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Suppose that $h:\mathbb{R}\to\mathbb{R}$ is infinitely differentiable. Define \begin{equation} k(w)=\left\{ \begin{array}{ll} \frac{d}{dw}\left(\frac{h(w)-h(0)}{w}\right)&w\neq 0,\\ 2h''(0)&w=0. \end{array} \right.\tag{*} \end{equation} Now fix $w\in\mathbb{R}$ and let $\phi$ denote the standard normal PDF. The claim is: $$ R\equiv\frac{1}{n}\int_{-\infty}^{\sqrt{n}w}k(v/\sqrt{n})\phi(v)dv $$ is $O(n^{-1})$.

I am not so sure about the importance of the functional form of $k$ as specified in (*). One certain thing is that $k$ is continuous.

I know that I should show that for there is an upperbound for $|\int_{-\infty}^{\sqrt{n}w}k(v/\sqrt{n})\phi(v)dv|$ for all large $n$. But how do I do this? Feel free to make any necessary assumption about $h$ and $k$.

Thank you!

p.s.

Reference: the above was stated as a remark without proof in Butler (2007). The expression $R$ above was the $\int pdq$ part in some integration by parts: $\int qdp=qp-\int pdq$.

Kim Jong Un
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Using the result here one can see that $k\in C^\infty(\mathbb R)$. But if this is all we have, the statement is false: take $k(w)=\exp(w^4)$, and then the integral is infinite.

You need some assumptions about the behavior of $k$ at infinity (or of $h$, from which $k$ was obtained). For example, if $k$ is bounded by $M$, then $$\int_{-\infty}^{\sqrt{n}w}k(v/\sqrt{n})\phi(v)\,dv \le M\int_{-\infty}^{\infty} \phi(v)dv = M$$

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  • Thank you Thursday! Your comment led me to checking the original reference from which Butler (2007) extracts the claim above. In the original paper (Temme (1982)), there was indeed some assumption on $h$. I'm feeling now that Butler (2007) has misled the readers by stating the claim above without an informative qualifying phrase. Even "under certain conditions" would at least warn the readers about the omission of certain details. – Kim Jong Un Aug 24 '14 at 02:41