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Instrumental Variable Extensions · Tutorial 563 of 972

Monotonicity Violations and Defiers

Learn how defiers are defined, why monotonicity excludes them, and how their presence can make the Wald ratio differ from the complier average treatment effect.

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What You'll Learn

  • Define compliers, always-takers, never-takers, and defiers using potential treatment uptake.
  • State the individual-level monotonicity condition and explain what it rules out.
  • Derive the reduced-form and first-stage contributions of compliers and defiers.
  • Calculate how defiers change the Wald ratio, including when the first stage is negative.
  • Explain why observed uptake rates cannot establish that there are no defiers.

Why Individual Responses to Encouragement Matter

An instrumental-variable interpretation can depend on how each person would respond to either level of an instrument. For example, an encouragement may increase treatment uptake for some people while causing others to avoid treatment. The latter response is not merely a weaker form of compliance: it is a distinct counterfactual pattern that changes what the usual Wald ratio represents.

To isolate this issue, consider a binary instrument \(Z\), a binary treatment \(D\), and an outcome \(Y\). Think of \(Z=1\) as receiving an encouragement and \(Z=0\) as not receiving it. Assume a fixed version of treatment, so the issue here is the direction of uptake response rather than differences between versions.

Definition: \(D(1)\) and \(D(0)\) are a person’s potential treatment uptake under instrument levels 1 and 0. A complier has \(D(1)=1,D(0)=0\); an always-taker has \(D(1)=1,D(0)=1\); a never-taker has \(D(1)=0,D(0)=0\); and a defier has \(D(1)=0,D(0)=1\). These four patterns exhaust the possibilities for binary \(D(1)\) and \(D(0)\).

A defier takes treatment without the encouragement but would not take it if encouraged. The label describes a comparison between two potential responses for the same person; it does not mean that the person’s observed behavior is irrational or that the instrument was assigned incorrectly.

Monotonicity and Its Consequences

The standard monotonicity assumption is an individual-level restriction on potential uptake:

$$ D(1) \ge D(0) \quad \text{for every person.} $$

Because uptake is binary, this condition permits always-takers, never-takers, and compliers, but it excludes the pattern \(D(1)=0,D(0)=1\). Monotonicity therefore rules out defiers by assumption. It is stronger than saying that encouragement increases uptake on average: a positive average increase allows some people’s uptake to move in the opposite direction.

For the usual Wald interpretation, monotonicity is considered alongside other instrumental-variable assumptions. The instrument is independent of potential uptake and potential outcomes, as under random assignment; the observed treatment satisfies \(D=D(Z)\); and the observed outcome satisfies \(Y=Y(D)\), the exclusion restriction that the instrument affects the outcome only through treatment. Consistency and no interference are also assumed. Finally, the instrument must change the treatment rate, so \(E[D\mid Z=1]-E[D\mid Z=0]\ne 0\). These conditions are stated here to clarify which part of the interpretation is at issue: defiers violate monotonicity, not necessarily random assignment or exclusion.

Key distinction: Monotonicity is not a claim that the encouragement raises uptake for everyone. It says that nobody’s uptake would fall under encouragement. People whose uptake does not change may still be always-takers or never-takers.

Let \(p_C\) and \(p_D\) be the population proportions of compliers and defiers. Under random assignment and consistency, the first stage—the difference in treatment uptake rates by instrument assignment—is

$$ E[D\mid Z=1]-E[D\mid Z=0]=p_C-p_D. $$

Always-takers and never-takers contribute no difference between the two assignment groups. Compliers contribute \(+1\) to their individual uptake difference, and defiers contribute \(-1\). Thus a positive first stage means \(p_C>p_D\), not \(p_D=0\).

How Defiers Change the Wald Ratio

Write \(Y(1)\) and \(Y(0)\) for the potential outcome under treatment and no treatment, and let \(\tau_C=E[Y(1)-Y(0)\mid C]\) and \(\tau_D=E[Y(1)-Y(0)\mid D_f]\) denote the average treatment effects among compliers and defiers. Here \(C\) and \(D_f\) denote the corresponding principal strata. Under independence, exclusion, and consistency, the reduced form is

$$ E[Y\mid Z=1]-E[Y\mid Z=0]=p_C\tau_C-p_D\tau_D. $$

A complier’s treatment changes from 0 to 1 as the instrument changes from 0 to 1, so the person contributes \(Y(1)-Y(0)\) to the outcome contrast. A defier’s treatment changes from 1 to 0, so the contribution is \(Y(0)-Y(1)\), the negative of that treatment effect. Always-takers and never-takers have unchanged treatment and contribute zero to this contrast under exclusion.

When the first stage is nonzero, dividing the reduced form by the first stage gives the Wald ratio:

$$ \frac{E[Y\mid Z=1]-E[Y\mid Z=0]} {E[D\mid Z=1]-E[D\mid Z=0]} = \frac{p_C\tau_C-p_D\tau_D}{p_C-p_D}. $$

If \(p_D=0\), this ratio equals \(\tau_C\), the complier average treatment effect. If \(p_D>0\), it is generally not that effect. When \(p_C>p_D\), it can be written as

$$ \frac{p_C}{p_C-p_D}\tau_C - \frac{p_D}{p_C-p_D}\tau_D. $$

The defier effect enters with a negative weight. Those weights do not form an ordinary nonnegative weighted average, so the ratio can fall outside the range of the two stratum-specific effects. If \(p_C=p_D\), the first stage is zero and the Wald ratio is undefined; the reduced form need not be zero. A zero denominator does not produce a complier effect.

Worked Example: A Positive First Stage with Defiers

Suppose a hypothetical randomized encouragement has four uptake types: 20% always-takers, 40% compliers, 30% never-takers, and 10% defiers. Assume the instrumental-variable conditions above hold. For illustration, suppose everyone has mean untreated outcome 10 points, and the treatment effects are 4 points among compliers and 1 point among defiers. Assume the treatment effects among always-takers and never-takers are 0 points; these groups’ effects do not enter the assignment contrasts because their uptake does not change.

Under \(Z=0\), always-takers and defiers take treatment. Under \(Z=1\), always-takers and compliers take treatment. Thus the treatment rates are \(0.20+0.10=0.30\) and \(0.20+0.40=0.60\), respectively. The first stage is \(0.60-0.30=0.30\).

The mean outcome under \(Z=0\) is \(0.20(10)+0.10(11)+0.30(10)+0.40(10)=10.1\) points. Under \(Z=1\), it is \(0.20(10)+0.40(14)+0.30(10)+0.10(10)=11.6\) points. The reduced form is \(11.6-10.1=1.5\) points.

Therefore the Wald ratio is \(1.5/0.30=5\) outcome points per treatment uptake. The stratum formula gives the same result: \(\{0.40(4)-0.10(1)\}/(0.40-0.10)=1.5/0.30=5\). This is not the complier effect of 4 points. The example satisfies the stated randomization, exclusion, and consistency conditions by construction, but it does not satisfy monotonicity because 10% are defiers.

Defiers Can Reverse or Distort the Interpretation

The sign and size of the Wald ratio depend on both the treatment effects and the balance of compliers and defiers. A negative Wald ratio does not necessarily mean that treatment harms compliers; it may result from a negative first stage combined with positive effects in both response groups.

Worked Example: A Negative First Stage

In a second hypothetical randomized study, suppose 20% of people are compliers, 35% are defiers, 25% are always-takers, and 20% are never-takers. Let the complier treatment effect be 5 outcome units and the defier treatment effect be 2 outcome units. Assume random assignment, exclusion, consistency, and no interference.

The treatment rate under \(Z=1\) is \(0.25+0.20=0.45\), because always-takers and compliers take treatment. Under \(Z=0\), the rate is \(0.25+0.35=0.60\), because always-takers and defiers take treatment. The first stage is \(0.45-0.60=-0.15\).

The reduced form is \(0.20(5)-0.35(2)=1.00-0.70=0.30\) outcome units. The Wald ratio is \(0.30/(-0.15)=-2\) outcome units per treatment uptake. It is neither the complier effect of \(5\) nor the defier effect of \(2\). The negative sign reflects the negative net uptake response as well as the outcome contrast; it cannot be read as evidence that treatment has a negative effect for compliers.

Even when the first stage is positive, defiers can make the ratio depart substantially from the complier effect. For instance, the first example’s ratio of 5 exceeds both stratum-specific effects, 4 and 1. This happens because, with \(p_C>p_D>0\), the defier effect is subtracted rather than added as a positive-weight component.

Worked Example: Uptake Rates Do Not Reveal the Defier Share

Suppose a hypothetical randomized encouragement produces treatment uptake rates of 30% for \(Z=0\) and 60% for \(Z=1\). The observed first stage is \(0.60-0.30=0.30\). These rates are compatible with a population containing no defiers: 30% compliers, 30% always-takers, and 40% never-takers.

They are also compatible with a population containing defiers: 45% compliers, 15% defiers, 15% always-takers, and 25% never-takers. In this second population, the rate under \(Z=0\) is \(0.15+0.15=0.30\), and the rate under \(Z=1\) is \(0.15+0.45=0.60\). Both populations therefore have the same uptake rates and the same first stage, since \(0.45-0.15=0.30\) in the second population.

The example satisfies the uptake-rate arithmetic under random assignment, but the rates alone do not determine the individual response types. Each person is observed under only one instrument level, so their other potential uptake is unobserved. A positive first stage is consistent with defiers being present; it does not establish monotonicity.

Justifying the Decomposition

The reduced-form identity follows by comparing potential outcomes across the two instrument assignments. Under random assignment, the mean outcome in each assignment group equals the corresponding population mean of \(Y(D(z))\). Exclusion and consistency let us write the observed outcome as \(Y(D(z))\). The difference is therefore

$$ E[Y(D(1))-Y(D(0))]. $$

Partition the population into its four uptake types. For always-takers, \(D(1)=D(0)=1\), so the individual outcome difference is \(Y(1)-Y(1)=0\). For never-takers, it is \(Y(0)-Y(0)=0\). For compliers, it is \(Y(1)-Y(0)\); averaging over their population share contributes \(p_C\tau_C\). For defiers, it is \(Y(0)-Y(1)\); averaging contributes \(-p_D\tau_D\). Adding the four contributions gives \(p_C\tau_C-p_D\tau_D\). The corresponding uptake difference is \(+1\) for each complier, \(-1\) for each defier, and 0 for the other types, yielding \(p_C-p_D\).

Common Mistake and Practical Meaning

A common mistake is to reason, “The encouragement increases treatment uptake overall, so nobody can be a defier; therefore the Wald ratio is the complier effect.” The first clause establishes only \(p_C-p_D>0\), not \(p_D=0\). For the second conclusion, one needs the individual-level monotonicity assumption, as well as the other instrumental-variable assumptions. The assumption should be defended using the mechanism and setting, not inferred from a positive average first stage alone.

For example, an encouragement might increase uptake for most people but provoke reactance in a smaller group, leading them to avoid treatment. Whether that mechanism is plausible depends on the specific intervention and population. Monotonicity may be a reasonable working assumption in some settings, but it is not a consequence of random assignment and cannot generally be verified by comparing observed uptake rates.

Key takeaway: Defiers have \(D(1)=0,D(0)=1\), and monotonicity \(D(1)\ge D(0)\) rules them out. If defiers are present, the reduced form and first stage contain opposing defier contributions, so the Wald ratio is generally not the complier average treatment effect.

Check Your Understanding

Use potential uptake and the Wald decomposition to answer these questions.

  1. What are \(D(1)\) and \(D(0)\) for a defier, and why does this type violate monotonicity?
  2. If \(p_C=0.50\) and \(p_D=0.10\), what is the first stage under random assignment? Does it establish that there are no defiers?
  3. Suppose \(p_C=0.30\), \(p_D=0.10\), \(\tau_C=6\), and \(\tau_D=2\). Calculate the reduced form and Wald ratio. Is the ratio the complier effect?
  4. Why can always-takers and never-takers have nonzero treatment effects without contributing directly to the reduced form across instrument assignments?
  5. Two studies have the same uptake rates under each instrument level. Why might their defier shares still differ?
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