I hope someone can clarify this for me:

When I am estimating treatment effects with a regression model, I can use interactions between the treatment indicator $W_i$ and covariates $X_i$ to model potential heterogeneity in treatment effects. While researching I came across this quote:

"Second, we can allow for a model with a full set of interactions: $Y_{i}^{\mathrm{obs}}=\alpha+\tau \cdot W_{i}+\beta^{\prime} {X}_{i}+\gamma^{\prime} {X}_{i} \cdot W_{i}+\varepsilon_{i}$

In general the least squares estimates based on these regression functions are not unbiased for the average treatment effects over the randomization distribution given the finite population. There is one exception. If the covariates are all indicators and they partition the population, and we estimate the model with a full set of interactions, Eq. (13), then the least squares estimate of $\tau$ is unbiased for the average treatment effect. " from "The Econometrics of Randomized Experiments, Athey and Imbens 2017, p.97"

Unfortunately, I don't understand how Athey and Imbens get to this conclusion. I know that I can't take $\hat{\tau}$ as the estimated average treatment effect when I have continuous covariates. The estimated ATE would be $\hat{\tau} + \hat{\gamma}*\mu_x$. Is that what they mean with not unbiased? Why would this bias go away with finite population ?

  • $\begingroup$ Why do you say this? I know that I can't take 𝜏̂ as the estimated average treatment effect when I have continuous covariates. $\endgroup$ Aug 14, 2023 at 17:56

1 Answer 1


The idea is simple: they are describing a saturated model. It helps to define the covariate levels: all possible combos of $X$ that you want to/can estimate; this is like a sample space for the design matrix. A saturated model has degrees of freedom equal to the number of unique covariate levels. For instance, if covariates of sex and smoking status are measured, there are 4 covariate levels: male nonsmoker, male smoker, female nonsmoker, and female smoker. Thus the saturated model has 4 degrees of freedom. One way to specify such a model with common statistical languages is to adjust for sex, smoking, and their interaction which, together with the intercept, is a 4 degree of freedom model. You can show that the 4 degree of freedom models all give the same predictions for any combination of covariates, hence they are equivalent in a broad sense.

When a model is saturated, it just predicts the average response for each covariate level rather than any complex weighted average. If I modeled systolic blood pressure, the effect of intercept + male sex + male smoker interaction would give me the average SBP for male smokers.

Consequently, if we add a randomized treatment assignment, there are now 8 levels (for each modify by "treated" and "untreated"). Treatment effect, therefore, can be estimated using quite general and sophisticated ways. You can, for instance, predict the stratum-specific treatment effect by, e.g. subtracting the predicted response from treated male smokers from untreated male smokers. You can also average up the stratum specific responses according to population frequencies (a process called marginal standardization) to estimate average treatment effects.

This is based on the idea that the average is unbiased, but ignores the fact that estimates can be biased by design as well. That is a separate discussion.

  • $\begingroup$ Thank you for your response. How do you see that they are talking about a saturated model? Does full set of interaction = saturated? $\endgroup$
    – Blo4d
    Jul 24, 2019 at 16:50
  • $\begingroup$ @Blo4d yes. In fact, I think their presentation would have been simpler by predefining model saturation and referring to it here, I can see no other possible meaning. We have to be careful about "full set of interactions". With 3 factors, you need all 2-way but also all 3-way interactions to achieve model saturation, and higher orders for more factors. There are analogues of model saturation for continuous covariates, so while not achievable we can get really close by thinking carefully about splines. $\endgroup$
    – AdamO
    Jul 24, 2019 at 17:31

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