Sign up ×
Theoretical Computer Science Stack Exchange is a question and answer site for theoretical computer scientists and researchers in related fields. It's 100% free, no registration required.

Is there an reverse Chernoff bound which bounds that the tail probability is at least so much.

i.e if $X_1,X_2,\ldots,X_n$ are independent binomial random variables and $\mu=\mathbb{E}[\sum_{i=1}^n X_i]$. Then can we prove $Pr[\sum_{i=1}^n X_i\geq (1+\delta)\mu]\geq f(\mu,\delta,n)$ for some function $f$.

share|cite|improve this question
Your example is asking too much: with $p=n^{-2/3}$, a standard Chernoff bound shows that $\Pr[|T\cap S_1| \geq \sqrt{1.1}n^{1/3}]$ and $\Pr[|T\cap S_2|\sqrt{1.1}\leq n^{1/3}]$ are at most $\exp(-cn^{1/3})$ for some $c$. – Colin McQuillan Nov 25 '12 at 21:19
You are right, I got confused about which term in chernoff bound has the square. I have changed the question to reflect a weaker bound. I don't think it will help me in my current application, but it might be interesting for other reasons. – Ashwinkumar B V Nov 25 '12 at 21:50

4 Answers 4

up vote 14 down vote accepted

Here is an explicit proof that a standard Chernoff bound is tight up to constant factors in the exponent for a particular range of the parameters. (In particular, whenever the variables are 0 or 1, and 1 with probability 1/2 or less, and $\epsilon\in(0,1/2)$, and the Chernoff upper bound is less than a constant.)

If you find a mistake, please let me know.

Lemma 1. (tightness of Chernoff bound) Let $X$ be the average of $k$ independent, 0/1 random variables (r.v.). For any $\epsilon\in(0,1/2]$ and $p\in(0,1/2]$, assuming $\epsilon^2 p k \ge 3$,

(i) If each r.v. is 1 with probability at most $p$, then $$\displaystyle \Pr[X\le (1-\epsilon)p] ~\ge~ \exp\big({-9\epsilon^2 pk}\big).$$

(ii) If each r.v. is 1 with probability at least $p$, then $$\displaystyle \Pr[X\ge (1+\epsilon)p] ~\ge~ \exp\big({-9\epsilon^2 pk}\big).$$

Proof. We use the following observation:

Claim 1. If $1\le \ell \le k-1$, then $\displaystyle {k \choose \ell} ~\ge~ \frac{1}{e\sqrt{2\pi\ell}} \Big(\frac{k}{\ell}\Big)^{\ell} \Big(\frac{k}{k-\ell}\Big)^{k-\ell}$

Proof of Claim 1. By Stirling's approximation, $i!=\sqrt{2\pi i}(i/e)^ie^\lambda$ where $\lambda\in[1/(12i+1),1/12i].$

Thus, $k\choose \ell$, which is $\frac{k!}{\ell! (k-\ell)!}$, is at least $$ \frac{\sqrt{2\pi k}\,(\frac{k}{e})^k} { \sqrt{2\pi \ell}\,(\frac{\ell}{e})^\ell ~~\sqrt{2\pi (k-\ell)}\,(\frac{k-\ell}{e})^{k-\ell} } \exp\Big(\frac{1}{12k+1} - \frac{1}{12\ell} - \frac{1}{12(k-\ell)}\Big)$$ $$ ~\ge~ \frac{1}{\sqrt{2\pi\ell}} \Big(\frac{k}{\ell}\Big)^{\ell} \Big(\frac{k}{k-\ell}\Big)^{k-\ell}e^{-1}. $$ QED

Proof of Lemma 1 Part (i). Without loss of generality assume each 0/1 random variable in the sum $X$ is 1 with probability exactly $p$. Note $\Pr[X\le (1-\epsilon)p]$ equals the sum $\sum_{i = 0}^{\lfloor(1-\epsilon)pk\rfloor} \Pr[X=i/k]$, and $\Pr[X=i/k] = {k \choose i} p^i (1-p)^{k-i}$.

Fix $\ell = \lfloor(1-2\epsilon)pk\rfloor+1$. The terms in the sum are increasing, so the terms with index $i\ge\ell$ each have value at least $\Pr[X=\ell/k]$, so their sum has total value at least $(\epsilon pk - 2) \Pr[X=\ell/k]$. To complete the proof, we show that $$(\epsilon pk - 2) \Pr[X=\ell/k] ~\ge~ \exp({-9\epsilon^2 pk}).$$

The assumptions $\epsilon^2pk\ge 3$ and $\epsilon\le 1/2$ give $\epsilon pk \ge 6$, so the left-hand side above is at least $\frac{2}{3}\epsilon pk\, {k \choose \ell} p^\ell(1-p)^{k-\ell}$. Using Claim 1, to bound $k\choose \ell$, this is in turn at least $A\, B$ where $A = \frac{2}{3e}\epsilon p k/ \sqrt{2\pi \ell}$ and $ B= \big(\frac{k}{\ell}\big)^\ell \big(\frac{k}{k-\ell}\big)^{k-\ell} p^\ell (1-p)^{k-\ell}. $

To finish we show $A\ge \exp(-\epsilon^2pk)$ and $B \ge \exp(-8\epsilon^2 pk)$.

Claim 2. $A \ge \exp({-\epsilon^2 pk})$

Proof of Claim 2. The assumptions $\epsilon^2 pk \ge 3$ and $\epsilon\le 1/2$ imply (i) $pk\ge 12$.

By definition, $\ell \le pk + 1$. By (i), $p k \ge 12$. Thus, (ii) $\ell \,\le\, 1.1 pk$.

Substituting the right-hand side of (ii) for $\ell$ in $A$ gives (iii) $A \ge \frac{2}{3e} \epsilon \sqrt{p k / 2.2\pi}$.

The assumption, $\epsilon^2 pk \ge 3$, implies $\epsilon\sqrt{ pk} \ge \sqrt 3$, which with (iii) gives (iv) $A \ge \frac{2}{3e}\sqrt{3/2.2\pi} \ge 0.1$.

From $\epsilon^2pk \ge 3$ it follows that (v) $\exp(-\epsilon^2pk) \le \exp(-3) \le 0.04$.

(iv) and (v) together give the claim. QED

Claim 3. $B\ge \exp({-8\epsilon^2 pk})$.

Proof of Claim 3. Fix $\delta$ such that $\ell=(1-\delta)pk$.
The choice of $\ell$ implies $\delta\le 2\epsilon$, so the claim will hold as long as $B \ge \exp(-2\delta^2pk)$. Taking each side of this latter inequality to the power $-1/\ell$ and simplifying, it is equivalent to $$ \frac{\ell}{p k} \Big(\frac{k-\ell}{(1-p) k}\Big)^{k/\ell-1} ~\le~ \exp\Big(\frac{2\delta^2 pk}{\ell}\Big). $$ Substituting $\ell= (1-\delta)pk$ and simplifying, it is equivalent to $$ (1-\delta) \Big(1+\frac{\delta p}{1-p}\Big)^{\displaystyle \frac{1}{(1-\delta)p}-1} ~\le~ \exp\Big(\frac{2\delta^2}{1-\delta}\Big). $$ Taking the logarithm of both sides and using $\ln(1+z)\le z$ twice, it will hold as long as $$ -\delta\, +\,\frac{\delta p}{1-p}\Big(\frac{1}{(1-\delta)p}-1\Big) ~\le~ \frac{2\delta^2}{1-\delta}. $$ The left-hand side above simplifies to $\delta^2/\,(1-p)(1-\delta)$, which is less than $2\delta^2/(1-\delta)$ because $p\le 1/2$. QED

Claims 2 and 3 imply $A B \ge \exp({-\epsilon^2pk})\exp({- 8\epsilon^2pk})$. This implies part (i) of the lemma.

Proof of Lemma 1 Part (ii). Without loss of generality assume each random variable is $1$ with probability exactly $p$.

Note $\Pr[X\ge (1+\epsilon)p] = \sum_{i = \lceil(1-\epsilon)pk\rceil}^n \Pr[X=i/k]$. Fix $\hat\ell = \lceil (1+2\epsilon)pk \rceil - 1$.

The last $\epsilon pk$ terms in the sum total at least $(\epsilon pk-2)\Pr[X=\hat\ell/k]$, which is at least $\exp({-9\epsilon^2 pk})$. (The proof of that is the same as for (i), except with $\ell$ replaced by $\hat\ell$ and $\delta$ replaced by $-\hat\delta$ such that $\hat\ell = (1+\hat\delta)pk$.) QED

share|cite|improve this answer

The Berry-Esseen theorem can give tail probability lower bounds, as long as they are higher than $n^{-1/2}$.

Another tool you can use is the Paley-Zygmund inequality. It implies that for any even integer $k$, and any real-valued random variable $X$,

$$ \Pr[|X| >= \frac{1}{2}(\mathbb{E}[X^k])^{1/k}] \geq \frac{\mathbb{E}[X^k]^2}{4\mathbb{E}[X^{2k}]} $$

Together with the multinomial theorem, for $X$ a sum of $n$ rademacher random variables Paley-Zygmund can get you pretty strong lower bounds. Also it works with bounded-independence random variables. For example you easily get that the sum of $n$ 4-wise independent $\pm 1$ random variables is $\Omega(\sqrt{n})$ with constant probability.

share|cite|improve this answer

The de Moivre-Laplace Theorem shows that variables like $|T\cap S_1|$, after being suitably normalised and under certain conditions, will converge in distribution to a normal distribution. That's enough if you want constant lower bounds.

For lower bounds like $n^{-c}$, you need a slightly finer tool. Here's one reference I know of (but only by accident - I've never had the opportunity to use such an inequality myself). Some explicit lower bounds on tail probabilities of binomial distributions are given as Theorem 1.5 the book Random graphs by Béla Bollobás, Cambridge, 2nd edition, where further references are given to An introduction to probability and its applications by Feller and Foundations of Probability by Rényi.

share|cite|improve this answer

If you are indeed okay with bounding sums of Bernoulli trials (and not, say, bounded random variables), the following is pretty tight.

Slud's Inequality*. Let $\{X_i\}_{i=1}^n$ be i.i.d. draws from a Bernoulli r.v. with $\mathbb{E}(X_1) = p$, and let integer $k\leq n$ be given. If either (a) $p\leq 1/4$ and $np \leq k$, or (b) $np \leq k \leq n(1-p)$, then $$\text{Pr}\big[\sum_i X_i \geq k\big] \geq 1 - \Phi\left(\frac{k-np}{\sqrt{np(1-p)}}\right),$$ where $\Phi$ is the cdf of a standard normal.

(Treating the argument to $\Phi$ as transforming the standard normal, this agrees exactly with what the CLT tells you; in fact, it tells us that Binomials satisfying the conditions of the theorem will dominate their corresponding Gaussians on upper tails.)

From here, you can use bounds on $\Phi$ to get something nicer. For instance, in Feller's first book, in the section on Gaussians, it is shown for every $z>0$ that $$(1-\varphi(z))(z^{-1} - z^{-3}) < 1-\Phi(z) < (1-\varphi(z))z^{-1},$$ where $\varphi$ is the density of a standard normal. There are similar bounds in the Wikipedia article for "Q-function" as well.

Other than that, and what other people have said, you can also try using the Binomial directly, perhaps with some Stirling.

(*) Some newer statements of Slud's inequality leave out some of these conditions; I've reproduced the one in Slud's paper.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.