proof of completeness under ucp convergence


Let (Ω,,(t)t+,) be a filtered probability space, be a sub-σ-algebra of (+), and S be a set of real valued functions on + which is closed (http://planetmath.org/Closed) under uniform convergenceMathworldPlanetmath on compactsPlanetmathPlanetmath. We show that both the set of -measurable processes and the set of jointly measurable processes with sample paths almost surely in S are completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmath (http://planetmath.org/Complete) under ucp convergence. The method used will be to show that we can pass to a subsequence which almost surely converges uniformly on compacts.

We start by writing out the metric generating the topology of uniform convergence on compacts (compact-open topologyMathworldPlanetmath) for functions +. This is the same as uniform convergence on each of the bounded intervals [0,n) for positive integers n,

d(X)n=12-nmin(1,supt<n|Xt|).

Then, the metric is (X,Y)d(X-Y). Convergence under the ucp topology is given by

Ducp(X)=𝔼[d(X)]

for any jointly measurable stochastic process X, with the (pseudo)metric being (X,Y)Ducp(X-Y).

Now, suppose that Xn is a sequence of jointly measurable processes such that Xn-Xmucp0 as m,n. Then, Ducp(Xn-Xm)0 and we may pass to a subsequence Xnk satisfying Ducp(Xnj-Xnk)2-j whenever k>j. So,

𝔼[kd(Xnk-Xnk+1)]=kDucp(Xnk-Xnk+1)k2-k=1.

In particular, this shows that kd(Xnk-Xnk+1) is almost surely finite and, therefore,

d(Xnj-Xnk)i=jk-1d(Xni-Xni+1)0

as k>j, with probability one.

So, the sequence Xnk is almost surely Cauchy (http://planetmath.org/CauchySequence), under the topology of uniform convergence on compacts. We set

Xt(ω){limkXtnk(ω),if the limit exists,0,otherwise.

As measurability of real valued functions is preserved under pointwise convergenceMathworldPlanetmath, it follows that if Xn are -measurable, then so is X. In particular, X is a jointly measurable process. Furthermore, since convergence is almost surely uniform on compacts, if Xn have sample paths in S with probability one then so does X.

It only remains to show that XnucpX. However, we have already shown that d(Xnk-X)0 with probability one, hence Ducp(Xnk-X)0.

Ducp(Xn-X)Ducp(Xn-Xnk)+Ducp(Xnk-X).

Letting k go to infinityMathworldPlanetmathPlanetmath, this is boundedPlanetmathPlanetmathPlanetmath by supm>nDucp(Xn-Xm), which goes to zero as n.

Title proof of completeness under ucp convergence
Canonical name ProofOfCompletenessUnderUcpConvergence
Date of creation 2013-03-22 18:40:35
Last modified on 2013-03-22 18:40:35
Owner gel (22282)
Last modified by gel (22282)
Numerical id 5
Author gel (22282)
Entry type Proof
Classification msc 60G07
Classification msc 60G05