Design and analysis of shortest two-sided confidence intervals for a probability under prior information

Abstract

Two-sided confidence intervals for a probability $p$ under a prescribed confidence level $\gamma$ are an elementary tool of statistical data analysis. A confidence interval has two basic quality characteristics: i) exactness, i. e., whether the actual coverage probability equals or exceeds the prescribed level $\gamma$; ii) inferential precision, measured by the length of the confidence interval. The interval provided by Clopper and Pearson (Biometrika 26:404–413, 1934) is the only exact interval actually used in statistical data analysis. Various authors have suggested shorter, i. e., more precise exact intervals. The present paper makes two contributions. i) We provide a general design scheme for minimum volume confidence regions under prior knowledge on the target parameter. ii) We apply the scheme to the problem of confidence intervals for a probability $p$ where prior knowledge is expressed in a flexible way by a beta distribution on a subset of the unit interval.

Appendices

Proofs related to Section 2

1.1 Proof of theorem 1

Proof of assertion b) of theorem 1

Let $A$ be an arbitrary level $\gamma$ MPS. Let $y\in R_2$ be fixed. We have

$$\begin{aligned}&\displaystyle \text{ P }_{y} (X\in A_y\cap A^{\star }_y)+ \text{ P }_{y} (X\in A_y{\setminus }A^{\star }_y)=\text{ P }_{y} (X\in A_y) \;\ge \; \gamma \ge \text{ P }_{y}\\&\displaystyle (X\in A^{\star }_y)=\text{ P }_{y} (X\in A_y\cap A^{\star }_y)+ \text{ P }_{y} (X\in A^{\star }_y{\setminus }A_y), \end{aligned}$$

and hence $\text{ P }_{y} (X\in A^{\star }_y{\setminus }A_y) \le \text{ P }_{y} (X\in A_y{\setminus }A^{\star }_y)$. We have $A^{\star }_y{\setminus }A_y \subset D_{\ge s_y} (y)$ and $A_y{\setminus }A^{\star }_y \subset D_{\le s_y} (y)$, thus $\inf _{x\in A^{\star }_y{\setminus }A_y} Q_y(x) \ge \sup _{x\in A_y{\setminus }A^{\star }_y} Q_y(x)$. If $\inf _{x\in A^{\star }_y{\setminus }A_y} Q_y(x)=0$, then $s_y =0$, hence $D_{> s_y}(y)=\{x\vert f_{X\vert Y=y}(x)>0\}$, and thus $G_y(s_y)=\text{ P }_{y} (X\in D_{> s_y} (y)) =1>\gamma$ contradictory to the assumptions of theorem 1. Hence $\inf _{x\in A^{\star }_y{\setminus }A_y} Q_y(x)>0$. We obtain the inequalities

$$\begin{aligned} \displaystyle \inf _{x\in A^{\star }_y{\setminus }A_y} Q_y(x) \int \limits _{A^{\star }_y{\setminus }A_y} f_{X}(x) \, \text{ d }\mu _1(x)&\le \int \limits _{A^{\star }_y{\setminus }A_y} Q_y(x)f_{X}(x) \, \text{ d }\mu _1(x) \\&= \displaystyle \int \limits _{A^{\star }_y{\setminus }A_y} f_{X\vert Y=y}(x) \, \text{ d }\mu _1(x) = \text{ P }_{y} (X\in A^{\star }_y{\setminus }A_y)\\&\le \text{ P }_{y} (X\in A_y{\setminus }A^{\star }_y) = \displaystyle \int \limits _{A_y{\setminus }A^{\star }_y} f_{X\vert Y=y}(x) \, \text{ d }\mu _1(x)\\&= \int \limits _{A_y{\setminus }A^{\star }_y} Q_y(x)f_{X}(x) \, \text{ d }\mu _1(x)\\&\le \displaystyle \sup _{x\in A_y{\setminus }A^{\star }_y} Q_y(x) \int \limits _{A_y{\setminus }A^{\star }_y} f_{X}(x) \, \text{ d }\mu _1(x)\\&\le \inf _{x\in A^{\star }_y{\setminus }A_y} Q_y(x) \int \limits _{A_y{\setminus }A^{\star }_y} f_{X}(x) \, \text{ d }\mu _1(x). \end{aligned}$$

Hence $\int _{A^{\star }_y{\setminus }A_y} f_{X}(x) \, \text{ d }\mu _1(x)\le \int _{A_y{\setminus }A^{\star }_y} f_{X}(x) \, \text{ d }\mu _1(x)$, and thus we have $\int _{A^{\star }_y} f_{X}(x) \, \text{ d }\mu _1 (x)\le \int _{A_y} f_{X}(x) \, \text{ d }\mu _1(x)$.

The latter inequality was proven for arbitrary $y\in R_2$. Hence by definition (3) we obtain $V(A^{\star }) \le V(A)$. This completes the proof of assertion b) of theorem 1.

Proof of assertion c) of theorem 1

Consider $y\in R_2$ where $D_{= s_y} (y)= \bigcup _I B_i$ is a countable union of intervals $B_i$ with disjoint interior. We can assume $G_y(s_y^-)- G_y(s_y)= \text{ P }_y (X\in D_{= s_y}(y))>0$. From properties of the Lebesgue–Borel integral it follows that for any $0\le \alpha _i\le \int _{B_i}f_X\, \text{ d }\mu _1$ there is a subinterval $B_{i,\alpha _i}\subset B_i$ with $\int _{B_{i,\alpha _i}}f_X\, \text{ d }\mu _1 =\alpha _i$. Let $\rho = [\gamma -G_y(s_y)]/[G_y(s_y^-)- G_y(s_y)]$. For $i\in I$, let $\alpha _i= \rho \int _{B_{i}}f_X\, \text{ d }\mu _1$. Let $E(y)= \bigcup _I B_{i,\alpha _i}$. Then

$$\begin{aligned} \text{ P }_y (X\in E(y)) = \sum \limits _I \int \limits _{B_{i,\alpha _i}}f_X\, \text{ d }\mu _1 = \rho \text{ P }_y (X\in D_{= s_y}(y)) = \gamma -G_y(s_y), \end{aligned}$$

hence $\text{ P }_{y} (X\in D_{> s_y}(y)\cup E(y)) = G_y(s_y) + \gamma -G_y(s_y)=\gamma$.

1.2 Proof of proposition 2

We prove assertion a) of proposition 2. The proof of assertion b) is completely analogous. The sets of conditions a.i) and a.ii) are completely symmetric. Hence it suffices to prove the implication from a.i) to a.ii). Assume that the conditions a.i.1), a.i.2), a.i.3) are valid.

To prove the validity of b.ii.1), assume $y\in R_2$ such that $A_y$ is not an interval. Then there are $x_1, x^{\prime }, x_2 \in A_y, x_1 < x^{\prime }< x_2$, with $x_1, x_2\in A_y, x^{\prime }\notin A_y$. Hence $y\notin A_{x^{\prime }}$, i. e., either $y < A_{x^{\prime }}$, or $y > A_{x^{\prime }}$. Consider the former case. By property i.2), $y<A_{x_2}$, hence $x_2\notin A_y$, contradictory to the assumptions on $x_2$. The second case $y > A_{x^{\prime }}$ is treated analogously. It follows that $A_y$ is an interval.

To prove the validity of b.ii.2), consider $y_1, y_2 \in R_2, y_1 < y_2, x\in R_1$ with $x < A_{y_1}$. Assume $x^{\prime }\in A_{y_2}$ with $x\ge x^{\prime }$. Then $x^{\prime } \le x < A_{y_1}$. Hence $x^{\prime } \in A_{y_2}{\setminus }A_{y_1}$. Since $x\notin A_{y_1}$ we have $y_1\notin A_{x}$. Because of the stipulated interval property of $A_x$, see assumption a.i.1), we have either $y_1< A_x$ or $y_1> A_x$. Assume $y_1> A_x$. Then by property a.i.3) also $y_1> A_{x^{\prime }}$, hence $y_2> A_{x^{\prime }}$, hence $x^{\prime }\notin A_{y_2}$, in contradiction to the above result $x^{\prime } \in A_{y_2}{\setminus }A_{y_1}$. Thus $y_1< A_x$. From $x^{\prime } \le x < A_{y_1}$ we obtain by property i.2) for all $x^{\prime \prime }\in A_{y_1}$ that $y_1< A_{x^{\prime \prime }}$, hence in particular $x^{\prime \prime }\notin A_{y_1}$, in contradiction to the assumption $x^{\prime \prime }\in A_{y_1}$. This final contradiction results from assuming an $x^{\prime }\in A_{y_2}$ with $x\ge x^{\prime }$. Hence $x < A_{y_2}$. This proves the validity of bii.2).

The proof of the validity of b.ii.3) proceeds analogously.

Proofs related to Section 5

1.1 Proof of proposition 5

Proof of assertion a) of proposition 5

Let $0<p_0 <p_1 <1$. We have

$$\begin{aligned} \frac{\text{ d }^{m}}{\text{ d } t^m} t^x (1-t)^{n-x} = \ln \left( \frac{t}{1-t}\right) ^m t^x (1-t)^{n-x} \quad \text{ for } p_0\le t \le p_1. \end{aligned}$$

$[p_0;p_1]\ni t \mapsto \ln (t/(1-t))^m$ is a continuous function on $[p_0; p_1]$. Hence

$$\begin{aligned} \sup _{p_0\le t\le p_1} \left| \ln \left( \frac{t}{1-t}\right) ^m t^x (1-t)^{n-x} \right| \le \sup _{p_0\le t\le p_1} \left| \ln \left( \frac{t}{1-t}\right) ^m \right| < +\infty . \end{aligned}$$

Hence, by applying the well-known theorem on differentiation under the integral sign to (13), we find that $v_{p_0,p_1,a,b}(x)$ can be differentiated with respect to $x$ with the result given by Eq. (16). By Eq. (16) we have $v^{\prime \prime }_{p_0,p_1,a,b}(x)>0$ on $[0;n]$. Hence $v^{\prime }_{p_0,p_1,a,b}$ is strictly increasing on $[0;n]$. The remainder of assertion a) follows obviously.

Proof of assertion b) of proposition 5

The first partial derivatives of the symmetric beta function are given by the equation $\frac{\partial }{\partial s} B(s,t) = B(s,t) \Big [\psi (s)-\psi (s+t)\Big ]$, see Abramowitz and Stegun (1972), equation 6.6.8, where $\psi$ is the digamma function. With Eqs. (9) and (13) we obtain the derivative (17). The digamma function is strictly increasing on $(0;+\infty )$, see equation 6.3.21 of Abramowitz and Stegun (1972), for instance. Since $B(x+a,n-x+b)>0$, the sign of $v^{\prime }_{0,1}(x)$ is determined by the sign of $\psi (x+1)-\psi (n-x+1)$. The integral representation 6.3.21 of Abramowitz and Stegun (1972) shows that $\psi$ is strictly increasing on $(0;+\infty )$, hence $[0;n] \ni x \mapsto \psi (x+a)-\psi (n-x+b)$ is strictly increasing.

Proof of assertion c) of proposition 5

Let $0<p_0 < 1$. For $p_0 <q \le 1, x\in [0;n]$ let

$$\begin{aligned} \displaystyle h_{p_0,q,a,b}(x) \!=\! \displaystyle \frac{1}{B(a, b)(1-p_0)} \int \limits _{p_0}^{q} \!\left( \frac{y\!-\!p_0}{1\!-\!p_0}\right) ^{a-1} \left[ 1\!-\! \frac{y\!-\! p_0}{1\!-\!p_0}\right] ^{b-1} y^x (1\!-\!y)^{n-x} \, \text{ d } y.\nonumber \\ \end{aligned}$$

(23)

Evidently, $\lim _{q\uparrow 1} h_{p_0,q,a,b}(x) = h_{p_0,1,a,b}(x) = v_{p_0,1,a,b}(x)$ $x\in [0;n]$. Analogously to the proof of a) of proposition 5 we find

$$\begin{aligned} \begin{array} {l} \displaystyle h^{(m)}_{p_0,q,a,b}(x)= \frac{1}{B(a, b)(1-p_0)} \; \cdot \\ \quad \displaystyle \int \limits _{p_0}^{q} \left( \frac{y-p_0}{1 -p_0}\right) ^{a-1} \left[ 1- \frac{y- p_0}{1-p_0}\right] ^{b-1} \ln \left( \frac{y}{1-y}\right) ^m y^x (1-q)^{n-x} \, \text{ d } y \end{array} \end{aligned}$$

(24)

for $m=0,1,\ldots$, and we see that there exists a value $0\le z_{p_0,q,a,b}\le n$ such that $h^{\prime }_{p_0,q,a,b}(x)<0$ for $0 < x < z_{p_0,q,a,b}$, and $h^{\prime }_{p_0,q,a,b}(x)>0$ for $z_{p_0,q,a,b}< x < n$. From Eq. (24) we obtain for fixed $x\in [0;n], m=0,1,\ldots$,

$$\begin{aligned} \displaystyle \frac{\small {\text{ d }}}{\small {\text{ d }} q} h_{p_0,q,a,b}^{(m)} (x) = {\ln \left( \frac{q}{1-q}\right) ^m \over B(a,b)(1-p_0)} \left( {q-p_0\over 1-p_0}\right) ^{a-1} \left[ 1- {q- p_0\over 1-p_0}\right] ^{b-1} \displaystyle q^x (1-q)^{n-x} \nonumber \\ \end{aligned}$$

(25)

for $p_0 <q < 1$, with the recursion

$$\begin{aligned} \frac{\small {\text{ d }}}{\small {d} q} h_{p_0,q,a,b}^{(m)} (x) = \ln \left( {q\over 1-q}\right) \frac{\small {\text{ d }}}{\small {\text{ d }} q} h_{p_0,q,a,b}^{(m-1)} (x) \quad \text{ for } p_0 <q < 1, \; m\ge 1. \end{aligned}$$

(26)

From (25) we find in particular for $m=1,2,\ldots$

$$\begin{aligned} \frac{\small {\text{ d }}}{\small {\text{ d }} q} h_{p_0,q,a,b}^{(m)} (x) \;>\;0 \quad \text{ for } x\in (0;n), \; p_0 <q < 1, \;q>0.5. \end{aligned}$$

(27)

Let $p_0 < q_1<q_2<\cdots <1$ with $\lim _k q_k=1, q_k>0.5$. From (27) we see that the sequence $(z_{p_0, q_k,a,b})_{k\in \mathbb{N }}$ is decreasing. Since $0\le z_{p_0, q_k,a,b} \le n, (z_{p_0, q_k,a,b})_{k\in \mathbb{N }}$ converges to a value $x_{p_0,1,a,b} \in [0;n]$. Let $0<u <w <x_{p_0,1,a,b}$. Then $0<u <w <z_{p_0, q_k,a,b}$ for all $k\in \mathbb{N }$, thus $h_{p_0,q_k,a,b}(u) > h_{p_0,q_k,a,b}(w)$ for all $k\in \mathbb{N }$, and hence by convergence $v_{p_0,1,a,b}(u) \ge v_{p_0, 1,a,b}(w)$. Let $x_{p_0,1,a,b} <u <w <n$. Then there is a $k_0\in \mathbb{N }$ with $z_{p_0, q_k,a,b} <u <w <n$ for all $k\ge k_0$. Thus $h_{p_0,q_k,a,b}(u) < h_{p_0,q_k,a,b}(w)$ for all $k\ge k_0$, and hence by convergence $v_{p_0,1,a,b}(u) \le v_{p_0, 1,a,b}(w)$.

The proof of assertion d) of proposition 5 is completely analogous to the proof of assertion c).

1.2 Proof of proposition 6

Elementary calculus provides the derivative (18) in assertion a).

Proof of assertion b) of proposition 6

Consider assertion b) with the case $0<p_0 <p_1 <1$. Let $0<x<n$ and let the measure $\mu _x$ on the Borel field in $[p_0;p_1]$ be defined by

$$\begin{aligned} \mu _x(D) = {1\over B(a,b)(p_1-p_0)} \int \limits _D \left( {t-p_0\over p_1-p_0}\right) ^{a-1} \left[ 1- {t- p_0\over p_1-p_0}\right] ^{b-1} t^x (1-t)^{n-x} \, \text{ d }t. \end{aligned}$$

For $x\in (0;n)$ we obtain from Eq. (13) $\mu _x([p_0;p_1])= v_{p_0,p_1,a,b}(x)$. With Eq. (16) we obtain for $x\in (0;n)$

$$\begin{aligned} \displaystyle \frac{\text{ d }}{\text{ d } x} {v^{\prime }_{p_0,p_1,a,b}(x) \over v_{p_0,p_1,a,b}(x)} = \displaystyle {\int \limits _{[p_0;p_1]} \ln \left( {t\over 1-t}\right) ^2 \, \text{ d }\mu _x(t) \mu _x([p_0;p_1]) \;-\; \left[ \int \limits _{[p_0;p_1]} \ln \left( {t\over 1-t}\right) \, \text{ d }\mu _x(t)\right] ^2 \over v_{p_0,p_1,a,b}(x)^2}. \end{aligned}$$

The strict Cauchy–Schwarz inequality $\;[\int _B \vert g_1 g_2\vert \, \text{ d }\mu _x]^2 < \int _B g_1^2 \, \text{ d }\mu _x(t) \int _B g_2^2 \, \text{ d }\mu _x(t)\;$ holds for $g_1(t)= \ln (t/(1-t)), g_2(t)=1$, since these functions are not linearly dependent, see Hewitt and Stromberg (1969). Hence

$$\begin{aligned} \left[ \int \limits _{[p_0;p_1]} \ln \left( {t\over 1-t}\right) \, \text{ d } \mu _x(t)\right] ^2&\le \left[ \int \limits _{[p_0;p_1]} \vert g_1(t) g_2(t)\vert \, \text{ d }\mu _x(t)\right] ^2 \\&< \int \limits _{[p_0;p_1]} g_1(t)^2 \, \text{ d }\mu _x(t) \int \limits _{[p_0;p_1]} g_2(t)^2 \, \text{ d }\mu _x(t) \\&= \int \limits _{[p_0;p_1]} \ln \left( {t\over 1-t}\right) ^2 \, \text{ d }\mu _x(t) \; \mu _x([p_0;p_1]) \end{aligned}$$

Hence ${{\text{ d }}\over {\text{ d }} x} {v^{\prime }_{p_0,p_1,a,b}(x) \over v_{p_0,p_1,a,b}(x)} >0$ on $[0;n]$, and $[0;n]\ni x \mapsto v^{\prime }_{p_0,p_1,a,b}(x) / v_{p_0,p_1,a,b}(x)$ is strictly increasing. The remainder of assertion b) follows from the Eq. (18) for the derivative $Q_{p_0,p_1,a,b,y}^{\prime }$ on $[0;n]$.

Consider assertion b) with the case $0=p_0, 1=p_1$. From Eq. (17) we obtain

$$\begin{aligned} {v^{\prime }_{0,1,a,b}(x) \over v_{0,1,a,b}(x)} = \psi (x+a)-\psi (n-x+b) \quad \text{ for } x \in [0;n]. \end{aligned}$$

The proof of assertion b) of proposition 5 demonstrates that $[0;n] \ni x \mapsto \psi (x+a)-\psi (n-x+b)$ has at most one change of sign on $[0;n]$ which is from $-$ to $+$, if existing. Again, the remainder of assertion b) follows from the Eq. (18) for the derivative $Q_{p_0,p_1,a,b,p}^{\prime }$ on $[0;n]$.

Proof of assertion c) of proposition 6

For $p_0 <q \le 1, x\in [0;n]$, consider $h_{p_0,q,a,b}(x)$ as defined by Eq. (23). Proceeding analogously to the proof of of assertion b) of proposition 6, we find that ${\text{ d } \over \text{ d } x} {h^{\prime }_{p_0,q,a,b}(x) \over h_{p_0,q,a,b}(x)} >0$ on $[0;n]$, and $[0;n]\ni x \mapsto h^{\prime }_{p_0,q,a,b}(x) / h_{p_0,q,a,b}(x)$ is strictly increasing for $p_0 < q< 1$. In analogy to the definition of $x_{p_0,q,a,b,y}$ in assertion b), define $0\le z_{p_0,q,a,b,y}\le n$ by comparing $h^{\prime }_{p_0,p_1,a,b}/ h_{p_0,p_1,a,b}$ with $\ln (y /[ 1-y])$. From Eqs. (23) and (24) we find

$$\begin{aligned} h^{(m)}_{p_0,q,a,b}(x) \;< \; \ln \left( {q\over 1-q}\right) ^m h_{p_0,q,a,b}(x) \quad \text{ for } x\in [0;n],\; p_0 <q < 1. \end{aligned}$$

(28)

With the recursion (26) and with the inequality (28) we obtain for fixed $x\in [0;n]$

$$\begin{aligned} \frac{\text{ d }}{\text{ d } q} \frac{h_{p_0,q,a,b}^{\prime } (x)}{h_{p_0,q,a,b} (x)} = \frac{{\frac{\text{ d }}{\text{ d }q}} h_{p_0,q,a,b} (x)}{h_{p_0,q,a,b} (x)^2} \left\{ {\ln \left( \frac{q}{1-q}\right) h_{p_0,q,a,b} (x) - h_{p_0,q,a,b}^{\prime } (x)}\right\} > 0 \nonumber \\ \end{aligned}$$

(29)

for $p_0 <q < 1, q>0.5$. Let $p_0 < q_1<q_2<\cdots <1$ with $\lim _k q_k=1, q_k>0.5$. From (29) we see that the sequence $(z_{p_0, q_k,a,b,y})_{k\in \mathbb{N }}$ is decreasing. Since $0\le z_{p_0, q_k,a,b,y} \le n, (z_{p_0, q_k,a,b,y})_{k\in \mathbb{N }}$ converges to a value $x_{p_0, 1,a,b, y} \in [0;n]$. The remainder of the proof proceeds analogously to the proof of assertion c) of proposition 5.

The proof of assertion d) of proposition 6 is completely analogous to the proof of assertion c).

Proofs related to Section 6

1.1 Proof of proposition 7

Consider the assumptions of assertion a). By proposition 2, $A_x =\{y \in [p_0;p_1] \vert c_L(y) \le x\le c_U(y)\}$ is an interval for all $x\in \{0,\ldots ,n\}$ with endpoints $\inf A_x, \sup A_x$ increasing in $x$. We have to prove that the endpoints $\inf A_x, \sup A_x$ are elements of $A_x$ for $x\in \{0,\ldots ,n\}$. Let $x\in \{0,\ldots ,n\}$. If $A_x$ is one-elementary, then clearly $\inf A_x, \sup A_x \in A_x$. Let $A_x$ have an open interior, and let $(y_l)$ be a sequence from $A_x$ with $y_1 > y_2 >\ldots , \lim _l y_l= \inf A_x$. Then $c_L(y_l)\le x \le c_U(y_l)$ for all $l$. Hence by convergence $c_L ({\inf A_x}) \le c_L ({\inf A_x}^+) \le x \le c_U ({\inf A_x}^+)=c_U ({\inf A_x})$, hence $\inf A_x \in A_x$, and thus $\inf A_x =\min A_x$. The proof of $\sup A_x =\max A_x \in A_x$ is completely analogous.

Assertion b) is an application of proposition 2.

1.2 Proof of proposition 8

The proof makes use of the subsequent proposition 9 on the binomial OC $L_{n,c}(y)$ defined by Eq. (22), see Uhlmann (1982).

Proposition 9

(Binomial OC) Let $n\in \mathbb{N }, c \in \mathbb{N _{0}}, c\le n$.

For $y \!\in \! (0;1)$ we have $\; L^{\prime }_{n, c} (y) \!=\! - n \left( \begin{array}{l} n\!-\!1\\ c \end{array}\right) y^c (1-y)^{n -c-1} \!=\! -(n -c) \left( \begin{array}{l} n\\ c \end{array}\right) y^c (1-y)^{n-c-1}$. In case of $c < n, L_{n,c}$ is strictly decreasing on the interval $[0;1]$ with $L_{n,c}(0)=1, L_{n,c}(1)=0$. In case of $c= n$ we have $L_{n,c}(y) = L_{n, n}(y) = 1$ for $y\in [0;1]$.

Assertion a) of proposition 8 follows directly from the definition of the quantities $p_{x,\gamma }, \widetilde{p}_{x, \gamma }$, and the monotonicity properties of the binomial OC explained by proposition 9.

For the proof of assertion b), let $\gamma \ge 0.5$. Let $0<x<n$. Then

$$\begin{aligned} L_{n,x}(\widetilde{p}_{x, \gamma }) = 1-\gamma \;\le \; \gamma = L_{n,x-1}(p_{x, \gamma }) \;<\; L_{n,x}(p_{x, \gamma }), \end{aligned}$$

hence $\widetilde{p}_{x, \gamma }> p_{x, \gamma }$ since $L_{n,x}$ is strictly decreasing on $[0;1]$. In the case of $x=0$, we have by definition $p_{x, \gamma }=0.0 <\widetilde{p}_{x, \gamma }$, and similarly in the case of $x=n$ by definition $p_{x, \gamma }<1.0=\widetilde{p}_{x, \gamma }$.

For the proof of assertion c), let $x\in \{0,\ldots ,n\}, y\in (p_{x, \gamma }; \widetilde{p}_{x, \gamma }) \cap [p_0;p_1]$. Assume $x\le c_L(y)-1$. Then we have $x\le n-1$ and we find

$$\begin{aligned} L_{n, c_{U}(y)}(y)- L_{n, c_{L }(y)-1}(y)&\le 1- L_{n, c_{1 }(y)-1}(y) \le 1- L_{n, x}(y)\\&< 1- L_{n, x}(\widetilde{p}_{x, \gamma }) =\gamma \end{aligned}$$

in contradiction to the property (21). Now assume $x\ge c_U(y)+1$. Then we have $x\ge 1$ and we find

$$\begin{aligned} L_{n, c_{U}(y)}(y)- L_{n, c_{L }(y)-1}(y)&\le L_{n, c_{U}(y)}(y) \le L_{n, x-1}(y) < L_{n, x-1}(p_{x, \gamma }) =\gamma \end{aligned}$$

in contradiction to the property (21). This proves $c_L(y) \le x\le c_U(y)$, and hence also $y_L(x)\le y \le y_U(x)$.