Sharp IV Bounds for Binary Potential Outcomes

Why Use Latent Response Types?

With a binary instrument, treatment, and outcome, each person has potential treatment choices under both instrument values and potential outcomes under both treatment values. The observed data reveal only the treatment and outcome corresponding to the instrument value actually assigned. The unobserved combinations leave room for different causal effects, even when the instrument is randomly assigned and has no direct effect on the outcome.

A linear program makes this missing information explicit. It assigns probabilities to possible combinations of potential responses, requires those probabilities to reproduce the observed data, and minimizes or maximizes the average treatment effect over all compatible assignments. The endpoints are useful only if they are sharp: each must be attainable by at least one data-generating process satisfying the assumptions.

Model and assumptions: 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 that would occur under instrument value \(z\), and \(Y(d)\) for the outcome under treatment value \(d\). Assume consistency, no interference between individuals, random assignment \(Z\perp (D(0),D(1),Y(0),Y(1))\), and exclusion: the outcome under instrument value \(z\) and treatment \(d\) is \(Y(d)\), with no additional dependence on \(z\). The target is the population average treatment effect \(E[Y(1)-Y(0)]\). No monotonicity restriction on treatment responses is imposed.

Encoding the Observed Distribution

There are \(2^4=16\) possible response types. A type specifies the tuple \((D(0),D(1),Y(0),Y(1))\). For example, \((0,1,0,1)\) describes a person who receives treatment only when \(Z=1\), has outcome 0 under control, and outcome 1 under treatment. Let \(q_{d_0d_1y_0y_1}\) be the population probability of that type. These probabilities are nonnegative and sum to one.

The observed data identify the conditional cell probabilities \(r_{zdy}=P(D=d,Y=y\mid Z=z)\), provided both instrument values occur with positive probability. For any given type and instrument value \(z\), the observed treatment is \(d_z\) and, by exclusion and consistency, the observed outcome is \(y_{d_z}\). Thus the cell probability is the sum of the probabilities of all types that produce that cell.

r_{zdy} = \sum_{d_0,d_1,y_0,y_1} q_{d_0d_1y_0y_1} \mathbf{1}\{d_z=d,\;y_d=y\}, \qquad z,d,y\in\{0,1\}.

Here \(\mathbf{1}\{\cdot\}\) equals 1 when the stated conditions hold and 0 otherwise. The eight cell equations (four for each instrument value) are linear in the 16 latent probabilities. Some equations are redundant because each set of four cells sums to one, but retaining them is convenient. The average treatment effect is also linear in those probabilities:

E[Y(1)-Y(0)] = \sum_{d_0,d_1,y_0,y_1} q_{d_0d_1y_0y_1}(y_1-y_0).

Definition: The sharp lower bound is the minimum of this linear objective over all nonnegative response-type probabilities that satisfy the observed-cell equations and sum to one. The sharp upper bound is the corresponding maximum. If no feasible probabilities exist, the observed distribution is incompatible with the assumptions.

This formulation allows all 16 response types unless the data or an additional, explicitly imposed restriction rule some out. In particular, equality of treatment rates across instrument groups does not by itself establish that every individual has the same treatment response under both instrument values. That is a statement about the unobserved joint response \((D(0),D(1))\), not just its marginal rates.

Why the Bounds Are Sharp

The linear program is not merely a way to obtain conservative limits. Its feasible set describes exactly the response-type distributions that reproduce the observed law under the stated assumptions. This gives a direct justification of sharpness.

Every admissible model gives a feasible point.
The model assigns each person one of the 16 potential-response types. Their population probabilities are nonnegative, sum to one, and reproduce each observed cell by consistency and exclusion. Random assignment makes those type probabilities the same in both instrument groups.

Every feasible point gives an admissible model.
Take any nonnegative probabilities satisfying the cell equations. Draw a response type according to those probabilities, assign \(Z\) independently with the observed assignment probability, and set \(D=D(Z)\) and \(Y=Y(D)\). This construction satisfies random assignment and exclusion and reproduces the observed conditional cells.

The objective is the model’s average treatment effect.
Under the constructed model, the average treatment effect is exactly the linear objective evaluated at the chosen probabilities. Therefore, a feasible minimizer and maximizer generate admissible models attaining the two endpoints.

The feasible set is a closed and bounded subset of the 16-dimensional probability simplex, so if it is nonempty, both extrema are attained. This completes the justification: no compatible model has an effect outside the interval, and models exist that attain each endpoint. The interval is an identified set, not a confidence interval. Sampling uncertainty in estimated cell probabilities is a separate issue.

Worked Example: Treatment Equals the Instrument

Suppose a hypothetical population has the following conditional outcome rates: when \(Z=0\), everyone has \(D=0\) and 20% have \(Y=1\); when \(Z=1\), everyone has \(D=1\) and 60% have \(Y=1\). The instrument is randomly assigned, and both instrument values occur.

Under \(Z=0\), the observed treatment rate is zero. Since random assignment gives the same distribution of response types in both instrument groups, \(D(0)=0\) for the whole population, apart from probability-zero types. Under \(Z=1\), everyone has \(D=1\), so \(D(1)=1\) for the whole population. Consequently, the first group reveals \(Y(0)\) for the population and the second reveals \(Y(1)\) for the population.

The observed rates give \(E[Y(0)]=0.20\) and \(E[Y(1)]=0.60\). Substituting into the estimand,

$$ E[Y(1)-Y(0)]=0.60-0.20=0.40. $$

The sharp lower and upper bounds both equal 0.40. In this hypothetical population, the average treatment effect is point-identified as an increase of 0.40 in the probability of a positive outcome. The joint association between \(Y(0)\) and \(Y(1)\) remains unknown, but it does not affect their difference in means.

A Partially Identified Example

A less complete instrument response can leave some potential outcomes unobserved for an entire subgroup. The next example shows how the cell equations identify the observed subgroup quantities while leaving a precise range for the population effect.

Worked Example: Half Never-Takers and Half Compliers

Consider the following hypothetical conditional cell probabilities. In the \(Z=0\) group, everyone has \(D=0\), with \(P(Y=1\mid Z=0)=0.40\). In the \(Z=1\) group, half have \(D=1\) and half have \(D=0\). Among those with \(D=1\), \(P(Y=1\mid D=1,Z=1)=0.60\); among those with \(D=0\), \(P(Y=1\mid D=0,Z=1)=0.20\). Assignment is random, and exclusion holds.

Because nobody is treated when \(Z=0\), random assignment rules out response types with \(D(0)=1\). In the \(Z=1\) group, half are treated, so the response types must consist of half with \((D(0),D(1))=(0,1)\) and half with \((0,0)\). Call these groups compliers and never-takers, respectively. This conclusion follows from the observed zero treatment rate at \(Z=0\) and the 0.50 treatment rate at \(Z=1\); no additional monotonicity assumption is needed for this particular data pattern.

The \(Z=1,D=0\) cell consists of never-takers, so it identifies their control-outcome mean as \(E[Y(0)\mid N]=0.20\). The \(Z=1,D=1\) cell consists of compliers and identifies \(E[Y(1)\mid C]=0.60\). The overall \(Z=0\) outcome rate is a mixture of control outcomes for both groups:

0.40 = 0.50E[Y(0)\mid C] + 0.50E[Y(0)\mid N].

Substituting \(E[Y(0)\mid N]=0.20\) gives

E[Y(0)\mid C] = \frac{0.40-(0.50)(0.20)}{0.50} = \frac{0.30}{0.50} = 0.60.

The complier average effect is therefore \(0.60-0.60=0\). For never-takers, \(Y(1)\) is never observed, so let \(\theta=E[Y(1)\mid N]\). Since the outcome is binary, \(0\leq\theta\leq1\). The population effect is

E[Y(1)-Y(0)] = 0.50(0.60-0.60)+0.50(\theta-0.20) = 0.50\theta-0.10.

As \(\theta\) ranges from 0 to 1, the effect ranges from \(-0.10\) to \(0.40\). Both endpoints are attainable: set never-takers’ \(Y(1)\) to zero for the lower endpoint and to one for the upper endpoint, while keeping the observed outcomes and the identified control-outcome means unchanged. The sharp ATE bounds are therefore \([-0.10,\,0.40]\), in outcome-probability units.

Using the Linear Program When Outcomes Are Missing

The previous example could be simplified by reading off the response groups from the zero treatment cell. In less structured cases, the 16-type linear program provides a systematic method. The following case illustrates how the same constraints give bounds even when the instrument does not change treatment for anyone.

Worked Example: No Treatment Under Either Instrument Value

Suppose a hypothetical population has \(D=0\) for everyone in both instrument groups, and \(P(Y=1\mid Z=0)=P(Y=1\mid Z=1)=0.30\). Assume random assignment and exclusion.

The observed treatment cells imply \(D(0)=D(1)=0\) for the population. Thus every feasible response type has \(d_0=d_1=0\). The observed outcome rate identifies \(E[Y(0)]=0.30\), but the data provide no observation of \(Y(1)\). Write \(\theta=E[Y(1)]\), where binary outcomes imply \(0\leq\theta\leq1\). The LP objective reduces to

E[Y(1)-Y(0)]=\theta-0.30.

The minimum is \(0-0.30=-0.30\), and the maximum is \(1-0.30=0.70\). To see sharpness directly, for the lower endpoint assign \(Y(1)=0\) to everyone; for the upper endpoint assign \(Y(1)=1\) to everyone. In both cases assign \(Y(0)=1\) to 30% of the population and \(Y(0)=0\) to the remaining 70%. These response-type distributions reproduce the observed data, so the sharp bounds are \([-0.30,\,0.70]\).

Interpretation and a Common Pitfall

Each endpoint answers an optimization question over a set of data-generating processes, not an uncertainty question about a parameter drawn at random. In the second example, every compatible model gives a population average effect between \(-0.10\) and \(0.40\), and compatible models attain both values. This does not mean there is a 95% probability that the effect lies in that range; no confidence level is attached to the identification bounds.

A common mistake is to treat random assignment and exclusion as if they revealed both potential outcomes for every person. Random assignment makes the response-type distribution comparable across instrument groups, and exclusion removes a direct effect of the instrument. Neither assumption observes the missing treatment outcome for people who never receive treatment. In the no-treatment example, \(Y(1)\) remains unrestricted even though assignment is random and exclusion holds. The correction is to encode the unobserved responses explicitly and optimize over all assignments consistent with the observed cells.

The resulting bounds depend on the assumptions used. Here, no monotonicity restriction was added, and exclusion was imposed exactly. Adding restrictions can change the feasible set and therefore the bounds; relaxing exclusion changes the model itself. The numerical endpoints should always be reported with the assumptions and the outcome scale in view.

Key takeaway: Under random assignment and exclusion, binary IV data impose linear constraints on the probabilities of 16 potential-response types. Minimizing and maximizing the average treatment effect over those constraints yields sharp bounds: every compatible model lies within them, and each endpoint is attained by a compatible model.

Check Your Understanding

Answer each question using the latent-response formulation.

How many response types are there when \(D(0)\), \(D(1)\), \(Y(0)\), and \(Y(1)\) are all binary?
In the cell equation, which potential outcome determines the observed outcome for a type with \(D(z)=1\)?
Why does random assignment allow the same response-type probabilities to be used in both instrument groups?
If everyone has \(D=0\) under both instrument values and the observed positive-outcome rate is 0.25, what are the sharp ATE bounds under these assumptions?
What additional fact must be shown to call an optimization interval sharp, rather than merely a valid outer bound?