The first stage an instrument needs is set by the violation nobody can see

The error each estimator converges on when the instrument has a direct effect of 0.05 on the outcome — a path the exclusion restriction asserts is zero and no sample can check. The instrument's error is δ/π exactly, so it is the reciprocal of the very quantity that made the method work: 1.0000 at a first stage of 0.05 and 0.0833 at 0.60. Least squares carries the confounding instead, at 0.3440 at a first stage of 0.30. The two cross at π = 0.1389, and the crossing is exactly δ times 2.7778 — the first stage an instrument needs is proportional to the violation it is assumed not to have, and below that line the method being corrected is the better estimator.

The counted mean bias of two-stage least squares at 4 instruments and 200 rows, over 2000 draws a setting, against the standard approximation and against the least-squares inconsistency the instrument was brought in to remove. At π = 0.02 the counted bias is 0.3220 ± 0.0142 where least squares is out by 0.3594 — 89.6% of the way back. At π = 0.3 it is 0.0118 against 0.2647. The approximation, the inconsistency over the population first-stage F, tracks the count at the weak end and sits above it in the middle: 0.968, 0.971, 0.918, 0.810, 0.740, 0.722, 0.846 as the ratio of counted to approximated bias.

Counted coverage of two nominal 95.0% intervals for the same causal effect, read off the same 2000 draws of 200 rows at each first stage. The exact Anderson–Rubin set covers 95.3% at every setting — flat, because the statistic it inverts is built from y − tβ, which contains no π at all, and is therefore the same number on the same draw whatever the instrument is worth. The conventional interval covers 99.1% at π = 0.02 and 95.6% at π = 0.6: it goes wrong at the weak end by covering too MUCH, at a median width of 7.320, because its standard error is computed from residuals taken at an estimate that has itself gone wrong. A weak instrument does not make this interval lie about its coverage; it makes it useless while telling the truth.

Four quantities in a population where the effect is not the same for everybody: 40.0% compliers, 25.0% always-takers and 35.0% never-takers, with the compliers carrying an effect 1.0000 larger than everybody else's. The population average effect is 0.5000. The compliers' average effect is 1.1000. A valid instrument converges on 1.1000 — the second of those, not the first, and the two differ by 0.6000. Comparing the treated with the untreated as they stand gives 1.4182, wrong by 0.9182 in the same direction, because always-takers start 1.5000 above never-takers before any treatment happens. The instrument removes the selection and changes the question at the same time, and only one of those is reported.

The overidentification test's rejection rate against the error the violation actually puts into the estimate, so the two rows are the same estimate being equally wrong. With the whole violation on one instrument the test keeps its size at 5.0% under the null and reaches 86.4% by an error of 0.800. With both instruments violating in the same ratio as their first stages the two Wald ratios are identical, the test has nothing to compare, and it rejects at 5.8% at that same error — its own size. Over 1000 draws of 300 rows at each setting, at a nominal 5.0%. The test is a comparison between instruments and it was never a check on either.

Coverage of three nominal 95.0% intervals on the same 1000 draws of 200 rows at each count, when a total concentration parameter of 8 is spread over 1 to 32 instruments. Two-stage least squares covers 97.2%, 96.4%, 94.0%, 86.7%, 73.2%, 51.5%. Building each row's fitted treatment from a first stage that never saw that row covers 97.1%, 97.2%, 98.3%, 97.8%, 97.9%, 98.7%. Limited-information maximum likelihood with its conventional standard error covers 97.2%, 96.7%, 95.7%, 90.9%, 85.2%, 79.0%. At one instrument the likelihood estimator is two-stage least squares exactly, which is why the first readings of those two agree to the last draw.