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Why do we fix $\omega$ in the definition of a sample path of a stochastic process? If $f:T \to \Omega$ is an arbitrary function, can't we define sample path to be equal to $\{X(t, f(t)): t \in T\}$?

Denisof
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    A sample path is usually function $T\to\mathbb R$ prescribed by $t\mapsto X_t(\omega)$. Every $\omega\in\Omega$ induces such a function. In that sense the label "sample path" is reserved. What is the use placing the label also on sets (so not functions) as described in your question? – drhab Nov 27 '18 at 13:35
  • I can rephrase my question in your notation, why can't we define sample path as $t \to X_t(f(t))$, where $f:T \to \Omega$ is an arbitrary function? I'm asking this, because it seems for me that these are exactly the same, though people use only one of them. – Denisof Nov 27 '18 at 14:07
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    They are not exactly the same. You can at most say that the function $t\mapsto X_t(\omega)$ is a special case of the (more general) $t\mapsto X_t(f(t))$. It arises if $f$ is a constant function. If you are interested in this more general case then it might be handy to find a suitable label for it. Labels that allready exist must not be used for that. – drhab Nov 27 '18 at 14:19
  • In case of i.i.d.'s they are the same, yes. Basically, I was wondering, why do people use this definition instead of the more general one. – Denisof Nov 27 '18 at 14:26
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    I don't understand what you are trying to say. What is the role of the concept "iid" in this? If $f$ is not a constant function then it is most unlikely that you can find an $\omega\in\Omega$ such that the functions $t\mapsto X_t(\omega)$ and $t\mapsto X_t(f(t))$ are the same. – drhab Nov 27 '18 at 14:35
  • Well, it's not so clear to me. In simple words, $X_\omega(t)$ is some random function, that is its values are random variables. For example, if all the random variables $X_t$ have standard normal distribution, is there any difference between my definition and the one you wrote? Where does this $\omega$ kick in? – Denisof Nov 27 '18 at 15:39
  • Check this answer if still interested. So as I understood, $\omega$ in the definition of a sample path is an element of $\Omega_{X_t}^T$. – Denisof Nov 28 '18 at 11:44
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    If we identify heads with $0$ and tails with $1$ then in that situation $\Omega={0,1}^T$ for $T=\mathbb N$. Then a stochastic process can be looked at as a function $X:\mathbb N\times\Omega\to\mathbb R$. Setting $X_n:=X(n,-):\Omega\to\mathbb R$ it can be looked at as collection of random variables: ${X_n\mid n\in\mathbb N}$. A sample path is then a function $\mathbb N\to\mathbb R$ prescribed by $n\mapsto X_n(\omega)$ or (if you like that more) a sequence $(X_1(\omega),X_2(\omega),\cdots)$ – drhab Nov 28 '18 at 13:43
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    You call $\omega$ an element of $\Omega^T_{X_t}$ in that context. I don't know what you mean by that. $\omega$ is an element of $\Omega$. In this case that means that any $\omega$ can be looked at as a sequence like $(0,0,1,0,1,1,0,\dots)$. – drhab Nov 28 '18 at 13:45
  • So $\Omega$ of random variable $X_n$ is ${\text{H}, \text{T}}$, but $\omega$ from the definition of a path is not an element of $\Omega$, it's instead an element of the set of all sequences $2^\mathbb{N}$. – Denisof Nov 28 '18 at 14:13
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    As said: $X_n:\Omega\to\mathbb R$ where $\Omega={0,1}^{\mathbb N}$ (not ${H,T}$) and every $\omega\in\Omega$ induces sample path which is a function $\mathbb N\to\mathbb R$. Every $X_n$ takes values in ${0,1}$ (or ${H,T}$ if you like) but that does not mean that ${0,1}$ is the "$\Omega$ of $X_n$". You can at most say that $X_n$ induces a distribution on its range ${0,1}$. – drhab Nov 28 '18 at 14:40
  • Thanks, now it's clear, I was confused, because in many sources it's usually written something like this: for any $n \in \mathbb{N}$, $X_n$ is a random variable with defined probability space $(\Omega, \mathcal{F}, P)$, and path is a function $X_\omega(n) = X(\omega, n)$, where $\omega \in \Omega$. I was confused by this notation. For example here https://en.wikipedia.org/wiki/Stochastic_process it's written so, it's not so clear that $\Omega$ in case of heads and tails is actually a set of functions from $T$ to ${0,1}$. – Denisof Nov 28 '18 at 15:45

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