# Representation of unconfoundedness of Rubin causal models on Pearl causal models

Disclaimer: I'm new to the potential outcome framework so this question may not make lot of sense, in case please let me understand where I am failing.

I was reading this question about the unconfoundedness assumption for Rubin's causal models that claims that the potential outcomes are independent of the treament given the observations (mathematically that is $$(Y(0),Y(1))\perp T|X$$) and, since I'm used to Pearl's view of causality, I was trying to figure out if such assumption could be translated visually using a DAG.

I don't think it is possible to do it without relying on a Richardson's single world intervention graph but I wanted to have a confirmation from somebody that is more used to both Rubin and Pearl worlds than me.

Yes, you can represent them in a DAG.

In the structural framework, the DAG is nothing but a visual representation of the functional arguments that enter the structural equations. For a quick review see the appendix of the Crash Course in Good and Bad controls.

As an example, consider the following structural causal model:

\begin{align} X &= f_x(U_x)\\ T &= f_t(X, U_t)\\ Y &= f_y(T, X, U_y) \end{align}

Where here the disturbances $$U$$ are mutually independent (this assumption is not necessary, and just for illustration purposes).

The model above has an associated DAG:

Where independent disturbances are usually omitted for clarity (they need not be).

Now let us represent the potential outcomes in a DAG.

Note the potential outcome $$Y(1)$$ is defined as the solution to the structural equations when we set $$T=1$$.

$$Y(1) = f_y(1, X, U_y)$$

Note the only parents of $$Y(1)$$ are $$X$$ and $$U_y$$. You can thus write the DAG of the variables $$T$$, $$X$$ and $$Y(1)$$ as:

In this DAG you can read that $$Y(1) \perp\!\!\!\perp T \mid X$$. The above logic is general, and with that you can write as many potential outcomes as you would like in your DAGs.

For instance, using the same logic, you can derive $$Y(0) = f_y(0, X, U_y)$$, and if you want to, you can represent both potential outcomes in the same DAG:

Where here we now do need to write the disturbance $$U_y$$, since it is common parent of $$Y(1)$$ and $$Y(0)$$.

This allows you to read the joint independence $$\{Y(1), Y(0)\} \perp\!\!\!\perp T \mid X$$.

• Hi Carlos, thanks a lot for the answer, very well written! However it is not totally clear to me why you didn't substitute 1 to T as well. Shouldn't we modify it as well? Commented Dec 27, 2022 at 20:55
• @DaSim You can represent as many random variables in the graph as you want, the original $T$, the original $Y$, potential outcomes etc. Here since you want to check whether $T$ is independent of $Y(1)$, you also want to keep the random variable $T$ in the graph, precisely to be able to check that independence. Commented Dec 28, 2022 at 8:29