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Below is actually a statement in textbook. But I don't have a good intuition of it.

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.

I hope someone can point out the essence of this argument to me. Also, Is there a continuous process satisfying the first two requirements?

Thanks!

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    $\begingroup$ Doesn't the process $W_t=0$ for all $t$ almost surely satisfy all the conditions and also have continuous paths? What book is this? $\endgroup$
    – Paul Tupper
    Commented Sep 24, 2011 at 16:15
  • $\begingroup$ @Paul: Oksendal's Stochastic Differential Equation. chp3. Certainly he is talking about nontrivial cases. $\endgroup$
    – pde_bk
    Commented Sep 24, 2011 at 16:25
  • $\begingroup$ @duomededuome: The process can't even be (jointly) measurable, which is a much weaker condition than continuity. $\endgroup$
    – George Lowther
    Commented Sep 24, 2011 at 17:14
  • $\begingroup$ After I got a relatively complete understanding(I think;-),let me comment a bit. The proof in spirit of BSteinhurst's post: Contradiction is got by working on $E[|W_s-W_t|^2]$($E[|W_s-W_t|]$ is not easy to work on). Certainly attention should be paid to classification of bounded and unbounded cases. There are other proofs, such as the one on page67 of Doob’s Stochastic Process and the one on page10 of Kallianpur’s Stochastic Filtering Theory, as well as Y.Sun's paper Michael pointed out in the answer. But seemingly they require mutual independence instead of only pair independence. $\endgroup$
    – pde_bk
    Commented Sep 26, 2011 at 21:41
  • $\begingroup$ However those other proofs mentioned above give stronger results as said by George and Michael in their posts. Thank everyone again! $\endgroup$
    – pde_bk
    Commented Sep 26, 2011 at 21:44

2 Answers 2

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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, The almost equivalence of pairwise and mutual independence and the duality with exchangeability, Probab. Theory Relat. Fields 112, 425- 456 (1998). The proof is not very complicated.

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.

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  • $\begingroup$ Ahh. A much more satisfactory answer than mine. Thank you Michael. $\endgroup$
    – BSteinhurst
    Commented Sep 24, 2011 at 18:02
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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.

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  • $\begingroup$ Both answer are instructive. Thank you Ben! $\endgroup$
    – pde_bk
    Commented Sep 24, 2011 at 17:34
  • $\begingroup$ You tried to say that I can construct $t_n$ s.t. $E|W_{t_n}-W_t|>\epsilon$, right? $\endgroup$
    – pde_bk
    Commented Sep 24, 2011 at 17:43
  • $\begingroup$ Yes, I did. Thank you for catching that. $\endgroup$
    – BSteinhurst
    Commented Sep 24, 2011 at 18:00

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