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Suppose I have a set of sample data from an unknown or complex distribution, and I want to perform some inference on a statistic $T$ of the data. My default inclination is to just generate a bunch of bootstrap samples with replacement, and calculate my statistic $T$ on each bootstrap sample to create an estimated distribution for $T$.

What are examples where this is a bad idea?

For example, one case where naively performing this bootstrap would fail is if I'm trying to use the bootstrap on time series data (say, to test whether I have significant autocorrelation). The naive bootstrap described above (generating the $i$th datapoint of the nth bootstrap sample series by sampling with replacement from my original series) would (I think) be ill-advised, since it ignores the structure in my original time series, and so we get fancier bootstrap techniques like the block bootstrap.

To put it another way, what is there to the bootstrap besides "sampling with replacement"?

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    $\begingroup$ If you want to do inference for the mean of i.i.d. data, the bootstrap is a great tool. Everything else is questionable, and requires case-by-case proof of weak convergence. $\endgroup$
    – StasK
    Apr 22, 2015 at 13:45

3 Answers 3

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If the quantity of interest, usually a functional of a distribution, is reasonably smooth and your data are i.i.d., you're usually in pretty safe territory. Of course, there are other circumstances when the bootstrap will work as well.

What it means for the bootstrap to "fail"

Broadly speaking, the purpose of the bootstrap is to construct an approximate sampling distribution for the statistic of interest. It's not about actual estimation of the parameter. So, if the statistic of interest (under some rescaling and centering) is $\newcommand{\Xhat}{\hat{X}_n}\Xhat$ and $\Xhat \to X_\infty$ in distribution, we'd like our bootstrap distribution to converge to the distribution of $X_\infty$. If we don't have this, then we can't trust the inferences made.

The canonical example of when the bootstrap can fail, even in an i.i.d. framework is when trying to approximate the sampling distribution of an extreme order statistic. Below is a brief discussion.

Maximum order statistic of a random sample from a $\;\mathcal{U}[0,\theta]$ distribution

Let $X_1, X_2, \ldots$ be a sequence of i.i.d. uniform random variables on $[0,\theta]$. Let $\newcommand{\Xmax}{X_{(n)}} \Xmax = \max_{1\leq k \leq n} X_k$. The distribution of $\Xmax$ is $$ \renewcommand{\Pr}{\mathbb{P}}\Pr(\Xmax \leq x) = (x/\theta)^n \>. $$ (Note that by a very simple argument, this actually also shows that $\Xmax \to \theta$ in probability, and even, almost surely, if the random variables are all defined on the same space.)

An elementary calculation yields $$ \Pr( n(\theta - \Xmax) \leq x ) = 1 - \Big(1 - \frac{x}{\theta n}\Big)^n \to 1 - e^{-x/\theta} \>, $$ or, in other words, $n(\theta - \Xmax)$ converges in distribution to an exponential random variable with mean $\theta$.

Now, we form a (naive) bootstrap estimate of the distribution of $n(\theta - \Xmax)$ by resampling $X_1, \ldots, X_n$ with replacement to get $X_1^\star,\ldots,X_n^\star$ and using the distribution of $n(\Xmax - \Xmax^\star)$ conditional on $X_1,\ldots,X_n$.

But, observe that $\Xmax^\star = \Xmax$ with probability $1 - (1-1/n)^n \to 1 - e^{-1}$, and so the bootstrap distribution has a point mass at zero even asymptotically despite the fact that the actual limiting distribution is continuous.

More explicitly, though the true limiting distribution is exponential with mean $\theta$, the limiting bootstrap distribution places a point mass at zero of size $1−e^{-1} \approx 0.632$ independent of the actual value of $\theta$. By taking $\theta$ sufficiently large, we can make the probability of the true limiting distribution arbitrary small for any fixed interval $[0,\varepsilon)$, yet the bootstrap will (still!) report that there is at least probability 0.632 in this interval! From this it should be clear that the bootstrap can behave arbitrarily badly in this setting.

In summary, the bootstrap fails (miserably) in this case. Things tend to go wrong when dealing with parameters at the edge of the parameter space.

An example from a sample of normal random variables

There are other similar examples of the failure of the bootstrap in surprisingly simple circumstances.

Consider a sample $X_1, X_2, \ldots$ from $\mathcal{N}(\mu,1)$ where the parameter space for $\mu$ is restricted to $[0,\infty)$. The MLE in this case is $\newcommand{\Xbar}{\bar{X}}\Xhat = \max(\bar{X},0)$. Again, we use the bootstrap estimate $\Xhat^\star = \max(\Xbar^\star, 0)$. Again, it can be shown that the distribution of $\sqrt{n}(\Xhat^\star - \Xhat)$ (conditional on the observed sample) does not converge to the same limiting distribution as $\sqrt{n}(\Xhat - \mu)$.

Exchangeable arrays

Perhaps one of the most dramatic examples is for an exchangeable array. Let $\newcommand{\bm}[1]{\mathbf{#1}}\bm{Y} = (Y_{ij})$ be an array of random variables such that, for every pair of permutation matrices $\bm{P}$ and $\bm{Q}$, the arrays $\bm{Y}$ and $\bm{P} \bm{Y} \bm{Q}$ have the same joint distribution. That is, permuting rows and columns of $\bm{Y}$ keeps the distribution invariant. (You can think of a two-way random effects model with one observation per cell as an example, though the model is much more general.)

Suppose we wish to estimate a confidence interval for the mean $\mu = \mathbb{E}(Y_{ij}) = \mathbb{E}(Y_{11})$ (due to the exchangeability assumption described above the means of all the cells must be the same).

McCullagh (2000) considered two different natural (i.e., naive) ways of bootstrapping such an array. Neither of them get the asymptotic variance for the sample mean correct. He also considers some examples of a one-way exchangeable array and linear regression.

References

Unfortunately, the subject matter is nontrivial, so none of these are particularly easy reads.

P. Bickel and D. Freedman, Some asymptotic theory for the bootstrap. Ann. Stat., vol. 9, no. 6 (1981), 1196–1217.

D. W. K. Andrews, Inconsistency of the bootstrap when a parameter is on the boundary of the parameter space, Econometrica, vol. 68, no. 2 (2000), 399–405.

P. McCullagh, Resampling and exchangeable arrays, Bernoulli, vol. 6, no. 2 (2000), 285–301.

E. L. Lehmann and J. P. Romano, Testing Statistical Hypotheses, 3rd. ed., Springer (2005). [Chapter 15: General Large Sample Methods]

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  • $\begingroup$ The behaviour of the order statistics bootstrap seems reasonable to me, given that the exponential distribution has a similar "point mass" at zero - The mode of an exponential distribution is 0, so it seems reasonable that the probability should be non-zero at the most likely value! The bootstrap would probably be something more like an geometric distribution which is a discrete analogue of the exponential. I wouldn't take this as a "failure" of the bootstrap here - for the estimated quantity of $\theta$ always lies in the appropriate interval $\theta\geq X_{(n)}$ $\endgroup$ Apr 19, 2011 at 6:31
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    $\begingroup$ @cardinal - the asymptotic distribution is not the appropriate benchmark - unless you have an infinite sample. The bootstrap distribution should be compared to the finite sample distribution that it was designed to approximate. What you want to show is that as number of bootstrap iterations goes to infinity, the bootstrap distribution converges to the finite sampling distribution. letting $n\to\infty$ is an approximate solution not an exact one. $\endgroup$ Apr 19, 2011 at 13:36
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    $\begingroup$ @cardinal +1, I've upvoted the question earlier, but I just want to thank for a very good answer, examples and links to the articles. $\endgroup$
    – mpiktas
    Apr 19, 2011 at 13:45
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    $\begingroup$ I'll try to post an example in a day or so. But, I think you're likely misunderstanding. The logic is that the coverage implied by the bootstrap distribution is wrong, even asymptotically. That leaves little hope for it to work in finite samples. You might reread the second-to-last paragraph of that section, which should make it clear that the bootstrap can behave arbitrarily badly in terms of the approximation of the sampling distribution. $\endgroup$
    – cardinal
    Apr 20, 2011 at 1:56
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    $\begingroup$ @probabilityislogic, at first, I only saw the the latter of your two most recent comments. To address the former, you can see the first two sentences of the section above with heading "What it means for the bootstrap to 'fail'", where this is addressed explicitly. The bootstrap is not about estimating the parameter. We assume we have a good way to estimate the desired parameter (in this case, $X_{(n)}$ works fine). The bootstrap is about knowing something about the distribution of the parameter so that we can do inference. Here, the bootstrap gets the distribution (very!) wrong. $\endgroup$
    – cardinal
    Apr 20, 2011 at 13:06
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The following book has a chapter (Ch.9) devoted to "When Bootstrapping Fails Along with Remedies for Failures":

M. R. Chernick, Bootstrap methods: A guide for practitioners and researchers, 2nd ed. Hoboken N.J.: Wiley-Interscience, 2008.

The topics are:

  1. Too Small of a Sample Size
  2. Distributions with Infinite Moments
  3. Estimating Extreme Values
  4. Survey Sampling
  5. Data Sequences that Are M-Dependent
  6. Unstable Autoregressive Processes
  7. Long-Range Dependence
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    $\begingroup$ Have you seen this comment to an answer in this thread? Incidentally, that comment links to an Amazon page for Chernick's book; the reader reviews are enlightening. $\endgroup$
    – whuber
    Dec 30, 2013 at 15:58
  • $\begingroup$ @whuber Well, I didn't notice that comment. Should I remove my answer? $\endgroup$
    – Sadeghd
    Dec 31, 2013 at 15:21
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    $\begingroup$ Because your answer is more detailed than the reference in the comment, it potentially has value: but in keeping with SE policies and aims, it would be nice to see it amplified with some explanation of why you are recommending this book or--even better--to include a summary of the information in it. Otherwise it adds little and should be deleted or converted into a comment to the question. $\endgroup$
    – whuber
    Dec 31, 2013 at 15:24
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The naive bootstrap depends on the sample size being large, so that the empirical CDF for the data are a good approximation to the "true" CDF. This ensures that sampling from the empirical CDF is very much like sampling from the "true" CDF. The extreme case is when you have only sampled one data point - bootstrapping achieves nothing here. It will become more and more useless as it approaches this degenerate case.

Bootstrapping naively will not necessarily fail in times series analysis (although it may be inefficient) - if you model the series using basis functions of continuous time (such a legendre polynomials) for a trend component, and sine and cosine functions of continuous time for cyclical components (plus normal noise error term). Then you just put in what-ever times you happen to have sampled into the likelihood function. No disaster for bootstrapping here.

Any auto-correlation or ARIMA model has a representation in this format above - this model is just easier to use and I think to understand and interpret (easy to understand cycles in sine and cosine functions, hard to understand coefficients of an ARIMA model). For example the auto-correlation function is the inverse Fourier transform of the power spectrum of a time series.

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    $\begingroup$ @probabilityislogic, for time-series processes the time plays important role, so distribution of vector $(X_t,X_{t+1})$ is different from $(X_{t+1},X_t)$. The resampling as done in naive bootstrap destroys this structure, so for example if you try to fit AR(1) model, after resampling you might get that you are trying to fit $Y_{10}$ as $\rho Y_{15}$, which is does not seem natural. If you google for "bootstrapping time series" the second article gives example of how estimate of variance of time series has... $\endgroup$
    – mpiktas
    Apr 19, 2011 at 8:51
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    $\begingroup$ @probabilityislogic, ok so assume that we have a sample of vectors $(X_1,1),(X_2,2),(X_3,3)$, which comes from AR(1) model $Y_t=\rho Y_{t-1}+\varepsilon_t$. Suppose we estimate $\rho$ using linear regression without the intercept. For the given sample the estimate will be $(X_2X_1+X_3X_2)/(X_1^2+X_2^2)$. Suppose now we have a bootstrap sample $(X_1,1),(X_1,1),(X_3,3)$, how would one define the the estimate? Since time structure is kept, I should take it into account, but how? I cannot pair $(X_3,3)$ with $(X_1,1)$, since then I will estimate AR(2) model $Y_t=\alpha Y_{t-2}+u_t$. $\endgroup$
    – mpiktas
    Apr 19, 2011 at 13:38
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    $\begingroup$ @probabilityislogic, would it be possible for you to demonstrate your idea in your answer for naive bootstrap estimate of $\rho$ in AR(1) model $Y_t=\rho Y_{t-1}+u_t$? I do not think that it is possible, hence the basic reason for downvote. I would be glad to get proven wrong. $\endgroup$
    – mpiktas
    Apr 19, 2011 at 13:41
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    $\begingroup$ @probabilityislogic, and? What will be the estimate of $rho$ in that case? I am sorry for pestering, but I genuinely do not see how can you show that naive bootstrap will not fail in this case. $\endgroup$
    – mpiktas
    Apr 19, 2011 at 14:00
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    $\begingroup$ My book here has a chapter on when the bootstrap fails and also a chapter on how the bootstrap is applied in time series. For time series the bootstrap can be applied to residuals from a model in the model based approach. The other nonparametric time domain approach is block bootstrap of which there are many types. $\endgroup$ Aug 28, 2012 at 16:29

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