Constructing Conditional Likelihood Ratio Confidence Sets

From a Conditional Test to a Confidence Set

A weak first stage can make conventional instrumental-variable confidence intervals unreliable. A confidence set obtained by inverting a test designed for weak instruments offers a different approach: test each possible value of the structural effect, then retain the values that the test does not reject. For the conditional likelihood ratio (CLR) test, this inversion can produce more than one interval, an unbounded set, or even an empty set. Its shape is part of the result, not a technical detail to hide.

Consider an outcome $Y_i$, an endogenous treatment $X_i$, included covariates $W_i$, and excluded instruments $Z_i$. The target is the coefficient $\beta$ in the structural equation. For clarity, write the model as

Y_i = \beta X_i + W_i'\gamma + \varepsilon_i, \qquad X_i = Z_i'\pi + W_i'\delta + v_i.

The CLR test evaluates a null hypothesis $H_0:\beta=\beta_0$ for a proposed value $\beta_0$. Its conditional likelihood ratio statistic compares the maximized likelihood under the null with the maximized likelihood under the alternative, conditional on a statistic that removes the nuisance dependence on the unknown first-stage coefficients. Write the statistic as $\mathrm{CLR}(\beta_0)$, and its conditional critical value as $c_{1-\alpha}\{t(\beta_0)\}$, where $t(\beta_0)$ denotes the conditioning statistic for that null. The test rejects when the statistic exceeds its conditional critical value. Equivalently, it rejects when its conditional p-value $p_{\mathrm{CLR}}(\beta_0)$ is below $\alpha$.

Definition: The $1-\alpha$ CLR confidence set is the set of all structural-effect values that the level-$\alpha$ CLR test does not reject: $$ \mathcal{C}_{1-\alpha} = \{\beta_0: p_{\mathrm{CLR}}(\beta_0)\geq \alpha\} = \{\beta_0: \mathrm{CLR}(\beta_0)\leq c_{1-\alpha}\{t(\beta_0)\}\}. $$

The conditioning statistic and hence the critical value can depend on $\beta_0$. Therefore, inversion does not generally mean comparing every candidate value to one fixed cutoff. A reliable implementation evaluates the appropriate conditional test separately at each proposed value.

Assumptions and the Coverage Argument

The usual CLR justification is developed for the linear IV reduced-form model with fixed or exogenous instruments, a correctly specified linear structural equation, and jointly normal reduced-form errors with a common covariance matrix across observations. The instrument and covariate design must have the required rank so the reduced forms can be estimated. The covariance matrix must be nonsingular. The instruments must satisfy the relevant exclusion and exogeneity conditions for the structural interpretation of $\beta$. Crucially, the CLR procedure is designed not to require the first-stage coefficient $\pi$ to be large.

These assumptions matter in different ways. The conditional test's reference distribution relies on its reduced-form distributional model; the causal interpretation additionally relies on instrument validity and the structural model. Weak-IV robustness does not repair invalid instruments, misspecified equations, or dependence and heteroskedasticity that the chosen test procedure does not accommodate. Use a version of the test whose variance and sampling assumptions match the design.

The inversion argument itself is short. Suppose the test has size at most $\alpha$ for every value of the nuisance parameters under $H_0:\beta=\beta_0$. At the true effect $\beta$, the event that $\beta$ belongs to the inverted set is exactly the event that the test does not reject that true-value null. Thus:

P_{\beta}\{\beta\in\mathcal{C}_{1-\alpha}\} = P_{\beta}\{p_{\mathrm{CLR}}(\beta)\geq\alpha\} \geq 1-\alpha.

If the test has exact size $\alpha$, the coverage probability is $1-\alpha$; if it is conservative, coverage is at least $1-\alpha$. This is a repeated-sampling statement about the procedure under its assumptions. It does not mean there is a $1-\alpha$ probability that a fixed parameter lies in the particular set computed from the observed data.

Worked Example: A Bounded CLR Confidence Set

Suppose a hypothetical analysis has treatment measured in hours and outcome measured in points. At $\alpha=0.05$, a CLR routine evaluates conditional p-values over candidate values of $\beta_0$. Selected hypothetical results are shown below. They are test outputs, not estimates from a real study.

Candidate $\beta_0$ (points per hour)	Conditional CLR p-value	Decision at 0.05
0	0.003	Reject
1	0.041	Reject
2	0.180	Do not reject
3	0.620	Do not reject
4	0.090	Do not reject
5	0.012	Reject

The candidate values with p-values at least 0.05 are retained. The endpoints lie between the adjacent grid values where the decision changes. To illustrate a linear interpolation of the p-value curve near the first crossing, use the values at 1 and 2:

\beta_{\mathrm{lower}} \approx 1+\frac{0.050-0.041}{0.180-0.041}(2-1) = 1+\frac{0.009}{0.139} \approx 1.065.

At the upper crossing, interpolate between 4 and 5:

\beta_{\mathrm{upper}} \approx 4+\frac{0.090-0.050}{0.090-0.012}(5-4) = 4+\frac{0.040}{0.078} \approx 4.513.

This gives an approximate confidence set from 1.065 to 4.513 points per hour. The interpolation is a numerical approximation, not part of the CLR theory; in practice, refine the grid or use a root-finding routine on the test's p-value or statistic-minus-critical-value function. The set does not establish that the true effect is inside with probability 0.95. It says the values in this range are not rejected by these level-0.05 tests.

Finding Endpoints Without Assuming a Single Interval

To compute a set, evaluate the test over a range of candidate $\beta_0$ values, identify every transition between rejection and non-rejection, and refine each transition. An endpoint ordinarily solves

p_{\mathrm{CLR}}(\beta_0)=\alpha, \qquad\text{or equivalently}\qquad \mathrm{CLR}(\beta_0)=c_{1-\alpha}\{t(\beta_0)\}.

This is a numerical root-finding problem. Bisection is useful once a crossing has been bracketed: if one endpoint of a bracket is rejected and the other is not, evaluate the midpoint, retain the half-bracket containing the transition, and repeat until the desired precision is reached. Since the p-value need not be monotone in $\beta_0$, search for all crossings rather than presuming there are exactly two.

Specify the target and level.
Set the structural parameter, its units, and the desired significance level, such as $\alpha=0.05$ for a nominal 95% set.

Evaluate the conditional test.
For each candidate value, compute the CLR statistic, the critical value appropriate to that null's conditioning statistic, and the conditional p-value.

Mark retained and rejected values.
Retain candidates with p-values at least $\alpha$, using the test's stated boundary convention.

Refine every boundary and inspect the tails.
Numerically locate all crossings, then check whether extreme candidate values are retained or rejected before reporting the set's shape.

A finite grid alone cannot establish that a set is bounded. If all candidates at the edge of the grid are retained, expand the search and assess tail behavior using the test function and appropriate numerical checks. If the set has separate retained stretches, report their union rather than filling in the gaps.

Worked Example: Disconnected and Unbounded Regions

Consider a second hypothetical CLR calculation at the 5% level, with $\beta_0$ again measured in points per hour. The evaluated p-values are 0.21 at $-8$, 0.09 at $-2$, 0.03 at $-1$, 0.01 at 0, 0.08 at 1, 0.19 at 2, and 0.41 at 8. Suppose further numerical tail checks establish that the p-value remains above 0.05 sufficiently far in both directions. That tail check is essential: the finite grid alone would not prove unboundedness.

The first crossing is between $-2$ and $-1$. Linear interpolation gives

\beta_1 \approx -2+\frac{0.090-0.050}{0.090-0.030}(-1-(-2)) = -2+\frac{0.040}{0.060} \approx -1.333.

The next crossing is between 0 and 1:

\beta_2 \approx 0+\frac{0.050-0.010}{0.080-0.010}(1-0) = \frac{0.040}{0.070} \approx 0.571.

The retained regions are approximately $(-\infty,-1.333]$ and $[0.571,\infty)$, with boundary inclusion understood according to the test convention. The rejected values form the gap between them. This result is not meaningfully summarized by one conventional “lower limit to upper limit” interval: doing so would include values that the test rejects. Both unbounded components should be reported, together with the units and confidence level.

Worked Example: An Empty Inverted Set

In a third hypothetical calculation, suppose a careful search over the full parameter range finds that the largest conditional CLR p-value is 0.043. At $\alpha=0.05$, every candidate value is rejected because $0.043<0.05$. The inverted 95% set is therefore empty:

\mathcal{C}_{0.95} = \{\beta_0:p_{\mathrm{CLR}}(\beta_0)\geq 0.05\} = \varnothing.

This is a possible outcome of test inversion; it is not a zero-width interval and should not be rewritten as one. It indicates that, under the test and model used, no candidate value survives at the chosen level. It can prompt checks of data coding, model specification, instrument validity, and numerical implementation. It does not by itself identify which assumption failed, and a finite grid whose largest p-value is 0.043 would not suffice unless the search also established that no unexamined region contains retained values.

Diagnostics, Comparisons, and Limitations

Several checks are especially important when constructing the set. First, confirm that each candidate's test uses its own conditional calculation; a single cutoff reused across values can change the result. Second, use a sufficiently fine grid and refine all crossings. Third, search both tails, particularly when the reported set appears unbounded. Fourth, verify that reported components include all non-rejected regions and exclude rejected gaps. Finally, check that the software's p-value and rejection conventions agree with the boundary rule used to report endpoints.

The methods differ in what they protect against and what they return. Conventional Wald intervals use an estimate and its standard error with an approximate normal reference distribution; under weak identification, that approximation can fail. Anderson–Rubin confidence sets are obtained by inverting Anderson–Rubin tests and are weak-IV-robust under their own assumptions, but can be conservative or wide. CLR inversion also gives weak-IV-robust inference under its model, and conditions to account for nuisance first-stage parameters. Both weak-IV-robust methods can produce irregular sets; neither guarantees a neat bounded interval. The choice must also respect the error structure and design, rather than being made solely from which set looks narrowest.

The CLR set is not automatically robust to heteroskedasticity, clustering, invalid instruments, or structural misspecification. Its usual likelihood-based calibration depends on the reduced-form distributional assumptions. A weak-IV-robust set also does not cure a lack of substantive identification: an unbounded set communicates that the data and assumptions do not rule out arbitrarily large values in that direction. Report the entire set, its construction level, the effect's units, and the assumptions used.

Common Mistake

A frequent mistake is to find two crossings and report the region between them as a confidence interval, assuming p-values must rise and then fall in a simple monotone pattern. CLR p-values need not have that shape. The correction is to evaluate the test over the parameter range, retain every non-rejected component, and check the tails before declaring endpoints finite. Also avoid interpreting the complement of the set as a probability statement about the true effect: it is the collection of values rejected by the specified tests, not a posterior distribution.

Key takeaway: A CLR confidence set is the inversion of candidate-specific conditional tests. Its boundaries solve the test's rejection-threshold equation, and its complete shape—including gaps and unbounded tails—must be preserved in the report.

Check Your Understanding

Answer each question, then compare with the brief solutions.

At level $\alpha=0.05$, what does a candidate value with a CLR p-value of 0.07 contribute to the inverted set?
A grid shows non-rejection at its largest positive candidate value. Is that enough to call the set unbounded above? Explain.
For a crossing bracketed by p-values 0.02 at $\beta_0=3$ and 0.08 at $\beta_0=4$, what is the linear-interpolation approximation to the 0.05 boundary?
Why can a disconnected CLR set not be reported as one interval spanning its smallest and largest values?

Solutions. (1) It is retained because $0.07\geq0.05$. (2) No. Expand the search and assess the tail; a finite grid cannot establish that all sufficiently large values are retained. (3) The interpolated boundary is $3+(0.05-0.02)/(0.08-0.02)=3+0.5=3.5$. (4) The span would include the rejected gap, falsely labeling rejected candidate values as part of the confidence set.