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Partial Identification and Bounds · Tutorial 411 of 699

When Outcome Bounds Are Unknown Or Unbounded

See why unrestricted worst-case bounds can span the real line, and how explicit support or moment restrictions yield finite, interpretable identified sets.

Advanced 11 min read

What You'll Learn

  • Express the average treatment effect using the two unobserved potential-outcome means.
  • Determine how the outcome support constrains those missing means.
  • Explain why nonnegative, unbounded outcomes can still leave the effect unbounded in both directions.
  • Derive finite effect bounds from known outcome limits.
  • Use conditional second-moment restrictions to bound missing means and the effect.
  • Diagnose unsupported outcome limits and tail assumptions.

Why Unknown Outcome Limits Matter

Worst-case bounds use restrictions on the possible values of outcomes to constrain missing potential outcomes. If an outcome is known to lie between fixed limits, those limits constrain the means of outcomes that were not observed. But a sample’s observed minimum and maximum do not, by themselves, establish limits for the population outcome or its potential outcomes. When no defensible limits or other restrictions are available, the bounds can be extremely wide or uninformative.

The issue is not merely that an outcome is continuous. A continuous outcome can be confined to a known interval, while an unbounded outcome may be restricted to nonnegative values. The missing means’ feasible ranges depend on the outcome’s support and on any additional assumptions about its tails. This tutorial makes those dependencies explicit.

Definition: Let \(D\in\{0,1\}\) denote treatment status, with potential outcomes \(Y(1)\) and \(Y(0)\), and observed outcome \(Y=D Y(1)+(1-D)Y(0)\). The average treatment effect is \(ATE=E[Y(1)-Y(0)]\). Write \(p=P(D=1)\), \(m_1=E[Y\mid D=1]\), and \(m_0=E[Y\mid D=0]\). For \(0<p<1\), define the missing potential-outcome means \(a=E[Y(0)\mid D=1]\) and \(b=E[Y(1)\mid D=0]\). Then \(ATE=p(m_1-a)+(1-p)(b-m_0)\).

The equality follows by splitting the population into treated and untreated groups. In the treated group, \(Y(1)\) is observed and its mean is \(m_1\), while the mean \(a\) of \(Y(0)\) is missing. In the untreated group, \(Y(0)\) is observed with mean \(m_0\), while the mean \(b\) of \(Y(1)\) is missing. Thus the observed data determine \(p,m_1,m_0\), but not generally \(a\) or \(b\).

Here the target is the population average effect, and the setup assumes consistency, well-defined treatment and potential outcomes, and finite observed group means. It does not assume that treatment is randomized or that potential outcomes are independent of treatment. To discuss both missing means, both treatment groups must have positive population probability: \(0<p<1\).

How Support Restrictions Determine the Bounds

Suppose the outcome is known to lie in \([L,U]\), with finite \(L\) and \(U\). Then both missing means must lie in that interval: \(a,b\in[L,U]\). Define the part of the ATE expression determined by observed quantities as \(C=p m_1-(1-p)m_0\). The remaining terms are \(-pa+(1-p)b\). Because \(p>0\), this expression decreases as \(a\) increases and increases as \(b\) increases. Its smallest value therefore uses \(a=U,b=L\); its largest uses \(a=L,b=U\).

$$ ATE=C-pa+(1-p)b,\qquad C=p m_1-(1-p)m_0 $$
$$ ATE\in\left[C-pU+(1-p)L,\ C-pL+(1-p)U\right] $$

These bounds are sharp under the stated support restriction alone: the lower endpoint is attainable by assigning the missing outcomes the allowed endpoint values \(a=U\) and \(b=L\), and the upper endpoint by assigning \(a=L\) and \(b=U\). More generally, if all that is known is that \(a,b\) belong to a feasible set of means \(\mathcal M\), the bounds depend on the infimum and supremum of that set. The support determines \(\mathcal M\); “unbounded” alone does not specify its lower endpoint or whether it extends in both directions.

Worked Example: A Defensible Bounded Outcome Scale

Suppose a hypothetical outcome is a score known by design to range from 0 to 20 points. In the population, \(p=0.40\), \(m_1=12\) points, and \(m_0=8\) points. No restriction is imposed on the missing potential-outcome means beyond the score’s known limits. Find the worst-case ATE bounds.

First calculate the observed-data component:

$$ C=p m_1-(1-p)m_0 =(0.40)(12)-(0.60)(8) =4.8-4.8=0\text{ points}. $$

Here \(L=0\) and \(U=20\). The lower bound assigns the largest possible missing mean to \(a\) and the smallest to \(b\):

$$ ATE_{\mathrm{lower}}=0-(0.40)(20)+(0.60)(0)=-8\text{ points}. $$

The upper bound reverses those assignments:

$$ ATE_{\mathrm{upper}}=0-(0.40)(0)+(0.60)(20)=12\text{ points}. $$

Thus the identified set is \([-8,12]\) points. These data and assumptions do not establish whether the average effect is beneficial or harmful: both negative and positive values are consistent with them. The calculations require the 0-to-20 scale to be a genuine population limit, not merely the range seen in a sample.

Unbounded Does Not Always Mean Unrestricted

If the outcome has no restrictions on its values—that is, its support allows any real value—then the missing means can each be any finite real number under the support restriction alone. With \(0<p<1\), the term \(-pa\) can be made arbitrarily large or small by varying \(a\); independently, \((1-p)b\) can also be made arbitrarily large or small by varying \(b\). The resulting ATE identified set is the whole real line. This conclusion requires unrestricted support in both directions; it does not follow merely from saying that an outcome is unbounded.

For example, a nonnegative outcome is unbounded above but cannot have negative values. Its missing means must be nonnegative, not arbitrary real numbers. Nevertheless, if both missing means can grow without limit and \(0<p<1\), the ATE can still take any real value: increasing \(a\) drives the effect downward, while increasing \(b\) drives it upward. The restrictions on each mean and the range of their weighted difference are distinct questions.

Worked Example: Nonnegative and Unbounded Outcomes

Suppose a hypothetical outcome is nonnegative and has no known upper limit. Let \(p=0.30\), \(m_1=15\) units, and \(m_0=10\) units. The support restriction implies \(a,b\geq0\), and permits each to be arbitrarily large. The observed component is

$$ C=(0.30)(15)-(0.70)(10)=4.5-7=-2.5\text{ units}. $$

If \(a=b=0\), the implied ATE is \(-2.5\) units. To produce an ATE of \(5\) units, set \(a=0\) and solve \(-2.5+0.70b=5\), giving \(b=7.5/0.70=10.714\) units. To produce an ATE of \(-8\) units, set \(b=0\) and solve \(-2.5-0.30a=-8\), giving \(a=5.5/0.30=18.333\) units.

In general, for any target \(t\geq-2.5\), choose \(a=0\) and \(b=(t+2.5)/0.70\). For any \(t<-2.5\), choose \(b=0\) and \(a=(-2.5-t)/0.30\). These choices are finite and nonnegative, and they yield \(ATE=t\). Therefore the identified set is \((-\infty,\infty)\), even though each missing mean is restricted to be nonnegative. This result relies on both groups being present and both missing means being free to grow without limit.

Other support shapes can produce other feasible mean ranges. For instance, if the outcome is bounded below but unbounded above, each missing mean must respect that lower bound; it cannot be assigned an arbitrarily small value. If the support is bounded above but not below, the corresponding restriction runs the other way. The ATE set must be calculated from those feasible ranges, rather than inferred from the word “unbounded.”

Tail Restrictions Can Tighten the Set

A tail restriction can make the identified set informative without asserting a hard upper limit for every individual outcome. The restriction must, however, constrain the missing potential-outcome means—not just describe the observed outcomes. A finite observed sample maximum, a finite observed mean, or a statement that outcomes have finite first moments does not give a uniform numerical bound on \(a\) and \(b\).

One possible assumption is a known upper bound on the second moment of each missing potential outcome within the group where it is unobserved. For a random variable \(Z\), the Cauchy–Schwarz inequality gives \(|E[Z]|\leq\sqrt{E[Z^2]}\). When \(Z\geq0\), this yields \(0\leq E[Z]\leq\sqrt{E[Z^2]}\). Such a restriction is substantive: it must be justified from subject-matter knowledge, design information, or suitable auxiliary evidence. It is not established just because the observed outcomes have modest values.

Worked Example: Bounding Missing Means With Second-Moment Limits

Consider again a hypothetical nonnegative outcome with \(p=0.30\), \(m_1=15\) units, and \(m_0=10\) units. Suppose subject-matter evidence justifies the restrictions \(E[Y(0)^2\mid D=1]\leq100\) square units and \(E[Y(1)^2\mid D=0]\leq144\) square units. These are restrictions on the missing potential outcomes in their respective groups, not on the observed outcomes alone.

By nonnegativity and Cauchy–Schwarz,

$$ 0\leq a\leq\sqrt{100}=10\text{ units}, \qquad 0\leq b\leq\sqrt{144}=12\text{ units}. $$

The observed component is \(C=-2.5\) units. The smallest ATE uses \(a=10\) and \(b=0\):

$$ ATE_{\mathrm{lower}}=-2.5-(0.30)(10)+(0.70)(0) =-5.5\text{ units}. $$

The largest uses \(a=0\) and \(b=12\):

$$ ATE_{\mathrm{upper}}=-2.5-(0.30)(0)+(0.70)(12) =5.9\text{ units}. $$

The resulting interval is \([-5.5,5.9]\) units. It is narrower than the whole real line because the assumed second-moment limits bound the missing means. The calculation does not show that those limits are true; the interval is conditional on them. Moreover, these mean bounds need not be attainable by distributions satisfying every possible additional constraint. They are valid outer bounds; sharpness requires checking that the endpoint means are feasible under all imposed restrictions.

Derivation, Diagnostics, and Alternatives

The endpoint calculation follows directly from the signs of the missing means’ coefficients. If \(a\in[a_L,a_U]\) and \(b\in[b_L,b_U]\), then, because \(p\) and \(1-p\) are positive, the smallest value uses \(a=a_U\) and \(b=b_L\), while the largest uses \(a=a_L\) and \(b=b_U\). Therefore

$$ ATE\in \left[C-pa_U+(1-p)b_L,\ C-pa_L+(1-p)b_U\right]. $$

The formula also makes clear why an unrestricted mean range \((-\infty,\infty)\) gives an unrestricted ATE range, and why nonnegative means with no upper limits still do so when both treatment groups have positive weight. If only one group has positive probability, this two-missing-mean argument does not apply; the missing contribution and feasible range must be reconsidered for that case.

A practical diagnostic is to write down the precise restrictions on \(a\) and \(b\) before calculating bounds. Then ask whether each restriction is a property of the outcome’s scale, a justified scientific assumption, or merely an extrapolation from the observed sample. Check that the observed group means are compatible with any claimed support limits. For moment restrictions, verify that the moments are stated for the correct treatment group and potential outcome, and that their units are squared when a second moment is used.

The main alternative is to report bounds under several defensible support or tail assumptions rather than choose one limit without explanation. A known physical or instrument-imposed scale can support a hard bound. A moment or tail restriction may be more suitable when extreme values are possible but their magnitude or frequency can be limited. A parametric model for the tail can also yield a narrower estimate, but its conclusions rely on the model rather than on worst-case support restrictions alone. Transforming or clipping the outcome changes the target or its scale; it does not, by itself, identify bounds for the original outcome’s ATE.

Common Mistakes and Limitations

  • Equating unbounded with unrestricted. An outcome can be nonnegative and unbounded above. Its missing means then cannot be negative, even if the ATE set can still be the whole real line.
  • Treating the observed range as a population limit. A sample maximum is not a guaranteed upper bound on unobserved potential outcomes. A hard limit needs a defensible basis.
  • Assuming finite observed means bound missing means. The observed means describe outcomes under the treatment actually received. They do not, without further assumptions, cap the unobserved counterfactual means.
  • Using a tail condition that does not bound the relevant mean. Saying that each distribution has a finite first moment, without a known numerical limit, still allows the mean to be arbitrarily large across compatible distributions.
  • Presenting assumption-dependent bounds as data-only conclusions. A narrow interval derived from a second-moment limit is conditional on that limit and should be reported as such.

Worst-case bounds are limited by what can be justified about the missing outcomes. If every scientifically defensible restriction leaves a broad set, that breadth is a finding about the available information, not a computational failure. Conversely, a narrow set can be misleading if it depends on an unsupported cap or tail assumption. The goal is to make both the identified set and the assumptions that produce it transparent.

Check Your Understanding

For each question, first identify the restrictions on the missing means, then determine what follows for the ATE set.

  1. With \(0<p<1\), the outcome can take any real value and there are no other restrictions on \(a\) and \(b\). What is the ATE identified set?
  2. The outcome is nonnegative and unbounded above. Are \(a\) and \(b\) allowed to be negative? Can the ATE nevertheless range over the whole real line?
  3. For \(p=0.25\), \(m_1=8\), \(m_0=4\), and known support \([0,10]\), calculate the worst-case ATE bounds.
  4. Why does knowing that observed outcomes have finite means not, by itself, provide finite bounds on \(a\) and \(b\)?
  5. Suppose a nonnegative missing potential outcome has conditional second moment at most 49 square units. What upper bound does Cauchy–Schwarz place on its conditional mean?

Solutions. (1) The set is \((-\infty,\infty)\), because \(a\) and \(b\) can vary freely and their coefficients \(-p\) and \(1-p\) have opposite signs. (2) Neither mean can be negative, but both can grow without limit. Increasing \(a\) makes the ATE arbitrarily negative, and increasing \(b\) makes it arbitrarily positive, so the ATE set is still the whole real line. (3) Here \(C=(0.25)(8)-(0.75)(4)=2-3=-1\). The lower bound is \(-1-(0.25)(10)+(0.75)(0)=-3.5\); the upper bound is \(-1-(0.25)(0)+(0.75)(10)=6.5\). The bounds are \([-3.5,6.5]\) outcome units. (4) Observed means concern the potential outcome revealed by treatment status; they do not constrain the missing potential-outcome distributions unless an additional assumption links them. (5) The conditional mean is at most \(\sqrt{49}=7\) units, and is at least 0 because the outcome is nonnegative.

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