Worst-Case Treatment-Effect Bounds Without Exclusion

Why Bound the Effect When Exclusion Is Not Assumed?

A randomized encouragement can change treatment uptake, but it may also affect the outcome through another route. When that direct effect is possible, the usual instrumental-variable interpretation cannot attribute the entire outcome difference between assignment groups to treatment. Randomization still makes assignment groups comparable, but it does not reveal the outcomes people would have had under a treatment level they did not receive.

For a binary outcome, the fact that outcomes lie between 0 and 1 gives useful worst-case bounds. These bounds use the observed joint probabilities of assignment, uptake, and outcome. They do not require treatment uptake to be randomized, the instrument to satisfy exclusion, or uptake to follow a monotonicity rule.

Estimand and Assumptions

Let $Z\in\{0,1\}$ be randomized instrument assignment, $D\in\{0,1\}$ treatment uptake, and $Y\in\{0,1\}$ the outcome. Because exclusion is not assumed, write $Y(d,z)$ for the outcome that would occur under treatment level $d$ and instrument assignment $z$. Also write $D(z)$ for the treatment that would be taken under assignment $z$.

For a fixed assignment level $z$, define the population average treatment effect $\tau_z=E[Y(1,z)-Y(0,z)]$. The overall target in this tutorial averages those effects using the instrument's randomization probabilities: $\tau=\sum_{z=0}^1 P(Z=z)\tau_z$. It is the average effect of changing treatment while holding assignment fixed, averaged over the assignment distribution. Defining the target this way matters: when exclusion is not assumed, there is no single outcome $Y(d)$ that necessarily applies regardless of $z$.

Assumptions: (1) $Z,D,Y$ are binary; (2) assignment is randomized, so $Z$ is independent of the collection of potential uptake and outcome values $\{D(0),D(1),Y(d,z):d,z\in\{0,1\}\}$; (3) consistency holds, so the observed uptake is $D=D(Z)$ and the observed outcome is $Y=Y(D,Z)$; and (4) there is no interference between individuals. Both assignment levels must have positive probability to estimate bounds for both $\tau_0$ and $\tau_1$. No exclusion restriction is imposed.

The randomization assumption lets observed probabilities conditional on $Z=z$ describe the corresponding potential uptake and outcome under assignment $z$ in the population. It does not make $D(z)$ independent of the potential outcomes. Within each assignment group, people who take treatment may differ from those who do not.

Worst-Case Bounds Within an Assignment Group

For each $z$, let $q_{dy,z}=P(D=d,Y=y\mid Z=z)$. These four observed probabilities sum to 1. Define the uptake probabilities $p_{1,z}=P(D=1\mid Z=z)=q_{10,z}+q_{11,z}$ and $p_{0,z}=P(D=0\mid Z=z)=q_{00,z}+q_{01,z}$.

Among people with $D=1$ under assignment $z$, the observed outcome is $Y(1,z)$; their $Y(0,z)$ is unobserved. Among people with $D=0$, the observed outcome is $Y(0,z)$; their $Y(1,z)$ is unobserved. Since $Y$ is binary, each missing potential outcome is bounded between 0 and 1. Thus the population means of the two potential outcomes satisfy:

q_{11,z}\leq E[Y(1,z)]\leq q_{11,z}+p_{0,z}, \qquad q_{01,z}\leq E[Y(0,z)]\leq q_{01,z}+p_{1,z}.

The lower bound for $E[Y(1,z)]$ counts observed successes among those treated and assigns zero to the unobserved potential outcomes of those untreated. Its upper bound assigns one to those missing outcomes. The corresponding reasoning for $E[Y(0,z)]$ uses observed successes among the untreated and bounds the missing outcomes among the treated.

Worst-case bounds: Subtracting the largest possible mean of $Y(0,z)$ from the smallest possible mean of $Y(1,z)$ gives the lower bound; subtracting the smallest possible mean of $Y(0,z)$ from the largest possible mean of $Y(1,z)$ gives the upper bound: $$ q_{11,z}-q_{01,z}-p_{1,z} \ \leq\ \tau_z\ \leq\ q_{11,z}+p_{0,z}-q_{01,z}. $$

These bounds are worst-case because they make no assumptions about the missing potential outcomes beyond their binary range. Each endpoint can be attained by assigning the missing potential outcomes the values that minimize or maximize the contrast. Consequently, the bounds cannot be narrowed using these observed probabilities alone and the stated assumptions.

Worked Examples

Worked Example: Bounding the Effect at One Assignment Level

Suppose hypothetical data under $Z=0$ yield the following conditional probabilities: $q_{11,0}=0.30$, $q_{10,0}=0.20$, $q_{01,0}=0.20$, and $q_{00,0}=0.30$. They sum to $1.00$, as required.

The treatment uptake probabilities are $p_{1,0}=q_{10,0}+q_{11,0}=0.20+0.30=0.50$ and $p_{0,0}=q_{00,0}+q_{01,0}=0.30+0.20=0.50$. The lower endpoint is $q_{11,0}-q_{01,0}-p_{1,0}=0.30-0.20-0.50=-0.40$. The upper endpoint is $q_{11,0}+p_{0,0}-q_{01,0}=0.30+0.50-0.20=0.60$.

Thus $-0.40\leq\tau_0\leq0.60$. Because $Y$ is binary, these effects are differences in outcome probabilities, or percentage-point differences. The data and assumptions allow effects ranging from a 40-percentage-point decrease to a 60-percentage-point increase under $Z=0$; they do not determine even the sign of the effect.

Worked Example: Averaging Bounds Across Randomized Assignments

Suppose a hypothetical randomized study assigns $Z=0$ and $Z=1$ with probability $0.50$ each. Under $Z=0$, use the probabilities from the preceding example, which give $-0.40\leq\tau_0\leq0.60$. Under $Z=1$, suppose $q_{11,1}=0.50$, $q_{10,1}=0.10$, $q_{01,1}=0.15$, and $q_{00,1}=0.25$. These probabilities also sum to 1.

For $Z=1$, $p_{1,1}=0.10+0.50=0.60$ and $p_{0,1}=0.25+0.15=0.40$. The lower endpoint is $0.50-0.15-0.60=-0.25$, and the upper endpoint is $0.50+0.40-0.15=0.75$. Therefore, $-0.25\leq\tau_1\leq0.75$.

The overall target averages the assignment-specific effects with their assignment probabilities. Its lower endpoint is $0.50(-0.40)+0.50(-0.25)=-0.325$; its upper endpoint is $0.50(0.60)+0.50(0.75)=0.675$. Hence $-0.325\leq\tau\leq0.675$, a range from a 32.5-percentage-point decrease to a 67.5-percentage-point increase in the outcome probability. The bounds are wide because uptake is not randomized within assignment groups and no restriction links treatment and outcome beyond the stated assumptions.

Worked Example: Showing That the Endpoints Can Be Attained

Return to the $Z=0$ probabilities in the first example. To attain the lower endpoint, leave all observed outcomes unchanged, set $Y(1,0)=0$ for the untreated group, and set $Y(0,0)=1$ for the treated group. The mean $E[Y(1,0)]$ is then $0.30$, the observed successes among the treated. The mean $E[Y(0,0)]$ is $0.20+0.50=0.70$, the observed successes among the untreated plus all treated people. The contrast is $0.30-0.70=-0.40$.

To attain the upper endpoint, instead set $Y(1,0)=1$ for the untreated group and $Y(0,0)=0$ for the treated group. The means become $E[Y(1,0)]=0.30+0.50=0.80$ and $E[Y(0,0)]=0.20$. The contrast is $0.80-0.20=0.60$. Both completions preserve every observed probability. This demonstrates why, without further assumptions, the observed data cannot exclude either endpoint.

Why the Randomized Instrument Does Not Identify the Effect

In the second example, the observed outcome probability is $0.50$ under $Z=0$ and $0.75$ under $Z=1$. The assignment-outcome difference is $0.75-0.50=0.25$, or 25 percentage points. Treatment uptake also changes, from $0.50$ to $0.60$, a 10-percentage-point difference. But the outcome difference may reflect both the change in treatment uptake and a direct effect of assignment on outcomes. Without exclusion, the outcome difference cannot be attributed solely to treatment or divided by the uptake difference to identify a treatment effect.

Randomization supports comparisons across $Z$, but it does not reveal either person's unobserved outcome under the other treatment level while holding $Z$ fixed. The worst-case bounds make this limitation explicit rather than replacing it with an unsupported point estimate.

Derivation and Scope of the Bounds

For a fixed $z$, consistency means that among those with $D=1$, the observed outcome equals $Y(1,z)$. Their contribution to the population mean $E[Y(1,z)]$ is $P(D=1,Y=1\mid Z=z)=q_{11,z}$. Among those with $D=0$, $Y(1,z)$ is missing and lies in $[0,1]$; this group has probability $p_{0,z}$. Adding the smallest and largest possible missing contributions gives $q_{11,z}\leq E[Y(1,z)]\leq q_{11,z}+p_{0,z}$. The same argument for $Y(0,z)$ gives $q_{01,z}\leq E[Y(0,z)]\leq q_{01,z}+p_{1,z}$.

For any two quantities in these intervals, the smallest possible difference uses the smallest first quantity and the largest second quantity. The largest possible difference uses the largest first quantity and the smallest second quantity. Therefore,

\begin{aligned} \tau_z &=E[Y(1,z)]-E[Y(0,z)]\\ &\geq q_{11,z}-(q_{01,z}+p_{1,z}) =q_{11,z}-q_{01,z}-p_{1,z},\\ \tau_z &\leq (q_{11,z}+p_{0,z})-q_{01,z} =q_{11,z}+p_{0,z}-q_{01,z}. \end{aligned}

The endpoint constructions in the worked example show attainability for one assignment level. The same choices can be made separately within each assignment level; averaging the resulting contrasts with the nonnegative weights $P(Z=z)$ gives the stated bounds for $\tau$. This argument requires no assumption that outcomes under different assignment levels are equal.

Common Mistakes

Treating random assignment as randomized treatment. Randomization applies to $Z$, not necessarily to $D$. The people who take treatment can differ from those who do not, even within an assignment group.
Using the assignment-outcome difference as the treatment effect. Without exclusion, assignment may affect the outcome directly. The assignment-outcome contrast is not, by itself, a treatment effect.
Forgetting to average assignment-specific effects using the assignment probabilities. The overall target is weighted by $P(Z=z)$; it is an unweighted average only when the assignment probabilities are equal.
Interpreting a bound as a confidence interval. These are identification bounds: they describe the range allowed by the population probabilities and assumptions. They are not, by themselves, a statement about sampling uncertainty.

Key takeaway: With a binary outcome, randomized assignment and observed uptake-outcome probabilities yield worst-case treatment-effect bounds even when exclusion is not assumed. The bounds are informative only to the extent that the observed data and assumptions restrict the missing potential outcomes.

Check Your Understanding

Use the definitions and formulas above to answer the following questions.

In a fixed assignment group, suppose $q_{11}=0.25$, $q_{10}=0.25$, $q_{01}=0.10$, and $q_{00}=0.40$. Calculate the lower and upper bounds for the assignment-specific treatment effect.
Why does randomization of $Z$ not imply that treatment uptake $D$ is independent of potential outcomes?
For a binary outcome, explain why the unobserved contribution to $E[Y(1,z)]$ among untreated individuals lies between zero and $P(D=0\mid Z=z)$.
If $P(Z=0)=0.70$ and $P(Z=1)=0.30$, how should assignment-specific lower bounds be combined to obtain the lower bound for the overall target?
What additional causal assumption would be needed before interpreting an assignment-outcome contrast as arising only through treatment uptake?