Bivariate additive-noise direction identifiability under Condition 19

Put $\xi =\log p_X$, $\nu =\log p_N$, and $\pi (x,y)=\log p(x,y)$. By [F2], $\pi (x,y)=\xi (x)+\nu (y-f(x))$ is a $C^3$ function. The chain rule gives ${\pi }_{xy}=-{\nu }^{\mathrm{\prime }\mathrm{\prime }}(y-f(x))f^{\mathrm{\prime }}(x),{\pi }_{xx}={\nu }^{\mathrm{\prime }\mathrm{\prime }}(y-f(x))(f^{\mathrm{\prime }}(x){)}^2-{\nu }^{\mathrm{\prime }}(y-f(x))f^{\mathrm{\prime }\mathrm{\prime }}(x)+{\xi }^{\mathrm{\prime }\mathrm{\prime }}(x)\mathrm{.}$ Thus both derivatives are defined, and ${\pi }_{xy}\mathrm{\ne }0$ exactly on the Condition 19 domain.

Suppose a reverse representation as in the Statement exists. By [F4] and [F2], $p_Y(y)={\int }_{\mathbb{R}}p(x,y)dx$ is positive and finite for almost every $y$. The sectionwise translation $m=x-g(y)$ gives the joint density $p_{M,Y}(m,y)=p(m+g(y),y)$ by [F5]; independence makes it equal to $p_M(m)p_Y(y)$ almost everywhere. Choose $y_0$ for which $0<p_Y(y_0)<\mathrm{\infty }$ and this factorization holds for almost every $m$. Then $p_M(m)=\frac{p(m+g(y_0),y_0)}{p_Y(y_0)}\text{for almost every }m\mathrm{.}$ The right side is a strictly positive $C^3$ function of $m$, so choose it as the density version of $M$ and put $\tilde{\nu }=\log p_M$. Consequently the forward and reverse density formulas agree almost everywhere: $p(x,y)=p_Y(y)p_M(x-g(y))\mathrm{.}$ For almost every $y$ with $0<p_Y(y)<\mathrm{\infty }$, the two sides as functions of $x$ are continuous and agree almost everywhere, hence agree for every $x$. Taking logarithmic $x$-derivatives gives ${\pi }_x(x,y)={\tilde{\nu }}^{\mathrm{\prime }}(x-g(y))$ for almost every $(x,y)$. Both sides are continuous on ${\mathbb{R}}^2$, so this identity holds everywhere. Differentiating in $x$ and $y$ gives ${\pi }_{xx}={\tilde{\nu }}^{\mathrm{\prime }\mathrm{\prime }}(x-g(y))$ and ${\pi }_{xy}=-{\tilde{\nu }}^{\mathrm{\prime }\mathrm{\prime }}(x-g(y))g^{\mathrm{\prime }}(y)$. Wherever ${\pi }_{xy}\mathrm{\ne }0$, it follows that $\frac{{\pi }_{xx}}{{\pi }_{xy}}=-\frac{1}{g^{\mathrm{\prime }}(y)},$ which is independent of $x$; hence ${\mathrm{\partial }}_x({\pi }_{xx}\mathrm{/}{\pi }_{xy})=0$.

From the forward expressions in step 1.1, on the same nonzero domain, $\frac{{\pi }_{xx}}{{\pi }_{xy}}=-f^{\mathrm{\prime }}(x)+\frac{{\nu }^{\mathrm{\prime }}(y-f(x))}{{\nu }^{\mathrm{\prime }\mathrm{\prime }}(y-f(x))}\frac{f^{\mathrm{\prime }\mathrm{\prime }}(x)}{f^{\mathrm{\prime }}(x)}-\frac{{\xi }^{\mathrm{\prime }\mathrm{\prime }}(x)}{{\nu }^{\mathrm{\prime }\mathrm{\prime }}(y-f(x))f^{\mathrm{\prime }}(x)}\mathrm{.}$ Differentiating this expression in $x$, using ${\mathrm{\partial }}_x{\nu }^{\mathrm{\prime }}=-f^{\mathrm{\prime }}{\nu }^{\mathrm{\prime }\mathrm{\prime }}$ and ${\mathrm{\partial }}_x{\nu }^{\mathrm{\prime }\mathrm{\prime }}=-f^{\mathrm{\prime }}{\nu }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}$, and setting the derivative to zero gives ${\xi }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(x)={\xi }^{\mathrm{\prime }\mathrm{\prime }}(x)(-\frac{{\nu }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(y-f(x))f^{\mathrm{\prime }}(x)}{{\nu }^{\mathrm{\prime }\mathrm{\prime }}(y-f(x))}+\frac{f^{\mathrm{\prime }\mathrm{\prime }}(x)}{f^{\mathrm{\prime }}(x)})-2{\nu }^{\mathrm{\prime }\mathrm{\prime }}(y-f(x))f^{\mathrm{\prime }\mathrm{\prime }}(x)f^{\mathrm{\prime }}(x)+{\nu }^{\mathrm{\prime }}(y-f(x))f^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(x)+\frac{{\nu }^{\mathrm{\prime }}(y-f(x)){\nu }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(y-f(x))f^{\mathrm{\prime }\mathrm{\prime }}(x)f^{\mathrm{\prime }}(x)}{{\nu }^{\mathrm{\prime }\mathrm{\prime }}(y-f(x))}-\frac{{\nu }^{\mathrm{\prime }}(y-f(x))(f^{\mathrm{\prime }\mathrm{\prime }}(x){)}^2}{f^{\mathrm{\prime }}(x)}\mathrm{.}$ The quotient rule is legitimate because both denominator factors are nonzero on this domain.

Step 3.1 forces Condition 19's differential equation at every point where ${\nu }^{\mathrm{\prime }\mathrm{\prime }}(y-f(x))f^{\mathrm{\prime }}(x)\mathrm{\ne }0$, contrary to [F3]. Therefore no reverse representation in the stated class exists.

The law cannot be independent: if $X$ and $Y$ were independent, positivity of $p_X$ and the forward density formula [F2] would make $p_N(y-f(x))$ equal $p_Y(y)$ for almost every $(x,y)$. By Fubini, for almost every pair $x,x^{\mathrm{\prime }}$, the translates $p_N(y-f(x))$ and $p_N(y-f(x^{\mathrm{\prime }}))$ agree for almost every $y$. Their probability laws are translates of the law of $N$; [F6] forces $f(x)=f(x^{\mathrm{\prime }})$ for almost every pair. Continuity of $f$ makes it constant everywhere, contradicting the hypothesis. Thus the empty graph is excluded as well, and the only smooth bivariate ANM direction in the class is $X\to Y$.