I am trying to prove that a stochastic process with the following properties cannot exist.

Let $\{X_t: 0 \leq t \leq 1 \}$ be a stochastic process such that

i) $X_s$ and $X_t$ are independent whenever $s\neq t$;

ii) Each $X_t$ has the same distribution and variance $1$;

iii) The path $t \rightarrow X_t(\omega)$ is continuous for almost every $\omega$.

Here is my attempt:

Let $\Omega_0$ be the set of full measure on which paths of $X_t(\omega)$ are continuous. I fix a $t \in (0,1)$ and choose a sequence $(t_n)$ such that $t_n \rightarrow t$. Continuity gives $X_{t_n}(\omega) \rightarrow X_t(\omega)$ for every $\omega \in \Omega_0$ (almost surely, that is) and hence in probability.

$$\forall{\varepsilon >0} \quad \lim_{n\rightarrow\infty} P\{\lvert X_{t_n}-X_t\rvert > \varepsilon\} = 0$$

Using the properties (i) and (ii) we can see that $Z_t := \lvert X_{t_n}-X_t\rvert$ has the same distribution for all $n\geq 1$. Therefore $$P\{\lvert X_{t_n}-X_t\rvert > \varepsilon\} =P\{\lvert X_{t_m}-X_t\rvert > \varepsilon\}$$ for any $n,m \geq 1$. This then implies $X_{t} = X_{s}$ almost surely. But this doesn't yet exclude a process like $X_t(\omega) = \omega$. At some point I need to show that $X_t(\omega) = C$ for some constant $C$ to get a contradiction with the variance property and this is where I am stuck. Could someone point out if my proof is OK so far and how I finish it? Thanks!

  • 1
    $\begingroup$ This question is related: math.stackexchange.com/q/1117168 $\endgroup$
    – saz
    Mar 26, 2015 at 14:35
  • $\begingroup$ You are welcome. Does this answer your question (about how to finish the proof)? $\endgroup$
    – saz
    Mar 26, 2015 at 15:13
  • $\begingroup$ @Saz It definitely does. However, now I need to study Kolmogorov's 0-1 Law, which we had not discussed in our lectures (I am doing a course on SDE's not on Measure theoretic probability per se so we skip some of these important results). But I will try to find an argument (inspired by Kolmogorov) which we did cover in class. I have an additional property for the process, namely the identical distribution of $(X_t)$. Maybe that helps me come up with something simpler. $\endgroup$
    – Calculon
    Mar 26, 2015 at 15:20
  • $\begingroup$ Yeah, there is in fact a simpler proof. See my answer below. $\endgroup$
    – saz
    Mar 26, 2015 at 19:43

1 Answer 1


You have already shown that $X_t = X_s$ almost surely for any $s,t \geq 0$. It follows from the continuity of the process, that there exists a random variable $Y$ such that $$X_t = Y \qquad \text{for all $t \geq 0$ almost surely}.$$ Since, by (i), $X_t$ and $X_s$ are independent for $s \neq t$, we get that $X_t = Y$ is independent of $X_s = Y$. Hence, $Y$ is constant (i.e. does not depend on $\omega$).

Remark: If we drop assumption (ii), then we get $X_t = f(t)$ for a (deterministic) function $f$; see this question.

  • $\begingroup$ That is exactly the kind of answer I was hoping to find. Thank you! $\endgroup$
    – Calculon
    Mar 26, 2015 at 20:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .