Moderate deviations for random fields and random complex zeroes

0.45 MB·English

by  Boris Tsirelson

Moderate deviations for random fields and
random complex zeroes
Boris Tsirelson
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Abstract
a
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Moderate deviations for random complex zeroes are deduced from
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a new theorem on moderate deviations for random fields.

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R
P Contents
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h
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a 1 A chain of inequalities 3
m
[ 2 Splittable random processes 17

1
v 3 Logarithm of a Gaussian process 26
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5
0 4 An example of a splittable process 33
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1 5 Splittable random fields 36
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6 An example of a splittable field 47
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7 Random complex zeroes 53
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r
a
Introduction
Recent results on moderate deviations for random complex zeroes [7] are
a challenge for probability theory, since they involve complex analysis, in
contrast to asymptotic normality obtained via random fields [8]. Taking up
the challenge, I deduce moderate deviations for random complex zeroes from
a new general theorem on moderate deviations for random fields. As a by-
product, the same general theorem gives the asymptotic normality, avoiding
1Thisresearchwassupportedbytheisraelsciencefoundation(grantNo.683/05).
1
thediagramtechniques of[8]. However, Iconsideronlysmoothtestfunctions,
leaving aside interesting effects of sharp boundary [7].
The random complex zeroes are singularities of a stationary random field
on the complex plane (the logarithm of the absolute value of a normalized
Gaussiananalyticfunction). Thisrandomfieldtranscendstheexistingtheory
of moderate deviations (see [3], [4] and references therein) in several aspects:
(a) it is a random field on the plane, not a random process on the line;
(b) it has some, but not all exponential moments;
(c) it is some function of a Gaussian random field, but the function
(z ln z ) is singular at 0, which entangles moderate deviations of the
7→ | |
non-Gaussian field with small deviations of the underlying Gaussian field;
(d) the underlying Gaussian field is non-stationary.
The first part (Sections 1–4) contains the general result in dimension one
(random processes on the line). Dimension two (random fields on the plane)
is treated in the second part (Sections 5–6), using the first part. The third
part (Section 7) deals with random complex zeros.
Main results of the first part and the third part are formulated below.
1 Definition. A stationary random process X = (Xt)t R is splittable, if
E exp X < , EX = 0, and there exists (on some pr∈obability space) a
0 0
| | ∞
triple of random processes X0,X ,X+ such that
−
(a) the two processes X ,X+ are independent;
−
(b) the four processes X,X0,X ,X+ are identically distributed;
−
(c) E exp 0 |Xt− −Xt0|dt+ 0∞|Xt+ −Xt0|dt < ∞.
−∞
2 Theorem.(cid:0)RFor every splittable Rstationary random(cid:1) process X there exists
σ [0, ) such that for every compactly supported continuous function
∈ ∞
f : R R,
→
1 t σ2
lim lnE expλ ∞ f X dt = f 2 .
r→∞,λ→0 rλ2 r t 2 k kL2(R)
λlogr→0 Z−∞ (cid:16) (cid:17)
That is, for every ε there exist R and δ such that the given expression
is ε-close to the right-hand side for all r R and all λ = 0 such that
≥ 6
λ logr δ.
| | ≤
3 Corollary. Let X, σ and f be as in Theorem 2, and σ = 0. Then
6
1 t 1
r→∞lim,c→∞ c2 lnP f r Xtdt ≥ cσkfkL2(R)√r = −2 .
(clogr)2/r→0 (cid:18)Z (cid:16) (cid:17) (cid:19)
Unfortunately, the region of moderate deviations (r , c , c2
→ ∞ → ∞ r →
0) is not covered. The condition (clogr)2 0 leaves a small gap between
r →
Corollary 3 and large deviations (c2 = const).
r
2
4 Corollary. Let X, σ and f be as in Theorem 2. Then the distribu-
tion of r 1/2 f t X dt converges (as r ) to the normal distribution
− r t → ∞
N(0,σ2 f 2).
k k R (cid:0) (cid:1)
Consider now the random entire function ψ : C C defined by
→
∞ ζ zk
k
ψ(z) = ,
√k!
k=0
X
where ζ ,ζ ,... are independent standard complex Gaussian random vari-
0 1
ables.
5 Theorem. There exists an absolute constant σ (0, ) such that for
∈ ∞
every compactly supported C2-function h : C R,
→
1 z r2 σ2
lim lnE expλr2 h hdm = ∆h 2dm;
r→∞,λ→0 r2λ2 r − π 2 | |
λlog2r→0 (cid:18)z:ψX(z)=0 (cid:16) (cid:17) R (cid:19) R
here m is the Lebesgue measure on C, and ∆h is the Laplacian of h.
6 Corollary. Let σ and h be as in Theorem 5. Then
1 z r2 cσ 1
lim lnP h hdm ∆h 2dm = .
r→∞,c→∞ c2 r − π ≥ r | | −2
(clog2r)/r→0 (cid:18)z:ψX(z)=0 (cid:16) (cid:17) R qR (cid:19)
The same holds for ( h), of course.
−
7 Corollary. Let σ and h be as in Theorem 5. Then the distribution of
z r2
r h hdm
r − π
(cid:18)z:ψX(z)=0 (cid:16) (cid:17) R (cid:19)
converges (as r ) to the normal distribution N(0,σ2 ∆h 2dm).
→ ∞ | |
This is the asymptotic normality established in [8] usRing moments and
diagrams.
1 A chain of inequalities
Main results of this section, formulated below, areTheorem 1.2 (used inSect.
2), Proposition 1.5 (also used in Sect. 2), and Proposition 1.6 (used in Sect.
5, see 5.10).
3
1.1 Definition. Let X,Y be random variables (possibly on different prob-
ability spaces) and C [0, ). We say that Y is a C-duplication of X, if
∈ ∞
there exist random variables X ,X ,Z (on some probability space) such that
1 2
X and X are independent,
1 2
X ,X ,X are identically distributed,
1 2
X +X +Z and Y are identically distributed,
1 2
lnE expλZ Cλ2 for all λ [ 1,1].
≤ ∈ −
1.2 Theorem. Let random variables X ,X ,... (possibly on different prob-
1 2
ability spaces) and numbers C ,C , [0, ) be such that
1 2
··· ∈ ∞
(a) X is a C -duplication of X (for all n = 1,2,...);
n+1 n n
(b) sup (2θ) nC < for some θ < 1;
n − n ∞
(c) E expε X < for some ε > 0;
1
(cid:0) | | ∞(cid:1)
(d) EX = 0.
1
Then the following limit exists:
1
lim lnE expλX .
n ,λn 02nλ2 n
→∞ →
Given a function f : (0, ) [0, ] and a number C [0, ), we define
∞ → ∞ ∈ ∞
another function f [C] : (0, ) [0, ] as follows:
+
∞ → ∞
2 p
f [C](λ) = inf f(pλ)+ Cλ2 for λ (0,1),
+
(1.3) p [1/(1 λ), ) p p 1 ∈
∈ − ∞ −
f [C](λ) = for λ [1, ).
+
∞ ∈ ∞
Further, we define recursively for n = 0,1,2,...
(1.4) f [C ,...,C ] = (f [C ,...,C ]) [C ].
+ 0 n+1 + 0 n + n+1
1.5 Proposition. For every ε,θ (0,1),
∈
1
limsup f [C ,...,C ](λ)
2n+1λ2 + 0 n ≤
n ,λn 0+
→∞ →
1 1 1 C
n
limsup f(λ)+ sup .
≤ 1 ε λ 0 λ2 2ε(1 √θ)2 n (2θ)n
− → −
1.6 Proposition. For every ε,θ (0,1), n and λ,
∈
(a) if 0 < λ εθn/2(1 √θ) then
≤ −
1 λ λ2 1 C
k
f [C ,...,C ](λ) (1 ε)f + max ;
2n+1 + 0 n ≤ − 1 ε 2ε(1 √θ)2 k=0,...,n (2θ)k
(cid:16) − (cid:17) −
4
(b) let m 0,1,...,n 1 be such that (n m)λ < 1 and µ εθm/2(1
∈ { − } − ≤ −
√θ), where µ = λ ; then
1 (n m)λ
− −
1
f [C ,...,C ](λ)
2n+1 + 0 n ≤
µ µ2 1 λ θm+1 C
k
(1 ε)f + + max .
≤ − 1 ε 2ε(1 √θ)2 21 θ k=0,...,n (2θ)k
(cid:16) − (cid:17) (cid:18) − − (cid:19)
The proof of Th. 1.2 uses Prop. 1.5 (whose proof uses Prop. 1.6). The
relevance of (1.3)-(1.4) and 1.5 to 1.2 stems from Lemmas 1.7, 1.8. The
proofs of Propositions 1.6 and 1.5 are given after Lemmas 1.9, 1.12. For the
proof of Th. 1.2 see the end of this section.
1.7 Lemma. For all random variables X,Y and all p (1, ),
∈ ∞
1 1
plnE exp X (p 1)lnE exp Y
p − − − p 1 ≤
(cid:16) − (cid:17)
1 p 1 p
lnE exp(X +Y) lnE exppX + − lnE exp Y .
≤ ≤ p p p 1
−
(In the lower bound we interpret as .)
∞−∞ −∞
Proof. By the H¨older inequality,
p (p 1)/p
E exp(X +Y) = E(expX expY) (E exppX)1/p E exp Y − ;
· ≤ p 1
(cid:16) − (cid:17)
the upper boundfollows. Weapply the upper bound to 1(X+Y) and 1Y
p −p
instead of X,Y:
(cid:0) (cid:1)
1 1 p 1 1
lnE exp X lnE exp(X +Y)+ − lnE exp Y ;
p ≤ p p − p 1
(cid:16) − (cid:17)
the lower bound follows.
1.8 Lemma. (a) Let random variables X,Y and a function f : (0, )
∞ →
[0, ] be such that
∞
Y is a C-duplication of X ,
lnE expλX f(λ) for all λ (0, ).
≤ ∈ ∞
Then
lnE expλY f [C](λ) for all λ (0, ).
+
≤ ∈ ∞
5
(b) Let random variables X,Y,Z and a function f : (0, ) [0, ] be
∞ → ∞
such that
X,Y are independent,
lnE expλX f(λ) and lnE expλY f(λ) for all λ (0, ),
≤ ≤ ∈ ∞
lnE expλZ Cλ2 for all λ [0,1].
≤ ∈
Then
lnE expλ(X +Y +Z) f [C](λ) for all λ (0, ).
+
≤ ∈ ∞
Proof. Item (a) is a special case of Item (b). Item (b) follows from Lemma
1.7 (the upper bound) applied to λ(X +Y) and λZ.
1.9 Lemma.
n
1 λ λ
f [C ,...,C ](λ) 1 (n+1)λ f + 2 kC
2n+1 + 0 n ≤ − 1 (n+1)λ 2 − k
(cid:0) (cid:1) (cid:16) − (cid:17) Xk=0
1
for 0 < λ < .
n+1
Proof. Induction in n. For n = 0, the needed inequality
1 λ λ
(1.10) f [C](λ) (1 λ)f + C for 0 < λ < 1
+
2 ≤ − 1 λ 2
(cid:16) − (cid:17)
follows from (1.3) for p = 1 . For n > 0, denoting f [C ,...,C ] by g,
1 λ + 0 n 1
the assumed inequality for n− 1 takes the form −
−
n 1
1 λ λ − 1
(1.11) g(λ) 1 nλ f + 2 kC for 0 < λ < ,
2n ≤ − 1 nλ 2 − k n
(cid:0) (cid:1) (cid:16) − (cid:17) Xk=0
while the needed inequality for n becomes
n
1 λ λ
g [C ](λ) 1 (n+1)λ f + 2 kC
2n+1 + n ≤ − 1 (n+1)λ 2 − k
(cid:0) (cid:1) (cid:16) − (cid:17) Xk=0
1
for 0 < λ < .
n+1
Let 0 < λ < 1 . By (1.10), 1g [C ](λ) (1 λ)g(µ) + λC , where
n+1 2 + n ≤ − 2 n
µ = λ . We note that 1 nµ = 1−(n+1)λ > 0, µ = λ and get from
1 λ − 1 λ 1 nµ 1 (n+1)λ
− − − −
(1.11)
n 1
1 1 (n+1)λ λ µ −
g(µ) − f + 2 kC .
− k
2n ≤ 1 λ 1 (n+1)λ 2
− (cid:16) − (cid:17) Xk=0
6
Thus,
1 1 λ λ
g [C ](λ) − g(µ)+ C
2n+1 + n ≤ 2n 2n+1 n ≤
n 1
λ λ − λ
1 (n+1)λ f + 2 kC + 2 nC .
− k − n
≤ − 1 (n+1)λ 2 2
(cid:0) (cid:1) (cid:16) − (cid:17) Xk=0
1.12 Lemma. For every ε (0,1),
∈
1 λ λ2 n 2
f [C ,...,C ](λ) (1 ε)f + 2 k/2 C
2n+1 + 0 n ≤ − 1 ε 2ε − k
(cid:16) − (cid:17) (cid:18)Xk=0 p (cid:19)
min 2 k/2√C
k=0,...,n − k
for 0 < λ ε .
≤ 2 k/2√C
k=0,...,n − k
Proof. Induction in n. For n = 0, the needed inequaliPty
1 λ λ2
f [C](λ) (1 ε)f + C for 0 < λ ε
+
2 ≤ − 1 ε 2ε ≤
(cid:16) − (cid:17)
follows from (1.3) for p = 1 . For n > 0, we write the assumed inequality
1 ε
for n 1 in the form −
−
1 λ λ2
(1.13) g(λ ) (1 ε )f old + old σ2
2n old ≤ − old 1 ε 2ε old
old old
(cid:16) − (cid:17) m
old
whenever 0 < λ ε and 0 < ε < 1;
old old old
≤ σ
old
here g = f+[C0,...,Cn 1], σold = kn=−012−k/2√Ck and mold =
min 2 k/2√C . We−have to prove that
k=0,...,n 1 − k
− P
1 λ λ2
(1.14) g [C ](λ) (1 ε)f + σ2
2n+1 + n ≤ − 1 ε 2ε
(cid:16) − (cid:17) m
whenever 0 < λ ε and 0 < ε < 1;
≤ σ
hereσ = n 2 k/2√C = σ +2 n/2√C andm = min 2 k/2√C =
k=0 − k old − n k=0,...,n − k
min(m ,2 n/2√C ).
old − n
P
By (1.3) for p = 1 ,
1 εnew
−
λ C λ2
n
g [C ](λ) 2(1 ε )g +
+ n new
≤ − 1 ε ε
new new
(cid:16) − (cid:17)
whenever 0 < λ ε and 0 < ε < 1.
new new
≤
7
We combine it with (1.13) for λ = λ :
old 1 εnew
−
1
g [C ](λ)
2n+1 + n ≤
λ λ2 C λ2
(1 ε ) (1 ε )f + σ2 + n .
− new − old (1 ε )(1 ε ) 2ε (1 ε )2 old 2n+1ε
(cid:18) (cid:16) − new − old (cid:17) old − new (cid:19) new
In order to get (1.14) it remains to find ε ,ε (0,1) such that (1
old new
∈ −
ε )(1 ε ) = 1 ε,
old new
− −
σ2 C σ2
(1.15) old + n
2ε (1 ε ) 2n+1ε ≤ 2ε
old new new
−
and λ ε , λ ε mold.
We≤definenwe 1−εnew ≤ oldσold
σ εx ε
old
x = , ε = , ε = .
old new
2 n/2√C x+1 ε x+1
− n −
Clearly, ε < 1 (since εx < x < x+1 ε) and ε < 1 (since ε < 1 < x+1).
old new
−
Also,
x+1 ε εx x+1 ε
(1 ε )(1 ε ) = − − − = 1 ε.
old new
− − x+1 ε · x+1 −
−
Taking into account that 2−n/2√Cn = σoxld and σoσld = x+x1 = (1−εneεw)εold we
get
m ε σ
λ ε = min m , =
old
≤ σ σ x+1
(cid:16) εm (cid:17)ε (1 ε )ε
old new old
= min , = min m − ,ε .
old new
σ x+1 · σ
old
(cid:16) (cid:17) (cid:16) (cid:17)
Finally, we check (1.15):
σ2 C σ2 x+1 1 σ 2
old + n = old + old =
2ε (1 ε ) 2n+1ε 2 · εx 2ε x
old new new new
− (cid:16) (cid:17)
σ2 (x+1) (x+1)σ2 σ2 (x+1)
= old + old = old (x+1) =
2εx 2εx2 2εx2
1 σ (x+1) 2 σ2
old
= = .
2ε x 2ε
(cid:16) (cid:17)
8
The reader may wonder, how did I found these formulas for ε ,ε in
old new
the proof of Lemma 1.12. In fact, I have minimized the left-hand side of
(1.15) in ε ,ε restricted by (1 ε )(1 ε ) = 1 ε. The expression
old new old new
− − −
2 k/2√C 2 have appeared afterwards.
− k
P(cid:0)Proof of Prop.(cid:1)1.6. Denotemax Ck by M. Without loss of generality
k=0,...,n (2θ)k
we assume that C = M(2θ)k for all k. Then
k
n √M
2 k/2 C and min 2 k/2 C = √Mθn/2.
− k − k
≤ 1 √θ k=0,...,n
Xk=0 p − p
Item (a) follows immediately from Lemma 1.12.
Proving Item (b), we use Lemma 1.12 for passing from f to g =
f [C ,...,C ], and Lemma 1.9 for passing from g to g [C ,...,C ] =
+ 0 m + m+1 n
f [C ,...,C ]. Namely,
+ 0 n
1 µ µ2 M
f [C ,...,C ](µ) (1 ε)f +
2m+1 + 0 m ≤ − 1 ε 2ε(1 √θ)2
(cid:16) − (cid:17) −
and
1
g [C ,...,C ](λ)
2n m + m+1 n ≤
−
n
λ λ
1 (n m)λ g + 2 (k m 1)C .
− − − k
≤ − − 1 (n m)λ 2
(cid:0) (cid:1) (cid:16) − − (cid:17) k=Xm+1
Proof of Prop. 1.5. Let sup Cn = M < (otherwise there is nothing to
n (2θ)n ∞
prove). We consider two overlapping cases separately.
The first case: λ θn/2.
≪
Prop. 1.6(a) gives eventually (when θ n/2λ is small enough),
−
1 λ λ2 M
f [C ,...,C ](λ) (1 ε)f + .
2n+1 + 0 n ≤ − 1 ε 2ε(1 √θ)2
(cid:16) − (cid:17) −
It remains to divide by λ2 and note that
1 λ 1 1
(1 ε)limsup f = limsup f(λ).
− λ2 1 ε 1 ε λ2
λ 0+ λ 0+
→ (cid:16) − (cid:17) − →
The second case: λ θn.
≫
9
We choose m 0,1,...,n 1 such that
∈ { − }
λ2 θm λ,
≪ ≪
apply Prop. 1.6(b), observe that µ λ, λθm+1 = o(λ2) and get
∼ 2 1 θ
−
1 µ µ2 1
f [C ,...,C ](λ) (1 ε)f + +o λ2 M .
2n+1 + 0 n ≤ − 1 ε 2ε(1 √θ)2
(cid:16) − (cid:17) (cid:18) − (cid:19)
(cid:0) (cid:1)
It remains to divide by λ2 and note that
1 µ 1 1
(1 ε)limsup f = limsup f(λ).
− λ2 1 ε 1 ε λ2
λ 0+
(cid:16) − (cid:17) − →
We are not yet in position to prove Theorem 1.2, since the chains of
inequalities considered above are based on the upper bound of Lemma 1.7.
We need similar results on lower bounds.
Given f : (0, ) [0, ] and C [0, ), we define f [C] : (0, )
∞ → ∞ ∈ ∞ − ∞ →
[0, ] by
∞
λ 1
(1.16) f [C](λ) = sup 2pf Cλ2 for λ (0, ).
− p [λ+1, ) p − p 1 ∈ ∞
∈ ∞ (cid:16) (cid:17) −
(The supremum is nonnegative, since the limit as p is nonnegative.)
→ ∞
Further, we define recursively for n = 0,1,2,...
(1.17) f [C ,...,C ] = (f [C ,...,C ]) [C ].
0 n+1 0 n n+1
− − −
Lemmas 1.18, 1.19, 1.22 and Propositions 1.26, 1.27 are lower-bound
counterparts of 1.8, 1.9, 1.12, 1.6 and 1.5. Proofs are quite similar, but many
small changes in formulas look unpredictable, especially in the proof of 1.22.
1.18 Lemma. (a) Let random variables X,Y and a function f : (0, )
∞ →
[0, ] be such that
∞
Y is a C-duplication of X ,
lnE expλX f(λ) for all λ (0, ).
≥ ∈ ∞
Then
lnE expλY f [C](λ) for all λ (0, ).
≥ − ∈ ∞
10