Presentation Quenched Central Limit Theorem
Sketch of Proof
Ti − Ti−1∞i=1 are independent under Pω. Lindberg-Feller⇒
limn→∞
Pω
(Tn − EωTn
σ√
n≤ x
)= Φ(x), P − a.s.
Define X ∗t := maxXn : n ≤ t. Then,
limn→∞
Pω
(X ∗n − nvP + Rn(ω,X ∗n )
v3/2P σ√
n≤ x
)= Φ(x), P − a.s.
Difficulty is to replace Rn(ω,X ∗n ) by Zn(ω), which only depends on theenvironment.
Jonathon Peterson 7/24/2008 10 / 35
Thesis Presentation Quenched Limits: s < 2
Traps
Define the potential of the environmnet
V (i) :=
∑i−1
k=0 log ρk , i > 00, i = 0∑−1
k=i − log ρk , i < 0
Trap: An atypical section of environment where the potential isincreasing.
Time to cross a trap is exponential in the height of the uphill.
Largest uphill of V (·) in [0,n] is ∼ 1s log n (Erdos & Renyi ’70).
⇒ scaling of n1/s in limit laws of Tn.
Jonathon Peterson 7/24/2008 11 / 35
Thesis Presentation Quenched Limits: s < 2
Traps
Define the potential of the environmnet
V (i) :=
∑i−1
k=0 log ρk , i > 00, i = 0∑−1
k=i − log ρk , i < 0
Trap: An atypical section of environment where the potential isincreasing.
Time to cross a trap is exponential in the height of the uphill.
Largest uphill of V (·) in [0,n] is ∼ 1s log n (Erdos & Renyi ’70).
⇒ scaling of n1/s in limit laws of Tn.
Jonathon Peterson 7/24/2008 11 / 35
Thesis Presentation Quenched Limits: s < 2
Blocks of the environment
Ladder locations νn defined by ν0 = 0,
νn := infi > νn−1 : V (i) < V (νn−1)ν−n := supj < ν−n+1 : V (k) > V (j) ∀k < j
##HH\\cc
XX JJcc@@HH
##cc
XXJJHHcc
TT
ccHH
ν1 ν2 ν3 ν4 ν5ν−1ν−3
ν−4ν−5ν−6
ν−2ν0
Define a new measure on environments
Q(·) = P ( · |V (i) > 0,∀i < 0)
Under Q, the environment is stationary under shifts of the νi .Jonathon Peterson 7/24/2008 12 / 35
Thesis Presentation Quenched Limits: s < 2
Blocks of the environment
Ladder locations νn defined by ν0 = 0,
νn := infi > νn−1 : V (i) < V (νn−1)ν−n := supj < ν−n+1 : V (k) > V (j) ∀k < j
##HH\\cc
XX JJcc@@HH
##cc
XXJJHHcc
TT
ccHH
ν1 ν2 ν3 ν4 ν5ν−1ν−3
ν−4ν−5ν−6
ν−2ν0
Define a new measure on environments
Q(·) = P ( · |V (i) > 0,∀i < 0)
Under Q, the environment is stationary under shifts of the νi .Jonathon Peterson 7/24/2008 12 / 35
Thesis Presentation Quenched Limits: s < 2
Heuristics of Quenched Limit Laws
Tνn =n∑
i=1
(Tνi − Tνi−1)Law≈
n∑i=1
exp(µi,ω)
where µi,ω = Eω(Tνi − Tνi−1) ≈√
Varω(Tνi − Tνi−1).
Quenched CLT? Only if
limn→∞
maxi≤n
µ2i,ω
VarωTνn
= 0, P − a.s.
Exponential limit if
limn→∞
maxi≤n
µ2i,ω
VarωTνn
= 1, P − a.s.
Jonathon Peterson 7/24/2008 13 / 35
Thesis Presentation Quenched Limits: s < 2
Heuristics of Quenched Limit Laws
Tνn =n∑
i=1
(Tνi − Tνi−1)Law≈
n∑i=1
exp(µi,ω)
where µi,ω = Eω(Tνi − Tνi−1) ≈√
Varω(Tνi − Tνi−1).
Quenched CLT? Only if
limn→∞
maxi≤n
µ2i,ω
VarωTνn
= 0, P − a.s.
Exponential limit if
limn→∞
maxi≤n
µ2i,ω
VarωTνn
= 1, P − a.s.
Jonathon Peterson 7/24/2008 13 / 35
Thesis Presentation Quenched Limits: s < 2
Heuristics of Quenched Limit Laws
Tνn =n∑
i=1
(Tνi − Tνi−1)Law≈
n∑i=1
exp(µi,ω)
where µi,ω = Eω(Tνi − Tνi−1) ≈√
Varω(Tνi − Tνi−1).
Quenched CLT? Only if
limn→∞
maxi≤n
µ2i,ω
VarωTνn
= 0, P − a.s.
Exponential limit if
limn→∞
maxi≤n
µ2i,ω
VarωTνn
= 1, P − a.s.
Jonathon Peterson 7/24/2008 13 / 35
Thesis Presentation Quenched Limits: s < 2
Heuristics of Quenched Limit Laws
Tνn =n∑
i=1
(Tνi − Tνi−1)Law≈
n∑i=1
exp(µi,ω)
where µi,ω = Eω(Tνi − Tνi−1) ≈√
Varω(Tνi − Tνi−1).
Quenched CLT? Only if
limn→∞
maxi≤n
µ2i,ω
VarωTνn
= 0, P − a.s.
Exponential limit if
limn→∞
maxi≤n
µ2i,ω
VarωTνn
= 1, P − a.s.
Jonathon Peterson 7/24/2008 13 / 35
Thesis Presentation Quenched Limits: s < 2
Theorem (P. ’07)Assume s < 2. Then ∃b > 0 s.t.
limn→∞
Q(
VarωTνn
n2/s ≤ x)
= L s2 ,b
(x).
α-stable process with α < 1 has jumps.This hints that when s < 2
lim infn→∞
Q
(maxi≤n
µ2i,ω
VarωTνn
< δ
)> 0
and
lim infn→∞
Q
(maxi≤n
µ2i,ω
VarωTνn
> 1− δ
)> 0
Jonathon Peterson 7/24/2008 14 / 35
Thesis Presentation Quenched Limits: s < 2
Theorem (P. ’07)Assume s < 2. Then ∃b > 0 s.t.
limn→∞
Q(
VarωTνn
n2/s ≤ x)
= L s2 ,b
(x).
α-stable process with α < 1 has jumps.This hints that when s < 2
lim infn→∞
Q
(maxi≤n
µ2i,ω
VarωTνn
< δ
)> 0
and
lim infn→∞
Q
(maxi≤n
µ2i,ω
VarωTνn
> 1− δ
)> 0
Jonathon Peterson 7/24/2008 14 / 35
Thesis Presentation Quenched Limits: s < 2
Quenched Limit Laws (sub-gaussian regime)
Theorem (P.’07)
If s < 2 then P − a.s. there exist random subsequences nk = nk (ω),and mk = mk (ω) such that
(a) limk→∞
Pω
(Tnk − EωTnk√
VarωTnk
≤ x
)= Φ(x)
(b) limk→∞
Pω
(Tmk − EωTmk√
VarωTmk
≤ x
)=
0 if x < −11− e−x−1 if x ≥ −1
Contrast with the annealed results:
s ∈ (0,1)⇒ limn→∞
P(
Tn
n1/s ≤ x)
= Ls,b(x)
s ∈ (1,2)⇒ limn→∞
P
(Tn − nv−1
Pn1/s ≤ x
)= Ls,b(x)
Jonathon Peterson 7/24/2008 15 / 35
Thesis Presentation Quenched Limits: s < 2
Quenched Limit Laws (sub-gaussian regime)
Theorem (P.’07)
If s < 2 then P − a.s. there exist random subsequences nk = nk (ω),and mk = mk (ω) such that
(a) limk→∞
Pω
(Tnk − EωTnk√
VarωTnk
≤ x
)= Φ(x)
(b) limk→∞
Pω
(Tmk − EωTmk√
VarωTmk
≤ x
)=
0 if x < −11− e−x−1 if x ≥ −1
Contrast with the annealed results:
s ∈ (0,1)⇒ limn→∞
P(
Tn
n1/s ≤ x)
= Ls,b(x)
s ∈ (1,2)⇒ limn→∞
P
(Tn − nv−1
Pn1/s ≤ x
)= Ls,b(x)
Jonathon Peterson 7/24/2008 15 / 35
Thesis Presentation Quenched Limits: s < 2
Quenched Limit Laws (ballistic, sub-gaussian regime)
Theorem (P.’07)
If s ∈ (1,2) then P − a.s. there exist random subsequences nk = nk (ω)and mk = mk (ω) such that
(a) limk→∞
Pω
(Xtk − nk
vP√
VarωTnk
≤ x
)= Φ(x)
(b) limk→∞
Pω
(Xt ′k−mk
vP√
VarωTmk
< x
)=
ex−1 if x < 11 if x ≥ 1
,
where tk = EωTnk and t ′k = EωTmk .
Contrast with
limn→∞
P(
Xn − nvP
n1/s ≤ x)
= 1− Ls,b(−x)
Jonathon Peterson 7/24/2008 16 / 35
Thesis Presentation Quenched Limits: s < 2
Quenched Limit Laws (ballistic, sub-gaussian regime)
Theorem (P.’07)
If s ∈ (1,2) then P − a.s. there exist random subsequences nk = nk (ω)and mk = mk (ω) such that
(a) limk→∞
Pω
(Xtk − nk
vP√
VarωTnk
≤ x
)= Φ(x)
(b) limk→∞
Pω
(Xt ′k−mk
vP√
VarωTmk
< x
)=
ex−1 if x < 11 if x ≥ 1
,
where tk = EωTnk and t ′k = EωTmk .
Contrast with
limn→∞
P(
Xn − nvP
n1/s ≤ x)
= 1− Ls,b(−x)
Jonathon Peterson 7/24/2008 16 / 35
Thesis Presentation Quenched Limits: s < 2
Quenched Limit Laws (Zero-Speed Regime)
Theorem (P., Zeitouni ’07)
If s ∈ (0,1), then P − a.s. there exist random subsequencesnk = nk (ω), mk = mk (ω), tk = tk (ω), and uk = uk (ω) s.t.
(a) limk→∞
Pω
(Xnk
mk≤ x
)=
0 x ≤ 012 0 < x <∞
and limk→∞
log mk
log nk= s
(b) limk→∞
Pω
(Xtk − uk
log2 tk∈ [−δ, δ]
)= 1, ∀δ > 0.
Contrast with
limn→∞
P(
Xn
ns ≤ x)
= 1− Ls,b(x−1/s)
Jonathon Peterson 7/24/2008 17 / 35
Thesis Presentation Quenched Limits: s < 2
Quenched Limit Laws (Zero-Speed Regime)
Theorem (P., Zeitouni ’07)
If s ∈ (0,1), then P − a.s. there exist random subsequencesnk = nk (ω), mk = mk (ω), tk = tk (ω), and uk = uk (ω) s.t.
(a) limk→∞
Pω
(Xnk
mk≤ x
)=
0 x ≤ 012 0 < x <∞
and limk→∞
log mk
log nk= s
(b) limk→∞
Pω
(Xtk − uk
log2 tk∈ [−δ, δ]
)= 1, ∀δ > 0.
Contrast with
limn→∞
P(
Xn
ns ≤ x)
= 1− Ls,b(x−1/s)
Jonathon Peterson 7/24/2008 17 / 35
Thesis Presentation Part II
Part II: Annealed LargeDeviations for Multidimensional
RWRE
Jonathon Peterson 7/24/2008 18 / 35
Thesis Presentation Large Deviations: Background
Large Deviations: Definitions
Rate function: A lower semi-continuous function h : Rd → [0,∞].
Good rate function: x : |h(x)| ≤ C compact ∀C <∞.
ξn ∈ Rd satisfy a large deviation principle (LDP) if:
− infx∈Γ
h(x) ≤ lim infn→∞
1n
log P (ξn ∈ Γ)
≤ lim supn→∞
1n
log P (ξn ∈ Γ) ≤ − infx∈Γ
h(x),
where h is a good rate function.That is
P(ξn ≈ x) ≈ e−nh(x).
Jonathon Peterson 7/24/2008 19 / 35
Thesis Presentation Large Deviations: Background
Large Deviations: Definitions
Rate function: A lower semi-continuous function h : Rd → [0,∞].
Good rate function: x : |h(x)| ≤ C compact ∀C <∞.
ξn ∈ Rd satisfy a large deviation principle (LDP) if:
− infx∈Γ
h(x) ≤ lim infn→∞
1n
log P (ξn ∈ Γ)
≤ lim supn→∞
1n
log P (ξn ∈ Γ) ≤ − infx∈Γ
h(x),
where h is a good rate function.That is
P(ξn ≈ x) ≈ e−nh(x).
Jonathon Peterson 7/24/2008 19 / 35
Thesis Presentation Large Deviations: Background
Large Deviations: Definitions
Rate function: A lower semi-continuous function h : Rd → [0,∞].
Good rate function: x : |h(x)| ≤ C compact ∀C <∞.
ξn ∈ Rd satisfy a large deviation principle (LDP) if:
− infx∈Γ
h(x) ≤ lim infn→∞
1n
log P (ξn ∈ Γ)
≤ lim supn→∞
1n
log P (ξn ∈ Γ) ≤ − infx∈Γ
h(x),
where h is a good rate function.That is
P(ξn ≈ x) ≈ e−nh(x).
Jonathon Peterson 7/24/2008 19 / 35
Thesis Presentation Large Deviations: Background
LLN for multidimensional RWRE?
No known LLN in general.(In fact no 0-1 law for transience in a given direction).
However, the random variable V := limn→∞Xnn exists, P− a.s.
(Due to results of Sznitman and Zerner)
Moreover, either1 V =: vP is P− a.s. constant.2 supp(V ) = v−, v+, with v− = cv+ for some c ≤ 0.
There are known conditions such that a V = vP is constant, P− a.s.
Jonathon Peterson 7/24/2008 20 / 35
Thesis Presentation Large Deviations: Background
LLN for multidimensional RWRE?
No known LLN in general.(In fact no 0-1 law for transience in a given direction).
However, the random variable V := limn→∞Xnn exists, P− a.s.
(Due to results of Sznitman and Zerner)
Moreover, either1 V =: vP is P− a.s. constant.2 supp(V ) = v−, v+, with v− = cv+ for some c ≤ 0.
There are known conditions such that a V = vP is constant, P− a.s.
Jonathon Peterson 7/24/2008 20 / 35
Thesis Presentation Large Deviations: Background
LLN for multidimensional RWRE?
No known LLN in general.(In fact no 0-1 law for transience in a given direction).
However, the random variable V := limn→∞Xnn exists, P− a.s.
(Due to results of Sznitman and Zerner)
Moreover, either1 V =: vP is P− a.s. constant.2 supp(V ) = v−, v+, with v− = cv+ for some c ≤ 0.
There are known conditions such that a V = vP is constant, P− a.s.
Jonathon Peterson 7/24/2008 20 / 35
Thesis Presentation Annealed Large Deviations
Annealed Large Deviations
Theorem (Varadhan ’03)
Let Xn be a uniformly elliptic, nearst neighbor RWRE on Zd . Then,there exists a convex good rate function H(v) such that Xn
n satisfies anannealed LDP with rate function H(v).
This implies
limδ→0
lim infn→∞
1n
log P(‖Xn − nv‖ < δ) = H(v).
limδ→0
lim supn→∞
1n
log P(‖Xn − nv‖ < δ) = H(v).
Jonathon Peterson 7/24/2008 21 / 35
Thesis Presentation Annealed Large Deviations
Annealed Large Deviations
Theorem (Varadhan ’03)
Let Xn be a uniformly elliptic, nearst neighbor RWRE on Zd . Then,there exists a convex good rate function H(v) such that Xn
n satisfies anannealed LDP with rate function H(v).
This implies
limδ→0
lim infn→∞
1n
log P(‖Xn − nv‖ < δ) = H(v).
limδ→0
lim supn→∞
1n
log P(‖Xn − nv‖ < δ) = H(v).
Jonathon Peterson 7/24/2008 21 / 35
Thesis Presentation Annealed Large Deviations
Zero Set of the Rate Function
Drift at the origin: d(ω) := EωX1.Possible drifts: K := conv (supp (d(ω))).Nestling: 0 ∈ K.Non-nestling: 0 /∈ K.
Theorem (Varadhan ’03)
The set Z := v : H(v) = 0 is either a single point or an intervalcontaining the origin.Non-nestling ⇒ Z = vP.Nestling, supp(V ) = vP ⇒ Z = [0, vP ].Nestling, supp(V ) = v−, v+ ⇒ Z = [v−, v+].
Jonathon Peterson 7/24/2008 22 / 35
Thesis Presentation Annealed Large Deviations
Varadhan’s proof
Xn is not a Markov chain (long term memory).Study the comets of the random walk:
Wn := (−Xn,−Xn + X1, . . . ,−Xn + Xn−1,0)
Wn is a Markov chain (on a horrible state space W ).Obtain a LDP for the empirical distribution process
Rn :=1n
n∑j=1
δWn
with rate function J (µ).Contract for LDP for Xn
n : H(v) = infm(µ)=v J (µ).
Jonathon Peterson 7/24/2008 23 / 35
Thesis Presentation Annealed Large Deviations
Properties of the Annealed Rate Function H(v)
Theorem (P., Zeitouni ’08)
Assume the law P is non-nestling. Then, H(v) is analytic in aneighborhood of vP .
Idea:1 Define a new function J(v), which is analytic near vP .2 Show H(v) = J(v) near vP .
Jonathon Peterson 7/24/2008 24 / 35
Thesis Presentation Annealed Large Deviations
Properties of the Annealed Rate Function H(v)
Theorem (P., Zeitouni ’08)
Assume the law P is non-nestling. Then, H(v) is analytic in aneighborhood of vP .
Idea:1 Define a new function J(v), which is analytic near vP .2 Show H(v) = J(v) near vP .
Jonathon Peterson 7/24/2008 24 / 35
Thesis Presentation Annealed Large Deviations
Regeneration Times
Let ` ∈ Rd with ‖`‖2 = 1.Regeneration times (in direction `):
t
τ4τ3
τ2
τ1
Xt · `Xτ1 · ` Xτ2 · ` Xτ3 · ` Xτ4 · `
Jonathon Peterson 7/24/2008 25 / 35
Thesis Presentation Annealed Large Deviations
Regeneration Times
Assume that P(limn→∞ Xn · ` = +∞) = 1.Define P(·) := P (·|Xn · ` ≥ 0, ∀n) .
(Xτ1 , τ1), (Xτ2 − Xτ1 , τ2 − τ1), (Xτ3 − Xτ2 , τ3 − τ2), . . .
independent sequence under Pi.i.d. under P
Moreover,
vP := limn→∞
Xn
n=
EXτ1
Eτ1
Jonathon Peterson 7/24/2008 26 / 35
Thesis Presentation Annealed Large Deviations
Regeneration Times
Assume that P(limn→∞ Xn · ` = +∞) = 1.Define P(·) := P (·|Xn · ` ≥ 0, ∀n) .
(Xτ1 , τ1), (Xτ2 − Xτ1 , τ2 − τ1), (Xτ3 − Xτ2 , τ3 − τ2), . . .
independent sequence under Pi.i.d. under P
Moreover,
vP := limn→∞
Xn
n=
EXτ1
Eτ1
Jonathon Peterson 7/24/2008 26 / 35
Thesis Presentation Annealed Large Deviations
Regeneration Times
Assume that P(limn→∞ Xn · ` = +∞) = 1.Define P(·) := P (·|Xn · ` ≥ 0, ∀n) .
(Xτ1 , τ1), (Xτ2 − Xτ1 , τ2 − τ1), (Xτ3 − Xτ2 , τ3 − τ2), . . .
independent sequence under Pi.i.d. under P
Moreover,
vP := limn→∞
Xn
n=
EXτ1
Eτ1
Jonathon Peterson 7/24/2008 26 / 35
Thesis Presentation Annealed Large Deviations
The function IDefine for λ ∈ Rd+1
Λ(λ) := log Eeλ·(Xτ1 ,τ1),
andI(x , t) := sup
λ∈Rd+1λ · (x , t)− Λ(λ).
Cramer’s Theorem:(
Xτkk , τk
k
)∈ Rd+1 satisfies a LDP under P with rate
function I.
I(x , t) is convex.I(Eτ1vP ,Eτ1) = 0.Λ(λ) is analytic in the interior of its domain and I(x , t) is convexand analytic in a neighborhood of (Eτ1vP ,Eτ1).
Jonathon Peterson 7/24/2008 27 / 35
Thesis Presentation Annealed Large Deviations
The function IDefine for λ ∈ Rd+1
Λ(λ) := log Eeλ·(Xτ1 ,τ1),
andI(x , t) := sup
λ∈Rd+1λ · (x , t)− Λ(λ).
Cramer’s Theorem:(
Xτkk , τk
k
)∈ Rd+1 satisfies a LDP under P with rate
function I.
I(x , t) is convex.I(Eτ1vP ,Eτ1) = 0.Λ(λ) is analytic in the interior of its domain and I(x , t) is convexand analytic in a neighborhood of (Eτ1vP ,Eτ1).
Jonathon Peterson 7/24/2008 27 / 35
Thesis Presentation Annealed Large Deviations
The function J
Let
J(v) := infr∈(0,1]
r I(
vr,1r
).
J(v) is convex.J(v) is analytic in a neighborhood of vP .
We want to show
limδ→∞
lim supn→∞
1n
log P(‖Xn
n− v‖ < δ
)≤ −J(v),
and
limδ→∞
lim infn→∞
1n
log P(‖Xn
n− v‖ < δ
)≥ −J(v).
For convenience we’ll work with P instead of P.
Jonathon Peterson 7/24/2008 28 / 35
Thesis Presentation Annealed Large Deviations
The function J
Let
J(v) := infr∈(0,1]
r I(
vr,1r
).
J(v) is convex.J(v) is analytic in a neighborhood of vP .
We want to show
limδ→∞
lim supn→∞
1n
log P(‖Xn
n− v‖ < δ
)≤ −J(v),
and
limδ→∞
lim infn→∞
1n
log P(‖Xn
n− v‖ < δ
)≥ −J(v).
For convenience we’ll work with P instead of P.
Jonathon Peterson 7/24/2008 28 / 35
Thesis Presentation Sketch of Proof
Sketch of the proof
Idea:
P(Xn ≈ nv) ≈ P(Xτk ≈ nv , τk ≈ n, k = rn)
≈ e−nrI( vr ,
1r )
Lower bound:Force Xτk ≈ nv and τk ≈ n for some k = rn.Choose optimal r .
Upper bound:Harder. Need to show that above strategy is optimal.That is, rule out long regeneration times.
Jonathon Peterson 7/24/2008 29 / 35
Thesis Presentation Sketch of Proof
Sketch of the proof
Idea:
P(Xn ≈ nv) ≈ P(Xτk ≈ nv , τk ≈ n, k = rn)
≈ e−nrI( vr ,
1r )
Lower bound:Force Xτk ≈ nv and τk ≈ n for some k = rn.Choose optimal r .
Upper bound:Harder. Need to show that above strategy is optimal.That is, rule out long regeneration times.
Jonathon Peterson 7/24/2008 29 / 35
Thesis Presentation Sketch of Proof
Sketch of the proof
Idea:
P(Xn ≈ nv) ≈ P(Xτk ≈ nv , τk ≈ n, k = rn)
≈ e−nrI( vr ,
1r )
Lower bound:Force Xτk ≈ nv and τk ≈ n for some k = rn.Choose optimal r .
Upper bound:Harder. Need to show that above strategy is optimal.That is, rule out long regeneration times.
Jonathon Peterson 7/24/2008 29 / 35
Thesis Presentation Sketch of Proof
Lower bound
Fix r ∈ (0,1], and let k = rn.
1n
log P(‖Xn − nv‖ < 2δn)
≥ 1n
log P(‖Xτk − nv‖ < δn, |τk − n| < δn)
=rk
log P(‖
Xτk
k− v
r‖ < δ
r, |τk
k− 1
r| < δ
r
)Limit as k →∞ and then δ → 0 : r I
( vr ,
1r
).
This lower bound is true for all r ∈ (0,1] and so
limδ→∞
lim infn→∞
1n
log P(‖Xn
n− v‖ < δ
)≥ − inf
r∈(0,1]r I(
vr,1r
).
Note: Lower bound holds for any v · ` > 0.
Jonathon Peterson 7/24/2008 30 / 35
Thesis Presentation Sketch of Proof
Lower bound
Fix r ∈ (0,1], and let k = rn.
1n
log P(‖Xn − nv‖ < 2δn)
≥ 1n
log P(‖Xτk − nv‖ < δn, |τk − n| < δn)
=rk
log P(‖
Xτk
k− v
r‖ < δ
r, |τk
k− 1
r| < δ
r
)Limit as k →∞ and then δ → 0 : r I
( vr ,
1r
).
This lower bound is true for all r ∈ (0,1] and so
limδ→∞
lim infn→∞
1n
log P(‖Xn
n− v‖ < δ
)≥ − inf
r∈(0,1]r I(
vr,1r
).
Note: Lower bound holds for any v · ` > 0.
Jonathon Peterson 7/24/2008 30 / 35
Thesis Presentation Sketch of Proof
Lower bound
Fix r ∈ (0,1], and let k = rn.
1n
log P(‖Xn − nv‖ < 2δn)
≥ 1n
log P(‖Xτk − nv‖ < δn, |τk − n| < δn)
=rk
log P(‖
Xτk
k− v
r‖ < δ
r, |τk
k− 1
r| < δ
r
)Limit as k →∞ and then δ → 0 : r I
( vr ,
1r
).
This lower bound is true for all r ∈ (0,1] and so
limδ→∞
lim infn→∞
1n
log P(‖Xn
n− v‖ < δ
)≥ − inf
r∈(0,1]r I(
vr,1r
).
Note: Lower bound holds for any v · ` > 0.
Jonathon Peterson 7/24/2008 30 / 35
Thesis Presentation Sketch of Proof
Upper bound
First, note that by Chebychev’s inequality
P(Xτk = x , τk = t) ≤ e−λ·(x ,t)Eeλ·(Xτk ,τk )
= e−λ·(x ,t)+kΛ(λ) = e−k(λ·( xk ,
tk )−Λ(λ)).
True for any λ ∈ Rd+1 thus
P(Xτk = x , τk = t) ≤ e−kI( xk ,
tk ) = e−t k
t I( xt
tk ,
tk ) ≤ e−tJ( x
t ).
Note: The final bound does not depend on k .
Would like to say that
P(Xn ≈ nv) ≤ C P(∃k : Xτk ≈ nv , τk ≈ n).
Jonathon Peterson 7/24/2008 31 / 35
Thesis Presentation Sketch of Proof
Upper bound
First, note that by Chebychev’s inequality
P(Xτk = x , τk = t) ≤ e−λ·(x ,t)Eeλ·(Xτk ,τk )
= e−λ·(x ,t)+kΛ(λ) = e−k(λ·( xk ,
tk )−Λ(λ)).
True for any λ ∈ Rd+1 thus
P(Xτk = x , τk = t) ≤ e−kI( xk ,
tk ) = e−t k
t I( xt
tk ,
tk ) ≤ e−tJ( x
t ).
Note: The final bound does not depend on k .
Would like to say that
P(Xn ≈ nv) ≤ C P(∃k : Xτk ≈ nv , τk ≈ n).
Jonathon Peterson 7/24/2008 31 / 35
Thesis Presentation Sketch of Proof
Upper bound
First, note that by Chebychev’s inequality
P(Xτk = x , τk = t) ≤ e−λ·(x ,t)Eeλ·(Xτk ,τk )
= e−λ·(x ,t)+kΛ(λ) = e−k(λ·( xk ,
tk )−Λ(λ)).
True for any λ ∈ Rd+1 thus
P(Xτk = x , τk = t) ≤ e−kI( xk ,
tk ) = e−t k
t I( xt
tk ,
tk ) ≤ e−tJ( x
t ).
Note: The final bound does not depend on k .
Would like to say that
P(Xn ≈ nv) ≤ C P(∃k : Xτk ≈ nv , τk ≈ n).
Jonathon Peterson 7/24/2008 31 / 35
Thesis Presentation Sketch of Proof
Upper bound
Since P is non-nestling, τ1 has exponential tails:
P(τ1 ≥ εn) ≤ Ce−Cεn.
Fix ε small.Since J(vP) = 0, J(v) < Cε in a neighborhood of vP .
Thus we may assume τk − τk−1 < εn for all k ≤ n.Need an upper bound for
P(∃k : τk ∈ (n − εn,n], ‖Xn − nv‖ < nδ, τk+1 > n).
Jonathon Peterson 7/24/2008 32 / 35
Thesis Presentation Sketch of Proof
The event τk ∈ (n − εn,n], ‖Xn − nv‖ < nδ, τk+1 > n impliesτk = (1− s)n for some s ∈ [0, ε)
‖Xτk − nv‖ < n(δ + s)
τk+1 − τk > ns
P(∃k : τk ∈ (n − εn,n], ‖Xn − nv‖ < nδ, τk+1 > n)
≤∑k≤n
∑s∈[0,ε)
∑‖x−v‖<δ+s
P(τk = (1− s)n, Xτk = xn)P(τ1 > ns)
≤ Cnd+2 sups∈[0,ε)
sup‖x−v‖<δ+s
e−n(1−s)J( x1−s )e−Csn
Claim: Since J(v) is quadratic near vP , for v near vP
infs∈[0,ε)
inf‖x−v‖<δ+s
(1− s)J(
x1− s
)+ Cs = inf
‖x−v‖<δJ(x).
Jonathon Peterson 7/24/2008 33 / 35
Thesis Presentation Sketch of Proof
The event τk ∈ (n − εn,n], ‖Xn − nv‖ < nδ, τk+1 > n impliesτk = (1− s)n for some s ∈ [0, ε)
‖Xτk − nv‖ < n(δ + s)
τk+1 − τk > ns
P(∃k : τk ∈ (n − εn,n], ‖Xn − nv‖ < nδ, τk+1 > n)
≤∑k≤n
∑s∈[0,ε)
∑‖x−v‖<δ+s
P(τk = (1− s)n, Xτk = xn)P(τ1 > ns)
≤ Cnd+2 sups∈[0,ε)
sup‖x−v‖<δ+s
e−n(1−s)J( x1−s )e−Csn
Claim: Since J(v) is quadratic near vP , for v near vP
infs∈[0,ε)
inf‖x−v‖<δ+s
(1− s)J(
x1− s
)+ Cs = inf
‖x−v‖<δJ(x).
Jonathon Peterson 7/24/2008 33 / 35
Thesis Presentation Sketch of Proof
The event τk ∈ (n − εn,n], ‖Xn − nv‖ < nδ, τk+1 > n impliesτk = (1− s)n for some s ∈ [0, ε)
‖Xτk − nv‖ < n(δ + s)
τk+1 − τk > ns
P(∃k : τk ∈ (n − εn,n], ‖Xn − nv‖ < nδ, τk+1 > n)
≤∑k≤n
∑s∈[0,ε)
∑‖x−v‖<δ+s
P(τk = (1− s)n, Xτk = xn)P(τ1 > ns)
≤ Cnd+2 sups∈[0,ε)
sup‖x−v‖<δ+s
e−n(1−s)J( x1−s )e−Csn
Claim: Since J(v) is quadratic near vP , for v near vP
infs∈[0,ε)
inf‖x−v‖<δ+s
(1− s)J(
x1− s
)+ Cs = inf
‖x−v‖<δJ(x).
Jonathon Peterson 7/24/2008 33 / 35
Thesis Presentation Sketch of Proof
Claim: Since J(v) is quadratic near vP , for v near vP
infs∈[0,ε)
inf‖x−v‖<δ+s
(1− s)J(
x1− s
)+ Cs = inf
‖x−v‖<δJ(x).
eeeeeeee
x vP
C
Jonathon Peterson 7/24/2008 34 / 35
Thesis Presentation Sketch of Proof
Claim: Since J(v) is quadratic near vP , for v near vP
infs∈[0,ε)
inf‖x−v‖<δ+s
(1− s)J(
x1− s
)+ Cs = inf
‖x−v‖<δJ(x).
eeeeeeee
x x1−s vP
C
Jonathon Peterson 7/24/2008 34 / 35
Thesis Presentation Sketch of Proof
Claim: Since J(v) is quadratic near vP , for v near vP
infs∈[0,ε)
inf‖x−v‖<δ+s
(1− s)J(
x1− s
)+ Cs = inf
‖x−v‖<δJ(x).
eeeeeeee
x x1−s vP
C
(1− s)J(
x1−s
)+ Cs
Jonathon Peterson 7/24/2008 34 / 35
Thesis Presentation Future Work:
Other Results and Future Work:
When d = 1, have shown H(v) = J(v) for all v > 0 (even innestling case).When d ≥ 2, does H(v) = J(v) for all v · ` > 0?Analytic behavior of H(v) for ”speedup” in nestling case?Can anything be done for v · ` < 0?
Jonathon Peterson 7/24/2008 35 / 35