Show that $\int_{-\infty}^{a_1} (a_1-x)^r f(x) \mathrm{d} x$ and $\int_{a_n}^\infty (x - a_n)^r f(x) \mathrm{d} x$ are of order $\mathcal{O}(n^{-r})$

Question

Suppose that $f(x)$ is a smooth probability density function on $\mathbb R$ and denote by $a_i$ the $\frac{2i-1}{2n}$-th quantile of $F$ for $1\leq i \leq n$, where $F(x)$ is the cumulative distribution function corresponding to the density $f(x)$. It is claimed in this monograph (without any proof unfortunately) that both $$\int_{-\infty}^{a_1} (a_1-x)^r f(x) \mathrm{d} x$$ and $$\int_{a_n}^\infty (x - a_n)^r f(x) \mathrm{d} x$$ are of order $\mathcal{O}(n^{-r})$. May I know how should we prove the aforementioned claim? You can safely assume that $f$ has finite moments up to order $r \geq 1$.

---

Remark: I forgot to mention the important condition/assumption on the density $f$. Indeed, it is required that the (smooth) density $f$ has some finite moment of order $q > r$. i.e., $$\int_{\mathbb R} |x|^q f(x) \mathrm{d} x < +\infty$$ for some $q > r$.

Answer 1

Let \begin{equation*} I_r=\int_{a_n}^{\infty}(x-a_n)^rf(x)\,\mathrm{d}x, \tag{1} \end{equation*} where $a_n$ denote the $\frac{2n-1}{2n}$-th quantile of $F$, $F$ is the cumulative function corresponding to density $f(x)$.

In the following, another expression of $I_r$ will be derived, it will be convenient to estimate the order of $I_r$ as $n\to\infty$.

Let \begin{equation*} \overline{F}_1(x)=1-F(x)=\int_{x}^{\infty}f(t)\,\mathrm{d}t,\qquad x\in\mathbb{R}, \tag{2} \end{equation*} Suppose \begin{equation*} \overline{F}_1(x)=O(x^{-(r+\delta)}), \quad \delta>0,\qquad \text{as }\; x\to+\infty, \tag{3} \end{equation*} then \begin{align*} I_r &= \int_{a_n}^{\infty}(x-a_n)^rf(x)\,\mathrm{d}x \\ &= -\int_{a_n}^{\infty}(x-a_n)^r\,\mathrm{d}\overline{F}_1(x)\\ &=-\big[(x-a_n)^r\overline{F}_1(x)\big]_{a_n}^\infty + r\int_{a_n}^{\infty}(x-a_n)^{r-1}\overline{F}_1(x)\,\mathrm{d}x\\ &=r\int_{a_n}^{\infty}(x-a_n)^{r-1}\overline{F}_1(x)\,\mathrm{d}x \end{align*} Now let \begin{equation*} \overline{F}_{k+1}(x)=\int_{x}^{\infty}\overline{F}_k(x)\,\mathrm{d}x, \qquad k\ge 1. \end{equation*} Then, inductively, \begin{align*} \overline{F}_k(x) & =O(x^{-(r+1-k+\delta)}), \quad 1\le k \le r+1, \qquad \text{as }\; x\to\infty, \\ I_r&=\frac{r!}{(r-k)!}\int_{a_n}^{\infty}(x-a_n)^{r-k}\overline{F}_k(x)\, \mathrm{d}x\\ &=-\frac{r!}{(r-k)!}\int_{a_n}^{\infty}(x-a_n)^{r-k}\, \mathrm{d}\overline{F}_{k+1}(x)\\ &=\frac{r!}{(r-k-1)!}\int_{a_n}^{\infty}(x-a_n)^{r-k-1}\overline{F}_{k+1}(x)\, \mathrm{d}x\\ &\cdots\cdots\\ &=r!\int_{a_n}^{\infty} \overline{F}_{r}(x)\,\mathrm{d}x\\ I_r&=r!\overline{F}_{r+1}(a_n). \tag{4} \end{align*} The (4) may be used to estimate the order of $I_r$.

Now suppose \begin{equation*} \lim_{x\to+\infty} x^{r+\delta}\overline{F}_1(x)=c>0, \end{equation*} also written it as \begin{equation*} \overline{F}_1(x)\sim x^{-(r+\delta)}, \end{equation*} then \begin{gather*} \overline{F}_{r+1}(x)\sim x^{-\delta},\\ \frac{1}{2n}=\overline{F}_1(a_n)\sim a_n^{-(r+\delta)},\\ a_n\sim n^{1/(r+\delta)},\\ I_r=r!\overline{F}_{r+1}(a_n)\sim n^{-\delta/(r+\delta)}. \end{gather*}

Exapmle Suppose $f(x)=e^{-x}1_{[0,\infty)}(x)$, then \begin{equation*} \overline{F}_k(x)=e^{-x}, \qquad a_n=\log(2n) \end{equation*} hence from (4) \begin{equation*} I_r=r!\overline{F}_{r+1}(a_n)=\frac{r!}{2n}. \end{equation*}