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Horvitz–Thompson exposure mean in a two-unit design

Example

The two units form one group, which is assigned one of the four binary vectors $(0, 0), (0, 1), (1, 0), (1, 1)$ , each with probability $1 / 4$ . For both units, take the exposure map $D_{i} = (Z_{1}, Z_{2})$ . Under exposure $d = (0, 1)$ , let $Y_{1} (d) = 4$ and $Y_{2} (d) = 10$ ; set both units' potential outcomes to zero at the other three exposure labels. The map is sufficient because the exposure is the full within-group assignment vector.

Verification

Given: The finite assignment law and exposure model above.

[F1] Exposure mappings and exposure consistency: exposure consistency gives a fixed $Y_{i} (d)$ on the assignments mapped to $d$ , and $π_{i} (d)$ is the sum of the design weights of those assignments.

[F2] Horvitz–Thompson exposure-specific mean is design-unbiased: the exposure-specific target is the finite average of the fixed $Y_{i} (d)$ values and the estimator divides each observed exposed outcome by its positive exposure probability.

1.1givenF1F2algebra

Only assignment $(0, 1)$ maps either unit to $d$ , so $π_{1} (d) = π_{2} (d) = 1 / 4$ . The target is $μ (d) = (4 + 10) / 2 = 7$ .

2.1step 1.1F2algebra∎

Under assignment $(0, 1)$ the estimator equals $\frac{1}{2} (4 / (1 / 4) + 10 / (1 / 4)) = 28$ . Under each of the other three assignments it is zero. Its design expectation is therefore $(28 + 0 + 0 + 0) / 4 = 7 = μ (d)$ , verifying the unbiasedness identity directly.

Depends on

Used by

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Sources

Peter M. Aronow and Cyrus Samii, Estimating Average Causal Effects Under General Interference, §4 and Lemma 4.1 (standard reference, not scraped)