Critical points of smooth Gaussian random fields Jonathan Taylor - - PowerPoint PPT Presentation

Critical points of smooth Gaussian random fields

Jonathan Taylor (Stanford) November 11, 2014

Two Stages

• A model for random sets.
• Critical points: Kac-Rice formula.
• Tube formulae.
• Gaussian integral geometry.
• Selective inference for critical points.

References

• Most of what I am going to say today can be found in

Random Fields and Geometry.

• Results built on top of earlier work of:
• Robert Adler
• Iain Johnstone
• Satoshi Kuriki
• David Siegmund
• Jiayang Sun
• Akimichi Takemura
• Keith Worsley
• Weyl, Hotelling
• Last part of talk related to selective inference (See

arxiv.org/1308.3020)

A model for random sets

• Our basic building blocks are R-valued Gaussian random fields
• n some n-dimensional manifold M (maybe with corners).
• We think of M as fixed, not large volume / high frequency

properties.

• The only asymptotic we look at is excursion above a high level.
• For most of the talk, we will assume:
• E(ft) = 0.
• E(f 2

t ) = 1.

• R(t, s) = E(ft · fs).
• (Put canonical picture / example on blackboard)
• Is Gaussian necessary?
• Only when we want to do some explicit computations.
• Heavy-tailed can of course have different behavior.

Excursion above 0

Excursion above 1

Excursion above 1.5

Excursion above 2

Excursion above 2.5

Excursion above 3

Why care?

• Integral geometric properties tell a nice geometric story.
• Each component of the excursion set contains a critical point
• f f.
• By Morse Theory, the Euler characteristic can be expressed in

terms of critical points of f|f−1[u+∞). . .

• Critical points / values are of fundamental importance here.
• One part of the story: for large u

E

• χ
• M ∩ f−1[u, +∞)
• ≈ P
• sup

t∈M

ft ≥ u

• .

Statistical motivation

• Signal detection in smooth noise. A natural test statistic for

1-sparse means H0 : µ ≡ 0, Ha : µ(·) = α · R(t0, ·).

• Nonregular likelihood ratio / score tests. Limiting distribution

is often of the form sup

t∈M

max(ft, 0)2.

Kac-Rice formula

• A fundamental tool for counting zeros.
• Kac was interested in number of zeros of random polynomials.
• Rice was interested in the number of upcrossings of a process

above a level.

• Suppose h : M → Rn is sufficiently smooth and

non-degenerate and g is continuous E (# {t ∈ M : ht = 0, gt ∈ O}) =

• M

E

• 1{gt∈O} · |Jht|
• ht = 0
• φht(0) dt.

Kac-Rice formula

• With a little imagination, this is roughly the same as

E  

• t∈M:ht=0

F(gt)   =

• M

E

• F(gt) · |Jh(t)|
• ht = 0
• φht(0) dt

for reasonable functions F : R → R.

Rough proof of Kac-Rice

2 = lim

ǫ↓0

1 2ǫ

• [0,T]

1[u−ǫ,u+ǫ](h(t)) |h′(t)| dt

Rough proof of Kac-Rice

Interchange ǫ ↓ 0 and expectation. . .

Tail probability

• Applying Kac-Rice to local maxima:

P

• sup

t∈M

ft ≥ u

• ≤ E
• #{t ∈ M : ∇ft = 0, ft ≥ u, ∇2ft < 0}
• =
• M

E

• det(−∇2ft)1{ft≥u,∇2ft<0}
• ∇ft = 0
• φ∇ft(0) dt

=

• M

Mn

t (1{ft≥u}) dt

• Above,

Mj

t(h) def

= E

• h · det(−∇2ft)1{index(∇2ft)=j}
• ∇ft = 0
• φ∇ft(0)

Tail probability

• The EC heuristic:

n

• j=0

Mj

t(1{ft≥u}) u→∞

≈ Mn

t (1{ft≥u}).

• Morse’s theorem, followed by integration over M yields

E

• χ
• M ∩ f−1[u, +∞)
• =
• M

n

• j=0

Mj

t(1{ft≥u}) dt

What does this tell you?

• Let M be a 2-manifold without boundary, then

E

• χ
• f−1[0, +∞)
• = 1

2χ(M) .

• If we allow boundary, then

E

• χ
• f−1[0, +∞)
• = 1

2χ(M) + 1 2π|∂M|.

• Lengths and areas are computed with respect to a Riemannian

metric from f: g(Xt, Yt) = E(Xtf · Ytf).

Expected EC is computable

• The EC stands out as being explicitly computable in wide
• generality. (Here and on last slide is where we use our centered Gaussian constant variance

assumption)

• Specifically, define

ρj(u) =

• 1 − Φ(u)

j = 0 Hj−1(u)e−u2/2(2π)(j+1)/2 j ≥ 1

• Then,

E

• χ
• M ∩ f−1[u, +∞)
• =

n

• j=0

Lj(M)ρj(u).

How good is this approximation?

• The expected EC heuristic does not assume Gaussianity (though

calculations would be difficult otherwise).

• However, if f is Gaussian and as assumed here, a careful

application of Kac-Rice yields Error(u) =

• P
• sup

t∈M

ft ≥ u

• − E
• χ
• M ∩ f−1[u, +∞)
• u→∞

= Oexp

• e

−u2/2

• 1+

1 σ2 c (f,M)

• Error is roughly the cost of having two critical points above

the level u.

Tube formulae

• For small r, the functionals Lj(M) are implicitly defined by

Steiner-Weyl formula for r ≤ rc(M) Hk

• x ∈ Rk : d(x, M) ≤ r
• =

k

• j=0

ωk−jrk−jLj(M)

• The quantity σ2

c(f, M) is completely analogous to the critical

radius of embedding of M in Hf, the RKHS of f: M ∋ t → R(t, ·) ∈ S(Hf).

The cube

H3 (Tube([0, a] × [0, b] × [0, c], r)) = abc + 2r · (ab + bc + ac) + (πr2) · (a + b + c) + 4πr3 3

How to compute volume of a tube

t t +r ·ηt

The Jacobean

• Most of the work (and all of the local information) is encoded

in the Jacobean of (t, ηt) → t + r · ηt, η2 = 1

This is what Weyl said anyone decent student of calculus could do.

• Some careful thought and / or more calculus shows that

det(−∇2ft) has a very similar structure to the above Jacobean.

Gaussian Kinematic Formula

• Let f = (f1, . . . , fk) be made of IID copies of our original

Gaussian field.

• Consider the additive functional on Rk that takes a rejection

region D → E

• χ
• M ∩ f−1D
• .
• For D that are rare under the marginal distribution

(γk ∼ N(0, Ik×k)) the expected EC heuristic says E

• χ
• M ∩ f−1D
• ≈ P
• M ∩ f−1D = ∅
• .
• How is M involved?

(We suspect through Lj(M).)

• How is D involved?

T random field

A simple cone

The rejection region for a t statistic T(x1, x2, x3) = x1

• (x2

2 + x2 3)/2

.

Inverse image

Gaussian Kinematic Formula

• Define additive functionals Mγ

j on Rk by

γk

• y ∈ Rk : d(y, D) ≤ r
• =
• j≥0

( √ 2πr)j j! Mγk

j (D)

• Then, the Gaussian Kinematic Formula asserts

E

• χ(M ∩ f−1D)
• =

n

• j=0

Lj(M)Mγk

j (D).

Gaussian Kinematic Formula

• Why do Mγk

j

arise?

Not clear beyond direct calculation.

• Can be proved by direct calculation with Kac-Rice.
• Alternate proof based on classical Kinematic Fundamental

Formula on S√

N(RN) =

• x ∈ RN : x2 =

√ N

• ,

N → ∞.

• Both proofs involve recognizing an integral as a coefficient in

Gaussian tube expansion.

• Because many canonical statistics are based on distance, it

turns out there are perhaps more explicit examples of Gaussian tube formulae than Steiner . . .

• Instead of examples I want to return to selective inference . . .

Selective inference

• The measures Mj

t, suitably normalized, can be interpreted as a

type of Palm distribution / Slepian model.

• Define the normalized measures

Qj

t(˜

h) = Mj

t(˜

h) Mj

t(1)

.

• Formally, by Kac-Rice

h → Qj

t(h(ft))

determines the law of ft given t is a critical point of f with index j.

Selective inference

• Can derive tests of

H0 : E(ft) ≡ 0 based on sup

t∈M

ft

• t∗ = argmaxt∈Mft.
• Or selective tests of H0 : ft∗ = 0.
• Let’s take a closer look at the structure of such a test.

A discrete Kac-Rice calculation

• Suppose Z ∼ N(µk×1, Ck×k) with diag(C) = 1.
• Set

i∗ = argmaxiZi

• A simple calculation yields

{i∗ = i} = {Zi > max

j=i Zj}

=

• Zi > max

j=i

Zj − Ci,jZi 1 − Ci,j

• Note that

Mi = max

j:j=i

Zj − Ci,jZi 1 − Ci,j is independent of Zi for each i.

A discrete Kac-Rice calculation

• We see

Pµ(Zi∗ > t) =

k

• i=1

Pµ(Zi > t, i∗ = i) =

k

• i=1

Pµ(Zi > t, Zi ≥ Mi) =

k

• i=1

Eµ(1 − Φ(max(t, Mi))) =

k

• i=1

Qi,µ(1{Zi≥t})P(i∗ = i) where Qi,µ(h) = Eµ(h|i∗ = i).

Choice of model

• Define

˜ Qi,µ(h) = Qi,µ(h(Zi)).

• The selective test is a test of µi = 0 constructed to control

selective type I error.

• Without any assumption on µ, there is a (k − 1) dimensional

nuisance parameter when we want to test µi = 0 under ˜ Qi,µ.

• Standard approach is to condition on Z − E(Z|Zi), yielding

1 − Φ(Zi∗) 1 − Φ(Mi∗)

H0:µi∗=0

∼ Unif(0, 1).

• If we assume that µ ≡ 0, then we can draw from ˜

Qi,µ (under µ ≡ 0) and construct the selective test.

• This test based on ˜

Qi,µ is provably more powerful. (See arxiv.org/1410.2597)

Thanks

NSF-DMS 1208857 and AFOSR-113039.