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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:

enter image description here

EDIT:

Problem 7 on page 662:

enter image description here

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  • $\begingroup$ For further context, can you state "Problem 7 on page 662"? $\endgroup$ Aug 28, 2019 at 5:02
  • $\begingroup$ Hi. I have added it now. Thanks! $\endgroup$ Aug 28, 2019 at 7:05
  • $\begingroup$ Great, thank you for that, I will spend some time on your question. $\endgroup$ Aug 28, 2019 at 7:13
  • $\begingroup$ I've spent time, and I must admit that I've also struggled. The least I can tell you is this : for the second part of the question, the reason why singletons are to be forced into $\mathcal B_2$ is so that the indicator of $\{g(x)\}$ is a measurable function and hence the Radon Nikodym derivative which is proposed is measurable. Other than that, I don't see any reason why it's there. Also, a Borel space does contain all singletons, however, in the corollary it is no assumed that $\mathcal B_2$ is Borel, so it may not contain singletons. $\endgroup$ Aug 29, 2019 at 6:53
  • $\begingroup$ I see. I have reached similar conclusions for the latter part of the question, in regards to the singletons. However how is the development looking like for the first portion of the question? Have you managed to make any progress? :) I must admit I am still stuck here too. $\endgroup$ Aug 31, 2019 at 8:18

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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 $\mathcal Y$.

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