Bounds From Known Population Outcome Means

How a Population Mean Constrains Missing Outcomes

For each unit, only one potential outcome is observed: the outcome under the treatment actually received. An external source can sometimes supply a valid population mean for one potential outcome, even though individual outcomes under the alternative treatment remain unobserved. That mean is not a substitute for observing every missing outcome. Instead, it constrains their average through the known proportions treated and untreated.

The resulting information can identify an average effect for one treatment group, while leaving the population average treatment effect (ATE) only partially identified. The bounds depend on the known outcome scale and on whether the supplied mean actually describes the target population. No assumption of treatment assignment independent of potential outcomes is required.

Definition: Let $D\in\{0,1\}$ indicate treatment receipt, and let $Y(1)$ and $Y(0)$ be the potential outcomes under treatment and control. The observed outcome is $Y=DY(1)+(1-D)Y(0)$. Write $p=P(D=1)$, $\mu_1=E[Y\mid D=1]$, and $\mu_0=E[Y\mid D=0]$. The ATE is $\tau=E[Y(1)-Y(0)]$. The treated-group effect is $E[Y(1)-Y(0)\mid D=1]$, and the untreated-group effect is $E[Y(1)-Y(0)\mid D=0]$.

Suppose both potential outcomes lie in a known interval $[L,U]$. The external source supplies $m_0=E[Y(0)]$, the target-population mean of the control potential outcome. This is a population quantity, not the observed control-group mean $\mu_0$. The distinction matters because $Y(0)$ is missing for treated units.

Assumptions: The potential outcomes and treatment indicator refer to the same target population; consistency holds, so the observed outcome equals the potential outcome for the treatment received; and potential outcomes lie in the stated support $[L,U]$. The treatment proportion satisfies $0<p<1$. The external value $m_0$ is valid for this population and outcome definition. These conditions do not imply random assignment, exchangeability, or no unmeasured confounding.

Recovering the Missing Control-Outcome Mean

The population mean of $Y(0)$ is a mixture of its mean among untreated units, where it is observed, and its mean among treated units, where it is missing. Denote the latter by $E[Y(0)\mid D=1]$. The law of total expectation gives the mixture equation:

m_0=(1-p)\mu_0+pE[Y(0)\mid D=1].

If $m_0$ is known and $p>0$, rearranging identifies the missing treated-group mean. This is a constraint on an average; it does not reveal each treated unit's control outcome or its distribution.

E[Y(0)\mid D=1]=\frac{m_0-(1-p)\mu_0}{p}.

Consequently, the treated-group effect is identified by subtracting this missing mean from the observed treated-group outcome mean. In contrast, the missing mean $E[Y(1)\mid D=0]$ is not determined by $m_0$. Its support restriction supplies an interval, which leads to ATE bounds.

\begin{aligned} \tau_{\mathrm{ATT}} &=E[Y(1)-Y(0)\mid D=1]\\ &=\mu_1-\frac{m_0-(1-p)\mu_0}{p},\\[4pt] \tau &=p\mu_1+(1-p)E[Y(1)\mid D=0]-m_0. \end{aligned}

Because $L\leq E[Y(1)\mid D=0]\leq U$, multiplying by the nonnegative share $1-p$ and adding the other terms gives the ATE bounds. The lower endpoint uses $L$ for the missing treated outcome under treatment among controls; the upper endpoint uses $U$.

Theorem: Given the assumptions above and a valid known $m_0=E[Y(0)]$, the ATE is bounded by $$ p\mu_1+(1-p)L-m_0 \ \leq\ \tau\ \leq\ p\mu_1+(1-p)U-m_0. $$ The treated-group effect is identified by $\tau_{\mathrm{ATT}}=\mu_1-[m_0-(1-p)\mu_0]/p$, provided the implied missing mean is feasible on $[L,U]$.

These ATE bounds are sharp under the stated restrictions: each endpoint can be attained by setting the missing $Y(1)$ for every untreated unit to $L$, or to $U$, respectively. The observed outcomes and the mean constraint on $Y(0)$ can remain as specified. Intermediate values of the missing mean produce intermediate ATE values. This sharpness claim uses no restrictions on how the two potential outcomes are paired within a unit.

Worked Applications

Worked Example: A Known Control Mean Identifies the Treated-Group Effect

Suppose a hypothetical binary-outcome population has $L=0$, $U=1$, treatment proportion $p=0.40$, observed mean $\mu_1=0.60$ among treated units, observed mean $\mu_0=0.30$ among controls, and a valid external control-potential-outcome mean $m_0=0.42$. Check the missing mean, calculate the treated-group effect, and bound the ATE.

The missing $Y(0)$ mean among treated units is $$ E[Y(0)\mid D=1]=\frac{0.42-(1-0.40)(0.30)}{0.40} =\frac{0.42-0.18}{0.40}=0.60. $$ This is in the binary outcome range $[0,1]$, so the supplied mean is compatible with the observed control mean and treatment share.

The treated-group effect is $0.60-0.60=0$, in probability-point units. For the ATE, the lower endpoint is $$ 0.40(0.60)+0.60(0)-0.42=0.24-0.42=-0.18, $$ and the upper endpoint is $$ 0.40(0.60)+0.60(1)-0.42=0.24+0.60-0.42=0.42. $$ Thus the ATE lies between $-0.18$ and $0.42$, or between $-18$ and $42$ percentage points. The treated-group effect being zero does not force the population effect to be zero: the effect among controls remains unknown.

Worked Example: A Bounded Continuous Outcome

Consider a hypothetical outcome measured in points on a known scale from 0 to 10. Suppose $p=0.30$, $\mu_1=6$, $\mu_0=4$, and a valid external source gives $m_0=5.5$. Find the missing control-outcome mean among treated units, the treated-group effect, and the ATE bounds.

The implied missing mean is $$ E[Y(0)\mid D=1]=\frac{5.5-(1-0.30)(4)}{0.30} =\frac{5.5-2.8}{0.30}=9. $$ A mean of 9 points is within the stated 0-to-10 scale. The treated-group effect is $6-9=-3$ points.

The ATE lower endpoint assigns the lowest allowable mean, 0, to the missing $Y(1)$ outcomes among controls: $$ 0.30(6)+0.70(0)-5.5=1.8-5.5=-3.7\text{ points}. $$ The upper endpoint uses 10 points: $$ 0.30(6)+0.70(10)-5.5=1.8+7-5.5=3.3\text{ points}. $$ Accordingly, these assumptions bound the population ATE between $-3.7$ and $3.3$ points. They do not establish that the effect is positive or negative.

The Symmetric Case: A Known Treated-Potential-Outcome Mean

If the external source instead supplies $m_1=E[Y(1)]$, the same mixture argument constrains the missing treated-potential-outcome mean among controls:

E[Y(1)\mid D=0]=\frac{m_1-p\mu_1}{1-p}.

This identifies the untreated-group effect, since $Y(0)$ is observed among controls. The missing $Y(0)$ mean among treated units can still range from $L$ to $U$. Writing the ATE as $m_1-E[Y(0)]$ therefore gives the following bounds.

Result: If $m_1=E[Y(1)]$ is known and the same support and population assumptions hold, then $$ m_1-\{(1-p)\mu_0+pU\} \ \leq\ \tau\ \leq\ m_1-\{(1-p)\mu_0+pL\}. $$ The untreated-group effect is $E[Y(1)\mid D=0]-\mu_0$, with the missing mean recovered from the mixture equation.

Worked Example: Using a Known Treated-Potential-Outcome Mean

In a hypothetical study, let the outcome be measured from 0 to 10 points. Suppose $p=0.50$, $\mu_1=8$, $\mu_0=5$, and an external source validly supplies $m_1=7.2$. Calculate the missing $Y(1)$ mean among controls and the ATE bounds.

The mixture equation implies $$ E[Y(1)\mid D=0]=\frac{7.2-(0.50)(8)}{0.50} =\frac{7.2-4}{0.50}=6.4\text{ points}. $$ The untreated-group effect is therefore $6.4-5=1.4$ points. For the ATE, the population mean of $Y(0)$ is $0.50(5)+0.50E[Y(0)\mid D=1]$. The missing treated-group mean can be anywhere from 0 to 10, so $$ 7.2-\{0.50(5)+0.50(10)\}=-0.30 \ \leq\ \tau\ \leq\ 7.2-\{0.50(5)+0.50(0)\}=4.70. $$ The ATE is bounded from $-0.30$ to $4.70$ points. The positive untreated-group effect does not settle the sign of the population effect.

Compatibility Checks, Comparisons, and Limitations

Before interpreting any bound, check that the known mean is feasible. With known $m_0$, the implied $E[Y(0)\mid D=1]$ must lie in $[L,U]$. Equivalently, the supplied mean must satisfy $$ (1-p)\mu_0+pL\leq m_0\leq(1-p)\mu_0+pU. $$ If this condition fails, the data summaries, outcome support, target population, or external mean cannot all be correct under the stated setup. The appropriate response is to investigate the mismatch, not to report the impossible implied mean as an estimate. For known $m_1$, check that $(m_1-p\mu_1)/(1-p)$ lies in $[L,U]$.

A known population mean is different from assuming exchangeability. Exchangeability would relate potential outcomes across treatment groups through an assignment assumption; the mean constraint instead supplies one marginal moment and leaves the other missing outcomes restricted only by support. It is also different from worst-case bounds without external information: the valid mean can identify a subgroup effect and alter the population-effect interval. If both $m_0$ and $m_1$ are validly known for the same population, then $\tau=m_1-m_0$ is point identified, though each supplied mean must still be compatible with the observed group mean and support.

A common mistake is to treat $m_0$ as though it were the observed control mean $\mu_0$, or to conclude that knowing one potential-outcome mean reveals all missing outcomes. The correction is to write the mixture equation first: the observed group contributes only its population share, and the remaining share's mean is inferred by subtraction and division. That yields a missing-group average, not individual counterfactuals. Likewise, an ATE interval is not a confidence interval: these are identification bounds under population restrictions. If $p$, observed means, or the external benchmark are estimated from samples, their sampling uncertainty must be addressed separately.

Align the information.
Confirm that the external mean and observed summaries refer to the same population, treatment definition, outcome, and measurement scale.

Write the population mixture.
Weight the observed potential-outcome mean and the missing subgroup mean by $1-p$ and $p$, or the reverse as appropriate.

Check feasibility.
Verify that the implied missing mean lies within $[L,U]$; otherwise the inputs are incompatible under the assumptions.

Bound the remaining missing mean.
Use its support endpoints in the ATE expression, and state the units and the population to which the resulting interval refers.

These bounds can remain wide when the treated share is small or when the outcome scale is broad. They are also only as credible as the benchmark's target-population relevance and the stated support. Additional valid restrictions may narrow the identified set, but a mean borrowed from a different population or defined using a different outcome cannot be inserted into the mixture equation without justification.

Check Your Understanding

Use the stated support and population assumptions in each question. Solutions are included below.

For $L=0$, $U=1$, $p=0.25$, and $\mu_0=0.70$, what is the smallest feasible value of $m_0$?
Let $p=0.40$, $\mu_1=5$, $L=0$, $U=10$, and $m_0=4$. Calculate the ATE bounds, in outcome units.
With $p=0.50$, $\mu_1=7$, $\mu_0=4$, and $m_1=6$, calculate the missing $Y(1)$ mean among controls and the untreated-group effect.
Does a known $m_0$ point identify the ATE? Explain what it identifies and what remains unknown.

Solutions. (1) The lower feasibility limit is $(1-p)\mu_0+pL=(0.75)(0.70)+(0.25)(0)=0.525$. If $m_0<0.525$, the implied treated-group mean is below 0. (2) The lower endpoint is $(0.40)(5)+(0.60)(0)-4=-2$; the upper endpoint is $(0.40)(5)+(0.60)(10)-4=4$. The bounds are $[-2,4]$ outcome units. (3) The missing mean is $[6-(0.50)(7)]/0.50=(6-3.5)/0.50=5$. The untreated-group effect is $5-4=1$ outcome unit. (4) Not in general. It identifies $E[Y(0)\mid D=1]$ and thus the treated-group effect, but $E[Y(1)\mid D=0]$ remains unknown within its support, so the ATE is generally bounded rather than point identified.