I am currently trying to get my head around some things concerning parametric bootstrap. Most things are probably trivial but I still think I may have missed something.

Suppose I want to get confidence intervals for data using a parametric bootstrap procedure.

So I have this sample and I assume its normally distributed. I would then estimate variance $\hat{v}$ and mean $\hat{m}$ and get my distribution estimate $\hat{P}$, which is obviously just $N(\hat{m},\hat{v})$.

Instead of sampling from that distribution I could just calculate the quantiles analytically and be done.

a) I conclude: in this trivial case, the parametric bootstrap would be the same as calculating things in a normal-distribution-assumption?

So theoretically this would be the case for all parametric bootstrap models, as long as I can handle the calculations.

b) I conclude: using the assumption of a certain distribution will bring me extra accuracy in the parametric bootstrap over the nonparametric one (if it is correct of course). But other than that, I just do it because I can't handle the analytic calculations und try to simulate my way out of it?

c) I would also use it if the calculations are "usually" done using some approximation because this would perhaps give me more accuracy...?

To me, the benefit of the (nonparametric) bootstrap seemed to lie in the fact that I don't need to assume any distribution. For the parametric bootstrap that advantage is gone - or are there things I've missed and where the parametric bootstrap provides a benefit over the things mentioned above?

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    $\begingroup$ You're basically correct - you are trading analytic error for monte carlo error. The parametric bootstrap is also an approximate posterior sample. $\endgroup$ Commented May 8, 2014 at 11:06
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    $\begingroup$ you mean approximate posterior sample as in bayesian? i still dont quite get the connection between bootstrapping and maximum likelihood estimation. but thats a different story. thank you for your answer! $\endgroup$ Commented May 9, 2014 at 15:15

1 Answer 1


Yes. You are right. But Parametric bootstrap shields better results when the assumptions hold. Think of it this way:

We have a random sample $X_1, \ldots, X_n$ from a distribution $F$. We estimate a parameter of interest $\theta$ as a function of the sample, $\hat{\theta} = h (X_1, \ldots, X_n)$. This estimate is a random variable, so it has a distribution we call $G$. This distribution is fully determined by $h$ and $F$ meaning $G=G(h,F)$. When doing any kind of bootstrap (parametric, non-parametric, re-sampling) what we are doing is to estimate $F$ with $\hat{F}$ in order to get an estimate of $G$, $\hat G = G(h,\hat{F})$. From $\hat G$ we estimate the properties of $\hat \theta$. What changes fom differents types of bootstrap is how we get $\hat{F}$.

If you can analytically calculate $\hat{G} = G(h,\hat{F})$ you should go for it, but in general, it is a rather hard thing to do. The magic of bootstrap is that we can generate samples with distribution $\hat G$. To do this, we generate random samples $X^b_1, \ldots, X^b_n$ with distribution $\hat F$ and calculate $\hat {\theta}^b = h(X^b_1, \ldots, X^b_n)$ which will follow the $\hat G$ distribution.

Once you think of it this way, the advantages of parametric bootstrap are obvious. $\hat{F}$ would be a better approximation of $F$, then $\hat{G}$ would be closer to $G$ and finally the estimations of $\hat{\theta}$'s properties would be better.

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    $\begingroup$ So if we put it in terms of higher order convergence we see that although parametric and nonparametric bootstrap are of the same order of convergence (i think thats whats written in van der vaarts asymptotic statistics), parametric is still better. but only in terms of some factor? $\endgroup$ Commented Aug 20, 2014 at 17:59

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