# In the superpopulation framework of causal inference, what is the necessity of assuming the outcomes, treatment and covariates are jointly iid?

In the textbook "Causal Inference for Statistics" by Rubin and Imbens, the following argument is made on pg. 39:

"In part of this text we view our sample of size N as a random sample from an infinite super-population. In that case we employ slightly different formulations of the restric- tions on the assignment mechanism. Sampling from the super-population generates a joint sampling distribution on the quadruple of unit-level variables (Yi(0), Yi(1), Wi, Xi), i = 1, . . . , N. More explicitly, we assume the (Yi(0), Yi(1), Wi, Xi) are independently and identically distributed draws from a distribution indexed by a global parameter."

I am wondering why the bolded portion is needed to be assumed. In the context of observational studies what happens if 1) we have dependence, and 2) if the tuple is not jointly identically distributed.

Is this a general assumption that can be relaxed? Where do we really need it? Specifically:

1) What happens if we have dependence?

2) What happens if they are NOT identically distributed?

• Here's some intuition, although probably not enough for a full answer: If the observations are not independent, the SUTVA assumption would definitely break down. It would be no longer possible to write $Y_i(0), Y_i(1)$ because the potential outcomes would now depend not only on whether unit $i$ receives the treatment or not but also on whether other individuals receive the treatment. On the other hand, being not identically distributed would imply that each of the observations come from a different data-generating process. So, you'd have no longer a single "average causal effect," so to say. – baruuum Mar 3 at 21:34
• As you say, $\tau = E[Y_i(1)] - E[Y_i(0)]$ is the assumption, i.e., if $\{Y_i(1), Y_i(0)\}_{i=1}^n$ come from the same distribution (that is, identically distributed or generated by the same DGP), then $E[Y_1(1)] = E[Y_2(1)] = ...$, so we can estimate $E[Y_i(1)]$ from $n^{-1}\sum_{i=1}^n y_i\mathbb I(z_i = 1)$, by the law of large numbers (if the treatment assignment is ignorable). And the same applies to $E[Y_i(0)]$. If they come from different distributions, we might have $E[Y_i(1)] \ne E[Y_{i'}(1)], i\ne i'$. – baruuum Mar 4 at 19:24
• So, our estimated "treatment effect" would be some kind of weighted average across the $E[Y_i(1)-Y_i(0)]$s. At least how I think about the problem. I'll leave these as comments, since I'm not an expert in causal inference; I hope there'll be other people who can give you better explanations. – baruuum Mar 4 at 19:26
• A simple example: Consider the potential outcomes $Y_i(Z_i)$ and $Y_j(Z_j)$, where $Z_i$ and $Z_j$ are the treatment assignment variables for unit $i$ and $j$, respectively. If the potential outcomes are not independent, then you'd have to write $Y_i(Z_i,Z_j)$ and $Y_j(Z_j, Z_i)$. As the notation suggests, the response of unit $i$ now depends not only on whether unit $i$ receives the treatment, $Z_i$, but also whether unit $j$ receives the treatment, $Z_j$. That you can write $Y_i(Z_i)$ and $Y_j(Z_j)$ already incorporates the assumption that the treatment of unit $j$ does not affect $Y_i(Z_i)$ – baruuum Mar 5 at 1:56
• Possible duplicate of Statistics and causal inference? – Michael Chernick Mar 12 at 1:36

Your question is not straightforward to answer because violations of i.i.d can happen in many different ways. In short, I would say that regarding your main doubt, the answers would be the following: i.i.d is invoked for estimation from samples, and violations of i.i.d can be dealt with, but this depends on the model and the query of interest.

In causal inference, when you obtain an identification result, this is will be a functional of a joint distribution $$P$$. For instance, assuming conditional ignorability $$Y_{x} \perp \!\!\! \perp X|W$$, the distribution of the potential outcome is identified with the functional $$P(y_{x})= \sum_{w}P(y|x, w)P(w)$$. Note this is an equality of population quantities.

If you had the exact population values of the distribution $$P$$, you would be done. But usually you have a sample of size $$n$$, which can be summarized by the empirical distribution $$P_{n}$$. When you assume to have an i.i.d sample of $$P(y,x,w)$$, you know you can estimate $$P$$ consistently given large enough data, for instance, by appealing to the Glivenko–Cantelli theorem or simpler parametric results if you assume a parametric model. That's why i.i.d sample is usually assumed.

Is this a general assumption that can be relaxed? What happens if we have dependence? What happens if they are NOT identically distributed?

Yes this assumption can be relaxed in some ways, but usually at the expense of another (parametric) assumption. As a simple example, suppose you have time series data and a linear structural model for $$Y$$,

$$Y_{t} = \tau D_{t} + \beta{Y_{t-1}} + \epsilon_{t}$$

With $$|\beta| < 1$$ and $$\epsilon_{t}$$ Gaussian and $$E(\epsilon_{t}|D_{t}, Y_{t-1}, ..., Y_{1})=0$$. Consider you only have one observation per $$t$$. Here each of the $$Y_{t}$$ are both dependent and not identically distributed. Yet, $$\tau$$ can be consistently estimated via OLS.