Inverting Anderson-Rubin Tests into Confidence Sets

Why Invert a Test?

A confidence set for an unknown coefficient can be constructed by testing each possible value of that coefficient and retaining the values that the test does not reject. This is particularly useful with weak instruments: rather than relying on a normal approximation for an instrumental-variables estimator, inversion carries the Anderson–Rubin (AR) test's protection against weak identification into the confidence set. The resulting set is not guaranteed to resemble a familiar estimate plus or minus a margin of error.

We work with the scalar instrumental-variables model $y_i=\beta x_i+u_i$, where $Z$ is an $n\times q$ matrix of excluded instruments. Assume $Z$ has full column rank $q$, with $n>q$. There are no additional regressors in the notation below. Write $P_Z=Z(Z'Z)^{-1}Z'$ for the projection onto the instrument columns and $M_Z=I-P_Z$ for the complementary projection.

Definition: For a candidate value $b$ of the structural coefficient, let $r(b)=y-bx$. The Anderson–Rubin statistic is $$ F_{\mathrm{AR}}(b)= \frac{r(b)'P_Zr(b)/q}{r(b)'M_Zr(b)/(n-q)}, $$ when the denominator is positive. At test level $\alpha$, the inverted AR confidence set is the collection of candidate values not rejected: $$ \mathcal C_{1-\alpha}=\{b:F_{\mathrm{AR}}(b)\leq c_{q,n-q}(1-\alpha)\}, $$ where $c_{q,n-q}(1-\alpha)$ is the $1-\alpha$ quantile of the $F_{q,n-q}$ distribution.

For a candidate $b$, the test asks whether the instruments explain the reduced-form outcome $y-bx$. Under the null hypothesis $b=\beta$, this outcome is the structural error $u$. That observation is the source of the test's weak-instrument robustness: the null distribution does not depend on how strongly the instruments predict $x$.

Exact Test and Coverage

For the classical finite-sample result, assume that, conditional on $Z$, the errors satisfy $u\sim N(0,\sigma^2I)$ for some $\sigma^2>0$, and that the instruments are exogenous. The regressor $x$ may be endogenous and may be weakly related to $Z$. For a fixed candidate $b$, use the cross-multiplied acceptance criterion $$ (n-q)r(b)'P_Zr(b)-q\,c_{q,n-q}(1-\alpha)\,r(b)'M_Zr(b)\leq 0. $$ When $r(b)'M_Zr(b)>0$, this is exactly $F_{\mathrm{AR}}(b)\leq c_{q,n-q}(1-\alpha)$. The criterion also specifies the boundary case of a zero denominator; under the stated continuous-error model, that case has probability zero at the true parameter.

Theorem: Under the stated rank and Gaussian-error assumptions, the inverted set $\mathcal C_{1-\alpha}$ contains the true coefficient $\beta$ with probability $1-\alpha$. This probability statement holds for every first-stage coefficient, including zero.

Proof. At the true value, $r(\beta)=y-\beta x=u$. Conditional on $Z$, the matrices $P_Z$ and $M_Z$ are symmetric, idempotent, orthogonal projections with ranks $q$ and $n-q$, respectively. Since $u\sim N(0,\sigma^2I)$, $$ \frac{u'P_Zu}{\sigma^2}\sim\chi^2_q, \qquad \frac{u'M_Zu}{\sigma^2}\sim\chi^2_{n-q}. $$ The two quadratic forms are independent because the projections are orthogonal: $P_ZM_Z=0$. Consequently, $$ \frac{u'P_Zu/q}{u'M_Zu/(n-q)}\sim F_{q,n-q}. $$ The test therefore rejects the true value with probability $\alpha$, conditional on $Z$, and hence also unconditionally. The true value belongs to the inverted set precisely when its test does not reject, giving coverage $1-\alpha$. Neither the distribution calculation nor the resulting coverage uses the strength of the first stage. $\square$

This is an exact finite-sample result under the specified Gaussian, homoskedastic model. Other error assumptions require an appropriate test calibration; the exact $F$ claim should not be transferred automatically to them. The essential construction remains inversion: the confidence set consists of all candidate values accepted by the chosen AR test.

The Shape of the Inverted Set

The acceptance condition is a quadratic inequality in $b$. To see this, define $N(b)=r(b)'P_Zr(b)$, $D(b)=r(b)'M_Zr(b)$, and let $c=c_{q,n-q}(1-\alpha)$. Since $r(b)=y-bx$, both $N(b)$ and $D(b)$ are quadratic functions. Thus the set is determined by $$ (n-q)N(b)-qcD(b)\leq 0. $$ Expanding gives a polynomial of degree at most two. Depending on its coefficients and roots, its solution can be a bounded interval, two unbounded rays, a half-line, a single point, the empty set, or the entire real line. In particular, there is no general requirement that an inverted confidence set be a single bounded interval.

Worked Example: A Bounded Confidence Set

Take $n=3$, $q=1$, and $Z=(1,0,0)'$. Use the cutoff $c=1$ to illustrate the geometry; this is a generic test cutoff, not a claim that it is the usual 5% critical value. Let $y=(0,1,0)'$ and $x=(-1,0,0)'$. For candidate $b$, $$ r(b)=y-bx=(b,1,0)'. $$ The instrument component is $b$, and the orthogonal residual component is $(1,0)'$. Therefore $N(b)=b^2$, $D(b)=1$, and $$ F_{\mathrm{AR}}(b)=\frac{b^2}{1/2}=2b^2. $$ The acceptance condition $2b^2\leq1$ is equivalent to $|b|\leq1/\sqrt{2}$. The inverted set is the bounded interval $[-1/\sqrt{2},1/\sqrt{2}]$.

Worked Example: An Unbounded, Disconnected Set

Keep $n=3$, $q=1$, $Z=(1,0,0)'$, and cutoff $c=1$. Let $y=(1,0,1)'$ and $x=(0,-1,0)'$. Then $$ r(b)=y-bx=(1,b,1)'. $$ The instrument component is $1$, while the orthogonal component is $(b,1)'$. Hence $N(b)=1$ and $D(b)=b^2+1$, which is positive for every $b$. The statistic and acceptance condition are $$ F_{\mathrm{AR}}(b)=\frac{1}{(b^2+1)/2}=\frac{2}{b^2+1}, \qquad \frac{2}{b^2+1}\leq1. $$ Multiplying by the positive quantity $b^2+1$ gives $b^2\geq1$. Therefore the set is $(-\infty,-1]\cup[1,\infty)$: it is disconnected and unbounded. There is no contradiction in a confidence set excluding the values between its two accepted regions; each value is assessed by its own test.

Worked Example: The Entire Parameter Space

Again take $n=3$, $q=1$, $Z=(1,0,0)'$, and $c=1$. Set $y=(0,1,0)'$ and $x=(0,0,-1)'$. For every candidate $b$, $$ r(b)=y-bx=(0,1,b)'. $$ There is no component in the instrument direction, so $N(b)=0$, while $D(b)=1+b^2>0$. Thus $F_{\mathrm{AR}}(b)=0$ for every real $b$, and every candidate is accepted: $$ \mathcal C=\mathbb R. $$ A confidence set equal to the entire parameter space may be uninformative about the coefficient, but it accurately reflects that these data and this test do not rule out any value.

What Weak Identification Changes—and What It Does Not

If the instruments provide little information about $x$, AR inversion can yield a very wide set, or a set with unbounded components. If the data do not reject any candidate, it can yield the entire parameter space. These outcomes are not failures of the inversion procedure. They signal that the test has not found evidence against those candidate values. Weak instruments can make precise identification difficult; they do not justify reporting a narrow set merely because a conventional estimator and standard error produce one.

A practical way to compute the set is to form the quadratic polynomial in the acceptance condition, find its real roots, and test the sign on the intervals between roots. If its leading coefficient is zero, the inequality may instead be linear or constant. This calculation also avoids assuming in advance that the answer has two endpoints.

Choose a candidate value.
For each $b$, construct the transformed outcome $r(b)=y-bx$.

Compute the AR statistic.
Compare its instrument-explained component with its residual component, using the chosen test calibration.

Retain non-rejected values.
The collection of all retained candidates is the inverted confidence set; do not impose interval shape or boundedness.

Common Pitfall

Wrong reasoning: “A confidence set must be an interval around an estimate, so if inversion produces two separated regions or an unbounded region, one region must be discarded.” This imports the shape of a familiar Wald interval into a different procedure. Correction: the AR confidence set is defined by test non-rejection, not by distance from an estimator. Every candidate satisfying the AR acceptance inequality belongs in the set. Disconnected or unbounded results can therefore be the correct report, especially when the data provide limited information about the structural coefficient.

Key takeaway: Inverting an AR test means retaining every parameter value that the test does not reject. Under the classical finite-sample assumptions, coverage does not depend on instrument strength, but the resulting set can be wide, unbounded, disconnected, or equal to the whole parameter space.

Check Your Understanding

Use the definitions and calculations above to check your understanding.

Why does the null distribution of the AR statistic at the true coefficient not depend on the first-stage coefficient?
For $q=1$ and $n=3$, write the acceptance inequality using $N(b)$, $D(b)$, and a cutoff $c$.
In the disconnected-set example, why is it valid to multiply the acceptance inequality by $b^2+1$?
What does an inverted set equal to the entire real line say about the candidate values tested?
Why is it incorrect to force every AR confidence set to be a bounded interval around an instrumental-variables estimate?