I want to understand this entry, but do not understand how the $\mathcal{O}\left(\frac{1}{n^2}\right)$ in the accepted answer comes into play.

I reproduce the question here: We have $x \sim \mathrm{Bin}_{p,n}$ and want to approximate $E[\ln(x+1)]$ with a Taylor series. And I'm only interested in the case when the Taylor series is convergent for all values of $x$. (This condition implies $p> \frac{1}{2} $). Thus, \begin{equation} E[T_{x_0=np}\left(\ln(x+1)\right)]=\ln(np + 1) - \sum_{i=1}\frac{(-1)^{i}}{i(np-1)^{i}}E[(x-np)^{i}] \end{equation} In the accepted solution, the suggestion is to use the following approximation: \begin{equation} E[\ln(x+1)]=\ln(np + 1) - \frac{np(1-p)}{2(np+1)^{2}} + \mathcal{O}\left(\frac{1}{n^2}\right) \end{equation} Where the second summand comes from the variance term $(i = 2)$. I have not found an upper bound for central moments of Binomial variates that could lead to the big Oh notation. Plugging in a closed form expression for the respective central moments from this paper does not lead me to this approximation.


2 Answers 2


There are many ways to bound the Binomial central moments. A good (very general) estimate is provided by the Marcinkiweicz-Zygmund inequality, in the sharp form due to Burkholder [1], Writing $X-np$ as a sum of $n$ i.i.d. mean zero variables $Y_i$ taking values $1-p$ and $-p$, observe that the square function S(X) defined on page 87 in [1] is bounded by $n^{1/2}$. Theorem 3.1 page 87 in [1] then gives $$E[|X-np|^k] \le \bigl((k-1)n^{1/2}\bigr)^k \,.$$
The case $k=3$ is best dealt with directly, and here we cannot afford the absolute value inside the expectation: $$E[(X-np)^3] =\sum_{i,j,\ell \le n} E[Y_iY_jY_\ell]=\sum_{i \le n} E[Y_i^3] \le n \,.$$

[1] Burkholder, Donald L. "Sharp inequalities for martingales and stochastic integrals." Astérisque 157, no. 158 (1988): 75-94. http://www.numdam.org/article/AST_1988__157-158__75_0.pdf

Remark: In my answer I focus on the question regarding Binomial central moments. for estimating the logarithm, see Iosif Pinelis' answer.


"And I'm only interested in the case when the Taylor series is convergent for all values of $x$." -- There is no such Taylor series for $\ln$ (which is only defined on $(0,\infty)$).

However, to get the desired $O(1/n^2)$, you do not need a convergent series; you do not need any series at all. Instead, you need the Taylor expansion \begin{equation} \ln(1+u)=u-u^2/2+u^3/3+O(u^4) \tag{1} \end{equation} for $u\ge-1/2$ (say), with a universal constant in $O(u^4)$.

Indeed, letting \begin{equation} U:=\frac{X-np}{np+a}, \tag{2} \end{equation} for all real $a>0$ you can write \begin{align*} &E\ln(X+a)-\ln(np+a) \\ &=E\ln(1+U) \\ &=E\ln(1+U)1(U\ge-1/2) \\ &+E\ln(1+U)1(U<-1/2). \tag{3} \end{align*} By (1),
\begin{align*} &E\ln(1+U)\,1(U\ge-1/2) \\ &=E(U-U^2/2+U^3/3+O(U^4))\,1(U\ge-1/2) \\ &=E(U-U^2/2+U^3/3)+O(EU^4) \\ &-E(U-U^2/2+U^3/3)1(U<-1/2). \tag{4} \end{align*} Now we are going to use (say) Rosenthal's inequality (see e.g. formula (7) in this paper), which implies \begin{equation} EU^4=O(1/n^2) \tag{5} \end{equation} and \begin{equation} EU^6=O(1/n^3); \tag{6} \end{equation} here in what follows, the constants in $O(\cdot)$ may depend only on $p$ (and, in (11) and (12), also on $a$). By (6), for $m=0,1,2,3$ we have $|U|^m\,1(U<-1/2)|\le U^6(1/2)^{m-6}$ and hence \begin{equation} |EU^m\,1(U<-1/2)|\le EU^6(1/2)^{m-6}=O(1/n^3). \tag{7} \end{equation} Also, \begin{equation} EU=0,\quad EU^2=\frac{npq}{(np+a)^2} \tag{8} \end{equation} (where $q:=1-p$), $E(X-np)^3=n(pq^3-qp^3)=O(n)$ and hence \begin{equation} EU^3=O(1/n^2). \tag{9} \end{equation} By (4), (5), (7), (8), (9), \begin{align*} &E\ln(1+U)\,1(U\ge-1/2) \\ &=E(U-U^2/2+U^3/3)+O(1/n^2) \\ &=-\frac{npq}{2(np+a)^2}+O(1/n^2). \tag{10} \end{align*} By (2) and the obvious inequality $X\ge0$, we have $1+U\ge\frac a{np+a}$. Also, if $U<-1/2$, then $1+U<1/2$, whence $\ln(1+U)<0$ and hence $|\ln(1+U)|\le \ln\frac{np+a}a$. So, \begin{align*} &E|\ln(1+U)\,1(U<-1/2)| \\ &\le\ln\frac{np+a}a E1(U<-1/2)=O(1/n^2), \tag{11} \end{align*} by (7) with $m=0$. Now (3) and (10) yield \begin{align*} &E\ln(X+a)=\ln(np+a)-\frac{npq}{2(np+a)^2}+O(1/n^2), \tag{12} \end{align*} as desired.

  • $\begingroup$ thank you for this amazing proof. I have two questions. 1, The Taylor expansion in eq. 1 is valid for $|u|<1$. Now $U \in (-1, \frac{n}{\alpha})$. How is it garanteed, that eq. 1 is applicable for all $U1(U\geq -\frac12)$? 2, I understand eq. 11, that it is assumed that $E|\ln(1+U)1(U<-\frac12)|\leq E|U|^m1(U<-\frac12)$ for $m \in \{0,1,2,3\}$, which is not true. Could you help me to understand it better? $\endgroup$
    – qwert
    Jun 14, 2021 at 14:13
  • $\begingroup$ @qwert : Concerning (1): Use the Taylor expansion for $|u|\le1/2$ and then use the boundedness of $(\ln(1+u)-(u-u^2/2+u^3/3))/u^4$ for $u\ge1/2$. Concerning (11): I have added details on that. (I did not use there what you thought I used). $\endgroup$ Jun 14, 2021 at 15:06

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