Given the causal graph ($Z$->$X$$(Z\to X$, $Z$->$Y$$Z\to Y$, $X$->$Y$)$X\to Y)$, according to Pearl’s intervention, the effect of intervening $X$ on $Y$ can be estimated as
$$P(Y=y|do(X=x)) = \sum_z P(Y=y|X=x,Z=z)P(Z=z)$$$$P(Y=y|\operatorname{do}(X=x)) = \sum_z P(Y=y|X=x,Z=z)\,P(Z=z).$$
Regarding this formula, I have 2 following questions:
- If for $Z=z^*$,$count(X=x,Z=z^*)=0$,$Z=z^*,$ $\text{count}(X=x,Z=z^*)=0,$ do we set $P(Y=y│X=x,Z=z^* )=0$ ?$P(Y=y│X=x,Z=z^*)=0?$
- In cases where $P(z^*)$ is large but there is only anone instance where $Z=z^*$ and $X=x$ simultaneously, i.e. $count(X=x,Z=z^* )=1$, hence, $P(Y=y│X=x,Z=z^* )=1$.$\text{count}(X=x,Z=z^* )=1,$ 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 $P(Y=y│do(X=x))$ ? Is that a problem in this estimation?
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). Have $$P(Y=1│do(X=1))=P(Y=1│X=1,Z=z_1)+P(Y=1│X=1,Z=z_2)+P(Y=1│X=1,Z=z_3 )$$ $$=1/4×5/10+ 0/1×3/10+\mathbf{0/0}×2/10$$ $$P(Y=1│do(X=0) )=P(Y=1│X=0,Z=z_1)+P(Y=1│X=0,Z=z_2)+P(Y=1│X=0,Z=z_3)$$: $$=\mathbf{1/1×5/10}+ 1/2×3/10+2/2×2/10$$\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*}
