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This question is a repost from Mathematics Stack Exchange, where it did not receive any answer.

Assume $(X_i)_{i=1}^{\infty}$ is a sequence of i.i.d. real-valued random variables such that $\mathbb E[X_i^2]<\infty$. Denote by $F_X(t) := \mathbb P(X\leq t)$ their common distribution function. The regular Donsker's Theorem states that

\begin{equation}\tag{1} n^{1/2}\left(\frac{\sum_{i=1}^n \mathbb 1_{\{X_i\leq t\}}}{n}-F_X(t)\right)\stackrel{\mathrm d}{\rightarrow} B_0(F(t)), \end{equation}

where $\stackrel{\mathrm d}{\rightarrow}$ denotes convergence in distribution, and $B_0(\cdot)$ is a Brownian Bridge on $[0,1]$. My question is related to a previous one, which dealt with a generalization of the above result.

Let us replace $t$ by $\bar t + t n^{-\alpha}$ in $(1)$, for some fixed $\bar t >0$ and $0<\alpha <1$. Accordingly, we replace the scaling constant $n^{1/2}$ by $n^{(1+\alpha)/2}$. We obtain

\begin{equation}\tag{2} n^{(1+\alpha)/2}\left(\frac{\sum_{i=1}^n \mathbb 1_{\{X_i\leq \bar t + t n^{-\alpha}\}}}{n}-F_X(\bar t + t n^{-\alpha})\right). \end{equation}

My question is: does $(2)$ converge in distribution to some process (necessarily defined on $(-\infty,\infty)$)? If so, is there an explicit characterization? Has this been studied before?

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  • $\begingroup$ At $t=0$ the limit in (2) is just like the limit in (1) at tbar $=0$ except multiplied by $n^{\frac {\alpha} 2} $and so it does not exist. You may want something based on the increments, $\sum_{i=1}^n \mathbb 1_{\{t < X_i\leq \bar t + t n^{-\alpha}\}}$ $\endgroup$ – user83457 Apr 25 '16 at 9:39
  • $\begingroup$ @michael You are correct, and indeed I am essentially looking for a way to re-normalize (2) so that it converges to "something". $\endgroup$ – Indigo Apr 25 '16 at 10:11
  • $\begingroup$ clarification question , $\mathbb{E} \lbrack X \rbrack$ is for the sum of $X_i$, or for every value of $i$? $\endgroup$ – Amir Sagiv Apr 25 '16 at 10:12
  • $\begingroup$ @Amir For every value of $i$. I edited the original post accordingly. $\endgroup$ – Indigo Apr 25 '16 at 10:30
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    $\begingroup$ The correct normalisation is again $n^{1/2}$ and the limit is the same as (1). Maybe you meant to ask a different question? $\endgroup$ – Martin Hairer Apr 25 '16 at 10:45

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