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I would like an explanation on the statement in bold below. At first glance, I'd think that a weak instrumental variable would yield a even bigger standard error estimate.

"When instruments are weak, however, two serious problems emerge for two-stage least squares. First is a problem of bias. Even though two-stage least squares coefficient estimates are consistent — so that they almost certainly approach the true value as the sample size approaches infinity — the estimates are always biased in finite samples. When the instrumental variable is weak, this bias can be large, even in very large samples. Second, when an instrumental variable is weak, two-stage least squares’ estimated standard errors become far too small. Thus, when instruments are weak, confidence intervals computed for two-stage least squares estimates can be very misleading because their mid-point is biased and their width is too narrow, which undermines hypothesis tests based on two-stage least squares."

Murray, Michael P.. Avoiding Invalid Instruments and Coping with Weak Instruments. Journal of Economic Perspectives, v. 20, pp. 111-132.

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  • $\begingroup$ Not only the point estimate is biased, but the standard error that's reported by default is biased, as well. If you are not surprised that the OLS-reported standard error is biased when the data violate the homoskedasticity assumption of OLS, you should not be surprised to hear that the standard errors are wrong when the strong instrument assumption of IV is violated. $\endgroup$ – StasK Mar 17 '14 at 20:55
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The main reference to answer this would be Dufour (1997) who was the first to analytically show what happens to test statistics and confidence intervals of the 2-stage least squares estimator when instruments are weak. I wouldn't recommend reading the paper because it is awfully technical.
In summary:

  • Wald-type statistics are not pivotal anymore (i.e. no unique large-sample distribution for the test can provide valid tests and confidence intervals based on the asymptotic distribution of the test)
  • Wald-type test statistics have distributions that depend strongly on the unobserved nuisance parameters
  • the confidence level of Wald-type confidence intervals is essentially zero regardless of the sample size, so they tend to be arbitrarily small (see Dufour, 2003)

When you read papers on weak instruments like Stock et al. (2002) you will notice that most of them will avoid details on the problems of hypothesis tests with weak instruments in IV models. This is mostly because the arguments are very involved and given that there are remedies for inference with weak identification (for instance, the Anderson-Rubin, Kleibergen or Moreira statistics, or the reduced-form procedure by Chernozhukov and Hansen, 2008) they usually just refer to the papers which derived the problems of the usual testing procedures like the Dufour papers.

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A quick non-technical answer: since the estimate is biased, it will fall outside of the confidence interval more often than the nominal proportion of times that is should fall outside of the confidence interval. As n goes to infinity, the estimate converges to the true parameter, so that it will indeed fall within the CI the advertised number of times. It basically arises when using an estimator that is biased but consistent. A lot of econometrics is based on asymptotics. This is all to say that IV is biased unless you've got a lot of data or a really good instrument.

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  • $\begingroup$ I agree with you about the bias explanation. Actually, I just edited the question to a more specific part of the citation. $\endgroup$ – Robson Mar 17 '14 at 2:30
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    $\begingroup$ My point was that the two things are connected. The too-small nature of the SE's is only a problem because the estimate is biased. They would be correct SE's if the estimator was unbiased. I'd need to sit down with pencil and paper to show this formally. $\endgroup$ – generic_user Mar 17 '14 at 2:48

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