concentration inequality for a weighted sum of independent but not identical binary variables

Question

Let $\alpha\in[0,1]$ be a fixed constant, and let $w,x\in[0,1]^n$ be two vectors such that $\sum_i w_i x_i=\alpha$.

Define $Y = \sum_i w_i X_i$, where $X_i \sim \operatorname{Bernoulli}(x_i)$, so it holds $\mathbb{E}[Y]=\alpha\leq 1$, and assume that the $X_i$'s are independent variables, .

Can we use some concentration inequality for $Y$ to obtain some bound of the form $\Pr[Y>1] \leq c(\alpha)$ for some constant $c(\alpha)\ll 1$? Of course, we know from Markov's inequality that $\Pr[Y>1]\leq\alpha$, but I wans hopping to optain better bounds using the particular structure of $Y$, in particular for $\alpha$ close (or even equal) to $1$.

What I tried so far

• The Chernoff-Hoeffding inequality gives me a bound of the form $$ \Pr[Y>1] \leq \exp\left(-\frac{2(1-\alpha)^2}{n}\right), $$ but this is useless when $n\to\infty$.

• We can also use Bernstein inequality, as $\mathbb{V}[Y]=\sum_i w_i^2 x_i (1-x_i)\leq\sum_i w_i x_i=\alpha$, to obtain: $$ \Pr[Y>1] \leq \exp\left(-\frac{(1-\alpha)^2}{2(\alpha+\frac{1-\alpha}{3})}\right). $$ This is better, but it does not beat the Markov bound $\Pr[Y>1]\leq \alpha$ neither!

Answer 1

Without further restrictions on $w,x$, you cannot beat the Markov bound by much for $\alpha$ close to $1$ (as in your post).

Indeed, let $a:=\alpha$. Let $x_i=a$ for all $i=1,\dots,n$, and let $w_1=1$ and $w_2=\cdots=w_n=0$. Then all the conditions in the OP on $w,x$ will hold, and \begin{equation} P(Y\ge1)=P(X_1=1)=a, \end{equation} which is the Markov bound on $P(Y\ge1)$.

For $P(Y>1)$ and $a$ close to $1$, you can get about half as good as this, as follows. Take any natural $n\ge2$ and any $a\in(0,1]$. For $t\in(0,a/2)$, let \begin{equation} w_1=\frac a2,\quad w_2=1+t-\frac a2,\quad w_3=\cdots=w_n=0, \end{equation} \begin{equation} x_1=\frac{a/2}{w_1}=1,\quad x_2=\frac{a/2}{w_2}=\frac a{2-a+t},\quad x_3=\cdots=x_n=0. \end{equation} Then all the conditions in the OP on $w,x$ will hold, and one will also have $w_1+w_2>1$, whence \begin{multline} P(Y>1)=P(w_1X_1+w_2X_2>1)\ge P(X_1=X_2=1) \\ =x_1x_2=\frac a{2-a+t}\underset{t\downarrow0}\longrightarrow \frac a{2-a} =1-(1-a)(2+o(1)) \end{multline} as $a\uparrow1$, which is indeed about about half as good for $a$ close to $1$ as the Markov bound $a=1-(1-a)$.