3

Suppose $X_t$ is a stochastic process of $t$ on $[0,\infty)$ with almost surely continuous sample path. Does $X_t$ have to be almost surely deterministic and almost surely continuous in $t$ so that $X_{t_1}$ and $X_{t_2}$ are independent of each other for arbitrary and distinct $t_1$ and $t_2$?

Hans
  • 9,804
  • 1
    Sure, let $X_t$ be a constant process, e.g. $X_t=0$ for all $t$. – Math1000 Jan 24 '15 at 01:40
  • @Math1000 Really ? – Olórin Jan 24 '15 at 01:57
  • Well, clearly any sample path $X_t(\omega)$ is continuous as it is a constant function, and the value of $X_t$ is independent of the value of $X_s$. It is a trivial example, but it works. – Math1000 Jan 24 '15 at 02:09
  • @Math1000 And why is the value of $X_t$ independent of the value of $X_s$ ? – Olórin Jan 24 '15 at 02:11
  • $\mathbb P(X_t = x | X_s = 0)$ is $1$ if $x=0$, and zero otherwise. I suppose you're saying you run into problems when conditioning on something like ${X_s\neq 0}$? – Math1000 Jan 24 '15 at 02:50
  • @Math1000 You could see this more easily via the $\sigma$-algebras generated by $X_s$ and $X_t$ : they will be trivial, and therefore trivially independent, and then by definition $X_s$ and $X_t$ will be independent. – Olórin Jan 24 '15 at 02:58
  • Yeah, I agree with that. I thought you were disagreeing with me so I was confused... – Math1000 Jan 24 '15 at 04:31
  • @RobertGreen: Why did you delete your answer? Was it not right? Math1000's answer is very clever. That would however lead me to modify my question to whether the constant is necessarily the only satisfying process. Robert Green's answer seemed to negate the last statement. I have a few questions regarding Robert Green's now deleted answer. Could you please put it back on? Thanks. – Hans Jan 24 '15 at 06:42
  • @Math1000: Very clever answer. Please also see my above comment. – Hans Jan 24 '15 at 06:43
  • I didn't think it was very clever but thanks :) – Math1000 Jan 24 '15 at 12:50

1 Answers1

4

Recall the following statement from probability theory:

Theorem: Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of pairwise independent random variables. Then $Y:=\limsup_{n \to \infty} Y_n$ is constant almost surely.

Fix $t \geq 0$ and a sequence $(t_n)_{n \in \mathbb{N}}$, $t_n \neq t$, such that $t_n \to t$. By the continuity of the process $(X_s)_{s \geq 0}$, we know that $$Y_n := X_{t_n} \to X_t =: Y \qquad \text{almost surely as $n \to \infty$.}$$ Appyling the above lemma, we get that $X_t$ is constant almost surely. Since this holds for all $t \geq 0$, we conclude from the continuity of the process that $X_t=f(t)$ almost surely for some continuous function $f:[0,\infty) \to \mathbb{R}$ (which does not depend on $\omega$).

Remark: Note that the situation is totally different if we assume that $(X_t)_{t \geq 0}$ has independent increments. This leads to so-called additive processes.

Proof of the theorem: We use the following version of the Borel-Cantelli lemma (for a proof see e.g. Kai Lai Chung: A Course in Probability Theory, Theorem 4.2.5 + Corollary):

Let $(A_n)_{n \in \mathbb{N}}$ be a sequence of pairwise independent events. Then $$\mathbb{P} \left( \limsup_{n \to \infty} A_n \right) \in \{0,1\}. \tag{1}$$

Fix $c \in \mathbb{R}$ and set

$$A_n := \{Y_n \geq c\}.$$

Since the sequence $(Y_n)_{n \in \mathbb{N}}$ is pairwise independent, the sequence $(A_n)_{n \in \mathbb{N}}$ is also pairwise independent. Moreover,

$$\begin{align*} \limsup_{n \to \infty} A_n &= \bigcap_{n \in \mathbb{N}} \bigcup_{k \geq n} \{Y_k \geq c\} = \left\{ \limsup_{n \to \infty} Y_n \geq c \right\}. \end{align*}$$

From $(1)$ we conclude

$$\mathbb{P} \left( \limsup_{n \to \infty} Y_n \geq c \right) \in \{0,1\}.$$

This means that $\limsup_{n \to \infty} Y_n$ is almost surely constant.

saz
  • 120,083
  • Thank you, saz. Could you please provide a reference or a link to the proof of the above lemma? – Hans Jan 24 '15 at 07:44
  • 1
    Hans: Kolmogorov zero-one law. – Did Jan 24 '15 at 08:13
  • 1
    @Hans See e.g. "Kai Lai Chung: A course in probability theory", Theorem 4.2.5+Corollary. – saz Jan 24 '15 at 08:56
  • Wait, saz and @Did. $Y$ is constant almost surely refers to being constant with respect to sample $\omega$. So $X_t$ is constant almost surely on sample space $\Omega$ for given $t$. Then the conclusion should be $X_t$ is a continuous function of $t$ only, but not necessarily a constant in $t$, right? – Hans Jan 24 '15 at 19:21
  • @Hans Yeah, sure, you are right. Obviosuly, any process of the form $X(t,\omega) := f(t)$ for some continuous $f$ satisfies the claim. – saz Jan 24 '15 at 19:31
  • Could you please elucidate another detail? How do you reckon the null set where discontinuity occurs and the one where $X_t$ is not constant with respect to $\omega$ and conclude that the sample subset where $X_t(\omega)$ is discontinuous and stochastic is null? – Hans Jan 24 '15 at 20:36
  • 1
    @Hans Since the process is (almost surely) continuous, it can be uniquely described by the sequence of random variables $(X_{t_n}){n \in \mathbb{N}}$ where $(t_n){n \in \mathbb{N}} \subseteq [0,\infty)$ is dense. This means that we have countable many exceptional null sets and since the union of countable many null sets is again a null set, the claim follows. – saz Jan 24 '15 at 20:45
  • Would you be so kind as to derive your lemma from "Kai Lai Chung: A course in probability theory", Theorem 4.2.5+Corollary? I have some vague ideas but have some confusion on specification of $E_n$ in Theorem 4.2.5. – Hans Jan 25 '15 at 21:45
  • @Hans I have edited my answer. – saz Jan 26 '15 at 14:03
  • Setting the constant $c$ and letting $\limsup\limits_{n\to\infty}>c$ was the key and was what I missed. Thank you, Saz. – Hans Jan 27 '15 at 06:13