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Weak-Instrument Diagnostics · Tutorial 537 of 699

Fieller Geometry for Weak-IV Confidence Sets

Learn how the geometry of a ratio explains weak-IV confidence sets and how to classify their shape from the coefficients of a quadratic inequality.

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What You'll Learn

  • Express an IV ratio confidence set as an inequality for a reduced-form contrast.
  • Derive the Fieller quadratic and connect it to lines meeting an ellipse.
  • Classify bounded, disconnected, half-line, and all-real confidence sets.
  • Calculate example confidence sets using estimates, standard errors, and covariance.
  • Diagnose weak-denominator problems and distinguish Fieller inference from a Wald interval.

Why a Ratio Can Have an Unusual Confidence Set

An instrumental-variables (IV) estimand in a just-identified linear model is a ratio: a reduced-form effect on the outcome divided by a first-stage effect on the treatment. When the first-stage estimate is imprecise, plausible denominator values can be close to zero or can lie on both sides of zero. In that case, uncertainty in the ratio is not well represented by a symmetric interval around the estimated ratio. Fieller’s method makes the geometry of this problem explicit.

The key question is not simply how far the ratio estimate lies from its target. It is whether, for a proposed ratio, the estimated reduced-form coefficients are consistent with the linear restriction that the outcome reduced form equals that ratio times the treatment first stage. Inverting tests of that restriction gives the Fieller confidence set. Depending on the strength and covariance of the two estimates, the set may be bounded, disconnected, unbounded, or all of the real line.

Definition: Let \(\hat\beta_{RF}\) be an estimate of the outcome reduced-form coefficient and \(\hat\pi_{RF}\) an estimate of the treatment first-stage coefficient. The just-identified IV ratio estimate is \(\hat\theta=\hat\beta_{RF}/\hat\pi_{RF}\), when \(\hat\pi_{RF}\ne0\). A Fieller confidence set for \(\theta\) contains the candidate values \(\theta_0\) that are not rejected by a confidence test of \(H_0:\beta_{RF}-\theta_0\pi_{RF}=0\).

Assume the pair \((\hat\beta_{RF},\hat\pi_{RF})\) is jointly normal, or sufficiently well approximated by a joint normal distribution, with estimated covariance matrix \(\begin{pmatrix} \widehat V_\beta & \widehat C\\ \widehat C & \widehat V_\pi \end{pmatrix}\). The variances and covariance must be estimated for the same sample and model. For a two-sided confidence level \(1-\alpha\), let \(c\) be the corresponding standard-normal critical value; for example, \(c=1.96\) for a 95% large-sample interval. A \(t\) critical value can be used when justified by the sampling procedure. The scalar contrast has estimated variance \(\widehat V_\beta-2\theta_0\widehat C+\theta_0^2\widehat V_\pi\).

From the Test to the Quadratic

For each proposed \(\theta_0\), the null is retained when the estimated contrast is no more than \(c\) estimated standard errors from zero. Squaring both sides yields the inversion rule. The quadratic form is the source of Fieller geometry; it is not an approximation made by dividing endpoints of separate intervals.

$$ (\hat\beta_{RF}-\theta_0\hat\pi_{RF})^2 \leq c^2(\widehat V_\beta-2\theta_0\widehat C+\theta_0^2\widehat V_\pi). $$

Rearranging gives \(A\theta_0^2+B\theta_0+C_0\leq0\), where

$$ A=\hat\pi_{RF}^{\,2}-c^2\widehat V_\pi,\qquad B=-2\hat\beta_{RF}\hat\pi_{RF}+2c^2\widehat C,\qquad C_0=\hat\beta_{RF}^{\,2}-c^2\widehat V_\beta. $$

The discriminant \(\Delta=B^2-4AC_0\) and the sign of \(A\) determine the set. If \(A>0\) and \(\Delta\geq0\), the quadratic opens upward and the set is the interval between its roots. If \(A<0\) and \(\Delta>0\), it opens downward and the set consists of the two exterior rays. If \(A<0\) and \(\Delta<0\), it is negative for every real value, so the set is all real numbers. When \(A=0\), the inequality is linear unless \(B=0\), and can produce a half-line. A zero discriminant gives a repeated root; the inequality must still be evaluated with its direction rather than assumed to be an ordinary interval.

Geometric interpretation: In the plane with horizontal coordinate \(\pi_{RF}\) and vertical coordinate \(\beta_{RF}\), each candidate ratio \(\theta_0\) defines a line through the origin, \(\beta_{RF}=\theta_0\pi_{RF}\). The Fieller inequality says that this line comes within the test’s covariance-scaled distance of the estimated coefficient pair. Equivalently, it intersects the corresponding acceptance ellipse. The accepted slopes can therefore include widely separated rays or every possible slope when the ellipse reaches the origin.

The ellipse in this interpretation is the contour defined by the contrast-test critical value. It should not automatically be described as a joint 95% confidence ellipse: a critical value chosen for a one-dimensional contrast does not in general give 95% joint coverage for the two coefficients. The confidence statement concerns each scalar ratio restriction, and the set follows by inverting those tests.

Worked Fieller Calculations

Worked Example: A Bounded Set When the First Stage Is Precise

Suppose a hypothetical analysis estimates an outcome reduced form of \(\hat\beta_{RF}=0.30\) outcome units per instrument unit and a first stage of \(\hat\pi_{RF}=0.20\) treatment units per instrument unit. Let \(SE(\hat\beta_{RF})=0.08\), \(SE(\hat\pi_{RF})=0.05\), and \(\widehat C=0.001\). Use \(c=1.96\), so \(c^2=3.8416\). The covariance matrix is admissible: its determinant is \(0.0064(0.0025)-0.001^2=0.000015>0\).

The variances are \(\widehat V_\beta=0.08^2=0.0064\) and \(\widehat V_\pi=0.05^2=0.0025\). The coefficients are

$$ A=0.20^2-3.8416(0.0025)=0.030396, $$
$$ B=-2(0.30)(0.20)+2(3.8416)(0.001)=-0.1123168, $$
$$ C_0=0.30^2-3.8416(0.0064)=0.06541376. $$

Here \(A>0\), and \(\Delta=(-0.1123168)^2-4(0.030396)(0.06541376)\approx0.0046618>0\). Thus the accepted values lie between the two real roots:

$$ \frac{-B\pm\sqrt{\Delta}}{2A} = \frac{0.1123168\pm\sqrt{0.0046618}}{0.060792} \approx 0.7244,\ 2.971. $$

The Fieller 95% set is approximately \([0.7244,2.971]\) outcome units per treatment unit. The ratio estimate itself is \(0.30/0.20=1.50\) in those units, but the interval is obtained by test inversion, not by treating that ratio as normally distributed.

Worked Example: A Disconnected Set With an Uncertain Denominator

Now suppose the hypothetical estimates are \(\hat\beta_{RF}=0.10\) outcome units per instrument unit and \(\hat\pi_{RF}=0.03\) treatment units per instrument unit, with standard errors \(0.04\) and \(0.05\), respectively, and covariance zero. Again take \(c=1.96\). The covariance matrix is positive definite because both variances are positive and the covariance is zero.

The variances are \(0.0016\) and \(0.0025\). Substitution gives

$$ A=0.03^2-3.8416(0.0025)=-0.008704, $$
$$ B=-2(0.10)(0.03)=-0.006, \qquad C_0=0.10^2-3.8416(0.0016)=0.00385344. $$

The discriminant is \(\Delta=(-0.006)^2-4(-0.008704)(0.00385344)\approx0.0001701614\), which is positive. The roots are

$$ \frac{-B\pm\sqrt{\Delta}}{2A} = \frac{0.006\pm\sqrt{0.0001701614}}{-0.017408} \approx -1.094,\ 0.4047. $$

Because \(A<0\), the inequality holds outside the roots: the confidence set is approximately \((-\infty,-1.094]\cup[0.4047,\infty)\), in outcome units per treatment unit. The point estimate is \(0.10/0.03\approx3.333\), which belongs to the set. The disconnected shape is not a computational defect: the first-stage estimate is small relative to its uncertainty, so ratios with large positive or negative magnitudes cannot be ruled out.

Worked Example: The Entire Real Line Is Retained

For a third hypothetical calculation, let \(\hat\beta_{RF}=0.02\), \(\hat\pi_{RF}=0.03\), both standard errors equal \(0.05\), and \(\widehat C=0\). Use \(c=1.96\). The covariance matrix is positive definite, and the estimates are both small relative to their standard errors.

The coefficient calculations are

$$ A=0.03^2-3.8416(0.05^2)=-0.008704,\qquad B=-2(0.02)(0.03)=-0.0012, $$
$$ C_0=0.02^2-3.8416(0.05^2)=-0.009204. $$

Then \(\Delta=(-0.0012)^2-4(-0.008704)(-0.009204)\approx-0.0003190065<0\). The quadratic opens downward and has no real roots, so it is negative for every real \(\theta_0\). Therefore every value is retained: the confidence set is \((-\infty,\infty)\) outcome units per treatment unit. This is a valid statement of how little the data identify the ratio, not evidence that every value is equally plausible.

Why the Shape Changes

The sign of \(A=\hat\pi_{RF}^{\,2}-c^2\widehat V_\pi\) has a direct diagnostic interpretation. If \(A>0\), the first-stage estimate is far enough from zero, on the chosen critical-value scale, to exclude zero in its corresponding two-sided interval. That condition produces an upward-opening quadratic and, when real roots exist, a bounded set. If \(A<0\), zero remains plausible for the denominator. Large-magnitude ratios can then satisfy the test, leading to an unbounded set. Whether the set is two rays or all real values depends on the discriminant and the other coefficient estimates.

This gives a short justification for why conventional ratio intervals can fail. A Wald interval approximates the ratio by a locally linear function of the two estimates; that approximation relies on the denominator being sufficiently separated from zero. The Fieller inversion does not require that separation. Its unusual shapes communicate the lack of denominator information rather than concealing it in an apparently ordinary interval.

In weak-IV analysis, Anderson–Rubin confidence sets also arise by inverting tests of candidate structural effects. Under the just-identified linear setup, the same reduced-form restriction explains the connection to the Fieller quadratic. With multiple instruments, covariates, or robust test constructions, the inverted set need not reduce to this particular two-coefficient quadratic. The appropriate test and variance estimator must match the design; the geometric lesson remains that weak denominator information can make a ratio set nonstandard.

Diagnostics, Alternatives, and Pitfalls

Before reporting a Fieller set, verify that the estimates are for the same specification and sample, that the covariance term is included with the correct sign, and that the covariance matrix is valid. Then compute \(A\), \(B\), \(C_0\), and \(\Delta\) from unrounded inputs. Check the roots by substituting them into the quadratic, and check the inequality on a point between or outside the roots to determine which region is accepted. If \(A\) is near zero, small numerical changes can change the apparent shape, so report adequate precision and avoid interpreting rounded coefficients as exact.

A common mistake is to divide the endpoints of a confidence interval for \(\hat\beta_{RF}\) by the endpoints of an interval for \(\hat\pi_{RF}\), or to report \(\hat\theta\pm1.96SE(\hat\theta)\) without assessing the denominator. The first procedure does not generally produce the Fieller set, and the second can be badly misleading when the first stage is weak. The correction is to test each candidate ratio using the contrast \(\hat\beta_{RF}-\theta_0\hat\pi_{RF}\), include its covariance-adjusted variance, and invert that test. An unbounded result should not be “repaired” by truncating it to plausible-looking values; any scientifically motivated restriction should be stated separately from the statistical confidence set.

Fieller geometry is most directly useful for understanding the shape and interpretation of a just-identified ratio confidence set. It does not by itself resolve instrument validity, model misspecification, or the choice of instruments, and its stated coverage depends on the normal or large-sample approximation and variance estimator used. It complements, rather than replaces, weak-IV-robust tests suited to the actual design.

Check Your Understanding

Use the quadratic rule and the geometry to answer each question. Solutions follow the questions.

  1. In the quadratic \(A\theta^2+B\theta+C_0\leq0\), what shape is obtained when \(A<0\) and there are two distinct real roots?
  2. Why does \(A<0\) indicate that the denominator estimate is not separated from zero at the chosen critical-value scale?
  3. For \(A=-2\), \(B=0\), and \(C_0=1\), determine the accepted set.
  4. For \(A=1\), \(B=0\), and \(C_0=1\), determine the accepted set and explain why.

Solutions. (1) The parabola opens downward, so it is nonpositive outside its two roots; the set is two exterior rays. (2) Since \(A=\hat\pi_{RF}^{\,2}-c^2\widehat V_\pi\), \(A<0\) means \(|\hat\pi_{RF}|<c\sqrt{\widehat V_\pi}\), so the corresponding two-sided interval for the first-stage coefficient includes zero. (3) The expression is \(-2\theta^2+1\leq0\), equivalent to \(\theta^2\geq1/2\), so the set is \((-\infty,-1/\sqrt{2}]\cup[1/\sqrt{2},\infty)\). (4) The expression is \(\theta^2+1\), which is strictly positive for every real \(\theta\); the accepted set is empty. Such coefficients cannot arise from a nonempty acceptance region under the usual valid Fieller setup, but the calculation illustrates why the inequality must be checked rather than inferred from the sign of \(A\) alone.

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