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

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.

My Answer So Far:

$(\to)$ 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?

Many thanks for your time!


1 Answer 1


$\newcommand{\doop}{\operatorname{do}}\newcommand{\op}[1]{\operatorname{#1}}$ 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..

  • $\begingroup$ 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. $\endgroup$ Commented Jun 8, 2020 at 18:12
  • $\begingroup$ In (ii), why does $$P(v_i|\operatorname{do}(x))=E_{\operatorname{pa}_i}\!\!\left[P(v_i|\operatorname{pa}_i,\operatorname{do}(x))\right]?$$ $\endgroup$ Commented Jun 10, 2020 at 20:23
  • $\begingroup$ 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 ? $\endgroup$
    – Pohoua
    Commented Jun 10, 2020 at 20:50

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