Let $\gamma>1$ be a real number and let $n\in \mathbb{N}$. Define $f\colon\mathbb{N}\to[0,1]$ $$ f(n_0) = \frac{\binom{n-n_0}{m}}{\binom{n}{m}}, $$ where $$ m = \Big\lfloor{\frac{n}{\lceil\gamma n_0 \rceil}}\Big\rfloor. $$ Remark: $f(n_0) = \mathbb{P}(X = 0 )$, where $X\sim\text{Hypergeometric}(n,n_0,m)$ (following the notation from Wikipedia)
Goal: I aim to find a non-trivial upper bound for $f(n_0)$. Specifically, I would like to show that there exists a constant $c=c(\gamma)<1$ such that $$ f(n_0) < c $$ for all $n$ sufficiently large, and $n_0 \leq n/(2\gamma)$.
Progress and Observations: Note that the following expression can be helpful $$ f(n_0) =\frac{(n-m)\ldots(n-m-n_0+1)}{n\dots(n-n_0+1)}. $$ Although $f$ is not necessarily increasing, I have shown that $$ f(n_0) \leq \frac{\binom{n-n_0}{ n/\lceil\gamma n_0\rceil -1 }}{\binom{n}{n/\lceil\gamma n_0\rceil -1}} =:\tilde f(n_0), $$ and $\tilde f$ appears to be increasing, although I have not been able to prove it. If I could establish that $\tilde f(n_0)$ is increasing, then for $n_0 \leq n/(2\gamma)$ (assuming that it is an integer), we would get the desired result $$ f(n_0) \leq \tilde f(n_0) \leq \tilde f(n/(2\gamma)) = 1 - \frac{1}{2\gamma}. $$
Remark: I have numerically checked and it seems that the bound is correct (and that $c$ exists)
Questions:
- Are there any known references or techniques that can help with this type of upper bound?
- Is there a way to prove that $\tilde f(n_0)$ is increasing?
