# Non-technical conditions for validity of nonparametric bootstrap confidence intervals

I am rather ignorant about nonparametric bootstrap. Assume the same context as in this question: ${(x_i)}_{i=1}^m$ and ${(y_i)}_{i=1}^n$ are independent data samples, the $x_i$ are independent replicates from a distribution with expectation $\mu_X$ and similarly the $y_i$ are independent replicates from a distribution with expectation $\mu_Y$. We estimate the ratio $\theta=\mu_X/\mu_Y$ by the ratio $\hat\theta = \bar x / \bar y$ of the two sample means. Under which conditions the quantiles of the boostrap samples of $\hat\theta$ provide a valid confidence interval about $\theta$ ? I am not interested in the finest required technical conditions (such as a $L^{1+\epsilon}$-integrability condition), but rather in more easy conditions which are reasonable to assume for common real datasets.

EDIT: I am not interested in trivial counter-examples too. For instance I assume the the $y_i$ cannot be negative (the support of their distribution is an interval of strictly positive numbers) and the unknown distributions are discrete or continuous distributions.

EDIT: Maybe what I expect to understand is more clear if I ask a different question : assuming the previous 'edit' and strong distributional assumptions (such as $L^2$), under which conditions have we a valid bootstrap confidence interval about $f(\mu_X,\mu_Y)$ for a suitable function $f$ when using the bootstrap samples of the estimate $f(\bar x, \bar y)$ ? Is the unbiasedness of this estimate a required condition ?

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It's easy to find some conditions where this is not reasonable. E.g., let $Y$ have an absolutely continuous distribution with mean $0$ so that $\theta$ is undefined. Then any sample of $Y$ has no zero a.s., whence all bootstrap samples of $\hat\theta$ are defined and finite, implying no confidence interval will cover the true value. – whuber Jun 7 '12 at 19:58
If the denominator has probability concentrated around 0 the ratio has very heavytails witness the Cauchy distribution as the ratio of 2 N(0,1) random variables. – Michael Chernick Jun 7 '12 at 20:04
@whuber> Thanks for this remark. I have just edited my post to discard such a counter-example (this discards @Michael's counter-example as well). – Stéphane Laurent Jun 7 '12 at 20:10
The real force of my example (and also, I suspect, @Michael Chernick's remark), Stéphane, was to indicate that your estimator potentially has such a strong bias (even under your edited assumptions) that you may have no right to expect bootstrapped quantiles to work well at all. A good use of bootstrapping here might be to estimate and compensate for that bias, rather than to produce (bad) confidence intervals. – whuber Jun 7 '12 at 20:12
Thanks @whuber, you help me to better formulate my question. I've just edited it again. – Stéphane Laurent Jun 7 '12 at 20:22
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## 1 Answer

In their book, An Introduction to the Bootstrap (Chapman and Hall, New York 1994), Efron and Tibshirani give an example of estimating a ratio with the patch data study. The ratio of the averages is a biased estimate for the average of the ratio. So Efron and Tibshirani use the patch data to show how the bootstrap can correct for this bias.

These data are based on a clinical trial where the effect of a hormonal patch produced at a new manufacturing plant is compared to the effect at the old (approved) manufacturing plant. The parameter of interest in this problem is

$$\theta = \frac{E(\text{new})-E(\text{placebo})}{E(\text{old})-E(\text{placebo})}.$$

$E(.)$ is the expected level of hormone in the bloodstream by "new" = patch produced at new plant, "old" = patch produced at the approved plant, and "placebo" = patch without hormone.

The FDA requirement for approval of the new manufacturing plant is that the new patch produces at least 80% of the effect that the old patch produced over placebo. So the idea is to estimate $\theta$ and show via a confidence interval (or hypothesis testing) that it is greater than 0.80; or, equivalently, that $1-\theta$ is less than 0.20.

In this problem the numerator and denominator both have distributions concentrated on the positive half of the real line and bounded away from 0. So the problem that Bill Huber mentioned does not occur. In the book, Efron and Tibshirani compute the biased point estimate along with its bootstrap estimate of bias. They show that the bias-corrected estimate of $1-\theta$ is well below 0.20. Bootstrap confidence intervals could be used for this by again separately bootstrapping the two samples and computing the ratio estimates, say $B$ times, by Monte Carlo.

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Thanks! It's late, your answer seems to be excellent but I will come back tomorrow to read it seriously. – Stéphane Laurent Jun 7 '12 at 20:37
You can find out more about the example in Efron-Tibshirani book or Chernick-LaBudde "An Introduction to Bootstrap Methods with Applications to R" pp 51-53. I have not attempted to address the question about more general functions of the two samples. – Michael Chernick Jun 7 '12 at 20:45
I notice that this answer hasn't received any upvotes yet. While it seems that it doesn't answer the OP's question directly ("under which conditions the quantiles of the boostrap samples of $\hat{\theta}$ provide a valid confidence interval about $\theta$?"), it seems like you provide a useful and highly relevant example. I think that the answer would benefit from better formatting though - it would be easier to read if you added some blank space and used $\LaTeX$ for equations and Greek symbols. You could add an Amazon link to the book as well (these help pay for the Stackexchange sites!). – MånsT Jun 8 '12 at 7:49
Thanks. Well at least maybe I should be grateful for not getting downvotes. I do think by mentioning the need for the denominator rv to be bounded away from 0 I do provide at least a partial answer to the question about providing "nontechnical conditions". Maybe linking to my own book is going a little too far toward publizing it. – Michael Chernick Jun 8 '12 at 9:46
When your own work is a suitable reference, @Michael, please don't hold back from mentioning it. It helps to acknowledge that it's a self-reference: this can counter any misperception that you're engaged in self promotion. – whuber Jun 8 '12 at 13:48
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