Additive-noise causal discovery and identifiability — Examples

Under AC [A1], treat the full Model A vector $(U,X,Y)$ as an ANM with noises $(N_U,{ϵ}_X,{ϵ}_Y)$. By [F4]–[F5], its density is $\varphi (u)\varphi (x-u)\varphi (y-u)$; marginalizing the latent $U$ gives $p_A(x,y)={\int }_{\mathbb{R}}\varphi (u)\varphi (x-u)\varphi (y-u)du\mathrm{.}$ Completing the square, $u^2+(x-u{)}^2+(y-u{)}^2=3{(u-\frac{x+y}{3})}^2+\frac{2}{3}(x^2-xy+y^2)\mathrm{.}$ Thus [F3] and the substitution $t=\sqrt{3\mathrm{/}2}(u-(x+y)\mathrm{/}3)$ give $p_A(x,y)=\frac{1}{2\pi \sqrt{3}}\exp \text{\hspace{0.17em}⁣}(-\frac{x^2-xy+y^2}{3})\mathrm{.}$

The scaled standard normal variables in Model B have densities $q_2(x)=\frac{1}{2\sqrt{\pi }}{e}^{-x^2\mathrm{/}4}$ and $q_{3\mathrm{/}2}(e)=\frac{1}{\sqrt{3\pi }}{e}^{-e^2\mathrm{/}3}$ by [F2]. The Model B equation is an ANM, so [F4] gives $p_B(x,y)=q_2(x)q_{3\mathrm{/}2}(y-x\mathrm{/}2)=\frac{1}{2\pi \sqrt{3}}\exp \text{\hspace{0.17em}⁣}(-\frac{x^2-xy+y^2}{3}),$ where the last equality follows by expanding $x^2\mathrm{/}4+(y-x\mathrm{/}2{)}^2\mathrm{/}3=(x^2-xy+y^2)\mathrm{/}3$. Hence $p_A=p_B$ pointwise.

Under $do(X=1)$ in Model A, [F1] replaces only the equation for $X$; the unchanged $Y$ equation remains $Y=U+{ϵ}_Y$. Its density is the convolution $\int \varphi (u)\varphi (y-u)du=\frac{1}{2\sqrt{\pi }}{e}^{-y^2\mathrm{/}4}$, by completing the square and [F3]. In Model B the intervention gives $Y=1\mathrm{/}2+E$, whose density is $\frac{1}{\sqrt{3\pi }}{e}^{-(y-1\mathrm{/}2{)}^2\mathrm{/}3}$ by [F2]. At $y=1\mathrm{/}2$ the second density is larger than the first by the factor $\frac{2}{\sqrt{3}}{e}^{1\mathrm{/}16}>1$, so the intervention laws differ although the observational densities agree.

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.

Under AC [A1], the equation is a bivariate additive-noise model with $f(x)=x^3$, independent noise, strictly positive $C^3$ input and noise densities, and nonconstant $C^3$ function. By [F1] and [F4], its joint density is $p(x,y)=\varphi (x)\varphi (y-x^3)$.

Put $z=y-f(x)$. The standard normal log densities give ${\xi }^{\mathrm{\prime }\mathrm{\prime }}(x)=-1$, ${\xi }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(x)=0$, ${\nu }^{\mathrm{\prime }}(z)=-z$, ${\nu }^{\mathrm{\prime }\mathrm{\prime }}(z)=-1$, and ${\nu }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(z)=0$. At $(x,y)=(1,0)$, $z=-1$, $f^{\mathrm{\prime }}=3$, $f^{\mathrm{\prime }\mathrm{\prime }}=6$, and $f^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=6$, so ${\nu }^{\mathrm{\prime }\mathrm{\prime }}(z)f^{\mathrm{\prime }}=-3\mathrm{\ne }0$.

Substitution into Condition 19's right side gives $(-1)\left(\frac{6}{3}\right)-2(-1)(6)(3)+(1)(6)+0-(1)\frac{6^2}{3}=-2+36+6-12=28,$ whereas ${\xi }^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(1)=0$. Hence Condition 19 holds. The theorem [F3] now excludes a reverse ANM in its stated class, proving the Example's identifiability assertion.

The input and noise densities, and the cubic function, satisfy the hypotheses in the Statement; the nonzero-domain witness in step 1.2 satisfies the Condition 19 inequality in step 2.1.