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I did a bootstrapping with a mixed model (several variables with interaction and one random variable). I got this result (only partial):

> boot_out

ORDINARY NONPARAMETRIC BOOTSTRAP

Call:
boot(data = a001a1, statistic = bootReg, R = 1000)

Bootstrap Statistics :
          original        bias     std. error
t1*   4.887383e+01 -1.677061e+00 4.362948e-01
t2*   3.066825e+01  1.264024e+00 5.328387e-01
t3*   8.105422e+01  2.368599e+00 6.789091e-01
t4*   1.620562e+02  4.908711e+00 1.779522e+00
......

Now, I wanted to get the confidence intervals for the intercept:

> boot.ci(boot_out,type=c("norm","basic","perc"), index=1)
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 1000 bootstrap replicates

CALL : 
boot.ci(boot.out = boot_out, type = c("norm", "basic", 
"perc"), index = 1)

Intervals : 
Level      Normal              Basic              Percentile     
95%   (49.70, 51.41 )   (49.70, 51.41 )   (46.34, 48.05 )  
Calculations and Intervals on Original Scale

The bias corrected estimated is:

48.873 -1.677
1 47.196

The problem I have is that the normal and basic CI are outside of the estimate (original and corrected). I just wonder how to cope with that.

Update 1:
Here is a similar questions with a lot of responses.

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    $\begingroup$ Just a comment: Efron & Tibshirani (1993) in the classic book were arguing rather against bias correction saying it is a "dangerous" and "problematic" practice that could lead to increased standard error. $\endgroup$
    – Tim
    Commented Feb 2, 2015 at 11:07
  • $\begingroup$ @Tim Thanks for your comment. I will have a look at the book. Maybe, a solution is to use the estimates and and the bootstrap se* to compute the confidence intervals. In my case, the bias affect the estimates only slightly. $\endgroup$
    – giordano
    Commented Feb 2, 2015 at 13:08

1 Answer 1

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The difficulty you are facing is from the implied mathematics. A center of location estimator, or an interval estimator, can be thought of as the minimization of a cost function over a distribution. The sample mean over the Gaussian minimizes quadratic loss, while the median minimizes the absolute linear loss function over the Gaussian. Even though in the population they are located at the same point, they are discovered using different cost functions.

We give you an algorithm and say "do this," but before the algorithm was developed someone solved an optimization problem.

You have applied four different cost functions giving you three intervals and a point estimator. Since the cost functions are different, they provide you different points and intervals. There is nothing to be done about it except to manually unify the methodology.

You need to find the underlying papers and look at the underlying code to understand which ones map to what types of problems.

Sorry to say this, but you were betrayed by the software. It did its job, and on average this works out great, but you got the sample where the software won't work. Or, rather, it is working perfectly and you need to actually work your way backward through the literature to determine what it is really doing.

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  • $\begingroup$ Thanks for It did its job, and on average this works out great, but you got the sample where the software won't work and the other insights. Indeed, the sample is strange that's why I wanted to use Bootsstrap to compute CI. Obviously, this method seems not so simple as it appears. $\endgroup$
    – giordano
    Commented Dec 7, 2016 at 9:31

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