Why a Confidence Set May Not Be an Interval
A confidence interval is often pictured as two finite endpoints with every value between them included. That picture is not guaranteed in instrumental-variables (IV) analysis. When an Anderson–Rubin (AR) test is inverted, the result is the set of all hypothesized effects the test does not reject. If the instrument carries little information about the treatment, that set can be very wide, unbounded, or even disconnected. Its shape is part of the inferential result, not a cosmetic detail.
The earlier material on inverting Anderson–Rubin tests established the basic operation: test each candidate value of the structural coefficient, then collect the values not rejected at level \(\alpha\). Here the emphasis is on interpreting the resulting set. A set with finite endpoints conveys different information from one that extends indefinitely, and neither should be summarized by an ordinary interval without explanation.
Assumptions and a Simple Acceptance Inequality
Consider a structural equation \(Y=\beta X+u\), with an excluded instrument \(Z\). With included covariates, the same reasoning applies after accounting for them in the specified model. A causal interpretation requires a defensible structural model and instrument conditions: the instrument must be exogenous for the outcome disturbance and excluded from the outcome equation except through \(X\). The covariance calculation and reference distribution used by the test must also be appropriate for the sampling design. For example, dependence within clusters calls for a cluster-robust covariance calculation.
Instrument relevance is important for identifying \(\beta\) and obtaining informative inference. But the AR test does not divide by an estimated first-stage coefficient. Under the null restriction, its validity does not require that the first stage be strong. This protection against weak-instrument distortion is why inversion can honestly produce a set that is very large or unbounded.
A useful illustration is the one-instrument, scalar reduced-form version. Let \(\hat\gamma_y\) and \(\hat\gamma_x\) be the estimated coefficients of the instrument in the reduced-form regressions for \(Y\) and \(X\). Under the null value \(\beta\), the restriction is \(\gamma_y-\beta\gamma_x=0\). Suppose their estimated covariance matrix has entries \(V_{yy},V_{yx},V_{xx}\). A one-degree-of-freedom asymptotic test uses the squared standardized restriction. At the 5% level, the critical value is approximately 3.841. The candidate value is accepted when
The right-hand variance is the variance of \(\hat\gamma_y-\beta\hat\gamma_x\). Moving all terms to the left gives a quadratic inequality \(A\beta^2+B\beta+C\leq 0\), when the covariance estimate is held fixed. Its roots and the sign of its leading coefficient determine the accepted values. This algebra is an illustration for the stated scalar asymptotic test; other AR implementations may require numerical inversion rather than this exact quadratic form.
How the Quadratic Determines the Shape
For \(Q(\beta)=A\beta^2+B\beta+C\), the accepted set in this illustration is \(\{\beta:Q(\beta)\leq0\}\). When \(A\) is positive and there are two real roots, the parabola opens upward and is nonpositive between its roots, giving a bounded interval. When \(A\) is negative and there are two real roots, it opens downward and is nonpositive outside the roots, giving two unbounded pieces. If the quadratic has no real roots, its sign is constant; it can yield an empty set or the whole real line. A linear inequality can produce a single unbounded ray. The equality points are included because the rule uses “does not reject,” including the test's critical boundary.
Use the test statistic and critical value for the specified level and reference distribution.
When the rule reduces to a quadratic, solve \(A\beta^2+B\beta+C=0\); otherwise find the acceptance boundaries numerically.
Test a point in each region separated by the boundaries, and verify whether equality is accepted under the test rule.
State all accepted components and any excluded gaps, as well as the confidence level and assumptions.
Worked Example: A Bounded Set
Suppose a hypothetical one-instrument analysis reports \(\hat\gamma_y=1.5\) outcome units per instrument unit and \(\hat\gamma_x=4\) treatment units per instrument unit. Let \(V_{yx}=0\), \(V_{yy}=2.09/3.841\), and \(V_{xx}=7.384/3.841\). These covariance values are positive, and the diagonal covariance matrix is positive definite. Use the 5% rule above.
Substituting into the acceptance inequality and multiplying out gives
The discriminant is \(D=(-12)^2-4(8.616)(0.16)=138.48576\). Thus the roots are
Since the coefficient of \(\beta^2\) is positive, the quadratic is nonpositive between the roots. The 95% confidence set is approximately \([0.01346,1.379]\) in the units of the structural coefficient. It is bounded in this example, but the interval is still broad; boundedness alone does not mean the estimate is precise. The displayed coefficients use the unrounded covariance values specified above.
Worked Example: A Disconnected Set
For a second hypothetical analysis, take \(\hat\gamma_y=-3\), \(\hat\gamma_x=1\), \(V_{yx}=0\), \(V_{yy}=11/3.841\), and \(V_{xx}=3/3.841\). Both variances are positive, so this diagonal covariance matrix is also positive definite. The same 5% acceptance rule becomes
The roots solve \(2\beta^2-6\beta+2=0\), so
The quadratic opens downward. It is positive between its roots and nonpositive outside them, so the confidence set is approximately \((-\infty,0.3820]\cup[2.618,\infty)\). For a direct check, \(Q(0)=-2\), which is accepted, while \(Q(1)=2\), which is rejected. Reporting only the convex hull \((-\infty,\infty)\) would conceal that the test rejects the entire middle region.
Worked Example: An Unbounded Set That Is the Whole Line
In a third hypothetical analysis, let \(\hat\gamma_y=0.2\), \(\hat\gamma_x=0.1\), \(V_{yx}=0\), \(V_{yy}=1\), and \(V_{xx}=4\). The covariance matrix is positive definite. The acceptance inequality is
The leading coefficient is negative, and the discriminant is \(D=(-0.04)^2-4(-15.354)(-3.801)<0\). There are no real roots. Because \(Q(0)=-3.801<0\), the quadratic is negative everywhere, so every real \(\beta\) is accepted. The 95% confidence set is \(\mathbb{R}\), the whole real line.
In this example, the estimated reduced-form coefficient for \(X\) is small relative to its standard error: \(\sqrt{V_{xx}}=2\), whereas \(\hat\gamma_x=0.1\). That is consistent with the set being uninformative about the structural coefficient. It does not establish that the instrument is invalid, nor does it show that every effect value is equally plausible.
What Shape Reveals—and What It Does Not
A bounded set means that the test rejects sufficiently extreme values in both directions. An unbounded ray or a disconnected set means that at least one direction is not bounded by the test's rejection region. The whole real line means the test rejects none of the candidate values at that level. These outcomes often arise when the data provide weak information about the ratio linking the outcome and treatment reduced forms, but the shape alone is not a formal test of instrument strength or validity.
A disconnected set deserves particular care. Values in the gap are rejected, so replacing the set by its convex hull includes values the test rules out. Conversely, choosing just one component because it is closer to a conventional point estimate discards the other accepted values. Report the actual set and explain its components.
| Set shape | What the inverted test says | Reporting caution |
|---|---|---|
| Bounded interval | Values outside finite boundaries are rejected at the stated level. | A finite set can still be wide and imprecise. |
| Unbounded ray or whole line | Some or all extreme values remain unrejected. | Do not invent a finite endpoint or present a conventional interval as if it were robust. |
| Disconnected components | Separated ranges are accepted, with a rejected gap between them. | Do not report only the endpoints' convex hull or omit an accepted component. |
| Empty set | Every candidate value is rejected by this procedure. | Investigate model specification, test calibration, and assumptions; do not interpret it as proof that no parameter exists. |
Conventional IV Wald intervals are based on an estimated coefficient and its standard error. They can look narrow even when the first stage is weak, which is why they may fail to reflect weak-instrument uncertainty. AR inversion instead tests the structural null without relying on division by the first-stage estimate, at the cost of potentially wide or irregular sets. First-stage diagnostics describe relevance, but they do not replace confidence-set inference about \(\beta\). The introductory material on Anderson–Rubin tests covers the test itself; the comparison here concerns the information conveyed by the inverted set.
Diagnostics and Common Pitfalls
For computed sets, evaluate the test statistic or acceptance inequality at several points: just inside and outside each reported boundary, in every gap, and far into each tail. A dense grid is useful for finding missed components, but it does not prove that no narrow component exists; use a reliable root-finding or inversion method and verify boundaries directly. Report the test version, covariance estimator, reference distribution, confidence level, and any covariates or clustering used. If a set is unbounded or unexpectedly empty, check calculations and assumptions as well as the first-stage evidence.
- Mistake: “The confidence set is unbounded, so the effect must be infinite.” Correction: it means the test does not reject arbitrarily large values in at least one direction; the data have not supplied a finite bound under this procedure.
- Mistake: “The endpoints give an ordinary interval, so all values between them are accepted.” Correction: that is false for a disconnected set. State every accepted component and the rejected gaps.
- Mistake: “The entire real line is accepted, so the instrument is proven invalid.” Correction: the set shows that this test does not exclude any candidate effect at the chosen level. It does not determine why the test is uninformative or establish an instrument violation.
- Mistake: “A bounded AR set guarantees a strong instrument.” Correction: boundedness describes this test inversion; assess instrument strength with relevant diagnostics and examine assumptions separately.
Check Your Understanding
Answer each question, then compare with the short solutions below.
- For \(Q(\beta)=2\beta^2-8\beta+6\leq0\), find the accepted set and classify its shape.
- For \(Q(\beta)=-\beta^2+4\beta-5\leq0\), determine whether the accepted set is empty, bounded, or the whole real line.
- Why is reporting the convex hull of a disconnected confidence set potentially misleading?
- What does an AR confidence set equal to the whole real line establish, and what does it not establish?
Solutions. (1) Factor \(Q(\beta)=2(\beta-1)(\beta-3)\). The upward-opening quadratic is nonpositive between its roots, so the set is \([1,3]\), bounded. (2) The discriminant is \(4^2-4(-1)(-5)=-4\). The quadratic opens downward and \(Q(0)=-5<0\), so it is negative everywhere; the set is the whole real line. (3) The convex hull fills in the gap, including values the test rejects. (4) It establishes that this test does not reject any real candidate value at the chosen level. It does not prove that all effects are equally plausible, that the instrument is invalid, or that the assumptions hold.