Multiplicative approximation for a negative moment of the binomial distribution

Question

Let $X$ be a binomial random variable with parameters $n,p$.

Define the function $f(n, p, t) = E\frac{1}{1 + t X}, $ where $t > 0$.

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Question: Can we find an elementary function $F(n, p, t)$ such that for some constant $C \in [1, \infty)$, we have $$ F(n, p, t) \leq f(n, p, t) \leq C \, F(n, p, t), $$ for all $t > 0$?

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Some (perhaps obvious) observations:

• Jensen's inequality gives $$f(n, p, t) \geq \frac{1}{1 + t n p},$$ but this is evidently not sharp (described more below).
• Consider $p_n = 1/n, t_n = 1$. Then, $$ (1 - o(1)) \frac{1}{e} = (1 - p_n)^n \leq f(n, p_n, t_n) \leq 1, \quad\qquad (1) $$ as $n \to \infty$. This shows that (1) isn't sharp: $(1 + t_n n p_n)^{-1} = (1+n)^{-1} \to 0$.
• Similarly, we see that if $p \leq \tfrac{1}{2n}$, then $$ \frac{1}{2} \leq (1-p)^n \leq f(n, p, t) \leq 1. $$ Thus, we can focus on the case that $p \geq \tfrac{1}{2n}$.

Answer 1

The answer is yes, we can find such an elementary function.

Indeed, in the case when $np\le1/2$ we have
\begin{equation} f(n,p,t)\ge(1-p)^n\ge(1-p)^{1/(2p)}\ge1/2, \end{equation} since $p\le1/2$ in this case, so that \begin{equation} 1/2\le f(n,p,t)\le1. \end{equation}

Consider now the case $np>1/2$. By Chebyshev's inequality,
\begin{equation} P(X\le np/2)\le\frac{np(1-p)}{(np/2-np)^2}\le\frac4{np}. \end{equation} So, \begin{equation} E\frac{1(X\le np/2)}{1+tX} =P(X=0)+E\frac{1(1\le X\le np/2)}{1+tX} \\ \le(1-p)^n+\frac4{np}\frac1{1+t} \le(1-p)^n+\frac8{1+npt}, \end{equation} since $np\ge1/2$, and \begin{equation} E\frac{1(X>np/2)}{1+tX} \le\frac1{1+npt/2}\le\frac2{1+npt}, \end{equation} so that \begin{equation} E\frac1{1+tX}\le(1-p)^n+\frac{10}{1+npt}. \end{equation} On the other hand, as you showed, \begin{equation} E\frac1{1+tX}\ge\max\Big((1-p)^n,\frac1{1+npt}\Big) \ge\frac12\Big((1-p)^n+\frac1{1+npt}\Big). \end{equation}

Thus, \begin{equation} F(n,p,t)\le f(n,p,t)\le CF(n,p,t), \end{equation} where $C:=20$ and \begin{equation} F(n,p,t):= \left\{ \begin{alignedat}{2} &\frac12&&\text{ if }np\le\frac12, \\ &\frac12\Big((1-p)^n+\frac1{1+npt}\Big) &&\text{ if }np>\frac12. \end{alignedat} \right. \end{equation}