# Bound on the MGF of the product of two independent binomial, one being centered

Consider the following: $$r_1,...,r_t$$ are iid symmetric signs taking value $$\pm1$$, independent of $$B\sim Binomial(p, q)$$ with integer $$p$$ and $$q=p^{-1.01}$$.

Question: Consider $$t$$ as a non-decreasing function of $$p$$. For which such non-decreasing function of $$p$$ does there exist a constant $$\lambda>0$$ independent of $$p$$ such that the following holds: $$E[\exp(\lambda B \sum_{u=1}^t \frac{r_u}{\sqrt t})] \le 1.05,$$ for $$p$$ large enough.

Remark: Note that $$\tilde Z = \sum_u\frac{r_u}{\sqrt t}$$ has variance 1 and is very close to $$N(0,1)$$, for instance by Tusnady's inequality. If $$\tilde Z$$ were replaced by 1 inside the exponential, we could proceed as follows: using the exact formula for the MGF of the binomial, and using $$(1+x/p)^p\le e^x$$, $$E[\exp(\lambda B)= (1-q + qe^\lambda)^p \le (1 + qe^\lambda)^p \le \exp(pq e^\lambda) \le \exp(p^{-0.01} e^{\lambda}),$$ so that for any constant $$\lambda$$ (say, $$\lambda=1$$), the RHS is bounded from above by 1.05 or any other constant strictly greater than 1. If $$t$$ is constant, not growing to $$\infty$$ simultaneously as $$p$$, then we can use this remark to answer the question.

Edit: Another remark: the previous argument shows $$E[\exp(\lambda \tilde Z B)= (1-q + qe^{\lambda\tilde Z})^p \le (1 + qe^{\lambda\tilde Z})^p \le \exp(pq e^{\lambda\tilde Z}) \le \exp(p^{-0.01} e^{\lambda\tilde Z})$$ and since $$\tilde Z\le \sqrt t$$ always holds, $$\sqrt t = \log(p^{0.009})\asymp \log p$$ with $$\lambda=0$$ works. The harder question is with $$t$$ varying more rapidly than this.

• Is "$t$ [...] a non-increasing function of $p$" or an "increasing function of $p$"? Mar 16, 2022 at 1:06
• Thanks. t is a non-decreasing function of $p$. For instance $t=p^c$ for some $c>0$, or $t=\log(p)^c$. Will fix the question. Mar 16, 2022 at 1:38
• I added a remark that show that $\sqrt t \asymp \log p$ would work, for some small enough multiplicative constants. I am wondering about more rapidly growing functions of $p$. Mar 16, 2022 at 1:53

$$\newcommand{\la}{\lambda}$$The estimate of $$t$$ that you got cannot be improved. Indeed, let $$\begin{equation} S_t:=\sum_{u=1}^t \frac{r_u}{\sqrt t}. \end{equation}$$ Suppose, as you did, that $$\begin{equation} E\exp(\la BS_t)\le1.05 \end{equation}$$ for some real $$\la>0$$. Then $$\begin{equation} 1.05\ge\frac{\la^{2p}}{(2p)!}\,EB^{2p}\,ES_t^{2p} \gg \frac{\la^{2p}}{p^{2p}}\,EB^{2p}\,ES_t^{2p}, \end{equation}$$ where $$a\gg b$$ means that $$a\ge c^p b$$ for some universal real constant $$c>0$$.
Using Corollaries 1 and 2 by Latała, we get $$\begin{equation} EB^{2p}\gg\Big(\frac p{\ln p}\Big)^{2p} \end{equation}$$ and $$\begin{equation} ES_t^{2p}\gg \min[t^p,p^p]. \end{equation}$$ Thus, $$\begin{equation} 1.05\gg \frac{\la^{2p}}{p^{2p}}\,\Big(\frac p{\ln p}\Big)^{2p}\,\min[t^p,p^p] =\la^{2p}\Big(\frac{\min[t,p]}{\ln^2 p}\Big)^p, \end{equation}$$ which implies $$\sqrt t=O(\ln p)$$, as claimed.