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ON THE KAC-RICE FORMULA
LIVIU I. NICOLAESCU
Abstract. This is an informal introduction to the Kac-Rice
formula and some of its (mostlyone-dimensional) applications.
Contents
1. Gaussian random functions 12. The one-dimensional Kac-Rice
formula 43. Some applications 94. Energy landscapes 15Appendix A. A
probabilistic dictionary 17References 19
1. Gaussian random functions
Let I be an interval of the real axis. Fix smooth functions
f0, f1, . . . , fN : I → R,and independent normal random
variables Xk, k = 0, 1, . . . , N , of mean 0 and variance vk >
0,defined on the same probability space (Ω,A,P ). Form the linear
combination
F (t) = Fω(t) =N∑k=0
Xk(ω)fk(t). (1.1)
This is an example of random function. The value of this random
function at each pointt ∈ I is a Gaussian random variable.
As probability space Ω we can take Ω = RN+1 equiped with the
Gaussian measure
P (dω0 · · · dωN ) =
N∏k=0
e−
ω2k2vk
√2πvk
dω0 · · · dωN .The random variables Xk are then the coordinate
functions
Xk(ω0, . . . , ωN ) = ωk, ∀k = 0, . . . , N.
The covariance kernel of this random function is the function K
: I × I → R defined by
K(s, t) = E(F (s) · F (t)
)=
N∑j,k=0
fj(s)fk(t)E(XjXk).
Date: Started September 16,2014. Completed on October 9, 2014.
Last revision October 10, 2014.Notes for a Graduate Seminar talk,
October 2014.
1
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2 LIVIU I. NICOLAESCU
Since E(X2k) = vk and the variables Xj , Xk, j 6= k, are
independent we have
E(XjXk) = E(Xj)E(Xk) = 0, ∀j 6= k.
We deduce
K(s, t) =N∑k=0
vkfk(s)fk(t). (1.2)
We denote by Ẑ(F, I) the expected number of zeros of F on
I.
Example 1.1 (Random polynomials). FixN independent normal random
variablesX0, . . . , XN ,with mean 0 and var(Xk) = vk, ∀k. Then
F (t) = X0 +X1tk + · · ·+XN tN .
is a polynomial of degree ≤ N whose coefficients are independent
random normal variables.Observe that
F (−t) =N∑k=0
(−1)kXktk.
The random variables (−1)kXk are independent normal variables
with mean 0 and variancesvk. Thus the random polynomials F (t) and
F
−(t) = F (−t) have the same statistics and wededuce
Ẑ(F, [0,∞)
)= Ẑ
(F, (−∞, 0]
).
Suppose additionally that the variances vk satisfy the symmetry
condition
vk = vN−k, ∀k = 0, 1, . . . , N. (1.3)
In this case the random polynomials F (t) and
F ∗(t) = tNF (t−1) =n∑k=0
XN−ktk
have the same statistics and we deduce that
Ẑ(F, (0, 1)
)= Ẑ
(F, (1,∞)
).
We deduce that if F satisfies the symmetry condition (1.3),
then
Ẑ(F,R
)= 4Ẑ
(F, (0, 1)
)= 4Ẑ
(F, (1,∞)
). (1.4)
Here are two famous examples of random polynomials satisfying
the symmetry condition(1.3).
The Kac statistics. In this case the random variables have the
same variance
v0 = v1 = · · · = vN = 1.
We deduce that in this case the covariance kernel is
K(s, t) =N∑k=0
(st)k =1− (st)N+1
1− st. (1.5)
The Kostlan statistics. In this case the variances are
vk =
(N
k
), k = 0, 1, . . . , n.
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ON THE KAC-RICE FORMULA 3
In this case the covariance kernel is
K(s, t) =N∑k=0
(N
k
)(st)k =
(1 + st
)N. (1.6)
The above construction can be generalized as follows. Consider a
family of polynomials
p0(t), p1(t), . . . , pN (t), deg pk(t) = k, ∀k = 0, 1, 2, . . .
N.Fix N independent normal random variables X0, . . . , XN ,
var(Xk) = vk, ∀k. Then
F (t) = X0p0(t) +X1p1(t) + · · ·+XNpN (t)random polynomial of
degree N .
The Legendre statistics. Recall that the Legendre polynomials
are obtained from thesequence of monomials (tk)k≥0 by applying the
Gramm-Schmidt procedure with respect tothe inner product in L2([−1,
1], dt). Concretely, the degree n Legendre polynomial is
pn(t) =
√n+
1
2`n(t), `n(t) =
1
2nn!
dn
dtn(t2 − 1)n. (1.7)
We can construct a random polynomial
FN (t) =N∑k=0
Xkpk(t),
where Xk are independent standard normal random variables, Xk ∈
N(1), ∀k. Using theChristoffel-Darboux theorem [22] we deduce that
its covariance kernel is given by
KN (s, t) =N + 1
2
`N+1(t)`N (s)− `N+1(s)`N (t)t− s
. ut
Example 1.2 (Random trigonometric polynomials). We assume I =
[0, 2π], N is even,N = 2m and
f0 = 1, f2k−1(t) = sin(kt), fk(t) = cos(kt).
Asumev2k−1 = v2k = 2rk > 0.
For uniformity we set r0 = v0. In this case we have
K(s, t) = r0 +m∑k=1
2r0(
cos(ks) cos(kt) + sin(ks) sin(kt))
= r0 + 2m∑k=1
rk cos k(t− s) = r0 +m∑k=1
rk(eik(s−t) + e−ik(t−s)
).
(1.8)
In the special case when r0 = r1 = · · · = rm = 1 we deduce
K(s, t) = 1 + 2
m∑k=1
cos k(t− s) =m∑
k=−meik(t−s)
=e
i2
(2m+1)(t−s) − e−i2
(2m+1)(t−s)
ei2
(t−s) − e−i2
(t−s)=
sin (2m+1)(t−s)2sin t−s2
.
(1.9)
ut
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4 LIVIU I. NICOLAESCU
For any interval I and any function F : I → R we denote by Z(F,
I) the number ofzeros of f . When F is a random function Z(F, I) is
a random variable and we would like toinvestigate some of its
statistical invariants. The simplest such invariants is its
expectation.The Kac-Rice formula gives a description of the
expected number of zeros of the randomfunction F in the interval
I.
2. The one-dimensional Kac-Rice formula
Consider the random function defined in the previous section
F : I → R, F (t) =N∑k=0
Xkfk(t).
We would like to compute the expected number of zeros of F in I.
The key to this computationis Kac’s counting formula. For any ε
> 0 define (see Figure 1)
ηε : R→ R, ηε(y) =
{12ε , |y| < ε,0, |y| ≥ ε.
Figure 1. Approximating the Dirac function.
Let us point out that the family ηε converges as ε → 0 to
Dirac’s δ-function, i.e., for anycontinuous function f : R→ R we
have
limε↘0
∫Rηε(t)f(t) = f(0).
For every C1-function F : I → R we set
Zε(F, I) :=
∫Iηε(F (t))|F ′(t)|dt.
Definition 2.1. (a) Fix a compact interval [a, b]. We say that a
C1-function f : [a, b] → Ris convenient if the following hold.
• f(a) · f(b) 6= 0.• All the zeros of f are nondegenerate, i.e.,
if f(t) = 0, then f ′(t) 6= 0.
(b) We say that a C1-function F : R → R is convenient if it is
proper and all its zeros arenondegernerate. ut
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ON THE KAC-RICE FORMULA 5
Lemma 2.2 (Kac’s counting formula). (a) Suppose that the
C1-function F : [a, b] → R isconvenient. Then
Z(F, [a, b]) = limε↘0
Zε(F, [a, b]).
(b) Suppose that the C1-function F : R→ R is convenient.
ThenZ(F,R) = lim
ε↘0Zε(F,R).
(c) Let I = [a, b] or I = R and suppose that F : I → R is a C1
convenient function with νzeros and κ 0 we have
Zε(F, I) ≤ ν + 2κ. (2.1)
Proof. (a) Since F is convenient, it has finitely many zeros τ1
< · · · < τν . The set C ofcritical points of F is compact
and disjoint from the zero set of F because F is
convenient.Thus
ε0 := minx∈C|F (x)| > 0.
Fix ε ∈ (0, ε0). Then ∫ baηε(F (t)
)|F ′(t)|dt = 1
2ε
∫{|F | 1 for |t| > a.The restriction of F to [−a, a] is
convenient and we have
Z(F, [−a, a]) = Z(F,R).Next observe that for ε ∈ (0, 1) we
have∫
Rηε(F (t))|F ′(t)|dt =
∫ a−aηε(F (t))|F ′(t)|dt.
Hence
limε↘0
∫Rηε(F (t))|F ′(t)|dt = lim
ε↘0
∫ a−aηε(F (t))| = F ′(t)|dt = Z(F, [−a, a]) = Z(F,R).
(c) Since F has only finitely many critical points we deduce
from Rolle’s theorem that forany c ∈ R the equation F (t) = c has
only finitely many solutions.
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6 LIVIU I. NICOLAESCU
Fix ε > 0.The connected components of the set {|F | < ε}
are bounded intervals (a, b) suchthat |F (a)| = |F (b)| = ε. Since
the equation |F (t)| = ε has only finitely many solutions wededuce
that {|F | < ε} has only finitely many components
J` = (a`, b`), ` = 1, . . . , L.
We have ∫Iηε(F (t))|F ′(t)|dt =
1
2ε
L∑`=1
∫J`
|F ′(t)|dt.
Denote by k` the number of turning points of F (t) in J`, i.e.,
the number of local pointswhere F ′(t) changes sign. Let observe
that if J` contains no turning points, then F is eitherincreasing
or decreasing on this interval so F (a`)F (b`) < 0 and thus J`
contains a uniquezero of F . In particular, if J` contains no
turning point, then
L∑`=1
∫J`
|F ′(t)|dt = 2ε.
We setL0 :=
{` = 1, . . . , `; J` contains no turning point
},
L1 :={` = 1, . . . , `; J` contains turning points
}.
The above discussion shows that
#L0 = ν, #L1 ≤ κ,and thus
1
2ε
L∑`=1
∫J`
|F ′(t)|dt = 12ε
∑`∈L0
∫J`
|F ′(t)|dt+ 12ε
∑`∈L1
∫J`
|F ′(t)|dt
= ν +1
2ε
∑`∈L1
∫J`
|F ′(t)|dt.
Let now ` ∈ L1 and supose that the turning points of F in J`
aret1 < · · · < tk` .
Then∫J`
|F ′(t)|dt = |F (a`)− F (t1)|+ |F (t1)− F (t2)|+ · · · |F (tk`)−
F (b`)| ≤ 2ε(k` + 1).
Hence1
2ε
∑`∈L1
∫J`
|F ′(t)|dt ≤∑`∈L1
(2k` + 1) ≤ κ+ #L1 ≤ 2κ.
ut
Let us apply Kac’s counting formula to the random function in
(1.1)
F = Fω : I → R, Fω(t) =N∑k=0
Xk(ω)fk(t), Xk : (Ω,A,P )→ R.
Here we assume that I is either a compact interval or I = R. We
make the followingassumptions about the random function.
The random function F is almost surely convenient. (A1)
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ON THE KAC-RICE FORMULA 7
∃C > 0 such that, almost surely, Z(F, I) + Z(F ′, I) < C.
(A2)Let us point out that the random polynomials in Example 1.1
satisfy these assumptions.As for the random trigonometric
polynomials in Example 1.2, assumption (A1) follows froma direct
application of Sard’s theorem, while assumption (A2) follows by
observing that atrigonometric polynomial of degree ≤ m has at most
2m zeros and at most 2m critical point.
Let Fω be convenient. Then Kac’s counting formula implies.
Z(Fω, I) = limε↘0
Zε(Fω, I)
Thus
Ẑ(F, I) = E(Z(F, I)
)=
∫ΩZ(Fω, I)P (dω) =
∫Ω
limε↘0
Zε(Fω, I)P (dω).
At this point we want to switch the order of operations∫
Ω and limε↘0. Using assumption(A2), Lemma 2.2(c) and Lebesque’s
dominated convergence theorem we deduce that∫
Ωlimε↘0
Zε(F, I)P (dω) = limε↘0
∫ΩZε(Fω, I)P (dω)
= limε↘0
E
(∫Iηε(F (t))|F ′(t)|dt
)= lim
ε↘0
∫IE(ηε(F (t)
)|F ′(t)|
)dt.
Hence
Ẑ(F, I) = limε↘0
∫IE(ηε(F (t)
)|F ′(t)|
)dt. (2.2)
To compute the expectation of ηε(F (t)
)|F ′(t)| we observe that for each t ∈ I the random
vector
Ω 3 ω 7→(F (t), F ′(t)
)∈ R2
is Gaussian and its covariance form is described by the
symmetric matrix
Ct =
[at btbt ct
],
where
at = E(F (t)2
), bt = E
(F (t)F ′(t)
), ct = E
(F ′(t)2
).
We can describe the entries of Ct in terms of the covariance
kernel K(s, t) = E(F (s)F (t)).More precisely, we have
at = K(t, t), bt = K′t(s, t)|s=t, ct = K ′′st(s, t)|s=t.
(2.3)
We set
∆t := detCt = atct − b2t , vt :=∆tat,
and we make another assumption on F (t), namely
∆t > 0, ∀t ∈ I. (A3)
This assumption is automatically satisfied for the random
functions in Examples 1.1 and 1.2provided that N is sufficiently
large.1 The Gaussian measure ΓC on R2 with covariance formC is
then
ΓCt(dxdy) =1
2π√
∆te− 1
2∆t(ctx2−2tbxy+aty2)dxdy.
1We need N ≥ 2 in Example 1.1 and m ≥ 2 in Example 1.2.
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8 LIVIU I. NICOLAESCU
We then have
E(ηε(F (t)
)|F ′(t)|
)=
1
2ε
∫|x|
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ON THE KAC-RICE FORMULA 9
polynomial of degree 2N − 2 and ∆t is an even polynomial of
degree ≤ 4N − 3. In particulardeg ∆t ≤ 4N − 4 and we deduce
√∆tat
+|bt|a2t
= O(t−2) as |t| → ∞.
We deduce
E(ηε(F (t)
)|F ′(t)|
)=
1
2ε
∫ ε−ε
Φt(x)dx ≤ µ(t).
Invoking (2.2), (2.4), (A4) and the dominated convergence
theorem we conclude that
Ẑ(F, I))
= limε↘0
∫IE(ηε(F (t)
)|F ′(t)|
)dt =
∫I
limε↘0
E(ηε(F (t)
)|F ′(t)|
)dt =
∫I
Φt(0).
Observe that
Φt(0) =2√
2πat
∫ ∞0γvt(y)dy =
1
π√atvt
∫ ∞0
ye− y
2
2vt dy =1
πρt, ρt :=
√∆tat
.
Observe that
ρ2t =atct − b2t
a2t=K(s, t)s=tK
′′st(s, t)s=t −K ′t(s, t)2s=tK(s, t)2s=t
= ∂2st logK(s, t)s=t.
We have thus proved the Kac-Rice formula.
Theorem 2.3. Let I = [a, b] or I = R. Suppose that f0, f1, . . .
, fN : I → R are smoothfunctions and Xk ∈ N(vk), k = 0, . . . , N
are independent normal random variables. Weform the random
function
F : I → R, F (t) =N∑k=0
Xkfk(t)
with covariance kernel
K(s, t) = E(F (s)F (t)
).
If the random function F (t) satisfies the assumptions A1, A2,
A3, A4, then the expectednumber of zeros of F in I is
Ẑ(F, I) =1
π
∫Iρtdt, ρt =
ö2st logK(s, t)s=t =
√K(s, t)K ′′st(s, t)−K ′(s, t)2
K(s, t)2
∣∣∣s=t
. (2.5)
Remark 2.4. The quantity 1πρt in (2.5) is the expected density
of zeros of F at t, i.e.,1πρtdt
is the expected number of zeros in the infinitesimal interval
[t, t+ dt]. ut
3. Some applications
We describe a few immediate consequences of Theorem 2.3.
Example 3.1 (The Kac statistics). Suppose that FN (t) is the
degree N random Kacpolynomial
FN (t) =
N∑k=0
Xktk, (3.1)
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10 LIVIU I. NICOLAESCU
where the random variables X are independent normal variables Xk
∈ N(1). In this casethe covariance kernel is
KN (s, t) =1− (st)N+1
1− st.
Then
∂2st logKN (s, t)s=t =1
(t2 − 1)2− (N + 1)
2t2N
(t2N+2 − 1)2=: fN (t)dt.
We deduce that
ZN := Ẑ(FN ,R) = 4Ẑ(FN , [1,∞)
)=
4
π
∫ ∞1
√fN (t)dt.
Theorem 3.2 (Kac, Edelman-Kostlan). As N →∞ we have
ZN =2
π
(logN + C
)+ o(1), (3.2)
where
C = log 2 +
∫ ∞0
(√1
x2− 1
sinh2 x− 1x+ 1
)dx ≈ 0.6257....
Proof. We follow the approach [7, §2.5]. We make the change in
variables t = 1 + xN and wededuce
ZN =2
π
∫ ∞0
√gN (x)dx,
where
gN (x) =
(1
x(1 + x/(2N))
)2︸ ︷︷ ︸
=AN (x)2
−
(2 (N+1)N
(1 + x/N
)N+1(1 + x/N)2N − 1
)2︸ ︷︷ ︸
=:BN (x)2
.
The function gN has an apparent pole at x = 0, but it has an
extension as a smooth functionon [0,∞). Note that
limN→∞
AN (x) = A∞(x) :=1
x, ∀x > 0. (3.3)
The equality
limN→∞
(1 + x/N)N = ex (3.4)
implies that
limN→∞
BN (x) = B∞(x) :=2ex
e2x − 1=
1
sinhx, ∀x > 0. (3.5)
The function
g∞(x) = A∞(x)2 −B∞(x)2 =
1
x2− 1
sinh2 x=
sinh2 x− x2
x2 sinh2 x, x > 0
extends by continuity at x = 0 because
sinh2 x− x2 = x4
3+O(x6) as x→ 0.
Moreover
limN→∞
gN (x) = g∞(x) uniformly for x ∈ [0, 1].
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ON THE KAC-RICE FORMULA 11
We have √gN (x)−AN (x) =
√An(x)2 −BN (x)2 −AN (x) = −
BN (x)√gN (x) +AN (x)
limN→∞
BN (x)√gN (x) +AN (x)
=B∞(x)√
g∞(x) +A∞(x).
The functions√gN (x)−AN (x) are integrable on [1,∞) and we
have2
limN→∞
∫ ∞1
(√gN (x)−AN (x)
)= −
∫ ∞1
B∞(x)√g∞(x) +A∞(x)
dx
=
∫ ∞1
(√g∞(x)−A∞(x)
)dx =
∫ ∞1
(√g∞(x)−
1
x
)dx.
We deduce that as N →∞ we have∫ ∞0
√gN (x)dx =
∫ ∞1
AN (x) +
∫ 10
√g∞(x)dx+
∫ ∞1
(√g∞(x)−
1
x
)dx+ o(1).
Now observe that
−∫ 1
0
1
x+ 1dx+
∫ ∞1
(1
x− 1x+ 1
)dx = − log 2 + log 2 = 0,
and ∫ ∞1
AN (x) =
∫ ∞1
(1
x− 1
2N + x
)dx = log(2N + 1).
We deduce ∫ ∞0
√gN (x)dx = log(2N + 1) +
∫ ∞0
(√g∞(x)−
1
x+ 1
)dx+ o(1)
= logN + log 2 +
∫ ∞0
(√g∞(x)−
1
x+ 1
)dx+ o(1).
ut
Remark 3.3. (a) The graph of
ρn(t) =
√1
(1− t2)2− (N + 1)
2t2N
(1− t2N+2)2
is depicted in Figure 2.It has two “peaks” at t = ±1 which
suggest that the real roots of a Kac random polynomial
tend to concentrate near t = ±1. This statement can be made much
more precise. In [11,§2, Lemma 1] it is shown that for any s ∈ (0,
1] the expected number of roots in the interval(−1 + (logN)−s, 1−
(logN)−s)) is O((logN)s log logN) as N →∞,
Ẑ(FKac,
(−1 + 1
(logN)s, 1− 1
(logN)s
))= O((logN)s log logN).
2The passage to limit can be justified by invoking the dominated
convergence theorem whose applicabilitycan be verified upon a
closer inspection of (3.3), (3.4) and (3.5).
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12 LIVIU I. NICOLAESCU
Figure 2. The density of zeros of a random Kac polynomial of
degree 25.
(b) The asymptotics of the variance VN of the number of roots of
a Kac random polynomialof large degree N was described by N.B.
Maslova, [13]. More precisely, she proved that
VN ∼1
4π
(1− 2
π
)logN as N →∞.
Using Chebyshev’s inequality we deduce that for any k > 0,
the probability that a randomKac polynomial has more than ( 2π + k)
logN real roots is smaller than
1k2√
logNas N →∞.
(c) Intuitively, one should expect that a random polynomial has,
on average many morecritical points than zeros. M. Das [6] has
shown that the expected number of critical points
of a random Kac polynomial of degree N is asymptotic to 1+√
3π logN . This confirms the
intuition, but it also shows that expected number of critical
points has the same order ofgrowth as the expected number of
zeros.
(d) A weaker version of the asymptotic estimate (3.2) is valid
for more general classes ofrandom polynomials. More precisely
Ibragimov and Maslova have shown in [11] that if Fn(t)is a random
degree N polynomial of the form
FN (t) =
N∑k=0
Xktk,
where (Xk)k≥0 are independent identically distributed L2-random
variables, then
Ẑ(FN ,R) ∼2
πlogN as N →∞. (3.6)
The proof in [11] is much more complicated and is based on ideas
developed by Erdös-Offord[8] where they discuss the special case
when Xk are Bernoulli variables taking values ±1 withequal
probability. Recently (2014) H. Nguyen, O. Nguyen and V. Vu proved
in [16] that ifXk ∈ Lp for some p > 2, then (3.6) can be refined
to
Ẑ(FN ,R) =2
πlogN +O(1) as N →∞. (3.7)
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ON THE KAC-RICE FORMULA 13
(e) Any polynomial of degree n has n complex roots. A degree N
random Kac polynomial(3.1) FN (t) = FN (ω, t) has with probability
1 N distinct complex roots. As n → ∞, theseroots tend to distribute
uniformly in annuli around the unit circle |z| = 1.
More precisely, if α, β ∈ [0, 2π), α < β and δ ∈ (0, 1), and
we denote by ZN (ω|α, β, δ) theexpected number of zeros of FN (ω,
t) in the region
α < arg z < β, 1− δ < |z| < 1 + δ,
then the random variable
ω 7→ 1NZN (ω|α, β, δ)
converges almost surely and in any Lp, p ∈ (0, 1) to β−α2π . For
proofs and details we refer to[2], [4, §8.2], [20]. ut
Example 3.4 (The Kostlan statistics). Consider the Kostlan
random polynomials
PN (t) =
N∑k=0
Xktk
where Xk are independent normal random variables with zero means
and variances
var(Xk) =
(N
k
).
In this case the covariance kernel is
K(s, t) = (1 + st)N .
We have
logK(s, t) = N log(1 + st), ∂t logK(s, t) =Ns
1 + st,
∂2st logK(s, t) = N(1 + st)− st
(1 + st)2=
N
(1 + st)2.
The density of zeros is then ö2st logK(s, t)|s=t =
√N
1 + t2.
The Kac-Rice formula implies that the expected number of zeros
is
2√N
π
∫ ∞0
1
1 + t2=√N.
We see that the Kostlan random polynomials have, on average,
more real zeros than the Kacrandom polynomials. One can show that
the expected number of critical points of a Kostlanrandom
polynomial is equal to 2
√3N − 2; see [17].
ut
Remark 3.5. Consider the complex analog of the Kostlan random
polynomial
FN (ω, z) =
N∑k=0
Xk(ω)zk,
-
14 LIVIU I. NICOLAESCU
where Xk : Ω → C are independent complex valued random variable
normally distributedwith mean zero and variances
varXk =
(N
k
).
The distribution of zeros of FN (ω, z) in the complex plane C
has many beautiful properties,[10, 21].
For any Borel set B ⊂ C we denote by volS2(B) the area of B
viewed as a subset of S2via the stereographic projection from the
North Pole to the plane that cuts the Equator. IfZN (B) denotes the
expected number of critical points of FN in B, then (see [10,
21])
ZN (B) =N
4πvolS2(B).
In other words, we expect the zeros of the random polynomial FN
(ω, z) to be uniformlydistributed with respect to the probability
measure 14π volS2 on C. As explained in [21], thishappens with high
probability as N →∞. ut
Example 3.6 (The Legendre statistics). Consider the random
polynomial
FN (t) =
N∑k=0
Xkpk(t),
where Xk are independent standard normal random variables, and
pk is the degree k Legendrepolynomial defined in (1.7). M. Das [5]
has shown that the expected number of zeros of FN (t)in [−1, 1] is
asymptotic to 1√
3N for large N . ut
Example 3.7 (Fáry-Milnor). Suppose that
[0, L] 3 s 7→ r(s) = (x(s), y(s), z(s)) ∈ R3
is an immersed curve smooth parametrized by arclength. We obtain
a random function onthe curve
F (s) = Ax(s) +By(s) + Cz(s)
where A,B,C are independent standard normal random variables
with mean zero and vari-ance 1. Its derivative
F ′(s) = Ax′(s) +By′(s) + Cz′(s)
has covariance kernel
K(s1, s2) = x′(s1)x
′(s2) + y′(s1)y
′(s2) + z′(s1)z
′(s2) = T (s1) • T (s2),where (T ,N) is the Frenet frame along
the curve and • denotes the standard inner productin R3. We
have
∂s2K(s1, s2) = T (s1) • T ′(s2) = κ(s2)T (s1) •N(s2),where κ
denotes the curvature of the curve. Similarly
∂2s1s2K(s1s2) = κ(s1)κ(s2)N(s1) •N(s2).We deduce
K(s, s) = 1, ∂s2K(s, s) = 0, ∂2s1s2K(s, s) = κ(s)
2.
The Kac-Rice formula implies that the expected number of
critical points of F (s) along thecurve is
1
π
∫ L0κ(s)ds =
1
π× the total curvature of the curve.
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ON THE KAC-RICE FORMULA 15
This result was first independently and by different methods by
I. Fáry [9] and J. Milnor[14]. ut
Example 3.8 (Trigonometric polynomials). Consider random
trigonometric polynomialsof the form
Fm(t) = A0 +√
2
m∑k=0
(Ak cos kt+Bk sin kt), t ∈ [0, 2π]
where A0, A1, . . . , Am, B1, . . . , Bm are independent
standard normal random variables. Thecovariance kernel of this
random function is described in (1.9),
K(s, t) = 1 +m∑k=1
cos k(t− s).
We have
∂tK(s, t) = −2m∑k=0
k sin k(t− s), ∂2stK(s, t) = 2m∑k=0
k2 cos k(t− s).
We deduce
K(t, t) = 2m = 1, ∂2stK(t, t) = 2
m∑k=0
k2 =m(m+ 1)(2m+ 1)
3.
We conclude that the density of zeros is
ρm(t) =1
π
√m(m+ 1)
3,
so that the expected number of zeros in [0, 2π] of this random
trigonometric polynomial is
2
√m(m+ 1)
3∼ 2m√
3as m→∞.
ut
4. Energy landscapes
In this last part I will discuss possible generalizations of the
example involving trigono-metric polynomials. We start with a few
simple observations.
A 2π-periodic function f(θ) can be viewed as a function on the
unit circle
S1 ={
(x, y) ∈ R2; x2 + y2}.
Moreover, a function f : S1 → R is a trigonometric polynomial of
degree ≤ n if and only ifthere exists a polynomial P : R2 → R such
that degP ≤ n and f = P |S1 , i.e.,
f(θ) = P (cos θ, sin θ), ∀θ ∈ [0, 2π].Consider the unit
sphere
S2 ={~x ∈ R3; |~x| = 1
},
equipped with the normalized area element 14πdA given in
spherical coordinates (θ, ϕ), θ ∈[0, 2π], ϕ ∈ [0, π] by
1
4πdA =
sinϕ
4πdϕdθ.
For each natural number n we denote by Un the subspace of C∞(S2)
consisting of the
restrictions to S2 of the polynomials P (x, y, z) of degree ≤ n.
Let us point out that two
-
16 LIVIU I. NICOLAESCU
Figure 3. The graph of the restriction to S2 of a random degree
7 polynomial.
different polynomials in the variables x, y, z can have
identical restrictions to S2. For examplethe polynomial F (~x) =
|~x|2 = x2 + y2 + z2 has the same restriction to S2 as the
constantpolynomial 1. In any case, one can show (see e.g. [15])
that dimUn = n
2.The zero set of a generic smooth function f : S2 → R is
1-dimensional 1-submanifold of
S2 and thus it is a disjoint union of circles. We can think of a
function f : S2 → R as analtitude function describing the altitude
f(p) of a point p ∈ S2 relative to the sea-level andwe can
visualize as defining relief on a spherical shaped planet; see
Figure 3. From this pointof view, the zero set describes the shore
lines of this relief; see Figure 4.
The (energy) landscape of a smooth function f : S2 → R concerns
the location of thecritical points and its critical values. Recall
that the critical points of f are the points where∂θf = ∂ϕf = 0. If
p is a critical point of f , then the value f(p) of f at p is
called a criticalvalue of f . The coordinates of the critical
points indicate the locations of the mountain tops,the mountain
passes and of the sea bottoms. The critical values record the
altitudes at thesespecial points.
The space Un is equipped with an inner product
(u,v) =1
4π
∫S2uvdA, u,v ∈ Un.
This defines a Gaussian probability measure Γn on Un given
by
Γn(du) = (2π)−n2/2e−
‖u‖22 λ(du),
where λ denotes the canonical Lebesgue measure on Un defined by
the above inner product.If we fix an orthonormal basis (ei) of Un,
then we can represent a random polynomial u in
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ON THE KAC-RICE FORMULA 17
Figure 4. The zero set of the restriction to S2 of a random
degree 7 polynomial.
the ensemble (Un,Γn) as a linear combination
u =∑i
Xiei
where Xi are independent standard normal random variables.For
each open set O ⊂ S2, any open interval I ⊂ R and any smooth
function f : S2 → R
we denote by C(f,O) the number of critical points of f in O, and
by D(f, I) the number ofcritical values of f in I. We obtain in
this fashion random variables
Un 3 u 7→ C(f,O), D(f, I) ∈ Z≥0 ∪ {∞}.
We denote by Cn(O) and respectively Dn(I) their
expectations.Using a higher dimensional version of the Kac-Rice
formula [1, 3] we proved in [18] that
Cn(O) ∼2n2
3√
3area (O) as n→∞. (4.1)
The behavior of Dn(I) is a bit more complicated, but in [19], we
described is large nbehavior by relating it to statistical
invariants of certain ensembles of random, symmetric3× 3
matrices.
Appendix A. A probabilistic dictionary
For v > 0 and m ∈ R we denote by γv,m the probability measure
on R described by
γv,m(dx) =1√2πv
e−(x−m)2
2v dx.
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18 LIVIU I. NICOLAESCU
We will refer to it as the Gaussian measure on R with mean m and
variance v. As v ↘ 0the measure γv,m converges weakly to the Dirac
measure concentrated at m for this reasonwe set
γ0,m := δm.
A probability measure µ on R is called Gaussian if∃v ≥ 0, m ∈ R
: µ = γv,m.
If above m = 0, then we say that µ is a centered gaussian
measure.Suppose that (Ω,A,P ) is a probability space and X : Ω → R
is a random variable, i.e.
a measurable function. The random variable X is called Gaussian
or normal if the pushforward X#P is a Gaussian measure γv,m on R.
We write this as X ∈ N(v,m). Whenm = 0 we say that X is a centered
Gaussian variable and we write this X ∈N(v).
If X ∈N(v,m), then m is the mean or expectation of X
m = E(X) :=
∫ΩX(ω)P (dω),
and v is the variance of X,
v = var(X) =
∫Ω
(X(ω)−m
)2P (dω).
Let us observe that if X1, X2 : (Ω,A,P ) → R are two independent
normal variables, Xi ∈N(vi,mi), i = 1, 2, and c1, c2 ∈ R, then
c1X1 + c2X2 ∈N(c21v1 + c
22v2, c1m1 + c2m2
).
Suppose that U is a finite dimensional real vector space. Denote
by BU the σ-algebra of Borelsubsets of U . A centered Gaussian
measure on U is a probability measure Γ on BU such thatany linear
function ξ : U → R is a centered Gaussian random variable with
variance v(ξ).
Such a measure defines a symmetric bilinear form on the dual
U∗
C = CΓ : U∗ × U∗ → R, C(ξ1, ξ2) = EΓ(ξ1ξ2) =
∫Uξ1(u)ξ2(u)Γ(du).
Clearly CΓ is nonnegative definite. The bilinear form CΓ is
called the covariance form ofthe Gaussian measure Γ. The centered
Gaussian measure Γ is completely determined by itscovariance
form.
Fix an inner product (−,−) on U with norm ‖ − ‖. Then we can
identify U∗ with Uand a symmetric, nonnegative definite bilinear
form C : U∗ × U∗ → R with a symmetric,nonnegative definite operator
C : U → U . The inner product determines a canonical
centeredGaussian measure ΓI on U defined by
ΓI(du) = (2π)− 1
2dimUe−
‖u‖22 du.
Its covariance form is given by the identity operator U → U .
For an arbitrary symmetricnonnegative definite operator C : U → U
we set
ΓC = C12#ΓI = the pushforward of ΓI via the linear map C
12 : U → U.
Then ΓC is the centered Gaussian measure on U with covariance
form C. If C is invertible,then
ΓC(du) =1√
det 2πCe−
12
(C−1u,u)du.
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ON THE KAC-RICE FORMULA 19
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Department of Mathematics, University of Notre Dame, Notre Dame,
IN 46556-4618.E-mail address: [email protected]:
http://www.nd.edu/~lnicolae/
http://arxiv.org/abs/1402.4628http://arxiv.org/pdf/1006.1267.pdfhttp://front.math.ucdavis.edu/1101.5990http://www.nd.edu/~lnicolae/RandCrVal.pdfhttp://www.nd.edu/~lnicolae/
1. Gaussian random functions2. The one-dimensional Kac-Rice
formula3. Some applications4. Energy landscapesAppendix A. A
probabilistic dictionaryReferences