# Bootstrapping and Subsampling: Part I

January 20, 2013
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(This article was originally published at Normal Deviate, and syndicated at StatsBlogs.)

BOOTSTRAPPING AND SUBSAMPLING: PART I

Bootstrapping and subsampling are in the “amazing” category in statistics. They seem much more popular in statistics than machine learning for some reason.

1. The Bootstrap

The bootstrap (a.k.a. the shotgun) was invented by Brad Efron. Here is how it works. We have data ${X_1,\ldots, X_n \sim P}$ and we want a confidence interval for ${\theta = T(P)}$. For example, ${\theta}$ could be the median of ${P}$ or the mean of ${P}$ or something more complicated like, the largest eigenvalue of the covariance matrix of ${P}$.

The ${1-\alpha}$ bootstrap confidence interval is

$\displaystyle C_n = \left[ \hat\theta - \frac{\hat t_{1-\alpha/2}}{\sqrt{n}},\ \hat\theta - \frac{\hat t_{\alpha/2}}{\sqrt{n}}\right]$

where ${\hat\theta}$ is an estimator of ${\theta}$ and ${\hat t_{\alpha/2}}$ and ${\hat t_{1-\alpha/2}}$ are sample bootstrap quantiles that I will describe below. Before I explain this in more detail, notice two things. First, there is a minus sign in both the lower and upper endpoint. Second, the ${\alpha/2}$ and ${1-\alpha/2}$ quantiles are in the upper and lower endpoints, the reverse of what you might expect. The reason for the strange looking interval will be clear when we derive the interval.

Now for some details. Think of the parameter of interest ${\theta}$ as a function of the unknown distribution, which is why we write it as ${\theta = T(P)}$. Let ${P_n}$ denote the empirical distribution:

$\displaystyle P_n(A) = \frac{1}{n}\sum_{i=1}^n I_A(X_i).$

In other words, ${P_n}$ is the distribution that puts mass ${1/n}$ at each ${X_i}$.

The estimator is just the function ${T}$ applied to ${P_n}$, that is, ${\hat\theta = T(P_n)}$. For example, if ${\theta}$ is the median of ${P}$ then ${\hat\theta}$ is the median of ${P_n}$ which is just the sample median.

Now let

$\displaystyle R_n = \sqrt{n}(\hat\theta - \theta).$

We use ${R_n}$ because typically it converges in distribution to some well-defined distribution (such as a Normal). Now let ${H_n}$ denote the (unknown) distribution of ${R_n}$:

$\displaystyle H_n(t) = \mathbb{P}(R_n \leq t).$

Suppose, for a moment, that we did know ${H_n}$. We could then find the ${\alpha/2}$ quantile ${t_{\alpha/2}}$ and the ${1-\alpha/2}$ quantile ${t_{1-\alpha/2}}$, namely,

$\displaystyle t_{\alpha/2} = H_n^{-1}(\alpha/2)\ \ \ \mbox{and}\ \ \ t_{1-\alpha/2} = H_n^{-1}(1-\alpha/2).$

It follows that

$\displaystyle \mathbb{P}( t_{\alpha/2} \leq R_n \leq t_{1-\alpha/2}) = 1- \alpha.$

Continuing with the fantasy that we know ${H_n}$, define

$\displaystyle \Omega_n = \left[ \hat\theta - \frac{t_{1-\alpha/2}}{\sqrt{n}},\ \hat\theta - \frac{t_{\alpha/2}}{\sqrt{n}}\right].$

Now I will show you that ${\Omega_n}$ is an exact ${1-\alpha}$ confidence interval. This follows since

$\displaystyle \mathbb{P}(\theta \in \Omega_n) = \mathbb{P}\left(\hat\theta - \frac{t_{1-\alpha/2}}{\sqrt{n}} \leq \theta \leq \hat\theta - \frac{t_{\alpha/2}}{\sqrt{n}}\right)= \mathbb{P}(t_{\alpha/2} \leq R_n \leq t_{1-\alpha/2}) = H_n(t_{1-\alpha/2}) - H_n(t_{\alpha/2})= (1-\alpha/2) - \alpha/2 = 1-\alpha.$

We engineered ${\Omega_n}$ so that the last line would be exactly ${1-\alpha}$. The strange form of ${\Omega_n}$ is explained by the fact that we really have a probability statement for ${R_n}$ which we then manipulate into the form of an interval for ${\theta}$. (You can check that if ${H_n}$ were standard Gaussian, then using the symmetry of the Gaussian, the interval could be re-written in the more familiar looking form ${\hat\theta\pm z_{\alpha/2}/\sqrt{n}}$ where ${z_{\alpha/2}}$ is the upper-tail quantile of a Normal.)

The problem is that we don’t know ${H_n}$ and hence we don’t know ${t_{\alpha/2}}$ or ${t_{1-\alpha/2}}$. The bootstrap is a method for approximating ${H_n}$. Let ${B}$ be a large number (for example, ${B=100,000}$.) Now do this:

1. Draw ${n}$ observations ${X_1^*,\ldots, X_n^*}$ from ${P_n}$ and compute ${\hat\theta^*}$ from these new data.
2. Repeat step 1 ${B}$ times yielding values ${\hat\theta_1^*,\ldots, \hat\theta_B^*}$.
3. Approximate ${H_n}$ with

$\displaystyle \hat H_n(t) = \frac{1}{B}\sum_{j=1}^B I\Bigl( \sqrt{n}(\hat\theta_j^* - \hat\theta)\leq t\Bigr)$

where ${I}$ denotes the indicator function.

4. Find the quantiles ${\hat t_{\alpha/2}}$ and ${\hat t_{1-\alpha/2}}$ of ${\hat H_n}$ and construct ${C_n}$ as defined earlier.

The interval ${C_n}$ is the same as ${\Omega_n}$ except we use the estimated quantiles for ${C_n}$. What we are doing here is estimating ${H_n}$ by using ${P_n}$ as an estimate of ${P}$. (That’s why we draw ${X_1^*,\ldots, X_n^*}$ from ${P_n}$.) If ${\hat H_n}$ is close to ${H_n}$ then ${\hat t_{\alpha/2} \approx t_{\alpha/2}}$ and ${\hat t_{1-\alpha/2} \approx t_{1-\alpha/2}}$ and then ${C_n \approx \Omega_n}$.

There are two sources of error. First we approximate

$\displaystyle H_n(t)=\mathbb{P}(\sqrt{n} (\hat\theta - \theta))$

with

$\displaystyle H_n^*(t)=\mathbb{P}_n(\sqrt{n} (\hat\theta^* - \hat\theta)).$

Essentially, we are replacing ${P}$ with ${P_n}$. Second, we are approximating ${H_n^*(t)}$ with

$\displaystyle \hat H_n(t) = \frac{1}{B}\sum_{j=1}^B I\Bigl( \sqrt{n}(\hat\theta_j^* - \hat\theta)\leq t\Bigr).$

This second source of error is negligible because we can make ${B}$ as large as we want.

Remark: A moment’s reflection should convince you that drawing a sample of size ${n}$ from ${P_n}$ is the same as drawing ${n}$ points with replacement from the original data. This is how the bootstrap is often described but I think it is clearer to describe it as drawing ${n}$ observations from ${P_n}$.

2. Why Does It Work?

If ${\hat H_n}$ is close to ${H_n}$ then the bootstrap confidence interval will have coverage close to ${1-\alpha}$. Formally, one has to show that

$\displaystyle \sup_t |\hat H_n(t) - H_n(t)| \stackrel{P}{\rightarrow} 0$

in which case

$\displaystyle \mathbb{P}(\theta\in C_n) \rightarrow 1-\alpha$

as ${n\rightarrow\infty}$.

It is non-trivial to show that ${\sup_t |\hat H_n(t) - H_n(t)| \stackrel{P}{\rightarrow} 0}$ but it has been shown in some generality. See Chapter 23 of van der Vaart (1998) for example.

3. Why Does It Fail?

The bootstrap does not always work. It can fail for a variety of reasons such as when the dimension is high or when ${T}$ is poorly behaved.

An example of a bootstrap failure is in the problem of estimating phylogenetic trees. The problem here is that ${T(P)}$ is an extremely complex object and the regularity conditions needed to make the bootstrap work are unlikely to hold.

In fact, this is a general problem with the bootstrap: it is most useful in complex situations, but these are often the situations where the theory breaks down.

4. What Do We Do?

So what do we do when the bootstrap fails? One answer is: subsampling. This is a variant of the bootstrap that works under much weaker conditions than the bootstrap. Interestingly, the theory behind subsampling is much simpler than the theory behind the bootstrap. The former involves little more than a simple concentration inequality while the latter uses high-powered techniques from empirical process theory.

So what is subsampling?

Stay tuned. I will describe it in my next post. In the meantime, I’d be interested to hear about your experiences with the bootstrap. Also, why do you think the bootstrap is not more popular in machine learning?

References

Efron, Bradley. (1979). Bootstrap methods: Another look at the jackknife. The Annals of Statistics, 1-26.

Efron, Bradley, and Tibshirani, R. (1994). An Introduction to the Bootstrap. Chapman and Hall.

Holmes, Susan. (2003). Bootstrapping phylogenetic trees: Theory and methods. Statistical Science, 241-255.

van der Vaart, A. (1996). Asymptotic Statistics. Cambridge.

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