## Abstract

Let $X_{(i\mid n)}$ be the $i$th smallest order statistic from a population with $\operatorname{pdf} f(x)$ and $\operatorname{cdf} F(x)$. When $\tilde{x}$ is the population median, $n$ is the sample size and $G(x) = F(x)(1 - F(x))$, the following are proved. \begin{equation*}\begin{split}E(X_{(n-i+1\mid n)}) + E(X_{(i\mid n)}) \\ = n(n - 2i + 1)\binom{n - 1}{i - 1} \sum^{\lbrack (n-2i+1)/2\rbrack}_{\nu = 0}\frac{(-1)^\nu}{n - 2i + 1 - \nu}\binom{n - 2i + 1 - \nu}{\nu} \\ \times \Psi_1(i - 1 + \nu),\end{split}\end{equation*} $E(X_{(n-i+1\mid n)}) - E(X_{(i\mid n)}) = n\binom{n - 1}{i - 1} \sum^{\lbrack (n-2i)/2\rbrack}_{\nu = 0}(-1)^\nu\binom{n-2i-\nu}{\nu}\Psi_2(i - 1 + \nu)$ for $i = 1,2,\cdots,\lbrack n/2 \rbrack$ and $E(X_{(i\mid n)}) = n\binom{n - 1}{i - 1}\Psi_1(i - 1)$ for $i = (n + 1)/2$ when $n$ is odd, where $\Psi_1(l) = \int^\infty_{-\infty} x f(x)\{G(x)\}^l dx$ and \begin{equation*}\begin{split}\Psi_2(l) = -\int^{\tilde{x}}_{-\infty} x f(x) \{G(x)\}^l(1 - 4G(x))^{\frac{1}{2}} dx \\ + \int^\infty_{\tilde{x}}x f(x) \{G(x)\}^l(1 - 4G(x))^{\frac{1}{2}} dx.\end{split}\end{equation*} Parallel formulas are obtained for the $k$th order moment.

## Citation

Kosaburo Hirakawa. "Moments of Order Statistics." Ann. Statist. 1 (2) 392 - 394, March, 1973. https://doi.org/10.1214/aos/1176342380

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