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Let $\pi$ be a probability measure on $\mathbb{R}^2$ with respective marginals $\mu$ and $\nu$ such that $(X,Y) \sim \pi$.

Notation: $\pi_{X=x}$ be the conditional distribution of $Y$ given $X=x$, $\pi_{Y=y}$ be the conditional distribution of $X$ given $Y=y$.

Question: If for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$, we have: $$ \nu\left( \left\{ y \in \mathbb{R} \mid \pi_{Y=y}(A) = 1 \right\} \right) > 0, $$ and for all $B \in \mathcal{B}(\mathbb{R})$ such that $\nu(B) > 0$, we have: $$ \mu\left( \left\{ x \in \mathbb{R} \mid \pi_{X=x}(B) = 1 \right\} \right) > 0, $$ then is $X$ necessarily measurable with respect to $Y$ (and vice-versa)?

My intuition is yes, but I have no idea how to prove it. Any idea/intuition/conter-example is welcome! A highly related question is the following: Existence of Dirac measures in the context of joint and marginal distributions.

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  • $\begingroup$ This is quite an interesting condition. Can you say anything about where the question came from? $\endgroup$
    – Anthony Quas
    Commented Oct 27 at 14:10
  • $\begingroup$ @AnthonyQuas Thanks you again for your interest! The question comes from another question: Does the assertion: for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$, we have: $$ \nu\left( \left\{ y \in \mathbb{R} : \pi_{Y=y}(A) = 1 \right\} \right) > 0 $$ imply the assertion that for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$, $$ \nu\left( \left\{ y \in D : \pi_{Y=y}(A) = 1 \right\} \right) > 0? $$ $\endgroup$
    – thibault jeannin
    Commented Nov 5 at 15:15
  • $\begingroup$ $$ D = \{ y : \pi_{Y=y} \text{ is a Dirac measure} \} $$ Using your approach, I demonstrated the following weaker result: If for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$, $B_A = \{ y \in \mathbb{R} \mid \pi_{Y=y}(A) = 1 \}$ satisfies $\nu(B_A) > 0$, then for all $A \in \mathcal{B}(\mathbb{R})$ such that $\mu(A) > 0$, $$ \operatorname{essinf}_{\mathbb{1}_{B_A}(y) \, \nu(dy)} V_X(y) = 0. $$ I asked this question there: [mathoverflow.net/questions/480583/… $\endgroup$
    – thibault jeannin
    Commented Nov 5 at 15:34
  • $\begingroup$ I guess I am interested in how you arrived at the earlier question also. These kind of conditions seem as though they could be of interest in ergodic theory. There one builds abstract joinings (invariant couplings) of a pair of measures, and it can be of interest to know when the joining is supported on the graph of a function. $\endgroup$
    – Anthony Quas
    Commented Nov 5 at 23:13

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I believe the answer is yes. The argument is a little bizarre (using contradiction in what feels, for now, like an essential way). I feel there should be a cleaner version, but this is what I have for now. By the way, I have replaced your $\pi_{Y=y}(\cdot)$ with the notation $\mathbb P(\cdot|Y=y)$ which feels a bit more familiar to me.

Let $V_Y(x)=\mathbb E(\tanh^2 Y|X=x)-\mathbb E(\tanh Y|X=x)^2$, that is the variance of $\tanh Y$ conditioned on $X=x$. Similarly, let $V_X(y)$ be the variance of $\tanh X$ conditioned on $Y=y$. The function $V_Y(x)$ is $\mu$-a.e. zero if and only if $Y$ is measurable with respect to $X$; and likewise $V_X(y)$ is $\nu$-a.e. zero if and only if $X$ is measurable with respect to $Y$. (The role of the hyperbolic tangent is to ensure that these quantities are finite).

We now prove your claim by contradiction. Suppose your conditions hold, but that $X$ is not measurable with respect to $Y$. Then $V_X$ is not $\nu$-a.e. zero. It follows that there exists $\delta>0$ and a set $B$ with $\nu(B)>0$ with the property that $V_X(y)>\delta^2$ for all $y\in B$.

We now apply your second condition: let $A=\{x\colon \mathbb P(Y\in B|X=x)=1\}$. By your condition, $\mu(A)>0$. Using countable additivity, let $A_1$ be a subset of $A$ with $\mu(A_1)>0$ of the form $A_1=\{x\in A\colon \tanh x\in (c,c+\delta)\}$.

Finally, let $B_1=\{y\colon \mathbb P(X\in A_1|Y=y)=1\}$. By the first condition, $\nu(B_1)>0$. We claim that $B_1\subset B$ up to a set of measure 0. To see this, notice that $\{Y\in B_1\}$ is a subset of $\{X\in A_1\}$ (up to a set of measure 0); and $\{X\in A_1\}$ is a subset of $\{X\in A\}$, which is a subset (up to measure 0) of $\{Y\in B\}$.

If $y\in B_1$, then $V_X(y)<\frac{\delta^2}4$ since conditioned on $Y=y$, $\tanh X\in (c,c+\delta)$. However since $y\in B$, we also have $V_X(y)>\delta^2$. This is a contradiction.

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  • $\begingroup$ Good evening Sir, thank you very much for this proof, which I find very elegant. Thank you so much! $\endgroup$
    – thibault jeannin
    Commented Oct 23 at 20:17

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