Anti-concentration for sums of geometric random variables

Question

Consider the random variable $Y = Y_1 + \dots + Y_k$, where each $Y_i$ is iid distributed as a geometric random variable with sucess probability $p$; here we should think of $p$ as being close to zero. The mean of $Y$ is $k (1-p) / p \approx k/p$. I would like to show that there is a large probability that $Y$ is above its mean. When $k = 1$, we know that $Y$ can become as large as $\log (1/\beta) / p$ with probability $\beta$; for $\beta$ small, this is much larger than the mean value $1/p$.

How can one show anti-concentration for $k > 1$? I would like to show that for small $\beta$ then $Y$ is significantly larger (more than a constant multiple) above its mean $k/p$, with probability at least $\beta$.

Answer 1

This is just the negative binomial distribution.

So you can work with the corresponding cdf, the regularized incomplete beta function.

Answer 2

This paper might help https://cs.uwaterloo.ca/~browndg/negbin.pdf

No concentration inequality would work for small $k$. The above paper finds a concentration inequality for sum of $n$ independent geometrical variable via negative binomial distribution.

The form looks like this $$ P\bigg(\sum_{i = 1}^k Y_i > rk(1-p)/p\bigg) \leq \exp\bigg\{\frac{-rk(1-1/r)^2}{2}\bigg\}, $$ where $r$ is any constant that is greater than 1.