Bounding the tail of an average using the the tail of individual members

Let $X_1,X_2,\ldots,X_n$ be an i.i.d. sequence of $n$ positive random variables with mean $E[X_1]=\mu_X<\infty$ and the second moment $E[X_1^2]=\infty$.

I am interested in upper-bounding $P\left(|\frac{1}{n}\sum_{i=1}^nX_i-\mu_X|\geq x\right)$ (though a lower bound on $P\left(\frac{1}{n}\sum_{i=1}^nX_i-\mu_X \geq x\right)$ would be great as well). My go-to method for this is Chebyshev's inequality, however, the second moment for $X_1$'s (and, hence, the variance) is infinite so that's of no use.

However, I do have an expression for the tail probability of individual member of this i.i.d. sequence, $P(X_1>x)$ in a (more-or-less) nice form. Can I use it to get the desired bound(s) on the sequence average?

• If you cannot use Chebyshev’s inequality (because of the infinite variance) you can always consider applying “truncated Chebyshev inequality” – see e.g. books.google.pl/… – Waldemar Nov 20 '13 at 8:18
• How about the following for a simple lower bound: $$P\left(\frac{1}{n}X_i - \mu_X > x\right)\geq \left[P(X_i - \mu_X > x)\right]^n.$$ This turns out to be sufficiently tight in certain cases, for instance, when $P(X_1 > x)$ goes down exponentially in $x$. – Skoro Nov 23 '13 at 17:45
• @Waldemar Been working with your suggestion before going away for a few days... Haven't gotten the result I need yet. Didn't know about the truncated Cheybyshev's inequality before, it's neat. Thanks for the pointer to the book -- I do realize it's one of the first results when one searches for "truncated Chebyshev's inequality" but it's quite good. – Bullmoose Nov 28 '13 at 4:07
• @Skoro Are there random variables whose tails decay exponentially that have infinite (or undefined) variance? – Bullmoose Nov 28 '13 at 4:08