4
$\begingroup$

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 $\beta$.

$\endgroup$
5
$\begingroup$

This is just the negative binomial distribution.

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

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.