Measure Concentration for Stable Laws with Index Close to 2

We give upper bounds for the probability P(|f (X) − Ef (X)| > x), where X is a stable random variable with index close to 2 and f is a Lipschitz function. While the optimal upper bound is known to be of order 1/x α for large x, we establish, for smaller x, an upper bound of order exp(−x α /2), which relates the result to the gaussian concentration.

For α close to 2, this roughly tells us that the natural (and optimal, up to a multiplicative constant) upper bound L/x α holds for x α of order LM (logM ) 2 .On the other hand, suppose that X is a 1-dimensional, stable random variable and let Y (1) be the infinitely divisible vector whose Lévy measure is the Lévy measure of X truncated at 1. Then it is easy to check that var(Y (1) ) = LM .This clearly indicates that one cannot hope to obtain any interesting inequality if x 2 is much smaller than LM .In fact, when x α is of order LM , another result in [3] gives an upper bound of order cLM/x α .However, comparing this with the bound cL/x α of Theorem 1, we see that there is an important discrepancy when M is large, and so it is natural to investigate the case when x α lies in the range [LM, LM (log M ) 2 ] for large M .Here is our result: Theorem 2 Using the same notations as in Theorem 1, we have: (i) Let a < 1 and a , ε > 0. Then if M is sufficiently large, for every x of the form Then if M is sufficiently large, for every x such that x α > aLM log M , As a consequence of (i), let X (α) be the stable law whose Lévy measure ν is the uniform measure on S d−1 with total mass 1/M .Then since LM = 1, (3) can be rewritten as for x smaller than (log M ) 1/α .When α → 2, X (α) converges in distribution to a standard gaussian variable X , for which we have the following classical bound [1,6], valid for all x > 0: So we see that (4) recovers the result for the gaussian concentration.
Remark that (ii) slightly improves Theorem 1 when the index α is close to 2 and x α is of order To some extent, the existence of two regimes (i) and (ii), depending on the order of magnitude of x with regard to (LM log M ) 1/α , is reminiscent of the famous Talagrand inequality: where U is an infinitely divisible random variable with Lévy measure given by and f is a Lipschitz function, a and b being related to the L 1 and L 2 norm of f , respectively (see [7] for a precise statement).We now proceed to the proof of Theorem 2.

Proof of the result
The proof essentially follows the lines of the proof to be found in [3], where the case x α < LM (log M ) 2 had been overlooked.We write X = Y (R) + Z (R) , where Y (R) , Z (R) are two independent, infinitely divisible random variables whose Lévy measures are the Lévy measure of X truncated, above and below respectively, at R > 0. We have Since Z (R) is a compound Poisson process, it is easy to check that On the other hand, Thus we have to compare Ef (X) and Ef (Y (R) ).For large R, these two quantities are very close, since Given x, we choose R so that which entails that x ≤ R. Therefore we can write Let b be the real such that x α = bLM .Let b be such that R α = b LM , which, according to (8), entails or, equivalently, When M is large, b can be made arbitrarily close to b.To estimate quantities of the type ) ≥ y), we use Theorem 1 in [2], which states that where Using the fact that for s ∈ (0, R), we get the following upper bound for h R (s): See [3] for details of computations.The idea is to compare the two terms in the right-hand side of (11).Typically, for small s, the first term is dominant while for large s, the second term is dominant.
Let us first prove (i).Fix ε, a > 0 and a < 1.If δ, s, R > 0 are three reals satisfying the inequality As a consequence, if y is such that the real s = s(y) defined by It is clear that if s(y) satisfies (12), then for every 0 < y < y, s(y ) also satisfies (12) with the same reals δ and R. Therefore one can integrate (13) and one has: whenever s(y) satisfies (12).If y has the form y α = ALM/(3 − α) with A/(3 − α) < a log M and if we take R = y, Condition (12) becomes For M sufficiently large, this holds whenever Set Given a > 0, if M is large enough, δ(A) > 0 for every A such that a /2 < A < log M , and thus (15) is fulfilled.In that case, since we take R = y, (14) becomes Using the expression of δ, For M large enough, this quantity is bounded by (1 + ε/4)e −b /2 .To sum up, given ε > 0 and a > 0, if M is large enough, then for every b satisfying a /2 < b < log M , writing R α = b LM , we have Remark that given a > 0 and a < 1, if a < b < a log M , then taking b as defined by (9), we have a /2 < b < log M for M large enough and we can apply (17).Hence if x has the form x α = bLM with a < b < a log M , setting R α = b LM , we have for M large enough, This provides an upper bound for the first term of the right-hand side of (5).
To bound the second term of the right-hand side of ( 5), recall (6) and remark that choosing Given a > 0 and a < 1, if b satisfies a < b < a log M , then for M large enough, using again This concludes the proof of (i).
To prove (ii), we shall decompose the integral (10).Fix a > 2, take x of the form x α = bLM log M with b ≥ a and let R = (b LM log M ) 1/α with b given by (9).First let Then for M large enough, the same arguments as for (14) give On the other hand, for M large enough, if sR ≥ log M + log log M , Hence using (11), we have for every u > u 1 , where Now let R = (b LM log M ) 1/α with b given by (9).Then for M sufficiently large, R > u 1 .In that case, we can integrate (19) and this gives where we denote For M large enough, this leads to Finally, since h −1 R is increasing, Together with (18),(20), ( 6) and (9), this yields (ii).