Fix $N$ to be a large prime. Let $A \subset \mathbb{Z}/N\mathbb{Z}$ be a random subset defined by $\mathbb{P}(a \in A) = p$, where $p = N^{-2/3 + \epsilon}$ for some fixed $\epsilon > 0$. My question is what kind of concentration inequalities do we have for the random variable $|A+A|$?



Does $|A+A|$ concentrate near its mean?
If $\epsilon>1/6$ then $|A+A|=N$ with high probability. Suppose that $\epsilon<1/6$. Call a pair $(x,y) \in {\mathbb Z}/N{\mathbb Z} \times {\mathbb Z}/N{\mathbb Z} $ "bad" if $(x,y) \in A \times A $ and there is another pair $(z,w) \in A \times A $ (also different from $(y,x)$), such that $z+w=x+y$. The probability that a particular pair $(x,y) \in {\mathbb Z}/N{\mathbb Z} \times {\mathbb Z}/N{\mathbb Z} $ will be "bad" is at most $Np^4$ so the expected number of "bad" pairs is at most $N^3 p^4$. Every good pair $(x,y)$ contributes one element to $|A+A|$, with $(y,x)$ yielding the same eklement. By Markov's inequality, 
$|A+A| \ge (|A|^2+|A|)/2-O(N^3 p^4)$ with high probability. Under the assumption 
 $\epsilon<1/6$, we have that $N^3p^4$ is negligible compared to $N^2p^2$
so $|A+A|$ will be well approximated by $|A|^2/2$, which is half the square of a Binomial variable.


