Assume $(\Omega,\mathcal{A},\mu)$ is given. We are interested in the space, $\mathcal{M}$ of measurable functions from $\Omega$ to a set $S=\{1,2,\ldots,n\}$ with the standard discrete topology. Also let $(p_1,p_2,\ldots,p_n)$ be given, where $\sum_{j=1}^n p_j=1$ and $0<p_j<1$.
What are the necessary conditions on $(\Omega,\mathcal{A},\mu)$ to conclude that there exists a sequence, $X_1, X_2, X_3, \ldots$, of $\mu$-independent (def. below) random variables in $\mathcal{M}$, with identical distributions $\mu(X_i^{-1}(\{j\}))=p_j$?
- $X$ and $Y$ are $\mu$-independent if for all $A\in\sigma(X)$ and $B\in\sigma(Y)$, $\mu(A\cap B)=\mu(A)\mu(B)$, where $\sigma(X)$ and $\sigma(Y)$ are the sub-sigma algebras of $\mathcal{A}$ generated by the preimages of measurable sets in $S$ (which means preimages of all subsets of $S$ since $S$ is discrete)
Examples:
- Any discrete $\Omega$ does not allow for an i.i.d sequence as described above.
Proof: The set $\{\sigma(X_i), i=1,\ldots,\infty\}$ is finite because $\mathcal{A}$ is finite. Thus, for some $i_1\ne i_2$, $\sigma(X_{i_1})=\sigma(X_{i_2})$. By independence, if $A=X_{i_1}^{-1}(\{j\})=X_{i_2}^{-1}(\{j\})$, then $\mu(A)=\mu(A\cap A)=\mu(A)\mu(A)$. This implies $\mu(A)\in\{0,1\}$, violating the assumption that $0<p_j<1$.
- If $n=2$, $p_1=2/3, p_2=1/3$, and $\Omega=\mathbb{R}$ with Lebesgue measure then there does exist an i.i.d. sequence with the given law.
Construction See image below. Basically, construct $X_1$ to have $X_1([0, 2/3))=1$ and $X_1([2/3, 1])=2$. Then for subsequent $X_j$ divide each of the intervals on the previous step in a self-similar fashion. There needs to be some rule for assigning endpoints like all intervals are of the form $[a,b)$ unless $b=1$.
