Regular Variation Talk

MotivationDefinition and basic properties

Abel-Tauber Theorems

A (Gentle) Introduction to Regular Variation

Daniel Tokarev

MASCOS

The University of Melbourne, 20 August, 2008

Daniel Tokarev A (Gentle) Introduction to Regular Variation



Relationship between function and transformExtreme Values

References:

1 Feller, An Introduction to Probability Theory and ItsApplications, Vol II.

2 Bingham, Goldie & Teugels, Regular Variation.

3 D. Tokarev, Growth of Integral Transforms and Extinction inCritical Galton-Watson Processes, JAP, 2008.





Question: Given a real-valued measurable function F andan operator (eg integral transform, nth iterate, etc), what isa relationship between the asymptotic behaviour of thefunction and the operator?Example 1: Laplace Transform: Let U(x) be a monotonefunction (Probabilists: think "distribution function" (df) or"measure") possessing a Laplace Transform (LT)

LU(t) :=

∫ ∞0

e−xtU{dx}

(=

∫ ∞0

e−xtu(x)dx , provided u(x) := U ′(x) exists)

Obvious Observation ]1: As t →∞ in the transform,what will most likely matter is the behaviour of U(x) asx → 0.






LU(t) :=

∫ ∞0

e−xtU{dx}

(=

∫ ∞0








LU(t) :=

∫ ∞0

e−xtU{dx}

(=

∫ ∞0







Example 2 (Feller II, pp. 277-8)Let X (n) := max{X1, . . . , Xn}, where Xi ’s are non-negativeiid rv’s. Do there exist normalising constants an, such that

limn→∞

X (n)

an= X , in distribution, for some non-degenerate

X?Let F denote the common df of Xi ’s, then X (n) has df F n

and the equation above becomes F (anx)n → G(x), whereG is the df of X .Taking log of both sides and using log(1− x) ∼ −x , whenx is close to 0, we obtain

−n(1− F (anx)) → log G(x)






limn→∞

X (n)










limn→∞

X (n)










limn→∞

X (n)









Continued from before....


On the LHS, we need to "cancel out" n, ie choose an togrow appropriately, but this choice has to be independentof x !For this it is sufficient (and also necessary! - exercise) if forall x > 0:

limt→∞

1− F (tx)

1− F (t)= h(x).

Since then 1− F (anx) ∼ (1− F (an))h(x) and we are atliberty to choose the appropriate an (ie an = F←(1− 1/n)).








limt→∞

1− F (tx)

1− F (t)= h(x).









limt→∞

1− F (tx)

1− F (t)= h(x).









limt→∞

1− F (tx)

1− F (t)= h(x).





DefinitionExamplesUniform convergenceRepresentation Theorem

Theorem (Karamata, 1930)Suppose that for a dense set of x > 0,

limt→∞

U(tx)

U(t)= h(x) < ∞ ∗

Then h(x) = xρ, for some ρ ∈ R and we say that U is regularlyvarying of index ρ.

Since U(x1x2t)U(t) = U(x1x2t)

U(x2t)U(x2t)U(t) , taking the limit x →∞ and

applying ∗, we obtain h(x1x2) = h(x1)h(x2) and henceh(x) = xρ

(see Feller I, p.459 for bounded on finite intervals version,and BGT p. 2-5 for measurable and Baire versions).






limt→∞

U(tx)

U(t)= h(x) < ∞ ∗











limt→∞

U(tx)

U(t)= h(x) < ∞ ∗










We now know G from before: From−n(1− F (anx)) = G(x), x comes out as xρ and we obtain:

− log G(x) = xρ hence G(x) = e−xρ.

Theorem (Gnedenko-Fisher)

Let X (n) = max{X1, . . . , Xn}, where Xi are non-negative iid rv’swith df F . Then there exists a sequence {an} such that X (n)/anconverges in distribution to a non-degenerate rv X with df G iff1− F is regularly varying, in which case G(x) = e−xρ

.

The case ρ = 0 , ie U(xt)/U(t) → 1 is called slowly varying(usually denoted by L(x)).We have U(x)/xρ is slowly varying and conversely xρL(x)is regularly varying with index ρ, ie U(x) = xρL(x).









.










.










.






Examples of regularly and slowly varying functions:Any polynomial is regularly varying; any rational function isregularly varying; log is slowly varying; iterates and powersof log are slowly varyingslowly varying functions can be unbounded and oscillateinfinitely often, eg L(x) = e(log x)1/3(cos x)1/3

Also if L1, L2 are slowly varying, then so are L1L2, L1 + L2and if L2 →∞ as x →∞, also L1(L2)

Obvious Observation ]2: Note that most of thetheory can be applied to an asymptote at 0, or any otherpoint x0 by transformation x → 1/(x0 − x)

































Theorem (Uniform Convergence, Csiszar & Erdös, 1967 (BGTp.9))

If L is slowly varying then L(tx)/L(t) → 1 as x →∞ uniformlyon each compact x-set in (0,∞).

Let h(x) := log L(ex), then the assumption is

h(x + u)− h(x) → 0, as x →∞ ∀u ∈ R ?Assertion is uniform convergence on u-sets in R.Suppose not, ie there exist ε > 0, xn →∞ and a boundedsequence un with |h(xn +un)−h(xn)| > 2ε, for n = 1, 2, . . ..Passing to a subsequence if necessary, we may assumethat un → u0 and that |un − u0| ≤ 1 for large enough n.From ?, for every y and for large enough n,|h(xn + y)− h(xn)| ≤ ε, hence |h(xn + un)− h(xn + y)| > εfor large enough n.





































So there are sequences xn →∞ and un → u0 such that∀y ∈ R and for large enough n, |h(xn + un)− h(xn + y)| > ε.

We want to find z s.t. |h(xn + un)−h(xn + un − z)| > ε ∗Consider

⋃∞k=1 Ik = [−1, 1], where

Ik := [−1, 1] ∩ {y : ∀n ≥ k , |h(xn + yn)− h(xn + y)| > ε}.

Each Ik is measurable, hence ∃K s.t. IK has +ve measure.

Zn := un − IK = {un − y : y ∈ IK} and let Z :=∞⋂

j=1

∞⋃n=1

Zn.

Since all Zn have measure |Ik | and we took |un − u0| ≤ 1,they are subsets of [u0 − 2, u0 + 2] and

|Z | = limj→∞

|∞⋃

n=j

Zn| ≥ |IK | > 0.

Now take z ∈ Z and we have a contradiction as in ∗!Daniel Tokarev A (Gentle) Introduction to Regular Variation






⋃∞k=1 Ik = [−1, 1], where




j=1

∞⋃n=1

Zn.


|Z | = limj→∞

|∞⋃

n=j

Zn| ≥ |IK | > 0.







⋃∞k=1 Ik = [−1, 1], where




j=1

∞⋃n=1

Zn.


|Z | = limj→∞

|∞⋃

n=j

Zn| ≥ |IK | > 0.







⋃∞k=1 Ik = [−1, 1], where




j=1

∞⋃n=1

Zn.


|Z | = limj→∞

|∞⋃

n=j

Zn| ≥ |IK | > 0.







⋃∞k=1 Ik = [−1, 1], where




j=1

∞⋃n=1

Zn.


|Z | = limj→∞

|∞⋃

n=j

Zn| ≥ |IK | > 0.





Theorem (Local Boundedness, Seneta,1973 (BGT, p. 12))If L is positive and slowly varying, then for large enough X, it isbounded on finite intervals [X , Y ], 0 < X < Y < ∞.

Let h(x) = log(L(ex)) as before. From UniformConvergence, we know that there exists X such that foru ∈ [0, 1] and x ≥ X : |h(u + x)− h(x)| < 1

Hence |h(x)| ≤ |h(X )|+ 1 on [X , X + 1] and by induction|h(x)| ≤ |h(X )|+ n on [X , X + n]. The conclusion for Lfollows.



















Theorem (Representation Theorem (Karamata, Korevaar))L is slowly varying iff

L(x) = c(x) exp(

∫ x

a

ε(u)

udu), or h(x) = d(x) +

∫ x

be(u) du,

for some c(x) → c; d(x) → d; ε(x), e(x) → 0; a, b < x.

Since h is integrable on finite intervals far enough to theright, we can write

h(x) =

∫ x+1

xh(x)−h(u) du+

∫ x

Xh(u+1)−h(u) du+

∫ X+1

Xh(u) du

Observe that the first term on the right is just∫ 10 h(x + 1)− h(u) du which must tend to 0 by Uniform

Convergence and the last term is constant.






L(x) = c(x) exp(

∫ x

a

ε(u)


∫ x

be(u) du,



h(x) =

∫ x+1

xh(x)−h(u) du+

∫ x

Xh(u+1)−h(u) du+

∫ X+1

Xh(u) du








L(x) = c(x) exp(

∫ x

a

ε(u)


∫ x

be(u) du,



h(x) =

∫ x+1

xh(x)−h(u) du+

∫ x

Xh(u+1)−h(u) du+

∫ X+1

Xh(u) du






Mellin-type transformsTauberian theorem for iterates

If U(λ) has LT LU(t) then U(λx) has LT LU(t/x)

U(λx1)U(x1)

U(λx2)U(x2)

· · · U(λxi )U(xi )

→ λρ, for xi →∞↓ ↓ ↓ ↓

LU(t/x1)U(x1)

LU(t/x2)U(x1)

· · · LU(t/x1)U(xi )

→ t−ρΓ(1 + ρ)

Need to use Extended Continuity of Laplace Transforms(provided the transforms remain bounded - Exercise)

This means LU(t/x)U(x) → t−ρΓ(1 + ρ) which for t = 1 gives the

relationship we alluded to at the start and in fact we obtain:

Theorem ((Karamata, Feller II, pp. 442-3))

Let U(x) be a measure with LT LU(t). U is regularly varying at∞ with index ρ iff LU is regularly varying at 0 with index −ρ andeach of these is equivalent to

LU(1/t)/U(t) → Γ(1 + ρ), as t →∞.






U(λx1)U(x1)

U(λx2)U(x2)

· · · U(λxi )U(xi )

→ λρ, for xi →∞↓ ↓ ↓ ↓

LU(t/x1)U(x1)

LU(t/x2)U(x1)

· · · LU(t/x1)U(xi )

→ t−ρΓ(1 + ρ)






LU(1/t)/U(t) → Γ(1 + ρ), as t →∞.






U(λx1)U(x1)

U(λx2)U(x2)

· · · U(λxi )U(xi )

→ λρ, for xi →∞↓ ↓ ↓ ↓

LU(t/x1)U(x1)

LU(t/x2)U(x1)

· · · LU(t/x1)U(xi )

→ t−ρΓ(1 + ρ)






LU(1/t)/U(t) → Γ(1 + ρ), as t →∞.






U(λx1)U(x1)

U(λx2)U(x2)

· · · U(λxi )U(xi )

→ λρ, for xi →∞↓ ↓ ↓ ↓

LU(t/x1)U(x1)

LU(t/x2)U(x1)

· · · LU(t/x1)U(xi )

→ t−ρΓ(1 + ρ)






LU(1/t)/U(t) → Γ(1 + ρ), as t →∞.






U(λx1)U(x1)

U(λx2)U(x2)

· · · U(λxi )U(xi )

→ λρ, for xi →∞↓ ↓ ↓ ↓

LU(t/x1)U(x1)

LU(t/x2)U(x1)

· · · LU(t/x1)U(xi )

→ t−ρΓ(1 + ρ)






LU(1/t)/U(t) → Γ(1 + ρ), as t →∞.





Theorem ((Tokarev, JAP 2008))

Let g, h : [0, 1) → R+, L a slowly varying function at 0, and letρ ∈ (0, 1) ((1) remains valid for ρ = 0). Define

G(k) :=

∫ 1

0skg(s) ds H(k) :=

∫ 1

0(1− sk )h(s) ds

Then, as x → 1 and k →∞1 g(x) ∼ (1− x)−ρL(1− x) ⇐⇒ G(k) ∼ Γ(1− ρ)kρ−1L

( 1k

)2 h(x) ∼ (1− x)−ρ−1L(1− x) ⇐⇒ H(k) ∼ Γ(1−ρ)

ρ kρL( 1

k

)Sketch of the proof of (1), (2) is analogous. Lets = 1− x

k , then G(k) becomes

kρ−1∫ k

0

(1− x

k

)kx−ρL(x/k) ds.







G(k) :=

∫ 1

0skg(s) ds H(k) :=

∫ 1

0(1− sk )h(s) ds


( 1k

)2 h(x) ∼ (1− x)−ρ−1L(1− x) ⇐⇒ H(k) ∼ Γ(1−ρ)

ρ kρL( 1

k



kρ−1∫ k

0

(1− x

k

)kx−ρL(x/k) ds.







G(k) :=

∫ 1

0skg(s) ds H(k) :=

∫ 1

0(1− sk )h(s) ds


( 1k

)2 h(x) ∼ (1− x)−ρ−1L(1− x) ⇐⇒ H(k) ∼ Γ(1−ρ)

ρ kρL( 1

k



kρ−1∫ k

0

(1− x

k

)kx−ρL(x/k) ds.





Let ε ∈ (0, 1) and N > 0 be a "large" number, divide G(k)by kρ−1L(1/k) and split the integral into two parts:

∫ N

0

(1− x

k

)kx−ρ L(x/k)

L(1/k)dx +

∫ k

N

(1− x

k

)kx−ρ L(x/k)

L(1/k)dx

=

∫ N

0

(1− x

k

)kx−ρ

(L(x/k)

L(1/k)− 1

)dx +

∫ N

0

(1− x

k

)kx−ρ dx

+

∫ k

N

(1− x

k

)kx−ρ L(x/k)

L(1/k)dx

We can pick N so that by Uniform Convergence the firstterm on the right is < ε/3, the second is within ε/3 ofΓ(1− ρ) and using Potter’s Thm (BGT p. 25), the last termis < ε/3.






∫ N

0

(1− x

k

)kx−ρ L(x/k)

L(1/k)dx +

∫ k

N

(1− x

k

)kx−ρ L(x/k)

L(1/k)dx

=

∫ N

0

(1− x

k

)kx−ρ

(L(x/k)

L(1/k)− 1

)dx +

∫ N

0

(1− x

k

)kx−ρ dx

+

∫ k

N

(1− x

k

)kx−ρ L(x/k)

L(1/k)dx







∫ N

0

(1− x

k

)kx−ρ L(x/k)

L(1/k)dx +

∫ k

N

(1− x

k

)kx−ρ L(x/k)

L(1/k)dx

=

∫ N

0

(1− x

k

)kx−ρ

(L(x/k)

L(1/k)− 1

)dx +

∫ N

0

(1− x

k

)kx−ρ dx

+

∫ k

N

(1− x

k

)kx−ρ L(x/k)

L(1/k)dx






The above result can be used to easily obtain a Tauberiantheorem for iterates.

Theorem (adapted from Tokarev, JAP 2008)Let f be a monotone function with a monotone derivative andf (1) = f ′(1) = 1 and let fn denote the nth functional iterate of f .Then f (s)− s = (1− s)1+ρL(1− s) iff 1− fn = n−1/ρL∗(n),where L is slowly varying at 0 and L∗ is slowly varying at ∞.

The proof connects f and fn via the quantity∑∞

i=0 1− fi(0)n

(in a special case this is the Expected Time to Extinction ofa Galton-Watson process started with k ancestors and aprobability generating function f ).








i=0 1− fi(0)n









i=0 1− fi(0)n






∑∞i=0 1− fi(0)n can be shown to be asymptotically

equivalent to∫ ∞0

1−U(s)k ds =

∫ 1

0U←(s1/k ) ds = k

∫ ∞0

sk−1U←(s) ds,

where U(s) = fbsc(0). Regular variation of fn is equivalentto regular variation of U and this can be inverted to obtainregular variation of U←. The integral is then of the form ofG of the above.∑∞

i=0 1− fi(0)n can also be shown to be asymptotically

equivalent to∫ 1

0

1− sk

f (s)− sds. This is of the form of H.

Obvious Observation ]3:And that’s probably enoughfor today!







1−U(s)k ds =

∫ 1

0U←(s1/k ) ds = k

∫ ∞0

sk−1U←(s) ds,



equivalent to∫ 1

0

1− sk









1−U(s)k ds =

∫ 1

0U←(s1/k ) ds = k

∫ ∞0

sk−1U←(s) ds,



equivalent to∫ 1

0

1− sk




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