Stochastic processes. Continuity and differentiability of sample …roustant/Documents/Sample...
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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
Introduction Sample functions properties in quadratic mean Sample functions properties References
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
Introduction Sample functions properties in quadratic mean Sample functions properties References
INTRODUCTIONSAMPLE FUNCTIONS PROPERTIES
Introduction Sample functions properties in quadratic mean Sample functions properties References
Example (from km.sim{DiceKriging})
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Introduction Sample functions properties in quadratic mean Sample functions properties References
FIRST PARTSAMPLE FUNCTIONS PROPERTIES in QUADRATIC MEAN
Introduction Sample functions properties in quadratic mean Sample functions properties References
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.
Introduction Sample functions properties in quadratic mean Sample functions properties References
Processes with finite second-order moments
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.
Introduction Sample functions properties in quadratic mean Sample functions properties References
Processes with finite second-order moments
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 <, >.
Introduction Sample functions properties in quadratic mean Sample functions properties References
Processes with finite second-order moments
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 >.
Introduction Sample functions properties in quadratic mean Sample functions properties References
Processes with finite second-order moments
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...
Introduction Sample functions properties in quadratic mean Sample functions properties References
2nd order stationary processes
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).
Introduction Sample functions properties in quadratic mean Sample functions properties References
2nd order stationary processes
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.
Introduction Sample functions properties in quadratic mean Sample functions properties References
2nd order stationary processes
A difficulty
Last results are easy to obtain. However :Continuity or differentiability in q.m. by no means necessarilyimplies sample function continuity or differentiability.
Introduction Sample functions properties in quadratic mean Sample functions properties References
SECOND PARTSAMPLE FUNCTIONS PROPERTIES
Introduction Sample functions properties in quadratic mean Sample functions properties References
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
Introduction Sample functions properties in quadratic mean Sample functions properties References
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
Introduction Sample functions properties in quadratic mean Sample functions properties References
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.
Introduction Sample functions properties in quadratic mean Sample functions properties References
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).
Introduction Sample functions properties in quadratic mean Sample functions properties References
Sample function continuity
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.
Introduction Sample functions properties in quadratic mean Sample functions properties References
Sample function continuity
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.
Introduction Sample functions properties in quadratic mean Sample functions properties References
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].
Introduction Sample functions properties in quadratic mean Sample functions properties References
Sample function differentiability
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.
Introduction Sample functions properties in quadratic mean Sample functions properties References
Sample function differentiability
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.
Introduction Sample functions properties in quadratic mean Sample functions properties References
Sample function differentiability
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.
Introduction Sample functions properties in quadratic mean Sample functions properties References
REFERENCES
Introduction Sample functions properties in quadratic mean Sample functions properties 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.