How to deal with data edge cases when applying Pearl intervention formula for estimation?

Question

Given the causal graph $(Z\to X$, $Z\to Y$, $X\to Y)$, according to Pearl’s intervention, the effect of intervening $X$ on $Y$ can be estimated as $$P(Y=y|\operatorname{do}(X=x)) = \sum_z P(Y=y|X=x,Z=z)\,P(Z=z).$$ Regarding this formula, I have 2 following questions:

• If for $Z=z^*,$ $\text{count}(X=x,Z=z^*)=0,$ do we set $P(Y=y│X=x,Z=z^*)=0?$
• In cases where $P(z^*)$ is large but there is only one instance where $Z=z^*$ and $X=x$ simultaneously, i.e. $\text{count}(X=x,Z=z^* )=1,$ do we write $P(Y=y│X=x,Z=z^*)=1?$
  

Will the term $P(Y=y│X=x,Z=z^*)\,P(Z=z^*)$ dominate the estimation of $P(Y=y│\operatorname{do}(X=x))?$ Is that a problem in this estimation?

The above questions are demonstrated with the following data. The terms related to the questions are bold in the equation. Within this data, $X\in\{0,1\},Y\in\{0,1\}$, and $Z\in\{z_1,z_2,z_3\}$.

To analyze the causal effect of $X$ on $Y$, we estimate the average causal effect (ACE): \begin{align*} P(Y=1│\operatorname{do}(X=1)) &=P(Y=1│X=1,Z=z_1)\cdot P(Z=z_1)\\ &\quad+P(Y=1│X=1,Z=z_2)\cdot P(Z=z_2)\\ &\quad+P(Y=1│X=1,Z=z_3)\cdot P(Z=z_3)\\ &=1/4\times 5/10+ 0/1\times 3/10+\mathbf{0/0}\times 2/10,\\ P(Y=1│\operatorname{do}(X=0)) &=P(Y=1│X=0,Z=z_1)\cdot P(Z=z_1)\\ &\quad+P(Y=1│X=0,Z=z_2)\cdot P(Z=z_2)\\ &\quad+P(Y=1│X=0,Z=z_3)\cdot P(Z=z_3)\\ &=\mathbf{1/1\times 5/10}+ 1/2\times 3/10+2/2\times 2/10. \end{align*}

Answer 1

If in your sample data all the rows are equi-probable, then your calculations seem correct to me, except that you never compute a probability as $0/0.$ That term should just be $0.$ If an event never happens, its probability is zero; the answer to your first question is "yes". The answer to your second question is also "yes". The probability $P(Z=z^*)$ is largely independent of the probability $P(Y=y|X=x,Z=z).$ Remember that in the New Causal Revolution, the concept of conditional probability has not changed meaning in the slightest. It's still defined as the probability an event will occur in some sort of "reduced world":

$$P(A|B)=\frac{P(A\cap B)}{P(B)}.$$

In this expression, we restrict our attention to the "world where $B$ has occurred." In that world, what is the probability that $A$ has occurred? That's what conditional probability is all about.

Whenever you compute some complex expression like the backdoor adjustment formula, it's of course always possible that a few terms will dominate others. I don't see that that's a problem as long as you do the computations correctly!

Answer 2

Since Pearl's intervention formula is probabilistic in nature based on graph surgery of causal graph DAGs, it's assuming true marginal and/or conditional distributions of all involved variables. And your problem here is only to estimate some probability mass of the effect $Y$ after two interventions on causal variable $A$ based entirely on very limited available data. And for your concerned edge cases of the empirical data distribution estimated based on frequency counting, if no instance available for some conditional case (i.e., conditioned on $X=1, Z=z_3$ in your question) as frequency's denominator then apparently any conditional frequency should be $0$ since the conditioned sample space is empty. Similarly if there's only one single example under a specific condition then you just count its corresponding frequency denominator as $1$ as you correctly did for the first term in your last line above.

As for one term's domination of empirical intervention estimation, you comment on an existing answer is incorrect, $P(Y=1│do(X=1))=1/4$ which is actually much smaller than $ P(Y=1│do(X=0))$, it seems you missed the $P(Z=z_1)$ term. Therefore this very limited data set is actually fairly reasonable since it's consistent with the expected effect of $Y=0$ (you may interpret it as bad effect while $Y=1$ as good effect) once we intervened as $A=1$ (you may interpret it as smoke, and you may further interpret $Z$ as health mindfulness). Under such interpretation, $Z=z_1$ could mean poor health mindfulness which may explain why you have no conditional instance for your above concerned case $X=1, Z=z_3$. A high health mindful person would most likely not smoke, so there's no available data in your given small dataset. In summary even for empirical estimation problems so long as your samples are representative such as from a simple random sampling, there's no theoretical dominance issue caused by certain term in the intervention formula as you concerned, the primary concern if any here is your data quality.