\subsection{Control of Absolute Error}

In many situations, it is desirable to construct an estimator for
$p$ with guaranteed absolute precision and confidence level.  For
this purpose, we have


\begin{theorem} \label{coverage_abs} Let $0 < \varepsilon < \frac{1}{2}, \; 0 < \delta < 1, \;
\zeta > 0$ and $\rho > 0$. Let $n_1 < n_2 < \cdots < n_s$ be the ascending
arrangement of all distinct elements of {\small $\left \{ \left \lceil \left
( \frac{24 \varepsilon - 16 \varepsilon^2}{9} \right )^{ 1 - \frac{i}{\tau} } \frac{\ln
\frac{1}{\zeta \delta}}{2 \varepsilon^2} \right \rceil : i = 0, 1, \cdots, \tau \right \}$}
with {\small $\tau = \left \lceil \frac{ \ln \frac{9}{24 \varepsilon - 16 \varepsilon^2} } {
\ln (1 + \rho) } \right \rceil$.}    For $\ell = 1, \cdots, s$, define $K_\ell
= \sum_{i = 1}^{n_\ell} X_i, \; \widehat{\boldsymbol{p}}_\ell = \frac{ K_\ell
}{n_\ell}$  and $\boldsymbol{D}_\ell$ such that $\boldsymbol{D}_\ell = 1$ if {\small
$\left ( \left | \widehat{\boldsymbol{p}}_\ell - \frac{1}{2} \right | - \frac{2 \varepsilon }{3} \right
)^2 \geq \frac{1}{4} + \frac{ \varepsilon^2 n_\ell } {2 \ln (\zeta \delta) }$}; and
$\boldsymbol{D}_\ell = 0$ otherwise.  Suppose the stopping rule is that
sampling is continued until $\boldsymbol{D}_\ell = 1$ for some $\ell \in
\{1, \cdots, s\}$. Define $\boldsymbol{\widehat{p}} = \frac{\sum_{i=1}^{\mathbf{n}}
X_i}{\mathbf{n}}$ where $\mathbf{n}$ is the sample size when the
sampling is terminated. Define {\small  \[ \mathscr{Q}^+ =
\bigcup_{\ell = 1}^s \left \{ \frac{k}{n_\ell} + \varepsilon \in \left (0,
\frac{1}{2} \right ) : k \in \mathbb{Z} \right \} \bigcup \left \{ \frac{1}{2} \right \},
\qquad \mathscr{Q}^- = \bigcup_{\ell = 1}^s \left \{ \frac{k}{n_\ell} - \varepsilon
\in \left (0, \frac{1}{2} \right ) : k \in \mathbb{Z} \right \} \bigcup \left \{
\frac{1}{2} \right \}.
\]}
Then, a sufficient condition to guarantee $\Pr \left \{ \left |
\boldsymbol{\widehat{p}} - p \right | < \varepsilon \mid p \right \}
> 1 - \delta$ for any $p \in (0, 1)$ is that {\small  \begin{eqnarray} &  &
\sum_{\ell = 1}^s \Pr \{ \widehat{\boldsymbol{p}}_\ell \geq p + \varepsilon, \;
\boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} < \frac{\delta}{2}
\qquad \forall p \in
\mathscr{Q}^-, \label{2D1}\\
&  & \sum_{\ell = 1}^s \Pr \{ \widehat{\boldsymbol{p}}_\ell  \leq p - \varepsilon, \;
\boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} < \frac{\delta}{2}
\qquad \forall p \in \mathscr{Q}^+ \label{2D2}
 \end{eqnarray}}
where both (\ref{2D1}) and (\ref{2D2}) are satisfied if $0 < \zeta <
\frac{1}{2(\tau + 1)}$.

\end{theorem}

\subsection{Control of Absolute and Relative Errors}

To construct an estimator satisfying a mixed criterion in terms of
absolute and relative errors with a prescribed confidence level, we
have

\begin{theorem} \label{coverage_mixed}  Let $0 < \delta < 1, \; \zeta > 0$ and $\rho >
0$. Let $\varepsilon_a$ and $\varepsilon_r$ be positive numbers such that $0 <
\varepsilon_a < \frac{3}{8}$ and {\small $\frac{6 \varepsilon_a}{3 - 2 \varepsilon_a } < \varepsilon_r
< 1$}.  Let $n_1 < n_2 < \cdots < n_s$ be the ascending arrangement of
all distinct elements of {\small $\left \{ \left \lceil \left [ \frac{3}{2} \left
( \frac{1}{\varepsilon_a} - \frac{1}{\varepsilon_r} - \frac{1}{3} \right ) \right ]^{ \frac{i } {
\tau } } \frac{ 4(3 + \varepsilon_r) } {9 \varepsilon_r } \ln \frac{1} { \zeta \delta } \right
\rceil: i = 0, 1, \cdots, \tau \right \}$ } with {\small $\tau = \left \lceil \frac{
\ln \left [ \frac{3}{2} \left ( \frac{1}{\varepsilon_a} - \frac{1}{\varepsilon_r} - \frac{1}{3}
\right ) \right ] } { \ln (1 + \rho) } \right \rceil$. }   For $\ell = 1, \cdots,
s$, define {\small $K_\ell = \sum_{i = 1}^{n_\ell} X_i, \;
\widehat{\boldsymbol{p}}_\ell = \frac{K_\ell}{n_\ell},$} {\small
\[ \boldsymbol{D}_\ell = \begin{cases} 0 & \mathrm{for} \; \frac{1}{2} - \frac{2}{3} \varepsilon_a -
\sqrt{ \frac{1}{4} + \frac{ n_\ell \varepsilon_a^2 } {2 \ln (\zeta \delta) } } <
\widehat{\boldsymbol{p}}_\ell < \frac{ 6(1 - \varepsilon_r) (3 - \varepsilon_r) \ln (\zeta
\delta) } { 2 (3 - \varepsilon_r)^2 \ln (\zeta \delta) - 9 n_\ell \varepsilon_r^2} \; \mathrm{or}\\
  &  \quad \;\; \frac{1}{2} + \frac{2}{3} \varepsilon_a - \sqrt{ \frac{1}{4} + \frac{
n_\ell \varepsilon_a^2 } {2 \ln (\zeta \delta) } } < \widehat{\boldsymbol{p}}_\ell < \frac{ 6(1
+ \varepsilon_r) (3 + \varepsilon_r) \ln (\zeta \delta) } { 2 (3 +
\varepsilon_r)^2 \ln (\zeta \delta) - 9 n_\ell  \varepsilon_r^2},\\
1 & \mathrm{else}
 \end{cases}
\]}
for $\ell = 1, \cdots, s - 1$ and $\boldsymbol{D}_s = 1$.  Suppose the stopping
rule is that sampling is continued until {\small $\boldsymbol{D}_\ell = 1$}
for some $\ell \in \{1, \cdots, s\}$. Let {\small $\widehat{\boldsymbol{p}} =
\frac{\sum_{i=1}^{\mathbf{n}} X_i} {\mathbf{n}}$} where $\mathbf{n}$ is
the sample size when the sampling is terminated. Define {\small
$p^\star = \frac{\varepsilon_a}{\varepsilon_r}$} and {\small
\[ \mathscr{Q}_a^+ = \bigcup_{\ell = 1}^s \left \{ \frac{k}{n_\ell} + \varepsilon_a
\in \left (0, p^\star \right ) : k \in \mathbb{Z} \right \} \cup \left \{ p^\star
\right \}, \qquad \mathscr{Q}_a^- = \bigcup_{\ell = 1}^s \left \{
\frac{k}{n_\ell} - \varepsilon_a \in \left (0, p^\star \right ) : k \in \mathbb{Z} \right
\} \cup \left \{ p^\star \right \},
\]
\[
\qquad \mathscr{Q}_r^+ = \bigcup_{\ell = 1}^s \left \{ \frac{k}{n_\ell (1 +
\varepsilon_r)} \in \left (p^\star, 1 \right ) : k \in \mathbb{Z} \right \}, \quad \qquad
\quad \mathscr{Q}_r^- = \bigcup_{\ell = 1}^s \left \{ \frac{k}{n_\ell (1 -
\varepsilon_r)} \in \left (p^\star, 1 \right ) : k \in \mathbb{Z} \right \}.\qquad \qquad
\qquad\qqu
\]}
Then, {\small $\Pr \left \{ \left | \widehat{\boldsymbol{p}} - p
 \right | < \varepsilon_a  \; \mathrm{or} \; \left |
\frac{\widehat{\boldsymbol{p}} - p } {p }
 \right | < \varepsilon_r \mid p \right \} > 1 - \delta$
 }
for any $p \in (0, 1)$  provided that {\small \begin{eqnarray} &   & \sum_{\ell
= 1}^s \Pr \{ \widehat{\boldsymbol{p}}_\ell \geq p + \varepsilon_a, \; \boldsymbol{D}_{\ell - 1}
= 0, \; \boldsymbol{D}_\ell = 1 \mid p \} < \frac{\delta}{2} \qquad \forall p \in
\mathscr{Q}_a^-, \label{mix1}\\
&   & \sum_{\ell = 1}^s \Pr \{ \widehat{\boldsymbol{p}}_\ell \leq p - \varepsilon_a, \;
\boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} < \frac{\delta}{2}
\qquad \forall p \in \mathscr{Q}_a^+, \label{mix2}\\
&   & \sum_{\ell = 1}^s \Pr \{ \widehat{\boldsymbol{p}}_\ell \geq p (1 + \varepsilon_r),
\; \boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} < \frac{\delta}{2}
\qquad \forall p \in \mathscr{Q}_r^+, \label{mix3}\\
&   &  \sum_{\ell = 1}^s \Pr \{ \widehat{\boldsymbol{p}}_\ell \leq p (1 -
\varepsilon_r), \; \boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} <
\frac{\delta}{2} \qquad \forall p \in \mathscr{Q}_r^- \label{mix4}
 \end{eqnarray}}  where these conditions are satisfied for $0 < \zeta < \frac{1}{2 (\tau + 1) }$ .  \end{theorem}

\subsection{Control of Relative Error}


In many situations, it is desirable to design a sampling scheme to
estimate $p$ such that the estimator satisfies a relative error
criterion with a prescribed confidence level.  By virtue of the
function {\small \[ g(\varepsilon, \gamma) = 1 - \sum_{i= 0}^{\gamma - 1}
\frac{1}{i!} \left ( \frac{\gamma}{ 1 + \varepsilon} \right )^i \exp \left ( - \frac{\gamma}{ 1 +
\varepsilon} \right ) + \sum_{i= 0}^{\gamma - 1} \frac{1}{i!}  \left ( \frac{\gamma}{ 1 -
\varepsilon} \right )^i \exp \left ( - \frac{\gamma}{ 1 - \varepsilon} \right ),
\]}
we have developed a simple sampling scheme as described by the
following theorem.

\begin{theorem} \label{coverage_rev_a} Let $0 < \varepsilon < 1, \; 0 < \delta < 1, \; \zeta >
0$ and $\rho > 0$. Let $\gamma_1 < \gamma_2 < \cdots < \gamma_s$ be the ascending
arrangement of all distinct elements of {\small $\left \{ \left \lceil \left
[  \frac{3}{2} \left ( \frac{1}{\varepsilon} + 1 \right ) \right ]^{ \frac{i} { \tau } } \frac{
4(3 + \varepsilon) } {9 \varepsilon } \ln \frac{1} { \zeta \delta } \right \rceil: i = 0, 1,
\cdots, \tau \right \}$ } with {\small $\tau = \left \lceil \frac{ \ln \left [
\frac{3}{2} \left ( \frac{1}{\varepsilon} + 1 \right )  \right ] } { \ln (1 + \rho) } \right
\rceil$.}   Let $\widehat{\boldsymbol{p}}_\ell = \frac{ \sum_{i = 1}^{\mathbf{n}_\ell}
X_i } { \mathbf{n}_\ell }$ where $\mathbf{n}_\ell$ is the minimum
number of samples such that $\sum_{i = 1}^{\mathbf{n}_\ell} X_i =
\gamma_\ell$. For $\ell = 1, \cdots, s$,  define $\boldsymbol{D}_\ell$ such that
$\boldsymbol{D}_\ell = 1$ if {\small $\widehat{\boldsymbol{p}}_\ell \geq 1 + \frac{ 2
\varepsilon}{3 + \varepsilon} + \frac{ 9 \varepsilon^2 \gamma_\ell } { 2 (3 + \varepsilon)^2 \ln (\zeta
\delta)}$}; and $\boldsymbol{D}_\ell = 0$ otherwise. Suppose the stopping rule
is that sampling is continued until $\boldsymbol{D}_\ell = 1$ for some $\ell
\in \{1, \cdots, s\}$. Define estimator $\widehat{\boldsymbol{p}} =
\frac{\sum_{i=1}^{\mathbf{n}} X_i}{\mathbf{n}}$ where $\mathbf{n}$ is
the sample size when the sampling is terminated. Then, {\small $\Pr
\left \{ \left | \frac{ \widehat{\boldsymbol{p}} - p } { p }  \right | \leq \varepsilon \mid p \right
\} \geq 1 - \delta$} for any $p \in (0, 1)$ provided that $\zeta > 0$ is
sufficiently small to guarantee $g(\varepsilon, \gamma_s) < \delta$ and
 {\small \begin{eqnarray} &  & \ln ( \zeta \delta )
< \left [ \frac{ \left ( 1 + \varepsilon  +  \sqrt{ 1 + 4 \varepsilon + \varepsilon^2 }  \right )^2}
{ 4 \varepsilon^2 } + \frac{1}{2} \right ] \left [ \frac{\varepsilon}{ 1 + \varepsilon}  - \ln (1 +
\varepsilon) \right ], \label{cona}\\
&  & \sum_{\ell = 1}^s \Pr \{ \widehat{\boldsymbol{p}}_\ell \leq (1 - \varepsilon) p, \;
\boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} \leq \frac{\delta}{2}
\qquad \forall p \in \mathscr{Q}_r^-, \label{rev1}\\
&  & \sum_{\ell = 1}^s \Pr \{ \widehat{\boldsymbol{p}}_\ell \geq (1 + \varepsilon) p, \;
\boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} \leq \frac{\delta}{2}
\qquad \forall p \in \mathscr{Q}_r^+  \label{rev2} \end{eqnarray}} where {\small
$\mathscr{Q}_r^+ = \bigcup_{\ell = 1}^s \left \{\frac{\gamma_\ell}{m (1 + \varepsilon)
} \in (p^*, 1) : m \in \mathbb{N} \right \} $} and {\small $\mathscr{Q}_r^- =
\bigcup_{\ell = 1}^s \left \{\frac{\gamma_\ell}{m (1 - \varepsilon) } \in (p^*, 1)
: m \in \mathbb{N} \right \}$} with $p^* \in (0, z_{s-1})$ denoting the
unique number satisfying \[ g(\varepsilon, \gamma_s)  + \sum_{\ell = 1}^{s -
1} \exp \left ( \frac{\gamma_\ell}{z_\ell} \frac{ (p^* - z_\ell)^2 } {2 \left (
\frac{2 p^*}{3} + \frac{z_\ell}{3} \right ) \left ( \frac{2 p^*}{3} +
\frac{z_\ell}{3} - 1 \right ) } \right ) = \delta \]
 where {\small $z_\ell =  1 + \frac{ 2 \varepsilon}{3 + \varepsilon} + \frac{ 9 \varepsilon^2 \gamma_\ell } { 2 (3
+ \varepsilon)^2 \ln (\zeta \delta)}$} for $\ell = 1, \cdots, s - 1$.
 \end{theorem}

 In this section, we have proposed a multistage inverse sampling plan
for estimating a binomial parameter, $p$, with relative precision.
In some situations, the cost of sampling operation may be high since
samples are obtained one by one when inverse sampling is involved.
In view of this fact, it is desirable to develop multistage
estimation methods without using inverse sampling.  For this
purpose, we have

\begin{theorem} \label{noinverse} Let $0 < \varepsilon < 1, \; 0 < \delta < 1$ and $\zeta
> 0 $.  Let $\tau$ be a positive integer. For $\ell = 1, 2, \cdots$, let $\widehat{\boldsymbol{p}}_\ell = \frac{ \sum_{i =
1}^{n_\ell} X_i } { n_\ell }$, where $n_\ell$ is deterministic and
stands for the sample size at the $\ell$-th stage. For $\ell = 1, 2,
\cdots$, define $\boldsymbol{D}_\ell$ such that $\boldsymbol{D}_\ell = 1$ if {\small
$\widehat{\boldsymbol{p}}_\ell \geq \frac{ 6(1 + \varepsilon) (3 + \varepsilon) \ln (\zeta \delta_\ell)
} { 2 (3 + \varepsilon)^2 \ln (\zeta \delta_\ell) - 9 n_\ell \varepsilon^2}$}; and
$\boldsymbol{D}_\ell = 0$ otherwise, where $\delta_\ell = \delta$ for $1 \leq \ell
\leq \tau$ and $\delta_\ell = \delta 2^{\tau - \ell}$ for $\ell > \tau$.
Suppose the stopping rule is that sampling is continued until
$\boldsymbol{D}_\ell = 1$ for some stage with index $\ell$. Define estimator
$\widehat{\boldsymbol{p}} = \widehat{\boldsymbol{p}}_{\boldsymbol{l}}$, where $\boldsymbol{l}$ is the index of
stage at which the sampling is terminated. Then, $\Pr \{ \boldsymbol{l} <
\infty \} = 1$ and {\small $\Pr \left \{ \left | \frac{ \widehat{\boldsymbol{p}} - p } { p
} \right | \leq \varepsilon \mid p \right \} \geq 1 - \delta$} for any $p \in (0,
1)$ provided that $2 (\tau + 1 ) \zeta \leq 1$ and $\inf_{\ell > 0}
\frac{n_{\ell + 1}}{n_\ell} > 0$.

\end{theorem}


\subsection{Fixed-width Confidence Intervals}

In some literature, the estimation of $p$ has been formulated as the
problem of constructing a fixed-width confidence interval $(\boldsymbol{L},
\boldsymbol{U})$ such that $\boldsymbol{U} - \boldsymbol{L} \leq 2 \varepsilon$ and that $\Pr \left
\{ \boldsymbol{L} < p < \boldsymbol{U} \mid p \right \}
> 1 - \delta$ for any $p \in (0, 1)$ with prescribed $\varepsilon \in (0, \frac{1}{2})$ and $\delta \in (0,
1)$.  For completeness, we shall develop multistage sampling schemes
in this setting.

Making use of the Clopper-Pearson confidence interval
\cite{Clopper}, we have established the following sampling scheme.
\begin{theorem} \label{FW1} For $\alpha \in (0, 1)$ and integers $0 \leq k \leq n$,
define
\[ \mathcal{L} (n, k, \alpha) = \left\{\begin{array}{ll}
   0 \;\;\;&  {\rm if}\; k=0\\
   \underline{p} \; \;\;\;&
   {\rm if}\; k > 0
\end{array} \right.\;\;\;{\rm and}\;\;\;
\mathcal{U} (n, k, \alpha) = \left\{\begin{array}{ll}
   1  &  {\rm if}\; k= n\\
   \overline{p} \;
   & {\rm if}\; k < n
\end{array} \right.
\]
with $\underline{p} \in (0,1)$ satisfying $\sum_{j=k}^n {n \choose
j} \underline{p}^j (1- \underline{p})^{n-j} = \frac{\alpha}{2}$  and
$\overline{p} \in (0,1)$ satisfying $\sum_{j=0}^{k} {n \choose j}
   \overline{p}^j (1- \overline{p})^{n-j} = \frac{\alpha}{2}$.  Let $\zeta > 0$ and $\rho > 0$. Let $n_1 < n_2 < \cdots < n_s$ be the
ascending arrangement of all distinct elements of {\small $ \left \{
\left \lceil \left ( \frac{ 2 \varepsilon^2} { \ln \frac{1}{1- 2 \varepsilon} } \right )^{ 1 -
\frac{i}{\tau} } \frac{ \ln \frac{1}{\zeta \delta} } { 2 \varepsilon^2 } \right \rceil : i = 0,
1, \cdots, \tau \right \}$} with {\small  $\tau = \left \lceil \frac{ \ln \left (
\frac{1}{ 2 \varepsilon^2} \ln \frac{1}{1- 2 \varepsilon} \right ) } { \ln (1 + \rho)} \right
\rceil$. }   For $\ell = 1, \cdots, s$, define $K_\ell = \sum_{i =
1}^{n_\ell} X_i$  and $\boldsymbol{D}_\ell$ such that $\boldsymbol{D}_\ell = 1$ if
$\mathcal{U} (n_\ell, K_\ell, \zeta \delta) - \mathcal{L} (n_\ell, K_\ell, \zeta
\delta) \leq 2 \varepsilon$; and $\boldsymbol{D}_\ell = 0$ otherwise.  Suppose the
stopping rule is that sampling is continued until $\boldsymbol{D}_\ell = 1$
for some $\ell \in \{1, \cdots, s\}$.  Define {\small $\boldsymbol{L} =
\mathcal{L} \left (\mathbf{n}, \sum_{i=1}^{\mathbf{n}} X_i, \zeta \delta \right
)$} and {\small $\boldsymbol{U} = \mathcal{U} \left (\mathbf{n},
\sum_{i=1}^{\mathbf{n}} X_i, \zeta \delta \right )$}, where $\mathbf{n}$ is
the sample size when the sampling is terminated. Define {\small  \[
\mathscr{Q}_L = \bigcup_{\ell = 1}^s \left \{ \mathcal{L} (n_\ell, k, \zeta
\delta)  \in \left ( 0,  1 \right ) : 0 \leq k \leq n_\ell \right \}, \qquad
\mathscr{Q}_U = \bigcup_{\ell = 1}^s \left \{ \mathcal{U} (n_\ell, k, \zeta
\delta)  \in \left ( 0,  1 \right ) : 0 \leq k \leq n_\ell \right \}.
\]}
Then, a sufficient condition to guarantee $\Pr \left \{ \boldsymbol{L} < p <
\boldsymbol{U} \mid p \right \}
> 1 - \delta$ for any $p \in (0, 1)$ is that {\small  \begin{eqnarray} &  &
\sum_{\ell = 1}^s \Pr \{ \mathcal{L} (n_\ell, K_\ell, \zeta \delta) \geq p,
\; \boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} < \frac{\delta}{2}
\qquad \forall p \in \mathscr{Q}_L, \label{2D1F}\\
&  & \sum_{\ell = 1}^s \Pr \{ \mathcal{U} (n_\ell, K_\ell, \zeta \delta)
\leq p, \; \boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} <
\frac{\delta}{2} \qquad \forall p \in \mathscr{Q}_U \label{2D2F}
 \end{eqnarray}}
where both (\ref{2D1F}) and (\ref{2D2F}) are satisfied if $0 < \zeta <
\frac{1}{2(\tau + 1)}$.

\end{theorem}

Making use of Chernoff-Hoeffding inequalities \cite{Chernoff,
Hoeffding}, we have established the following sampling scheme.


\begin{theorem} \label{FW2} For $\alpha \in (0, 1)$ and integers $0 \leq k \leq n$,
define
\[
\mathcal{L} (n, k, \alpha) = \left \{\begin{array}{ll}
   \underline{p}  \;\;\;&  {\rm for}\; 0 < k < n,\\
   \left( \frac{\alpha}{2} \right)^{\frac{1}{n}} \; \;\;\;&
   {\rm for} \; k = n,\\
   0 \; \;\;\;&
   {\rm for} \; k = 0
\end{array} \right .  \qquad \mathcal{U} (n, k, \alpha) = \left \{\begin{array}{ll}
   \overline{p}  \;\;\;&  {\rm for}\; 0 < k < n,\\
   1 - \left( \frac{\alpha}{2} \right)^{\frac{1}{n}} \; \;\;\;&
   {\rm for} \; k = 0,\\
   1 \; \;\;\;&
   {\rm for} \; k = n
\end{array} \right .
\]
with $\underline{p} \in (0, \frac{k}{n})$ satisfying $\mathscr{M}_{\mathrm{B}} \left
( \frac{k}{n}, \underline{p} \right )  = \frac{ \ln (\zeta \delta) } { n}$ and $\overline{p}
\in (\frac{k}{n}, 1)$ satisfying $\mathscr{M}_{\mathrm{B}} \left ( \frac{k}{n},
\overline{p} \right )  = \frac{ \ln (\zeta \delta) } { n}$, where
$\mathscr{M}_{\mathrm{B}} (.,.)$ is a function such that
$\mathscr{M}_{\mathrm{B}} (z,\theta) = z \ln \frac{\theta}{z} + (1 - z) \ln \frac{1 -
\theta}{1 - z}$ for $z \in (0,1)$ and $\theta \in (0, 1)$. Let $\zeta > 0$
and $\rho
> 0$. Let $n_1 < n_2 < \cdots < n_s$ be the ascending arrangement of
all distinct elements of {\small $ \left \{ \left \lceil \left ( \frac{ 2
\varepsilon^2} { \ln \frac{1}{1- 2 \varepsilon} } \right )^{ 1 - \frac{i}{\tau} } \frac{ \ln
\frac{1}{\zeta \delta} } { 2 \varepsilon^2 } \right \rceil : i = 0, 1, \cdots, \tau \right \}$}
with {\small  $\tau = \left \lceil \frac{ \ln \left ( \frac{1}{ 2 \varepsilon^2} \ln
\frac{1}{1- 2 \varepsilon} \right ) } { \ln (1 + \rho)} \right \rceil$. }   For $\ell =
1, \cdots, s$, define $K_\ell = \sum_{i = 1}^{n_\ell} X_i$  and
$\boldsymbol{D}_\ell$ such that $\boldsymbol{D}_\ell = 1$ if $\mathcal{U} (n_\ell,
K_\ell, \zeta \delta) - \mathcal{L} (n_\ell, K_\ell, \zeta \delta) \leq 2 \varepsilon$;
and $\boldsymbol{D}_\ell = 0$ otherwise.  Suppose the stopping rule is that
sampling is continued until $\boldsymbol{D}_\ell = 1$ for some $\ell \in
\{1, \cdots, s\}$.  Define {\small $\boldsymbol{L} = \mathcal{L} \left (\mathbf{n},
\sum_{i=1}^{\mathbf{n}} X_i, \zeta \delta \right )$} and {\small $\boldsymbol{U} =
\mathcal{U} \left (\mathbf{n}, \sum_{i=1}^{\mathbf{n}} X_i, \zeta \delta \right
)$}, where $\mathbf{n}$ is the sample size when the sampling is
terminated. Define {\small  \[ \mathscr{Q}_L = \bigcup_{\ell = 1}^s \left
\{ \mathcal{L} (n_\ell, k, \zeta \delta)  \in \left ( 0,  1 \right ) : 0 \leq k
\leq n_\ell \right \}, \qquad \mathscr{Q}_U = \bigcup_{\ell = 1}^s \left \{
\mathcal{U} (n_\ell, k, \zeta \delta)  \in \left ( 0,  1 \right ) : 0 \leq k \leq
n_\ell \right \}.
\]}
Then, a sufficient condition to guarantee $\Pr \left \{ \boldsymbol{L} < p <
\boldsymbol{U} \mid p \right \}
> 1 - \delta$ for any $p \in (0, 1)$ is that {\small  \begin{eqnarray} &  &
\sum_{\ell = 1}^s \Pr \{ \mathcal{L} (n_\ell, K_\ell, \zeta \delta) \geq p,
\; \boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} < \frac{\delta}{2}
\qquad \forall p \in \mathscr{Q}_L, \label{2D1FC}\\
&  & \sum_{\ell = 1}^s \Pr \{ \mathcal{U} (n_\ell, K_\ell, \zeta \delta)
\leq p, \; \boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} <
\frac{\delta}{2} \qquad \forall p \in \mathscr{Q}_U \label{2D2FC}
 \end{eqnarray}}
where both (\ref{2D1FC}) and (\ref{2D2FC}) are satisfied if $0 < \zeta
< \frac{1}{2(\tau + 1)}$.

\end{theorem}

Making use of Massart's inequality \cite{Massart:90}, we have
established the following sampling scheme.

 \begin{theorem}
\label{FW3} For $\alpha \in (0, 1)$ and integers $0 \leq k \leq n$, define
{\small \[  \mathcal{L} (n, k, \alpha) = \max \left \{ 0, \; \frac{k}{n} +
\frac{3}{4} \; \frac{ 1 - \frac{2k}{n} - \sqrt{ 1 + \frac{9}{ 2 \ln
\frac{2}{\alpha} } \; k ( 1- \frac{k}{n}) } } {1 + \frac{9 n}{ 8 \ln
\frac{2}{\alpha} } } \right \},
\]} and {\small \[ \mathcal{U} (n, k, \alpha) = \min \left \{ 1, \; \frac{k}{n} +
\frac{3}{4} \; \frac{ 1 - \frac{2k}{n} + \sqrt{ 1 + \frac{9}{ 2 \ln
\frac{2}{\alpha} } \; k ( 1- \frac{k}{n}) } } {1 + \frac{9 n}{ 8 \ln
\frac{2}{\alpha} } } \right \}.
\]}  Let $\zeta > 0$ and $\rho
> 0$. Let $n_1 < n_2 < \cdots < n_s$ be the ascending arrangement of
all distinct elements of {\small $ \left \{ \left \lceil \frac{8}{9} \left (
\frac{3}{4 \varepsilon} + 1  \right )^{ \frac{i}{\tau} } \left ( \frac{3}{4 \varepsilon} -
1  \right ) \ln \frac{1}{\zeta \delta} \right \rceil : i = 0, 1, \cdots, \tau \right
\}$} with {\small  $\tau = \left \lceil \frac{ \ln \left ( \frac{3}{4 \varepsilon} +
1 \right ) } { \ln (1 + \rho)} \right \rceil$. } For $\ell = 1, \cdots, s$,
define $K_\ell = \sum_{i = 1}^{n_\ell} X_i$ and $\boldsymbol{D}_\ell$ such
that $\boldsymbol{D}_\ell = 1$ if \[
1 - \frac{9}{ 2 \ln (\zeta \delta) } \;
K_\ell \left ( 1- \frac{K_\ell }{n_\ell} \right ) \leq \varepsilon^2 \left [
\frac{4}{3} - \frac{3 n_\ell}{ 2 \ln (\zeta \delta) } \right ]^2, \] and
$\boldsymbol{D}_\ell = 0$ otherwise.  Suppose the stopping rule is that
sampling is continued until $\boldsymbol{D}_\ell = 1$ for some $\ell \in
\{1, \cdots, s\}$. Define {\small $\boldsymbol{L} = \mathcal{L} \left (\mathbf{n},
\sum_{i=1}^{\mathbf{n}} X_i, \zeta \delta \right )$} and {\small $\boldsymbol{U} =
\mathcal{U} \left (\mathbf{n}, \sum_{i=1}^{\mathbf{n}} X_i, \zeta \delta \right
)$}, where $\mathbf{n}$ is the sample size when the sampling is
terminated. Define {\small  \[ \mathscr{Q}_L = \bigcup_{\ell = 1}^s \left
\{ \mathcal{L} (n_\ell, k, \zeta \delta)  \in \left ( 0,  1 \right ) : 0 \leq k
\leq n_\ell \right \}, \qquad \mathscr{Q}_U = \bigcup_{\ell = 1}^s \left \{
\mathcal{U} (n_\ell, k, \zeta \delta)  \in \left ( 0,  1 \right ) : 0 \leq k \leq
n_\ell \right \}.
\]}
Then, a sufficient condition to guarantee $\Pr \left \{ \boldsymbol{L} < p <
\boldsymbol{U} \mid p \right \}
> 1 - \delta$ for any $p \in (0, 1)$ is that {\small  \begin{eqnarray} &  &
\sum_{\ell = 1}^s \Pr \{ \mathcal{L} (n_\ell, K_\ell, \zeta \delta) \geq p,
\; \boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} < \frac{\delta}{2}
\qquad \forall p \in \mathscr{Q}_L, \label{2D1mF}\\
&  & \sum_{\ell = 1}^s \Pr \{ \mathcal{U} (n_\ell, K_\ell, \zeta \delta)
\leq p, \; \boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_\ell = 1 \mid p \} <
\frac{\delta}{2} \qquad \forall p \in \mathscr{Q}_U \label{2D2mF}
 \end{eqnarray}}
where both (\ref{2D1mF}) and (\ref{2D2mF}) are satisfied if $0 < \zeta
< \frac{1}{2(\tau + 1)}$.

\end{theorem}

\bigskip

It should be noted that the interval estimation methods described in
Theorems \ref{FW1}--\ref{FW3} can be made less conservative by using
tight bounds of $C (p, \varepsilon) = 1 - \Pr \{ \boldsymbol{L} < p < \boldsymbol{U}  \mid
p \}$ for $p \in [a, b] \subseteq \Theta$ in Theorem \ref{bbsplit}.
Based on such bounds, a branch-and-bound type strategy described in
section 2.8 of \cite{Chen_EST} can be used to facilitate the search
of an appropriate value of $\zeta$ such that the coverage probability
associated with interval $(\boldsymbol{L}, \boldsymbol{U})$ is no less than $1 -
\delta$.


\begin{theorem}  \label{bbsplit} Let $\mathcal{L}_\ell = \mathcal{L} (\widehat{\boldsymbol{p}}_\ell,
\zeta, \delta)$ and $\mathcal{U}_\ell = \mathcal{U} (\widehat{\boldsymbol{p}}_\ell, \zeta,
\delta)$ for $\ell = 1, \cdots, s$.   Then, \begin{eqnarray*} C (p, \varepsilon) & \leq & \Pr
\{  \boldsymbol{L} \geq a \mid b \} + \Pr \{  \boldsymbol{U} \leq b \mid a \} \\
& \leq & \sum_{\ell = 1}^s \Pr \{ \mathcal{L}_\ell \geq a, \;
\boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_{\ell } = 1 \mid b \} + \sum_{\ell
= 1}^s \Pr \{ \mathcal{U}_\ell \leq b, \; \boldsymbol{D}_{\ell - 1} = 0, \;
\boldsymbol{D}_{\ell } = 1 \mid a \}, \end{eqnarray*} \begin{eqnarray*} C (p, \varepsilon) & \geq & \Pr \{
\boldsymbol{L} \geq b \mid a \} + \Pr \{  \boldsymbol{U} \leq a \mid b \} \\
& \geq & \sum_{\ell = 1}^s \Pr \{ \mathcal{L}_\ell \geq b, \;
\boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_{\ell } = 1 \mid a \} + \sum_{\ell
= 1}^s \Pr \{ \mathcal{U}_\ell \leq a, \; \boldsymbol{D}_{\ell - 1} = 0, \;
\boldsymbol{D}_{\ell } = 1 \mid b \} \end{eqnarray*} for any $p \in [a, b]$. Moreover,
if the open interval $(a, b)$ contains no element of the supports of
$\boldsymbol{L}$ and $\boldsymbol{U}$, then \begin{eqnarray*} C (p, \varepsilon) & \leq & \Pr
\{  \boldsymbol{L} \geq b \mid b \} + \Pr \{  \boldsymbol{U} \leq a \mid a \} \\
& \leq & \sum_{\ell = 1}^s \Pr \{ \mathcal{L}_\ell \geq b, \;
\boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_{\ell } = 1 \mid b \} + \sum_{\ell
= 1}^s \Pr \{ \mathcal{U}_\ell \leq a, \; \boldsymbol{D}_{\ell - 1} = 0, \;
\boldsymbol{D}_{\ell } = 1 \mid a \}, \end{eqnarray*} \begin{eqnarray*} C (p, \varepsilon) & \geq & \Pr
\{  \boldsymbol{L} > a \mid a \} + \Pr \{  \boldsymbol{U} < b \mid b \} \\
& \geq  & \sum_{\ell = 1}^s \Pr \{ \mathcal{L}_\ell > a, \;
\boldsymbol{D}_{\ell - 1} = 0, \; \boldsymbol{D}_{\ell } = 1 \mid a \} + \sum_{\ell
= 1}^s \Pr \{ \mathcal{U}_\ell < b, \; \boldsymbol{D}_{\ell - 1} = 0, \;
\boldsymbol{D}_{\ell } = 1 \mid b \} \end{eqnarray*} for any $p \in (a, b)$. \end{theorem}

\bigskip

We would like to note that Theorems 1 and 2 of \cite{Chen_EST} play
important roles in the establishment of the theorems in this
section. As can be seen from Theorems 1--6, the confidence
requirements can be satisfied by choosing $\zeta$ to be sufficiently
small.  The application of the double-decision-variable method and
the single-decision-variable method is obvious.  To determine $\zeta$
as large as possible and thus make the sampling schemes most
efficient, the computational techniques such as bisection confidence
tuning, domain truncation, triangular partition developed in
\cite{Chen_EST} can be applied.

With regard to the tightness of the double-decision-variable method,
we can develop results similar to Theorems 13, 18 and 23 of
\cite{Chen_EST}.

With regard to the asymptotic performance of our sampling schemes,
we can develop results similar to Theorems 14, 19 and 24 of
\cite{Chen_EST}.

\section{Estimation of Bounded-variable Means}

 The method proposed for estimating binomial parameters can be
 generalized for estimating means of random variables bounded in interval $[0, 1]$.
 Formally, let $Z \in [0, 1]$ be a random variable with expectation $\mu =
 \mathbb{E} [Z]$. We can estimate $\mu$ based on i.i.d. random samples
 $Z_1, Z_2, \cdots$ of $Z$ by virtue of the following results.

 \begin{theorem} \label{coverage_abs} Let $0 < \varepsilon < \frac{1}{2}$ and $0 < \delta < 1$.
Let $n_1 < n_2 < \cdots < n_s$ be a sequence of sample sizes such that
{\small $n_s \geq \frac{ \ln \frac{2 s} { \delta} } { 2 \varepsilon^2 }$}.  Define
$\widehat{\boldsymbol{\mu}}_\ell = \frac{ \sum_{i=1}^{n_\ell} Z_i }{n_\ell}$  for
$\ell = 1, \cdots, s$.  Suppose the stopping rule is that sampling is
continued until {\small $\left ( \left | \widehat{\boldsymbol{\mu}}_\ell - \frac{1}{2}
\right | - \frac{2 \varepsilon }{3} \right )^2 \geq \frac{1}{4} - \frac{ \varepsilon^2 n_\ell }
{2 \ln (2 s \slash \delta) }$} for some $\ell \in \{1, \cdots, s \}$. Define
$\boldsymbol{\widehat{\mu}} = \frac{\sum_{i=1}^{\mathbf{n}} Z_i}{\mathbf{n}}$ where
$\mathbf{n}$ is the sample size when the sampling is terminated.
 Then, $\Pr \left \{ \left | \boldsymbol{\widehat{\mu}} - \mu \right | < \varepsilon \right \}
\geq 1 - \delta$. \end{theorem}

\bigskip

This theorem can be shown by a variation of the argument for Theorem
1.


\begin{theorem} \label{coverage_mix_bound} Let $0 < \delta < 1, \; 0 < \varepsilon_a <
\frac{3}{8}$ and {\small $\frac{6 \varepsilon_a}{3 - 2 \varepsilon_a } < \varepsilon_r < 1$}.
Let $n_1 < n_2 < \cdots < n_s$ be a sequence of sample sizes such that
{\small $n_s \geq 2  \left ( \frac{1}{ \varepsilon_r } + \frac{1} {3} \right ) \left
( \frac{1}{ \varepsilon_a } - \frac{1}{ \varepsilon_r } - \frac{1} {3} \right ) \ln \left (
\frac{2s}{\delta} \right )$}.  Define $\widehat{\boldsymbol{\mu}}_\ell = \frac{
\sum_{i=1}^{n_\ell} Z_i }{n_\ell}$  for $\ell = 1, \cdots, s$.  Define
{\small \[ \boldsymbol{D}_\ell = \begin{cases} 0 & \mathrm{for} \; \frac{1}{2} - \frac{2}{3}
\varepsilon_a - \sqrt{ \frac{1}{4} + \frac{ n_\ell \varepsilon_a^2 } {2 \ln (\zeta \delta) } }
< \widehat{\boldsymbol{\mu}}_\ell < \frac{ 6(1 - \varepsilon_r) (3 - \varepsilon_r) \ln (\zeta
\delta) } { 2 (3 - \varepsilon_r)^2 \ln (\zeta \delta) - 9 n_\ell \varepsilon_r^2} \; \mathrm{or}\\
  &  \quad \;\; \frac{1}{2} + \frac{2}{3} \varepsilon_a - \sqrt{ \frac{1}{4} + \frac{
n_\ell \varepsilon_a^2 } {2 \ln (\zeta \delta) } } < \widehat{\boldsymbol{\mu}}_\ell < \frac{
6(1 + \varepsilon_r) (3 + \varepsilon_r) \ln (\zeta \delta) } { 2 (3 +
\varepsilon_r)^2 \ln (\zeta \delta) - 9 n_\ell  \varepsilon_r^2},\\
1 & \mathrm{else}
 \end{cases}
\]}
for $\ell = 1, \cdots, s - 1$ and $\boldsymbol{D}_s = 1$.  Suppose the stopping
rule is that sampling is continued until $\boldsymbol{D}_\ell = 1$ for some
$\ell \in \{1, \cdots, s \}$.  Define $\boldsymbol{\widehat{\mu}} =
\frac{\sum_{i=1}^{\mathbf{n}} Z_i}{\mathbf{n}}$ where $\mathbf{n}$ is
the sample size when the sampling is terminated.
 Then, $\Pr \left \{ \left | \boldsymbol{\widehat{\mu}} - \mu \right | < \varepsilon_a \; \text{or} \; \left | \boldsymbol{\widehat{\mu}} - \mu \right | < \varepsilon_r \mu \right \}
\geq 1 - \delta$. \end{theorem}

\bigskip

This theorem can be shown by a variation of the argument for Theorem
2.  In the general case that $Z$ is a random variable bounded in
$[a, b]$, it is useful to estimate the mean $\mu = \mathbb{E} [ Z]$
based on i.i.d. samples of $Z$ with a mixed criterion.  For this
purpose,  we shall introduce the function
\[
\mathcal{M}(z, \mu) = \begin{cases}  \frac{ (\mu - z)^2 } {2 \left ( \frac{2 \mu}{3} +
\frac{z}{3} \right ) \left ( \frac{2 \mu}{3} + \frac{z}{3} - 1 \right ) } & \text{for}
\; 0 \leq z \leq 1 \; \text{and} \;  \mu \in (0, 1),\\
- \infty & \text{for} \; 0 \leq z \leq 1 \; \text{and} \; \mu \notin (0, 1)
\end{cases}
\]
and propose the following multistage estimation method.

\begin{theorem} \label{coverage_mix_general} Let $0 < \delta < 1, \; \varepsilon_a > 0$ and
$0 < \varepsilon_r < 1$. Let $n_1 < n_2 < \cdots < n_s$ be a sequence of
sample sizes such that {\small $n_s \geq  \frac{(b - a)^2}{ 2 \varepsilon_a^2}
\ln \left ( \frac{2s}{\delta} \right )$}. Define $\widehat{\boldsymbol{\mu}}_\ell = \frac{
\sum_{i=1}^{n_\ell} Z_i }{n_\ell}, \; \widetilde{\boldsymbol{\mu}}_\ell  = a +
\frac{1}{b - a} \widehat{\boldsymbol{\mu}}_\ell$,
\[
 \underline{\boldsymbol{\mu}}_\ell = a +
\frac{1}{b - a}  \min \left \{ \widehat{\boldsymbol{\mu}}_\ell - \varepsilon_a, \;
\frac{\widehat{\boldsymbol{\mu}}_\ell}{ 1 + \mathrm{sgn} (\widehat{\boldsymbol{\mu}}_\ell) \varepsilon_r }
\right \}, \qquad \overline{\boldsymbol{\mu}}_\ell = a + \frac{1}{b - a}  \max \left \{
\widehat{\boldsymbol{\mu}}_\ell + \varepsilon_a, \; \frac{\widehat{\boldsymbol{\mu}}_\ell}{ 1 -
\mathrm{sgn} (\widehat{\boldsymbol{\mu}}_\ell) \varepsilon_r } \right \}
\]
for $\ell = 1, \cdots, s$.  Suppose the stopping rule is that sampling
is continued until $\mathcal{M}(\widetilde{\boldsymbol{\mu}}_\ell,
\underline{\boldsymbol{\mu}}_\ell) \leq \frac{1}{n_\ell} \ln \frac{\delta}{2s}$  and
$\mathcal{M}(\widetilde{\boldsymbol{\mu}}_\ell, \overline{\boldsymbol{\mu}}_\ell) \leq
\frac{1}{n_\ell} \ln \frac{\delta}{2s}$ for some $\ell \in \{1, \cdots, s \}$.
Define $\boldsymbol{\widehat{\mu}} = \frac{\sum_{i=1}^{\mathbf{n}} Z_i}{\mathbf{n}}$
where $\mathbf{n}$ is the sample size when the sampling is
terminated.
 Then, $\Pr \left \{ \left | \boldsymbol{\widehat{\mu}} - \mu \right | < \varepsilon_a \; \text{or} \; \left | \boldsymbol{\widehat{\mu}} - \mu \right | < \varepsilon_r |\mu| \right \}
\geq 1 - \delta$. \end{theorem}

 \section{A Link between
Binomial and Bounded Variables}

There exists an inherent connection between a binomial parameter and
the mean of a bounded variable.  In this regard, we have

\begin{theorem} \label{Link} Let $Z$ be a random variable bounded in $[0, 1]$. Let
$U$ a random variable uniformly distributed over $[0, 1]$. Suppose
$Z$ and $U$ are independent. Then,
\[ \mathbb{E} [Z] = \Pr \{ Z \geq U \}.
\] \end{theorem}

\begin{pf} Let $F_{Z, U}$ be the joint distribution of $Z$ and $U$.  Let
$F_Z$ be the cumulative distribution function of $Z$. Since $Z$ and
$U$ are independent, using Riemann-Stieltjes integration, we have
\[
\Pr \{ Z \geq U \} = \int_{z = 0}^1 \int_{u = 0}^z d F_{Z, U} =
\int_{z = 0}^1 \int_{u = 0}^z d u \; d F_Z =  \int_{z = 0}^1 z \; d
F_Z = \mathbb{E} [Z].
\]
\end{pf}

To see why Theorem \ref{Link} reveals a relationship between the
mean of a bounded variable and a binomial parameter, we define
\[
X = \begin{cases} 1 & \text{for} \; Z \geq U,\\
0 & \text{otherwise}. \end{cases}
\]
Then, by Theorem \ref{Link}, we have $\Pr \{ X = 1 \} = 1 - \Pr \{ X
= 0 \} = \mathbb{E} [Z]$. This implies that $X$ is a Bernoulli random
variable and $\mathbb{E} [Z]$ is actually a binomial parameter. As a
consequence, the techniques of estimating a binomial parameter can
be useful for estimating the mean of a bounded variable. Specially,
for a sequence of i.i.d. random samples $Z_1, Z_2, \cdots$ of bounded
variable $Z$ and a sequence of i.i.d. random samples $U_1, U_2, \cdots$
of uniform variable $U$ such that that $Z_i$ is independent with
$U_i$ for all $i$, we can define a sequence of i.i.d. random samples
$X_1, X_2, \cdots$  of Bernoulli random variable $X$ by
\[
X_i = \begin{cases} 1 & \text{for} \; Z_i \geq U_i,\\
0 & \text{otherwise}. \end{cases}
\]


\section{Conclusion}

We have established a new multistage approach for estimating the
mean of a bounded variable.  Our approach can provide an estimator
for the unknown mean which rigorously guarantees prescribed levels
of precision and confidence. Our approach is also very flexible in
the sense that the precision can be expressed in terms of different
types of margins of errors.

\bigskip

