Why LATE Is an Average
An instrumental-variable estimate is often summarized by one number. That number need not be the effect for every person whose treatment changes with the instrument: individual effects can differ. The important question is what the single number averages, and how the people contributing to that average are weighted.
Under the instrumental-variable assumptions, the Wald estimand is the local average treatment effect (LATE): the average individual treatment effect in the population of compliers. This tutorial makes that averaging statement explicit, derives it, and shows how different treatment-effect patterns among compliers change the average.
The effect \(\tau_i\) is individual-specific, but the conditional expectation \(E[\tau_i\mid D_i(1)>D_i(0)]\) averages those effects over the complier population. Each complier contributes according to their share of that population. A subgroup containing more compliers therefore receives more weight than a smaller subgroup; the average is not generally the simple mean of subgroup effects.
Assumptions and the Averaging Result
Write \(Z\) for a binary instrument and \(D\) for treatment. The Wald estimand is the effect of the instrument on the outcome divided by its effect on treatment. To interpret that ratio as an average of individual treatment effects, use the standard binary-instrument assumptions:
- Independence: \(Z\) is independent of the potential outcomes and potential treatment statuses, \((Y(1),Y(0),D(1),D(0))\).
- Exclusion: the outcome depends on the instrument only through treatment. In potential-outcome notation, \(Y(z,d)=Y(d)\).
- Monotonicity: \(D(1)\geq D(0)\) for every individual, so there are no defiers.
- Relevance: the instrument changes treatment uptake, so \(E[D\mid Z=1]-E[D\mid Z=0]\ne 0\).
Consistency links observed outcomes and treatment to the potential outcomes under the values actually received. As usual, the interpretation also presumes that one individual's instrument or treatment does not change another individual's outcome. Under monotonicity, individuals are never-takers, always-takers, or compliers. Only compliers change treatment when the instrument changes.
Here is the algebra that establishes the result. Independence allows assignment groups to identify averages of the corresponding potential outcomes and treatment statuses. By exclusion and consistency, the outcome observed under instrument value \(z\) is \(Y(D(z))\). Thus the numerator can be written as \(E[Y(D(1))-Y(D(0))]\), and the denominator as \(E[D(1)-D(0)]\).
For the second equality, if \(D(1)=D(0)\), the two potential outcomes in the first line are the same and the contribution is zero. If \(D(1)=1\) and \(D(0)=0\), the contribution is \(Y(1)-Y(0)=\tau\). Monotonicity rules out the remaining case, \(D(1)=0,D(0)=1\). The denominator follows by the same type-by-type argument:
The denominator is positive by relevance and monotonicity. Dividing the numerator by it gives the conditional mean among compliers, since \(E[\tau\,\mathbf{1}\{C\}]=P(C)E[\tau\mid C]\), where \(C\) denotes complier status. This completes the justification: the numerator adds complier effects in proportion to their population shares, while the first stage is the total share of compliers.
Worked Examples of Complier Averaging
Worked Example: Weighting Two Complier Subgroups
Consider a hypothetical complier population with 100 people. In a low-severity subgroup, 75 people each have a treatment effect of 1 outcome unit. In a high-severity subgroup, 25 people each have a treatment effect of 7 outcome units. Find the complier average treatment effect.
The subgroup shares among compliers are \(75/100=0.75\) and \(25/100=0.25\). The weighted average is
The same calculation using totals gives \(\{75(1)+25(7)\}/100=(75+175)/100=2.50\) outcome units. The high-severity subgroup has a larger effect, but it is only one-quarter of the complier population. The result is an average across all 100 compliers, not the effect for each of them.
Worked Example: Unequal Subgroup Sizes and Effects of Opposite Signs
In another hypothetical complier population, 12 people have an individual treatment effect of \(-2\) outcome units, while 8 people have an effect of \(5\) outcome units. Calculate the average and explain why simply averaging the two subgroup effects is inappropriate.
There are \(12+8=20\) compliers. The first subgroup contributes \(12(-2)=-24\) total effect units and the second contributes \(8(5)=40\). Therefore,
The simple mean of the two subgroup effects, \(\{-2+5\}/2=1.50\), would give each subgroup equal weight despite their different sizes. The relevant weights are \(12/20=0.60\) and \(8/20=0.40\), so the correct weighted calculation is \(0.60(-2)+0.40(5)=0.800\) outcome units. Negative and positive effects partly offset; the positive average does not mean every complier benefits.
Worked Example: Checking the Wald Ratio Against Complier Effects
Consider a hypothetical population with four principal groups. Always-takers make up 20%, compliers with an effect of 2 outcome units make up 30%, compliers with an effect of 6 units make up 20%, and never-takers make up 30%. Suppose the always-takers have \(Y(0)=10,Y(1)=12\); the first complier subgroup has \(Y(0)=20,Y(1)=22\); the second has \(Y(0)=30,Y(1)=36\); and the never-takers have \(Y(0)=8,Y(1)=11\). The instrument is independent of these potential values, exclusion holds, and there are no defiers. Calculate both the complier average and the Wald ratio.
The complier share is \(0.30+0.20=0.50\). Their average treatment effect is
To check the Wald ratio directly, under \(Z=1\), always-takers and compliers receive treatment, whereas never-takers do not. The mean outcome is
Under \(Z=0\), always-takers still receive treatment, compliers do not, and never-takers do not:
The reduced-form difference is \(18.60-16.80=1.80\) outcome units. Treatment uptake is \(0.70\) under \(Z=1\) and \(0.20\) under \(Z=0\), so the first stage is \(0.70-0.20=0.50\). The Wald ratio is therefore
The ratio matches the weighted average of the two complier effects. Always-takers and never-takers affect the observed outcome means, but their treatment status does not change with the instrument; under the assumptions, their contributions cancel in the reduced-form difference.
What the Average Does—and Does Not—Say
A LATE is an average over a particular population: people whose treatment status is changed by the specified instrument. If their effects are heterogeneous, the average depends on how those effects are distributed among those compliers. The weighted calculations above make this composition visible. A subgroup can have a large effect yet contribute modestly if it represents a small share of compliers.
The average does not reveal each complier's effect, nor does it identify which specific individuals are compliers. Even when the assumptions identify the population average, they do not make the individual effects equal. The estimate also does not, by itself, describe the average effect for people who would always take treatment, people who would never take it, or the full population.
Common Mistake
A frequent error is to say, “The instrument identifies the treatment effect for compliers,” as if the ratio were an effect shared by every complier. The derivation identifies \(E[\tau\mid C]\), not each \(\tau_i\). With heterogeneous effects, one complier might benefit greatly, another slightly, and another be harmed, while the reported LATE is their population average. Another error is to average subgroup effects without weighting by subgroup shares; that calculation is valid only when the subgroups have equal shares among compliers.
Check Your Understanding
Use the definitions and calculations above to answer these questions.
- In a complier population, 40% have an effect of 3 units and 60% have an effect of 1 unit. What is the complier average treatment effect?
- Why does the Wald numerator equal the complier share multiplied by the complier average treatment effect under the stated assumptions?
- A subgroup with 10 compliers has an effect of 4 units, and another subgroup with 30 compliers has an effect of 2 units. What is the weighted average effect, and why is the unweighted subgroup mean different?
- Does a LATE of 2 outcome units imply that every complier's treatment effect is 2 units? Explain.
- Which assumption rules out a group whose treatment status changes from treated under \(Z=0\) to untreated under \(Z=1\), and why does that matter for the averaging result?