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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?

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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.

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  • $\begingroup$ You can check what happens in some case where the large deviation behavior is known exactly, say the binomial distribution. $\endgroup$ – Yuval Filmus Apr 26 '15 at 6:53

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