What is the Probability Space Underlying a Markov Chain?

Question

Consider the random walk on $\mathbb{Z}$. The position at step $n$ can be described as a Markov chain $(X_n)$. We know that each $X_n$ is a random variable, so its domain is a probability space, so my question is, is that space the heads/tail space {H,T} or the product heads tail space $\Pi_{\mathbb{N}}\{H,T\}$ in which case $X_n$ are the projection to $n$-th component?

If the underlying probability space is the product space, how exactly do we assign probability to the events given each outcome is an infinite sequence of heads and tails?

Answer 1

If $X_n$ is the position of the random walk at time $n$ then the underlying probability space is $\prod_{n \in \mathbb N} \mathbb Z$ ie. the set of $\mathbb Z$-valued sequences.

The trick is to assign a value not to individual sequences but instead to define the probability of events of the form $$ X_0 = x_0, X_1 = x_1, \ldots, X_N = x_N $$ for every $N \in \mathbb N$ and every $(x_0, \ldots, x_N) \in \mathbb Z^{N+1}$.

The Kolmogorov extension theorem then tells you that this uniquely defines a probability on $\prod_{n \in \mathbb N} \mathbb Z$.

Answer 2

You can take the sequence space $\Pi_{\Bbb N}\{H,T\}$ as the underlying sample space. In this case, for a sample point $\omega =(\omega_1,\omega_2,\ldots )$ the step taken at time $k$ is $$ \xi_k(\omega) = \cases{1,&if $\omega_k=H$,\cr -1,&if $\omega_k=T$,\cr} $$ and the position of the walker after $n$ steps is $$ X_n(\omega) =\xi_1(\omega)+\cdots+\xi_n(\omega). $$ A complete assignment of probabilities requires something like the Kolmogorov theorem as noted by Leo in his answer. Simple events an be handled directly; for example, $$ P(\xi_1=\xi_2=1,\xi_3=-1) =1/8. $$