Setup: Let $S_n = n^{-1} \sum_{i=1}^n X_n$ denote a sample mean and let $S_n^*$ denote a stationary bootstrap re-sample of $S_n$. Let $F_n(x)$ denote the cumulative distribution function of $\sqrt{n} S_n$ and let $F_n^*(x)$ denote the cdf of $\sqrt{n} (S_n^* - S_n)$; thus $F_n^*(x)$ is conditional on $X_1, ..., X_n$. Assume $X_n$ is near-epoch-dependent on strong mixing base; assume $\mathbb{E} |X_n|^{2+\delta}$ is finite for some $\delta > 0$; and assume $\mathbb{E} X_n = 0, \forall n$. Then the standard stationary bootstrap result holds, ie:

\begin{equation} \sup_{x \in \mathbb{R}} |F_n(x) - F_n^*(x)| \overset{\mathbb{P}}{\rightarrow} 0, \: \mathrm{as} \: n \rightarrow \infty \end{equation}

Question: What (if any) additional conditions are needed for:

\begin{equation} \left| \int_a^b x^{1+\alpha} dF_n^*(x) - \int_a^b x^{1+\alpha} dF_n(x) \right| \overset{\mathbb{P}}{\rightarrow} 0, \: \mathrm{as} \: n \rightarrow \infty , \end{equation}

for some $\alpha > 0$, and some arbitrary choice of $a < b$ (possibly $a = -\infty$ and $b = \infty$)?

In words, given uniform convergence in probability of two cdfs (one of them conditional), what additional conditions are necessary to be certain that arbitrary moments also converge in probability?

Additional Information 1: I asked a very closely related question on CrossValidated here, and was informed (I think) that the moments do converge. However, the answer did not offer any proof, other than an oblique reference to the uniform convergence theorem in the comments, which actually raised more questions for me than it answered.

Additional Information 2: I'm fairly sure the result follows if the convergence in the setup is strengthened to almost sure, see Xiong and Li (2008) "Some Results on the Convergence of Conditional Distributions". However, I'm specifically interested in the case of convergence in probability.

  • $\begingroup$ I do not understand what you are trying to prove since the moment of your 'limit' random variable depends on n. I assume you mean to do this with absolute values of differences of the moments going to zero. I have not worked this out, but my guess is that it is necessary and sufficient that the sequence of moments is uniformly integrable. $\endgroup$ – Chris Janjigian Dec 10 '13 at 12:59
  • $\begingroup$ @ChrisJanjigian Yes, you are correct - I've fixed the second equation (sorry, the way I had it written before was sloppy). Yes, I suspect uniform integrability might be sufficient, but I'm not confident enough in my knowledge in this area to be sure. I'll offer a bounty in an hour or two to try and elicit a response with proof. :-) ps, thanks for your comment. $\endgroup$ – Colin T Bowers Dec 11 '13 at 0:40
  • $\begingroup$ In my opinion, there are some real conceptual problems with this question. I suggest carefully examining standard definitions to ensure you're asking the question you want answered. (1) Random variables converge in probability, functions (or cdfs) don't. So I'm not sure what you mean by the $\overset{\mathbb{P}}{\rightarrow}$ notation. (2) How's a conditional distribution differ from, well, any other distribution? One possibility (I really doubt this is what you mean) is that the conditional distribution is a random distribution that depends on the conditioning information. Please clarify. $\endgroup$ – Will Nelson Dec 11 '13 at 6:49
  • $\begingroup$ @WillNelson The convergence in probability notation is appropriate. Are you familiar with bootstrap methods in statistics? That is the motivation for the question. In that literature, the bootstrap cumulative distribution function (evaluated at a point) is conditional on the original sample of data and so is itself a random variable. Hence the convergence in probability. See eg Politis and Romano (1994) "The Stationary Bootstrap" equation 12, or the reference I've given in the section "Additional Information 2" for more detail. $\endgroup$ – Colin T Bowers Dec 11 '13 at 8:32
  • $\begingroup$ It seems you are assuming much more setup and structure to this problem than is evident in your Setup section. Why not try to boil the setup down to its essential components and clearly state them in your question? For example, is $F_n$ fixed or random? And $F_n^*$? And I don't think most probability theorists would have any idea how to interpret $F_n^*$ described as a "conditional cdf" without knowing the random variable being conditioned and the $\sigma$-algebra (or filtration) against which it's being conditioned... $\endgroup$ – Will Nelson Dec 11 '13 at 9:08

This addresses the case where $a$ and $b$ are fixed and finite.

Throughout, let $V(\cdot)$ denote the total variation of a function. The domain should be clear from context.

Lemma: Suppose $f$,$F$, and $G$ are functions of bounded variation on $[a,b]$ and suppose $|F(x)-G(x)|\le\epsilon$ for all $x\in[a,b]$. Let $M= \sup_{x\in[a,b]} |f(x)|$. Then $$ \left|\int_a^b f(x) dF(x) - \int_a^b f(x) dG(x)\right|\le \left(2M + V(f)\right)\epsilon. $$

Proof Integrate by parts: \begin{eqnarray} \int_a^b f(x) dF(x) - \int_a^b f(x) dG(x) &=& f(b)\left(F(b)-G(b)\right) - f(a)\left(F(a)-G(a)\right) \\ & & \ \ - \int_a^b F(x) - G(x)\ df(x). \end{eqnarray} Bounding each of the three terms on the right separately gives the result.

Corollary: Let $f:\mathbb{R}\to\mathbb{R}$ be a function of bounded variation. Suppose $F_n$ is a sequence of cdfs and $F_n^*$ is a sequence of random cdfs (on some underlying probability space such that the expressions below make sense). If $$ \sup_{x\in\mathbb{R}} |F_n(x)-F_n^*(x)| \overset{\mathbb{P}}{\rightarrow} 0, $$ then $$ \int f(x) dF_n(x) - \int f(x) dF_n^*(x) \overset{\mathbb{P}}{\rightarrow} 0. $$

Proof: Since $f$ has bounded variation, it is bounded. Let $M$ be such that $|f(x)|\le M$ for all $x$. Fix $\epsilon>0$. Let $N$ be such that $n>N$ implies $$ \mathbb{P}\left(\sup_{x\in\mathbb{R}} |F_n(x) - F_n^*(x)|<\frac{\epsilon}{2M+V(f)}\right) > 1-\epsilon. $$ From the lemma, for all $n>N$, $$ \mathbb{P}\left(\left|\int f(x)\ dF_n(x) - \int f(x)\ dF_n^*(x)\right|<\epsilon\right) > 1-\epsilon. $$ The result follows.

Comment: The integrals here are over $\mathbb{R}$, but one can take $$ f(x) = \begin{cases} x^{1+\alpha} \text{ if $x\in[a,b]$}\\ 0 \text{ if $x\notin[a,b]$} \end{cases} $$ for the original question.

  • $\begingroup$ This is very helpful, thanks Will. Next week I'll see if I can extend the ideas here to indefinite integrals. Bounty awarded. $\endgroup$ – Colin T Bowers Dec 17 '13 at 11:46

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