CIS Primer Question 2.5.1

Here are my solutions to question 2.5.1 of Causal Inference in Statistics: a Primer (CISP).

We’ll use the following simulated dataset to verify our answers. The coefficients are chosen so that each variable has approximately unit variance.

Part a and b

The book is not so clear on the definition of equivalent causal DAGs, and there are no explicit examples as far as I can tell. I’ll assume the following definition:

Causal homomorphism
Let $$G$$, $$H$$ be causal DAGs with skeletons $$G_s$$, $$H_s$$, respectively, and let $$f : G_s \to H_s$$ be a graph homomorphism. Then $$f$$ is said to be a causal homomorphism if each collider in $$G$$ is mapped to a collider in $$H$$; that is, if $$(x, z, y)$$ is a collider in $$G$$, then $$(f(x), f(z), f(y))$$ is a collider in $$H$$. Moreover, $$f$$ is a causal isomorphism if it is bijective with an inverse that is also a causal homomorphism.

One implication of this is that the degree of any node is invariant under a causal isomorphism. In particular, $$f(Z_3) = Z_3$$, since $$Z_3$$ is the only node of degree 4. Furthermore, $$f(X) \in \{ X, Y\}$$, so $$f(W) = W$$. Thus, the only non-trivial causal isomorphism maps $$f(X) = Y$$, $$f(Z_1) = Z_2$$. In other words, the following causal DAG is the only causal DAG that is equivalent to figure 2.9 with different causal implications.

The above graph is the only one that is observationally equivalent to figure 2.9

Part c

The directionality between $$\{X, W\}$$ and between $$\{W, Y\}$$ cannot be determined from nonexperimental data.

Part d

From part a of the previous question, we know that $$\{W, Z_1, Z_3\}$$ d-separates $$\{Y, X\}$$. This implies that the causal model from figure 2.9 is wrong is the coefficient of $$X$$ in the regression $$Y \sim 1 + X + W + Z_1 + Z_3$$ can be shown to be non-zero.

Part e

Again from part a of the previous question, we know that $$X$$ d-separates $$\{Z_3, W\}$$. This implies that the causal model from figure 2.9 is wrong is the coefficient of $$W$$ in the regression $$Z_3 \sim 1 + X + W$$ can be shown to be non-zero.

Part f

The only non-adjacent node to $$Z_3$$ is $$W$$. There is only one unblocked path between $$Z_3$$ and $$W$$: $$Z_3 \rightarrow X \rightarrow W$$. The only way to block this path is by conditioning on $$X$$, which wouldn’t be possible if $$X$$ were unobservable. Therefore, it seems like it’s not possible to find an equation that proves figure 2.9 wrong via a non-zero coefficient in a regression model of $$Z_3$$ with $$X$$ unobserved.

Part g

The causal DAG tells us we can decompose the joint probability as:

$\mathbb P(W, X, Y, Z_1, Z_2, Z_3) = \\ \mathbb P (Z_1) \cdot \mathbb P (Z_2) \cdot \mathbb P (Z_3 \mid Z_1, Z_2) \cdot \mathbb P (X \mid Z_1, Z_3) \cdot \mathbb P (W \mid X) \cdot \mathbb P (Y \mid W, Z_2, Z_3) .$

I’m unsure of the connection between this decomposition and implied vanishing partial regression coefficients. The 7 d-separation statements from 2.4.1 part a seem like they would be sufficient to fully test the model in this way.