Tutorials › Propensity Score Matching and Instrumental Variables › Measuring Strength With Multiple Instruments

Weak-Instrument Diagnostics · Tutorial 515 of 699

Measuring Strength With Multiple Instruments

Compute the joint first-stage F statistic for multiple excluded instruments and understand what it does—and does not—say about strength.

Advanced 11 min read

What You'll Learn

  • Specify the first-stage regression with included covariates and multiple excluded instruments.
  • Formulate the joint null that all excluded-instrument coefficients are zero.
  • Calculate the conventional F statistic from restricted and unrestricted residual sums of squares.
  • Relate the joint F statistic to partial R-squared and its numerator and denominator degrees of freedom.
  • Distinguish joint relevance from individual t-statistics, instrument validity, and practical strength.
  • Recognize when the conventional F reference distribution may not be appropriate.

Why Strength Is a Joint Question

When several excluded instruments are used for one endogenous regressor, their first-stage contribution is assessed as a group. The instruments may each have a small estimated coefficient, or their estimated coefficients may be imprecise when considered separately, while the instruments jointly explain meaningful variation in the endogenous regressor. Conversely, a small p-value for one coefficient does not by itself summarize the contribution of the full instrument set.

The relevant classical diagnostic is a joint test in the first-stage regression. Its result depends on the number of excluded instruments, the residual variation after fitting the model, and the covariance among the coefficient estimates. It is therefore not generally recoverable by simply counting which individual t-tests are significant.

Definition: Let \(D_i\) be one endogenous regressor, \(Z_i\) a vector of \(q\) excluded instruments, and \(X_i\) a vector of included exogenous regressors (including an intercept, if used). The first stage is $$ D_i = X_i'\gamma + Z_i'\pi + v_i. $$ The excluded instruments are jointly relevant in this linear first stage when their coefficient vector \(\pi\) is not zero. The joint first-stage null is \(H_0:\pi=0\); the alternative is \(H_a:\pi\ne 0\).

The null is about the excluded-instrument coefficients conditional on the included regressors. It is not a test of whether the instruments are exogenous, satisfy the exclusion restriction, or identify a causal effect. Those are distinct questions. Nor does rejecting the null automatically establish that the instruments are sufficiently strong for reliable IV inference.

The Conventional Joint First-Stage F Statistic

Fit the unrestricted first stage with both \(X_i\) and \(Z_i\), and the restricted first stage with \(X_i\) alone. Let \(SSR_U\) and \(SSR_R\) be their residual sums of squares, respectively. The unrestricted model estimates \(k_U\) parameters in total, including the intercept if present. Thus its residual degrees of freedom are \(df=n-k_U\). The joint test imposes \(q\) restrictions, one for each excluded-instrument coefficient.

Definition: Under the conventional homoskedastic linear regression assumptions, the partial F statistic for testing \(H_0:\pi=0\) is $$ F = \frac{(SSR_R-SSR_U)/q}{SSR_U/(n-k_U)}. $$ Under the null and normally distributed first-stage errors, \(F\) has an \(F_{q,n-k_U}\) distribution. Without normal errors, the same reference distribution is commonly used as a large-sample approximation under suitable conditions. The numerator degrees of freedom are \(q\), and the denominator degrees of freedom are \(n-k_U\).

The numerator measures the reduction in residual sum of squares from adding the excluded instruments, averaged per restriction. The denominator estimates the residual variance in the unrestricted first stage. The statistic is large when the instruments collectively reduce residual variation by a substantial amount relative to the remaining unexplained variation.

The same test can be expressed using a partial \(R^2\), which measures the fraction of the restricted model's residual sum of squares removed by adding the excluded instruments: $$ R^2_{\text{partial}} = \frac{SSR_R-SSR_U}{SSR_R}. $$ Substituting this expression into the F formula gives $$ F = \frac{R^2_{\text{partial}}/q}{(1-R^2_{\text{partial}})/(n-k_U)}. $$ This identity shows why partial \(R^2\) alone does not determine the F statistic: the number of instruments and the residual degrees of freedom matter too. The earlier material on the first-stage F statistic and partial R-squared develops those measures; here the emphasis is on their joint interpretation when \(q\) is greater than one.

Worked Tests With Multiple Instruments

Worked Example: Several Modest Individual Tests, a Stronger Joint Test

Suppose a hypothetical study estimates a first stage with three excluded instruments and included covariates. The unrestricted regression has residual sum of squares \(SSR_U=1080\), the restricted regression has \(SSR_R=1200\), and the unrestricted residual degrees of freedom are 196. The three individual coefficient t-statistics are 1.5, 1.2, and 1.7.

There are \(q=3\) restrictions. The joint statistic is $$ F = \frac{(1200-1080)/3}{1080/196} = \frac{120/3}{5.5102} = \frac{40}{5.5102} = 7.259. $$

This is an \(F_{3,196}\) statistic, with a p-value below 0.001. At a 5% significance level, reject \(H_0:\pi=0\): the instruments jointly predict the endogenous regressor in this first stage. The individual t-statistics do not contradict this result. A joint test uses the covariance among coefficient estimates as well as their individual standard errors, so its result cannot be inferred by counting individually significant coefficients. This finding is evidence of joint relevance, not proof of adequate strength or instrument validity.

For an equivalent calculation based on a regression output, the joint Wald form is $$ F = \frac{1}{q}(R\hat{\beta}-r)'\left[R\widehat{\operatorname{Var}}(\hat{\beta})R'\right]^{-1}(R\hat{\beta}-r), $$ where \(\hat{\beta}\) collects the unrestricted first-stage estimates and \(R\hat{\beta}=r\) encodes the restrictions being tested. For \(H_0:\pi=0\), \(R\) selects the \(q\) excluded-instrument coefficients and \(r=0\). With the usual homoskedastic OLS covariance estimate, this gives the same partial F test as the residual-sum-of-squares formula. The covariance matrix matters: separate coefficient estimates and standard errors are not enough to reconstruct the joint test.

Worked Example: Calculating the Test From Partial R-Squared

In a hypothetical first stage, two excluded instruments are added to a model containing the included covariates. Suppose \(SSR_R=500\), \(SSR_U=450\), and \(n-k_U=97\).

First calculate the partial \(R^2\): $$ R^2_{\text{partial}} = \frac{500-450}{500} = \frac{50}{500}=0.1000. $$ Then use \(q=2\) and the partial-\(R^2\) formula: $$ F = \frac{0.1000/2}{(1-0.1000)/97} = \frac{0.05000}{0.009278} = 5.389. $$

The test has \(2\) numerator and \(97\) denominator degrees of freedom, with a p-value of approximately 0.006. At \(\alpha=0.05\), reject the joint null. In this example, the instruments remove 10.0% of the variation left over after fitting the included covariates. That descriptive amount and the joint test result provide related but distinct information: neither, by itself, is a universal guarantee that conventional IV inference will perform well.

Degrees of Freedom, Controls, and Interpretation

The denominator degrees of freedom come from the unrestricted first stage, not from the restricted model. Suppose \(n=160\), the model includes an intercept and two covariates, and it has four excluded instruments. Then \(k_U=1+2+4=7\), so \(df=160-7=153\). The joint test has \(q=4\) numerator degrees of freedom. Counting the restrictions and parameters carefully prevents a common arithmetic error: the F statistic is not calculated by dividing the reduction in residual sum of squares by the total number of fitted parameters.

Worked Example: A Joint Test With Included Covariates

A hypothetical first stage uses \(n=160\) observations, an intercept, two included covariates, and four excluded instruments. It reports \(SSR_R=820\) and \(SSR_U=760\). There are \(q=4\) restrictions, and \(df=160-(1+2+4)=153\).

The statistic is $$ F = \frac{(820-760)/4}{760/153} = \frac{15}{4.9673} = 3.020. $$

For an \(F_{4,153}\) distribution, the p-value is approximately 0.020. At \(\alpha=0.05\), reject the joint null that all four excluded-instrument coefficients are zero, conditional on the included covariates. The test does not establish that each instrument is individually useful: it addresses their joint contribution. It also does not turn the p-value into a measure of the probability that the instruments are valid or strong.

The calculation assumes that the restricted and unrestricted regressions use the same observations and the same included covariates, with the unrestricted regression adding the \(q\) excluded instruments. For the stated exact F reference distribution, the first-stage errors must be independent, normally distributed with mean zero and constant variance, conditional on the regressors. In standard large samples, a conventional F test may be used as an approximation under weaker conditions, but its covariance assumptions still matter. If errors are heteroskedastic or dependent in ways that affect the variance calculation, the usual homoskedastic F reference may not be appropriate.

What the Joint F Does Not Tell You

A joint test answers a narrow question: whether the excluded-instrument coefficients are collectively distinguishable from zero under the test's assumptions. A low p-value may result with a large sample even when the improvement in prediction is modest. The p-value therefore measures evidence against the zero-coefficient null, not the practical size of the first-stage contribution. Conversely, a non-significant result is not proof that every instrument is irrelevant; it indicates that this sample and test did not provide sufficient evidence against the joint null.

The number of excluded instruments also matters. The F statistic averages the reduction in residual sum of squares across \(q\) restrictions, while its denominator uses the unrestricted residual variance. Two analyses with the same partial \(R^2\) can therefore have different F statistics if they use different numbers of instruments or residual degrees of freedom. The F-statistic rule of ten is not a universal threshold; its limitations depend on the design and the inferential goal.

A useful comparison is between the joint test and three other summaries:

SummaryQuestion it addressesWhat it does not establish
Individual t-statisticIs one coefficient distinguishable from zero, considered through its own standard error?The joint contribution of all excluded instruments
Joint first-stage FDo the excluded-instrument coefficients collectively differ from zero?Instrument validity or a universal measure of adequate strength
Partial \(R^2\)What fraction of restricted-model residual variation is removed by adding the instruments?Statistical significance without considering sample size and degrees of freedom

These measures are complementary. Report the joint statistic with its numerator and denominator degrees of freedom, the p-value, and, where useful, the partial \(R^2\). Describe the result as a first-stage diagnostic, not as a finding that the instruments are valid or that the causal effect is identified without qualification.

Diagnostics and Common Pitfalls

Before interpreting a reported joint F statistic, verify that the test corresponds to the intended first stage. Check that the excluded instruments are tested jointly, that included covariates remain in both restricted and unrestricted models, that the restriction count equals the number of excluded-instrument coefficients being tested, and that the residual degrees of freedom match the unrestricted model. Confirm that both regressions use the same observations. A mismatch in any of these can produce a statistic that answers a different question.

  • Counting individual significant coefficients. The wrong reasoning is that two significant t-tests mean the instrument set is jointly strong, or that no significant t-test means it is jointly irrelevant. The correction is to test the full set of coefficient restrictions using the joint covariance information.
  • Calling a small p-value “strong instruments.” The wrong reasoning treats statistical significance as a complete measure of predictive strength. The correction is to interpret the test alongside the partial \(R^2\), sample size, number of instruments, and the assumptions behind the statistic.
  • Using a causal interpretation for a first-stage test. The wrong reasoning is that a significant first stage proves the instruments are valid. The correction is that this test concerns prediction of \(D_i\); it does not test the exclusion restriction or instrument exogeneity.
  • Applying the conventional formula under any error structure. The wrong reasoning assumes the classical F reference applies regardless of heteroskedasticity or dependence. The correction is to ensure the test's variance assumptions fit the data and design.
Key takeaway: With multiple excluded instruments and one endogenous regressor, assess their conventional first-stage relevance by jointly testing \(H_0:\pi=0\). The F statistic summarizes their collective contribution conditional on included covariates; individual t-statistics, statistical significance, instrument validity, and adequate strength are different questions.

Check Your Understanding

Try each problem, then check the solutions below.

  1. A first stage has three excluded instruments. What is the joint null, and what are the numerator degrees of freedom?
  2. In a hypothetical model, \(SSR_R=360\), \(SSR_U=330\), \(q=2\), and the unrestricted residual degrees of freedom are 88. Calculate the F statistic.
  3. Why can a joint test reject even when no individual coefficient t-test rejects at the 5% level?
  4. A joint first-stage test has a small p-value. Does that establish instrument validity? Explain.
  5. For a model with \(n=120\), an intercept, three included covariates, and two excluded instruments, find the unrestricted residual degrees of freedom.

Solutions. (1) The null is that all three excluded-instrument coefficients are zero conditional on the included covariates; the numerator degrees of freedom are \(q=3\). (2) \(F=((360-330)/2)/(330/88)=15/3.75=4.000\), with \(2\) and \(88\) degrees of freedom. (3) The joint test uses the covariance among the coefficient estimates as well as their individual standard errors; the collective evidence need not match the separate t-test decisions. (4) No. The joint test assesses first-stage prediction, not exogeneity or the exclusion restriction. (5) The unrestricted model estimates \(1+3+2=6\) parameters, so the residual degrees of freedom are \(120-6=114\).

Clinical Biostats