Given a discrete $P(X,Y,Z)$ let's call $\Omega$ the set of all deterministic functions $f: XYZ \rightarrow W$ and $\Omega'$ the set of all deterministic functions $f': XY \rightarrow V$. Is it correct that $\Omega' \subseteq \Omega$? I am thinking that a deterministic function of a discrete random variable is essentially a partition of the outcomes of the random variable, thus the inclusion should be correct, but am not sure.
$\begingroup$
$\endgroup$
Add a comment
|
$\begingroup$
$\endgroup$
1
Technically, no, because the domain of the functions in the two sets is not the same. However, with a slight alteration, you can obtain the subset relation you want. If you let $\Omega$ be the class of all (measureable) functions of $(X,Y,Z)$ then you can define the subclass of functions:
$$\Omega' \equiv \{ f \in \Omega | f(x,y,z) = f(x,y,z') \text{ for all } z, z' \in \mathcal{Z} \},$$
where $\mathcal{Z}$ denotes the range of possible values of $z$. This gives you a subclass $\Omega' \subseteq \Omega$ containing functions that have the same domain as the functions in $\Omega$, but where all the functions in the subclass are invariant to the input value $z$.
(Incidentally, this question really has nothing to do with random variables --- it is just a question about functions. In the context of random variable inputs we would require the functions to be measureable, but that is the only difference.)
$\begingroup$
$\endgroup$
I think, in a more general sense, X,Y,Z don't have to be random variables let alone discrete. Intuitively, what you write resembles the following question:
Can we define all deterministic functions $f(x,y)$ in the form $f(x,y,z)$ instead?
This is yes. However, rigorously, the answer to your question is false because the domains of the functions are different.
