Sensitivity to Exclusion-Restriction Violations

Why Relax the Exclusion Restriction?

An instrumental variable is useful when it changes treatment take-up but is otherwise unrelated to the outcome. The exclusion restriction formalizes the “otherwise” part: holding treatment fixed, changing the instrument does not change the outcome. In some applications, that restriction is difficult to defend. A reminder, eligibility letter, or encouragement might affect an outcome through attention, information, or other channels in addition to its effect on treatment.

Instead of treating exclusion as all-or-nothing, we can specify how much direct influence the instrument may have and ask which treatment effects remain compatible with the data. This tutorial gives a precise version for a binary instrument, binary treatment, and binary outcome. It uses a latent-type linear program to calculate bounds; it does not assume that the instrument’s direct effect is zero.

Model and Violation Allowance

Let $Z\in\{0,1\}$ be the instrument, $D\in\{0,1\}$ the treatment, and $Y\in\{0,1\}$ the outcome. Write $D(z)$ for the treatment a unit would take under instrument value $z$, and $Y(d,z)$ for the outcome under treatment $d$ and instrument value $z$. The estimand here is the average treatment effect when the instrument is fixed at its reference value $z=0$:

\theta_0=E\!\left[Y(1,0)-Y(0,0)\right].

This is a population average, measured in outcome-probability units because $Y$ is binary. It is not necessarily the average treatment effect at $Z=1$: with exclusion violations, treatment effects may differ across instrument levels.

Definition: For binary outcomes, the direct-effect violation allowance $\delta\in[0,1]$ places the following restrictions on the fraction of units whose outcome would change when the instrument changes but treatment is held fixed: $$ E\!\left[|Y(d,1)-Y(d,0)|\right]\leq \delta,\qquad d\in\{0,1\}. $$ Because the potential outcomes are binary, the absolute difference is 1 exactly when the two potential outcomes disagree, and 0 otherwise. Thus, $\delta=0.10$ allows at most 10% of the population to have an instrument-induced outcome change at either fixed treatment level.

This is a bound on the average absolute direct change, not merely on the difference between the two average outcomes. For example, positive and negative individual changes could cancel in an average difference while many units’ outcomes still change. The allowance used here rules out that cancellation as a way of hiding a large share of direct effects.

The other assumptions are: (1) the instrument is randomly assigned, so $Z$ is independent of all potential treatments and outcomes; (2) consistency holds, so the observed treatment and outcome equal their potential values under the realized instrument and treatment; (3) there is no interference between units; and (4) treatment is monotone in the instrument, $D(1)\geq D(0)$. The last assumption rules out defiers, units whose treatment would move from 1 to 0 when the instrument changes from 0 to 1. Both instrument levels must occur with positive probability. These assumptions and the violation restrictions define the model for the bounds below.

Sharp Bounds as a Linear Program

A latent type records the potential treatment pair and all four potential outcomes:

t=(d_0,d_1,y_{00},y_{01},y_{10},y_{11}),

where $d_z=D(z)$ and $y_{dz}=Y(d,z)$. Each component is binary, and monotonicity permits only the treatment pairs $(0,0)$, $(0,1)$, and $(1,1)$. Let $\pi_t$ be the population share of type $t$. The shares are nonnegative and sum to one. Random assignment means this same population distribution of types applies in both instrument groups.

For each instrument value $z$, the data reveal the conditional probabilities of the four possible treatment-outcome pairs. The model requires the latent shares to reproduce those probabilities:

P(D=d,Y=y\mid Z=z) = \sum_t \pi_t\, \mathbf{1}\{d_z=d,\ y_{d,z}=y\}, \qquad d,y,z\in\{0,1\}.

Here $\mathbf{1}\{\cdot\}$ equals 1 when its condition holds and 0 otherwise. The violation restrictions become two more linear constraints:

\sum_t\pi_t |y_{d1}-y_{d0}|\leq\delta, \qquad d\in\{0,1\}.

To obtain the lower endpoint of the identified set, minimize the target across all type shares satisfying these constraints. To obtain the upper endpoint, maximize it:

\underline{\theta}_0(\delta) = \min_{\pi}\sum_t\pi_t(y_{10}-y_{00}), \qquad \overline{\theta}_0(\delta) = \max_{\pi}\sum_t\pi_t(y_{10}-y_{00}).

These are finite linear programs: the type list is finite, the data-matching equations are linear in $\pi_t$, the violation restrictions are linear in $\pi_t$, and the objective is linear. The resulting interval is sharp for the stated model: every value in it is attained by at least one compatible distribution of latent types, and no value outside it is compatible.

Justification of sharpness and widening: The feasible type shares form a bounded convex set, since they are nonnegative, sum to one, and satisfy linear equalities and inequalities. The target is a linear function of those shares, so its attainable values form a closed interval. If $\delta_1\leq\delta_2$, every type distribution allowed at $\delta_1$ is also allowed at $\delta_2$. Therefore the identified set at $\delta_1$ is contained in the set at $\delta_2$; its lower endpoint cannot increase and its upper endpoint cannot decrease. At $\delta=0$, the restrictions impose exact exclusion. At $\delta=1$, they impose no additional restriction, since binary outcomes can differ by at most 1.

Worked Examples

Worked Example: All Units Comply with the Instrument

Suppose hypothetical population data show $D=0$ for everyone when $Z=0$, and $D=1$ for everyone when $Z=1$. Among the $Z=0$ group, 30% have $Y=1$; among the $Z=1$ group, 55% have $Y=1$. Find the sharp bounds for $\theta_0$ when $\delta=0.10$.

The observed treatment pattern implies $D(0)=0$ and $D(1)=1$ for every unit. Thus the observed outcome rate at $Z=0$ identifies $E[Y(0,0)]=0.30$, while the rate at $Z=1$ identifies $E[Y(1,1)]=0.55$. Let $q=E[Y(1,0)]$, which is not observed. Since $Y(1,0)$ and $Y(1,1)$ are binary, the violation restriction for $d=1$ implies

|q-0.55| = \left|E[Y(1,0)-Y(1,1)]\right| \leq E[|Y(1,0)-Y(1,1)|] \leq 0.10.

Therefore $0.45\leq q\leq0.65$. Every value in this range is feasible: couple the two binary potential outcomes so that the fraction that disagree is $|q-0.55|$. For $d=0$, choose $Y(0,1)=Y(0,0)$, satisfying that violation restriction without changing the observed data. The treatment effect is $q-0.30$, so

\theta_0\in[0.45-0.30,\ 0.65-0.30]=[0.150,\ 0.350].

Under these assumptions and this allowance, the identified average treatment effect at $Z=0$ ranges from 0.150 to 0.350, or 15 to 35 percentage points. These are bounds, not a point estimate.

Worked Example: When Does the Sign Become Uncertain?

Use the same hypothetical data, but vary $\delta$. The restriction on $Y(1,0)$ gives $q\in[\max(0,0.55-\delta),\min(1,0.55+\delta)]$. For $\delta\leq0.45$, this yields

\theta_0\in[0.55-\delta-0.30,\ 0.55+\delta-0.30] =[0.25-\delta,\ 0.25+\delta].

At $\delta=0.20$, the interval is $[0.050,0.450]$, so every compatible treatment effect is positive. At $\delta=0.25$, it is $[0,0.500]$: zero is compatible with the data and assumptions. For $\delta=0.30$, it is $[-0.050,0.550]$, which includes both negative and positive effects. The threshold follows by solving $0.25-\delta=0$, giving $\delta=0.25$. In this example, allowing a direct outcome change for 25% of the population is enough for the identified set to include no average treatment effect.

This is a sensitivity calculation for the identified set, not a claim about sampling uncertainty. Its conclusion depends on the chosen definition of $\delta$, as well as on random assignment, monotonicity, and the other model restrictions.

Worked Example: No Treatment Take-Up Response

Consider a different hypothetical population. Everyone has $D=0$ under both instrument values. The outcome prevalence is 0.40 in both instrument groups. What can be learned about $\theta_0$, even when $\delta=0$?

The data identify $E[Y(0,0)]=0.40$, but never reveal an outcome under treatment. Set $Y(0,1)=Y(0,0)$ to satisfy exact exclusion at $d=0$. For the unobserved treated outcomes, set $Y(1,1)=Y(1,0)$; this satisfies exact exclusion at $d=1$ for any chosen distribution of $Y(1,0)$. Thus $E[Y(1,0)]$ can be any number from 0 to 1 without changing the observed data.

Consequently, even with $\delta=0$,

\theta_0=E[Y(1,0)]-0.40\in[0-0.40,\ 1-0.40]=[-0.400,\ 0.600].

The interval remains this wide for larger $\delta$, because the treatment outcome is unobserved and the instrument does not shift treatment. A small exclusion allowance cannot compensate for a lack of treatment variation.

How to Read the Bounds

The allowance $\delta$ has a direct interpretation in this binary-outcome model: it limits the population share whose outcome changes with the instrument while treatment is fixed. It is a sensitivity parameter, not something the observed treatment-outcome table generally identifies by itself. Choosing a range of plausible allowances makes explicit how much the substantive conclusion relies on exclusion.

For a given $\delta$, the sharp interval includes all and only average effects compatible with the observed conditional distributions and the maintained assumptions. If the whole interval is above zero, the model implies a positive average effect at $Z=0$; if it crosses zero, the sign is not determined. Increasing $\delta$ cannot make the interval narrower, although an endpoint may stay unchanged for some range of allowances.

A practical calculation uses the observed conditional cell probabilities in the data-matching equations, enumerates the allowed latent types, and solves the two linear programs. If no feasible type shares exist, then the observed probabilities conflict with at least one maintained assumption at that value of $\delta$. Sampling uncertainty is a separate issue: bounds computed from estimated cell probabilities are not, by themselves, confidence intervals.

Common Mistake

Wrong reasoning: “The instrument’s average outcome difference is small, so its direct effects must be small.” Individual direct effects can offset one another, leaving a small average difference even when many units’ outcomes change.
Correction: Here $\delta$ bounds $E[|Y(d,1)-Y(d,0)|]$, which for binary outcomes is the share of units with discordant potential outcomes at fixed treatment. Check which direct-effect quantity a sensitivity assumption actually limits before interpreting its allowance.
Wrong reasoning: “A larger allowance gives more information because it makes the model more realistic.”
Correction: A larger allowance relaxes a restriction, so the compatible set can only stay the same or widen. It may improve credibility, but it cannot sharpen identification in this model.

Key takeaway: Sensitivity to exclusion violations is made operational by specifying a direct-effect allowance and optimizing the target over all latent distributions consistent with the data and assumptions. The resulting sharp interval shows exactly which treatment effects remain possible at each allowance.

Check Your Understanding

Answer using the definitions and assumptions in this tutorial.

For binary outcomes, what population quantity does $\delta$ bound in this model?
Why does the identified set at $\delta=0.10$ have to be contained in the set at $\delta=0.20$?
In the all-complier example, calculate the bound endpoints when $\delta=0.05$.
Why can a small exclusion allowance still leave wide treatment-effect bounds when the instrument does not change treatment take-up?
What does it mean for the linear-program bounds to be sharp?