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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 $n=N-n_1$.

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1 Answer 1

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

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  • $\begingroup$ 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 $\sigma=1$. $\endgroup$
    – John_zyj
    Commented Aug 1 at 13:34
  • $\begingroup$ Please check another attempt, with another answer. I am not certain about proofs, but at least one of two answers is correct. $\endgroup$
    – Fedor Petrov
    Commented Aug 1 at 15:16
  • $\begingroup$ Very concise and elegant proof! Thank you! $\endgroup$
    – John_zyj
    Commented Aug 1 at 18:35
  • $\begingroup$ This problem can be generalized, the details are described here (mathoverflow.net/questions/484304/…). I turn to you for assistance. Thank you! $\endgroup$
    – John_zyj
    Commented 20 hours ago

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