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 $\beta$.



Anti-concentration for sums of geometric random variables
This is just the [negative binomial distribution](https://en.wikipedia.org/wiki/Negative_binomial_distribution#Sum_of_geometric_distributions).


So you can work with the corresponding cdf, [the regularized incomplete beta function](https://en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function).


