I recently came across this statement in page 627 of "Theory of Statistics", by Schervish.
If $Y = g(X)$, then the joint distribution of (X,Y) is not dominated by a product measure, even if the distribution of $X$ is dominated.
I have tried to pour some time into understanding this but:
- I currently have no intuition as to why this should be so? How come conventional product measures would not work on this joint space? How come a joint density written with these constraints ($Y = g(X)$), is not dominated even if $X$ is dominated?
In addition to this, he states to consider a measurable space:
$(\mathcal{Y},\mathcal{B}_2)$ a measurable space such that $\mathcal{B}_2$ contains all singletons
- What is implied by this? If a measurable space is Borel does it not already contain the singleton elements? Or does he mean a space with exclusively singleton elements? What does he mean to capture by saying this so explicitly.
For context, the following is the extract of the text, with the two parts which I would like to better understand:
EDIT:
Problem 7 on page 662:
Regarding your first question, it may become more apparent what the author is trying to convey by specializing to a more concrete situation first. Let $X$ be uniformly distributed on $[0,1]$, so that its law is (the restriction of) Lebesgue measure. Then the relevant product measure would be Lebesgue measure on the unit square. However, since $Y=g(X)$ we see that the support of the law of $(X,Y)$ is precisely the graph of the function $g(x)$, comprising the set $$ \bigl\{\bigl(x,g(x)\bigr)\colon x\in [0,1]\bigr\}. $$ When $g$ is a sufficiently smooth function, this set is $1$-dimensional and is therefore null with respect to the Lebesgue measure on $[0,1]^2$ (which is $2$-dimensional). Thus, any measure supported on it - in particular, the joint law of $(X,Y)$ - must be singular with respect to the Lebesgue measure on $[0,1]^2$. This is the point that the author is making.
Regarding your second question, note that the author is stating the result in great generality and that is the reason for the strange looking condition regarding containment of singletons. There is no topological structure mentioned at all for $(\mathcal Y,\mathcal B_2)$, and hence there cannot be any assumption of a Borel property - since it is only defined in the presence of a topological space, which is not given for $\mathcal Y$.