6
$\begingroup$

I need to prove the validity of the bootstrap for finding confidence sets defined by quantiles of a certain statistic of a general distribution function. I have spent several days reading proofs of the consistency of several bootstrap method and I find myself extremely confused. I keep going in circles because there are some fundamental issues that keep me from making the link to my particular setup.

This is my setup:

Suppose $X_1,\dots, X_n$, $i=1,\dots,n$ are an iid sample of an $\mathbb{R}^d$-valued random variable with unknown distribution function $F$. We assume that $F$ admits a density with respect to the Lebesgue measure. Define $\theta^*:=\mathbb{E}(X)$ and its plug-in estimate $\hat{\theta}=\frac{1}{n}\sum_{i=1}^nX_i$. I need to proof the validity of bootstrap confidence intervals for the following statistic $T(F)$

$$ T(F)=\|\hat{\theta} - \theta^*\|_2^2 $$

I am considering a weighted/multiplier bootstrap with independent weights $w_i \sim \mathcal{N}(1,1)$. (Or maybe i should consider normalized weights?) Therefore, define the measure $F^b:=\sum_{i=1}^{n}w_i\delta_{X_i}$ and $\theta^b:= \sum_{i=1}^{n}w_iX_i$ Hence, the bootstrap version $T(F^b)$ of $T$

$$ T(F^b)=\|\theta^b-\hat{\theta}\|_2^2 $$

Also, define for $z>0$ the set

$$ \mathcal{E}(z):=\{\theta: \|\hat{\theta}-\theta\|_2^2\leq z\} $$

We say that $\mathcal{E}(z_{\alpha})$ is a $100(1-\alpha)$ confidence set if

$$ P(\mathcal{E}(z_{\alpha}) \not\owns \theta^*)=1-\alpha $$

The bootstrap counterpart of $\mathcal{E}(z)$ is

$$ \mathcal{E}^b(z):=\{\hat{\theta}: \|\theta^b-\hat{\theta}\|_2^2\leq z\} $$ Since $T$ is expressed via the mean, I would assume this goes under Bootstrapping the mean, although I don't know if that's the right way to put it.

I need to prove that the above bootstrap procedure is valid for the construction of the above confidence sets. Or, more precisely, I need to find the least restrictive conditions for which this, or any other multiplier bootstrap procedure with iid weights is valid and need to proof it

Let's first define validity

Validity and metrics The least thing I have to expect from a bootstrap procedure is that it is asymptotically valid. That means that the distribution of the statistic $T$ under the bootstrap measure mimics well the distribution of $T$ under the true measure as $n$ grows to infinity. Therefore, define the following quantities $$ G_b(x):=P_{F^b}(T(F^b) \leq x) $$ and the distribution function of $T(F)$ as $$ G_n(x):=P(T(F) \leq x) $$

The following definition of consistency is taken from these lecture notes

Let $\rho(\cdot,\cdot)$ be a metric on the space of CDFs. $G_b$ is weakly consistent under $\rho$ for $T$ if $\rho(G_b, G_n) \xrightarrow[\mathbb{P}] 0$ strongly, if $\rho(G_b, G_n) \xrightarrow[a.s] 0$

The two metrics that are most frequently used in the literature are the following two. Let $F$ and $G$ be two distribution functions.

  1. Kolmogorov metric

$$ K(F,G)= sup_{x}\|F(x)-G(x)\| $$ It appears to me that there are many more Theorems that give conditions for the consistency of the Bootstrap for the Kolmogorov metric. It seems very natural to me, since it gives a uniform comparison of the distribution functions.

Also, there are a couple of results that seem useful to me, for example the fact that weak convergence for distributions where the distribution function of the limit is continuous implies convergence in the Kolmogorov metric. (see Lemma 2.11. in van der Vaart, Asymptotic statistics)

  1. The Mallows or Wasserstein or Monge-Kantorovich Metric

$$ W_2=inf_{\Gamma_{2,F,G}} \mathbb{E}(\|Y-X\|^2)^{1/2} $$

where $\Gamma_{2,F,G}$ is the class of all joint distributions of $(X,Y)$ with marginals $F$ and $G$, each with finite second moment.

This metric is used much less often then the Kantorovich metric. It has the interesting property that convergence in $W_2$ is equivalent to weak convergence and convergence in the second moment.

Now, there are $3$ very basic questions that I kept on asking myself

1. Focus on the quantity inside the norm or on the whole statistic - I figure that maybe it is the right thing to focus on showing that the distribution of $\theta^b- \hat{\theta}$ mimics the distribution $\hat{\theta}-\theta^*$. For example, if I could bound the total variation between these two distribution, I would have the desired result. - On the other hand, the more direct way would be to somehow show directly that, $G^b$ converges to the same limit as $G_n$, and then use the above fact

2. How to get the asymptotics in there? I want to proof asymptotic consistency and basically everything I read about bootstrapping the mean considers test statistic scaled and centered test statistics like. $$ \sqrt{n}(\hat{\theta}-\theta) $$ And then construct asymptotic confidence intervals from there. Do i need to do that as well? Do I actually need to consider the asymptotic distribution of something like $$ \sqrt{n}T(F) $$

3. How would my result differ if I chose to show convergence in Mallows distance? The proofs I have seen on convergence in Wasserstein distance seemd a little bit less technical than the ones for Kolmogorov. Also, in the above lecture notes it is stated that.

because the bootstrap is primarily used for estimating the CDF, mean and variance of a statistic, consistency in $W_2$ is just the right result.

Could anyone enlighten me, on what the author means by that.

If I could get answers to any of these questions, it would considerably improve my overall life satisfaction.

$\endgroup$

closed as unclear what you're asking by kjetil b halvorsen, Michael Chernick, mdewey, whuber Sep 10 '18 at 17:23

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ You lost me at the second paragraph, because $T$ is not a statistic: it depends on the unknown distribution function $F$. The concept of confidence interval therefore doesn't even apply. What are you actually trying to do? $\endgroup$ – whuber Dec 28 '17 at 21:59