When Stratum-Specific LATEs Are Aggregated

Why Aggregation Depends on Complier Mass

A conditional LATE describes the average treatment effect among compliers in a particular covariate stratum. It does not, by itself, say how much that stratum contributes to the effect among all compliers in the population. The relevant contribution depends on how many compliers the stratum contains, not just on its size or on its conditional LATE.

This tutorial derives that aggregation rule and distinguishes it from a simple average of stratum-specific effects. Throughout, $S$ denotes a pre-instrument stratum, $Z$ the instrument, $D$ treatment receipt, and $Y$ the outcome. Write $D(z)$ for treatment receipt under instrument value $z$, and $Y(d)$ for the outcome under treatment value $d$. The four compliance types and the conditional LATE are as defined in the earlier material on those subjects.

Definition: For stratum $k$, let $p_k=P(S=k)$, and let $\pi_k=P(D=1\mid Z=1,S=k)-P(D=1\mid Z=0,S=k)$ be the stratum-specific first stage. Under conditional instrument independence and monotonicity, $\pi_k=P(C\mid S=k)$, where $C$ denotes being a complier. When $\pi_k>0$, define $\tau_k=E[Y(1)-Y(0)\mid C,S=k]$. The stratum’s population complier mass is $p_k\pi_k$.

Assumptions and the Aggregation Result

The aggregation identity relies on the assumptions that identify the conditional LATEs. Within each stratum, the instrument must be independent of potential treatment receipt and potential outcomes; the exclusion restriction must rule out a direct effect of $Z$ on $Y$; and monotonicity must rule out defiers, so $D(1)\geq D(0)$. Consistency links observed treatment and outcomes to their potential values. Both instrument values must occur with positive probability in any stratum where a conditional contrast is estimated. The instrument direction is coded so that the first stage is nonnegative.

These assumptions imply that the conditional first stage equals the share of compliers in that stratum. To see this, conditional instrument independence allows the observed treatment contrast to be written as a contrast in potential treatment receipt. Under monotonicity, the individual difference $D(1)-D(0)$ is one for compliers and zero for always-takers and never-takers. Therefore:

\pi_k =E[D(1)-D(0)\mid S=k] =P(C\mid S=k).

For the outcome contrast, exclusion means that changing $Z$ affects the outcome only through treatment receipt. Under monotonicity, always-takers and never-takers do not change treatment receipt when $Z$ changes, so their expected outcome contrast is zero. Compliers do change receipt, and their mean outcome contrast is $\tau_k$. Thus, when $\pi_k>0$, the conditional reduced-form contrast is $\pi_k\tau_k$. This is the conditional Wald identification result expressed as a numerator and denominator.

E[Y\mid Z=1,S=k]-E[Y\mid Z=0,S=k] =\pi_k\tau_k.

Now condition on being a complier in the population. A complier is in stratum $k$ with probability proportional to the stratum’s population share times its within-stratum complier share. Let $q=P(C)$ be the total population complier share. The aggregation result applies only if $q>0$.

Theorem: If $q=\sum_k p_k\pi_k>0$, then the population average treatment effect among compliers is $$ \tau_C=E[Y(1)-Y(0)\mid C] =\sum_{k:\pi_k>0}w_k\tau_k, \qquad w_k=P(S=k\mid C) =\frac{p_k\pi_k}{\sum_j p_j\pi_j}. $$ The weights are nonnegative and sum to one. A stratum with $\pi_k=0$ has no complier mass and contributes no term. If $q=0$, there are no population compliers, so $\tau_C$ and these normalized weights are undefined.

The formula follows by partitioning the event $C$ according to stratum. In stratum $k$, the joint probability of being a complier and being in that stratum is $P(S=k,C)=p_kP(C\mid S=k)=p_k\pi_k$. Dividing by $P(C)=q$ gives $P(S=k\mid C)$. The conditional expectation of the treatment effect given $C$ is then the sum of the stratum-specific conditional means, each multiplied by this conditional stratum probability.

E[Y(1)-Y(0)\mid C] =\sum_{k:\pi_k>0} E[Y(1)-Y(0)\mid C,S=k]P(S=k\mid C).

Worked Calculations

Worked Example: Calculating Complier-Mass Weights

Consider a hypothetical population with two strata. Stratum 1 contains 60% of the population and has a first stage of 0.25. Stratum 2 contains 40% and has a first stage of 0.50. Suppose the conditional LATE is 2 outcome points in stratum 1 and 5 outcome points in stratum 2. Assume the conditional IV assumptions above hold in both strata.

Calculate each complier mass. The masses are $p_1\pi_1=0.60(0.25)=0.15$ and $p_2\pi_2=0.40(0.50)=0.20$. Thus the total complier share is $q=0.15+0.20=0.35$, which is positive.

Normalize the masses. The weights are $w_1=0.15/0.35=0.4286$ and $w_2=0.20/0.35=0.5714$. They sum to 1.0000.

Aggregate the effects. $ \tau_C=0.4286(2)+0.5714(5)=0.8572+2.857=3.714 $ outcome points (using unrounded weights gives $3.714$). The population complier effect is closer to 5 than to 2 because stratum 2 supplies the larger share of all compliers.

Worked Example: Why the Unweighted Average Can Be Wrong

Suppose a hypothetical population has $p_1=0.80$ and $p_2=0.20$. The stratum first stages are $\pi_1=0.10$ and $\pi_2=0.60$, and the respective conditional LATEs are 1 and 4 outcome points. Assume the conditional IV assumptions hold in both strata.

The complier masses are $0.80(0.10)=0.08$ and $0.20(0.60)=0.12$, so $q=0.20>0$. The normalized weights are $w_1=0.08/0.20=0.40$ and $w_2=0.12/0.20=0.60$. Hence $ \tau_C=0.40(1)+0.60(4)=0.40+2.40=2.80 $ outcome points.

The ordinary mean of the two conditional effects is $(1+4)/2=2.50$ points, not 2.80. Equal weighting gives each stratum half of the influence even though stratum 2 contains 60% of the population’s compliers. Equal complier masses guarantee that the ordinary mean equals the weighted mean; they are not necessary. For example, if both conditional LATEs were 3 points, their weighted mean and their ordinary mean would both be 3, whatever the weights.

Worked Example: A Stratum With No Compliers

In another hypothetical population, let $p_1=0.70$, $p_2=0.30$, $\pi_1=0.20$, and $\pi_2=0$. Suppose the conditional LATE in stratum 1 is 3 outcome points. Assume conditional independence and monotonicity, so the first stages equal the within-stratum complier shares.

The masses are $p_1\pi_1=0.70(0.20)=0.14$ and $p_2\pi_2=0.30(0)=0$. Since $q=0.14>0$, the overall effect is defined. The first stratum’s weight is $0.14/0.14=1$; the second has no compliers and receives no weight. Therefore $\tau_C=1(3)=3$ outcome points. There is no $\tau_2$ to estimate as a complier effect in stratum 2 because $P(C\mid S=2)=0$.

If instead both first stages were zero, then $q=0$. The formula would have a zero denominator, and it would be incorrect to report an overall complier effect or normalized weights. The assumptions do not guarantee that any compliers exist.

Connection to Standardized Wald Contrasts

The same aggregation can be seen through stratum-specific outcome and treatment contrasts. Define the standardized outcome contrast by averaging the within-stratum reduced forms using the population stratum shares, and define the standardized first stage the same way. If $q>0$, their ratio equals the complier-weighted effect.

\frac{\sum_k p_k\{E[Y\mid Z=1,S=k]-E[Y\mid Z=0,S=k]\}} {\sum_k p_k\{E[D\mid Z=1,S=k]-E[D\mid Z=0,S=k]\}} = \frac{\sum_k p_k\pi_k\tau_k}{\sum_k p_k\pi_k} =\tau_C.

Terms with $\pi_k=0$ contribute zero to both sums; a $\tau_k$ need not be defined in such a stratum. The ratio is a standardized contrast: it combines within-stratum contrasts using a common distribution of $S$. A raw comparison of all $Z=1$ units with all $Z=0$ units need not equal this expression if the instrument groups have different stratum composition. Conditional independence within strata alone does not ensure those raw groups have the same composition.

Diagnostics, Alternatives, and Limitations

In an applied analysis, the weights depend on both the population stratum shares and the first stages. Check that the estimated stratum shares are appropriate for the target population, that first stages have the intended direction, and that the calculations use a common set of strata. A tiny first stage means a small complier mass when the stratum share is fixed, but it does not necessarily mean a small normalized weight: the weight depends on that mass relative to the total mass elsewhere. A small first stage can also make the conditional Wald estimate imprecise because the ratio divides by a small denominator.

Sparse strata, near-zero first stages, or unstable estimated weights are warnings that the numerical decomposition may be unreliable. A zero first stage under the identifying assumptions implies zero complier share, whereas a noisy estimated first stage near zero does not establish that the population share is exactly zero. Estimation and uncertainty for these quantities require appropriate methods; treating estimated weights and conditional effects as known can understate uncertainty.

The main alternative to the complier-weighted aggregate is the unweighted mean of stratum-specific LATEs. That answers a different question unless the weights happen to give the same result. Equal complier masses guarantee equality, and equal conditional LATEs also guarantee it; other coincidences are possible. Another approach is to form standardized reduced-form and first-stage contrasts and take their ratio, which is algebraically equivalent under the stated assumptions. Neither approach removes the need for conditional instrument validity, exclusion, or monotonicity. Nor does the aggregate automatically describe every person in the population: it is an effect among the population’s compliers.

Common Mistake

Wrong reasoning: “There are two strata, so the overall LATE is the average of their two LATEs.” This implicitly assigns equal weight to each stratum, regardless of how many compliers each contains. Correction: calculate $p_k\pi_k$ for each stratum, divide each mass by the positive total $q=\sum_k p_k\pi_k$, and use those normalized weights. If $q=0$, stop: the population complier effect is undefined. Also, do not infer that a stratum’s normalized weight must be small just because its first stage is small; compare its complier mass with the total mass.

Check Your Understanding

Try each question, then check the solutions below.

Two strata have shares $p_1=0.50$, $p_2=0.50$, first stages $\pi_1=0.20$, $\pi_2=0.40$, and conditional LATEs 2 and 5 points. Find the complier weights and overall complier effect.
When is an ordinary average of two conditional LATEs guaranteed to equal the complier-weighted average? Is that condition necessary?
If one stratum has zero first stage and another has positive complier mass, what is the zero-first-stage stratum’s contribution to the overall effect?
If every stratum has zero first stage under conditional independence and monotonicity, can the overall complier effect be calculated?

Solutions. (1) The masses are $0.50(0.20)=0.10$ and $0.50(0.40)=0.20$, totaling $0.30$. The weights are $0.10/0.30=1/3$ and $0.20/0.30=2/3$. The effect is $(1/3)(2)+(2/3)(5)=4$ points. (2) Equal complier masses guarantee equality, but this is not necessary: equal conditional LATEs also make both averages identical for any weights. (3) It contributes zero complier mass and receives weight zero; its conditional LATE among compliers is undefined if its complier share is zero. (4) No. The total complier share is zero, so the population conditional effect and normalized weights are undefined.