[proofplan] We construct two binary recursive structural causal models explicitly. In the first model, $A$ directly causes $Y$ through the structural equation $Y=A$. In the second model, there is no arrow from $A$ to $Y$; instead, an unobserved Bernoulli variable $U$ determines both $A$ and $Y$. The two models give the same observational diagonal law on $(A,Y)$, but intervention on $A$ changes $Y$ in the first model and leaves $Y$ unchanged in the second. [/proofplan]

[step:Construct a direct-cause model with diagonal observational law] Let $\Omega_1:=\{0,1\}$ and let $\mathcal F_1:=2^{\{0,1\}}$. Define the probability measure $\mathbb P_1$ on $(\Omega_1,\mathcal F_1)$ by \begin{align*} \mathbb P_1(\{0\})=\mathbb P_1(\{1\})=\frac{1}{2}. \end{align*} Define the exogenous [random variable](/page/Random%20Variable) $U_1:(\Omega_1,\mathcal F_1)\to(\{0,1\},2^{\{0,1\}})$ by $U_1(\omega):=\omega$. Let $M_1$ be the recursive structural causal model with observed endogenous variables $A_1$ and $Y_1$, both valued in $\{0,1\}$, and structural assignments \begin{align*} A_1:=U_1 \end{align*} and \begin{align*} Y_1:=A_1. \end{align*} Equivalently, the structural map for $A_1$ is the identity map on $\{0,1\}$ applied to $U_1$, and the structural map for $Y_1$ is \begin{align*} f_{Y,1}:\{0,1\}\to\{0,1\},\qquad a\mapsto a. \end{align*} For every $\omega\in\Omega_1$, one has $A_1(\omega)=\omega$ and $Y_1(\omega)=\omega$. Hence the observational law of $(A_1,Y_1)$ is \begin{align*} \mathbb P_1(A_1=0,Y_1=0)=\frac{1}{2}, \end{align*} \begin{align*} \mathbb P_1(A_1=1,Y_1=1)=\frac{1}{2}, \end{align*} and \begin{align*} \mathbb P_1(A_1=0,Y_1=1)=\mathbb P_1(A_1=1,Y_1=0)=0. \end{align*} [/step]

[step:Compute the intervention in the direct-cause model] Let $M_1^{do(A=1)}$ denote the intervened model obtained from $M_1$ by replacing the structural assignment for $A_1$ with the constant value $1$ and leaving the structural assignment for $Y_1$ unchanged. In this step, $\mathbb P_{M_1}(Y=1\mid do(A=1))$ denotes the probability of the event $\{Y_1^{do(A=1)}=1\}$ under the same exogenous [probability space](/page/Probability%20Space) $(\Omega_1,\mathcal F_1,\mathbb P_1)$. In this intervened model, \begin{align*} A_1^{do(A=1)}=1 \end{align*} and therefore \begin{align*} Y_1^{do(A=1)}=A_1^{do(A=1)}=1. \end{align*} Thus \begin{align*} \mathbb P_{M_1}(Y=1\mid do(A=1))=1. \end{align*} [/step]

[step:Construct a hidden-common-cause model with the same observational law]Let $\Omega_2:=\{0,1\}$ and let $\mathcal F_2:=2^{\{0,1\}}$. Define the probability measure $\mathbb P_2$ on $(\Omega_2,\mathcal F_2)$ by \begin{align*} \mathbb P_2(\{0\})=\mathbb P_2(\{1\})=\frac{1}{2}. \end{align*} Define the unobserved exogenous random variable $U_2:(\Omega_2,\mathcal F_2)\to(\{0,1\},2^{\{0,1\}})$ by $U_2(\omega):=\omega$. Let $M_2$ be the recursive structural causal model with observed endogenous variables $A_2$ and $Y_2$, both valued in $\{0,1\}$, and structural assignments \begin{align*} A_2:=U_2 \end{align*} and \begin{align*} Y_2:=U_2. \end{align*} Here $U_2$ is not observed in the observational distribution of interest, and it is a common cause of both observed variables $A_2$ and $Y_2$. For every $\omega\in\Omega_2$, one has $A_2(\omega)=\omega$ and $Y_2(\omega)=\omega$. Hence \begin{align*} \mathbb P_2(A_2=0,Y_2=0)=\frac{1}{2}, \end{align*} \begin{align*} \mathbb P_2(A_2=1,Y_2=1)=\frac{1}{2}, \end{align*} and \begin{align*} \mathbb P_2(A_2=0,Y_2=1)=\mathbb P_2(A_2=1,Y_2=0)=0. \end{align*} Comparing with the computation for $M_1$, we obtain \begin{align*} \mathbb P_{M_1}(A=a,Y=y)=\mathbb P_{M_2}(A=a,Y=y) \end{align*} for every $(a,y)\in\{0,1\}^2$.[/step]

[guided]We now build the second model so that the observed association between $A$ and $Y$ is the same as in the first model, but its causal explanation is different. The sample space is again $\Omega_2:=\{0,1\}$ with $\mathcal F_2:=2^{\{0,1\}}$, with probability measure $\mathbb P_2$ satisfying \begin{align*} \mathbb P_2(\{0\})=\mathbb P_2(\{1\})=\frac{1}{2}. \end{align*} Define the unobserved exogenous random variable $U_2:(\Omega_2,\mathcal F_2)\to(\{0,1\},2^{\{0,1\}})$ by $U_2(\omega):=\omega$. The model $M_2$ has observed endogenous variables $A_2$ and $Y_2$, but both are generated from the same hidden variable: \begin{align*} A_2:=U_2. \end{align*} \begin{align*} Y_2:=U_2. \end{align*} Thus there is no structural arrow from $A_2$ to $Y_2$; the dependence between the observed variables is induced by the common unobserved cause $U_2$. We compute the observational distribution directly. If $\omega=0$, then $A_2(\omega)=0$ and $Y_2(\omega)=0$. If $\omega=1$, then $A_2(\omega)=1$ and $Y_2(\omega)=1$. Therefore \begin{align*} \mathbb P_2(A_2=0,Y_2=0)=\mathbb P_2(\{0\})=\frac{1}{2}, \end{align*} \begin{align*} \mathbb P_2(A_2=1,Y_2=1)=\mathbb P_2(\{1\})=\frac{1}{2}, \end{align*} and the off-diagonal events are empty: \begin{align*} \{A_2=0,Y_2=1\}=\varnothing,\qquad \{A_2=1,Y_2=0\}=\varnothing. \end{align*} Hence \begin{align*} \mathbb P_2(A_2=0,Y_2=1)=\mathbb P_2(A_2=1,Y_2=0)=0. \end{align*} For comparison, in the direct-cause model we have $A_1=U_1$ and $Y_1=A_1$, so $A_1(\omega)=Y_1(\omega)=\omega$ for every $\omega\in\Omega_1$. Therefore the observational law of $(A_1,Y_1)$ is \begin{align*} \mathbb P_1(A_1=0,Y_1=0)=\frac{1}{2}, \end{align*} \begin{align*} \mathbb P_1(A_1=1,Y_1=1)=\frac{1}{2}, \end{align*} and \begin{align*} \mathbb P_1(A_1=0,Y_1=1)=\mathbb P_1(A_1=1,Y_1=0)=0. \end{align*} The computation above shows that the observational law of $(A_2,Y_2)$ assigns exactly the same four masses. Hence, for every $(a,y)\in\{0,1\}^2$, \begin{align*} \mathbb P_{M_1}(A=a,Y=y)=\mathbb P_{M_2}(A=a,Y=y). \end{align*} The important point is that the same joint distribution of the observed pair $(A,Y)$ has two different causal explanations: direct causation in $M_1$ and hidden common causation in $M_2$.[/guided]

[step:Compute the intervention in the hidden-common-cause model] Let $M_2^{do(A=1)}$ denote the intervened model obtained from $M_2$ by replacing the structural assignment for $A_2$ with the constant value $1$ and leaving the structural assignment for $Y_2$ unchanged. In this step, $\mathbb P_{M_2}(Y=1\mid do(A=1))$ denotes the probability of the event $\{Y_2^{do(A=1)}=1\}$ under the same exogenous probability space $(\Omega_2,\mathcal F_2,\mathbb P_2)$. Thus \begin{align*} A_2^{do(A=1)}=1 \end{align*} while \begin{align*} Y_2^{do(A=1)}=U_2. \end{align*} Since $\mathbb P_2(U_2=1)=1/2$, it follows that \begin{align*} \mathbb P_{M_2}(Y=1\mid do(A=1))=\frac{1}{2}. \end{align*} [/step]

[step:Compare the two interventional laws] The observational laws of $(A,Y)$ under $M_1$ and $M_2$ are identical, since both assign mass $1/2$ to $(0,0)$, mass $1/2$ to $(1,1)$, and mass $0$ to the two off-diagonal points. However, the interventional distributions under $do(A=1)$ differ: \begin{align*} \mathbb P_{M_1}(Y=1\mid do(A=1))=1 \end{align*} whereas \begin{align*} \mathbb P_{M_2}(Y=1\mid do(A=1))=\frac{1}{2}. \end{align*} Thus the interventional distribution of $Y$ under intervention on $A$ is not determined by the observational distribution of $(A,Y)$ alone. This proves the claimed nonidentifiability under hidden confounding. [/step]