Let $(\Xi^{N},\mathcal{E}^{N})$ product space of the measurable space $(\Xi,\mathcal{E})$ where $N>0$. Let $\lambda$ be a probability measure on $(\Xi^{N},\mathcal{E}^{N})$.
The question: Is $\lambda$ be a product measure?, that is, are there $\mu_{i}$, $i=1,\ldots,N$, probability measures on $(\Xi,\mathcal{E})$ such that $\lambda=\otimes_{i=1}^{N}\mu_{i}$?
Another way of looking at my question is considering $\mathcal{M(\Xi)}$ the set of all probability measures supported on $\Xi$ and $\mathcal{M}(\Xi^{N})$ the set of all probability measures supported on $\Xi^{N}$, Can we identify $\mathcal{M}(\Xi^{N})$=$\mathcal{M}(\Xi)^{N}$?
If the answer is no, under what conditions is this identification possible?
Best Answer
No. The simplest counterexample is $\Xi = \{0,1\}$, with the discrete $\sigma$-algebra, and $N=2$.
Let $\lambda\{(0,0)\}=1-p$, $\lambda\{(1,1)\}=p$, and $\lambda\{(0,1)\}=\lambda\{(1,0)\}=0$. This is a probability measure on $(\Xi^N, \mathcal{E}^N)$, but is not a product measure.
In the language of random variables, this can be summarized as "for a joint distribution on $(X,Y)$, it is not necessarily true that $X$ and $Y$ are independent."