Is it possible to upper bound the probability bellow using Chernoff bound?
$$ \mathbb{P}\left\lbrace \sum\limits_{\substack{i}}^{n} \dfrac{1}{\mathbf{N(x_{i})}}> \dfrac{tn}{4} +\dfrac{n}{a_{n}} \right\rbrace, $$
where, $t$ and $a_{n}$ are constants. Also, $\mathbf{N}(x_{i})=Y_{i}+1$ is a random variable with $Y_{i} \sim Bin(n-1, r_{n}).$ The terms $Y_{i}$ are mutually dependent.
The problem is that I do not know the distribution of $\sum\limits_{\substack{i}}^{n} \dfrac{1}{\mathbf{N(x_{i})}}$. I only know the distribution of $ \sum\limits_{\substack{i}}^{n} \mathbf{N(x_{i})}$ but I could not have this term in the probability.
I could also do,
$$ \mathbb{P}\left\lbrace \sum\limits_{\substack{i}}^{n} \dfrac{1}{\mathbf{N(x_{i})}}> \dfrac{tn}{4} +\dfrac{n}{a_{n}} \right\rbrace \leq n \max_{i} \mathbb{P}\left\lbrace \dfrac{1}{\mathbf{N(x_{i})}}> \dfrac{t}{4} +\dfrac{1}{a_{n}} \right\rbrace$$
Then, I use the Chernoff bound but the obtained bound is not tight enough. So, I want to use the distribution of term $ \sum\limits_{\substack{i}}^{n} \mathbf{N(x_{i})}$.