An inequality about binomial distribution (generalization)

Question

Statement

Assume that $\sigma\in (1,+\infty)$, $N\in[3,+\infty)\cap \mathbb{Z}$, $p,t\in (0,1)$, $n_1\in\{0,1,2,\cdots,N-1\}$. Prove or disprove that $$[B(n_1)]^\frac{1}{\sigma} - [B(n_1+1)]^\frac{1}{\sigma} < 1 .$$ Here, the function $B$ is defined as $$ B(n_1):=\sum_{n_2=0}^{N-n_1} p(n_2 \mid n_1) \cdot \left( n_2+(N-n_1-n_2)t \right)^\sigma ,\ 0\le n_1\le N.$$

Here, $p(n_2 \mid n_1)$ is a conditional probability distribution satisfying $$ \begin{cases} \displaystyle\sum_{n_2=0}^{N-n_1} p(n_2\mid n_1) = 1 \\ \displaystyle\sum_{n_2=0}^{N-n_1} n_2\cdot p(n_2\mid n_1)=p\cdot (N-n_1) \end{cases} ,\ 0\le n_1\le N. $$

Motivation

Actually, this proposition is a generalization of An inequality about binomial distribution, which has been solved by @FedorPetrov. When $p(n_2\mid n_1) = \binom{N-n_1}{n_2}p^{n_2}(1-p)^{N-n_1-n_2}$, it is transformed into that problem.

My attempts

Assume that the random variables $N_1$ and $N_2$ such that the conditional probability mass function of $N_2$ given $N_1 = n_1$ is $\operatorname{P}(N_2 = n_2\mid N_1 = n_1) = p(n_2\mid n_1) $. Then we have $\mathbb{E}(N_2\mid N_1) = p\cdot (N-N_1)$, and $B(n_1)=\mathbb{E}( ( N_2 + (N-N_1-N_2)t )^\sigma \mid N_1 = n_1)$. But then I got stuck.

So how to prove or disprove this inequality? Thanks in advance!

Answer 1

$\newcommand\si\sigma$This is not true in general. Indeed, $$B(n_1)=B_\si(n_1):=E(((1-t)N_2+t(N-n_1))^\si|N_1=n_1)$$ for random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$.

If the inequality in question were true for all real $\si>1$, then we would have $$d_t(n_1):=M_\infty(n_1)-M_\infty(n_1+1)\le1 \tag{1}\label{1}$$ for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, where $$M_\infty(n_1):=\lim_{\si\to\infty}B_\si(n_1)^{1/\si}=(1-t)m(n_1)+t(N-n_1), $$ where in turn $m(n_1)$ is the maximum of the support set of the conditional distribution of $N_2$ given $N_1=n_1$. So, $$d_t(n_1)=(1-t)(m(n_1)-m(n_1+1))+t.$$ So, if \eqref{1} were true for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, then we would have $$d_0(n_1)=m(n_1)-m(n_1+1)\le1 \tag{2}\label{2}$$ for all $n_1\in\{0,\dots,N-1\}$.

However, it is easy to construct random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$ while \eqref{2} fails to hold for some $n_1\in\{0,\dots,N-1\}$.

E.g., let $N=3$, $p=1/2$, $P(N_2=1|N_1=1)=1$, and $P(N_2=n_2|N_1=n_1)=\frac1{4-n_1}$ for $n_1\in\{0,2,3\}$ and $n_2\in\{0,\dots,3-n_1\}$. Then all the conditions on $N_1$ and $N_2$ will hold, while $d_0(0)=m(0)-m(1)=3-1=2\not\le1$, so that \eqref{2} fails to hold.

Actually, for these random variables $N_1$ and $N_2$, $t=0$, and any $\si>1$ we have $$D(\si):=B_\si(0)^{1/\si}-B_\si(1)^{1/\si} =4^{-1/\sigma } \left(3^{\sigma }+2^{\sigma }+1\right)^{1/\si}-1>1$$ already for $\si\ge2.57$. $\quad\Box$

For an illustration, here is the graph $\{(\si,D(\si))\colon1<\si<12\}$ (solid):