3
$\begingroup$

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.
$\endgroup$
1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.