The Kolmogorov continuity theorem, Hölder continuity, and the Kolmogorov-Chentsov theorem

Jordan Bell
June 11, 2015

1 Modifications

Let (Ω,,P) be a probability space, let I be a nonempty set, and let (E,) be a measurable space. A stochastic process with index set I and state space E is a family (Xt)tI of random variables Xt:(Ω,)(E,). If X and Y are stochastic processes, we say that X is a modification of Y if for each tI,

P{ωΩ:Xt(ω)=Yt(ω)}=1.
Lemma 1.

If X is a modification of Y, then X and Y have the same finite-dimensional distributions.

Proof.

For t1,,tnI, let Ai for each 1in, and let

A=i=1nXti-1(Ai),B=i=1nYti-1(Ai).

If ωAB then there is some i for which ωYti-1(Ai), and ωXti-1(Ai) so Xti(ω)Yti(ω). Therefore

ABi=1n{ωΩ:Xti(ω)Yti(ω)}.

Because X is a modification of Y, the right-hand side is a union of finitely many P-null sets, hence is itself a P-null set. A and B each belong to , so P(AB)=0.11 1 We have not assumed that (Ω,,P) is a complete measure space, so we must verify that a set is measurable before speaking about its measure. Because P(AB)=0, P(A)=P(B), i.e.

P(Xt1A1,,XtnAn)=P(Yt1A1,,YtnAn).

This implies that22 2 http://individual.utoronto.ca/jordanbell/notes/finitedimdistributions.pdf

P*(Xt1Xtn)=P*(Yt1Ytn),

namely, X and Y have the same finite-dimensional distributions. ∎

2 Continuous modifications

Let E be a Polish space with Borel σ-algebra . A stochastic process (Xt)t0 is called continuous if for each ωΩ, the path tXt(ω) is continuous 0E.

A dyadic rational is an element of

D=i=02-i.

The Kolmogorov continuity theorem gives conditions under which a stochastic process whose state space is a Polish space has a continuous modification.33 3 Heinz Bauer, Probability Theory, p. 335, Theorem 39.3. It was only after working through the proof given by Bauer that I realized that the statement is true when the state space is a Polish space rather than merely d. In the proof I do not use that || is a norm on d, and only use that d(x,y)=|x-y| is a metric on d, so it is straightforward to rewrite the proof. This is like the Sobolev lemma,44 4 Walter Rudin, Functional Analysis, second ed., p. 202, Theorem 7.25. which states that if fHs(d) and s>k+d2, then there is some ϕCk(d) such that f=ϕ almost everywhere. It does not make sense to say that an element of a Sobolev space is itself Ck, because elements of Sobolev spaces are equivalence classes of functions, but it does make sense to say that there is a Ck version of this element.

Theorem 2 (Kolmogorov continuity theorem).

Suppose that (Ω,F,P,(Xt)tR0) is a stochastic process with state space Rd. If there are α,β,c>0 such that

E(|Xt-Xs|α)c|t-s|1+β,s,t0, (1)

then the stochastic process has a continuous modification that itself satisfies (1).

Proof.

Let 0<γ<βα and let

δ=β-αγ>0.

For m1, let Sm be the set of all pairs (s,t) with

s,t{j2-m:0j2m},

and |s-t|=2-m. There are 22m such pairs, i.e. |Sm|=22m. Let

Am=(s,t)Sm{|Xs-Xt|2-γm}.

For (s,t)Sm, using Chebyshev’s inequality and (1) we get

P(|Xt-Xs|2-γm) (2γm)αE(|Xt-Xs|α)
2αγmc|t-s|1+β
=c2αγm2-m(1+β)
<c2-m-δm.

Hence

P(Am)(s,t)SmP{|Xs-Xt|2-γm}<(s,t)Smc2-m-δm=2c2-δm.

Because mP(Am)<, the Borel-Cantelli lemma tells us that

P(n=1m=nAm)=P(N0)=0,

where for each ωΩN0 there is some m0(ω) such that ωAm when mm0(ω). That is, for ωΩN0 there is some m0(ω) such that

|Xt(ω)-Xs(ω)|<2-γm,mm0(ω),(s,t)Sm. (2)

Now let ωΩN0 and let s,t[0,1] be dyadic rationals satisfying

0<|s-t|2-m0(ω).

Let m=m(s,t) be the greatest integer such that |s-t|2-m:

2-m-1<|s-t|2-m, (3)

which implies that mm0(ω). There are some i0,j0{0,1,2,3,,2m} such that

s0=i02-ms<(i0+1)2-m,t0=j02-mt<(j0+1)2-m.

As 0s-s0<2-m and 0t-t0<2-m, there are sequences σj,τj{0,1}, j>m, each of which have cofinitely many zero entries, such that

s=s0+j>mσj2-j,t=t0+j>mτj2-j.

Because 0s-s0<2-m and t-t0<2-m,

2-m>|(s-s0)-(t-t0)|=|(s-t)-(s0-t0)||s0-t0|-|s-t|,

and with (3),

|s0-t0|<2-m+|s-t|2-m+2-m=2-m+1.

Thus |i0-j0|<2, so |i0-j0|{0,1} and so either s0=t0 or (s0,t0)Sm. In the first case, |Xt0(ω)-Xs0(ω)|=0. In the second case, since mm0(ω), by (2) we have

|Xt0(ω)-Xs0(ω)|<2-γm. (4)

Define by induction

sn=sn-1+σm+n2-(m+n),n1,

i.e.

sn=s0+m<jm+nσj2-j.

For each n1, sn-sn-1{0,2-(m+n)}, so either sn=sn-1 or (sn-1,sn)Sm+n, and because m+nm+1>m0(ω), applying (2) yields

|Xsn(ω)-Xsn-1(ω)|<2-γ(m+n).

Because the sequence σj is eventually equal to 0, the sequence sn is eventually equal to s. Thus

n=1(Xsn(ω)-Xsn-1(ω))=Xs(ω)-Xs0(ω),

whence

|Xs(ω)-Xs0(ω)|n=1|Xsn(ω)-Xsn-1(ω)|<n=12-γ(m+n)=2-γ(m+1)1-2-γ.

By the same reasoning we get

|Xt(ω)-Xt0(ω)|<2-γ(m+1)1-2-γ.

Using these and (4) yields

|Xt(ω)-Xs(ω)| |Xt(ω)-Xt0(ω)|+|Xt0(ω)-Xs0(ω)|+|Xs(ω)-Xs0(ω)|
<2-γ(m+1)1-2-γ+2-γm+2-γ(m+1)1-2-γ
=C2-γ(m+1),

for C=2γ+21-2-γ. By (3), 2-(m+1)<|t-s|, hence

|Xt(ω)-Xs(ω)|C|t-s|γ. (5)

This is true for all dyadic rationals s,t[0,1] with |s-t|2-m0(ω); when |s-t|=0 it is immediate.

For k1, let Xtk=Xk+t, which satisfies (1). By what we have worked out, there is a P-null set N1 such that for each ωΩN1 there is some m1(ω) such that mm1(ω) and (s,t)Sm imply that |Xt1(ω)-Xs1(ω)|<2-γm. Let N1=N0N1, which is P-null, and for ωΩN1 let m1(ω)=max{m0(ω),m1(ω)}. For s,tD[0,1] with |s-t|2-m1(ω), what we have worked out yields

|Xt(ω)-Xs(ω)|C|t-s|γ,|Xt1(ω)-Xs1(ω)|C|t-s|γ.

By induction, we get that for each k1 there are P-null sets N0N1Nk and for each ωΩNk there is some mk(ω) such that for s,tD[0,1] with |s-t|2-mk(ω),

|Xt(ω)-Xs(ω)| C|t-s|γ
|Xt1(ω)-Xs1(ω)| C|t-s|γ
|Xtk(ω)-Xsk(ω)| C|t-s|γ.

Let

Nγ=k1Nk,

which is an increasing sequence of sets whose union is P-null. For ωΩNγ, there is a nondecreasing sequence mk(ω) such that when 0jk and s,tD[j,j+1] with |s-t|2-mk(ω), it is the case that |Xt(ω)-Xs(ω)|C|t-s|γ. For s,tD[0,k+1] with |s-t|2-mk(ω), because |s-t|12, either there is some 0jk for which s,t[j,j+1] or there is some 1jk for which, say, s<j<t. In the first case, |Xt(ω)-Xs(ω)|C|t-s|γ. In the second case, because |j-s|<|t-s|2-mk(ω) and |t-j|<|t-s|2-mk(ω), we get, because s,jD[j-1,j] and j,tD[j,j+1],

|Xt(ω)-Xs(ω)| |Xt(ω)-Xj(ω)|+|Xj(ω)-Xs(ω)|
C|t-j|γ+C|j-s|γ
<2C|t-s|γ.

Thus for

Cγ=2C=2γ+1+41-2-γ,

we have established that for ωΩNγ, k1, and s,tD[0,k+1] satisfying |t-s|2-mk(ω), it is the case that

|Xt(ω)-Xs(ω)|Cγ|t-s|γ. (6)

This implies that for each ωΩNγ and for k1, the mapping tXt(ω) is uniformly continuous on D[0,k+1]. For t0 and ωΩNγ, define

Yt(ω)=limsDstXs(ω). (7)

For each k0, because tXt(ω) is uniformly continuous D[0,k+1]d, where D[0,k+1] is dense in [0,k+1] and d is a complete metric space, the map tYt(ω) is uniformly continuous [0,k+1]d.55 5 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 77, Lemma 3.11. Then tYt(ω) is continuous 0d. For ωNγ, we define

Yt(ω)=0,t0.

Then for each ωΩ, tYt(ω) is continuous 0d. For t0, ωYt(ω) is the pointwise limit of the sequence of mappings ωXs(ω) as st, sD. For each sD, ωXs(ω) is measurable d, which implies that ωYt(ω) is itself measurable d.66 6 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 142, Lemma 4.29. Namely, (Yt)t0 is a continuous stochastic process.

We must show that Y is a modification of X. For sD, for all ωΩNγ we have Ys(ω)=Xs(ω). For t0, there is a sequence snD tending to t, and then for all ωΩNγ by (7) we have Xsn(ω)Yt(ω). P(Nγ)=0, namely, Xsn converges to Yt almost surely. Because Xsn converges to Yt almost surely and P is a probability measure, Xsn converges in measure to Yt.77 7 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 479, Theorem 13.37. On the other hand, for η>0, by Chebyshev’s inequality and (1),

P{|Xsn-Xt|η}η-αE(|Xsn-Xt|α)η-αc|sn-t|1+β,

and because this is true for each η>0, this shows that Xsn converges in measure to Xt. Hence, the limits Yt and Xt are equal as equivalence classes of measurable functions Ωd.88 8 http://individual.utoronto.ca/jordanbell/notes/L0.pdf That is, P{Yt=Xt}=1. This is true for each t0, showing that Y is a modification of X, completing the proof. ∎

3 Hölder continuity

Let (X,d) and (Y,ρ) be metric spaces, let 0<γ<1, and let ϕ:XY be a function. For x0X, we say that ϕ is γ-Hölder continuous at x0 if there is some 0<ϵx0<1 and some Cx0 such that when d(x,x0)<ϵx0,

ρ(ϕ(x),ϕ(x0))Cx0d(x,x0)γ.

We say that ϕ is locally γ-Hölder continuous if for each x0X there is some 0<ϵx0<1 and some Cx0 such that when d(x,x0)<ϵx0 and d(y,x0)<ϵx0,

ρ(ϕ(x),ϕ(y))Cx0d(x,y)γ.

We say that ϕ is uniformly γ-Hölder continuous if there is some C such that for all x,yX,

ρ(ϕ(x),ϕ(y))Cd(x,y)γ.

We establish properties of Hölder continuous functions in the following.99 9 Achim Klenke, Probability Theory: A Comprehensive Course, p. 448, Lemma 21.3.

Lemma 3.

Let V be a nonempty subset of R0, let 0<γ<1, and let f:VRd be locally γ-Hölder continuous.

  1. 1.

    If 0<γ<γ then f is locally γ-Hölder continuous.

  2. 2.

    If V is compact, then f is uniformly γ-Hölder continuous.

  3. 3.

    If V is an interval of length T>0 and there is some ϵ>0 and some C such that for all s,tV with |t-s|ϵ we have

    |f(t)-f(s)|C|t-s|γ, (8)

    then

    |f(t)-f(s)|CTϵ1-γ|t-s|γ,s,tV.
Proof.

For t00, there is some 0<ϵt0<1 and some Ct0 such that when |t-t0|<ϵt0,

|f(t)-f(t0)|Ct0|t-t0|γCt0|t-t0|γ,

showing that f is locally γ-Hölder continuous.

With the metric inherited from 0, V is a compact metric space. For tV and ϵ>0, write

Bϵ(t)={vV:|v-t|<ϵ},

which is an open subset of V. Because f is locally γ-Hölder continuous, for each tV there is some 0<ϵt<1 and some Ct such that for all u,vBϵt(t),

|f(u)-f(v)|Ct|u-v|γ. (9)

Write Ut=Bϵt(t). Because tUt, {Ut:tV} is an open cover of V, and because V is compact there are t1,,tnV such that 𝔘={Ut1,,Utn} is an open cover of V. Because V is a compact metric space, there is a Lebesgue number δ>0 of the open cover 𝔘:1010 10 Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitchhiker’s Guide, third ed., p. 85, Lemma 3.27. for each tV, there is some 1in such that Bδ(t)Uti. Let

C=max{Ct1,,Ctn,2fuδ-γ},

For s,tV with |t-s|<δ, i.e. sBδ(t), there is some 1in with s,tUti. By (9),

|f(s)-f(t)|Cti|s-t|γC|s-t|γ.

On the other hand, for s,tV with |t-s|δ,

|f(s)-f(t)|2fu2fu(|s-t|δ)γ=2fuδ-γ|s-t|γC|s-t|γ.

Thus, for all s,tV,

|f(s)-f(t)|C|s-t|γ,

showing that f is uniformly γ-Hölder continuous.

Let n=Tϵ. For s,tV, because V is an interval of length T, |s-t|Tϵn, and then applying (8), because |t-s|nϵ,

|f(t)-f(s)| =|k=1nf(s+(t-s)kn)-f(s+(t-s)k-1n)|
k=1n|f(s+(t-s)kn)-f(s+(t-s)k-1n)|
k=1nC|t-sn|γ
=Cn1-γ|t-s|γ.

The following theorem does not speak about a version of a stochastic process. Rather, it shows what can be said about a stochastic process that satisfies (1) when almost all of its sample paths are continuous.1111 11 Heinz Bauer, Probability Theory, p. 338, Theorem 39.4.

Theorem 4.

If a stochastic process (Xt)tR0 with state space Rd satisfies (1) and for almost every ωΩ the map tXt(ω) is continuous R0Rd, then for almost every ωΩ, for every 0<γ<βα, the map tXt(ω) is locally γ-Hölder continuous.

Proof.

There is a P-null set N such that for ωΩN, the map tXt(ω) is continuous 0d. For each 0<γ<βα, we have established in (6) that there is a P-null set Nγ such that for k1 there is some mk(ω) such that when s,tD[0,k+1] and |t-s|2-mk(ω),

|Xt(ω)-Xs(ω)|Cγ|t-s|γ, (10)

where Cγ=2γ+1+41-2-γ. Write δ(k,ω)=2-mk(ω), and let Mγ=NγN. For ωΩMγ, the map tXt(ω) is continuous 0d. For k1 and for s,t[0,k+1] satisfying |s-t|δ(k,ω), say with st, let m=t-s2 and let ssnt be a sequence of dyadic rationals decreasing to s and let stnt be a sequence of dyadic rationals inceasing to t. Then sn,tnD[0,k+1] and |sn-tn||s-t|δ(k,ω), so by (10),

|Xtn(ω)-Xsn(ω)|Cγ|tn-sn|γ.

Because ωΩN, Xtn(ω)Xt(ω) and Xsn(ω)Xs(ω), so

|Xt(ω)-Xs(ω)| |Xt(ω)-Xtn(ω)|+|Xtn(ω)-Xsn(ω)|+|Xs(ω)-Xsn(ω)|
|Xt(ω)-Xtn(ω)|+Cγ|tn-sn|γ+|Xs(ω)-Xsn(ω)|
Cγ|t-s|γ,

thus

|Xt(ω)-Xs(ω)|Cγ|t-s|γ,

showing that for 0<γ<βα and ωΩMγ, the map tXt(ω) is locally γ-Hölder continuous.

Let 0<γn<βα be a sequence increasing to βα and let

M=n1Mγn,

which is a P-null set. Let 0<γ<βα and let n be such that γnγ. For ωΩM, the map tXt(ω) is locally γn-Hölder continuous, and because γγn this implies that the map is locally γ-Hölder continuous, completing the proof. ∎

Bauer attributes the following theorem to Kolgmorov and Chentsov.1212 12 Nikolai Nikolaevich Chentsov, 1930–1993, obituary in Russian Math. Surveys 48 (1993), no. 2, 161–166. It does not merely state that for any 0<γ<βα there is a modification that is locally γ-Hölder continuous, but that there is a modification that for all 0<γ<βα is locally γ-Hölder continuous.1313 13 Heinz Bauer, Probability Theory, p. 339, Corollary 39.5.

Theorem 5 (Kolmogorov-Chentsov theorem).

If a stochastic process (Xt)tR0 with state space Rd satisfies (1), then X has a modification Y such that for all ωΩ and 0<γ<βα, the path tYt(ω) is locally γ-Hölder continuous.

Proof.

Applying the Kolmogorov continuity theorem, there is a continuous modification Z of X that also satisfies (1). By Theorem 4, there is a P-null set M such that for ωΩM and 0<γ<βα, the map tZt(ω) is locally γ-Hölder continuous. For t0, define

Yt(ω)={Zt(ω)ωΩM0ωM,

i.e. Yt=1ΩMZt, which is measurable d, and so (Yt)t0 is a stochastic process. For every ωΩ and 0<γ<βα, the map tYt(ω) is locally γ-Hölder continuous. For t0,

{XtYt}={XtYt,Xt=Zt}{XtYt,XtZt}{YtZt}{XtZt}.

Because P(YtZt)=P(M)=0 and P(XtZt)=0, since Z is a modification of X, we get P(XtYt)=0, namely, Y is a modification of X. ∎