Problem 4.15 — the two claims behind "no ANM from $Y$ to $X$"

Setting (Definition 4.4, p. 67): an ANM from $Y$ to $X$ means there exist a measurable $g:\mathbb{R}\to \mathbb{R}$ and a random variable $M_X$ with

$$X=g(Y)+M_X,M_X\perp \perp Y.$$

Everything below assumes such $(g,M_X)$ exists and derives a contradiction.

What “conditional distribution given $Y=y$” means

For continuous $Y$ the event $\{Y=y\}$ has probability $0$, so $P(X\in A\mid Y=y)$ cannot be defined by division. Instead, a regular conditional distribution of $X$ given $Y$ is a family of probability measures $\{{\kappa }_y{\}}_{y\in \mathbb{R}}$ on $\mathbb{R}$ (measurable in $y$) such that

$$P(X\in A,Y\in B)={\int }_B{\kappa }_y(A)P_Y(dy)\text{for all Borel }A,B.\text{\hspace{1em}(∗)}$$

Such a family exists, and it is unique up to $P_Y$-null sets of $y$: if $\{{\kappa }_y\}$ and $\{{\kappa }_y^{\prime }\}$ both satisfy $(\ast )$ then ${\kappa }_y={\kappa }_y^{\prime }$ for $P_Y$-a.e. $y$. We write $P_{X\mid Y=y}:={\kappa }_y$.

So the sentence ”$P_{X\mid Y=y}$ is the law of $g(y)+M_X$” means: the family ${\kappa }_y(A):=P(g(y)+M_X\in A)$ satisfies $(\ast )$.

Claim 1 (independence $\Rightarrow$ conditionals are shifts of one law)

\begin{claim} If $X=g(Y)+M_X$ with $M_X\perp \perp Y$, then ${\kappa }_y(A):=P(g(y)+M_X\in A)$ is a regular conditional distribution of $X$ given $Y$. Equivalently: for $P_Y$-a.e.\ $y$, $P_{X\mid Y=y}=P_{M_X}(\cdot -g(y))$, the law of $M_X$ translated by $g(y)$. \end{claim}

Proof. Fix Borel $A,B$. Because $M_X\perp \perp Y$, the joint law of $(Y,M_X)$ is the product $P_Y\otimes P_{M_X}$. Then

$$P(X\in A,Y\in B)=P(g(Y)+M_X\in A,Y\in B)=\int \int 1_A(g(y)+m)1_B(y)P_{M_X}(dm)P_Y(dy).$$

The inner integral is $P(g(y)+M_X\in A)={\kappa }_y(A)$, so the right side is ${\int }_B{\kappa }_y(A)P_Y(dy)$, i.e. $(\ast )$ holds. $\square$

(This is the only place Fubini/Tonelli is used: writing the probability under a product measure as an iterated integral. Independence is what makes the joint law a product.)

Claim 2 (a shift-invariant statistic must be constant)

Let $T$ be any map from probability measures on $\mathbb{R}$ to $[0,\infty ]$ that is translation invariant: $T(\mu (\cdot -c))=T(\mu )$ for all $c\in \mathbb{R}$. Examples: $\mathrm{Var}(\mu )$; the Lebesgue measure of the support of $\mu$; the interquartile range.

\begin{claim} Under the hypothesis of Claim 1, $y\mapsto T(P_{X\mid Y=y})$ is $P_Y$-a.e.\ equal to the constant $T(P_{M_X})$. \end{claim}

Proof. By Claim 1, for $P_Y$-a.e. $y$, $P_{X\mid Y=y}=P_{M_X}(\cdot -g(y))$, hence $T(P_{X\mid Y=y})=T(P_{M_X})$ by translation invariance. $\square$

Contradiction for Problem 4.15(a)

From the joint density (uniform, value $\frac{1}{2}$, on the parallelogram $\{1\le x\le 3,|y-2x|\le \frac{1}{2}\}$),

$$P_{X\mid Y=y}=\mathcal{U}\left[\max (1,\frac{y}{2}-\frac{1}{4}),\min (3,\frac{y}{2}+\frac{1}{4})\right]\text{for }y\in [\frac{3}{2},\frac{13}{2}].$$

(This is a genuine version of the conditional distribution: one checks $(\ast )$ by integrating the joint density; it is unique a.e. by the uniqueness statement above.)

Take $T=$ support length (or $T=\mathrm{Var}$, which is ${\text{length}}^2/12$ for a uniform law):

• $y\in (\frac{5}{2},\frac{11}{2})$: length $\frac{1}{2}$;
• $y\in (\frac{3}{2},\frac{5}{2})$: length $\frac{y}{2}-\frac{3}{4}<\frac{1}{2}$.

Both intervals have positive $P_Y$-mass ($p_Y(y)=\frac{1}{2}\cdot \text{length}>0$ there). So $y\mapsto T(P_{X\mid Y=y})$ is not a.e. constant, contradicting Claim 2. Hence no $(g,M_X)$ with $M_X\perp \perp Y$ exists: $P_{X,Y}$ admits no ANM from $Y$ to $X$.

Remarks

• Fubini is not “needed for the idea”; it is needed only to make Claim 1 a theorem rather than a picture, because conditioning on a null event is defined through $(\ast )$.
• Variance is not needed either; any translation-invariant $T$ works. Support length is closest to the “draw the support” hint in the exercise.
• Theorem 4.5 (pp. 67–68) is not applicable: it needs strictly positive, three-times differentiable densities; the uniform noise here has neither.