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Assume $\Omega $={head, tail}, let T=$\mathbb N$ and $X_t$ $t\in T$ be a collection of i.i.d random variables following Bernoulli distribution. Since a stochastic process is a function of two variables. When $\omega$ is fixed, we get a sample path, so when fix $\omega$={head}, Is the sample path of $X_t(\omega)$ be the constant 1? But a realization of coin tossing can't be always head . So where is my fault?

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The error is that $\Omega=$ {heads,tails}$^T$, where {heads,tails}$^T$ is the set of all functions from $T$ to {heads,tails}. So a typical $\omega$ is an infinite sequences whose entries are either heads or tails.

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  • $\begingroup$ So the space of brownian motion is $\mathbb C(R^+,R)$, I see. $\endgroup$
    – user148857
    May 10 '14 at 8:47
  • $\begingroup$ What does $\Omega=\{heads,tails\}^T$ mean? Thanks! $\endgroup$ Aug 29 '16 at 19:09
  • $\begingroup$ @XianjinYang: I edited the question to answer your comment. $\endgroup$ Sep 5 '16 at 13:58
  • $\begingroup$ @SergioParreiras How to define a measure on a $\Omega=\{heads, tails\}^T$ if $\omega\in \Omega$ is a function? $\endgroup$ Sep 5 '16 at 14:11
  • $\begingroup$ @XianjinYang: if you don't like the word "function", think of it as a "sequence". You can use The Kolmogorov Extension Theorem to define a measures in this space, en.wikipedia.org/wiki/Kolmogorov_extension_theorem $\endgroup$ Sep 5 '16 at 14:21

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