support vs state space In one of my examples, I saw that the state space for a binomial distribution is $({0,1,...,n})$. I then thought that for $X_1, X_2,...,X_n$ iid RVs, where $f_\theta(x_i)=\theta x_i^{\theta -1}$ for $x \in (0,1)$, and $0$ otherwise, the state space is $(0,1)^n$. However, it was said to be $\Bbb R^n$, meaning other values outside $(0,1)$ are also counted in the state space (which would make sense, since the state space is the set of values assumed by our RV), and that $(0,1)^n$ is rather the support. But, following this logic, how comes the state space for the binomial distribution isn't $\Bbb N$, where for all $N>n,$ the probability that the RV is equal to that value is zero? Or ss it because the binomial distribution isn't defined for $n+1$ onwards?
 A: I don't know a formal definition of "state space" for distributions that are not discrete, but for discrete distributions, in some cases the support is bigger than the state space. The support is always a closed set.
The support of the distribution taking values in $(0,1)^n$ that you describe is the closed set $[0,1]^n.$
Here is the definition: The support of a probability distribution taking values in $\mathbb R^n$ is the set of points whose every open neighborhood has positive probability.
Suppose a discrete probability distribution assigns positive probability to each of $1,\,1/2,\, 1/3,\,1/4,\,\ldots\,.$ Then $0$ is a member of the support since every open neighborhood of $0$ has positive probability.
Now suppose
\begin{align}
& \Pr(X=x)>0 \text{ for } x = 1,2,3,\ldots \text{ and } \sum_{x=1}^\infty \Pr(X=x) = 1 \\[8pt]
\text{and } & \Pr(Y=y\mid X=x) = \frac 1 {x-1} \text{ for } y = 1,\ldots,x-1
\end{align}
and consider the discrete probability distribution of the random variable $Y/X \in(0,1).$ The state space in $\mathbb Q\cap[0,1]$ and the support is the larger set $[0,1].$
