Problem 4.15 (ANMs)

Problem 4.15 (ANMs)

1. Consider the SCM

$$\begin{array}{rl}X& :=N_X\\ Y& :=2X+N_Y\end{array}$$

with $N_X$ uniformly distributed between $1$ and $3$ and $N_Y$ uniformly distributed between $-0.5$ and $0.5$ and independent of $N_X$. The distribution $P_{X,Y}$ admits an ANM from $X$ to $Y$. Draw the support of the joint distribution of $X,Y$ and convince yourself that $P_{X,Y}$ does not admit an ANM from $Y$ to $X$, that is, there is no function $g$ and independent noise variables $M_X$ and $M_Y$ such that

$$\begin{array}{rl}X& =g(Y)+M_X\\ Y& =M_Y\end{array}$$

with $M_X$ independent of $M_Y$.

Solution Clearly $P\{X<1\vee X>3\}=0.$ Also clearly $\mathrm{supp}Y=[1.5,6.5]$

$$p(x)=\{\begin{array}{l}1/2,\text{ if }x\in [1,3]\\ 0,\text{ if }x\notin [1,3]\end{array}$$ $$p(n)=\{\begin{array}{l}1,\text{ if }x\in [-1/2,1/2]\\ 0,\text{ if }x\notin [-1/2,1/2]\end{array}$$

Recall convolution $p(y)=p_n(n)+{\int }_{-\infty }^{\infty }{p}_x(\frac{y-n}{2})dn$ or $p(y)=\int p_x(x)p_n(y-2x)dx$

So we get

$$p(y\mid x)=p_n(y-2x)=\{\begin{array}{ll}1,& \text{if }y-2x\in [-1/2,1/2]\\ 0,& \text{if }y-2x\notin [-1/2,1/2]\end{array}$$

Multiplying the marginal by the conditional gives the joint density

$$p(x,y)=p(x)p(y\mid x)=\{\begin{array}{ll}1/2,& \text{if }x\in [1,3]\text{ and }|y-2x|\le 1/2\\ 0,& \text{otherwise}\end{array}$$

so the support is the slanted band

$$\mathrm{supp}{P}_{X,Y}=\left\{(x,y):1\le x\le 3,2x-\frac{1}{2}\le y\le 2x+\frac{1}{2}\right\},$$

a parallelogram with vertices $(1,\frac{3}{2}),(1,\frac{5}{2}),(3,\frac{11}{2}),(3,\frac{13}{2})$.

Now read the band the other way. Conditioning on $Y=y$,

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

For $y\in [\frac{5}{2},\frac{11}{2}]$ neither bound binds and the interval is $\left[\frac{y}{2}-\frac{1}{4},\frac{y}{2}+\frac{1}{4}\right]$: width $\frac{1}{2}$, a pure translate of a fixed set, which is exactly what an ANM needs. But for $y\in [\frac{3}{2},\frac{5}{2})$ the constraint $x\ge 1$ truncates it to $[1,\frac{y}{2}+\frac{1}{4}]$, of width $\frac{y}{2}-\frac{3}{4}<\frac{1}{2}$, and symmetrically near $y=\frac{13}{2}$.

Why this rules out an ANM (Definition 4.4, p. 67). An ANM from $Y$ to $X$ means $X=g(Y)+M_X$ with $M_X\perp \perp Y$. Independence gives, for bounded measurable $h$ and Borel $B$, by Fubini,

$$\mathbb{E}[h(X)1_B(Y)]={\int }_B\mathbb{E}[h(g(y)+M_X)]P_Y(dy),$$

so for $P_Y$-a.e. $y$ the conditional law $P_{X\mid Y=y}$ is the law of $g(y)+M_X$. Hence $\mathrm{Var}(X\mid Y=y)=\mathrm{Var}(M_X)$ is a.e. constant in $y$. But from the conditional above, $\mathrm{Var}(X\mid Y=y)=\frac{1}{48}$ for $y\in (\frac{5}{2},\frac{11}{2})$ and $=\frac{1}{12}(\frac{y}{2}-\frac{3}{4}{)}^2<\frac{1}{48}$ for $y\in (\frac{3}{2},\frac{5}{2})$, and both sets have positive $P_Y$-mass. Contradiction: there is no ANM from $Y$ to $X$. (Theorem 4.5, p. 67–68, does not apply: it requires strictly positive, smooth densities.)

1. Similarly as in part a), consider the SCM

$$\begin{array}{rl}X& :=N_X\\ Y& :=X^2+N_Y\end{array}$$

with $N_X$ uniformly distributed between $1$ and $3$ and $N_Y$ uniformly distributed between $-0.5$ and $0.5$ and independent of $N_X$. Again, draw the support of $P_{X,Y}$ and convince yourself that there is no ANM from $Y$ to $X$.