\documentclass[fleqn,11pt]{article} \usepackage{amsmath} \renewcommand{\baselinestretch}{1.3} \setlength{\textwidth}{15.3cm} \setlength{\textheight}{20cm} \setlength{\oddsidemargin}{1cm} \setlength{\evensidemargin}{1cm} %\setlength{\mathindent}{0.1in} %equations intended by one inch %\setlength{\jot}{.2in} %increases space b/w lines of eqnarray by .2in \newtheorem{theorem}{\bf Theorem} \newtheorem{proposition}[theorem]{\bf Proposition} \newtheorem{lemma}{\bf Lemma} \newtheorem{corollary}[theorem]{\bf Corollary} \newtheorem{definition}[theorem]{\bf Definition} \newtheorem{exam}{\bf Example} \newenvironment{proof}% {\par \noindent {\it Proof. }}% {\par \indent} \newenvironment{counterexample}% {\par \noindent {\bf Counterexample:}}% {\par \indent} % the remarks \newenvironment{remark}% {\par \noindent {\bf Remark: }}% {\par \indent} % abbreviations of existing commands \newcommand{\be}{\begin{equation}} \newcommand{\bes}{\begin{displaymath}} \newcommand{\ee}{\end{equation}} \newcommand{\ees}{\end{displaymath}} \newcommand{\ra}{\rightarrow} \newcommand{\Var}{\mbox{Var}} % frequently used text pieces and formulas \newcommand{\rap}{\stackrel{P}{\rightarrow}} \newcommand{\raL}{\stackrel{\mathcal{L}}{\longrightarrow}} \newcommand{\deq}{\stackrel{d}{=}} % other customized commands \newcommand{\mathbold}[1]{\mbox{\boldmath $\bf#1$}} \newcommand{\bin}[2]{% \renewcommand{\arraystretch}{.5} \left({\setlength{\arraycolsep}{0mm} \begin{array}{c} #1 \\ #2 \end{array}} \right) \renewcommand{\arraystretch}{1} } \newcommand{\gdots}{\cdot \ldots \cdot} \usepackage{latexsym}% necessary for end of proof symbol according to % Jim \usepackage{rotate,epsfig}% necessary for figures %*********************************************** \begin{document} \title{Prediction intervals for future records and order statistics coming from two parameter exponential distribution} \author{ J. Ahmadi\footnote{ Corresponding author.\newline E-mail: ahmadi-j@um.ac.ir (J. Ahmadi), ta$_{-}$mi182@stu-mail.um.ac.ir (S. M. T. K. MirMostafaee). %{ The first author is a member Ordered and Spatial Data Center of Excellence of Ferdowsi University of Mashhad.} } ~ and S. M. T. K. MirMostafaee\\ {\small {Department of Statistics, School of Mathematical Sciences, }}\vspace{-0.1cm}\\ {\small { Ferdowsi University of Mashhad, P. O. Box 91775-1159, Mashhad,\ \ Iran}}} %\date{} \maketitle \begin{abstract} In this paper, we study the problem of predicting the $s$th record in the future sequence based on the order statistics from two parameter exponential distribution. The prediction intervals for the $s$th order statistic and also for the total lifetime in a future sample of size $m$ from two parameter exponential distribution are obtained on the basis of the first $n$ records coming from the same distribution. Finally, numerical computations are given for illustration of the proposed procedure. \end{abstract} \noindent {\it Key Words}: Gamma distribution; Negative binomial; Pivotal quantity; Two-sided prediction interval.\\ \noindent\textbf{Mathematics Subject Classification}~ Primary 62G30; Secondary 62E15. \section{ Introduction } Prediction of future events on the basis of the past and present knowledge is a fundamental problem of statistics, arising in many contexts and producing varied solutions. As in estimation, a predictor can be either a point or an interval predictor. Parametric and nonparametric predictions have been considered in literature. In many practical data-analytic situations we are interested in using the observations from an initial sample to construct an interval that will have a present probability of containing some statistic based on a future sample of observations from the same underlying distribution. Such an interval is called a {\it{prediction interval}} for the statistic of interest. \\ Suppose that $X_1, \cdots, X_n$ are independent and identically distributed (iid) observations from an absolutely continuous cumulative distribution function (cdf) $F(x)$ and probability density function (pdf) $f(x)$. The order statistics of the sample are defined by the arrangement of $X_1, \cdots, X_n$ from the smallest to the largest, denoted as $X_{1:n}\leq X_{2:n}\leq \cdots\leq X_{n:n}$. These statistics have been used in a wide range of problems, including robust statistical estimation, detection of outliers, characterization of probability distributions and goodness-of-fit tests, entropy estimation, analysis of censored samples, reliability analysis, quality control and strength of materials; for more details, see Arnold \textit{et al. }(1992), David and Nagaraja (2003) and references therein. Several authors have considered prediction intervals for future order statistics based on observed order statistics in parametric and nonparametric setting. See for example, Kaminsky and Nelson (1998), Lawless (1977), Nagaraja (1984), Kaminsky and Rhodin (1985), Chou (1988), Nagaraja (1995), Raqab and Nagaraja (1995), Hsieh (1997), Abdel-Aty et al. (2007) and so on. Let $X_1,X_2,...$ be a sequence of iid random variables having an absolutely continuous cdf $F(x)$ and pdf $f(x)$. An observation $X_j$ is called an {\it upper record} value if its value exceeds that of all previous observations. Thus, $X_j$ is an upper record if $X_j>X_i$ for every $i0. \end{equation} We study the problem of prediction of the $s$th record value in a future sample from a two parameter exponential distribution on the basis of order statistics from the same underlying distribution. The results are presented in Section 2. In Sections 3 and 4, the prediction intervals for the $s$th order statistic and also for the total lifetime in a future sample of size $m$ from $Exp(\sigma, \mu)$ based on the first $n$ records are obtained. Finally, in Section 5 numerical computations are given for illustration of the proposed procedure. Some tables are provided which can help us to construct appropriate prediction intervals for order statistics and record values. \section{Prediction intervals for record value } Let $X_{1:n}\leq X_{2:n} \leq ... \leq X_{r:n}$ be the $r$ smallest observations in a sample of size $n$ from $Exp(\sigma, \mu)$. Let $R_s$ be the $s$th upper record value in a future sequence from the same distribution. In this section, we are interested in using the $X_{i:n}, 1\leq i\leq r\leq n$ to get information about the value of $R_s$ before actually obtaining the future sequence, i.e., to construct prediction intervals for $R_s$. That is, if $L(X_{1:n}, X_{2:n}, ..., X_{r:n})$ and $U(X_{1:n}, X_{2:n}, ..., X_{r:n})$ are statistics based on the initial sample, then we say that $(L, U)$ is a $100(1-\alpha)\%$ prediction interval for $R_s$ if $ P(L\leq R_s \leq U)=1-\alpha$. To do this, we consider the following pivotal quantity, and use the characteristic that the distribution of this pivotal quantity does not depend on parameters, to give the lower and upper bounds of prediction intervals. Let \begin{equation} \label{eq:quantity-P-record} Z=\frac{n(R_s-X_{1:n})}{T}, \end{equation} where $T=\sum_{i=1}^r (X_{i:n}-X_{1:n})+(n-r)(X_{r:n}-X_{1:n})$. Let $T_s=\sigma(R_s-\mu)$, then it follows that $T_s$ is the $s$th upper record corresponding to standard exponential distribution. By \eqref{eq:pdf-upper-record} and \eqref{eq:pdf-exp}, $T_s$ has a gamma distribution with shape parameter $s$ and scale parameter 1. Also take $T_1=\sigma(X_{1:n}-\mu)$ and $S_1=\sigma T$, then $nT_1$ has a standard exponential distribution whereas $2S_1$ has a chi-square distribution with $2r-2$ degrees of freedom, moreover $T_1$ and $S_1$ are stochastically independent. So from \eqref{eq:quantity-P-record}, $Z$ can be expressed in terms of $T_1, T_s$ and $S_1$ as $Z=\frac{n(T_s-T_1)}{S_1}$. In the next result, we obtained the marginal cdf of $Z$. \begin{lemma} \label{lem-p-record1} Let $X_{1:n}, X_{2:n}, ..., X_{r:n}$ be the $r, (r\geq 2)$ smallest observations in a sample of size $n$ from $Exp(\sigma, \mu)$ and $R_s$ be the $s$th upper record value in a future sequence from the same distribution. Then the cdf of $Z$, as defined in \eqref{eq:quantity-P-record}, is \begin{eqnarray*}Pr(Z\leq z) =\left\{\begin{array}{ll}{\frac{(1-z)^{1-r}}{(n+1)^{s}}} ~&{\rm if }~ z\leq 0, \vspace{.2cm}\\ NB(r-1, p; s) +\sum_{i=0}^{s-1} {{i+r-2} \atopwithdelims ()i}q^i p^{r-1}(n+1)^{i-s} &{\rm if }~ z > 0, \\ \end{array}\right .\end{eqnarray*} where $p=\frac{n}{z+n}$, $q=1-p$ and $NB( r-1, p; s)$ stands for the tail probability at point $s$ of a negative binomial random variable with parameters $r-1$ and $p$ which equals $\sum_{i=s}^\infty {{i+r-2} \atopwithdelims ()i}q^i p^{r-1}$. \end{lemma} \noindent \textbf{Proof.} First we find the distribution of the numerator of $Z$, i.e., $V=n(T_s-T_1)$. Since $T_1$ and $T_s$ are independent, we have \begin{eqnarray*} Pr(V\leq v)&=&\int_0^\infty Pr(n(T_s-T_1)\leq v|T_s=t)dF_{T_s}(t)\\&=&\int_{0}^\infty Pr(nT_1\geq nt-v)dF_{T_s}(t). \end{eqnarray*} By assumptions $nT_1\sim Exp(0,1)$, so $$Pr(nT_1\geq nt-v) =\left\{\begin{array}{ll}{1} ~~~~~&{\rm if }~~~ nt-v \leq 0, \\ e^{-(nt-v)} ~~~~~&{\rm if }~~~ nt-v> 0. \\ \end{array}\right .$$ Therefore by substituting we get \begin{equation} \label{eqv} Pr(V\leq v)=Pr(nT_s\leq v)+\int_{\frac{v}{n}}^\infty e^{-(nt-v)}f_{T_s}(t)dt. \end{equation} The following equality \begin{equation} \label{eq:identity} \int_t^\infty \frac{\sigma^n x^{n-1}}{(n-1)!}e^{-\sigma x}dx=\sum_{j=0}^{n-1}e^{-\sigma t}\frac{(\sigma t)^j}{j!}, \end{equation} represents the relationships between the incomplete gamma function and sum of Poisson probabilities. So for $v>0$, using \eqref{eq:identity} and the fact that $T_s\sim \Gamma(s,1)$, \eqref{eqv} can be rewritten as \begin{eqnarray}\label{eq1} Pr(V\leq v)=e^{-\frac{v}{n}}\sum_{i=s}^\infty \frac{\left(\frac{v}{n}\right)^i}{i!}+e^{-\frac{v}{n}}\sum_{i=0}^{s-1} \frac{\left(\frac{v}{n}\right)^i}{i!}(n+1)^{i-s}, \end{eqnarray} For $v\leq 0$ from \eqref{eqv} we have \begin{eqnarray}\label{eq02} Pr(V\leq v)=\frac{e^v}{(n+1)^{s}}. \end{eqnarray} Now we find the cdf of $Z$; using conditional argument we have \begin{eqnarray}\label{eq3}\nonumber Pr(Z\leq z)&=&\int_0^\infty Pr(Z\leq z|S_1=u)d G_{S_1}(u)\\&=&\int_0^\infty Pr(V\leq z u) \frac{u^{r-2}}{(r-2)!}e^{-u}du, \end{eqnarray} where the second equality is obtained by noting that $V$ and $S_1$ are stochastically independent. By substituting (\ref{eq1}) and (\ref{eq02}) in (\ref{eq3}) the result follows. \hfill{$\Box$}\\ In Lemma \ref{lem-p-record1}, we have proved that the distribution of $Z=\frac{n(R_s-X_{1:n})}{T}$ does not depend on $\mu$ and $\sigma$ and depends only on $r, n$ and $s$. Thus, we can construct prediction intervals for $R_s$ which are free of $\mu$ and $\sigma$. Let $z_{\gamma}(r,n,s)$ be the $\gamma$th right quantile of $Z$, i.e., $Pr(Z\leq z_{\gamma}(r,n,s))=\gamma$, then we have \begin{equation} \label{prediction-order} Pr\left[ z_{\alpha/2}(r,n,s)\leq \frac{n(R_s-X_{1:n})}{T}\leq z_{1-\alpha/2}(r,n,s) \right]=1-\alpha. \end{equation} Hence $[X_{1:n}+z_{\alpha/2}(r,n,s)\frac{T}{n},~X_{1:n}+z_{1-\alpha/2}(r,n,s)\frac{T}{n}]$ is a two-sided equal tail $100(1-\alpha)\%$ prediction interval for $s$th upper record value in a future sample from $Exp(\mu,\sigma)$ based on $X_{1:n}, ..., X_{r:n}$. This is stated in the following Theorem. \begin{theorem} \label{the-p-record1} Under the assumptions of Lemma \ref{lem-p-record1}, $[X_{1:n}+z_{\alpha}(r,n,s)\frac{T}{n}, X_{1:n}+z_{\beta}(r,n,s)\frac{T}{n}]$ is a prediction interval for $R_s$ whose confidence coefficient is free of $\mu$ and $\sigma$ and is given by $\beta-\alpha$. \end{theorem} We have computed $z_{\gamma}(r,n,s)$ for $\gamma=0.025, 0.05, 0.95$ and $0.975$ and some selected values of $r, n$ and $s$. These are presented in Tables 1--2, which can help us to construct prediction intervals for $R_s$. \section{Prediction intervals for order statistic } Let $ R_1, R_2, ..., R_n$ be the first $n$ record values from $Exp(\mu,\sigma)$. Let $Y_{j:m}$ be the $j$th smallest observation in a future sample of size $m$ from the same distribution. In this section we wish, on the basis of observed $R_1, R_2, ..., R_n$, to construct prediction intervals for $Y_{j:m}$. We seek some pivotal quantity to find this prediction intervals. To do this, consider the quantity \begin{equation}\label{eq:pivot2}W=\frac{Y_{j:m}-R_1}{R_n-R_1}, \end{equation} which does not depend on $\mu$ and $\sigma$. Let $T_{j:m}=\sigma(Y_{j:m}-\mu), ~T_2=\sigma(R_1-\mu)$ and $S_2=\sigma (R_n-R_1)$. It is well-known that $T_2$ has a standard exponential distribution, that $2S_2$ has a chi-square distribution with $2n-2$ degrees of freedom and that $T_2$ and $S_2$ are independent (see Arnold {\it et al.}, 1998). Also $T_{j:m}$ is the $j$th order statistic from a sample of size $m$ from the standard exponential distribution and its pdf is given by \begin{eqnarray*} f_{T_{j:m}}(t)=j {m \choose j} (1-e^{-t})^{j-1}e^{-t(m-j+1)}, \ \ t>0. \end{eqnarray*} So $W$ can be expressed as $W=\frac{T_{j:m}-T_2}{S_2}$. We obtained its cdf, which does not depend on $\mu$ and $\sigma$, in the next result. \begin{lemma} \label{lem-p-order1} Let $R_1, R_2, ..., R_n$ be the first $n,(n\geq 2)$ record values from $Exp(\mu,\sigma)$ and $Y_{j:m}$ be the $j$th smallest observation in a future sample of size $m$ from the same distribution. Then the cdf of $W$, as defined in \eqref{eq:pivot2}, is $$Pr(W\leq w) =\left\{\begin{array}{ll}{\frac{(m-j+1)(1-w)^{1-n}}{(m+1)}} ~&{\rm for }~ w\leq 0, \vspace{.2cm}\\ 1-j{m\atopwithdelims () j}\sum_{i=0}^{j-1}{{j-1}\atopwithdelims () i}(-1)^i\frac{[1+w(m-j+i+1)]^{1-n}}{(m-j+i+1)(m-j+i+2)} &{\rm for }~ w > 0. \\ \end{array}\right .$$ \end{lemma} \noindent \textbf{Proof.} By proceeding the same line of the proof of Lemma \ref{lem-p-record1}, first we find the distribution of $V=T_{j:m}-T_2$. We have \begin{eqnarray*} Pr(V\leq v)&=&\int_0^\infty Pr(T_{j:m}-T_2\leq v|T_{j:m}=t)dF_{T_{j:m}}(t)\\&=&\int_{0}^\infty Pr(T_2\geq t-v)dF_{T_j}(t), \end{eqnarray*} since $T_{j:m}$ and $T_2$ are independent. By assumptions $T_2$ has a standard exponential distribution, so by substituting we obtain $$Pr(V\leq v)=Pr(T_j-v\leq 0)+\int_{v}^\infty e^{-(t-v)}f_{T_j}(t)dt.$$ For $v>0$ we get \begin{eqnarray}\label{eq:or1}Pr(V\leq v)=1-j {m\atopwithdelims () j}\sum_{i=0}^{j-1}\frac{{{j-1}\atopwithdelims () i}(-1)^i\exp[-v(m-j+i+1)]}{(m-j+i+1)(m-j+i+2)} \end{eqnarray} and for $v\leq 0$ we have \begin{eqnarray}\label{eq:or2}Pr(V\leq v)=\frac{e^v(m-j+1)}{m+1}. \end{eqnarray} Let $g_{S_2}(s)$ denote the pdf of $S_2$, using the conditional argument we have \begin{eqnarray}\label{eq:03} Pr(W\leq w)=\int_0^\infty Pr(V\leq w s)g_{S_2}(s)ds, \end{eqnarray} where \eqref{eq:03} is obtained by the fact that $V$ and $S_2$ are independent. Substituting (\ref{eq:or1}) and (\ref{eq:or2}) in (\ref{eq:03}), the result follows by algebraical computations. \hfill{$\Box$}\\ In Lemma \ref{lem-p-order1}, we have shown that the distribution of $W=\frac{Y_{j:m}-R_1}{R_n-R_1}$ does not depend on the parameters and depends only on $n,m$ and $j$. Thus, we can easily construct prediction intervals for $Y_{j:m}$ which are free of $\mu$ and $\sigma$. Let $w_{\gamma}(n,m,j)$ be the $\gamma$th quantile of $W$, i.e., $Pr(W\leq w_\gamma({n,m,j})=\gamma$ and $T_n=R_n-R_1$, then \begin{equation} \label{eq:order1} \left[R_1+ w_{\frac{\alpha}{2}}(n,m,j)T_n,~ R_1+w_{1-\frac{\alpha}{2}}(n,m,j)T_n\right] \end{equation} is a two-sided $100(1-\alpha)\%$ prediction interval for the $j$th smallest observation in a future sample of size $m$ from $Exp(\mu,\sigma)$ on the basis of $R_1, R_2, ..., R_n$. As a consequence, we have the following result. \begin{theorem} \label{the-p-order1} Under the assumptions of Lemma \ref{lem-p-order1}, $[R_1+w_\alpha(n,m,j) T_n, R_1+w_\beta(n,m,j) T_n]$ is a prediction interval for $Y_{j:m}$ whose confidence coefficient is free of $\mu$ and $\sigma$ and is given by $\beta-\alpha$. \end{theorem} We have computed $w_{\gamma}(n,m,j)$ for $\gamma=0.025, 0.05, 0.95$ and $0.975$ and some selected values of $n,m$ and $j$. These are presented in Tables 3 and 4, which can help us to construct prediction intervals for $Y_{j:m}$. \section{Prediction intervals for total time } Let $R_1, R_2, ..., R_n$ be the first $n$ record values from $Exp(\mu,\sigma)$. Let $\bar{Y}_m=\frac{1}{m}\sum_{i=1}^m Y_{i:m}$ be the mean of a future sample of size $m$ from the same distribution. We wish, on the basis of observed $R_1, R_2, ..., R_n$ to construct prediction intervals for $\bar{Y}_m$. To do this, consider the quantity \begin{equation}\label{eq:mean01}U=\frac{\bar{Y}_m-R_1}{R_n-R_1}. \end{equation} Let $T_u=\sigma(\bar{Y}_m-\mu), T_3=\sigma(R_1-\mu)$ and $S_3=\sigma (R_n-R_1)$. As mentioned before, $T_3$ has a standard exponential distribution, $2S_3$ has a chi-square distribution with $2n-2$ degrees of freedom and $T_3$ and $S_3$ are independent. Also $mT_u$ has a gamma distribution with shape parameter $m$ and scale parameter 1. So we can express $U$ in terms of $T_u, T_3$ and $S_3$ as $U=\frac{T_u-T_3}{S_3}$, whose distribution is obtained in the next result. \begin{lemma} \label{lem-p-mean1} Let $R_1, R_2, ..., R_n$ be the first $n,(n\geq 2)$ record values from $Exp(\mu,\sigma)$ and ${\bar{Y}}_m$ be the mean of a future sample of size $m$ from the same distribution. Then the cdf of $U$, as defined in \eqref{eq:mean01}, is $$Pr(U\leq u) =\left\{\begin{array}{ll}{\frac{(1-u)^{1-n}}{(1+m^{-1})^m}} ~&{\rm for }~ u\leq 0, \vspace{.2cm}\\ NB( n-1, p; m)+\sum_{i=0}^{m-1} {{n+i-2} \atopwithdelims ()i}q^i p^{n-1}\left(\frac{m}{m+1}\right)^{m-i} &{\rm for }~ u > 0, \\ \end{array}\right .$$ where $p=\frac{1}{mu+1}$, $q=1-p$ and $NB( n-1, p; m)$ is as defined in Lemma \ref{lem-p-record1}. \end{lemma} \noindent \textbf{Proof.} Proof is similar to that of Lemma \ref{lem-p-record1}. \hfill{$\Box$} \vskip 3mm According to Lemma \ref{lem-p-mean1}, the distribution of $U=\frac{\bar{Y}_m-R_1}{R_n-R_1}$ does not depend on the parameters and depends only on $n$ and $m$. Therefore, we can construct prediction intervals for $\bar{Y}_{m}$ needless of knowing the values of $\mu$ and $\sigma$. Let $u_{\gamma}(n,m)$ be the $\gamma$th quantile of $U$, i.e., $Pr(U\leq u_{\gamma}(n,m))=\gamma$ and $T_n=R_n-R_1$, then $$[R_1+ u_{\alpha/2}(n,m)T_n,~ R_1+u_{1-\alpha/2}(n,m)T_n]$$ is a two-sided $100(1-\alpha)\%$ prediction interval for the mean of a future sample of size $m$ from $E(\mu,\sigma)$, i.e. $\bar{Y}_m$, based on $R_1, R_2, ..., R_n$. Thus, we have the following result. \begin{theorem} \label{the-p-mean1} Under the assumptions of Lemma \ref{lem-p-mean1}, $[R_1+ u_\alpha(n,m) T_n,~ R_1+u_\beta(n,m) T_n]$ is a prediction interval for $\bar{Y}_m$ whose confidence coefficient is free of $\mu$ and $\sigma$ and is given by $\beta-\alpha$. \end{theorem} We have computed $u_{\gamma}(n,m)$ for $\gamma=0.025, 0.05, 0.95$ and $0.975$ and some selected values of $n$ and $m$. These are presented in Table 5, which enable us to construct prediction intervals for $\bar{Y}_m$. \section{Illustrative examples} In this section, we illustrate the proposed procedure of this paper by considering two real data sets. In the first example two schemes prediction intervals are made for future records on the basis of observed order statistics. In the second one the prediction intervals for the future order statistics and the mean of total time are derived based on observed records. \begin{exam} \label{example-records} Prediction intervals for records\vspace{-0.2cm} \end{exam} We consider the following data set due to Grubbs (1971): {\small \begin{center} \begin{tabular}{llllllllll} 162,& 200,&271,&302,&393,&508,&539,&629,&706,&777,\\ 884,&1008,&1101,&1182,&1463,&1603,&1984,&2355,&2880, \end{tabular} \end{center}} which were nineteen military personnel carriers failed in service for one reason or the other at the mileages. The author assumed that these data follow a two parameter exponential distribution. Lawless (1977) also used the above data set to construct prediction intervals for future order statistics. Here, we have a complete sample with $r=n=19$ from a two parameter exponential distribution, and the observed values of $T$ and $X_{1:19}$ given by $t=\sum_{i=1}^{19}(x_{i:19}-x_{1:19})=15869$ and $x_{1:19}=162$, respectively. \begin{itemize} \item \textbf{One-side prediction intervals:} Suppose we wish to construct 90\% and 95\% lower prediction bounds for the second record value, $R_2$, in a future sample. Using Lemma \ref{lem-p-record1}, for $n=r=19$ and $s=2$ we have $Pr(Z\geq 0.4989)~=0.9$ and $ Pr(Z\geq 0.3124)=0.95$. So using \eqref{eq:quantity-P-record}, $(578.6865, ~\infty)$ and $(422.9198, ~\infty)$ are 90\% and 95\% one-sided prediction intervals for the second record value, $R_2$, respectively. \item \textbf{Two-side prediction intervals:} For constructing 90\% and 95\% two-sided equal-tailed prediction intervals for $R_2$, in a future sample. Again by using Lemma \ref{lem-p-record1}, for $n=r=19$ and $s=2$, we have \begin{center} \begin{tabular}{c|cccc} $\gamma$&0.025&0.050&0.950&0.975\\ \hline $z_{\gamma}(19, 19, 2)$&0.3124&0.1926&5.4930&6.6158 \end{tabular} \end{center} So using \eqref{prediction-order}, $(422.9198,~ 4749.8114)$ and $(322.8615,~ 5687.5858)$ are 90\% and 95\% two-sided equal-tailed prediction intervals for the second record value, $R_2$, respectively. \end{itemize} \begin{exam} \label{example-order} Prediction intervals for order statistics \vspace{-0.2cm} \end{exam} A rock crushing machine has to be reset if, at any operation, the size of rock being crushed is larger than any that has been crushed before. The following data given by Dunsmore (1983) are the sizes dealt with up to the third time that the machine has been reset: 9.3, 0.6, 24.4, 18.1, 6.6, 9.0, 14.3, 6.6, 13.0, 2.4, 5.6, 33.8.\\ The record values were the sizes at the operation when resetting was necessary. Clearly $R_1=9.3$ and $R_3-R_1=24.5$. Dunsmore (1983) assumed that these data follow an $Exp(\sigma, 0)$ distribution. Based on these information, \eqref{eq:order1} and Lemma \ref{lem-p-order1}, we can construct prediction intervals for order statistics in a future sample from the same parent. For example $(23.7329,~312.0931)$ and $(11.9950, ~184.6220)$ are 90\% two-sided equal-tailed prediction intervals for $Y_{100:100}$ and $Y_{97:100}$ in a future sample of size $m=100$, respectively. \section{Concluding remark} In this paper, we have studied the prediction problems in \textit{two samples} and suggested some pivotal quantities to develop a prediction interval for the future record value (order statistics) based on observed order statistics (record values). The results do not depend on the parameters of the underlying distribution. The proposed procedure can be considered for other distributions. 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(2008), Prediction intervals for future records \textit{Statistics \& Probability Letters, } in press. \end{enumerate} %\newpage \begin{table} \caption{Values of $z_{\gamma}(r,10,s)$ for some selected values of $r, s$ and $\gamma$. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \label{tab:3} \begin{tabular}{|l|lc|l|l|l|l|l|} \hline &&s& 2 & 3 & 4& 5& 6\\ r& $\gamma$ & &&&&&\\ \hline 2&0.025& & 1.0820&3.5699 &6.1394 &8.7517 &11.3915 \\ &0.050& & 2.1004&5.2365 &8.4587 &11.7188&15.0015 \\ &0.950& & 365.3210&560.2135&755.2039 &950.1867 &1145.1615 \\ &0.975& & 745.5175&1140.2865&1535.2846&1930.2885&2325.2862 \\ \hline 4&0.025& & 0.3983&1.4481 &2.6288 &3.8753 &5.1601 \\ &0.050& & 0.7694&2.0631 &3.4780 &4.9512 &6.4572 \\ &0.950& & 29.2538& 41.8065& 54.2213 & 66.5716 &78.8863 \\ &0.975& & 40.2733&56.8677 &73.2771 &89.6005 &105.8766 \\ \hline 6&0.025& & 0.2450&0.9180 &1.6989 &2.5381 &3.4126 \\ &0.050& & 0.4734&1.2992 &2.2243 &3.2008 &4.2079 \\ &0.950& & 13.5285& 18.8961& 24.1507 & 29.3485 &34.5132 \\ &0.975& & 17.4268&23.9555 &30.3409 &36.6544 &42.9261 \\ \hline 7&0.025& & 0.2055&0.7766 &1.4455 &2.1685 &2.9248 \\ &0.050& & 0.3972&1.0972 &1.8871 &2.7249 &3.5914 \\ &0.950& & 10.5770& 14.6774& 18.6759 & 22.6220 &26.5369 \\ &0.975& & 13.4115&18.2945 &23.0500 &27.7401 &32.3912 \\ \hline 9&0.025& & 0.1555&0.5940 &1.1144 &1.6815 &2.2782 \\ &0.050& & 0.3005&0.8374 &1.4495 &2.1029 &2.7820 \\ &0.950& & 7.3300& 10.0832& 12.7520 & 15.3759 &17.9722 \\ &0.975& & 9.1173&12.3093 &15.3975 &18.4303 &21.4293 \\ \hline 10&0.025& & 0.1386&0.5316 &1.0002 &1.5123 &1.0524 \\ &0.050& & 0.2679&0.7489 &1.2992 &1.8882 &2.5014 \\ &0.950& & 6.3473& 8.7047& 10.9845 & 13.2224 &15.4344 \\ &0.975& & 7.8455&10.5540 &13.1675 &15.7297 &18.2603 \\ \hline \end{tabular} \end{table} \begin{table} \caption{Values of $z_{\gamma}(r,20,s)$ for some selected values of $r, s$ and $\gamma$. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \label{tab:4} \begin{tabular}{|l|lc|l|l|l|l|l|} \hline &&s& 2 & 3 & 4& 5& 6\\ r& $\gamma$ & &&&&&\\ \hline 5&0.025& & 0.8805&2.4763 &4.3273 &6.3133 &8.3791 \\ &0.050& & 1.4253&3.4112 &5.6262 &7.9581 &10.3581 \\ &0.950& & 37.8055&53.1025 & 68.1344 & 83.0364 &97.8643 \\ &0.975& & 49.8347&69.0378 &87.8976 &106.5893 &125.1852 \\ \hline 10&0.025&& 0.4057&1.1786 &2.1066 &3.1247 &4.2002 \\ &0.050&& 0.6544&1.6085 &2.7030 &3.8763 &5.0992 \\ &0.950&& 12.8503&17.5731 & 22.1385 & 26.6187 &31.0464 \\ &0.975& & 15.8605&21.2874 &26.5216 &31.6516 &36.7173 \\ \hline 15&0.025&& 0.2637&0.7745 &1.3956 &2.0833 &2.8149 \\ &0.050&& 0.4249&1.0540 &1.7831 &2.5707 &3.3964 \\ &0.950& & 7.6638&10.3860 & 12.9969 & 15.5454 &18.0539 \\ &0.975& & 9.2935&12.3403 &15.4121 &18.0104 &20.8795 \\ \hline 17&0.025& & 0.2313&0.6812 &1.2298 &1.8387 &2.4876 \\ &0.050& & 0.3727&0.9263 &1.5698 &2.2661 &2.9972 \\ &0.950& & 6.5942& 8.9173 & 11.1410 & 13.3081 &16.4388 \\ &0.975& & 7.9653&10.5499 &13.0150 &15.4121 &17.1654 \\ \hline 19&0.025& & 0.2060&0.6079 &1.0993 &1.6456 &2.2288 \\ &0.050& & 0.3319&0.8263 &1.4021 &2.0262 &2.6822 \\ &0.950& & 5.7856& 7.8106 & 9.7456 & 11.6292 &13.4792 \\ &0.975& & 6.9676&9.2099 &11.3444 &13.4171 &15.4496 \\ \hline 20&0.025& & 0.1954&0.5769 &1.0439 &1.5636 &1.1186 \\ &0.050& & 0.3147&0.7840 &1.3310 &1.9244 &2.5484 \\ &0.950& & 5.4511& 7.3638 & 9.1705 & 10.9379 &12.6733 \\ &0.975& & 6.5565&8.6591 &10.6590 &12.5996 &14.5018 \\ \hline \end{tabular} \end{table} \begin{table} \caption{Values of $w_{\gamma}(n,10,j)$ for some selected values of $n, j$ and $\gamma$. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \label{tab:5} \begin{tabular}{|l|lc|l|l|l|l|l|} \hline &&n& 2 & 3 & 4& 5& 6\\ j& $\gamma$& &&&&&\\ \hline 5&0.025& & -20.8182&-3.6710 &-1.7943 &-1.1612 &-0.8525 \\ &0.050& & -9.9091&-2.3029 &-1.2178 &-0.8174 &-0.6127 \\ &0.950& & 3.5015&0.8186 &0.4290 &0.2858 &0.2130 \\ &0.975&& 7.3254&1.2780 &0.6121 &0.3903 &0.2834 \\ \hline 6&0.025& & -17.1818&-3.2640 &-1.6296 &-1.0649 &-0.7862 \\ &0.050& & -8.9091&-2.0151 &-1.0871 &-0.7364 &-0.5550 \\ &0.950&& 5.5977&1.2029 &0.6131 &0.4026 &0.2975 \\ &0.975&& 11.6037&1.8506 &0.8604 &0.5403 &0.3888 \\ \hline 7&0.025&& -13.5454&-2.8139 &-1.4410 &-0.9529 &-0.7082 \\ &0.050&& -6.2727&-1.6968 &-0.9375 &-0.6422 &-0.4871 \\ &0.950& & 8.6679&1.7305 &0.8659 &0.5581 &0.4092 \\ &0.975& & 17.8560&2.6347 &1.1935 &0.7397 &0.5279 \\ \hline 8&0.025& & -9.9091&-2.3029 &-1.2178 &-0.8174 &-0.6127 \\ &0.050& & -4.4545&-1.3355 &-0.7603 &-0.5282 &-0.4039 \\ &0.950& & 13.3591&2.4976 &1.2151 &0.7792 &0.5674 \\ &0.975& & 27.3962&3.7733 &1.6705 &1.0233 &0.7251 \\ \hline 9&0.025& & -6.2727&-1.6968 &-0.9375 &-0.6422 &-0.4871 \\ &0.050& & -2.6364&-0.9069 &-0.5378 &-0.3809 &-0.2946 \\ &0.950& & 21.2859&3.7513 &1.7896 &1.1364 &0.8224 \\ &0.975& & 43.5054&5.6348 &2.4451 &1.4830 &1.0446 \\ \hline 10&0.025& & -2.6364&-0.9069 &-0.5378 &-0.3809 &-0.2946 \\ &0.050& & -0.8182&-0.3484 &-0.2205 &-0.1612 &-0.1270 \\ &0.950& & 38.8805&6.5176 &3.0637 &1.9323 &1.3934 \\ &0.975& & 79.2830&9.7606 &4.1772 &2.5194 &1.7701 \\ \hline \end{tabular} \end{table} \begin{table} \caption{Values of $w_{\gamma}(n,20,j)$ for some selected values of $n, j$ and $\gamma$. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \label{tab:6} \begin{tabular}{|l|lc|l|l|l|l|l|} \hline &&n& 2 & 3 & 4& 5& 6\\ j& $\gamma$ & &&&&&\\ \hline 12&0.025& &-16.1429&-3.1404 &-1.5785 &-1.0348 &-0.7653 \\ &0.050& & -7.5714&-1.9277 &-1.0465 &-0.7110 &-0.5368 \\ &0.950& & 5.7979&1.1992 &0.6012 &0.3910 &0.2870 \\ &0.975& & 11.9694&1.8270 &0.8323 &0.5162 &0.3682 \\ \hline 15&0.025& & -10.4286&-2.3806 &-1.2525 &-0.8386 &-0.6278 \\ &0.050& & -4.7143&-1.3905 &-0.7878 &-0.5461 &-0.4171 \\ &0.950& & 11.4467&2.1227 &1.0253 &0.6539 &0.4741 \\ &0.975& & 23.4521&3.1934 &1.3996 &0.8508 &0.5992 \\ \hline 17&0.025& & -6.6190&-1.7603 &-0.9677 &-0.6614 &-0.5016 \\ &0.050& & -2.8095&-0.9518 &-0.5618 &-0.3971 &-0.3067 \\ &0.950& & 18.3374&3.1883 &1.5063 &0.9497 &0.6836 \\ &0.975& & 37.4386&4.7674 &2.0425 &1.2270 &0.8580 \\ \hline 18&0.025& & -4.7143&-1.3905 &-0.7878 &-0.5461 &-0.4171 \\ &0.050& & -1.8571&-0.6903 &-0.4190 &-0.3001 &-0.2336 \\ &0.950& & 23.8908&4.0297 &1.8842 &1.1816 &0.8478 \\ &0.975& & 48.7069&6.0106 &2.5482 &1.5228 &1.0614 \\ \hline 19&0.025& & -2.8095&-0.9518 &-0.5618 &-0.3971 &-0.3067 \\ &0.050& & -0.9048&-0.3801 &-0.2396 &-0.1748 &-0.1375 \\ &0.950& & 32.6810&5.3560 &2.4808 &1.5486 &1.1080 \\ &0.975& & 66.5453&7.9732 &3.3490 &1.9927 &1.3856 \\ \hline 20&0.025&& -0.9048&-0.3801 &-0.2396 &-0.1748 &-0.1375 \\ &0.050& & 0.0476& 0.0241 & 0.0161 & 0.0121 & 0.0097 \\ &0.950&& 51.1467&8.1884 &3.7714 &2.3500 &1.6806 \\ &0.975& & 104.0601&12.1860&5.0968 &3.0316 &2.1098 \\ \hline \end{tabular} \end{table} \begin{table} \caption{Values of $u_{\gamma}(n,m)$ for some selected values of $n, m$ and $\gamma$. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \label{tab:7} \begin{tabular}{|l|lc|l|l|l|l|l|} \hline &&n& 2 & 3 & 4& 5& 6\\ m& $\gamma$ & &&&&&\\ \hline 5&0.025& & -15.0751&-3.0094 &-1.5238 &-1.0023 &-0.7427 \\ &0.050& & -7.0376 &-1.8351 &-1.0031 &-0.6838 &-0.5171 \\ &0.950& & 7.5395 &1.5606 &0.7866 &0.5139 &0.3786 \\ &0.975& & 15.5796 &2.3907 &1.0992 &0.6869 &0.4928 \\ \hline 10&0.025& & -14.4217&-2.9271 &-1.4891 &-0.9817 &-0.7283 \\ &0.050& & -6.7109&-1.7768 &-0.9756 &-0.6664 &-0.5046 \\ &0.950& & 7.2792&1.4592 &0.7244 &0.4689 &0.3432 \\ &0.975& & 14.9927&2.2159 &0.9998 &0.6173 &0.4391 \\ \hline 15&0.025& & -14.1925&-2.8978 &-1.4767 &-0.9743 &-0.7232 \\ &0.050& & -6.5962&-1.7561 &-0.9658 &-0.6602 &-0.5001 \\ &0.950& & 7.1869&1.4233 &0.7023 &0.4528 &0.3305 \\ &0.975& & 14.7858&2.1546 &0.9648 &0.5926 &0.4200 \\ \hline 17&0.025& & -14.1377&-2.8907 &-1.4737 &-0.9725 &-0.7219 \\ &0.050& & -6.5690&-1.7512 &-0.9634 &-0.6587 &-0.4990 \\ &0.950& & 7.1648&1.4148 &0.6970 &0.4489 &0.3274 \\ &0.975& & 14.7362&2.1399 &0.9564 &0.5867 &0.4154 \\ \hline 20&0.025& & -14.0756&-2.8827 &-1.4703 &-0.9705 &-0.7205 \\ &0.050& & -6.5378&-1.7455 &-0.9607 &-0.6570 &-0.4978 \\ &0.950& & 7.1397&1.4050 &0.6910 &0.4445 &0.3239 \\ &0.975& & 14.6801&2.1233 &0.9469 &0.5800 &0.4101 \\ \hline 25&0.025& & -14.0047&-2.8736 &-1.4665 &-0.9681 &-0.7189 \\ &0.050& & -6.5023&-1.7390 &-0.9576 &-0.6550 &-0.4964 \\ &0.950& & 7.1110&1.3939 &0.6841 &0.4394 &0.3199 \\ &0.975& & 14.6160&2.1044 &0.9360 &0.5722 &0.4041 \\ \hline 30&0.025& & -13.9571&-2.8674 &-1.4639 &-0.9666 &-0.7178 \\ &0.050& & -6.4785&-1.7347 &-0.9556 &-0.6537 &-0.4954 \\ &0.950& & 7.0918&1.3864 &0.6795 &0.4361 &0.3172 \\ &0.975& & 14.5729&2.0916 &0.9287 &0.5671 &0.4001 \\ \hline \end{tabular} \end{table} \end{document} \begin{eqnarray} \nonumber f_{R_s,\dots,R_n,Y_{j:m}}(r_s,\dots,r_n,y;\sigma)&=&\frac{(r_s-\mu)^{s-1}}{(s-1)!\sigma^{n+1}}e^{-\frac{r_n-\mu}{\sigma}} j{m \choose j}\left[1-e^{-\frac{y-\mu}{\sigma}}\right]^{j-1}{e^{\frac{-(m-j+1)(y-\mu)}{\sigma}}}, \\&& \ \ \ \mu A(j;m). \end{eqnarray*}