Consider an outcome $Y$, treatment $D$, and set of covariates $X$. The outcome is real-valued, the treatment has support $\{0,1\}$, and the set of covariates is a collection of binary variables with support $\{0,1\}$. Furthermore, let $Y(d)$ denote potential outcome of $Y$ if $D=d$ for $d\in\{0,1\}$, and note that $Y=Y(1)D+Y(0)(1-D)$. Now, assume the following.

Assumption 1 (Mean Independence). $E(Y(d)\mid X,D)=E(Y(d)\mid X)$ for $d\in\{0,1\}$.

In this context, how strong is the following assumption?

Assumption 2. $$E(Y(d)\mid X)=\alpha_d+\beta_d\cdot p(X)$$ for $d\in\{0,1\}$, where $\{(\alpha_1,\beta_1),(\alpha_0,\beta_0)\}$ are pairs of real numbers (i.e., scalars) and $p(X)=P(D=1\mid X)$ is the propensity score.

In other words, how strong is the assumption that the conditional means of the potential outcomes are linear in the propensity score? I am wondering if there are some easy-to-understand assumptions that imply Assumption 2. This is an important assumption that is used in the causal inference literature for adjustment on covariates (see, e.g., corollary 4.3 in [1]). Or is the assumption guaranteed already? For example, can it be derived from the simple fact that $E(Y(d)\mid X)$ and $p(X)$ are linear in $X$ because $X$ is a collection of binary variables (meaning that the model is saturated)?

  1. Rosenbaum, P. R., & Rubin, D. B. (1983). The central role of the propensity score in observational studies for causal effects. Biometrika, 70(1), 41-55.

2 Answers 2


This is a fairly strong assumption and one not likely to be true in practice. That assumption being true justifies the use of a linear regression of $Y$ on $D$ and $p(X)$ for estimating the causal effect of $D$ on $Y$. In practice, because this is assumption is rarely true, this method is not a valid way to reduce the bias due to confounding by $X$. Matching, weighting, and stratification on the propensity score do not require this assumption and therefore are better used for estimating the effect of $D$. If using regression adjustment on the propensity score, one should use a flexible model of $Y$ on $D$ and $p(X)$, such as a spline model, which is more likely to capture the true relationship.

Here's a simple example where the assumption is false:

Let $$E[Y(0)|X] = E[Y(1)|X] = E[Y|X] = .2X_1 + .3X_2$$ (i.e., no treatment effect). Also let $$p(X)=.3X_1+.2X_2$$ There are no scalars $\alpha$ and $\beta$ such that $$E[Y|X] = .2X_1 + .3X_2 = \alpha + \beta(.3X_1+.2X_2) = \alpha + \beta \cdot p(X)$$So, the assumption fails.


The expectation of $Y$ given $X$ is a function of $X.$ Hence if $X$ takes value in $\{0,1\}$ then there always exists $\alpha$ and $\beta$ such that $E[Y|X]=\alpha+\beta X$ for the mere reason that you can alway draw a line between two given points. Hence if $X$ is univariate, both $p(X)$ and $E[Y|X]$ are linear functions fo $X.$

Now if $X$ is multivariate each components belonging to $\{0,1\}$ again things get a little more messy. Consider $X$ is a 2-dimensional vector. You can write $$ E[Y|X]=E[Y|X_1=0,X_2=0]+ X_1E[Y|X_1=1,X_2=0]+ X_2E[Y|X_1=1,X_2=0]+ X_1X_2E[Y|X_1=1,X_2=1]. $$ While the first three terms of the above RHS are linear function of $X$, the last one is not.

In the complete general case, your "linear" representation needs cross effects -which can be expressed as products of components of $X$- to hold true.


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