Short Memo About Inverse Probability Weighting

Introduction

I want to briefly have on paper the logic behind using inverse probability-weighted estimators of the ATT (average treatment effect on the treated) in a causal inference context.

Setup

• Let \(Y(1)\) and \(Y(0)\) denote the potential outcomes under treatment and no treatment, respectively.

• The ATT is defined as:

$$ ATT=\mathbb{E}[Y(1)−Y(0)∣T=1] $$

where \(T \in \{0,1\}\) indicates treatment status.

We assume that conditional on covariates \(X\), the treatment is independent of \(Y(0)\), an assumption known as unconfoundedness.

The IPW estimator uses weights to adjust for the selection bias in treatment assignment. The weights are derived from the propensity score, $$e(X)=P(T=1∣X)$$ which is the probability of receiving treatment given covariates \(X\).

Idea

The first part of the \(ATT\), \(\mathbb{E}[Y(1) | T=1]\) is identified. The idea is to estimate \(\mathbb{E}[Y(0)|T=1]\) using the observed outcomes from the untreated group, for which \(T=0\). The IPW Estimator for \(\mathbb{E}[Y(0)|T=1]\) is:

$$ \frac{\mathbb{E}[Y \times 1(T=0) \times \frac{e(X)}{1−e(X)}]}{\mathbb{E}[1(T=0) \times \frac{e(X)}{1−e(X)}]}. $$

where \(\mathbb{E}(.)\) are now in-sample operations.

Unbiasedness Proof

I evaluate this estimator and show that the expected value equals \(\mathbb{E}[Y(0)|T=1]\).

1. For \(T=0\), the observed outcome \(Y\) equals \(Y(0)\). Thus, the numerator of the IPW estimator becomes:

$$ \mathbb{E}\left[Y \times 1(T=0)\times \frac{e(X)}{1−e(X)}\right]=\mathbb{E}\left[Y(0)\times 1(T=0)\times \frac{e(X)}{1−e(X)}\right]. $$

Using the law of iterated expectations, and making use of independence between \(Y(0)\) and \(T\) conditional on \(X\), we condition on \(X\):

$$ \mathbb{E} \left[Y(0) \times 1(T=0) \times \frac{e(X)}{1−e(X)}\right]=\mathbb{E}[\mathbb{E}[Y(0)∣X] \times \mathbb{E}[1(T=0) \times \frac{e(X)}{1−e(X)}∣X]]. $$

The indicator \(1(T=0)\) ensures \(P(T=0∣X)=1−e(X)\). Substituting the following:

\(\mathbb{E}[1(T=0) \times \frac{e(X)}{1−e(X)}∣X]=e(X)\)

into the numerator makes it that the numerator simplifies to:

$$ \mathbb{E}[Y(0) \cdot e(X)]. $$

2. Using the law of iterated expectation again (while conditioning again on \(X\)), the denominator of the IPW estimator normalizes the weights:

$$ \mathbb{E}[1(T=0)\times \frac{e(X)}{1−e(X)}]=E[e(X)]. $$

3. Combining Terms now makes it that the IPW estimator becomes:

$$ \frac{\mathbb{E}[Y(0) \times e(X)]}{E[e(X)]} $$

Now note that the denominator is \(P(T=1)\) and the nominator is the event that \(P(T=1) \cap Y(0)\). Thus, by the definition of conditional probability*, this equals \(\mathbb{E}[Y(0)|T=1]\).

*: For an explicit derivation of this, see below

Conclusion

This derivation showed that the inverse probability-weighted estimator is an unbiased estimator for the counterfactual outcome \(Y(0)\) for the treated group. The derivation is fundamental and pops up a lot in the treatment evaluation literature. I have loosely based it on Imbens and Rubin (2015).

Appendix: Explicit Deriviation

Here is an explicit derivation why the estimator \( \frac{\mathbb{E}[Y(0) \times e(X)]}{E[e(X)]} \) recovers \( \mathbb{E}[Y(0) \mid T = 1] \).

1. Using the Law of Total Probability:

The expectation \( \mathbb{E}[Y(0) \mid T = 1] \) can be expressed using the law of total probability:

$$ \mathbb{E}[Y(0) \mid T = 1] = \int \mathbb{E}[Y(0) \mid X] \cdot P(X \mid T = 1) dX. $$

Here:\( \mathbb{E}[Y(0) \mid X] \) is the expected potential outcome for untreated individuals at a given \( X \).

\( P(X \mid T = 1) \) is the distribution of \( X \) in the treated group.

2. Reweighting the Untreated Group to Match \( P(X \mid T = 1) \):

We can rewrite \( P(X \mid T = 1) \) in terms of \( P(X) \) and \( e(X) \) using Bayes’ rule:

$$ P(X \mid T = 1) = \frac{P(T = 1 \mid X) \cdot P(X)}{P(T = 1)} = \frac{e(X) \cdot P(X)}{\mathbb{E}[e(X)]}. $$

Substituting this into the formula for \( \mathbb{E}[Y(0) \mid T = 1] \), we have:

$$ \mathbb{E}[Y(0) \mid T = 1] = \int \mathbb{E}[Y(0) \mid X] \cdot \frac{e(X) \cdot P(X)}{\mathbb{E}[e(X)]} dX. $$

3. Expressing as a Weighted Expectation:

Recognizing that the integral over \( P(X) \) represents the total expectation over the population distribution of \( X \), we have:

$$ \mathbb{E}[Y(0) \mid T = 1] = \frac{\int \mathbb{E}[Y(0) \mid X] \cdot e(X) \cdot P(X) dX}{\mathbb{E}[e(X)]}. $$

The numerator,\( \int \mathbb{E}[Y(0) \mid X] \cdot e(X) \cdot P(X) \, dX \), is equivalent to \( \mathbb{E}[Y(0) \cdot e(X)] \), the expectation of \( Y(0) \cdot e(X) \) over the population. Hence:

$$ \mathbb{E}[Y(0) \mid T = 1] = \frac{\mathbb{E}[Y(0) \cdot e(X)]}{\mathbb{E}[e(X)]}. $$