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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 $t_2$?

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    $\begingroup$ Sure, let $X_t$ be a constant process, e.g. $X_t=0$ for all $t$. $\endgroup$
    – Math1000
    Jan 24, 2015 at 1:40
  • $\begingroup$ @Math1000 Really ? $\endgroup$
    – Olórin
    Jan 24, 2015 at 1:57
  • $\begingroup$ 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. $\endgroup$
    – Math1000
    Jan 24, 2015 at 2:09
  • $\begingroup$ @Math1000 And why is the value of $X_t$ independent of the value of $X_s$ ? $\endgroup$
    – Olórin
    Jan 24, 2015 at 2:11
  • $\begingroup$ $\mathbb P(X_t = x | X_s = 0)$ is $1$ if $x=0$, and zero otherwise. I suppose you're saying you run into problems when conditioning on something like $\{X_s\neq 0\}$? $\endgroup$
    – Math1000
    Jan 24, 2015 at 2:50

1 Answer 1

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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.

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  • $\begingroup$ Thank you, saz. Could you please provide a reference or a link to the proof of the above lemma? $\endgroup$
    – Hans
    Jan 24, 2015 at 7:44
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    $\begingroup$ Hans: Kolmogorov zero-one law. $\endgroup$
    – Did
    Jan 24, 2015 at 8:13
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    $\begingroup$ @Hans See e.g. "Kai Lai Chung: A course in probability theory", Theorem 4.2.5+Corollary. $\endgroup$
    – saz
    Jan 24, 2015 at 8:56
  • $\begingroup$ 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? $\endgroup$
    – Hans
    Jan 24, 2015 at 19:21
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    $\begingroup$ @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. $\endgroup$
    – saz
    Jan 24, 2015 at 20:45

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