Contents Gaussian random functionslnicolae/Kac_Rice.pdfWe denote by Zb(F;I) the expected number of zeros of Fon I. Example 1.1 (Random polynomials). Fix Nindependent normal random

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

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.

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


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


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.


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)


∃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.


We then have

E(ηε(F (t)

)|F ′(t)|

)=

1

2ε

∫|x|


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 =

√∂2st 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)


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].


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).


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)


(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 √∂2st 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,


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.


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


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


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.


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.


References

[1] R. Adler, R.J.E. Taylor: Random Fields and Geometry, Springer Monographs in Mathematics,Springer Verlag, 2007.

[2] L. Arnold: Über die Nullstellenverteilung zufälliger Polynome, Math. Z. 92 1966 1218.[3] J.-M. Azäıs, M. Wschebor: Level Sets and Extrema of Random Processes, John Wiley & Sons,

2009.[4] A.T. Barucha-Reid, M. Sambandhan: Random Polynomials, Academic Press, 1986.[5] M. Das: Real zeros of a random sum of orthogonal polynomials, Proc. A.M.S., 27(1971), 147-153.[6] M. Das: Real zeros of a class of random algebraic polynomials, J. Indian Math. Soc. 36(1972),

53-63.[7] A. Edelman, E. Kostlan: How many zeros of a random polynomial are real?, Bull. A.M.S. (New

Series), 32(1995), No.1, 1-37.[8] P. Erdös, A.C. Offord: On the number of real roots of a random algebraic equation, Proc. London

Math. Soc. 6(1956), 139-160.[9] I. Fáry: Sur la courbure totale d’une courbe gauche faisant un nord, Bull. Soc. Math. France,

77(1949), 128-138.[10] P.J. Forrester, G. Honner: Exact statistical properties of the zeros of complex random polynomials,

J.Phys. A: Math. Gen. 32(1999), 2961-2981.[11] I.A. Ibragimov, N.B Maslova: On the expected number of real zeros of random polynomials I.

Coefficients with zero means, Theory Prob. Appl., 16(1971), 228-248.[12] M. Kac: On the average number of real roots of a random algebraic equation, Bull. A.M.S.,

49(1943), 314-320.[13] N. B. Maslova: On the variance of the number of real roots of random polynomials, Theory Probab.

Appl., 19(1974), 35-52.[14] J.W. Milnor: On the total curvature of knots, Ann. Math., 52(1950), 248-237.[15] C. Müller: Analysis of Spherical Symmetries in Euclidean Spaces, Appl. Math. Sci. vol. 129,

Springer Verlag, 1998.[16] H. Nguyen, O. Nguyen, V. Vu: On the number of real roots of random polynomials, arXiv: 1402.4628[17] L.I. Nicolaescu: Statistics of linear families of smooth functions on knots, arXiv 1006.1267.[18] : Critical sets of random smooth functions functions on compact manifolds, arXiv: 1101.5990[19] Complexity of random smooth functions on compact manifolds, Indiana Univ. Math. J.,

63(2014), 1037-1065.[20] D.I. Šparo, M.G.Šur: On the distributions of roots of random polynomials, Vestn. Mosk. Univ. Ser.

1 Mat. Mekh, 1962, no.3, 40-43.[21] M. Sodin, B. Tsirelson: Random complex zeroes, I. Asymptotic normality, Israel J. Math.,

144(2004), 125-149.[22] G. Szegö: Orthogonal polynomials, Amer. Math. Soc., Colloquium Publications, vol.23, 1939. (2003

Reprint).

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

Documents

Contents Gaussian random functionslnicolae/Kac_Rice.pdfWe denote by Zb(F;I) the expected number of zeros of Fon I. Example 1.1 (Random polynomials). Fix Nindependent normal random