I'm working on a problem where I need information on the size of $E_n=|S_n-n\mu|$, where $S_n=X_1+\ldots+X_n$ is a sum of i.i.d. random variables and $\mu=\mathbb EX_1$. For this to make sense, the $(X_i)$ have to be integrable. In that case, the weak law of large numbers says $E_n/n$ converges to 0 in probability, while the strong law says $E_n=o(n)$ almost surely.

If $X_1$ is square integrable, then we get the (stronger) result $E_n/(n^{1/2+\epsilon})$ converges to 0 in probablility. What I am looking for is a result for the case $\mathbb E|X_1|^\alpha<\infty$ for some $1<\alpha<2$. Back-of-the-envelope calculations suggest a result of the form $E_n=O(n^{1/\alpha+\epsilon})$ holds. I suspect a relationship with $\alpha$-stable laws and have tracked down Gnedenko and Kolmogorov's book, but find (1) their notation is quite hard to read; (2) they seem to mainly care about something that's not so important for me; namely a result of the Central Limit Theorem type. They impose extra conditions under which suitably scaled and translated versions of $S_n$ converge to a non-trivial distribution. I don't want to assume anything beyond an $\alpha$th moment inequality, but am looking for a correspondingly weaker conclusion giving upper bounds on $E_n$.

Can anyone point me to some results about deviation from the mean for sums of iid random variables with $\alpha$th moments ($1<\alpha<2$)?
In a similar vein, what if the random variables fail to have an $\alpha$th moment for some $\alpha<1$. Here I'd expect to see some kind of result telling me that if $S_n/a_n$ converges in probability to a constant for some sequence $(a_n)$, then that constant must be 0.

Edit: Let me give a precise Chebyshev-like statement I would really like to be true. Let $(X_n)$ be iid; $\mathbb EX_1=0$ and $\mathbb E|X_1|^\alpha<\infty$ for some $1<\alpha<2$. Then $\mathbb P(|S_n|>t)<C_\alpha n\mathbb E|X_1|^\alpha/t^\alpha$, where $C_\alpha$ is a constant that only depends on $\alpha$ (I have guessed the bound based on "symbol-bashing" with the $\alpha$-stable law and believe it's at least "dimensionally correct".)

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    $\begingroup$ Maybe I'm confused, but doesn't a CLT-type result give you exactly what you want? In the finite-variance case, the classical CLT says that $(S_n - \mu n)/n^{1/2}$ converges weakly, so as a result $E_n/n^{1/2 + \epsilon} \to 0$ weakly and hence in probability (standard result). So a CLT-type result telling you that $(S_n - \mu n)/ n^{1/\alpha}$ converges in distribution would imply that $E_n / n^{1/\alpha + \epsilon} \to 0$ in probability. $\endgroup$ Apr 16, 2015 at 1:55
  • $\begingroup$ @Nate: Don't you need a second moment condition to apply CLT? $\endgroup$ Apr 16, 2015 at 2:05
  • $\begingroup$ Right. In the infinite second moment case, by "CLT-type result" I'm talking about a Kolmogorov-Gnedenko style stable law limit theorem giving the weak convergence of $(S_n - \mu n)/n^{1/\alpha}$ (typically the weak limit is a stable law, not the normal distribution). $\endgroup$ Apr 16, 2015 at 2:06
  • $\begingroup$ Sorry @NateEldredge for the misunderstanding. So yes. I hope that a K-G stable law limit theorem would do the trick. It's just that I did not find a reference for such a theorem that didn't have a bunch of extra conditions (about slowly varying densities etc). If you know where I can find a statement of the type you're mentioning, that will probably resolve (at least the first part of) my issue. $\endgroup$ Apr 16, 2015 at 2:19
  • $\begingroup$ I don't know an exact reference either, but I suspect that with a little work you can show that a distribution satisfying $E|X|^\alpha < \infty$ will satisfy the K-G hypotheses. $\endgroup$ Apr 16, 2015 at 2:21

4 Answers 4


An early occurence of such bounds is in the theorem of Theorem of vonBahr and Eseen

vonBahr, B., Esseen C.-G.: Inequalities for the rth absolute moment of a sum of random variables, $1\leq r \leq 2$. Ann. Math. Statist. 36, No.1, 299-393 (1965).

Theorem: Let $X_i$ be independent (not necessarily i.i.d.) zero mean random variables. Then, for $\alpha\in [1,2]$, $$E|\sum_{i=1}^n X_i|^\alpha\leq C_\alpha \sum_{i=1}^n E|X_i|^\alpha\,.$$ ($C_\alpha$ is explicit)

Applying Markov's inequality in your setup gives $$P(|S_n|>t)\leq C_\alpha n E|X_1|^\alpha t^{-\alpha} $$ as you wanted.

Note: the moment condition certainly does not imply convergence to $\alpha$-stable - one would need some regularly varying conditions on the tail for that.

Note2: A good summary (up to late 70s) of estimates of this kind (mostly from the Russian school) are in Nagaev's Annals of Probability paper (1979). Petrov's 1975 book is also a good resource - in particular the VonBahr-Essen bound is mentioned there (on page 60).


If we assume that the $X_i$ are iid in $L^p$ for some $p\in ]1,2[$, then we have:

$$ a.e. \ \ {1 \over n} \sum_{k=0}^{n−1} X_k=E(X_0)+o(n^{1/p−1})$$

This follows from the Kolmogorov three series theorem. This is done in the book of Durrett, probability, theory and examples, theorem 2.5.8 (now theorem 2.5.12 with the 5th version of the book).

Note that if you are not interested by the exact exponent, then the standard quick $L^2$ proof gives you such an estimate.

  • $\begingroup$ Thanks - but isn't Var$(Y_i)$ infinite? $\endgroup$ Apr 16, 2015 at 6:34
  • $\begingroup$ OK. Thanks a lot. I'll look in Durrett tomorrow. $\endgroup$ Apr 16, 2015 at 6:46
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    $\begingroup$ @quas the book is on his webpage, I have provided a link. Define $Y_i = X_i {\bf 1}_{|X_i|\leq i^{1/p}} / i^{1/p}$, check that $\sum Var(Y_i)$ is finite, this gives a.e. convergence of $\sum Y_i$, then apply the Kronecker lemma and Borel-Cantelli. $\endgroup$
    – coudy
    Apr 16, 2015 at 6:49
  • $\begingroup$ So I did look at Durrett. I have a paper copy that does not seem to include Thm 2.5.8, but the online version is as you say; and also includes an answer to the other question I asked. Thanks again. $\endgroup$ Apr 16, 2015 at 19:13
  • $\begingroup$ I realize this question is a number of years old, now, but do either of you know if it is possible to get a uniform constant for the little-o? $\endgroup$ Feb 23 at 17:59

Your exact question seems to be answered by Lenny Baum and Melvin Katz:

  • $\begingroup$ Thanks @Igor. This does seem very close to my question. I'll see if I can extract a Chebyshev-like result (which is what I really want) from this. $\endgroup$ Apr 16, 2015 at 4:52

I guess what you are looking for is a concentration inequality for heavy-tail distributions. How about this one:

Barthe, F., Cattiaux, P., & Roberto, C. (2005). Concentration for independent random variables with heavy tails. Applied Mathematics Research eXpress, 2005(2), 39-60.

Also, please take a look at the Bernstein inequalities.

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    $\begingroup$ Thanks @Taha. So I think Bernstein inequalities are based on exponential moments (which diverge in my setting); I was unable to pick out useful results from the paper by the French authors either. $\endgroup$ Apr 16, 2015 at 0:29
  • $\begingroup$ Well, I am pretty sure someone should have derived the concentration inequalities for heavy-tail distributions. I tried to look at this: Robust Estimation of High-Dimensional Mean Regression, but their lemmas require me to read the entire paper to understand. I'll look into the robust statistic literature to see what I can find. $\endgroup$
    – Taha
    Apr 16, 2015 at 4:22

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