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Introduction Sample functions properties in quadratic mean Sample functions properties References

Stochastic processes. Continuity anddifferentiability of sample functions

O. Roustant

École Nationale Supérieure des Mines de Saint-Étienne

12th of March 2009


Outline

1 Introduction

2 Sample functions properties in quadratic meanProcesses with finite second-order moments2nd order stationary processes

3 Sample functions propertiesSample function continuitySample function differentiability

4 References


INTRODUCTIONSAMPLE FUNCTIONS PROPERTIES


Example (from km.sim{DiceKriging})

0.0 1.0 2.0 3.0

0.00.2

0.40.6

0.81.0

distance

covariance

expmatern3_2matern5_2gauss

0.0 1.0 2.0 3.0

-2-1

01

2

input, x

outpu

t, f(x)


FIRST PARTSAMPLE FUNCTIONS PROPERTIES in QUADRATIC MEAN


Processes with finite second-order moments

Definitions and preliminary remarksLet Xt be a stochastic process.

Xt has finite second-order moments if ∀t , E(X 2t ) <∞

By Cauchy-Schwartz inequality this implies that∀u, v , E(Xu) and E(XuXv ) are well defined.As usual, L2 denotes the set of random variables with finitesecond order moments. Let us recall that L2 is a Hilbertspace, with < X , Y >= E(XY )

In the following, we will assume that E(Xt ) = 0 (if it not thecase, just consider Xt − E(Xt ))

We denote C(u, v) =< Xu, Xv > the covariance function.



Convergence in quadratic mean

DefinitionLet X1, X2, ..., Xn, ... and Y be r.v. with finite variances.Xn → Y in quadratic mean (q.m.) if ‖Xn − Y‖L2 → 0

Loeve criterionLet X1, X2, ..., Xn be n random variables with finite variances.Then Xn converges in q.m. iif < Xn, Xm > converges to a finitelimit c when n, m tend independently to infinity.

Proof hints.

For the "if" part, show that <, > is continuous.For the "iif" part, show that Xn is a Cauchy sequence.

Exercice. Find a counter example if n, m are not chosenindependently.



Continuity

Definition - Continuity in quadratic meanXt is continuous in q.m. at t = t0 if Xt → Xt0 in q.m.

Proposition1 Xt is continuous in q.m. at t = t0 iif the covariance function

C(u, v) is continuous at the diagonal point (t0, t0).2 If C(u, v) is continuous at every diagonal point (t , t), then it

is continuous everywhere.

Proof hints.1 For the "if" part, develop the expression E(Xt+h − Xt )

2

For the "iif" part, use the equality :C(t + h, t + k)− C(t , t) =< Xt+h − Xt , Xt+k − Xt >+ < Xt+h − Xt , Xt > + < Xt+k − Xt , Xt >

2 Use (1) and continuity of <, >.



Differentiability

Definition - Differentiability in quadratic mean

Xt is differentiable in q.m. at t if Xt+h−Xth converges in q.m.

Proposition

1 If ∂2C∂u∂v exists at (t , t), then Xt is differentiable in q.m. at t .

2 If ∂2C∂u∂v exists for every diagonal point (t , t), then ∂C

∂u (u, v)

and ∂2C∂u∂v (u, v) exist everywhere and we have :

cov(X ′u, Xv ) = ∂C∂u (u, v) and cov(X ′u, X ′v ) = ∂2C

∂u∂v (u, v)

Proof hint.1 Apply Loeve criterion to Yn =

Xt+hn−Xthn

for any seq. hn → 0.2 For the 1st derivative, use (1) and develop <

Xu+h−Xuh , Xv >.

Then develop <Xu+h−Xu

h ,Xv+k−Xv

k >.



Differentiability - Higher orders

Exercice

Show that if ∂4C∂u2∂v2 exists at (t , t), then Xt is diff. twice in q.m. at

t . In addition, if ∂4C∂u2∂v2 exists at every diagonal point, then all

derivatives written below exist everywhere and we have :

cov(X ′′u , Xv ) = ∂2C∂u2 (u, v)

cov(X ′′u , X ′v ) = ∂3C∂u2∂v (u, v)

cov(X ′′u , X ′′v ) = ∂4C∂u2∂v2 (u, v)

Generalize to any order.

Proof hint.For 2nd derivative existence, apply last proposition to X ′tand C′(u, v) := cov(X ′u, X ′v ).To prove the formulas, develop relevant expressions...


2nd order stationary processes


Definition

A stochastic process (Xt ) is 2nd order stationary if for any t , h,E(Xt ) and cov(Xt , Xt−h) do not depend on t .

Recall that we restrict to centered stochastic processes.Thus (Xt ) is stationary if C(u, v) is a function of u − v .We denote : C(h) = C(t , t − h).



Analytic properties of sample functions

PropositionLet Xt be a stationary stochastic process.

1 Xt is continuous in q.m. at t = t0 iif C(h) is continuous at 0.In this case, Xt is continuous everywhere.

2 If C(2k)(h) exists in an open set containing 0, then Xt isdifferentiable in q.m. at order k everywhere.

Proof hint.Show that the local properties of C(h) at 0 imply the sameproperties to C(u, v) at the diagonal points.



A difficulty

Last results are easy to obtain. However :Continuity or differentiability in q.m. by no means necessarilyimplies sample function continuity or differentiability.


SECOND PARTSAMPLE FUNCTIONS PROPERTIES


A difficulty

Exercice 1Let Xt and Yt two stochastic processes defined over [0, 1] by :Xt (ω) = 0 ∀t , ωYt (ω) = 1 if t = ω and 0 otherwiseThen Xt and Yt have the same finite-dimensional distributionsbut :P({Xt is continuous in [0, 1]}) = 1P({Yt is continuous in [0, 1]}) = 0


A difficulty (following)

Exercice 2Let Xt and Yt two stochastic processes defined over [0, 1] by :Xt (ω) = 0 ∀t , ωYt (ω) = 1 if t =

1+ωn

n+1 and 0 otherwiseThen Xt and Yt have the same finite-dimensional distributions,but :P({Xt is continuous at 0}) = 1P({Yt is continuous at 0}) = 0


Equivalent processes

DefinitionWe say that Xt and Yt are equivalent if they have the samefinite dimensional distributions :

∀t , P({Xt = Yt}) = 1

Remarks.

This implies that two equivalent processes have the samefamily of finite-dimensional distributions.The examples above show that two equivalent processesdo NOT have the same sample functions properties.


Sample function continuity

Sample functions continuity - Kolmogorov theoremLet Xt be a stochastic process defined over [0, 1]. Suppose thatfor all t , t + h in [0, 1],

P({|Xt+h − Xt | ≥ g(h)}) ≤ q(h)

where g and q are even functions of h, non increasing as h ↓ 0,and such that

∞∑n=1

g(2−n) <∞ and∞∑

n=1

2nq(2−n) <∞.

Then there exists an equivalent stochastic process Yt whosesample functions are, with probability one, continuous on [0, 1].

Proof. See (Cramér, Leadbetter, 1967).



CorollaryIf with the notation above we have

E |Xt+h − Xt |p ≤K |h|

|log|h||1+r

where p < r and K are positive constants, the conclusion of thetheorem holds.

Proof. Apply Markov inequality : P(|X | ≥ a) ≤ |X |p

ap , and takeg(h) := |log|h||−b with 1 < b < r/p.



Stochastic processes with finite 2nd-order moments

Let Xt be a stochastic process with finite 2nd order moments. Iffor all t and t + h in [a, b] the difference

∆2hC(t , t) := C(t + h, t + h)− C(t + h, t)− C(t , t + h)− C(t , t)

satisfies an inequality of the form ∆2hC(t , t) < K |h|

|log|h||q , withq > 3 and K > 0, then Xt is equivalent to a stochastic processwhich, with probability one, is sample continuous.

Stationary processesLet Xt be a stationary stochastic process. If C′′(0) exists, thenXt is equivalent to a stochastic process which, with probabilityone, is sample continuous.

Proof. Apply corollary with p = 2.


Sample function differentiability

Sample function differentiability - TheoremLet Xt be a stoc. process defined over [0, 1]. Suppose that thehypothesis of Kolmogorov theorem hold, and that, moreover, forall t − h, t , t + h in [0, 1],

P({|Xt+h + Xt−h − 2Xt | ≥ g1(h)}) ≤ q1(h)

where g1 and q1 are even functions of h, non increasing ash ↓ 0, and such that

∞∑n=1

2ng1(2−n) <∞ and∞∑

n=1

2nq1(2−n) <∞.

Then Xt is equivalent to a process which, with probability one,has continuous sample functions derivatives in [0, 1].



CorollaryIf the conditions of Kolmogorov theorem’s corollary are satisfiedand if, moreover

E |Xt+h + Xt−h − 2Xt |p ≤K |h|1+p

|log|h||1+r

where p < r and K are positive constants, the conclusion of thetheorem holds.

Proofs. For the theorem, see (Cramér, Leadbetter, 1967). Forthe corollary, apply again Markov inequality.



Stochastic processes with finite 2nd-order moments

Let Xt be a stochastic process with finite 2nd order moments. Iffor all t , h the fourth difference ∆4

hC(t , t) satisfies an inequality

of the form ∆4hC(t , t) < K |h|3

|log|h||q , with q > 3 and K > 0, then Xt isequivalent to a process which, with probability one, hascontinuous sample function derivatives.

Stationary processes

Let Xt be a stationary stochastic process. If C(4)(0) exists, thenXt is equivalent to a process which, with probability one, issample continuous.

Proof. Again appply corollary with p = 2.



Differentiability - Higher ordersThere are analogous results. In particular, if Xt is a stationarystochastic process and if C(2k+2)(0) exists, then Xt isequivalent to a process which, with probability one, has Ck

sample functions.


REFERENCES


ReferencesCramér H., Leadbetter M.R. (1967), Stationary andRelated Stochastic Processes - Sample FunctionProperties and Their Applications, Wiley.Rasmussen C.E., Williams C.K.I. (2006), GaussianProcesses for Machine Learning, the MIT Press.