A reversible linear-Gaussian additive-noise model

Under AC [A1], [F1]–[F2] and independence give the density $\varphi (x)\varphi (n)=(2\pi {)}^{-1}{e}^{-(x^2+n^2)\mathrm{/}2}$ for $(X,N)$. The forward equation $Y=X+N$ is an additive-noise equation by [F4].

The linear map $(x,n)\mapsto (y,m)=(x+n,(x-n)\mathrm{/}2)$ has determinant $-1$ and inverse $(y,m)\mapsto (x,n)=(y\mathrm{/}2+m,y\mathrm{/}2-m)$, whose determinant also has absolute value $1$. By [F3], substituting the inverse into the density from step 1.1 gives $p_{Y,M}(y,m)=\frac{1}{2\pi }\exp \text{\hspace{0.17em}⁣}(-\frac{(y\mathrm{/}2+m{)}^2+(y\mathrm{/}2-m{)}^2}{2})=\frac{1}{2\pi }{e}^{-y^2\mathrm{/}4-m^2}\mathrm{.}$

The last density factors as $p_Y(y)p_M(m)$ with $p_Y(y)=(2\sqrt{\pi }{)}^{-1}{e}^{-y^2\mathrm{/}4}$ and $p_M(m)={\pi }^{-1\mathrm{/}2}{e}^{-m^2}$. Each factor integrates to one by the affine normal-law description [F1] and normalization [F2], so the joint law is the product of its marginals and $M\perp Y$. Since $M=X-Y\mathrm{/}2$, rearranging gives $X=Y\mathrm{/}2+M$.

For $f(x)=x$ and standard normal input and noise, $\xi (x)=\nu (x)=-x^2\mathrm{/}2-\frac{1}{2}\log (2\pi )$, $f^{\mathrm{\prime }}=1$, $f^{\mathrm{\prime }\mathrm{\prime }}=f^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=0$, ${\xi }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}={\nu }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=0$, and ${\nu }^{\mathrm{\prime }\mathrm{\prime }}=-1$. The Condition 19 domain is all $(x,y)$ because ${\nu }^{\mathrm{\prime }\mathrm{\prime }}(y-f(x))f^{\mathrm{\prime }}(x)=-1\mathrm{\ne }0$; both sides of its equation equal zero there. Thus its required inequality never holds, so Condition 19 fails.