currently I am studying the model of potential outcomes and am still confused after reading the other answers on the other threads but also books treating this model.

How can the potential outcomes $(Y_{0i},Y_{1i})$ be independent of the treatment variable $T$?

Formally: $(Y_ {0i},Y_{1i}) \perp\!\!\!\perp T$ whereby $i$ is the $i$-th unit.

I know that we make this assumption to be able to eliminate the selection bias: $E[Y_{0i} | T_i=1] - E[Y_{0i} | T_i=0]$.
But how can the potential outcomes be independent of the treatment variable if we want to check for a causal relationship between $Y$ and $T$?

Looking at the switching equation: $Y_i=Y_{0i}+(Y_{1i}-Y_{0i})\cdot T_i$ with $T_i \in \{0,1\}$

$Y_i=Y_{0i}$ for $T_i=0$,
$Y_i=Y_{1i}$ for $T_i=1$

So I can not see why we can assume that the potential outcomes are independent of $T$ if the observed outcome we get is $Y_{0i}$ or $Y_{1i}$ depending on the treatment variable $T$.

The switching equations says that the potential outcome that will be assigned to our observed result $Y_i$ is dependent on the treatment variable $T$.
But assuming $(Y_ {0i},Y_{1i}) \perp\!\!\!\perp T$ while looking at the switching equation seems like a contradiction to me.

It looks like I am missing something here.
Would be happy if someone could explain me where my misconception lies.

  • $\begingroup$ See stats.stackexchange.com/a/182259/7071 $\endgroup$
    – dimitriy
    Commented Jul 15, 2023 at 17:02
  • $\begingroup$ @dimitriy read it yesterday. Did not clarify anything for me. $\endgroup$
    – lastwave
    Commented Jul 15, 2023 at 17:04
  • $\begingroup$ The switching equation has the observed outcome on the left, so T determines which potential outcome you observe. The orthogonality condition applies to the potential outcomes only, and not the observed outcome. It's as simple as that. $\endgroup$
    – dimitriy
    Commented Jul 15, 2023 at 17:13
  • $\begingroup$ @dimitriy How can it only apply for the potential outcomes if the observed outcome is assigned one of the potential outcomes? It does not make sense to me. Even if we would say that it applies only to the potential outcomes it does not change the fact that the observed outcome is one of the observed outcomes and therefore it has to also affect our observed outcome. $\endgroup$
    – lastwave
    Commented Jul 15, 2023 at 17:19
  • $\begingroup$ The counterfactual thought experiment is about pulling pairs of potential outcomes out if a hat. Orthogonality means the hat for treated is the same as the hat for untreated. Call the observed outcome Z to make the difference clear. The Ys are orthogonal to T, but Z is not orthogonal to T since Z is a function of Ys and T (assuming the treatment does something). $\endgroup$
    – dimitriy
    Commented Jul 15, 2023 at 17:31

1 Answer 1


Here is an example of an effective treatment assignment that satisfies PO orthogonality:

$$ \begin{array}{|l|c|c|c|c|} \hline \text{Person} & Y_0 & Y_1 & T & Y \\ \hline\hline \text{Alice} & 3 & 5 & 1 & 5 \\ \hline \text{Bob} & 3 & 5 & 0 & 3 \\ \hline \text{Cathy} & 3 & 5 & 1 & 5 \\ \hline \text{David} & 3 & 5 & 0 & 3 \\ \hline \end{array} $$

Above, the effect is 2 and $(Y_ {0},Y_{1}) \perp\!\!\!\perp T$ since the potential outcomes $(3,5)$ are the same when $T=0$ and $T=1$.

Below is an example where $(Y_ {0},Y_{1}) \not\perp\!\!\!\!\!\perp T$. Suppose the cost of treatment is $\$2.25$, and people get to choose if they are treated. Alice and Cathy opt in since $(5-2.25) = 2.75 > 0$. Bob and David do not since $(5-2.25)=2.75 < 3$. The potential outcomes are $(0,5)$ when $T=1$ and $(3,5)$ when $T=0$, so they are not orthogonal.

$$ \begin{array}{|l|c|c|c|c|} \hline \text{Person} & Y_0 & Y_1 & T & Y \\ \hline\hline \text{Alice} & 0 & 5 & 1 & 5 \\ \hline \text{Bob} & 3 & 5 & 0 & 3 \\ \hline \text{Cathy} & 0 & 5 & 1 & 5 \\ \hline \text{David} & 3 & 5 & 0 & 3 \\ \hline \end{array} $$

Here the actual effect is 3.5, and the measured effect is only 2 because you have self-selection out of treatment by folks with high $Y_0$s and in from people with low $Y_0$s. This biases the measured effect from the naive T bs C comparison down.

In both examples, $Y\not\perp\!\!\!\!\!\perp T$ since the treatment is effective.


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