Lee Bounds for Outcomes Under Monotone Attrition

Why Monotone Attrition Can Be Informative

When outcome observation depends on treatment, a comparison of observed outcomes may compare different groups of people. For example, treatment may make follow-up more likely, so the treated group with recorded outcomes includes some people who would not have been observed without treatment. The difference in observed means then need not equal an average treatment effect for a common population.

Lee bounds address this problem with a directional restriction: treatment weakly increases, or weakly decreases, the probability that an outcome is observed for each individual. This restriction does not reveal who would have been observed in both treatment states. It does, however, determine the share of the observed group who must belong to that common population. Trimming the observed outcome distribution by the implied share gives bounds on the treatment effect for that population.

Setup and Target Effect

Let $D\in\{0,1\}$ denote treatment assignment. For each individual, $Y(1)$ and $Y(0)$ are the potential outcomes under treatment and control. Let $S(1)$ and $S(0)$ indicate whether the corresponding outcome would be observed, with value 1 for observed and 0 for missing. The realized outcome is available only when $S=1$. Assume consistency: when $D=d$, the observed response indicator is $S(d)$, and if that response indicator is 1, the recorded outcome is $Y(d)$.

Assume treatment is randomly assigned, so $D$ is independent of $(Y(1),Y(0),S(1),S(0))$, and both assignment probabilities are positive. Also assume the potential outcomes have finite means in the populations used below. The target is not necessarily the average treatment effect for everyone. It is the average treatment effect for individuals whose outcome would be observed under either assignment.

Definition: The always-observed population consists of individuals with $S(1)=S(0)=1$. Its average treatment effect is $$ \tau_{\mathrm{AO}}=E\!\left[Y(1)-Y(0)\mid S(1)=S(0)=1\right]. $$ Lee bounds are lower and upper bounds on this effect under random assignment and a monotonicity restriction on observation.

The monotonicity restriction considered first is $S(1)\geq S(0)$ for every individual. It says treatment cannot cause someone who would otherwise have an observed outcome to have a missing outcome. It does not say treatment increases observation for everyone: some individuals may have $S(1)=S(0)=1$, while others may have $S(1)=S(0)=0$.

Write $p_d=P(S=1\mid D=d)$ for the outcome observation rate in assignment group $d$. Under random assignment, these rates equal the population probabilities $P(S(d)=1)$. With $S(1)\geq S(0)$, $p_1\geq p_0$. If $p_0>0$, the fraction of treated individuals with observed outcomes who must be always-observed is $q=p_0/p_1$.

Trimming the Treated Outcome Distribution

Under $S(1)\geq S(0)$, anyone observed in the control group must be always-observed: $S(0)=1$ implies $S(1)=1$. Consequently, the control observed outcome distribution is the distribution of $Y(0)$ in the always-observed population. The treated observed group is different. It contains all always-observed individuals and may also contain individuals with $S(1)=1,S(0)=0$.

The always-observed individuals therefore make up the fraction $q=p_0/p_1$ of treated individuals with observed outcomes. Their treated potential outcome mean is not generally identified, because the data do not identify which treated respondents are always-observed. But among all possible ways of selecting a fraction $q$ from the treated observed outcome distribution, the smallest possible selected mean comes from taking its lowest outcomes, and the largest comes from taking its highest outcomes.

Definition: Let $F_1$ be the distribution of observed outcomes among treated individuals, and let $Q_1(u)=\inf\{y:F_1(y)\geq u\}$ be its quantile function. For $0<q\leq1$, define the lower- and upper-trimmed means by $$ \mu_{1,L}(q)=\frac{1}{q}\int_0^q Q_1(u)\,du, \qquad \mu_{1,U}(q)=\frac{1}{q}\int_{1-q}^1 Q_1(u)\,du. $$ These formulas allow fractional trimming at a quantile when the distribution has an atom.

Let $\mu_{0,\mathrm{obs}}=E[Y\mid D=0,S=1]$, the mean outcome among observed controls. Under the assumptions above, this equals the mean of $Y(0)$ in the always-observed population. Subtracting that common control mean from the two possible extremes for the treated mean gives the Lee bounds.

Theorem: Under random assignment, consistency, finite means, $S(1)\geq S(0)$, and $p_1\geq p_0>0$, the sharp bounds for the always-observed average treatment effect are $$ \mu_{1,L}(q)-\mu_{0,\mathrm{obs}} \ \leq\ \tau_{\mathrm{AO}}\ \leq\ \mu_{1,U}(q)-\mu_{0,\mathrm{obs}}, \qquad q=\frac{p_0}{p_1}. $$ “Sharp” means that, without additional restrictions, the observed data and these assumptions permit values at both endpoints and values between them.

Why the Bounds Hold

Identify who is always observed among controls.
Monotonicity $S(1)\geq S(0)$ implies $S(0)=1$ only for people with $S(1)=1$. Thus observed controls are precisely the always-observed individuals, and their observed outcome mean is the target control potential-outcome mean.

Find the always-observed share among treated respondents.
The always-observed population has share $p_0$ in the full population, while treated respondents have share $p_1$. Therefore their share among treated respondents is $p_0/p_1=q$.

Bound the treated mean by rearrangement.
Among all subsets with share $q$ of a distribution, the lowest possible mean selects the lowest $q$ fraction and the highest possible mean selects the highest $q$ fraction. The treated always-observed mean must lie between these trimmed means.

Subtract the identified control mean.
The treatment effect for the always-observed population is its treated mean minus its control mean. Subtracting the same $\mu_{0,\mathrm{obs}}$ from both extremes yields the stated interval.

The rearrangement claim can be seen from the quantile representation. Represent the treated observed outcomes by $Q_1(U)$, where $U$ is uniform on $(0,1)$. Selecting the lowest $q$ fraction corresponds to $0<U<q$, and selecting the highest $q$ fraction corresponds to $1-q<U<1$. Any other selection of the same share replaces some lower-ranked values with values at least as large for the maximum, or replaces some higher-ranked values with values no larger for the minimum. Thus no subset of share $q$ can have a mean below $\frac{1}{q}\int_0^q Q_1(u)\,du$ or above $\frac{1}{q}\int_{1-q}^1Q_1(u)\,du$. With no further information about which treated respondents are always-observed, either extreme assignment of membership is compatible with the observed treated distribution and the known share. This also explains the sharpness claim.

Worked Example: Trimming a Discrete Treated Distribution

Suppose a hypothetical randomized experiment has observation rates $p_0=0.60$ and $p_1=0.80$, and assume $S(1)\geq S(0)$. Among treated respondents, suppose the outcome distribution has probability 0.25 at 10 points, 0.50 at 20 points, and 0.25 at 40 points. The observed control mean is 15 points. Find the Lee bounds.

The always-observed share among treated respondents is $q=p_0/p_1=0.60/0.80=0.75$. Thus we retain 75% of the treated observed distribution. For the lower trimmed mean, retain the lowest 75%: all 25% at 10 and all 50% at 20.

\mu_{1,L}(0.75) =\frac{0.25(10)+0.50(20)}{0.75} =\frac{12.5}{0.75} =16.67\text{ points}.

For the upper trimmed mean, retain the highest 75%: all 50% at 20 and all 25% at 40. The assumptions are consistent with the rates because $p_1\geq p_0$; the given outcome distribution supplies the observed treated outcomes, and the control mean is for respondents who, by monotonicity, are always-observed.

\mu_{1,U}(0.75) =\frac{0.50(20)+0.25(40)}{0.75} =\frac{20}{0.75} =26.67\text{ points}.

Subtract the control mean of 15 points from each endpoint. The effect for the always-observed population is bounded by $16.67-15=1.67$ points and $26.67-15=11.67$ points. The interval does not identify a single effect, but it rules out effects outside those bounds under the stated assumptions.

Worked Example: Trimming a Continuous Distribution

In a second hypothetical randomized study, suppose $p_0=0.50$, $p_1=0.80$, and treatment weakly increases observation for every individual. Suppose treated respondents’ outcomes are uniformly distributed from 0 to 10 hours, and the observed control mean is 4 hours. Calculate the bounds.

The retained share is $q=0.50/0.80=0.625$. For a uniform distribution on $[0,10]$, the quantile function is $Q_1(u)=10u$, for $0<u<1$. The lower trimmed mean is the average of the lowest 62.5% of this distribution:

\mu_{1,L}(0.625) =\frac{1}{0.625}\int_0^{0.625}10u\,du =\frac{5(0.625)^2}{0.625} =3.125\text{ hours}.

The upper 62.5% starts at $1-q=0.375$. Its mean is

\mu_{1,U}(0.625) =\frac{1}{0.625}\int_{0.375}^1 10u\,du =\frac{5(1-0.375^2)}{0.625} =6.875\text{ hours}.

Subtracting the observed control mean gives $[3.125-4,\ 6.875-4]=[-0.875,\ 2.875]$ hours. The data and monotonicity assumption bound the always-observed treatment effect between a decrease of 0.875 hours and an increase of 2.875 hours. They do not establish that the effect is positive or negative.

When Treatment Reduces Observation

The direction of monotonicity determines which group must be trimmed. Suppose instead that $S(1)\leq S(0)$ for every individual, so treatment weakly reduces observation. Now every treated respondent is always-observed, while the observed controls include the always-observed group plus people observed only under control. The treated observed mean identifies the always-observed treated mean, and the control distribution must be trimmed to retain the share $q=p_1/p_0$.

Because the treatment effect is the treated mean minus the control mean, the largest trimmed control mean gives the lower effect bound. The smallest trimmed control mean gives the upper effect bound.

Worked Example: Treatment Weakly Reduces Observation

Consider a hypothetical randomized experiment with $p_1=0.50$, $p_0=0.75$, and $S(1)\leq S(0)$. Suppose the treated observed mean is 25 points. Among observed controls, outcomes have equal probability at 10, 20, and 40 points. Find the bounds.

The always-observed share among controls is $q=p_1/p_0=0.50/0.75=2/3$. The lower trimmed control mean retains the lowest two thirds, the outcomes 10 and 20, so it is $(10+20)/2=15$ points. The upper trimmed control mean retains 20 and 40, so it is $(20+40)/2=30$ points. Here every treated respondent is always-observed, so their mean of 25 points is the treated mean for the target population.

25-30 \ \leq\ \tau_{\mathrm{AO}}\ \leq\ 25-15, \qquad\text{or}\qquad -5\text{ points}\ \leq\ \tau_{\mathrm{AO}}\ \leq\ 10\text{ points}.

The interval includes both negative and positive effects. Reversing the trimming order would give incorrect endpoints because a larger control mean implies a smaller treated-minus-control effect.

Interpretation and Common Pitfalls

Lee bounds concern the average treatment effect for a specific latent subgroup: people whose outcome would be observed under either assignment. They are not automatically bounds on the average treatment effect for the entire randomized population. The monotonicity assumption is also individual-level: $S(1)\geq S(0)$ must hold for each individual, not merely as an average pattern or as a difference in response rates. The rate inequality $p_1\geq p_0$ is necessary under that assumption, but does not establish it.

A common mistake is to trim the treated distribution and then interpret the trimmed treated mean as a point estimate of the always-observed treated mean. The identities of the always-observed treated respondents are unknown. Trimming gives the smallest and largest means compatible with the known share, not a way to identify membership. Another mistake is to discard the upper tail regardless of the monotonicity direction. The correct tails depend on which group contains the additional respondents: under $S(1)\geq S(0)$, trim treated outcomes; under $S(1)\leq S(0)$, trim control outcomes.

The response rates and outcome distributions are population quantities in the derivation. In an empirical analysis, they are estimated from data, and uncertainty in estimated endpoints requires appropriate inference. The bounds themselves do not remove uncertainty caused by sampling, nor do they justify a causal interpretation if treatment assignment is confounded. They show what the observed outcome distributions and the stated selection restriction imply about the target effect.

Key takeaway: With monotone observation, the response-rate ratio determines the share to retain. Trim the outcome distribution of the group that contains additional respondents, calculate both tail means, and subtract in the order required by the treatment effect.

Check Your Understanding

Use the assumptions and definitions in this tutorial to answer the following questions.

Under $S(1)\geq S(0)$, why does an observed control outcome come from the always-observed population?
If $p_0=0.45$ and $p_1=0.75$ under $S(1)\geq S(0)$, what share of treated respondents must be retained when forming the Lee bounds?
Under $S(1)\geq S(0)$, which tail of the treated observed outcome distribution gives the lower bound on the treated mean for the always-observed population?
Under $S(1)\leq S(0)$, explain why the largest trimmed control mean produces the lower treatment-effect bound.
Does $p_1>p_0$ by itself prove that $S(1)\geq S(0)$ for every individual? Explain.