# Causality: Models, Reasoning and Inference, by Judea Pearl: Causal Bayesian Networks and the Truncated Factorization

Background: $$\newcommand{\doop}{\operatorname{do}}\newcommand{\op}{\operatorname{#1}}$$

Definition 1.2.2 (Markov Compatibility) If a probability function $$P$$ admits the factorization of $$P(x_1,\dots,x_n)=\prod_i P(x_i|\operatorname{pa}_i)$$ relative to the Directed Acyclic Graph (DAG) $$G,$$ we say that $$G$$ represents $$P,$$ that $$G$$ and $$P$$ are compatible, or that $$P$$ is Markov relative to $$G.$$

Here the $$\operatorname{pa}_i$$ are the parents of $$x_i.$$

Definition 1.3.1 (Causal Bayesian Network) Let $$P(v)$$ be a probability distribution on a set $$V$$ of variables, and let $$P(v|\doop(x))$$ denote the distribution resulting from the intervention $$\doop(X=x)$$ that sets a subset $$X$$ of variables to constants $$x.$$ Denote by $$P_*$$ the set of all interventional distributions $$P(v|\doop(x)), X\subseteq V,$$ including $$P(v),$$ which represents no intervention (i.e., $$X=\varnothing$$). A DAG $$G$$ is said to be a causal Bayesian network compatible with $$P_*$$ if and only if the following three conditions hold for every interventional $$P\in P_*:$$

1. $$P(v|\doop(x))$$ is Markov relative to $$G;$$
2. $$P(v_i|\doop(x))=1$$ for all $$V_i\in X$$ whenever $$v_i$$ is consistent with $$X=x;$$
3. $$P(v_i|\operatorname{pa}_i,\doop(x))=P(v_i|\operatorname{pa}_i)$$ for all $$V_i\not\in X$$ whenever $$\operatorname{pa}_i$$ is consistent with $$X=x,$$ i.e., each $$P(v_i|\operatorname{pa}_i)$$ remains invariant to interventions not involving $$V_i.$$

Truncated Factorization: $$P(v|\doop(x))=\prod_{i|V_i\not\in X}P(v_i|\operatorname{pa}_i)\qquad\text{for all } v \text{ consistent with }x.$$

Problem Statement: Prove that the three conditions of Definition 1.3.1 (Causal Bayesian Network) are necessary and sufficient for the truncated factorization.

$$(\to)$$ Assume the three conditions of Definition 1.3.1 hold. We know by (i) that we can write $$P(v|\doop(x))=\prod_iP(v_i|\op{pa}_i,\doop(x)).$$ Then we factor into two products, depending on where the $$v_i$$ are: \begin{align*} P(v|\doop(x)) &=\prod_iP(v_i|\op{pa}_i,\doop(x))\\ &=\left[\prod_{i, v_i\in X}P(v_i|\op{pa}_i,\doop(x))\right]\left[\prod_{i, v_i\not\in X}P(v_i|\op{pa}_i,\doop(x))\right]. \end{align*} According to (ii), the first product is $$1,$$ yielding $$P(v|\doop(x))=\prod_{i, v_i\not\in X}P(v_i|\op{pa}_i,\doop(x)).$$ Finally, we argue that \begin{align*} P(v|\doop(x)) &=\prod_{i, v_i\not\in X}P(v_i|\op{pa}_i,\doop(x))\\ &=\prod_{i, v_i\not\in X}P(v_i|\op{pa}_i) \end{align*} by invoking (iii), since we assumed that $$P(v_i|\op{pa}_i)$$ are invariant to interventions involving $$X.$$ Am I correct so far?

$$(\leftarrow)$$ Other than the obvious first step of assuming we can write the truncated factorization, I don't have any ideas on this one. How can I proceed? Would the steps in the $$(\to)$$ direction all be reversible?

$$\newcommand{\doop}{\operatorname{do}}\newcommand{\op}{\operatorname{#1}}$$ I think your proof of the forward implication is correct. For the backward implication I may have something.

Suppose the Truncated Factorization: for all $$v$$ consistent with $$x$$, $$P(v\mid \mathrm{do}(x))=\prod_{i\mid Vi\notin X}P(v_i \mid \mathrm{pa}_i)$$ for a non cyclic oriented graph $$G$$.

## Proof that condition 3 is verified

Let be $$i$$, $$v_i$$, and an intervention $$X = x$$ be such that $$V_i \notin X$$ and a realization of $$\mathrm{pa}_i$$ is compatible with $$X = x$$. We need to prove that: $$P(v_i|\mathrm{pa}_i, \mathrm{do}(x)) = P(v_i |\mathrm{pa}_i).$$ To do so, let's take an intervention $$X' = x'$$ such that:

• $$X \subset X'.$$
• $$X = x$$ and $$X' = x'$$ are compatible.
• $$V_i \notin X'$$ and $$\forall j\neq i, V_j\in X'.$$
• The realization of $$\mathrm{pa_i}$$ considered is compatible with $$X'.$$

Intuitively, we are fixing everything but $$V_i$$ by an intervention, without contradicting the intervention $$X = x$$ nor the considered realization of $$\mathrm{pa}_i$$.

Then, using the factorization, $$P(v|\mathrm{do}(x')) = P(v_i| \mathrm{pa}_i)$$ since only index $$i$$ is left in the product, and thus $$P(v|\mathrm{do}(x'), \mathrm{do}(x)) = P(v_i|\mathrm{pa}_i, \mathrm{do}(x)).$$ But as $$X = x$$ is included in $$X' = x'$$, $$P(v|\mathrm{do}(x'), \mathrm{do}(x)) = P(v|\mathrm{do}(x'))$$. So we have that: $$P(v_i|\mathrm{pa}_i, \mathrm{do}(x)) = P(v_i|\mathrm{pa}_i),$$ which is what we wanted.

## Proof that condition 1 is verified

If we use the truncated factorization on a null intervention, we obtain that $$G$$ and $$P$$ are Markov compatible: $$P(v) = \prod_i P(v_i|\mathrm{pa}_i).$$ Conditioning the last equation on an intervention $$X = x$$, we get that $$P(v|\mathrm{do}(x)) = \prod_i P(v_i|\mathrm{pa}_i, \mathrm{do}(x)),$$ which is that $$P(v|\mathrm{do}(x))$$ and $$G$$ are Markov compatible.

## Proof that condition 2 is verified

Let's consider an intervention $$X =x$$. Using condition 1, we have: \begin{align*} P(v|\doop(x)) &=\prod_i P(v_i|\op{pa}_i, \doop(x))\\ &=\prod_{i|V_i\in X}P(v_i|\op{pa}_i,\doop(x))\cdot\prod_{i|V_i\not\in X}P(v_i|\op{pa}_i,\doop(x))\\ &=\prod_{i|V_i\in X}P(v_i|\op{pa}_i,\doop(x))\cdot\prod_{i|V_i\not\in X}P(v_i|\op{pa}_i), \end{align*} using condition 3. As $$P(v|\mathrm{do}(x))$$ can also be expressed with the truncated factorization, we get that: $$\prod_{i|V_i \notin X}P(v_i|\mathrm{pa}_i) = \prod_{i|V_i \in X}P(v_i|\mathrm{pa}_i, \mathrm{do}(x))\prod_{i|V_i \notin X}P(v_i| \mathrm{pa}_i)$$ and so, simplifying by dividing out $$P(v_i|\mathrm{pa}_i)$$: $$\prod_{i|V_i \in X}P(v_i|\mathrm{pa}_i) = 1 .$$ (To be allowed to make this simplification, we need to suppose that $$P(v_i|\mathrm{pa}_i) \neq 0$$, which is necessarily the case if we suppose $$P(v|\mathrm{do}(x)) \neq 0$$ for instance.)

In the end we have that $$P(v_i|\mathrm{pa}_i, \mathrm{do}(x)) = 1$$ for all $$i$$ such that $$i \in X$$ (since their product is $$1$$). To get to condition 2, write $$P(v_i|\mathrm{do}(x)) = \mathbb{E}_{\mathrm{pa}_i}\left[P(v_i| \mathrm{pa}_i, \mathrm{do}(x))\right] = \mathbb{E}_{\mathrm{pa}_i}\left[1\right] = 1.$$

I hope this is understandable, correct and helping..

• Thanks very much! I'll go ahead and award you the bounty, but I need to digest it a bit more before I accept it as the answer. – Adrian Keister Jun 8 at 18:12
• In (ii), why does $$P(v_i|\operatorname{do}(x))=E_{\operatorname{pa}_i}\!\!\left[P(v_i|\operatorname{pa}_i,\operatorname{do}(x))\right]?$$ – Adrian Keister Jun 10 at 20:23
• We start from the equality $P(v_i) = E_{pa_i}[P(v_i\mid pa_i)]$ and then condition on the intervention $do(x)$. Is this correct ? – Pohoua Jun 10 at 20:50