Objective Bayesian analysis for a capture–recapture model

Abstract

In this paper, we study a special capture–recapture model, the $M_t$ model, using objective Bayesian methods. The challenge is to find a justified objective prior for an unknown population size $N$. We develop an asymptotic objective prior for the discrete parameter $N$ and the Jeffreys’ prior for the capture probabilities $\varvec{\theta }$. Simulation studies are conducted and the results show that the reference prior has advantages over ad-hoc non-informative priors. In the end, two real data examples are presented.

Appendices

Appendix A: Proof of Theorem 1

The likelihood function (2) can be written as the product of the probability mass functions of iid random vectors ${{\varvec{e}}}_j$’s, $j=1,\ldots ,N$, where ${{\varvec{e}}}_j=(e_{j0}, e_{j1}, \cdots , e_{jL})$ and

$$\begin{aligned} {{\varvec{e}}}_j\mid \varvec{\theta }\sim \text{ Multi }(1;p_0(\varvec{\theta }),\ldots ,p_L(\varvec{\theta })). \end{aligned}$$

(36)

The logarithm of the density of ${{\varvec{e}}}_j$ is

$$\begin{aligned} \log p({{\varvec{e}}}_j\mid \varvec{\theta })=\sum _{i=0}^L e_{ji}\log p_i(\varvec{\theta }). \end{aligned}$$

(37)

For convenience, define

$$\begin{aligned} C_j\doteq \Big \{s\in \{0,1,\ldots ,L=2^k\}: \text{ the } j\mathrm{th} \;\text{ binary } \text{ digit } \text{(from } \text{ left } \text{ to } \text{ right) } \text{ of }\;s\;\text{ is } \text{1 }\Big \} \end{aligned}$$

For any $r=1,\ldots ,k$, we have

$$\begin{aligned}&\frac{\partial ^2}{\partial \theta _r^2}\sum _{i=0}^L e_{ji}\log p_i(\varvec{\theta }) \nonumber \\&\quad =\sum _{i=0}^L e_{ij} \sum _{j=1}^k \frac{\partial ^2}{\partial \theta _r^2} \left( b_{ij}\log \theta _j + (1-b_{ij})\log (1-\theta _j)\right) \nonumber \\&\quad = -\sum _{i=0}^L e_{ij} \left( \frac{b_{ir}}{\theta _r^2} + \frac{1-b_{ir}}{(1-\theta _r)^2}\right) . \end{aligned}$$

(38)

Since $\text{ E }(e_{ji})=p_i(\varvec{\theta })$, we have

$$\begin{aligned} \text{ E }\left( \frac{\partial ^2}{\partial \theta _r^2}\sum _{i=0}^Le_{ji}\log p_i(\varvec{\theta }) \right) = -\sum _{i=0}^L \left( \frac{p_i(\varvec{\theta })b_{ir}}{\theta _r^2}+\frac{p_i(\varvec{\theta })(1-b_{ir})}{(1-\theta _r)^2}\right) . \end{aligned}$$

(39)

Let $Q_{ir}=\frac{p_i(\varvec{\theta })b_{ir}}{\theta _r^2}+\frac{p_i(\varvec{\theta })(1-b_{ir})}{(1-\theta _r)^2}$. If $b_{ir}=0$,

$$\begin{aligned} Q_{ir}=\frac{p_i(\varvec{\theta })}{(1-\theta _r)^2}=(1-\theta _r)^{-1} \prod _{j\ne r}\theta _j^{b_{ij}}(1-\theta _j)^{1-b_{ij}}. \end{aligned}$$

(40)

If $b_{ir}=1$,

$$\begin{aligned} Q_{ir}=\frac{p_i(\varvec{\theta })}{\theta _r^2}=\theta _r^{-1} \prod _{j\ne r}\theta _j^{b_{ij}}(1-\theta _j)^{1-b_{ij}}. \end{aligned}$$

(41)

Notice that $\prod _{j\ne r}\theta _j^{b_{ij}}(1-\theta _j)^{1-b_{ij}}$ does not depend on $b_{ir}$ and equals $1$. Therefore, we have

$$\begin{aligned} \text{ E }\left( \frac{\partial ^2}{\partial \theta _r^2}\sum _{i=0}^Le_{ji}\log p_i(\varvec{\theta }) \right) =\frac{1}{\theta _r(1-\theta _r)}. \end{aligned}$$

(42)

Also, we have

$$\begin{aligned} \frac{\partial ^2}{\partial \theta _s\theta _t}\sum _{i=0}^Le_{ji}\log p_i(\varvec{\theta })=0. \end{aligned}$$

(43)

Therefore, the Fisher information matrix for $\varvec{\theta }$ is Diag $\big \{\theta _1^{-1}(1-\theta _1)^{-1},\ldots , \theta _k^{-1} (1-\theta _k)^{-1}\big \}$, which yields the result. $\square$

Appendix B: Proof of Lemma 1

1.1 B.1 Proof of (15)

We start from studying $J_1(N)$ to prove (15). Since $\frac{1}{(N\;+\;a\;+\;b\;+\;i)^2}$ is decreasing in $i$, for $i=0, 1, \cdots$,

$$\begin{aligned} \frac{k}{(N+a+b+i+1)^2} < \int \nolimits _i^{i+1}\frac{k}{(N+a+b+x)^2}dx < \frac{k}{(N+a+b+i)^2}. \end{aligned}$$

Adding up the inequality from $i=0$ to infinity, we get

$$\begin{aligned} 0\le \sum _{i=0}^\infty \frac{k}{(N+a+b+i)^2}-\int \nolimits _0^\infty \frac{k}{(N+a+b+x)^2}dx \le \frac{k}{(N+a+b)^2}.\qquad \end{aligned}$$

(44)

The integral in (44) is $\frac{k}{N+a+b}$. Therefore,

$$\begin{aligned} \frac{k}{N+a+b} \le \sum _{i=0}^\infty \frac{k}{(N+a+b+i)^2} \le \frac{k}{N+a+b}+\frac{k}{(N+a+b)^2}. \end{aligned}$$

(45)

Apply (45) for $k=1$ and $a+b=1$, we have

$$\begin{aligned} \frac{1}{N+1} \le \sum _{i=0}^\infty \frac{1}{(N+1+i)^2} \le \frac{1}{N+1}+\frac{k}{(N+1)^2}. \end{aligned}$$

(46)

Combing (45) and (46), we can bound $J_1(N)$ by

$$\begin{aligned} \frac{k}{N+a+b}-\frac{1}{N+1}-\frac{1}{(N+1)^2}\le J_1(N)\nonumber \\ \le \frac{k}{N+a+b} - \frac{1}{N+1}+ \frac{k}{(N+a+b)^2}. \end{aligned}$$

(47)

$J_1(N)$ does not depend on the data, so $\text{ E }(J_1(N))=J_1(N)$. So (15) follows from (47) immediately.

1.2 B.2 Proofs of (17) and (19)

The asymptotic order of $\text{ E }(J_2(N))$ depends on the value of $b$ and is summarized in (17) and (19). To prove these equations, we need the following result from Chao and Strawderman (1972). For $X\sim Bin(N,p)$,

$$\begin{aligned} \text{ E }\left( \frac{1}{X+1}\right) = \frac{1-(1-p)^{N+1}}{(N+1)p}. \end{aligned}$$

For all $1\le l\le k$, since $m_l\mid N,\theta _l \sim \text{ Bin }(N,1-\theta _l)$, we can apply the above result and get

$$\begin{aligned} \frac{1}{k}\text{ E }(J_2(N))&= \text{ E }\left[ \text{ E }\left( \frac{1}{N-m_1+1}\mid \theta _1\right) \right] \nonumber \\&= \int \nolimits _0^1 \frac{1-\theta _1^{N+1}}{(N+1)(1-\theta _1)}\frac{\varGamma (a+b)}{\varGamma (a)\varGamma (b)}\theta _1^{a-1}(1-\theta _1)^{b-1}d\theta _1 \nonumber \\&= \frac{\varGamma (a+b)}{\varGamma (a)\Gamma (b)(N+1)}\sum _{j=0}^N\int \nolimits _0^1 \theta _1^{j+a-1}(1-\theta _1)^{b-1}d\theta _1 \nonumber \\&= \frac{\varGamma (a+b)}{\varGamma (a)(N+1)}\sum _{j=0}^N\frac{\varGamma (j+a)}{\varGamma (j+a+b)}. \end{aligned}$$

(48)

Therefore, we have the following decomposition which will be used for several times throughout the proof of Lemma 1.

$$\begin{aligned} \frac{1}{k}\text{ E }\left( J_{2}(N)\right) = J_{21}(N)+J_{22}(N)+J_{23}(N), \end{aligned}$$

(49)

where

$$\begin{aligned} J_{21}(N)&= \frac{\varGamma (a+b)}{\varGamma (a)(N+1)}\int \nolimits _0^N \frac{1}{(x+a)^b}dx, \end{aligned}$$

(50)

$$\begin{aligned} J_{22}(N)&= \frac{\varGamma (a+b)}{\varGamma (a)(N+1)}\left( \sum _{j=0}^N \frac{1}{(j+a)^b}-\int \nolimits _0^N \frac{1}{(x+a)^b}dx\right) , \end{aligned}$$

(51)

$$\begin{aligned} J_{23}(N)&= \frac{\varGamma (a+b)}{\varGamma (a)(N+1)}\sum _{j=0}^N\left( \frac{\varGamma (j+a)}{\varGamma (j+a+b)}-\frac{1}{(j+a)^b}\right) . \end{aligned}$$

(52)

Next, we will study the limiting behaviors of these three terms. The order of $J_{21}(N)$ can be found out straightforwardly. When $0< b <1$, $J_{21}(N)$ can be written as

$$\begin{aligned} J_{21}(N)=\frac{\varGamma (a+b)}{(1-b)\varGamma (a)(N+1)}\left( (N+a)^{1-b} - a^{1-b}\right) \sim \frac{1}{N^b}. \end{aligned}$$

(53)

When $b=1$,

$$\begin{aligned} J_{21}(N) = \frac{\varGamma (a+b)}{(1-b)\varGamma (a)(N+1)}\log \left( \frac{N+a}{a}\right) \sim \frac{\log (N+a)}{N+1}. \end{aligned}$$

(54)

For the second term $J_{22}(N)$, we show that it is asymptotically smaller than $N^{-1}$. For any $0<a,b\le 1$, due to the monotonicity of $(x+a)^{-b}$, we can bound the integral as

$$\begin{aligned} \sum _{j=1}^N\frac{1}{(j+a)^b} < \int \nolimits _0^N \frac{1}{(x+a)^b}dx < \sum _{j=0}^{N-1}\frac{1}{(j+a)^b}, \end{aligned}$$

(55)

which is equivalent to

$$\begin{aligned} \frac{1}{(N+a)^b}<\sum _{j=0}^N\frac{1}{(j+a)^b}-\int \nolimits _0^N \frac{1}{(x+a)^b}dx < \frac{1}{a^b}. \end{aligned}$$

(56)

Therefore, $J_{22}(N)$ can be bounded as

$$\begin{aligned} \frac{\varGamma (a+b)}{\varGamma (a)(N+1)(N+a)^b}<J_{22}(N)<\frac{\varGamma (a+b)}{ \varGamma (a)a^b(N+1)}, \end{aligned}$$

(57)

where the upper bound is in the order of $1/N$.

Lastly, we prove that third term $J_{23}(N)$ approaches $1/N$ as $N\rightarrow \infty$ for any $0<a,b\le 1$. It follows from Erdelyi and Tricomi (1951) that for fixed $\alpha , \beta \ge 0$, as $z\rightarrow \infty$,

$$\begin{aligned} \frac{\varGamma (z+\alpha )}{\varGamma (z+\beta )}=z^{\alpha -\beta }\left( 1+\frac{(\alpha -\beta )(\alpha +\beta -1)}{2z}+O\left( \frac{1}{z^2}\right) \right) . \end{aligned}$$

(58)

Applying this expansion with $z=j+a$, $\alpha =0$, and $\beta =b$,

$$\begin{aligned} \frac{\varGamma (j+a)}{\varGamma (j+a+b)}=\frac{1}{(j+a)^b}\left( 1+ \frac{b(1-b)}{2(j+a)}+J_{23}^*(j)\right) , \end{aligned}$$

(59)

where $J_{23}^*(j)\sim O\left( (j+a)^{-2}\right)$, as $j\rightarrow \infty$. Thus, we can rewrite $J_{23}(N)$ as

$$\begin{aligned} J_{23}(N)=\frac{\varGamma (a+b)}{\varGamma (a)(N+1)}\left( \frac{b(1-b)}{2} \sum _{j=0}^N \frac{1}{(j+a)^{1+b}} +\sum _{j=0}^N\frac{J_{23}^*(j)}{(j+a)^b}\right) . \end{aligned}$$

The facts that

$$\begin{aligned} \sum _{j=0}^\infty \frac{1}{(j+a)^{1+b}}<\infty \quad \text{ and } \quad \sum _{j=0}^\infty \frac{J_{23}^*(j)}{(j+a)^b}<\infty \end{aligned}$$

ensure that as $N\rightarrow \infty$,

$$\begin{aligned} J_{23}(N) \sim \frac{1}{N}. \end{aligned}$$

(60)

Applying (53), (54), (57), and (60) to (49) will prove (17) and (19). $\square$

1.3 B.3 Proofs of (18) and (20)

Note that $J_3(N)$ can be bounded from both sides. On one hand,

$$\begin{aligned} J_{3}(N)&\le \sum _{l=1}^k\left[ \frac{1}{(N-m_l+b)^2}+\int \nolimits _0^\infty \frac{1}{(N-m_l+b+x)^2}dx\right] -\sum _{l=1}^k\frac{1}{N-m_l+1}\nonumber \\&= \sum _{l=1}^k\frac{1}{(N-m_l+b)^2}+\sum _{l=1}^k\left( \frac{1}{N-m_l+b}-\frac{1}{N-m_l+1}\right) \nonumber \\&= \sum _{l=1}^k\frac{1}{(N-m_l+b)^2}+\sum _{l=1}^k\frac{1-b}{(N-m_l+b)(N-m_l+1)} \nonumber \\&\le \sum _{l=1}^k\frac{1}{(N-m_l+b)^2}+\sum _{l=1}^k\frac{1-b}{(N-m_l+b)^2}\nonumber \\&= \sum _{l=1}^k\frac{2-b}{(N-m_l+b)^2}. \end{aligned}$$

(61)

On the other hand,

$$\begin{aligned} J_{3}(N)&\ge \sum _{l=1}^k\left( \int \nolimits _0^\infty \frac{1}{(N-m_l+b+x)^2}dx-\frac{1}{N-m_l+1}\right) \nonumber \\&= \sum _{l=1}^k\frac{1-b}{(N-m_l+b)(N-m_l+1)} \ge \sum _{l=1}^k\frac{b(1-b)}{(N-m_l+b)^2}. \end{aligned}$$

(62)

Let

$$\begin{aligned} J^*_{3}(N)=\text{ E }\left( \frac{1}{(N-m_1+b)^2}\right) . \end{aligned}$$

Inequalities (61) and (62) imply that $\text{ E }(J_3(N))/J^*_3(N)$ can be controlled by positive constants.

$$\begin{aligned} b(1-b)J^*_{3}(N)\le \text{ E }(J_{3}(N))\le (2-b)J^*_{3}(N). \end{aligned}$$

(63)

It suffices to find out the asymptotic order of $J^*_{3}(N)$. In fact,

$$\begin{aligned} J^*_{3}(N)&= \frac{\varGamma (a+b)}{\varGamma (a)\Gamma (b)}\int \nolimits _0^1 \frac{1}{(N-m_l+b)^2}\sum _{m_l=0}^N {N\atopwithdelims ()m_l} \theta _l^{m_l+a-1}(1-\theta _l)^{N-m_l+b-1} d\theta _l \nonumber \\&= \frac{\varGamma (a+b)}{\varGamma (a)\Gamma (b)}\sum _{i=0}^N\frac{\varGamma (N+1)\Gamma (i+a)\varGamma (N-i+b)}{(N-i+b)^2\varGamma (i+1) \Gamma (N-i+1)\varGamma (N+a+b)},\qquad \quad \end{aligned}$$

(64)

where $m_l$ is replaced by $i$ for notation simplicity. Since $J^*_{3}(N)$ is a finite summation of positive components, it can be bounded from below by its last term,

$$\begin{aligned} J^*_{3}(N)\ge \frac{\varGamma (a+b)\Gamma (N+a)}{b^2\varGamma (a)\Gamma (N+a+b)} \approx \frac{\varGamma (a+b)}{b^2\varGamma (a)(N+a)^b}. \end{aligned}$$

(65)

Now, we bound $J^*_{3}(N)$ from above. Applying (17) and (19) , we have for any $0<b\le 1$, as $N\rightarrow \infty$,

$$\begin{aligned} J^*_{3}(N)\le \frac{1}{b^2}\text{ E }\left( \frac{1}{(N-m_1+1)}\right) \sim \frac{\log ^{[b]}N}{(N+1)^b}. \end{aligned}$$

(66)

It follows from (65) and (66) that, as $N\rightarrow \infty$, when $0<b<1$,

$$\begin{aligned} J^*_{3}(N) \sim N^{-b}. \end{aligned}$$

(67)

When $b=1$,

$$\begin{aligned} \frac{a}{N+a}\le J^*_{3}(N)\le \frac{C_1\log (N+a)}{N+1}. \end{aligned}$$

(68)

Combining (63), (67), and (68), we completed the proof of (18) and (20). $\square$

1.4 B.4 Proof of (16)–Basis: $k=2, ~0<b<1$

$\text{ E }(J_4(N))$ is the leading term. Finding out its asymptotic order is the most challenging part of this proof. We will use induction method to study $\text{ E }(J_4(N))$ and prove (16). In subsections B.4 and B.5, we will focus on the base case (basis) when $k=2$. Specifically, in B.4, we will show that (16) holds when $k=2, ~0<b<1$. In B.5, we will prove (16) when $k=2, ~b=1$ with a different approach. Finally, we will prove (16) for any arbitrary $k\ge 3$ (the inductive step) in B.6.

When $k=2$, $n$ follows a Binomial distribution:

$$\begin{aligned} n\mid N,\theta _1,\theta _2 \sim \text{ Bin }\left( N,1-(1-\theta _1)(1-\theta _2)\right) . \end{aligned}$$

Let $\delta _i=1-\theta _i$, $i=1,2$. For $0<b<1$, the expectation of $J_4(N)$ when $k=2$, denoted by $\text{ E }_2(J_4(N))$, can be expressed as

$$\begin{aligned}&\text{ E }_2(J_4(N))=\sum _{n=0}^N\left( \sum _{i=0}^\infty \frac{1}{(N-n+1+i)^2}\right) \frac{\varGamma (N+1)}{\varGamma (n+1)\Gamma (N-n+1)} \nonumber \\&\qquad \int \nolimits _0^1\int \nolimits _0^1(1{-}\delta _1\delta _2)^n(\delta _1 \delta _2)^{N{-}n}B^{-2}(a,b) (\delta _1\delta _2)^{b-1} (1-\delta _1)^{a-1}(1-\delta _2)^{a-1} d\delta _1d\delta _2 \nonumber \\&\quad =\sum _{n=0}^N\left( \sum _{i=0}^\infty \frac{1}{(N-n+1+i)^2}\right) \frac{B^{-2}(a,b)\varGamma (N+1)}{\varGamma (n+1)\Gamma (N-n+1)}\nonumber \\&\qquad \int \nolimits _0^1\int \nolimits _0^1(1-\delta _2+\delta _2(1-\delta _1))^n (\delta _2 \delta _2)^{N-n+b-1} (1-\delta _1)^{a-1}(1-\delta _2)^{a-1} d\delta _1d\delta _2 \nonumber \\&\quad =\sum _{n=0}^N\left( \sum _{i=0}^\infty \frac{1}{(N-n+1+i)^2}\right) \!\frac{B^{-2}(a,b)\varGamma (N+1)}{\varGamma (n+1)\Gamma (N-n+1)}\!\sum _{j=0}^n\!\frac{\varGamma (n+1)}{\varGamma (j+1) \Gamma (n-j+1)}\nonumber \\&\qquad \int \nolimits _0^1\int \nolimits _0^1\delta _1^{N-n+b-1}(1-\delta _1)^{n-j+a-1}\delta _2^{N-j+b-1} (1-\delta _2)^{j+a-1} d\delta _1d\delta _2 \nonumber \\&\quad = J_{41}+J_{42}+J_{43}, \end{aligned}$$

(69)

where for $l=1,2,3$,

$$\begin{aligned} J_{4l}(N)=\frac{B^{-2}(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\sum _{n=0}^N \sum _{i=0}^\infty \frac{1}{(N-n+1+i)^2} \frac{\varGamma (N-n+b)}{\varGamma (N-n+1)}J_{4l}^*(n),\nonumber \\ \end{aligned}$$

(70)

and

$$\begin{aligned} J^*_{41}(n)&= \int \nolimits _0^n\frac{1}{(x+1)^{1-a}(n-x+1)^{1-a}(N-x+1)^a}dx,\end{aligned}$$

(71)

$$\begin{aligned} J^*_{42}(n)&= \sum _{j=0}^n \frac{1}{(j+1)^{1-a}(n-j+1)^{1-a}(N-j+1)^a} - J^*_{41}(n), \end{aligned}$$

(72)

$$\begin{aligned} J^*_{43}(n)&= \sum _{j=0}^n \frac{\varGamma (j+a)\Gamma (n-j+a)\varGamma (N-j+b)}{\varGamma (j+1) \Gamma (n-j+1)\varGamma (N-j+a+b)}- J^*_{41}(n) -J^*_{42}(n).\qquad \end{aligned}$$

(73)

Notice that the forms of $J^*_{42}(n), J^*_{43}(n)$ are similar with those of $J_{22}(N), J_{23}(N)$. Therefore, the asymptotic bounds of $J^*_{42}(n), J^*_{43}(n)$ can be obtained by adopting the similar approaches in the proof of (51) and (52). We omit the details and present the results below:

$$\begin{aligned} J_{4l}(N) \le \frac{1}{(N+1)^b} , \quad l=2,3. \end{aligned}$$

(74)

The term $J_{41}(N)$ can be decomposed into two partial sums,

$$\begin{aligned} J_{41}(N)=J_{411}(N)+J_{412}(N), \end{aligned}$$

(75)

where

$$\begin{aligned}&J_{411}(N)=\dfrac{B^{-2}(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\displaystyle \sum \limits _{n=0}^{N-\log N-1} \sum \limits _{i=0}^\infty \frac{1}{(N-n+1+i)^2} \frac{\varGamma (N-n+b)}{\varGamma (N-n+1)}J_{41}^*(n),\end{aligned}$$

(76)

$$\begin{aligned}&J_{412}(N)=\dfrac{B^{-2}(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\displaystyle \sum \limits _{n=N-\log N}^N \sum \limits _{i=0}^\infty \frac{1}{(N-n+1+i)^2} \frac{\varGamma (N-n+b)}{\varGamma (N-n+1)}J_{41}^*(n). \end{aligned}$$

(77)

Interestingly, although $J_{412}(N)$ has much less terms than $J_{411}(N)$, it is the leading term. In fact, for $k=2$ and $0<b<1$, there exists positive constants $C_2$, $C_3$ such that as $N\rightarrow \infty$,

$$\begin{aligned} J_{411}(N)&\le \frac{C_2\log ^b N}{(N+1)^b}, \end{aligned}$$

(78)

$$\begin{aligned} J_{412}(N)&\sim \frac{C_3B^{-2}(a,b)\log N}{(N+1)^b}. \end{aligned}$$

(79)

For $J_{411}(N)$, let $y=x/n$ and we can bound $J_{41}^*(n)$ by

$$\begin{aligned} J_{41}^*(n)&=\frac{1}{n^{1-a}}\int \nolimits _0^1\frac{1}{(y+\frac{1}{n})^{1-a}(1-y+\frac{1}{n})^{1-a} (1-y+\frac{1}{n}+\frac{N-n}{n})^a}dy \nonumber \\&\le \frac{1}{n^{1-a}}\int \nolimits _0^1\frac{1}{(y+\frac{1}{n})^{1-a}(1-y+\frac{1}{n})}dy. \end{aligned}$$

(80)

The integral can be bounded as following. Notice that

$$\begin{aligned} 0\!&\le \!\int \nolimits _0^{1/2}\frac{1}{(y+1/n)^{1-a}(1-y+1/n)}dy \nonumber \\&\le \frac{1}{1/2+1/n}\int \nolimits _0^{1/2}\frac{1}{(y+1/n)^{1-a}}dy \le \frac{2}{a}\left( \frac{1}{2}+\frac{1}{n}\right) ^a \rightarrow \frac{2^{1-a}}{a}. \end{aligned}$$

(81)

Similarly, we have

$$\begin{aligned} \log n/2 \le \int \nolimits _{1/2}^1\frac{1}{(y+1/n)^{1-a}(1-y+1/n)}dy \le 2^{1-a}\log (1+n/2). \end{aligned}$$

(82)

It follows from (80), (81), (82) that,

$$\begin{aligned} J_{41}^*(n) \le \frac{2^{1-a}\log (n+1)}{(n+1)^{1-a}}. \end{aligned}$$

(83)

and consequently, applying the same decomposition and arguments as in (50), (51), (52), we have

$$\begin{aligned} J_{411}(N)&\le \frac{2B^{-2}(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\sum _{n=0}^{N-\log N} \frac{\varGamma (N-n+b)}{\varGamma (N-n+2)}\frac{2^{1-a}\log (n+1)}{(n+1)^{1-a}}\nonumber \\&\le \frac{2^{4-2a}B^{-2}(a,b)\log ^b N}{(1-b)N^b}. \end{aligned}$$

(84)

Let $C_2=\frac{2^{4-2a}B^{-2}(a,b)}{1-b}$ and we proved (78).

For $J_{412}(N)$, note that $J^*_{41}(n)\le \frac{\log (n+1)}{(n+1)^{1-a}}$, we get

$$\begin{aligned} J_{412}(N)&\le \frac{B^{-2}(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\sum _{n=N-\log N}^N \left( \sum _{i=0}^\infty \frac{1}{(N-n+1+i)^2}\right) \nonumber \\&\times \frac{\log (n+1)\varGamma (N-n+b)}{(n+a)^{1-a}\varGamma (N-n+1)}\nonumber \\&\le \frac{B^{-2}(a,b)\log (N+1)\varGamma (N+1)}{(N-\log N+1)^{1-a}\varGamma (N+a+b)}H(N), \end{aligned}$$

(85)

where

$$\begin{aligned} H(N)=\sum _{n=N-\log N}^N \left( \sum _{i=0}^\infty \frac{1}{(N-n+1+i)^2}\right) \frac{\varGamma (N-n+b)}{\varGamma (N-n+1)} . \end{aligned}$$

(86)

On the other hand, notice that

$$\begin{aligned} J^*_{41}(N)&\ge \frac{1}{n^{1-a}}\int \nolimits _0^1 \frac{1}{\left( y+\frac{1}{n}+\frac{N-n}{n} \right) ^{1-a}\left( 1-y+\frac{1}{n}+\frac{N-n}{n}\right) }dy \\&\ge \frac{1}{n^{1-a}\left( 1+\frac{1}{n}+\frac{N-n}{n}\right) ^{1-a}} \int \nolimits _0^1 \frac{1}{ \left( 1-y+\frac{1}{n}+\frac{N-n}{n}\right) }dy \\&= \frac{1}{(N+1)^{1-a}} \left( \log n-\log (N-n+1)\right) . \end{aligned}$$

Since $n<N-\log N$ and $N-n > \log N$, we have

$$\begin{aligned} J^*_{41}(N) \ge \frac{1}{(N+1)^{1-a}} \left( \log (N-\log N)-\log (1+\log N)\right) . \end{aligned}$$

(87)

Consequently, we obtain a lower bound of $J_{412}(N)$ by

$$\begin{aligned} J_{412}(N) \ge B^{-2}(a,b)\frac{[\log (N-\log N)-\log (1+\log N)] \varGamma (N+1)}{(N+1)^{1-a}\varGamma (N+a+b)}H(N).\quad \end{aligned}$$

(88)

Let $m=N-n$ and we have

$$\begin{aligned} H(N)&= \sum _{m=0}^{\log N}\left( \sum _{i=0}^\infty \frac{1}{(m+1+i)^2}\right) \frac{\varGamma (m+b)}{\varGamma (m+1)}\nonumber \\&\le \sum _{m=0}^{\log N}\frac{2\varGamma (m+b)}{(m+1)\varGamma (m+1)} \le \sum _{m=0}^{\infty }\frac{2\varGamma (m+b)}{\varGamma (m+2)} \sim \sum _{m=0}^\infty \frac{1}{m^{2-b}} < \infty .\quad \end{aligned}$$

(89)

Since $H(N)$ is an increasing function of $N$, the limit of $H(N)$ as $N\rightarrow \infty$ exists and we assume $\lim _{N\rightarrow \infty } H(N) = C_4$, where $C_4$ is a positive constant. Combining (85) and (88), we have

$$\begin{aligned}&\frac{C_4B^{-2}(a,b)[\log (N-\log N)-\log (1+\log N)] \varGamma (N+1)}{(N+1)^{1-a}\varGamma (N+a+b)}\nonumber \\&\quad \le J_{412}(N) ~\le ~ \frac{C_4B^{-2}(a,b)\log (N+1)\varGamma (N+1)}{(N-\log N+1)^{1-a}\varGamma (N+a+b)}, \end{aligned}$$

(90)

which yields (79) immediately.

It follows from (75), (78), (79) that as $N\rightarrow \infty$,

$$\begin{aligned} J_{41}(N) \sim \frac{C_3B^{-2}(a,b)\log N}{(N+1)^b}. \end{aligned}$$

(91)

Combining with (69) and (74), it is clear that $J_{41}(N)$ is the leading term of $\text{ E }_2(J_4(N))$, hence (16) holds for $k=2, ~0<b<1$. $\square$

1.5 B.5 Proof of (16)–Basis: $k=2, b=1$

To show (16) for $k=2,~b=1$, a different approach is required. Because $J_4(N)$ can be bounded by

$$\begin{aligned} \frac{1}{N-n+1}\le J_4(N)\le \frac{1}{N-n+1} +\frac{1}{(N-n+1)^2}, \end{aligned}$$

(92)

it suffices to obtain the asymptotic orders of $\text{ E }\frac{1}{N-n+1}$ and $\text{ E }\frac{1}{(N-n+1)^2}$. Applying Chao and Strawderman (1972)’s result, the former expectation can be decomposed as

$$\begin{aligned} \text{ E }_2\Big (\frac{1}{N-n+1}\Big )&= \text{ E }\Big (\text{ E }\Big (\frac{1}{N-n+1}\mid \varvec{\theta }\Big ) \Big )\nonumber \\&= B^{-2}(a,1)\int \nolimits _0^1\int \nolimits _0^1 \frac{1-\left( 1-\prod _{i=1}^2(1-\theta _i)\right) ^{N+1}}{(N+1)\prod _{i=1}^2(1-\theta _i)}\prod _{i=1}^2\theta _i^{a-1} d\theta _1 d\theta _2 \nonumber \\&= B^{-2}(a,1)\int \nolimits _0^1 \int \nolimits _0^1 \frac{1-\left( 1-\prod _{i=1}^2 \delta _i\right) ^{N+1}}{(N+1)\prod _{i=1}^2\delta _i}\prod _{i=1}^2 (1-\delta _i)^{a-1}d\delta _1 d\delta _2 \nonumber \\&= \frac{B^{-2}(a,1)}{N\!+\!1}\!\sum _{j=0}^N \int \nolimits _0^1 \int \nolimits _0^1 \!\left( 1\!-\!\delta _{2}\!+\!\delta _{2}(1\!-\!\delta _1)\right) ^j \!\prod _{i=1}^{2} (1-\delta _i)^{a-1}d\delta _1 d\delta _2 \nonumber \\&= \frac{B^{-2}(a,1)}{N+1}\sum _{j=0}^N \frac{\varGamma (j+1)}{\varGamma (j+a+1)}\sum _{i=0}^j \frac{\varGamma (i+a)\Gamma (j-i+a)}{\varGamma (i+1)\Gamma (j-i+a+1)} \nonumber \\&= J_{44}(N)+J_{45}(N)+J_{46}(N), \end{aligned}$$

(93)

where for $l=4,5,6$,

$$\begin{aligned} J_{4l}(N)=\frac{B^{-2}(a,1)}{N+1}\sum _{j=0}^N \frac{\varGamma (j+1)}{\varGamma (j+a+1)}J_{4l}^*(j), \end{aligned}$$

(94)

and

$$\begin{aligned} J^*_{44}(j)&= \int \nolimits _0^j\frac{1}{(x+a)^{1-a}(j-x+a)}dx, \end{aligned}$$

(95)

$$\begin{aligned} J^*_{45}(j)&= \sum _{i=0}^j\frac{1}{(i+a)^ {1-a}(j-i+a)} -\int \nolimits _0^j\frac{1}{(x+a)^{1-a}(j-x+a)}dx, \end{aligned}$$

(96)

$$\begin{aligned} J^*_{46}(j)&= \sum _{i=0}^j\frac{ \varGamma (i+a)\Gamma (j-i+a)}{\varGamma (i+1) \Gamma (j\!-\!i\!+\!a\!+\!1)} - \sum _{i=0}^j\frac{1}{(i+a)^{1-a}(j\!-\!i\!+\!a)}. \end{aligned}$$

(97)

The asymptotic orders or bounds of $J_{4l}(N)$, $l=4,5,6$, and $\text{ E }\left( \frac{1}{(N-n+1)^2}\right)$ are summarized below. For fixed $0<a\le 1$ and $b=1$, sending $N$ to infinity, we have

$$\begin{aligned} \dfrac{B^{-2}(a,1)\log ^2 N}{2(N+1)}&\le J_{44}(N) \le \dfrac{B^{-2}(a,1)\log ^2 N}{2(N+1)}+ \dfrac{B^{-2}(a,1)(\log N)^{3/2}}{a(N+1)},\qquad \end{aligned}$$

(98)

$$\begin{aligned} J_{4l}(N)&\le \dfrac{B^{-2}(a,1)\log N}{N+1}, \quad l=5,6, \end{aligned}$$

(99)

$$\begin{aligned} \text{ E }\left( \frac{1}{(N-n+1)^2}\right)&\le \dfrac{2B^{-2}(a,1)\log (N+1)}{N} . \end{aligned}$$

(100)

Due to the similarity of the proofs in nature, we will only prove (98), which gives the asymptotic order of the leading term. In fact, we let $y=x/j$ and write $J^*_{44}(j)$ as

$$\begin{aligned} J^*_{44}(j)=\frac{1}{j^{1-a}}\int \nolimits _0^1\frac{1}{(y+a/j)^{1-a}(1-y+a/j)}dy. \end{aligned}$$

(101)

This integral can be shown to have an order of $\log j$ for $j$ large enough. Let $\epsilon =(\log j)^{-1/2}$. First, we integrate from $0$ to $1-\epsilon$.

$$\begin{aligned}&\int \nolimits _0^{1-\epsilon }\frac{1}{(y+a/j)^{1-a}(1-y+a/j)}dy \le \frac{1}{\epsilon +a/j}\int \nolimits _0^{1-\epsilon } \frac{1}{(y+a/j)^{1-a}}dy \nonumber \\&\qquad =\frac{1}{a(\epsilon +a/j)}\left[ (1-\epsilon +a/j)^a-(a/j)^a\right] \le \frac{(1-\epsilon +a/j)^a}{a(\epsilon +a/j)}. \end{aligned}$$

(102)

Second, we integrate from $1-\epsilon$ to $1$.

$$\begin{aligned}&\int \nolimits _{1-\epsilon }^1\frac{1}{(y+a/j)^{1-a}(1-y+a/j)}dy \nonumber \\&\quad \le \frac{1}{(1-\epsilon +a/j)^{1-a}}\int \nolimits _{1-\epsilon }^1 \frac{1}{1-y+a/j}dy = \frac{1}{(1-\epsilon +a/j)^{1-a}}\log \frac{\epsilon j+a}{a}.\nonumber \\\quad \end{aligned}$$

(103)

On the other hand,

$$\begin{aligned}&\int \nolimits _{1-\epsilon }^1\frac{1}{(y+a/j)^{1-a}(1-y+a/j)}dy \nonumber \\&\quad \ge \frac{1}{(1+a/j)^{1-a}}\int \nolimits _{1-\epsilon }^1 \frac{1}{1-a+a/j}dy = \frac{1}{(1+a/j)^{1-a}}\log \frac{\epsilon j+a}{a}.\qquad \end{aligned}$$

(104)

It follows from (81), (103), and (104) that

$$\begin{aligned}&\frac{1}{(1+a/j)^{1-a}}\log \frac{\epsilon j+a}{a} \le \int \nolimits _0^1\frac{1}{(y+a/j)^{1-a}(1-y+a/j)}dy \nonumber \\&\quad \le \frac{1}{(1-\epsilon +a/j)^{1-a}}\log \frac{\epsilon j+a}{a} + \frac{(1-\epsilon +a/j)^a}{\epsilon (\epsilon +a/j)}. \end{aligned}$$

(105)

Clearly,

$$\begin{aligned} \lim _{j\rightarrow \infty }\frac{1}{(1+a/j)^{1-a}}\frac{\log (\epsilon j+a)/a}{\log j}&= 1, \end{aligned}$$

(106)

$$\begin{aligned} \lim _{j\rightarrow \infty }\frac{1}{(1-\epsilon +a/j)^{1-a}} \frac{\log (\epsilon j+a)/a}{\log j}&= 1. \end{aligned}$$

(107)

Also, by the definition of $\epsilon$, we have

$$\begin{aligned} \frac{(1-\epsilon +a/j)^a}{a(\epsilon +a/j)\log j} \sim \frac{1}{a(\log j)^{1/2}}. \end{aligned}$$

(108)

It follows from (105), (106), (107), and (108) that

$$\begin{aligned} \log j\le \int \nolimits _0^1\frac{1}{(y+a/j)^{1-a}(1-y+a/j)}dy \le \log j+\frac{1}{a}(\log j)^{1/2}. \end{aligned}$$

(109)

Combining (109) and (101), we have

$$\begin{aligned} \frac{\log j}{j^{1-a}}\le J^*_{44}(j) \le \frac{\log j}{j^{1-a}} + \frac{(\log j)^{1/2}}{aj^{1-a}} . \end{aligned}$$

(110)

From (110), we can apply the same decomposition as in (50), (51), (52) and prove (16) for $k=2,~b=1$. The details of the decomposition are omitted to reduce the redundancy of the proof. $\square$

1.6 B.6 Proof of (16)–Inductive step

Here, we prove (16) for general $k\ge 3, ~0<b\le 1$. Based on the discussions in B.4 and B.5, the asymptotic order of $\text{ E }_2(J_4(N))$ can be summarized as the following. For $k=2$ and $0<a\le 1$,

• (1) when $0<b<1$,

  $$\begin{aligned} \!\frac{C_2B^{-2}(a,b)\log (N+1)}{N^b}\!\le \text{ E }_2(J_4(N)) \!\le \frac{C_2B^{-2}(a,b)\log (N+1)}{N^b} + \frac{\log ^b N}{N^b};\nonumber \\ \end{aligned}$$

  (111)

• (2) when $b=1$,

  $$\begin{aligned} \frac{B^{-2}(a,1)\log ^2 N}{2(N+1)}\le \text{ E }_2(J_4(N)) \le \frac{B^{-2}(a,1)\log ^2 N}{2(N+1)} + \frac{B^{-2}(a,1)\log ^{3/2} N}{a(N+1)}.\nonumber \\ \end{aligned}$$

  (112)

In general, we assume the capture–recapture experiment consists of $k+1$ sampling occasions. So

$$\begin{aligned} n\mid N,\varvec{\theta }\sim \text{ Bin }\Big (N, 1-\prod _{i=1}^{k+1}(1-\theta _i)\Big ). \end{aligned}$$

For any fixed $0<a,b\le 1$ and $u=1,\cdots ,k+1$, we let $\delta _u=1-\theta _u$, $r=n-j$, and $m=N-j$. The expected value of $J_4(N)$ can be expressed as

$$\begin{aligned}&\text{ E }_{k+1}(J_4(N))\\&\quad =\sum _{n=0}^N\sum _{i=0}^\infty \frac{1}{(N-n+1+i)^2} \frac{N!}{n!(N-n)!}\int \nolimits _0^1\cdots \int \nolimits _0^1\left( 1-\prod _{u=1}^{k+1} \delta _u\right) ^n\left( \prod _{u=1}^{k+1} x_u\right) ^{N-n} \\&B^{-(k+1)}(a,b)\prod _{u=1}^{k+1}\delta _u^{b-1}\prod _{u=1}^{k+1} (1-\delta _u)^{a-1}d\delta _1\cdots d\delta _{k+1} \\&\quad = B^{-(k+1)}(a,b)\sum _{n=0}^N\sum _{i=0}^\infty \frac{1}{(N\!-\!n\!+\!1\!+\!i)^2} \frac{\varGamma (N\!+\!1)}{\varGamma (n\!+\!1)\Gamma (N\!-\!n\!+\!1)}\\&\qquad \times \sum _{j=0}^n\frac{\varGamma (n+1)}{\varGamma (j+1)\!\Gamma (n-j+1)}\\&\quad \int \nolimits _0^1\cdots \!\int \nolimits _0^1 (1-\delta _{k+1})^j \delta _{k+1}^ {n-j} \!\left( 1-\prod _{u=1}^k\delta _u\right) ^{n-j} \delta _u^ {N-n+b-1}\\&\qquad \times \prod _{u=1}^{k+1} (1-\delta _u)^{a-1}d\delta _1\cdots d\delta _{k+1}. \end{aligned}$$

In this way, we can integrate out $\delta _{k+1}$ and get the recursive formula.

$$\begin{aligned}&\text{ RHS }=\frac{B^{-(k+1)}(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\sum _{n=0}^N\sum _{i=0}^\infty \frac{1}{(N-n+1+i)^2} \frac{1}{\varGamma (N-n+1)} \\&\qquad \times \sum _{j=0}^n\frac{\varGamma (j+a)\Gamma (N-j+b)}{\varGamma (j+1)\Gamma (n-j+1)}\\&\qquad \quad \int \nolimits _0^1\cdots \int \nolimits _0^1\left( 1-\prod _{u=1}^k\delta _u \right) ^{n-j}\prod _{u=1}^k\delta _u^{N-n+b-1}\prod _{u=1}^k(1-\delta _u)^{a-1} d\delta _1\cdots d\delta _k \\&\quad = \frac{B(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\sum _{m=0}^N \frac{\varGamma (N-m+a)\varGamma (m+b)}{\varGamma (N-m+1)\varGamma (m+1)} \sum _{r=0}^{m}\left( \sum _{i=0}^\infty \frac{1}{(m-r+1+i)^2}\right) \\&\qquad \quad \int \nolimits _0^1\cdots \int \nolimits _0^1 \frac{B^{-k}(a,b)\varGamma (m+1)}{\varGamma (r+1) \Gamma (m-r+1)}\left( 1-\prod _{u=1}^k\delta _u \right) ^r\\&\qquad \times \prod _{u=1}^k\delta _u^{m-r+b-1}\prod _{u=1}^k(1-\delta _u)^{a-1} d\delta _1\cdots d\delta _k \\&\quad =\frac{B(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\sum _{m=0}^N \frac{\varGamma (m+b)\Gamma (N-m+a)}{\varGamma (m+1)\Gamma (N-m+1)} \text{ E }_k(J_4(m)), \end{aligned}$$

where $\text{ E }_k(J_4(N))$ denotes the expectation of $J_4(N)$ when there are $k$ sampling occasions. For fixed $k\ge 3$, $0<a\le 1$, and $0<b<1$, we assume that for some constants $C_5(k), C_6(k)$ depending on $k$, such that as $m\rightarrow \infty$,

$$\begin{aligned} \left| \text{ E }_k(J_4(m)) - \frac{C_2B^{-k}(a,b)\log ^{k-1}(m)}{m^b}\right| \le \frac{C_5(k)\log ^{k-2+b}(m)}{m^b}. \end{aligned}$$

(113)

For $b=1$, we assume that

$$\begin{aligned} \left| \text{ E }_k(J_4(m)) - \frac{B^{-k}(a,1)\log ^{k}(m)}{m}\right| \le \frac{C_6(k)\log ^{k-1/2}(m)}{m}. \end{aligned}$$

(114)

Clearly, assumptions (113) and (114) hold for $k=2$.

For $k\ge 3$, we will show the following results that directly lead to (16). Assuming $k+1$ sampling occasions and assumptions (113) and (114), we have for $0<b<1$,

$$\begin{aligned} \left| \text{ E }_{k+1}(J_4(N)) - \frac{C_2B^{-(k+1)}(a,b)\log ^{k}(N)}{N^b}\right| \le \frac{C_5(k+1)\log ^{k-1+b}(N)}{N^b}. \end{aligned}$$

(115)

For $b=1$,

$$\begin{aligned} \left| \text{ E }_{k+1}(J_4(N)) - \frac{B^{-(k+1)}(a,1)\log ^{k+1}(N)}{N}\right| \le \frac{C_6(k+1)\log ^{k+1/2}(N)}{N}. \end{aligned}$$

(116)

We focus on the case when $0<b<1$ and prove (115). $\text{ E }_{k+1}(J_4(N))$ can be decomposed after removing the first $M+1$ terms ($M$ is a fixed integer).

$$\begin{aligned} \!\text{ E }_{k+1}(J_4(N))\!&= \! \frac{C_2B^{-(k+1)}(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\!\sum _{m=M}^N \!\frac{\varGamma (m+b)\Gamma (N-m+a)\log ^{k-1} m}{m^b\varGamma (m+1)\Gamma (N-m+1)} \nonumber \\&+\frac{B(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\sum _{m=M}^N \frac{\varGamma (m+b)\Gamma (N-m+a)}{\varGamma (m+1)\Gamma (N-m+1)}\nonumber \\&\times \left( \text{ E }_k(J_4(m))-\frac{C_2B^{-k}(a,b) \log ^{k-1} m}{m^b}\right) \nonumber \\&\equiv J_{4k1}(N) + J_{4k2}(N). \end{aligned}$$

(117)

By the similar approach in the proof of (50), (51), (52), we can approximate $J_{4k1}(N)$ by

$$\begin{aligned} J_{4k1}(N) \approx \frac{C_2B^{-(k+1)}(a,b)\varGamma (N+1)}{\varGamma (N+a+b)} I_k(N), \end{aligned}$$

(118)

where

$$\begin{aligned} I_k(N)= \int \nolimits _M^N\frac{\log ^{k-1} x}{(x+1)(N-x+1)^{1-a}} dx. \end{aligned}$$

(119)

Below, we show that for any $0<b<1$ and $k\ge 2$, there exist positive constants $C_7$ such that

$$\begin{aligned} \frac{\log ^k N}{N^{1-a}}- \frac{C_7\log ^{k-1}N}{N^{1-a}} \le I_k(N) \le \frac{\log ^k N}{N^{1-a}}. \end{aligned}$$

(120)

To prove (120), we first find the order of $I_k(N)$ for $k=2$. Let $y=N-x$,

$$\begin{aligned} I_2(N)&=\int \nolimits _M^N\frac{\log x}{(x+1)(N-x+1)^{1-a}} dx \nonumber \\&= \frac{1}{N^{1-a}}\int \nolimits _{M/N}^1 \frac{\log y+\log N}{(y+1/N)(1-y+1/N)^{1-a}}dy \nonumber \\&\sim \frac{\log ^2 N}{N^{1-a}}+ \frac{1}{N^{1-a}}\int \nolimits _{M/N}^1 \frac{\log y}{(y+1/N)(1-y+1/N)^{1-a}}dy \nonumber \\&\le \frac{\log ^2 N}{N^{1-a}}. \end{aligned}$$

(121)

Now, we bound $I_2(N)$ from below. Notice that

$$\begin{aligned}&\frac{1}{N^{1-a}}\int \nolimits _{M/N}^{1/2} \frac{\log y}{(y+1/N)(1-y+1/N)^{1-a}}dy \nonumber \\&\quad \ge \frac{1}{(1/2+1/N)N^{1-a}}\int \nolimits _{M/N}^{1/2}\frac{\log y}{(1-y+1/N)^{1-a}}dy \nonumber \\&\quad \ge \frac{\log M-\log N}{(1/2+1/N)N^{1-a}}\int \nolimits _{M/N}^{1/2}\frac{1}{(1-y+1/N)^{1-a}}dy \nonumber \\&\quad =\frac{\log M-\log N}{a(1/2+1/N)N^{1-a}}\left[ (1-M/N+1/N)^a-(1/2+1/N)^a\right] \nonumber \\&\quad \approx \frac{\log M-\log N}{aN^{1-a}}. \end{aligned}$$

(122)

We also have

$$\begin{aligned}&\frac{1}{N^{1-a}}\int \nolimits _{1/2}^1 \frac{\log y}{(y+1/N)(1-y+1/N)^{1-a}}dy \ge -\frac{\log 2}{N^{1-a}}\int \nolimits _{1/2}^1 \frac{1}{(1-y+1/N)^{1-a}}dy \nonumber \\&\qquad = -\frac{\log 2}{a(1/2+1/N)N^{1-a}}\left[ (1/2+1/N)^a-(1/N)^a\right] \approx -\frac{\log 2}{aN^{1-a}}. \end{aligned}$$

(123)

It follows from (121), (122), and (123) that

$$\begin{aligned} \frac{\log ^2 N}{N^{1-a}}+\frac{\log M-\log N}{aN^{1-a}}-\frac{\log 2}{aN^{1-a}} \le I_2(N) \le \frac{\log ^2 N}{N^{1-a}}. \end{aligned}$$

(124)

For any fixed $k\ge 2$, we obtain the upper bound of $I_{k+1}(N)$.

$$\begin{aligned} I_{k+1}(N)&= \int \nolimits _M^N\frac{\log ^k x}{(x+1)(N-x+1)^{1-a}}dx \nonumber \\&= \frac{1}{N^{1-a}}\int \nolimits _{M/N}^1 \frac{(\log y+\log N)^k}{(y+1/N)(1-y+1/N)^{1-a}}dy \nonumber \\&= \frac{1}{N^{1-a}}\int \nolimits _{M/N}^1 \frac{(\log y+\log N)^{k-1}(\log y+\log N)}{(y+1/N)(1-y+1/N)^{1-a}}dy \nonumber \\&= I_{k}(N)\log N + \frac{1}{N^{1-a}}\int \nolimits _{M/N}^1 \frac{(\log y+\log N)^{k-1}\log y}{(y+1/N)(1-y+1/N)^{1-a}}dy \nonumber \\&\le I_{k}(N)\log N . \end{aligned}$$

(125)

Next, we seek the lower bound of $I_{k+1}(N)$. First of all,

$$\begin{aligned}&\frac{1}{N^{1-a}}\int \nolimits _{M/N}^{1/2} \frac{(\log y+\log N)^{k-1}\log y}{(y+1/N)(1-y+1/N)^{1-a}}dy \nonumber \\&\quad \ge \frac{(\log M-\log N)\log ^{k-1}M}{N^{1-a}}\int \nolimits _{M/N}^{1/2} \frac{1}{(y+1/N)(1-y+1/N)^{1-a}}dy \nonumber \\&\quad \ge \frac{(\log M-\log N)\log ^{k-1}M}{(1/2+1/N)N^{1-a}}\int \nolimits _{M/N}^{1/2}\frac{1}{(1-y+1/N)^{1-a}}dy \nonumber \\&\quad = \frac{(\log M-\log N)\log ^{k-1}M}{a(1/2+1/N)N^{1-a}} \left[ (1-M/N+1/N)^a-(1/2+1/N)^a \right] \nonumber \\&\quad \approx \frac{2(1-2^{-a})(\log M-\log N)\log ^{k-1}M}{aN^{1-a}}. \end{aligned}$$

(126)

Similarly,

$$\begin{aligned}&\frac{1}{N^{1-a}}\int \nolimits _{1/2}^1 \frac{(\log y+\log N)^{k-1}\log y}{(y+1/N)(1-y+1/N)^{1-a}}dy \nonumber \\&\quad \ge -\frac{(\log N-\log 2)^{k-1}\log 2}{a(1/2+1/N)N^{1-a}} \left[ (1/2+1/N)^a -(1/N)^a\right] \nonumber \\&\quad \approx -\frac{2^{1-a}\log 2 (\log N-\log 2)^{k-1}}{aN^{1-a}} . \end{aligned}$$

(127)

It follows from (125), (126), and (127) that

$$\begin{aligned} 0&\le I_k(N)\log N-I_{k+1}(N)\nonumber \\&\le \frac{2(1-2^{-a})(\log N-\log M)\log ^{k-1}M}{aN^{1-a}} +\frac{2^{1-a}\log 2 (\log N-\log 2)^{k-1}}{aN^{1-a}}.\qquad \quad \end{aligned}$$

(128)

Combining (124) and (128), we have proved (120). It follows from (118) and (120) that

$$\begin{aligned} \left| J_{4k1}(N)-\frac{C_2B^{-(k+1)}(a,b)\log ^k N}{N^b}\right| \le \frac{C_2C_7B^{-(k+1)}(a,b)\log ^{k-1} N}{N^b}. \end{aligned}$$

(129)

Similarly, we can show that

$$\begin{aligned} |J_{4k2}(N)|&\le \frac{B(a,b)\varGamma (N+1)}{\varGamma (N+a+b)}\sum _{m=M}^N \frac{\varGamma (m+b)\Gamma (N-m+a)\log ^{k-2+b}N}{m^b\varGamma (m+1)\Gamma (N-m+1)}\nonumber \\&\le \frac{C_8\log ^{k-1+b} N}{N^b}, \end{aligned}$$

(130)

where $C_8$ is a positive constant. Combining (117), (129), (130), we get (115). In the case of $b=1$, (116) can be proved in a similar manner. (115) and (116) imply (16) in Lemma 1.$\square$