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Latent confounding can mimic an observed ANM direction

Example

Assume the Axiom of Choice. Let $U, ϵ_{X}, ϵ_{Y}$ be mutually independent standard normal random variables. In Model A, set $U = N_{U}$ , $X = U + ϵ_{X}$ , and $Y = U + ϵ_{Y}$ , with $U$ latent and $X, Y$ observed. In Model B, let $Z_{1}, Z_{2}$ be independent standard normal variables, set $X = \sqrt{2} Z_{1}$ , $E = \sqrt{3 / 2} Z_{2}$ , and $Y = X / 2 + E$ , with both vertices observed. In each model the observed vector $(X, Y)$ has density $p (x, y) = \frac{1}{2 π \sqrt{3}} \exp ⁣ (- \frac{x^{2} - x y + y^{2}}{3}) .$ Under $d o (X = 1)$ , Model A gives $Y = U + ϵ_{Y}$ with density $\frac{1}{2 \sqrt{π}} e^{- y^{2} / 4}$ , while Model B gives $Y = 1 / 2 + E$ with density $\frac{1}{\sqrt{3 π}} e^{- (y - 1 / 2)^{2} / 3}$ . The two intervention densities differ at $y = 1 / 2$ , so the same observational law is compatible with different causal effects when a latent common cause is allowed.

Example references

Latent vertices and intervention semantics

Observed additive-noise model in Model B

Joint density of the observed ANM in Model B

Peters et al., JMLR (2014), latent-variable discussion

Assaad, Devijver, and Gaussier, course notes (2023), slide 23

Peters, Janzing, and Schölkopf, Elements of Causal Inference, §7.2.3

Remarks

In the diagram, circles denote variables and arrows denote direct structural inputs. The dashed vertex $U$ is latent in Model A, so only $X$ and $Y$ are observed there. Model B has the observed edge $X \to Y$ . The diagrams show the two model graphs; the density and intervention calculations below prove the claimed observational and interventional comparisons.

\begin{tikzpicture}[>=stealth]
  \node[draw,circle,dashed] (u) at (0,0) {$U$};
  \node[draw,circle] (xa) at (2,1) {$X$};
  \node[draw,circle] (ya) at (2,-1) {$Y$};
  \draw[->] (u) -- (xa);
  \draw[->] (u) -- (ya);
  \node at (1,-2) {Model A};
  \node[draw,circle] (xb) at (4,0) {$X$};
  \node[draw,circle] (yb) at (6,0) {$Y$};
  \draw[->] (xb) -- (yb);
  \node at (5,-1) {Model B};
\end{tikzpicture}

Facts & Assumptions

Given: Assume AC. Use the standard normal density $ϕ (t) = (2 π)^{- 1 / 2} e^{- t^{2} / 2}$ and the independent noises specified in the Example.

[A1]

Assume AC for the standard-normal law; AC implies countable choice (AC $_{ω}$ ), which is the additional interface used by normalization and Lebesgue-change suppliers. The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ), AC implies DC implies countable choice

[F1]

The finite recursive SCM convention defines latent vertices, observed marginals, and $d o (X_{I} = a)$ by replacing only the intervened equations. Recursive structural causal models on finite DAGs

[F2]

A standard normal has density $ϕ$ ; its affine images have the corresponding scaled density. Standard normal and normal laws, The standard normal density has total mass one, A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not

[F3]

The Gaussian integral is $\int_{R} e^{- t^{2}} d t = \sqrt{π}$ . The Gaussian integral $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$

[F4]

For a finite ANM on fully observed vertices, the joint density is the product of its shifted noise densities. Continuous additive-noise models on finite DAGs, Positive joint density of a finite additive-noise model

[F5]

Nonnegative product integrals may be marginalized in either coordinate order. Tonelli's theorem for nonnegative measurable functions on a sigma-finite product

Verification

Given: The two SCMs and standard normal densities in the Example.

1.1givenA1F3F4F5algebraconstruct

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 $ϕ (u) ϕ (x - u) ϕ (y - u)$ ; marginalizing the latent $U$ gives $p_{A} (x, y) = \int_{R} ϕ (u) ϕ (x - u) ϕ (y - u) d u .$ Completing the square, $u^{2} + (x - u)^{2} + (y - u)^{2} = 3 {(u - \frac{x + y}{3})}^{2} + \frac{2}{3} (x^{2} - x y + y^{2}) .$ Thus [F3] and the substitution $t = \sqrt{3 / 2} (u - (x + y) / 3)$ give $p_{A} (x, y) = \frac{1}{2 π \sqrt{3}} \exp ⁣ (- \frac{x^{2} - x y + y^{2}}{3}) .$

1.2givenA1F2F4algebra

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

2.1givenA1F1F2F3step 1.1step 1.2algebra∎

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

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Sources

Jonas Peters, Joris Mooij, Dominik Janzing, and Bernhard Schölkopf, Causal Discovery with Continuous Additive Noise Models, JMLR 15 (2014), latent-variable discussion (standard reference, not scraped)
K. Assaad, E. Devijver, and G. Gaussier, Causal discovery: noise-based methods, ENS Lyon course notes (2023), slide 23 (standard reference, not scraped)
Jonas Peters, Dominik Janzing, and Bernhard Schölkopf, Elements of Causal Inference, §7.2.3 (standard reference, not scraped)