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So, I try to understand the confusing terminology of stochastics - especially that of stochastic processes.

Let $(\Omega,\Sigma,P)$ be a probability space and $(Z,\Sigma')$ a measure space with $\Sigma\Sigma'$-measureable functions $X_t$, i.e. stochastic variables at times t. The stochastic process X is a family of these stochastic variables $X=(X_t)_{t \in T}$ with the index set $T$.
$\omega \in \Omega$ denotes the possible outcomes.

Then, the following maps are defined:
$X:\Omega \times T\longrightarrow Z, (\omega,t) \mapsto X_t(\omega)$

$X(\omega,\cdot):T \longrightarrow Z, t \mapsto X_t(\omega)$

$X_t:\Omega \longrightarrow Z, \omega \mapsto X_t(\omega)$

$\\$

My question is how the sample space $\Omega$ looks like in a stochastic process.

For that let us consider a trivial example: At two times $t \in T=\{1,2\}$ a coin is flipped to either heads or tails. For heads a player loses 100 dollars, for tails he wins 100. Let the stochastic variable X be the won amount of money.

Is the sample space:

A) $\Omega=\{(heads,heads),(heads,tails),(tails,heads),(tails,tails)\}$

or

B) $\Omega=\{heads,tails\}$ ?

$\\\\$

My assumption is that A) is the right answer as I think it would be consistent with the above defined functions $X, X(\omega,\cdot), X_t$: For example $X_t$ would give out the amount won at t, like $X_1((heads,tails))=-100$ or $X_2((heads,tails))=-100+100=0$.

The stochastic process gives out the won amount for a given outcome $\omega$ and time t, like $X(((heads,tails),1))=X_1((heads,tails))=-100$ ,where $((heads,tails),1) \in \Omega \times T$

The trajectory $X(\omega,\cdot)$ gives out the won amount for a fixed outcome $\omega$ at different times t.

In this case the state space would look like this: $Z=\{-200,-100,0,100,,200\}$

It makes perfectly sense to me and seems consistent. Nonetheless Im a bit confused whether my understanding of stochastic processes and variables is even right or I may have some misunderstanding and B) is the right answer. What makes me doubt of my A) is that the size of the sample space would vary with the times the coin is flipped. This wouldn't be the case if B) is correct.

Thank you very much.

S. Den
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    $\Omega$ is what you define it to be. You could also say $\omega \in \Omega$ is one of infinitely many universe where at some point in time there will be two dices beeing tossed.

    The aim of you random variables is to remove the unecessary information in $\Omega$ and reduce it only to the important ones.

    So $\Omega$ must atleast be $A)$ because those are all necessary scenarios and if you where to reduce it even further you wouldnt have enough information anymore to distinguish the four scenarios.

     – crush3dice Aug 21 '20 at 11:58
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    There is no such thing as the sample space for a random variable or a stochastic process. There is no uniqueness for $(\Omega, \mathcal F, P)$. – Kavi Rama Murthy Aug 21 '20 at 12:02
  • Thank you very much. Your approach of explaining Ω enables a whole different view on the subject. So the sample space should mirror the possible of all iterations of the random experiment (A) not only the ones of the single iteration (B), right? – S. Den Aug 21 '20 at 12:23

1 Answers1

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As written in the comments, there can be many choices of $(\Omega, \mathcal{F}, \mathbb{P})$ that allow you to construct some random variable (or process) $X$.

(A) is one choice of $\Omega$ for modeling the scenario of flipping two coins. In the case of flipping an infinite number of coins, one choice of $\Omega$ is the Cantor Space, which is the space of infinite binary strings $2^\mathbb{N}$. You can then, for example, express the outcome of only flipping the first $N$ coins as the set of elements that have a certain prefix of length $N$.

oswinso
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