Why Remove Monotonicity?
The familiar binary-instrument argument for a complier effect often assumes that the instrument cannot make anyone less likely to receive treatment. That restriction rules out defiers: individuals who would take treatment when the instrument is off and not take it when the instrument is on. Without monotonicity, observed changes in treatment can combine compliers and defiers. The usual Wald ratio therefore need not identify a complier average treatment effect.
Removing monotonicity does not make causal analysis impossible. If the instrument is independent of the relevant potential variables and satisfies exclusion, the observed instrument-group distributions still restrict the possible average treatment effect (ATE). They may identify it exactly, or they may imply only a range. This tutorial derives that range for binary treatment and outcome.
The assumptions used here are instrument independence, \(Z\mathrel{\perp}(D(0),D(1),Y(0),Y(1))\); exclusion, so the instrument affects the outcome only through treatment; and consistency, so observed treatment and outcome are \(D=D(Z)\) and \(Y=Y(D)\). Both instrument values must occur with positive probability. No monotonicity assumption is made: either \(D(1)\geq D(0)\) for everyone or its reverse is not required.
Compliance Types and What Treatment Rates Reveal
The pair \((D(0),D(1))\) defines four principal strata. Always-takers have pair \((1,1)\), never-takers \((0,0)\), compliers \((0,1)\), and defiers \((1,0)\). Let their population probabilities be \(a,n,c,f\), respectively. These probabilities are nonnegative and sum to one.
To obtain this range, substitute \(f=q_0-a\), \(c=q_1-a\), and \(n=1-q_0-q_1+a\). Nonnegativity gives \(a\leq q_0\), \(a\leq q_1\), and \(a\geq q_0+q_1-1\); also \(a\geq0\). These restrictions yield the displayed interval. They are restrictions on the latent type shares, not assumptions about which types exist. In particular, when the interval has positive width, the observed treatment rates alone do not separate compliers from defiers.
Worked Example: A Range of Possible Compliance Shares
Suppose a hypothetical study observes treatment rates \(q_0=0.35\) when the instrument is off and \(q_1=0.55\) when it is on. Find the possible type shares implied by these rates.
The always-taker share obeys \(a\in[\max(0,0.35+0.55-1),\min(0.35,0.55)]=[0,0.35]\). For any choice of \(a\) in this interval, the remaining shares are \(f=0.35-a\), \(c=0.55-a\), and \(n=0.10+a\).
For example, \(a=0.10\) gives \(f=0.25\), \(c=0.45\), and \(n=0.20\). These shares sum to one and reproduce the treatment rates: \(a+f=0.35\) and \(a+c=0.55\). Another permissible value, \(a=0\), gives no always-takers and a different mixture of compliers and defiers. The treatment rates do not choose between these mixtures.
Constructing Sharp ATE Bounds
Treatment rates alone are not enough to use the outcome information. The full observed data are the conditional cell probabilities \(r_{zdy}=P(D=d,Y=y\mid Z=z)\), for each \(z,d,y\in\{0,1\}\). To describe all latent configurations consistent with these data, let \(\pi_{d_0,d_1,y_0,y_1}=P(D(0)=d_0,D(1)=d_1,Y(0)=y_0,Y(1)=y_1)\). There are 16 such probabilities, one for each combination of the four binary potential variables.
This is a finite linear program: its unknowns are the 16 latent type probabilities, its constraints are linear, and its objective is the ATE. The indicator in the cell equation selects people whose treatment under instrument value \(z\) is \(d\) and whose outcome under that treatment is \(y\). The equation follows directly from consistency and exclusion. Instrument independence makes the latent probabilities the same in both instrument groups, so the same \(\pi\) must reproduce the cells for \(Z=0\) and \(Z=1\).
The bounds are sharp, not merely conservative. Every latent distribution satisfying the constraints can be paired with an instrument independent of the potential variables and generates exactly the observed cells. Its ATE is the objective value for that distribution. Conversely, any population satisfying the assumptions and observed data yields one such feasible distribution. Thus no value inside the feasible objective range can be ruled out using only these assumptions and data, and no value outside it is feasible.
For each instrument group, obtain the probabilities of each treatment-outcome pair. Check that the cells sum to one within each group.
Use the 16 probabilities for \((D(0),D(1),Y(0),Y(1))\); require each to be nonnegative and their sum to equal one.
For both values of \(Z\), sum the latent probabilities that imply each observed \((D,Y)\) cell.
Minimize and maximize the weighted sum of \(Y(1)-Y(0)\). The two optima are the sharp lower and upper bounds.
Worked Example: All Observed Outcomes Are Zero
Suppose a hypothetical instrument group has treatment rate 0.50 under each instrument value, and every observed outcome is zero. Find the sharp ATE bounds without assuming monotonicity.
The treatment-rate equations give \(q_0=q_1=0.50\), so \(0\leq a\leq0.50\), \(c=0.50-a\), \(f=0.50-a\), and \(n=a\). Since every observed outcome is zero, anyone observed treated must have \(Y(1)=0\), and anyone observed untreated must have \(Y(0)=0\). An always-taker is treated under either instrument value: its \(Y(1)\) is observed as zero, while its \(Y(0)\) is unobserved. Its contribution to the population ATE can therefore range from \(-1\) to 0 per person. A never-taker has observed \(Y(0)=0\), but unobserved \(Y(1)\), so its contribution ranges from 0 to 1 per person. Compliers and defiers are observed both untreated and treated across the two instrument values, so both their potential outcomes equal zero.
For a fixed \(a\), these facts give \(-a\leq\text{ATE}\leq n=a\). Since \(a\) can range from 0 to 0.50, the overall bounds are \([-0.50,0.50]\). They are attainable: at \(a=n=0.50\), set \(Y(0)=1\) for the always-takers to attain \(-0.50\), or set \(Y(1)=1\) for the never-takers to attain \(0.50\). In either construction, the unobserved potential outcomes change the ATE but not any observed outcome.
Worked Example: Treatment Rates at Their Extremes
In another hypothetical population, \(q_0=0\) and \(q_1=1\). The observed outcome mean is 0.30 in the \(Z=0\) group and 0.60 in the \(Z=1\) group. Find the ATE.
The compliance-share formula gives \(a\in[\max(0,0+1-1),\min(0,1)]=[0,0]\). Thus \(a=0\), \(c=1\), and \(n=f=0\): everyone is a complier, as implied by the extreme treatment rates. With \(Z=0\), everyone receives treatment zero, so independence and consistency identify \(E[Y(0)]=0.30\). With \(Z=1\), everyone receives treatment one, identifying \(E[Y(1)]=0.60\). Therefore \[ \text{ATE}=E[Y(1)]-E[Y(0)]=0.60-0.30=0.30. \] The outcome is binary, so the result is a 0.30 difference in outcome probability, or 30 percentage points. Here the assumptions and observed rates yield point identification even though monotonicity was never imposed.
Worked Example: Everyone Is Treated Under Either Instrument Value
Suppose a hypothetical population has \(q_0=q_1=1\), and the observed outcome rate is 0.70 in each instrument group. Find the ATE bounds.
The type-share restrictions force \(a=1\), with \(c=f=n=0\): everyone is an always-taker. The observed outcome is \(Y(1)\), so independence identifies \(E[Y(1)]=0.70\). The population mean \(E[Y(0)]\) is not observed and, because \(Y(0)\) is binary, can be anywhere from zero to one. Hence \[ \text{ATE}=0.70-E[Y(0)]\in[0.70-1,\ 0.70-0]=[-0.30,\ 0.70]. \] Both endpoints are feasible: set \(Y(0)=1\) for everyone to attain \(-0.30\), or \(Y(0)=0\) for everyone to attain \(0.70\). These choices leave the observed treatment and outcome distributions unchanged. The instrument has no information about the untreated outcome for this population.
Interpretation and a Common Pitfall
An identified interval is a statement about which population ATE values are compatible with the observed distribution and the assumptions. If the interval is wide, the data and assumptions leave substantial uncertainty about the causal effect; if its endpoints coincide, the ATE is point identified. These are identification bounds, not confidence intervals. In applications, sampling uncertainty in the estimated cell probabilities must also be considered before making inferential claims about the population bounds.
- Pitfall: silently reinstating monotonicity. The incorrect reasoning is to call every increase in the treatment rate a complier share and set the defier share to zero. Without monotonicity, \(q_1-q_0=c-f\), not necessarily \(c\). The correction is to retain both compliers and defiers in the latent model and optimize over every feasible type distribution.
- Pitfall: reporting the Wald ratio as the ATE. Without monotonicity, the ratio does not generally identify the complier effect, and that effect would not automatically equal the population ATE even if identified. Use the observed-cell restrictions and the target-specific optimization instead.
- Pitfall: treating bounds as probabilities about the parameter. An identified interval does not mean there is a probability that the fixed ATE lies at a particular point in it. It describes the set of ATE values not ruled out by the maintained assumptions and observed distribution.
Check Your Understanding
Answer each question using the assumptions and definitions in this tutorial.
- If \(q_0=0.20\) and \(q_1=0.60\), what is the feasible interval for the always-taker share?
- Why does \(q_1-q_0\) equal the complier share only under an additional restriction?
- What does it mean for the linear-programming bounds to be sharp?
- If everyone is an always-taker and the observed treated-outcome rate is 0.40, what are the ATE bounds for a binary outcome?
- Are the bounds obtained from population cell probabilities a confidence interval? Explain the distinction.