Let $X_1,\dots, X_n\sim\operatorname{Bern}(\frac{1}{2})$ be independent, identically distributed random variables, and $\alpha=(\alpha_1,\dots,\alpha_n)\in[0,1]^n$ a vector of non-negative weights satisfying $\sum_{k=1}^n \alpha_k = 1$.

Define the random variable $X\stackrel{\rm def}{=} \sum_{k=1}^n \alpha_k X_k$. Given parameter $\delta > 0$, what is the best concentration bounds I get on $X$ in terms of the $p$-'norms' (allowing $p\in(0,1)$) of $\alpha$, i.e. $$ \mathbb{P}\{ X \leq \delta\} \leq \varphi(\delta, (\lVert\alpha\rVert_p)_{p>0}) $$ ?

I was thinking of using Berry—Esseen-like approximations to reduce this to studying a Gaussian random variable (leading to a bound in terms of $\frac{1}{\lVert \alpha\rVert_2^2}$, but this falls short of what I want (as, if I am not mistaken, this requires the vector $\alpha$ to be more or less uniform). Using Chernoff/Hoeffding bounds will get me a bound expressed in terms of $\frac{1}{\lVert \alpha\rVert_\infty}$, but it feels very loose.

Edit: using the conversion Bernoulli/Rademacher and the bounds for those, I get an upper bound of $e^{-\frac{(1-2\delta)^2}{\lVert \alpha\rVert_2^2}}$, similar to what the Gaussian approximation would result in. But I still don't really know if this is tight, that is if the "right" parameter for this is indeed $\lVert \alpha\rVert_2$, or something more "exotic" that just happens to coincides with $\lVert \alpha\rVert_2$ in the "uniform-weights regime."

  • 1
    $\begingroup$ What is the role of $\delta$? It sounds like you should be using Littlewood-Offord theory, see e.g. arxiv.org/abs/math/0703503 $\endgroup$
    – Terry Tao
    Jun 9, 2016 at 22:51
  • $\begingroup$ Oh, shoot. $\delta$ is the parameter $\varepsilon$ that got renamed by mistake, a small constant. I'll edit this. $\endgroup$
    – Clement C.
    Jun 9, 2016 at 22:52
  • 1
    $\begingroup$ Ah, I had thought your $X_i$ were centred around the origin, rather than around 1/2. In which case you're right, Littlewood-Offord theory will not be of much use. So you're probably best off with variants of the Chernoff/Hoeffding bounds that you've already tried. $\endgroup$
    – Terry Tao
    Jun 10, 2016 at 0:52
  • 3
    $\begingroup$ One can also work with the model case when $\alpha_k = 1/m$ for $m$ values of $k$ and $\alpha_k=0$ otherwise, for some parameter $1 \leq m \leq n$, in which case everything can be computed explicitly. This can be used as a test case to see how sharp the inequalities you already have are. $\endgroup$
    – Terry Tao
    Jun 10, 2016 at 0:56
  • 2
    $\begingroup$ Good point -- I somehow forgot to include the basic sanity checks. So, for any αα which is uniform on a subset of $m$ elements, then the inverse of the squared $2$-norm will be the right parameter (it's basically the same as the fully uniform case), i.e. the bound will be tight. I'm going to play around for some other simple examples. (Thank you!) $\endgroup$
    – Clement C.
    Jun 10, 2016 at 2:58

1 Answer 1


An almost complete answer. First of all, indeed the $\lVert \alpha\rVert_2$-based bound mentioned in the question can be shown to be tight for many "simple" $\alpha$'s, such as balanced, or uniform/balanced on a subset of $m$ coordinates. However, it is not tight in general, and the right answer appears to be captured by a quantity, the $K$-functional between $\ell_1$ and $\ell_2$.

Converting the problem to that of looking at a weighted sum $Y\stackrel{\rm def}{=}\sum_{k=1}^n \alpha_k Y_k$ of Rademacher random variables $Y_1,\dots, Y_n$ (where $Y_k = 1-2X_k$), $$ \mathbb{P}\{ X \leq \varepsilon \} = \mathbb{P}\{ Y \geq 1-2\varepsilon \}. $$ In Montgomery-Smith [1], the following is shown:

Theorem. For $c\stackrel{\rm def}{=} \frac{1}{4\ln 12}\simeq \frac{1}{10}$, we have $$\begin{align} \mathbb{P}\{ Y > K_{1,2}(\alpha,t) \} &\leq e^{-\frac{t^2}{2}}, \tag{1} \\ \mathbb{P}\{ Y > cK_{1,2}(\alpha,t) \} &\geq ce^{-\frac{t^2}{c}} \tag{2} \end{align}$$ where $K_{1,2} (\alpha,t) \stackrel{\rm def}{=} \inf\{\lVert \alpha'\rVert_1+t\lVert \alpha''\rVert_2 \ \colon\ \alpha',\alpha'' \in \ell_2,\ \alpha'+\alpha'' = \alpha\}$.

Moreover, the following bound holds for $K_{1,2}(\alpha,t)$, ([1], citing Theorem 4.2 of [2]): assuming wlog that $\alpha_1 \geq \alpha_2 \geq \dots\geq \alpha_n$, $$ \gamma K_{1,2}(\alpha,t) \leq \sum_{k=1}^{\lfloor t^2\rfloor} \alpha_k + \left(\sum_{k=\lfloor t^2\rfloor+1}^n \alpha_k^2\right)^{\frac{1}{2}} \leq K_{1,2}(\alpha,t) \tag{3} $$ for some absolute constant $\gamma \in (0,1)$.

This gives the following corollary, that can be found in [1]:

Corollary. With $c$ as above, we have that for all $\alpha\in\ell_2$ (which is trivial in our case as $\alpha$ has finite support) and $0 < t \leq c\frac{\lVert \alpha\rVert_2^2}{\lVert \alpha\rVert_\infty}$, $$\begin{align} ce^{-\frac{t^2}{c^3\lVert \alpha\rVert_2^2}} \leq \mathbb{P}\{ Y >t \} \leq e^{-\frac{t^2}{2\lVert \alpha\rVert_2^2}}, \tag{4} \end{align}$$ where the upper bound holds for all $t>0$.

Since the range of interest is $t\in(0,1]$, the above shows that the $e^{-\Theta\left(\frac{t^2}{\lVert \alpha\rVert_2^2}\right)}$ dependence is tight as long as $\lVert \alpha\rVert_2^2$ and $\lVert \alpha\rVert_\infty$ are within constant factors, which happens to be the case for $\alpha$ roughly balanced or uniform on a subset of coordinates, for instance.

Now, for a counter example, one can consider (a variant of) the "truncated random Harmonic series," where we set, for $1\leq k\leq n$, $$ \alpha_k \stackrel{\rm def}{=} \frac{1}{k H_n} \operatorname*{\sim}_{n\to \infty} \frac{1}{k\ln n}. $$

Then, using techniques similar as [1], one can show that $$ \mathbb{P}\{ Y > t \} \leq e^{-\Theta(n^t)} $$ for $t\in(0,1)$. However, $ \lVert \alpha\rVert_2^2 {\displaystyle\operatorname*{\sim}_{n\to\infty}} \frac{\pi^2}{6\ln^2 n} $, so (4) would only yield $$ \mathbb{P}\{ Y > t \} \leq e^{-\Theta( t^2 \ln^2 n )}. $$

Note that (sanity check) in this case, $\lVert \alpha\rVert_\infty = \frac{1}{H_n}$, so the lower bound of (4) does not apply: $\frac{\lVert \alpha\rVert_2^2}{\lVert \alpha\rVert_\infty} {\displaystyle\operatorname*{\sim}_{n\to\infty}} \frac{1}{\ln n} = o(1)$.

Edit: based on results from [3] (adapting and generalizing the proof of Theorem 2.2), one can get the following refinement of (2) above:

Theorem. Fix any $c>0$. Then, for any $\alpha\in\ell_1+\ell_2$, we have that for all $t\geq 1$, $$ \mathbb{P}\left\{ Y \geq \frac{1}{1+c}K_{1,2}(\alpha,t) \right\} \geq e^{-\left( \frac{2}{c}\ln\frac{\sqrt{6}(1+c)}{c}\right) (t+c)^2 }. $$ In particular, $$ \mathbb{P}\left\{ Y \geq \frac{1}{2}K_{1,2}(\alpha,t) \right\} \geq e^{-\left( \ln 24 \right) (t+1)^2 } \geq e^{-\left( 4\ln 24 \right)t^2 }. $$

[1] Montgomery-Smith S.J. (1990) The distribution of Rademacher sums. Proc. Amer. Math. Soc. 109:517522

[2] Holmstedt, Tord. (1970) Interpolation of quasi-normed spaces. Math. Scand. 26, 177–199.

[3] Astashkin, S.V. (2010) Rademacher functions in symmetric spaces Journal of Mathematical Sciences, 169(6):725–886


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