# An inequality about binomial distribution

Statement

Assume that $$\sigma,R\in (1,+\infty)$$, $$N\in\mathbb{N}^*$$, $$p\in (0,1)$$, $$n_1\in\{0,1,2,\cdots,N-1\}$$. Prove or disprove that $$B^\frac{1}{\sigma}(n_1)-B^\frac{1}{\sigma}(n_1+1)<1 .$$ Here, the function $$B$$ is defined as $$B(n_1):=\sum_{n_2=0}^{N-n_1}\binom{N-n_1}{n_2}p^{n_2}(1-p)^{N-n_1-n_2}\left(n_2+(N-n_1-n_2)R^{\frac{1}{\sigma}-1}\right)^\sigma .$$

My attempts

First I tried using the inequality $$B^\frac{1}{\sigma}(n_1)-B^\frac{1}{\sigma}(n_1+1)<\left(B(n_1)-B(n_1+1)\right)^\frac{1}{\sigma}$$ since $$\sigma>1$$, but $$\left(B(n_1)-B(n_1+1)\right)^\frac{1}{\sigma}$$ may be greater than 1 (concluded by computer). So this approach may not work.Then, I tried using $$\binom{N-n_1}{n_2}=\binom{N-n_1-1}{n_2}+\binom{N-n_1-1}{n_2-1}$$, but I still haven't proven it.

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

Update

Thanks to a user who answered earlier (he may have deleted his answer because of a slight error in his argument), I can transform the left side of the inequality to be proved into a form that is more probabilistic.

Equivalent proposition: Assume that the random variables $$X_1, X_2,\cdots , X_n (n\ge 1)$$ are independent and identically distributed, with $$\operatorname{P}(X_k=1)=p,\operatorname{P}(X_{k}=t)=1-p$$, where $$p,t\in (0,1)$$. Real number $$\sigma > 1$$. Prove that $$\operatorname{E}^{\frac{1}{\sigma}}\left[\left(\sum_{k=1}^{n}X_k\right)^\sigma\right] - \operatorname{E}^{\frac{1}{\sigma}}\left[\left(\sum_{k=1}^{n-1}X_k\right)^\sigma\right] < 1 .$$ P.S. Actually, $$t=R^{\frac{1}{\sigma}-1}$$ and $$n=N-n_1$$.

Another try, now I claim that the inequality holds.

For a random variable $$X$$, denote $$\|X\|=(\mathbb E |X|^\sigma)^{1/\sigma}$$. It is indeed a norm.

Then $$\|X_1+\ldots+X_n\|\leqslant \|X_1+\ldots+X_{n-1}\|+\|X_n\|\leqslant \|X_1+\ldots+X_{n-1}\|+1.$$

• Thank you for your reply! But I think $T$ should be less than $1$. Since $\frac{1}{\sigma}<1$, so $\frac{1}{\sigma}-1<0$, and therefore $T=R^{\frac{1}{\sigma}-1}<1$. So the inequality to be proven is true when $\sigma=1$. Commented Aug 1 at 13:34
• Please check another attempt, with another answer. I am not certain about proofs, but at least one of two answers is correct. Commented Aug 1 at 15:16
• Very concise and elegant proof! Thank you! Commented Aug 1 at 18:35