# State Space and Sigma Algebra for stochastic Process

Let $X_t$ be a random variable on $(\Omega,\mathcal{F} ,\rho)$ for all non-negative $t$.

From what I understand, the state-space for the stochastic process is the infinite tuple $(\omega_1, \omega_2,\ldots)$

What will be the sigma algebra for this? How is it related to the sigma algebra of the original random variables?

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The state space is simply $\Omega$. – Michael Greinecker Feb 13 '13 at 18:23
Thanks for a quick reply! However then we will severely constrain the state space of the process. We can only analyze those sample paths, in which each variable has taken the same state-space variable. What if X_1 and X_2 take different omegas? – MMM Feb 13 '13 at 18:28
All your $X_t$ are defined on $\Omega$, so there are no different omegas. – Michael Greinecker Feb 13 '13 at 18:46
No, I mean $X_1$ takes $\omega_1$ and $X_2$ takes $\omega_2$. In this case wouldn't we need $\omega$ = $(\omega_1,\omega_2,....)$ for capturing all sample paths. If we say \Omega is the sample space for it, the only possible sample paths will be $(\omega,\omega,....)$ For example in Markov processes we take $X_1$ = i, and $X_2$ = j. Both random variables are taking two different statespace variables there. – MMM Feb 13 '13 at 18:58
Nothing prevents $\Omega$ from being already a space of sequences. But if $X_t$ is a random variable on $\Omega$, than it can only depend on elements of $\Omega$. "e^\cdot" and $\cdot+2$ are both functions defined on $\mathbb{R}$. That these are two different functions does not mean that they have to be somehow be defined on $\mathbb{R}^2$. – Michael Greinecker Feb 13 '13 at 19:10

I think what the OP really wants to know is the following. Suppose $X_1$, $X_2$, $\dots$ take values in a state space $\mathfrak{X}$. In this case, the state space for the stochastic process is $\Omega = \mathfrak{X}^\mathbb{N}$. Now $\mathfrak{X}$ itself may be a measurable space with a sigma-algebra $\mathcal{E}$. I believe the OP wants to know how the sigma-algebra $\mathcal{F}$ for $\Omega$ can be obtained from the sigma-algebra $\mathcal{E}$ for $\mathfrak{X}$.
More abstractly, if $(X,\mathcal{F}_X)$ and $(Y,\mathcal{F}_y)$ are measurable spaces, then is there a "natural" sigma-algebra $\mathcal{F}$ on the set $X^Y$ that makes $(X^Y,\mathcal{F})$ into a measurable space? That is, does the category of measurable spaces allow exponential objects? I don't know the answer, and would be very interested to know it myself.
No, there is no natural $\sigma$-algebra of this kind. – Michael Greinecker Jan 28 '14 at 18:50