Skip to main content
added 9 characters in body
Source Link
Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

$\newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=}$ Let $U_1,U_2,\dots$ be iid random variables, each uniformly distributed on $[0,1]$. For a fixed natural $k$, let $Y_{n,k}$ be the $k$th largest value among $U_1,\dots,U_n$. For a fixed $p\in(0,1)$, let $q_{n,k}(p)$ be the $p$-quantile of $Y_{n,k}$. The problem then is to find \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)). \end{equation}

Note that $Y_{n,k}$ has the beta distribution with parameters $n+1-k,k$. So (see Sections Derived_from other distributions and Summation), \begin{equation} 1-Y_{n,k}\D\frac{S_k}{S_{n+1}}, \end{equation} where $\D$ denotes the equality in distribution, $S_j:=X_1+\dots+X_j$, and $X_1,X_2,\dots$ are iid standard exponential r.v.'s. So, by the law of large numbers, \begin{equation} S_{n,k}:=n(1-Y_{n,k})\eD S_k, \end{equation} where $\eD$ denotes the convergence in distribution.

(Note also that $S_k$ has the gamma distribution with parameters $k$ and $1$.)

So, $n(1-q_{n,k}(p))=\tilde q_{n,k}(1-p)\to \tilde q_k(1-p)$, where $\tilde q_{n,k}(1-p)$ and $\tilde q_k(1-p)$ denote the $(1-p)$-quantiles of $S_{n,k}$ and $S_k$, respectively. Hence,
\begin{align} n(1-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) &=\E(S_{n,k}|Y_{n,k}\le q_{n,k}(p)) \\ &=\E(S_{n,k}|S_{n,k}\ge \tilde q_{n,k}(1-p)) \\ &\to\E(S_k|S_k\ge \tilde q_k(1-p)), \end{align} by an appropriate uniform integrability involving (say) second moments. Thus, \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) =\E(S_k|S_k\ge \tilde q_k(1-p))-\tilde q_k(1-p). \end{equation}

$\newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=}$ Let $U_1,U_2,\dots$ be iid random variables, each uniformly distributed on $[0,1]$. For a fixed natural $k$, let $Y_{n,k}$ be the $k$th largest value among $U_1,\dots,U_n$. For a fixed $p\in(0,1)$, let $q_{n,k}(p)$ be the $p$-quantile of $Y_{n,k}$. The problem then is to find \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)). \end{equation}

Note that $Y_{n,k}$ has the beta distribution with parameters $n+1-k,k$. So (see Derived_from other distributions and Summation), \begin{equation} 1-Y_{n,k}\D\frac{S_k}{S_{n+1}}, \end{equation} where $\D$ denotes the equality in distribution, $S_j:=X_1+\dots+X_j$, and $X_1,X_2,\dots$ are iid standard exponential r.v.'s. So, by the law of large numbers, \begin{equation} S_{n,k}:=n(1-Y_{n,k})\eD S_k, \end{equation} where $\eD$ denotes the convergence in distribution.

(Note also that $S_k$ has the gamma distribution with parameters $k$ and $1$.)

So, $n(1-q_{n,k}(p))=\tilde q_{n,k}(1-p)\to \tilde q_k(1-p)$, where $\tilde q_{n,k}(1-p)$ and $\tilde q_k(1-p)$ denote the $(1-p)$-quantiles of $S_{n,k}$ and $S_k$, respectively. Hence,
\begin{align} n(1-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) &=\E(S_{n,k}|Y_{n,k}\le q_{n,k}(p)) \\ &=\E(S_{n,k}|S_{n,k}\ge \tilde q_{n,k}(1-p)) \\ &\to\E(S_k|S_k\ge \tilde q_k(1-p)), \end{align} by an appropriate uniform integrability involving (say) second moments. Thus, \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) =\E(S_k|S_k\ge \tilde q_k(1-p))-\tilde q_k(1-p). \end{equation}

$\newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=}$ Let $U_1,U_2,\dots$ be iid random variables, each uniformly distributed on $[0,1]$. For a fixed natural $k$, let $Y_{n,k}$ be the $k$th largest value among $U_1,\dots,U_n$. For a fixed $p\in(0,1)$, let $q_{n,k}(p)$ be the $p$-quantile of $Y_{n,k}$. The problem then is to find \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)). \end{equation}

Note that $Y_{n,k}$ has the beta distribution with parameters $n+1-k,k$. So (see Sections Derived_from other distributions and Summation), \begin{equation} 1-Y_{n,k}\D\frac{S_k}{S_{n+1}}, \end{equation} where $\D$ denotes the equality in distribution, $S_j:=X_1+\dots+X_j$, and $X_1,X_2,\dots$ are iid standard exponential r.v.'s. So, by the law of large numbers, \begin{equation} S_{n,k}:=n(1-Y_{n,k})\eD S_k, \end{equation} where $\eD$ denotes the convergence in distribution.

(Note also that $S_k$ has the gamma distribution with parameters $k$ and $1$.)

So, $n(1-q_{n,k}(p))=\tilde q_{n,k}(1-p)\to \tilde q_k(1-p)$, where $\tilde q_{n,k}(1-p)$ and $\tilde q_k(1-p)$ denote the $(1-p)$-quantiles of $S_{n,k}$ and $S_k$, respectively. Hence,
\begin{align} n(1-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) &=\E(S_{n,k}|Y_{n,k}\le q_{n,k}(p)) \\ &=\E(S_{n,k}|S_{n,k}\ge \tilde q_{n,k}(1-p)) \\ &\to\E(S_k|S_k\ge \tilde q_k(1-p)), \end{align} by an appropriate uniform integrability involving (say) second moments. Thus, \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) =\E(S_k|S_k\ge \tilde q_k(1-p))-\tilde q_k(1-p). \end{equation}

added 223 characters in body
Source Link
Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

$\newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=}$ Let $U_1,U_2,\dots$ be iid random variables, each uniformly distributed on $[0,1]$. For a fixed natural $k$, let $Y_{n,k}$ be the $k$th largest value among $U_1,\dots,U_n$. For a fixed $p\in(0,1)$, let $q_{n,k}(p)$ be the $p$-quantile of $Y_{n,k}$. The problem then is to find \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)). \end{equation}

Note that $Y_{n,k}$ has the beta distribution with parameters $n+1-k,k$. So (see Derived_from other distributions and Summation), \begin{equation} 1-Y_{n,k}\D\frac{S_k}{S_{n+1}}, \end{equation} where $\D$ denotes the equality in distribution, $S_j:=X_1+\dots+X_j$, and $X_1,X_2,\dots$ are iid standard exponential r.v.'s. So, by the law of large numbers, \begin{equation} S_{n,k}:=n(1-Y_{n,k})\eD S_k, \end{equation} where $\eD$ denotes the convergence in distribution.

(Note also that $S_k$ has the gamma distribution with parameters $k$ and $1$.)

So, $n(1-q_{n,k}(p))=\tilde q_{n,k}(1-p)\to \tilde q_k(1-p)$, where $\tilde q_{n,k}(1-p)$ and $\tilde q_k(1-p)$ denote the $(1-p)$-quantiles of $S_{n,k}$ and $S_k$, respectively. Hence,
\begin{align} n(1-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) &=\E(S_{n,k}|Y_{n,k}\le q_{n,k}(p)) \\ &=\E(S_{n,k}|S_{n,k}\ge \tilde q_{n,k}(1-p)) \\ &\to\E(S_k|S_k\ge \tilde q_k(1-p)), \end{align} by an appropriate uniform integrability involving (say) second moments. Thus, \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) =\E(S_k|S_k\ge \tilde q_k(1-p))-\tilde q_k(1-p). \end{equation}

$\newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=}$ Let $U_1,U_2,\dots$ be iid random variables, each uniformly distributed on $[0,1]$. For a fixed natural $k$, let $Y_{n,k}$ be the $k$th largest value among $U_1,\dots,U_n$. For a fixed $p\in(0,1)$, let $q_{n,k}(p)$ be the $p$-quantile of $Y_{n,k}$. The problem then is to find \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)). \end{equation}

Note that $Y_{n,k}$ has the beta distribution with parameters $n+1-k,k$. So, \begin{equation} 1-Y_{n,k}\D\frac{S_k}{S_{n+1}}, \end{equation} where $\D$ denotes the equality in distribution, $S_j:=X_1+\dots+X_j$, and $X_1,X_2,\dots$ are iid standard exponential r.v.'s. So, by the law of large numbers, \begin{equation} S_{n,k}:=n(1-Y_{n,k})\eD S_k, \end{equation} where $\eD$ denotes the convergence in distribution.

(Note also that $S_k$ has the gamma distribution with parameters $k$ and $1$.)

So, $n(1-q_{n,k}(p))=\tilde q_{n,k}(1-p)\to \tilde q_k(1-p)$, where $\tilde q_{n,k}(1-p)$ and $\tilde q_k(1-p)$ denote the $(1-p)$-quantiles of $S_{n,k}$ and $S_k$, respectively. Hence,
\begin{align} n(1-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) &=\E(S_{n,k}|Y_{n,k}\le q_{n,k}(p)) \\ &=\E(S_{n,k}|S_{n,k}\ge \tilde q_{n,k}(1-p)) \\ &\to\E(S_k|S_k\ge \tilde q_k(1-p)), \end{align} by an appropriate uniform integrability involving (say) second moments. Thus, \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) =\E(S_k|S_k\ge \tilde q_k(1-p))-\tilde q_k(1-p). \end{equation}

$\newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=}$ Let $U_1,U_2,\dots$ be iid random variables, each uniformly distributed on $[0,1]$. For a fixed natural $k$, let $Y_{n,k}$ be the $k$th largest value among $U_1,\dots,U_n$. For a fixed $p\in(0,1)$, let $q_{n,k}(p)$ be the $p$-quantile of $Y_{n,k}$. The problem then is to find \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)). \end{equation}

Note that $Y_{n,k}$ has the beta distribution with parameters $n+1-k,k$. So (see Derived_from other distributions and Summation), \begin{equation} 1-Y_{n,k}\D\frac{S_k}{S_{n+1}}, \end{equation} where $\D$ denotes the equality in distribution, $S_j:=X_1+\dots+X_j$, and $X_1,X_2,\dots$ are iid standard exponential r.v.'s. So, by the law of large numbers, \begin{equation} S_{n,k}:=n(1-Y_{n,k})\eD S_k, \end{equation} where $\eD$ denotes the convergence in distribution.

(Note also that $S_k$ has the gamma distribution with parameters $k$ and $1$.)

So, $n(1-q_{n,k}(p))=\tilde q_{n,k}(1-p)\to \tilde q_k(1-p)$, where $\tilde q_{n,k}(1-p)$ and $\tilde q_k(1-p)$ denote the $(1-p)$-quantiles of $S_{n,k}$ and $S_k$, respectively. Hence,
\begin{align} n(1-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) &=\E(S_{n,k}|Y_{n,k}\le q_{n,k}(p)) \\ &=\E(S_{n,k}|S_{n,k}\ge \tilde q_{n,k}(1-p)) \\ &\to\E(S_k|S_k\ge \tilde q_k(1-p)), \end{align} by an appropriate uniform integrability involving (say) second moments. Thus, \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) =\E(S_k|S_k\ge \tilde q_k(1-p))-\tilde q_k(1-p). \end{equation}

Source Link
Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

$\newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\eD}{\overset{\text{D}}\to} \newcommand{\D}{\overset{\text{D}}=}$ Let $U_1,U_2,\dots$ be iid random variables, each uniformly distributed on $[0,1]$. For a fixed natural $k$, let $Y_{n,k}$ be the $k$th largest value among $U_1,\dots,U_n$. For a fixed $p\in(0,1)$, let $q_{n,k}(p)$ be the $p$-quantile of $Y_{n,k}$. The problem then is to find \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)). \end{equation}

Note that $Y_{n,k}$ has the beta distribution with parameters $n+1-k,k$. So, \begin{equation} 1-Y_{n,k}\D\frac{S_k}{S_{n+1}}, \end{equation} where $\D$ denotes the equality in distribution, $S_j:=X_1+\dots+X_j$, and $X_1,X_2,\dots$ are iid standard exponential r.v.'s. So, by the law of large numbers, \begin{equation} S_{n,k}:=n(1-Y_{n,k})\eD S_k, \end{equation} where $\eD$ denotes the convergence in distribution.

(Note also that $S_k$ has the gamma distribution with parameters $k$ and $1$.)

So, $n(1-q_{n,k}(p))=\tilde q_{n,k}(1-p)\to \tilde q_k(1-p)$, where $\tilde q_{n,k}(1-p)$ and $\tilde q_k(1-p)$ denote the $(1-p)$-quantiles of $S_{n,k}$ and $S_k$, respectively. Hence,
\begin{align} n(1-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) &=\E(S_{n,k}|Y_{n,k}\le q_{n,k}(p)) \\ &=\E(S_{n,k}|S_{n,k}\ge \tilde q_{n,k}(1-p)) \\ &\to\E(S_k|S_k\ge \tilde q_k(1-p)), \end{align} by an appropriate uniform integrability involving (say) second moments. Thus, \begin{equation} \lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)) =\E(S_k|S_k\ge \tilde q_k(1-p))-\tilde q_k(1-p). \end{equation}