From Wikipedia:

The above formula is often unwieldy in practice, so the following looser but more convenient bounds are often used:

(i) $Pr(X\geq (1+\delta)\mu)\leq e^{-\frac{\delta^2\mu}{3}}, 0<\delta<1$

(ii) $Pr(X\leq (1-\delta)\mu)\leq e^{-\frac{\delta^2\mu}{2}}, 0<\delta<1$

The assumption they use is $E[X]=\mu$.

Would (i) still hold if we only assume $E[X]\leq \mu$? Would (ii) still hold if we only assume $E[X]\geq\mu$?

If not, what "practical forms" do we have in these cases?


1 Answer 1


It is more practical just because it has

the advantage of being easier to state and compute with in many situations (quoted from [1]).

I am not sure whether (i) still holds for $\mu' = E[X] \le \mu$. At least, we cannot obtain that directly from the following computation:

$$Pr[X \ge (1 + \delta) \mu] \le Pr[X \ge (1 + \delta) \mu'] \le e^{-\frac{\delta^2 \mu'}{3}} \ge e^{-\frac{\delta^2 \mu}{3}}$$

[1] Mitzenmacher, Michael and Upfal, Eli (2005). Probability and Computing: Randomized Algorithms and Probabilistic Analysis. Cambridge University Press. ISBN 0-521-83540-2.

  • $\begingroup$ You can check what happens in some case where the large deviation behavior is known exactly, say the binomial distribution. $\endgroup$ Apr 26, 2015 at 6:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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