A concentration problem Consider $N$ variables $X_1,X_2,\ldots,X_N$ with $\Pr(X_i=a_i)=\Pr(X_i=-a_i)=1/2$. Here $a_1,a_2,\ldots,a_N\in [0,1]$.
Does there exist some concentration result about $\sum_{i=1}^N X_i$?
 A: Let $S(a,\epsilon)=\sum_{k=1}^N \epsilon_i a_i$ with $\epsilon=(\epsilon_1,\ldots,\epsilon_N)$ and  $\epsilon_i\in\{\pm 1\}$ independent and uniformly distributed, and $a_i\in [0,1].$ We can actually assume $a_i\leq a_{i+1}$ (since the signs range over all of $\{\pm 1\}^n,$) and take $a_i\in (0,1]$ since otherwise we can reduce the dimension $N$ and reconsider the problem for $N'<N.$
Clearly $S(a,\epsilon)=-S(a,-\epsilon)$ so the distribution of $S(a,\epsilon)$ is symmetric around zero. Let any lower bound $B$ on $$\mathbb{P}\left\{|S((1,\ldots,1),\epsilon)|\leq 1\right\}\geq B$$ for the case $a=(1,1,\ldots,1)$ for all $1\leq i\leq N$ be given. Recall that we can assume $0<a_1\leq a_2\leq \cdots \leq a_n \leq 1.$
Define the all one vector as $a^{(0)}$ and now consider the vector 
$$a^{(1)}=(a_{i^{\ast}},\ldots a_{i^{\ast}},1,\ldots,1)$$
where we define $a^{(1)}_i=a_{i^{\ast}}$ for all $i\leq i^{\ast}.$
Lemma: Let $Z$ be any symmetric random variable, with support on the integers. Then $Z+\epsilon a$ where $\epsilon=\pm 1$ with equal probability
satisfies the inequality
$$\mathbb{P}\left\{|Z+\epsilon a|\leq 1\right\}\geq \mathbb{P}\left\{|Z+\epsilon |\leq 1\right\}$$ whenever $a\in (0,1].$ This can be seen by the fact that the probability mass function of $Z+\epsilon a$ is the sum $$\frac{q(z-a)+q(z+a)}{2}$$ while that of $Z+\epsilon$ is the sum $$\frac{q(z-1)+q(z+1)}{2}.$$ Here $q(z)$ is the probability mass function of $Z.$
This is because $a_{i^{\ast}}<1$ and the random sums $S(a^{(1)},\epsilon)$ are ``more central'' than the sums $S(a^{(0)},\epsilon)$ since their distribution result from at least one convolution with dominated random variables on $\{-a_{i^{\ast}}, a_{i^{\ast}}\}$ 
instead of random variables on $\{-1,1\}.$
The Lemma's argument can then be repeated using induction and applying it to compare
$$S(a^{(0)},\epsilon)\textrm{ with } S(a^{(1)},\epsilon),$$
and then moving from $a^{(1)}$ to $$a^{(2)}=(a_{i^{\ast}-1},\ldots a_{i^{\ast}-1}, a_{i^{\ast}},1,\ldots,1)$$
and applying the argument to obtain that
$$\mathbb{P}\left\{|S(a^{(2)},\epsilon)|\leq 1\right\}\geq B$$
still holds. Then use $a^{(3)},$ etc to finish the argument.
A: If $a_i$ are deterministic, than you can think about new random variables  $Y_i=\frac{X_i}{a_i}$ and you already have the lower bound for this.
