# Can I apply Chernoff bound to an arbitrary positive random variable?

I have $n$ independent random variables $X_i$. Can I use Chernoff's bound to find a bound on $P(X>\epsilon)$ where $X=\sum_i X_i$?

Assume that I know $E[X_i]$ and $0<X_i<U_i, \forall i$, but I don't know the distribution of $X_i$.

Is there any constraint on $n$? (I don't have any control on $n$. it can be small $n<10$ or large $n>100$.)

[EDIT] [This]1 was my motivation for choosing Chernoff. In page 1, it discusses using Chernoff's bound for any distribution where $0\leq X_i\leq 1$. Later in page 3, it talks about when $X_i>1$. But how can I apply that for my case? and what will be its implication? Is the bound get looser as $U_i$s get larger?

• Assuming $X_i$'s are iid, you can apply Chebychev's inequality to bound: $$\mathbb{P}\left[ |\frac{1}{n} X - \mathbb{E}[X]| \leq \epsilon \right] \geq \frac{\mathbb{V}[X]}{n\epsilon^2}$$ Commented Sep 9, 2013 at 22:33
• Just to clarify, both inequalities should infact be reversed for the above expression to be valid. Commented Sep 9, 2013 at 22:43
• Thanks Dave. But I don't know the variance and the variables are not identically distributed. Actually their mean can be different. Commented Sep 9, 2013 at 22:52

There is no constraint on $n$; as usual, the concentration rate is exponential in terms of $n$.
• Also the Hoeffding's inequality bound does not depend on $\mu$ but Chernoff's does. For my case $\frac{U_i}{\mu_i}$ is about 1000 which is much larger than $n$. Commented Sep 10, 2013 at 3:37