Concentration inequality without variance

Question

Let $X$ be a positive random variable with $\Bbb E X \leq M$. I would like to compute the expectation using Monte-Carlo method, so I am looking for the bounds on $\Bbb P(|\bar X_n - \Bbb E X| \geq \epsilon)$ where $\bar X_n$ is a sample average of $n$ i.i.d. samples of $X$. Chernoff inequality provides such bounds in case I can upped bound the variance, however I only can upper bound expectation with $M$. Is there a version of Chernoff inequality for that case?

Answer 1

There cannot be such a bound, since the constraint is compatible with arbitrarily high variance and hence, by the central limit theorem, high deviation probabilities for arbitrarily high $n$. The variance of $X$ with $P(X=0)=1-\epsilon$ and $P(X=M/\epsilon)=\epsilon$, with expectation $M$, is

$$ (1-\epsilon)M^2+\epsilon M^2\left(\frac1\epsilon-1\right)^2=M^2\left(\frac1\epsilon+O(1)\right)\;. $$

(This $X$ is merely non-negative, but the modificiation for positive $X$ is trivial.)