# Moment problem for order statistics

Problem:

Let $X_{1},\cdots,X_{n}$ be samples from $X$. If $E\left|X\right|^{a}<\infty$ for some $a>0,$ and $n,k,r$ satisfies $r\leq a\cdot\min\left(k,n-k+1\right),$ then $E\left|X_{\left(k\right)}\right|^{r}<\infty$.

Attempted Solution

Since $E\left|X\right|^{a}<\infty,$$E\left[X\right]^{r}<\infty$ for any $r\in\left[0,a\right].$ Note that \begin{align*} +\infty>E\left|X\right|^{a}=\int_{\mathbb{R}}\left|t\right|^{r}dF\geq\int_{x}^{+\infty}\left|t\right|^{r}dF=\int_{\mathbb{R}}\left|t\right|^{r}1_{[x,+\infty)}\left(t\right)dF. \end{align*} Since $\left|t\right|^{r}1_{[x,+\infty)}\left(t\right)\rightarrow0$, it follows from DCT that $\int_{x}^{+\infty}\left|t\right|^{r}dF\rightarrow0.$ On the other hand, we have \begin{align*} \int_{x}^{\infty}\left|t\right|^{r}dF & \geq\left|x\right|^{r}P\left(X>x\right)\geq0. \end{align*} Take the limit and we get \begin{align*} 0=\lim_{x\rightarrow\infty}\int_{x}^{+\infty}\left|t\right|^{r}dF\geq\lim_{x}\left|x\right|^{r}P\left(X>x\right)\geq0, \end{align*} which implies \begin{align*} \lim_{x}\left|x\right|^{r}P\left(X>x\right)=\lim_{x}\left|x\right|^{r}\left(1-F\left(x\right)\right)=0,\ \left|x\right|^{r}\left(1-F\left(x\right)\right)<\infty. \end{align*} Next by integration by parts we have ($G$ is the survival function) \begin{align*} \infty> & \int_{0}^{\infty}x^{a}dF=\int_{0}^{\infty}x^{a}dP\left(X\leq x\right)\\ = & \int_{0}^{\infty}x^{a}d\left[1-P\left(X>x\right)\right]\\ = & -\int_{0}^{\infty}x^{a}dP\left(X>x\right)\\ = & -\left[x^{a}P\left(X>x\right)\right]_{0}^{\infty}+a\int_{0}^{\infty}x^{a-1}P\left(X>x\right)dx\\ = & a\int_{0}^{\infty}x^{a-1}\left[1-F\left(x\right)\right]dx. \end{align*} Hence, we have $a\int_{0}^{+\infty}x^{a-1}\left[1-F\left(x\right)\right]dx<\infty.$ Next, let $G_{k}$ be the distribution function of $X_{\left(k\right)}.$ It is easy to see that \begin{align*} G_{k}\left(x\right)= & P\left(X_{\left(k\right)}<x\right)=P\left(\left(X_{1},\cdots,X_{n}\right)\mbox{ has at least }k\mbox{ elements }\leq x\right)\\ = & \sum_{i=k}^{n}\binom{n}{i}\left[F\left(x\right)\right]^{i}\left(1-F\left(x\right)\right)^{n-i}. \end{align*} Note that we can write \begin{align*} E\left|X_{\left(k\right)}\right|^{r}= & \int_{-\infty}^{0}\left|x\right|^{r}dG_{k}+\int_{0}^{+\infty}x^{r}dG_{k}. \end{align*} We talk about the second integral first. Note using similar method as before, we get \begin{align*} \int_{0}^{+\infty}x^{r}dG_{k}= & -\int_{0}^{\infty}x^{r}dP\left(X_{\left(k\right)}>x\right)\\ = & -\left[x^{r}P\left(X_{\left(k\right)}>x\right)\right]_{0}^{+\infty}+r\int_{0}^{\infty}x^{r-1}P\left(X_{\left(k\right)}>x\right)dx. \end{align*} Since \begin{align*} x^{r}P\left(X_{\left(k\right)}>x\right)= & x^{r}P\left(\left(X_{1},\cdots,X_{k}\right)\mbox{ has at most }\left(k-1\right)\ \leq x\right)\\ = & x^{r}\left(\sum_{i=0}^{k-1}\binom{n}{i}\left[F\left(x\right)\right]^{i}\left[1-F\left(x\right)\right]^{n-i}\right)\\ = & \sum_{i=1}^{k-1}\binom{n}{i}x^{r}\left[F\left(x\right)\right]^{i}\left[1-F\left(x\right)\right]^{n-i}\\ \leq & \sum_{i=1}^{k-1}\binom{n}{i}x^{r}\left[1-F\left(x\right)\right]\rightarrow0, \end{align*} where the last convergence results follows from previous argument, then it follows that \begin{align*} \int_{0}^{+\infty}x^{r}dG_{k}= & r\int_{0}^{+\infty}x^{r-1}P\left(X_{\left(k\right)}>x\right)dx\\ = & r\int_{0}^{+\infty}x^{r-1}\left(\sum_{i=0}^{k-1}\binom{n}{i}\left[F\left(x\right)^{i}\right]\left[1-F\left(x\right)\right]^{n-i}\right)dx\\ = & r\sum_{i=0}^{k-1}\binom{n}{i}\int_{0}^{+\infty}\left\{ x^{a}\left[1-F\left(x\right)\right]\right\} ^{\frac{r-a}{a}}\left[1-F\left(x\right)\right]^{n-i-\frac{r}{a}}\left[F\left(x\right)\right]^{i}x^{a-1}\left[1-F\left(x\right)\right]dx.\tag{1} \end{align*} Note that

(1)Since $x^{a}\left[1-F\left(x\right)\right]<\infty$ and $\lim_{x}x^{a}\left[1-F\left(x\right)\right]\rightarrow0,$ $\left\{ x^{a}\left[1-F\left(x\right)\right]\right\} ^{\frac{r-a}{a}}\leq M.$

(2) If $n-i-\frac{r}{a}\geq0,$ then $\left[1-F\left(x\right)\right]^{n-i-\frac{r}{a}}$ is bounded by 1. Since $i\in\{0,\cdots,k-1\},$ we have $r\leq a\left(n-i\right),\forall i\iff r\leq a\left(n-k+1\right).$

So if $r\leq a\left(n-k+1\right)$, then \begin{align*} \left(1\right)\leq & M\int_{0}^{\infty}x^{a-1}\left[1-F\left(x\right)\right]dx<\infty, \end{align*} as desired.

Question I am stuck at showing the another part of the integral. Namely, $\int_{-\infty}^0|x|^rdG_k<\infty$ if $r<ak$