Non-existence of such a continuous stochastic process - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:58:28Z http://mathoverflow.net/feeds/question/76274 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76274/non-existence-of-such-a-continuous-stochastic-process Non-existence of such a continuous stochastic process pde_bk 2011-09-24T15:00:53Z 2011-09-24T17:14:32Z <p>Below is actually a statement in textbook. But I don't have a good intuition of it.</p> <p>If we want a stochastic process $W_t$ to satisfy i). $s\neq t$ implies $W_s$ and $W_t$ are independent, ii). ${W_t}$ is stationary, iii). $E[W_t]=0$ for all t, then $W_t$ cannot have continuous paths.</p> <p>I hope someone can point out the essence of this argument to me. Also, Is there a continuous process satisfying the first two requirements?</p> <p>Thanks!</p> http://mathoverflow.net/questions/76274/non-existence-of-such-a-continuous-stochastic-process/76276#76276 Answer by BSteinhurst for Non-existence of such a continuous stochastic process BSteinhurst 2011-09-24T16:39:46Z 2011-09-24T16:39:46Z <p>Suppose you had such a process that is not trivial. Suppose you have $W_s \neq W_0$. For $t >s$ we have assumed that $W_s$ and $W_t$ are independent, have mean zero, and the same distribution. Now choose a sequence $t_n \downarrow s$ such that $|W_{t_n} - W_s| > \epsilon$. This sequence exists for some $\epsilon >0$ since $W_t$ is independent of $W_s$, but has the same distribution as $W_s$. Once you has this sequence you can quickly see that the process cannot be continuous. </p> http://mathoverflow.net/questions/76274/non-existence-of-such-a-continuous-stochastic-process/76277#76277 Answer by Michael Greinecker for Non-existence of such a continuous stochastic process Michael Greinecker 2011-09-24T17:14:32Z 2011-09-24T17:14:32Z <p>Something much stronger holds. One can actually show that no nontrivial such process has measurable sample parths. No assumption on the mean and no stationarity assumption is necessary. This is Proposition 1.1. in Y. Sun, <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.42.2689&amp;rep=rep1&amp;type=pdf" rel="nofollow">The almost equivalence of pairwise and mutual independence and the duality with exchangeability</a>, Probab. Theory Relat. Fields 112, 425- 456 (1998). The proof is not very complicated.</p> <p>In terms of the intuition, continuity tells you that you can understand the behavior at a point by observing points close by- independence tells you you cannot learn anything from such points.</p>