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I'm studying the stochastic analysis material, and stuck with a problem which states below:


Suppose $\{X_t, 0 \leq t \leq 1 \}$ is a real-valued stochastic process that satisfies

(a) $X_s$ and $X_t$ are independent if $s \neq t$

(b) Each $X_t$ has the same distribution, and variance 1

(c) the Path $t \mapsto X_t(\omega)$ is continuous for almost every $\omega$.

Show such processes does NOT exist.


My thinking is trying to show by contradiction. So first by (b) the variance = 1, I have $E\left[ {{{\left( {{X_t} - E\left[ {{X_t}} \right]} \right)}^2}} \right] = 1 \Rightarrow E\left[ {{X_t}^2} \right] - {\left( {E\left[ {{X_t}} \right]} \right)^2} = 1$. and since for each $X_t$ have same distribution, I could write

$ P(X_t \in B) = P(X_s \in B)$, for all $B$ in Borel set of real line.

and by (a), I know $E[X_s X_t] = E[X_s] E[X_t]$. But I have no clue about how to combine these result to get the contradiction.

Any thought is appreciated.

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    $\begingroup$ Should (a) read "$X_s$ and $X_t$"? $\endgroup$ – Juho Kokkala Sep 18 '14 at 14:30
  • $\begingroup$ C.f. my stronger, dropping the identical distribution condition, question math.stackexchange.com/q/1117168/64809. $\endgroup$ – Hans Sep 30 '18 at 23:25
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Here's a suggestion. Assume (a) and (b) hold.

  1. Let $\mu = E[X_0]$. Show that $P(X_0 > \mu) > 0$. (If not, then $X_0 \le E[X_0]$ almost surely. Conclude that $X_0$ is constant, contradicting the assumption in (b) that it has variance 1.) Then using continuity from below, show that there is some $\epsilon$ such that $P(X_0 > \mu + \epsilon) > 0$. Likewise, by making $\epsilon$ smaller if necessary, $P(X_0 < \mu - \epsilon) > 0$.

  2. Let $t_n$ be a sequence with $t_n \downarrow 0$ (such as $t_n = 1/n$). Use the second Borel-Cantelli lemma (or a similar argument) to show that, almost surely, $X_{t_n} > \mu + \epsilon$ for infinitely many $n$. Likewise, show that almost surely, $X_{t_n} < \mu - \epsilon$ for infinitely many $n$.

  3. Conclude that almost surely, $\lim_{n \to \infty} X_{t_n}$ does not exist, and hence (c) does not hold.

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  • $\begingroup$ Thanks, but I still confused with the following two points: 1. How to find such an open interval (a,b) in the beginning? Is this because the $X_t$ is defined on [0,1] and by (c), it attains the maximum and minimum? 2. I know how to conclude $X_{t_n} \in (a,b)$ for infinitely many $n$ (by independence of assumption $X_{t_n}$), but how to conclude $X_{t_n} \notin (a,b)$ for infinitely many $n$? $\endgroup$ – Chhsieh Sep 18 '14 at 7:19
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    $\begingroup$ I altered the hint a little bit. For your question 1, essentially you can take $(a,b) = (E[X_0], +\infty)$. Note that I am not assuming that (c) holds (indeed I am showing it does not). For your question 2, with my revised approach the symmetry should be more obvious. $\endgroup$ – Nate Eldredge Sep 18 '14 at 14:26
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    $\begingroup$ @Nate I guess you would need $X_{t_n}>\mu+\epsilon$ and $X_{t_n}<\mu-\epsilon$ infinitely often to preclude convergence; $\epsilon=0$ doesn't quite do it. (I know you know that, but maybe not the OP) :) $\endgroup$ – user940 Sep 18 '14 at 17:16
  • $\begingroup$ @ByronSchmuland: Oops, I know it now that you mention it. :-) Fixed now. $\endgroup$ – Nate Eldredge Sep 18 '14 at 17:48
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    $\begingroup$ Yes, if $P(X_0 \le \mu) = 1$ then $X_0$ is an almost sure constant. In particular, $P(X_0 = \mu) = 1$. (Note that $\mu - X_0$ would be an almost surely nonnegative random variable with expected value zero.) $\endgroup$ – Nate Eldredge Sep 19 '14 at 3:59

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