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.
Here's a suggestion. Assume (a) and (b) hold.
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, $P(X_0 < \mu - \epsilon) > 0$.
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 $n$.
Conclude that almost surely, $\lim_{n \to \infty} X_{t_n}$ does not exist, and hence (c) does not hold.