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As from this post as @Ben Bolker pointed out that :

.. note that these (as often pointed out by Doug Bates) the Wald standard errors are often very poor estimates of the uncertainty of variances, because the likelihood profiles are often far from quadratic on the variance scale ...

I have the following questions :

(1) If the likelihood profiles are often far from quadratic on the variance scale , does it imply the Wald confidence interval has a very poor chance to contain the true parameter value ?

(2) Why are the Wald standard errors often very poor estimates of the variances ? (Don't know whether there is any technical difference between "estimates of the variances" and "estimates of the uncertainty of variances") .

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    $\begingroup$ Minor correction: I suggest calling the standard errors "standard errors from the information matrix" and not "Wald standard errors". The name Wald is more attached to test statistics that use MLEs and the information matrix. Someone may know the history better than me though. $\endgroup$ Jul 30, 2015 at 15:06

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(1) When the true sampling distributions are asymmetric, then applying a symmetric confidence interval based on the mean and the variance of that distribution leads to uneven coverage. Instead of missing 5% on each tail for a 90% CI, you would get 3% on one tail and 12% on the other tail, with the actual coverage lower than the nominal.

(2) All asymptotic statistics have finite sample biases. Unfortunately for the likelihood based methods, these biases tend to go on the side of lowering the standard errors relative for what they would need to be. The area of statistics that studies these subtleties is called "higher order asymptotics" or "second order asymptotics" (the first order being the asymptotic normal distribution that many ML estimates follow under "standard" conditions).

Maybe this could help, too: Obtaining any Wald statistic you want.

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  • $\begingroup$ In point (1) , 3% in one tail and 12% in another tail , that is in total 15% type one error in 90% confidence interval ? But there needs to be 10% type 1 error in 90% confidence interval . Should it be 3% in one tail and 7% in another tail ? Am I doing mistake ? $\endgroup$
    – user81411
    Aug 2, 2015 at 14:56
  • $\begingroup$ @Munira, asymmetric and incorrect coverage is very typical for nonlinear statistics. My choice of numbers was deliberate. $\endgroup$
    – StasK
    Aug 7, 2015 at 11:16

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