SCM, DAGs, adjustment — Examples

In the confounding model, $P(A=1)=P(U=0,E=1)+P(U=1,E=0)=\frac{1}{2}\cdot \frac{1}{4}+\frac{1}{2}\cdot \frac{3}{4}=\frac{1}{2}$. Also $P(U=1,A=1)=\frac{1}{2}\cdot \frac{3}{4}=\frac{3}{8}$ and $P(U=1,A=0)=\frac{1}{2}\cdot \frac{1}{4}=\frac{1}{8}$. Thus $P(Y=1\mid A=1)=3\mathrm{/}4$ and $P(Y=1\mid A=0)=1\mathrm{/}4$, although under $do(A=a)$ the variable $Y=U$ remains Bernoulli$(1\mathrm{/}2)$. The difference between the observational conditionals and the intervention law is caused by the open backdoor path through $U$; conditioning on $U$ blocks that path and the adjusted mean is ${\sum }_{u\in \{0,1\}}P(Y=1\mid A=a,U=u)P(U=u)=1\mathrm{/}2$.

For the collider model, $P(B=1,C=1)=1\mathrm{/}4=P(B=1)P(C=1)$ by [F1], so the variables are independent before conditioning. But $D=1$ on exactly the three equiprobable pairs $(1,1),(1,0),(0,1)$. Hence $P(B=1\mid D=1)=P(C=1\mid D=1)=2\mathrm{/}3$ while $P(B=1,C=1\mid D=1)=1\mathrm{/}3\mathrm{\ne }4\mathrm{/}9$. Conditioning on the collider $D$ therefore creates dependence, exactly as the active-path rule [F2] predicts.

In either model, $A=Y=U$, so the observed joint law assigns probability $1\mathrm{/}2$ to $(A,Y)=(0,0)$ and $1\mathrm{/}2$ to $(A,Y)=(1,1)$, and zero to the other pairs. The parameter values representing the two SCMs are observationally equivalent by [F1].

Under $do(A=a)$ in SCM 1, the equation $Y=A$ remains and gives $Y=a$ with probability one. Under $do(A=a)$ in SCM 2, the equation $Y=U$ remains and $U$ is unchanged, so $Y$ is Bernoulli$(1\mathrm{/}2)$. Thus for $a=0$ the interventional success probabilities are respectively $0$ and $1\mathrm{/}2$; the intervention-law estimand is not constant on the observational-equivalence class in step 1.1. It is therefore not identifiable on the two-model statistical model by [F2].

Since $P(A=1)=1\mathrm{/}2$, Bayes' rule gives $P(U=1\mid A=1)=3\mathrm{/}4$ and $P(U=1\mid A=0)=1\mathrm{/}4$. Since $E_M$ is independent of $(U,A)$, $P(M=1\mid A=1)=4\mathrm{/}5$ and $P(M=1\mid A=0)=1\mathrm{/}5$. Thus for each $a\in \{0,1\}$ both mediator outcomes have positive conditional probability; the nested-law support condition in [F2] holds.

If $M=1$, then $Y=1$. If $M=0$, then $Y=U$, so $E(Y\mid M=0,A=1)=3\mathrm{/}4$ and $E(Y\mid M=0,A=0)=1\mathrm{/}4$. Averaging over the marginal law of $A$ gives $\int E(Y\mid M=0,A=a^{\mathrm{\prime }})P_A(d{a}^{\mathrm{\prime }})=1\mathrm{/}2$, while the same inner average for $M=1$ is $1$. The front-door formula [F2] therefore gives $\frac{4}{5}\cdot 1+\frac{1}{5}\cdot \frac{1}{2}=\frac{9}{10}\mathrm{.}$

Under $do(A=1)$, $M=1$ with probability $4\mathrm{/}5$ and $M=0$ with probability $1\mathrm{/}5$. The exogenous $U$ remains Bernoulli$(1\mathrm{/}2)$ and independent of $E_M$, so $E(Y\mid do(A=1),M=1)=1$ and $E(Y\mid do(A=1),M=0)=1\mathrm{/}2$. Hence $E[Y^1]=\frac{4}{5}\cdot 1+\frac{1}{5}\cdot \frac{1}{2}=9\mathrm{/}10$, agreeing with the observed functional.