Moment inequalities for sums of certain independent symmetric random variables

by

P. HITCZENKO* (Raleigh, NC), S. J. MONTGOMERY-SMITH^{†} (Columbia, MO) and K. OLESZKIEWICZ ^{‡} (Warszawa)

Abstract

This paper gives upper and lower bounds for moments of sums of independent random variables (X_{k}) which satisfy the condition that P (|X|_{k} ≥ t) = exp(−N_{k}(t)), where N_{k} are concave functions. As a consequence we obtain precise information about the tail probabilities of linear combinations of independent random variables for which N(t) = |t|^{r} for some fixed 0 < r ≤ 1. This complements work of Gluskin and Kwapień who have done the same for convex functions N.

1. Introduction.

Let X_{1}, X_{2} … be a sequence of independent random variables. In this paper we will be interested in obtaining estimates on the L_{p}-norm of sums of (X_{k}), i.e. we will be interested in the quantity,

∥∑ X_{k}∥_{p} = (E |∑ X_{k}|^{p})^{1/p} , 2 ≤ p < ∞.

Although using standard symmetrization arguments our results carry over to more general cases, we will assume for the sake of simplicity that X_{k} have symmetric distributions, i.e. that P (X_{k} ≤ t) = P (−X_{k} ≤ t) for all t ∈ R.

Let us start by considering the case of linear combinations of identically distributed, independent (i.i.d.) random variables, that is X_{k} = a_{k}Y_{k}, where Y, Y_{1}, … are i.i.d. We can assume without loss of generality that all a_{k}’s are nonnegative. Also, since the Y_{k}’s are i.i.d., we can rearrange the terms of (a_{k}) arbitrarily without affecting the sum ∑ a_{k}Y_{k}. Therefore, for notational convenience, we will adopt the following convention throughout this paper; whenever we are dealing with a sequence of real numbers we will always assume that its terms are nonnegative and form a nonincreasing sequence. In other words we identify a sequence (a_{k}) with the decreasing rearrangement of (|a_{k}|).

Of course, a huge number of inequalities concerning the L_{p}-norm of a sum ∑ a_{k}Y_{k} are known. Let us recall two of them. The first, called Khintchine’s inequality, deals with the Rademacher sequence, i.e. Y = ϵ, where ϵ takes values ±1, each with probability 1/2.

AMS 1991 Subject Classification: 60E15, 60G50

* Supported in part by NSF grant DMS 9401345 † Supported in part by NSF grant DMS 9424396 ‡ Supported in part by Foundation for Polish Science and KBN Grant 2 P301 022 07 For p ≥ 2 it can be formulated as follows (see e.g. Ledoux and Talagrand (1991, Lemma 4.1)); there exists an absolute constant K such that

(∑ a_{k}^{2})^{1/2} ≤ ∥∑ a_{k}ϵ_{k}∥_{p} ≤ K\sqrt{p} (∑ a_{k}^{2})^{1/2} .

The value of the smallest constant that can be put in place of K\sqrt{p} is known (see Haagerup (1982)). The second inequality, was proved by Rosenthal (1970), and in our generality says that for 2 ≤ p < ∞ there exists a constant B_{p} such that for a sequence a = (a_{k}) ∈ ℓ_{2},

max{∥a∥_{2} ∥Y ∥_{2} , ∥a∥_{p} ∥Y ∥_{p}} ≤ ∥∑ a_{k}Y_{k}∥_{p} ≤ B_{p} max{∥a∥_{2} ∥Y ∥_{2} , ∥a∥_{p} ∥Y ∥_{p}},

where ∥a∥_{s} denotes the ℓ_{s}-norm of a sequence a. The best possible constant B_{p} is known (cf. Utev (1985) for p ≥ 4, and Figiel, Hitczenko, Johnson, Schechtman and Zinn (1995) for 2 < p < 4). We refer the reader to the latter paper and references therein for more information on the best constants in Rosenthal’s inequality. For our purpose, we only note that Johnson, Schechtman and Zinn (1983) showed that B_{p} ≤ Kp/ log p for some absolute constant K, and that up to the value of K, this bound cannot be further improved. This, and other related results, were extended in various directions by Pinelis (1994). When specialized to our generality, his inequality reads as

max\{∥a∥_{2} ∥Y ∥_{2} , ∥a∥_{p} ∥Y ∥_{p}\} ≤ ∥∑ a_{k}Y_{k}∥_{p} ≤ inf_{1≤c≤p} max\{\sqrt{c}e^{p/c} ∥a∥_{2} ∥Y ∥_{2} , c ∥a∥_{p} ∥Y ∥_{p}\}.

A common feature of these two inequalities is that they express the L_{p}-norm of a sum ∑ a_{k}Y_{k} in terms of norms of individual terms a_{k}Y_{k}. This is very useful and both inequalities found numerous applications. However, upper and lower bounds in these inequalities differ by a factor that depends on p, and sometimes are quite insensitive to the structure of the coefficient sequence. (Rosenthal’s inequality is also insensitive to the common distribution of Y_{k}’s.) Consider, for example coefficient sequence a_{k} = 1/k, k ≥ 1. Then, the only information on ∥∑ a_{k}ϵ_{k}∥_{p} given by Khintchine’s inequality is that it is essentially between 1 and \sqrt{p}. Practically the same conclusion is given for two quite different sequences, namely a_{k} = 2^{−k}, k ≥ 1, and a_{k} = 1/\sqrt{n} or 0 according to whether k ≤ n or k > n, n ∈ N. (The “true values” of ∥∑ a_{k}ϵ_{k}∥_{p} are rather different in each of these three cases, and are of order log(1 + p), 1, and \sqrt{p ∧ n}, respectively.)

¿From this point of view, it is natural to ask whether more precise information on the size of ∥∑ a_{k}Y_{k}∥_{p} can be obtained. Although, in general the answer to this question may be difficult, there are cases for which there is a satisfactory answer. First of all, if Y_{k} is a standard Gaussian random variable, then ∥∑ a_{k}Y_{k}∥_{p} = ∥a∥_{2} ∥Y ∥_{p} so that

c\sqrt{p} ∥a∥_{2} ≤ ∥∑ a_{k}Y_{k}∥_{p} ≤ C\sqrt{p} ∥a∥_{2} ,

for some absolute constants c and C. Next consider the case when (Y_{k}) = (ϵ_{k}), a Rademacher sequence. We have:

c \{∑_{k≤p} a_{k} + \sqrt{p} (∑_{k>p} a_{k}^{2})^{1/2}\} ≤ ∥∑ a_{k}ϵ_{k}∥_{p} ≤ C \{∑_{k≤p} a_{k} + \sqrt{p} (∑_{k>p} a_{k}^{2})^{1/2}\} .

(Recall, that according to our convention, a_{1} ≥ a_{2} ≥ … ≥ 0.) The above inequality has been established in Hitczenko (1993), although the proof drew heavily on a technique from Montgomery-Smith (1990). (In fact, the inequality for Rademacher variables can be deduced from the results obtained in the latter paper.) The next step was done by Gluskin and Kwapień (1995) who dealt with the case of random variables with logarithmically concave tails. To describe their result precisely, suppose that Y is a symmetric random variable such that for t > 0 one has P (|Y | ≥ t) = exp(−N(t)), where N is an Orlicz function (i.e. convex, nondecreasing , and N(0) = 0). Recall, that if M is an Orlicz function and (a_{k}) is a sequence of scalars then the Orlicz norm of (a_{k}) is defined by ∥(a_{k})∥_{M} = inf{u > 0 : ∑ M(a_{k}/u) ≤ 1}. Let N′ be a function conjugate to N i.e. N′(t) = sup{st − N(s) : s > 0}, and put M_{p}(t) = N′(pt)/p. Then Gluskin and Kwapień (1995) proved that

c \{∥(a_{k})_{k≤p}∥_{M_{p}} + \sqrt{p} ∥(a_{k})_{k>p}∥_{2}\} ≤ ∥∑ a_{k}Y_{k}∥_{p} ≤ C \{∥(a_{k})_{k≤p}∥_{M_{p}} + \sqrt{p} ∥(a_{k})_{k>p}∥_{2}\} .

In the special case N(t) = |t|^{r}, r ≥ 1, this gives

\begin{matrix}c\{ ∥(a_{k})_{k≤p}∥_{r′} ∥Y ∥_{p} + & \sqrt{p} ∥(a_{k})_{k>p}∥_{2} ∥Y ∥_{2} \} ≤ ∥∑ a_{k}Y_{k}∥_{p} \\ ≤ & C\{ ∥(a_{k})_{k≤p}∥_{r′} ∥Y ∥_{p} + \sqrt{p} ∥(a_{k})_{k>p}∥_{2} ∥Y ∥_{2} \}, \end{matrix}

where 1/r′ + 1/r = 1, and c, C are absolute constants. From now on, we will refer to random variables corresponding to N(t) = |t|^{r} as symmetric Weibull random variables with parameter r.

Now let us describe the results of this paper. We will say that a symmetric random variable has a logarithmically convex tail if P (|X| ≥ t) = exp(−N(t)) for t ≥ 0, where N : R_{+} → R_{+} is a concave function with N(0) = 0. We will show the following result.

Theorem 1.1. There exist absolute constants c and K such that if (X_{k}) is a sequence of independent random variables with logarithmically convex tails, then for each p ≥ 2,

\begin{matrix}c\{(∑ ∥X_{k}∥_{p}^{p})^{1/p}+ & \sqrt{p}(∑ ∥X_{k}∥_{2}^{2})^{1/2}\} \\ ≤ & ∥ ∑ X_{k}∥_{p} ≤ K\{(∑ ∥X_{k}∥_{p}^{p})^{1/p} + \sqrt{p}(∑ ∥X_{k}∥_{2}^{2})^{1/2}\}. \end{matrix}

This result includes the special case of linear combinations of symmetric Weibull random variables with fixed parameter r ≤ 1, and in this case we are also able to obtain information about the tail distributions of the sums. In fact, we will give a second proof of the moment inequalities in this special case. We include this second proof, because we believe that the methods will be of great interest to the specialists.

Let us mention that the main difficulty is to obtain tight upper bounds. Once this is done the lower bounds are rather easy to prove. For this reason, the main emphasis of this paper is on the proofs of upper bounds.

We will now outline the main steps in both proofs of the upper bounds, and at the same time describe the organization of this paper. In Section 2, we describe a weaker version of the result that depends upon a certain moment condition. In particular, if we specialise to linear combinations of symmetric Weibull random variables, we obtain constants that become unbounded as the parameter r tends to zero (but are still universal in p ≥ 2). The proof is based on hypercontractive methods that were suggested to us by Kwapień. In Section 3 we will show that moments of sums of symmetric random variables with logarithmeically convex tails are dominated by the linear combinations of suitably chosen multiples of standard exponential random variables by independent symmetric three valued random variables. We are then able to obtain the main result.

The main idea of the second proof is to reduce the problem to the situation to the case of an i.i.d. sum. This is done in the Section 4. In Section 5 we deal with the problem of finding an upper bound for the L_{p}-norm of a sum of i.i.d. random variables. We accomplish this by estimating from above a decreasing rearrangement of the sum in terms of a decreasing rearrangement of an individual summand. In Section 6 we apply the results from the preceding two sections to obtain an upper bound on the L_{p}-norm of ∑ a_{k}X_{k}, where (X_{k}) are i.i.d. symmetric with P (|X| > t) = exp(−t^{r}), 0 < r ≤ 1.

2. Random variables satisfying moment condition. In this section we will prove a variant of Theorem 1.1, where the random variables satisfy a certain moment condition. We use hypercontractive methods, similar to those in Kwapień and Szulga (1991). This approach was suggested to us by S. Kwapień. It should be emphasized that if the result is specialized to symmetric Weibull random variables then the constant B in Theorem 2.2 below depends on r. Let us begin with the following.

Definition 2.1. We say that a random variable X satisfies moment condition if it is symmetric, all moments of X are finite and there exist positive constants b, c such that for all even natural numbers n ≥ k ≥ 2

\frac{1}{b} \frac{n}{k} ≤ \frac{∥X∥_{n}}{∥X∥_{k}} ≤ c^{n−k}.

It is easy to check that the existence of such a c is equivalent to the finiteness of

sup_{n∈N}(E|X|^{n})^{1/n^{2}}

or

lim sup_{t→∞}(ln t)^{−2} ln(P (|X| > t) < 0.

Here is the main result of this section.

Theorem 2.2. If independent real random variables X_{1}, X_{2}, … satisfy moment condition with the same constants b and c then there exist positive constants A and B such that for any p ≥ 2, any natural number m, and S = ∑_{i=1}^{m} X_{i}, the following inequalities hold:

A(\sqrt{p}∥S∥_{2} + ( ∑_{k=1}^{m} ∥X_{k}∥_{p}^{p})^{1/p}) ≤ ∥S∥_{p} ≤ B(\sqrt{p}∥S∥_{2} + ( ∑_{k=1}^{m} ∥X_{k}∥_{p}^{p})^{1/p}).

We will prove a lemma first.

Lemma 2.3. If X satisfies moment condition then there exists positive constant K such that for any q ≥ 1 and any real x the following inequality holds true:

(E|x + X|^{2q})^{1/q} ≤ 4eEX^{2}q + (|x|^{2q} + K^{2q}E|X|^{2q})^{1/q}.

Proof: Put K = \sqrt{2}ebc^{4} + 1. Let n be the least natural number greater than q. We consider two cases:

(i) |x| ≤ (K − 1)∥X∥_{2q}. Then ∥x + X∥_{2q} ≤ |x| + ∥X∥_{2q} ≤ K∥X∥_{2q}. (ii) |x| ≥ (K − 1)∥X∥_{2q} ≥ \frac{K−1}{c^{2}}∥X∥_{2n}. Then

\begin{matrix}E(x + X)^{2n} − x^{2n} = & ∑_{k=1}^{n} \frac{2n(2n − 1) … (2n − 2k + 1)}{(2k)!}x^{2n−2k}EX^{2k} \\ ≤ & 4n^{2}x^{2n−2} ∑_{k=1}^{n} \frac{(2n)^{2k−2}∥X∥_{2k}^{2k}}{(2k)!x^{2k−2}} \\ ≤ & 4n^{2}x^{2n−2} ∑_{k=1}^{n} ∥X∥_{2k}^{2} \frac{(2n)^{2k−2}}{(2k)!(\frac{K−1}{c^{2}})^{2k−2}} (\frac{∥X∥_{2k}}{∥X∥_{2n}})^{2k−2} \\ ≤ & 4n^{2}x^{2n−2} ∑_{k=1}^{n}(∥X∥_{2}c^{2k−2})^{2} \frac{(2n)^{2k−2}(\frac{bk}{n})^{2k−2}}{(2k)!(\frac{K−1}{c^{2}})^{2k−2}} \\ = & 4n^{2}x^{2n−2} ∑_{k=1}^{n} \frac{EX^{2}}{(2k)!} (\frac{2bc^{4}k}{K − 1})^{2k−2} \\ ≤ & n^{2}x^{2n−2}eEX^{2} ∑_{k=1}^{n} (\frac{ebc^{4}}{K − 1})^{2k−2} \\ ≤ & 2eEX^{2}n^{2}x^{2n−2}. \end{matrix}

Hence

∥x + X∥_{2q} ≤ ∥x + X∥_{2n} ≤ \sqrt{x^{2} + 2eEX^{2}n} ≤ \sqrt{x^{2} + 4eEX^{2}q}.

This completes the proof of Lemma 2.3.

Proof of Theorem 2.2: We begin with the left hand side inequality. The inequality

∥S∥_{p} ≥ (∑_{k=1}^{m} E|X_{k}|^{p})^{1/p}

follows from Rosenthal’s inequality (Rosenthal (1970)), or by an easy induction argument and the inequality 2(|x|^{p} + |y|^{p}) ≤ |x + y|^{p} + |x − y|^{p}. To complete the proof of this part we will show that if X is a random variable satisfying the left hand side inequality in Definition 2.1 with constant b then

∥S∥_{p} ≥ \frac{1}{2\sqrt{2}b}∥S∥_{2}\sqrt{p}.

To this end let G be a standard Gaussian random variable. Then, for every even natural number n we have

\frac{∥X∥_{2}}{2b} ∥G∥_{n} = ((n − 1)!!)^{1/n} \frac{∥X∥_{2}}{2b} ≤ n\frac{∥X∥_{2}}{2b} ≤ ∥X∥_{n}.

Hence, by the binomial formula, for every even natural number n and any real number x

E|x + \frac{∥X∥_{2}}{2b} G|^{n} ≤ E|x + X|^{n}.

If G_{k}’s are i.i.d. copies of G then, by an easy induction argument, we get that

E|x + \frac{∥S∥_{2}}{2b} G|^{n} = E|x + ∑ \frac{∥X_{k}∥_{2}}{2b} G_{k}|^{n} ≤ E|x + S|^{n}.

Letting n be the largest even number not exceeding p, and putting x = 0 we obtain

∥S∥_{p} ≥ ∥S∥_{n} ≥ \frac{∥S∥_{2}}{2b} ∥G∥_{n} ≥ \frac{∥S∥_{2}}{2b} \sqrt{n} ≥ \frac{1}{2\sqrt{2}b}∥S∥_{2}\sqrt{p},

which completes the proof of the first inequality of Theorem 2.2.

To prove the second inequality we will proceed by induction. For N = 1, 2, …, m, let S_{N} = ∑_{k=1}^{N} X_{k}, h_{N} = K(∑_{k=1}^{N} E|X_{k}|^{p})^{1/p}, and q = p/2. We will show that for any real x

(E|x + S_{N}|^{2q})^{1/q} ≤ 4eqES_{N}^{2} + (|x|^{2q} + h_{N}^{2q})^{1/q}.

For N = 1 this is just Lemma 2.3. Assume that the above inequality is satisfied for N < m. Let E′ = E( · |X_{1}, X_{2}, …, X_{N}), E′′ = E( · |X_{N+1}). Then

(E|x + S_{N+1}|^{2q})^{1/q} = (E′′E′|(x + S_{N}) + X_{N+1}|^{2q})^{1/q} ≤

by Lemma 2.3

\begin{matrix}≤ & (E′(4eqEX_{N+1}^{2} + (|x + S_{N}|^{2q} + K^{2q}E|X_{N+1}|^{2q})^{1/q})^{q})^{1/q} \\ ≤ & 4eqEX_{N+1}^{2} + (E′|x + S_{N}|^{2q} + K^{2q}E|X_{N+1}|^{2q})^{1/q} \end{matrix}

by the inductive assumption

\begin{matrix}≤ 4eqEX_{N+1}^{2}+ & ((4eqES_{N}^{2} + (|x|^{2q} + h_{N}^{2q})^{1/q})^{q} + K^{2q}E|X_{N+1}|^{2q})^{1/q} \\ = & 4eqEX_{N+1}^{2} + ∥(4eqES_{N}^{2} + (|x|^{2q} + h_{N}^{2q})^{1/q}, K^{2}(E|X_{N+1}|^{2q})^{1/q})∥_{q} \\ ≤ & 4eqEX_{N+1}^{2} + ∥(4eqES_{N}^{2}, 0)∥_{q} + ∥((|x|^{2q} + h_{N}^{2q})^{1/q}, K^{2}(E|X_{N+1}|^{2q})^{1/q})∥_{q} \\ = & 4eqES_{N+1}^{2} + (|x|^{2q} + h_{N+1}^{2q})^{1/q}. \end{matrix}

Our induction is finished. Taking N = m and x = 0 we have

∥S∥_{p} ≤ \sqrt{2epES^{2} + h_{m}^{2}} ≤ \sqrt{2e}∥S∥_{2}\sqrt{p} + K(∑_{k=1}^{m} E|X_{k}|^{p})^{1/p}.

This completes the proof of Theorem 2.2.

3. Random variables with logarithmically convex tails. The aim of this section is to prove the main result, Theorem 1.1. Although random variables with logarithmically convex tails do not have to satisfy moment condition we have the following.

Proposition 3.1. Let Γ be an exponential random variable with density e^{−x} for x ≥ 0. If X is a symmetric random variable with logarithmically convex tails then for any p ≥ q > 0 the following inequality is satisfied:

\frac{∥X∥_{p}}{∥Γ∥_{p}} ≥ \frac{∥X∥_{q}}{∥Γ∥_{q}} .

In particular, random variables with logarithmically convex tails satisfy the first inequality of Definition 2.1 with constant b = e/\sqrt{2}.

Proof: Let F = N^{−1}. Then |X| has the same distribution as F (Γ). Since N is a concave function and N(0) = 0, it follows that N(x)/x is nonincreasing. Therefore, f(x) = F (x)/x is nondecreasing. By a standard inequality for a measure x^{p}e^{−x}dx we have

(\frac{∫_{0}^{∞} f^{p}(x)x^{p}e^{−x}dx}{∫_{0}^{∞} x^{p}e^{−x}dx})^{1/p} ≥ (\frac{∫_{0}^{∞} f^{q}(x)x^{p}e^{−x}dx}{∫_{0}^{∞} x^{p}e^{−x}dx})^{1/q},

so it suffices to prove that

\frac{∫_{0}^{∞} f(x)^{q}x^{p}e^{−x}dx}{∫_{0}^{∞} x^{p}e^{−x}dx} ≥ \frac{∫_{0}^{∞} f(x)^{q}x^{q}e^{−x}dx}{∫_{0}^{∞} x^{q}e^{−x}dx}.

Since f^{q}(x) is a nondecreasing function, it is enough to show that

\frac{∫_{a}^{∞} x^{p}e^{−x}dx}{Γ(p + 1)} ≥ \frac{∫_{a}^{∞} x^{q}e^{−x}dx}{Γ(q + 1)}, a ≥ 0.

Let

h(a) = \frac{∫_{a}^{∞} x^{p}e^{−x}dx}{Γ(p + 1)} − \frac{∫_{a}^{∞} x^{q}e^{−x}dx}{Γ(q + 1)}.

Then, h(0) = 0 and lim_{a→∞} h(a) = 0. Since h′(a) is positive on (0, a_{0}), and negative on (a_{0}, ∞), where a_{0} = (\frac{Γ(p+1)}{Γ(q+1)})^{1/(p−q)}, we conclude that h(a) ≥ 0. To see that b = e/\sqrt{2}, notice that the sequence (n/∥Γ∥_{n}) = ((n^{n}/n!)^{1/n}) is increasing to e. Therefore, since 2/∥Γ∥_{2} = \sqrt{2},

\frac{∥X∥_{n}}{∥X∥_{k}} ≥ \frac{∥Γ∥_{n}}{∥Γ∥_{k}} ≥ \frac{n}{k} \frac{\sqrt{2}}{e}.

This completes the proof.

Since the left hand side inequality in Theorem 1.1 follows by exactly the same argument as in Theorem 2.3, we will concentrate on the right hand side inequality. We first establish a comparison result which may be of its own interest. Let Γ be an exponential random variable with mean 1. For a random variable X with logarithmically convex tails, and p > 2, we denote by E_{p}(X) a random variable distributed like aΘΓ, where Θ and Γ are independent, Θ is symmetric, 3-valued (i.e. P (Θ = ±1) = α/2, P (Θ = 0) = 1 − α), and a, α are chosen so that ∥X∥_{2} = ∥E_{p}(X)∥_{2} and ∥X∥_{p} = ∥E_{p}(X)∥_{p}. Proposition 3.1. guarantees that such a and α exist. We begin with the following lemma.

Lemma 3.2. There exists an absolute constant c such that for any random variable with logarithmically convex tails X and for every 2 ≤ q < p < ∞ one has

∥X∥_{q} ≤ c∥E_{p}(X)∥_{q}.

Proof: Replacing X by X/a we can assume a = 1. Assume first that 4 ≤ q ≤ p/2. We have

∥X∥_{q}^{q} = q ∫_{0}^{∞} t^{q−1}e^{−N(t)}dt = q( ∫_{0}^{2} + ∫_{2}^{p} + ∫_{p}^{∞} )t^{q−1}e^{−N(t)}dt.

We will estimate each integral separately. By Markov’s inequality, for every t > 0 we have

(∗)  t^{p}e^{−N(t)} ≤ ∥X∥_{p}^{p} = ∥Θ∥_{p}^{p}∥Γ∥_{p}^{p}.

Hence,

∫_{p}^{∞} t^{q−1}e^{−N(t)}dt ≤∥Θ∥_{p}^{p}∥Γ∥_{p}^{p} ∫_{p}^{∞} t^{q−1−p}dt ≤ ∥Θ∥_{p}^{p}Γ(p + 1)\frac{p^{q−p}}{p − q}
≤∥Θ∥_{q}^{q} \frac{2^{p}p^{q}}{e^{p}} ≤ K^{q}∥Θ∥_{q}^{q}q^{q} ≤ K^{q}∥Θ∥_{q}^{q}∥Γ∥_{q}^{q}.

We estimate the first integral in a similar way; since for every t > 0

(_{∗}^{∗})  t^{2}e^{−N(t)} ≤ ∥Θ∥_{2}^{2}∥Γ∥_{2}^{2},

we get

∫_{0}^{2} t^{q−1}e^{−N(t)}dt ≤ ∥Θ∥_{2}^{2}∥Γ∥_{2}^{2} ∫_{0}^{2} t^{q−3}dt ≤ C^{q}∥Θ∥_{2}^{2}∥Γ∥_{q}^{q} = C^{q}∥Θ∥_{q}^{q}∥Γ∥_{q}^{q}.

It remains to estimate the middle integral. Notice, that (∗) with t = p and (_{∗}^{∗}) with t = 2 imply that

N(p) ≥ p ln(p/∥Γ∥_{p}) − ln α ≥ p ln(e/2) − ln α,

and

N(2) ≥ 2 ln(e/2) − ln α.

Hence, by concavity of N, we infer that N(t) ≥ t ln(e/2)−ln α, for 2 ≤ t ≤ p. Consequently,

∫_{2}^{p} t^{q−1}e^{−N(t)}dt ≤ α ∫_{0}^{∞} t^{q−1}e^{−t ln(e/2)}dt ≤ K^{q}∥Θ∥_{q}^{q}∥Γ∥_{q}^{q}.

Now consider the case 2 ≤ q < 4. Since for q < p, ∥Γ∥_{q} ≥ kq/p∥Γ∥_{p} for some absolute constant k > 0, and for Θ Hölder’s inequality is in fact equality, we obtain the following. If 0 < s < 1 is chosen so that 1/q = s/(2q) + (1 − s)/2, then

∥ΘΓ∥_{q} =∥Θ∥_{2q}^{s}∥Γ∥_{q}^{s}∥Θ∥_{2}^{1−s}∥Γ∥_{q}^{1−s} ≥ ∥Θ∥_{2q}^{s}(\frac{k}{2} ∥Γ∥_{2q})^{s}∥Θ∥_{2}^{1−s}∥ G∥_{2}^{1−s}
≥\frac{k}{2} ∥ΘΓ∥_{2q}^{s}∥ΘΓ∥_{2}^{1−s} ≥ \frac{k}{2} (∥X∥_{2q}/c_{1})^{s}∥X∥_{2}^{1−s}
≥\frac{k}{2} c_{1}∥X∥_{q}.

Here c_{1} denotes an absolute constant obtained in the first part of proof (as 2q ≥ 4). Since the similar argument (with 1/q = s/p + (1 − s)/2) works for p/2 ≤ q ≤ p, the proof is completed.

In the remainder of this paper we will assume that if (X_{k}) are independent, then the corresponding variables E_{k}’s, are independent, too. We have the following.

Proposition 3.3. There exists an absolute constant K such that if (X_{k}) is a sequence of independent random variables with logarithmically convex tails and for p > 2 (E_{k}) = (E_{p}(X_{k})) is the corresponding sequence of (Θ_{k}Γ_{k}), then

∥ ∑ X_{k}∥_{p} ≤ K∥ ∑ E_{k}∥_{p}.

Proof: Let q ∈ N be an integer, p/2 ≤ 2q < p. It follows from Kwapień and Woyczyński (1992, Proposition 1.4.2) (cf. Hitczenko (1994, Proposition 4.1) for details) that if 2q ≤ p, and (Z_{k}) is any sequence of independent, symmetric random variables, then

∥ ∑ Z_{k}∥_{p} ≤ K \frac{p}{2q} \{∥ ∑ Z_{k}∥_{2q} + ∥ max |Z_{k}|∥_{p}\} ≤ K \frac{p}{2q} \{∥ ∑ Z_{k}∥_{2q} + (∑ ∥Z_{k}∥_{p}^{p})^{1/p}\}.

Therefore,

∥ ∑ X_{k}∥_{p} ≤ K{∥ ∑ X_{k}∥_{2q} + ( ∑ ∥X_{k}∥_{p}^{p})^{1/p}}.

It follows from the binomial formula and Lemma 3.2 that

∥ ∑ X_{k}∥_{2q} ≤ c∥ ∑ E_{k}∥_{2q}.

Hence we obtain,

∥ ∑ X_{k}∥_{p} ≤ K{∥ ∑ E_{k}∥_{2q} + ( ∑ ∥E_{k}∥_{p}^{p})^{1/p}} ≤ K∥ ∑ E_{k}∥_{p},

as required.

Remark 3.4. The above proposition is similar in spirit to a result of Utev (1985), who proved that if p > 4, (Y_{k}) is a sequence of independent symmetric random variables, and (Z_{k}) is a sequence of independent symmetric three-valued random variables, whose second and pth moments are the same as Y_{k}’s, then

∥ ∑ Y_{k}∥_{p} ≤ ∥ ∑ Z_{k}∥_{p}.

We do not know what is the best constant K in Proposition 3.3. We do know, however, that if p is an even number, then K ≤ e.

In order to finish the proof of Theorem 1.1 we need one more lemma.

Lemma 3.5. There exists a constant K, such that if Θ is a symmetric 3-valued random variable with P (|Θ| = 1) = δ = 1 − P (ξ = 0), independent of Γ, then for every q ≥ 1 and x ∈ R we have

(E|x + ΘΓ|^{2q})^{1/q} ≤ eqE(ΘΓ)^{2} + (|x|^{2q} + K^{2q}E|ΘΓ|^{2q})^{1/q}.

Proof: We need to prove that

(|x|^{2q} + δ(E|x + Γ|^{2q} − |x|^{2q}))^{1/q} ≤ 2eqδ + (|x|^{2q} + δK^{2q}EΓ^{2q})^{1/q}.

Set for simplicity A = E|x + Γ|^{2q} − |x|^{2q} and B = K^{2q}EΓ^{2q}. Since the case A ≤ B is trivial, assume A > B. It suffices to show that

\frac{ϕ(Aδ) − ϕ(Bδ)}{δ} ≤ 2eq,

where ϕ(t) = (x^{2q} + t)^{1/q}. Since ϕ is a concave function, the left hand side above is decreasing in δ, so it suffices to check that

2eq ≥ lim_{δ→0} \frac{ϕ(Aδ) − ϕ(Bδ)}{δ} = Aϕ′(0) − Bϕ′(0) = \frac{1}{q}(A − B)x^{2−2q}.

But this equivalent to

E|x + Γ|^{2q} − x^{2q} ≤ 2eq^{2}x^{2q−2} + K^{2q}EΓ^{2q} = eq^{2}x^{2q−2}EΓ^{2} + K^{2q}EΓ^{2q}.

The latter inequality follows from the proof of Lemma 2.3 since exponential variable satisfies moment condition. (A direct proof, giving K = e^{2} can be given, too.) The proof is completed.

Proof of Theorem 1.1 is now very easy. By Proposition 3.3 it suffices to prove the result for the sequence (E_{k}). But, in view of the previous Lemma, and the proof of Theorem 2.2 we get that

∥ ∑ E_{k}∥_{p} ≤ K\{( ∑ ∥E_{k}∥_{p}^{p})^{1/p} + \sqrt{p}( ∑ ∥E_{k}∥_{2}^{2})^{1/2}\},

which completes the proof.

4. L_{p}-domination of linear combinations by i.i.d. sums.

The aim of this section is to prove the following

Theorem 4.1. Let (Y_{k}) be i.i.d. symmetric random variables. For a sequence of scalars (a_{k}) ∈ ℓ_{2}, we let m = ⌈(∥a∥_{2}/∥a∥_{p})^{2p/(p−2)}⌉, where ⌈x⌉ denotes the smallest integer no less than x. Then we have

∥ ∑ a_{k}Y_{k}∥_{p} ≤ K \frac{∥a∥_{p}}{m^{1/p}} ∥ ∑_{k=1}^{m∨p} Y_{k}∥_{p},

for some absolute constant K.

Remark 4.2.

(i) This result is related to an inequality obtained by Figiel, Iwaniec and Pełczyński (1984, Proposition 2.2′). They proved that if (a_{k}) is a sequence of scalars, and (Y_{k}) are symmetrically exchangeable then

∥ ∑_{k=1}^{n} a_{k}Y_{k}∥_{p} ≤ \frac{(∑_{k=1}^{n} |a_{k}|^{p})^{1/p}}{n^{1/p}} ∥ ∑_{k=1}^{n} Y_{k}∥_{p}.

Although we do not get constant 1, our m is generally smaller than n. Also, (cf.Marshall and Olkin (1979, Chapter 12.G)) for certain random variables (Y_{k}), including the Rademacher sequence, one has

∥ ∑_{k=1}^{n} a_{k}Y_{k}∥_{p} ≤ \frac{(∑_{k=1}^{n} |a_{k}|^{2})^{1/2}}{\sqrt{n}} ∥ ∑_{k=1}^{n} Y_{k}∥_{p}.

Our m is chosen so that the ℓ_{2} and ℓ_{p} norms of a new coefficient sequence are essentially the same as those of the original one.

(ii) If, roughly, the first p coefficients in the sequence (a_{k}) are equal, then automatically m ≥ cp, and thus m ∨ p in the upper limit of summation can be replaced by m. This follows from the fact that if a_{1} ≥ a_{2} ≥, …, ≥ 0, then the ratio

\frac{∑_{j=1}^{k} a_{j}^{2}}{(∑_{j=1}^{k} a_{j}^{p})^{2/p}}

is increasing in k. This observation will be important in Section 6.

We will break up the proof of Theorem 4.1 into several propositions. Recall that for a random variable Z ∈ L_{2q}, q ∈ N, EZ^{2q} = (−1)^{q}ϕ_{Z}^{(2q)}(0), where ϕ_{Z} is a characteristic function of Z and ϕ_{Z}^{(2q)} its 2qth derivative.

Proposition 4.3. Let (Y_{k}) be a sequence of i.i.d. symmetric random variables such that

(−1)^{l}(ln ϕ_{Y})^{(2l)}(0) ≥ 0, for l = 1, …, q.

Suppose that a and b are two sequences of real numbers such that ∥a∥_{2l} ≤ ∥b∥_{2l} for l = 1, …, q. Then

E(∑_{k=1}^{n} a_{k}Y_{k})^{2q} ≤ E(∑_{k=1}^{n} b_{k}Y_{k})^{2q}.

Proof: Let S = ∑_{k=1}^{n} a_{k}Y_{k}. We will show that (−1)^{q}(ϕ_{S})^{(2q)}(0) is an increasing function of ∥a∥_{2l}, l = 1, …, q. We have

ϕ_{S}(t) = exp{ln ϕ_{S}(t)} = exp{∑_{k} ln ϕ_{a_{k}Y} (t)}.

Differentiating once and then 2q − 1 times using Leibnitz formula we get

ϕ_{S}^{(2q)} = ∑_{k}(ϕ_{S} ln′ ϕ_{a_{k}Y})^{(2q−1)} = ∑_{k} ∑_{j=0}^{2q−1} \binom{2q − 1}{j}(ln ϕ_{a_{k}Y})^{(j+1)}ϕ_{S}^{(2q−1−j)}.

Hence, evaluating at zero, and using ϕ_{S}^{(j)}(0) = 0, for odd j’s, we get

\begin{matrix}ϕ_{S}^{(2q)}(0) = & ∑_{k} ∑_{j=0}^{2q−1} \binom{2q − 1}{j}(ln ϕ_{a_{k}Y})^{(j+1)}(0)ϕ_{S}^{(2q−1−j)}(0) \\ = & ∑_{k} ∑_{j=1}^{q} \binom{2q − 1}{2j − 1}a_{k}^{2j}(ln ϕ_{Y})^{(2j)}(0)ϕ_{S}^{(2(q−j))}(0), \end{matrix}

and it follows that

∥S∥_{2q}^{2q} = (−1)^{q}ϕ_{S}^{(2q)}(0) = ∑_{j=1}^{q} ∥a∥_{2j}^{2j}\binom{2q − 1}{2j − 1}(−1)^{j}(ln ϕ_{Y})^{(2j)}(0)∥S∥_{2(q−j)}^{2(q−j)}.

Since we assume that (−1)^{j}(ln ϕ_{Y})^{(2j)}(0) ≥ 0, the result follows by induction on q.

Proposition 4.4. Let p ≥ 2. Suppose that a and b are two sequences of real numbers such that ∥a∥_{s} ≤ ∥b∥_{s} for 2 ≤ s ≤ p. Let (Y_{k}) be i.i.d. symmetric such that (−1)^{l}(ln ϕ_{Y})^{(2l)}(0) ≥ 0, for all l ≤ p. Then

∥ ∑_{k=1}^{n} a_{k}Y_{k}∥_{p} ≤ K∥ ∑_{k=1}^{n} b_{k}Y_{k}∥_{p},

for some absolute constant K.

Proof: It follows from Kwapień and Woyczyński (1992, Proposition 1.4.2) (cf. Hitczenko (1994, Proposition 4.1) for details) that if 2q ≤ p, and (Z_{k}) is any sequence of independent, symmetric random variables, then

∥ ∑ Z_{k}∥_{p} ≤ K \frac{p}{2q} \{∥ ∑ Z_{k}∥_{2q} + ∥ max |Z_{k}|∥_{p}\} ≤ K \frac{p}{2q} \{∥ ∑ Z_{k}∥_{2q} + (∑ ∥Z_{k}∥_{p}^{p})^{1/p}\}.

Applying this inequality to Z_{k} = a_{k}Y_{k}, using Proposition 4.3, and the inequalities

∥ ∑ Z_{k}∥_{2q} ≤ ∥ ∑ Z_{k}∥_{p}  and  (∑ ∥Z∥_{p}^{p})^{1/p} ≤ ∥ ∑ Z_{k}∥_{p},

we get the desired result, provided p/q ≤ C.

Remark 4.5. It is natural to ask whether the assumption (−1)^{l}(ln ϕ_{Y})^{(2l)}(0) ≥ 0 can be dropped. In general it cannot. To see this let (ϵ_{k}) be a Rademacher sequence, and for p ∈ N let a_{k} = 1 or 0 according to whether k ≤ p or k > p. If b_{1} = \sqrt{p} and b_{k} = 0 for k ≥ 2, then ∥a∥_{s} ≤ ∥b∥_{s} for 2 ≤ s ≤ p, but ∥ ∑ a_{k}ϵ_{k}∥_{p} ≈ p and ∥ ∑ b_{k}ϵ_{k}∥_{p} = \sqrt{p}.

Before we proceed, we need some more notation. For a random variable Z we let Pois(Z) = ∑_{k=1}^{N} Z_{k}, where N, Z_{1}, Z_{2}, …, are independent random variables, N is a Poisson random variable with parameter 1, and (Z_{k}) are i.i.d. copies of Z. Moreover, if the Z_{k} are independent, then Pois(Z_{k}) will always be chosen so that they are independent. Since P (|Y | > t) ≤ (1/2)(1 − e^{−1})P (|Pois(Y)| > t), the next proposition is a consequence of the contraction principle.

Proposition 4.6. Let (Y_{k}) be a sequence of independent symmetric random variables. Then

∥ ∑ Y_{k}∥_{p} ≤ C∥ ∑ Pois(Y_{k})∥_{p},

for some absolute constant C.

Next we note that, for arbitrary random variable Z

ϕ_{Pois(Z)} = exp{ϕ_{Z} − 1},

and it follows, in particular, that if Z is symmetric, then (−1)^{k}(ln ϕ_{Pois(Z)})^{(2k)}(0) ≥ 0 for all k and thus, by the above discussion, if (Z_{k}) are i.i.d. copies of Z, then

∥ ∑ a_{k}Pois(Y_{k})∥_{p} ≤ C∥ ∑ b_{k}Pois(Y_{k})∥_{p},

whenever ∥a∥_{s} ≤ ∥b∥_{s}, for 2 ≤ s ≤ p.

Now fix a sequence a, and let m = ⌈(∥a∥_{2}/∥a∥_{p})^{2p/(p−2)}⌉ as in Theorem 4.1. Define a sequence b by

b_{k} = \{ \begin{matrix}β, & if k ≤ m; \\ 0, & otherwise, \end{matrix}

where β = ∥a∥_{p}/m^{1/p}. Note that ∥b∥_{2} ≥ ∥a∥_{2}, and ∥b∥_{p} = ∥a∥_{p}. Hence by Hölder’s inequality ∥a∥_{s} ≤ ∥b∥_{s} for all 2 ≤ s ≤ p. Therefore, for an arbitrary sequence of i.i.d. symmetric random variables (Y_{k}) ⊂ L_{p} we have

∥ ∑ a_{k}Y_{k}∥_{p} ≤ K \frac{∥a∥_{p}}{m^{1/p}} ∥ ∑_{k=1}^{m} Pois(Y_{k})∥_{p} = K \frac{∥a∥_{p}}{m^{1/p}} ∥ ∑_{k=1}^{N_{m}} Y_{k}∥_{p},

where N_{m} is a Poisson random variable with parameter m, independent of the sequence (Y_{k}).

Our final step is to estimate the quantity ∥ ∑_{k=1}^{N_{m}} Y_{k}∥_{p}. The following computation is rather straightforward; a very similar calculation can be found, for example, in Klass (1976, proof of Proposition 1).

Proposition 4.7. For m ∈ N, and (Y_{k}) and N_{m} as above, we have

∥ ∑_{k=1}^{N_{m}} Y_{k}∥_{p} ≤ K∥ ∑_{k=1}^{m∨p} Y_{k}∥_{p}.

Proof: For j ≥ 1, let S_{j} = ∑_{k=1}^{j} Y_{k}. Note that, since (S_{j}/j) is a reversed martingale (or by an application of the triangle inequality), ∥S_{j}/j∥_{p} is decreasing in j. Therefore, for j_{0} to be specified later, we have

\begin{matrix}∥ ∑_{k=1}^{N_{m}} Y_{k}∥_{p}^{p} = & ∑_{j=1}^{∞} ∥S_{j}∥_{p}^{p} \frac{e^{−m}m^{j}}{j!} ≤ ∥S_{j_{0}}∥_{p}^{p} ∑_{j=1}^{j_{0}} \frac{e^{−m}m^{j}}{j!} + \frac{∥S_{j_{0}}∥_{p}^{p}}{j_{0}^{p}} ∑_{j>j_{0}} \frac{j^{p}e^{−m}m^{j}}{j!} \\ ≤ & ∥S_{j_{0}}∥_{p}^{p} (1 + \frac{e^{−m}}{j_{0}^{p}} ∑_{j>j_{0}} \frac{j^{p}m^{j}}{j!} ) . \end{matrix}

The ratio of two consecutive terms in the last sum is equal to

(\frac{j + 1}{j})^{p} \frac{m}{j + 1} ≤ \frac{e^{p/j}m}{j + 1} ≤ \frac{1}{2},

whenever j ≥ max{2em, p}. Therefore, choosing j_{0} ≈ max{2em, p}, we obtain

∥ ∑_{k=1}^{N_{m}} Y_{k}∥_{p}^{p} ≤ ∥S_{j_{0}}∥_{p}^{p} (1 + \frac{2e^{−m}}{j_{0}^{p}} j_{0}^{p}(\frac{m}{j_{0}})^{j_{0}}) ≤ 2∥S_{j_{0}}∥_{p}^{p} ≤ K^{p}∥S_{m∨p}∥_{p}^{p}.

This completes the proof of Proposition 4.7.

Theorem 4.1 now follows immediately from the above results.

5. Distribution of a sum of i.i.d. random variables. In this section, Y_{1}, …, Y_{n} are independent copies of a symmetric random variable Y. We fix n, and let S = S_{n} = ∑_{i=1}^{n} Y_{i}, and M = M_{n} = sup_{1≤i≤n} Y_{i}. Our aim is to calculate ∥S∥_{p}, and as we will see below, this is equivalent to finding ∥S∥_{p,∞} + ∥M∥_{p}. Let us recall, that for a random variable Z, ∥Z∥_{p,∞} is defined by

∥Z∥_{p,∞} = sup_{s>0} s^{1/p}Z^{∗}(s),

where Z^{∗} denotes the decreasing rearrangement of the distribution function of Z i.e.

Z^{∗}(s) = sup{t : P (|Z| > t) > s} 0 < s < 1.

The following simple observation will be useful

Lemma 5.1. Let p ≥ 2. For any sequence of independent, symmetric random variables (Z_{k}) ⊂ L_{p} we have

\frac{1}{4}\{ ∥∑ Z_{k}∥_{p,∞} + ∥max |Z_{k}|∥_{p} \} ≤ ∥∑ Z_{k}∥_{p} ≤ K\{ ∥∑ Z_{k}∥_{p,∞} + ∥max |Z_{k}|∥_{p} \},

for some absolute constant K.

Proof: Since for p > q ≥ 1 and for any random variable W, ∥W ∥_{q,∞} ≤ ∥W ∥_{q} ≤

(\frac{p}{p−q} )^{1/q} ∥W ∥_{p,∞}, the first inequality is trivial. For the second inequality we first use a result of Kwapień and Woyczyński (see proof of Proposition 4.4 above)

∥∑ Z_{k}∥_{p} ≤ K \frac{p}{q}\{ ∥∑ Z_{k}∥_{q} + ∥max |Z_{k}|∥_{p} \},

and then the right estimate above with q = p/2. This completes the proof.

Throughout the rest of this section, by f(x) ⪯ g(x) we mean that f(cx) ≤ Cg(x) for some constants c, C. We will write f(x) ≈ g(x) if f(x) ⪯ g(x) and g(x) ⪯ f(x). With this convention we have

Theorem 5.2. For 0 ≤ θ ≤ 1 we have

S^{∗}(θ) ⪯ T_{1}(θ) + T_{2}(θ),

where

T_{1}(θ) = \frac{log(1/θ)}{\sqrt{θ^{−1/n} − 1}} ∥Y^{∗}|_{[θ/n∨(θ^{−1/n}−1),1]}∥_{2} ,

and

T_{2}(θ) = log(1/θ) sup_{θ/n≤t≤θ^{−1/n}−1} \frac{Y^{∗}(t)}{log (1 + \frac{θ^{−1/n}−1}{t})}.

Proof: Let t > 0. Following the argument in Hahn and Klass (1995), for a > 0 we have

\begin{matrix}P (S ≥ t) ≤ & P (M ≥ a) + inf_{λ>0} E(e^{λ(S−t)}|M < a)P (M < a) \\ = & 1 − P^{n}(Y < a) + ( inf_{λ>0} E(e^{λ(Y −t/n)}|Y < a)P (Y < a))^{n}. \end{matrix}

If we choose a so that these two terms are equal, and then we let

θ = 1 − P^{n}(Y < a) = ( inf_{λ>0} E(e^{λ(Y −t/n)}|Y < a)P (Y < a))^{n},

then, by the above computation

P (S ≥ t) ≤ 2θ.

The whole idea now is to invert the above relationship, i.e. starting with θ we will find a relatively small t so that S^{∗}(θ) ⪯ t. Since the values of S^{∗}(θ) are of no importance for θ close to 1 we can assume that θ is bounded away from 1, θ ≤ 1/10, say. Then P (Y < a) = (1 − θ)^{1/n} ≈ 1 − θ/n, so that P (Y ≥ a) ≈ θ/n which, in turn, means that Y^{∗}(θ/n) ≈ a.

In order to find a t, we will use the relation

θ = inf_{λ} E (e^{λ(Y −t/n)}|Y < a)^{n} = inf_{λ} (Ee^{λ(Ỹ −t/n)})^{n} ,

where Ỹ denotes Y I_{Y <a}. Then, for λ > 0 we have

θ ≤ e^{−λt} (Ee^{λỸ} )^{n} ,

and taking logarithms on both sides we get

t ≤ \frac{1}{λ} ln(1/θ) + n ln (Ee^{λỸ} )^{1/λ} .

Hence, we can take

t = inf_{λ>0} \{\frac{1}{λ} ln(1/θ) + n ln (Ee^{λỸ} )^{1/λ} \}.

Note that the second term, being equal to n ln ∥eỸ ∥_{λ}, is an increasing function of λ. Since the first term is decreasing in λ, it follows that, up to a factor of 2, infimum is attained when both terms are equal. Thus t ≈ (1/λ) ln(1/θ), where λ is chosen so that ln(1/θ) = n ln (Ee^{λỸ} ), i.e. Ee^{λỸ} = θ^{−1/n}. Since Ỹ is symmetric, Ee^{λỸ} = E cosh(λ|Ỹ |), and the above equality is equivalent to

E (\frac{cosh(λỸ) − 1}{θ^{−1/n} − 1}) = 1.

But this simply means that

t ≈ ∥Ỹ ∥_{Φ} ln(1/θ), where Φ(s) = \frac{cosh(s) − 1}{θ^{−1/n} − 1}.

(Recall that if Ψ is an Orlicz function and Z is a random variable, then ∥Z∥_{Ψ} = inf{u > 0 : EΨ(Z/u) ≤ 1}.)

In order to complete the proof, it suffices to show that, with Φ as above, we have

∥Ỹ ∥_{Φ} ≈ sup_{θ/n≤t≤θ^{−1/n}−1} \frac{Y^{∗}(t)}{log (1 + \frac{θ^{−1/n}−1}{t})} + \frac{1}{\sqrt{θ^{−1/n} − 1}} ∥Y^{∗}|_{[θ/n∨(θ^{−1/n}−1),1]}∥_{2} .

We will need the following two observations. For an Orlicz function Ψ and a random variable Z, define the weak Orlicz norm by

∥Z∥_{Ψ,∞} = sup_{0<x<1} \{\frac{1}{Ψ^{−1}(1/x)}Z^{∗}(x)\} .

Lemma 5.3. Let ψ(s) = e^{s} − 1. Then,

∥Z∥_{ψ,∞} ≤ ∥Z∥_{ψ} ≤ 3 ∥Z∥_{ψ,∞} .

Proof: Only the second inequality needs justification. Suppose ∥Z∥_{ψ,∞} ≤ 1. This means that Z^{∗}(x) ≤ ln(1 + 1/x), for each 0 < x < 1. Hence

Eψ(Z/u) = Ee^{\{Z^{∗}/u\}} −1 ≤ ∫_{0}^{1} e^{(1/u) ln(1+1/x)}dx−1 ≤ ∫_{0}^{1}(2/x)^{1/u}dx−1 = \frac{2^{1/u}u}{u − 1} −1 ≤ 1,

whenever u ≥ 3. Remark 5.4. The above relationship is true for more general Orlicz functions, as long as they do grow fast enough; see Theorem 4.6 in Montgomery-Smith (1992) for more details.

Lemma 5.5. Suppose that Φ_{1}, Φ_{2} : [0, ∞) → [0, ∞) are nondecreasing continuous functions that map zero to zero, and such that there exists a constant c > 1 such that Φ_{i}(x/c) ≤ Φ_{i}(x)/2 (i = 1, 2). Suppose further that there are numbers a, A > 0 such that Φ_{1}(a) = Φ_{2}(a) = A. Define Φ : [0, ∞) → [0, ∞) by

Φ(x) = \{ \begin{matrix}Φ_{1}(x) & if x ≤ a \\ Φ_{2}(x) & if x > a . \end{matrix}

Then for any measurable scalar valued function f, we have that

∥f∥_{Φ} ≈ ∥f^{∗}|_{[0,A^{−1}]}∥_{Φ_{2}} + ∥f^{∗}|_{[A^{−1},∞)}∥_{Φ_{1}}.

The constants of approximation depend only upon c.

Proof: Notice that Φ(x/c^{2}) ≤ Φ(x)/2. This is clear if both x/c^{2} and x lie on the same side of a, and otherwise it follows by considering the cases whether x/c is greater or less than a.

Let us first suppose that the right hand side is less than 1. Then

∫_{0}^{A^{−1}} Φ_{2}(f^{∗}(x)) dx ≤ 1 and ∫_{A^{−1}}^{∞} Φ_{1}(f^{∗}(x)) dx ≤ 1.

¿From the first inequality, we see that A^{−1}Φ_{2}(f^{∗}(A^{−1})) ≤ 1, from which it follows that f^{∗}(A^{−1}) ≤ a. Let B be such that f^{∗}(B^{+}) ≤ a ≤ f^{∗}(B^{−}), so that B ≤ A. (Here, and in the rest of the proof, we define f(A^{+}) = lim_{x↘A} f(x), and f(A^{−}) = lim_{x↗A} f(x).) Then

∫_{0}^{B} Φ(f^{∗}(x)) dx = ∫_{0}^{B} Φ_{2}(f^{∗}(x)) dx ≤ 1,

and

∫_{B}^{A^{−1}} Φ(f^{∗}(x)) dx ≤ A^{−1}Φ(f^{∗}(B^{+})) ≤ 1,

and

∫_{A^{−1}}^{∞} Φ(f^{∗}(x)) dx = ∫_{A^{−1}}^{∞} Φ_{1}(f^{∗}(x)) dx ≤ 1.

Hence

∫_{0}^{∞} Φ(f^{∗}(x)) dx ≤ 3,

and so

∫_{0}^{∞} Φ(f^{∗}(x)/c^{4}) dx ≤ 1,

that is,

∥f∥_{Φ} ≤ c^{4}.

Now suppose that ∥f∥_{Φ} ≤ 1, so that

∫_{0}^{∞} Φ(f^{∗}(x)) dx ≤ 1.

Then A^{−1}Φ(f^{∗}(A^{−1})) ≤ 1, which implies that f^{∗}(A^{−1}) ≤ a. Let B be such that f^{∗}(B^{+}) ≤ a ≤ f^{∗}(B^{−}), so that B ≤ A. Then

∫_{0}^{B} Φ_{2}(f^{∗}(x)) dx = ∫_{0}^{B} Φ(f^{∗}(x)) dx ≤ 1,

and

∫_{B}^{A^{−1}} Φ_{2}(f^{∗}(x)) dx ≤ A^{−1}Φ_{2}(f^{∗}(B^{+})) ≤ 1.

Hence

∫_{0}^{A^{−1}} Φ_{2}(f^{∗}(x)) dx ≤ 2,

from which it follows that ∥f|_{[0,A^{−1}]}∥_{Φ_{2}} ≤ c. Also

∫_{A^{−1}}^{∞} Φ_{1}(f^{∗}(x)) dx = ∫_{A^{−1}}^{∞} Φ(f^{∗}(x)) dx ≤ 1.

and hence ∥f|_{[A^{−1},∞)}∥_{Φ_{1}} ≤ 1. This completes the proof of Lemma 5.5

Now, in order to finish the proof of Theorem 5.2, let Φ(s) = \frac{cosh(s)−1}{θ^{−1/n}−1}. Then,

Φ(s) = \{ \begin{matrix}Φ_{1}(s), & if 0 ≤ s ≤ 1; \\ Φ_{2}(s), & if s > 1; \end{matrix}

where

Φ_{1}(s) ≈ \frac{s^{2}}{θ^{−1/n} − 1} and Φ_{2}(s) ≈ \frac{e^{s} − 1}{θ^{−1/n} − 1}.

Since Ỹ^{∗} = (Y I_{Y <a})^{∗} ≈ (Y^{∗}|_{[θ/n,1]})^{∗}, by Lemma 5.5 we obtain

∥Ỹ ∥_{Φ} ≈ ∥Y^{∗}|_{[0,θ^{−1/θ}−1]}∥_{Φ_{2}} + ∥Y^{∗}|_{[θ^{−1/θ}−1,1]}∥_{Φ_{1}}
≈ ∥Y^{∗}|_{[0,θ^{−1/θ}−1]}∥_{Φ_{2},∞} + \frac{1}{\sqrt{θ^{−1/θ} − 1}} ∥Y^{∗}|_{[θ^{−1/θ}−1,1]}∥_{2}
≈ sup_{θ/n≤t≤θ^{−1/n}−1} \frac{Y^{∗}(t)}{log (1 + \frac{θ^{−1/n}−1}{t})} + \frac{1}{\sqrt{θ^{−1/n} − 1}} ∥Y^{∗}|_{[θ/n∨(θ^{−1/n}−1),1]}∥_{2} .

This completes the proof of Theorem 5.2.

6. Moments of linear combination of symmetric Weibull random variables.

In this section we will apply our methods to obtain an upper bound for the L_{p}-norm of a linear combination of i.i.d. symmetric Weibull random variables with parameter 0 < r < 1. We will also show that this upper bound is tight. A random variable X will be called symmetric Weibull with parameter r, 0 < r < ∞ if X is symmetric and |X| has density given by

f_{|X|}(t) = rt^{r−1}e^{−t^{r}}, t > 0.

We refer to Johnson and Kotz (1970, Chapter 20) for more information on Weibull random variables. Here we only note that by change of variables

∥X∥_{p}^{p} = Γ(1 + \frac{p}{r}), p > 0,

so that, using Stirling formula and elementary estimates we have the following.

Lemma 6.1. If X is a symmetric Weibull random variable with parameter 0 < r < 1, and p ≥ 2, then

p∥X∥_{2} ≤ C∥X∥_{p},

where C is a constant not depending on p or r.

As we mentioned in the Introduction, Gluskin and Kwapień (1995) established a two sided inequality for the L_{p}-norm of a linear combinations of i.i.d. symmetric random variables with logarithmically concave tails. In the special case where P (|ξ| > t) = exp(−t^{r}), r ≥ 1, their result reads as follows

\begin{matrix}c\{(∑_{k≤p} a_{k}^{r′})^{1/r′} ∥ξ∥_{p} + & \sqrt{p} ∥ξ∥_{2} (∑_{k>p} a_{k}^{2})^{1/2}\} ≤ ∥ ∑ a_{k}ξ_{k}∥_{p} \\ ≤ & C\{(∑_{k≤p} a_{k}^{r′})^{1/r′} ∥ξ∥_{p} + \sqrt{p} ∥ξ∥_{2} (∑_{k>p} a_{k}^{2})^{1/2}\}, \end{matrix}

where r′ is the exponent conjugate to r, i.e. 1/r + 1/r′ = 1, and c and C are absolute constants. In this section we complement the result of Gluskin and Kwapień, by treating the case r < 1. Here is the main result of this section.

Theorem 6.2. There exist absolute constants c, C such that if (X_{i}) is a sequence of i.i.d. symmetric Weibull random variables with parameter r, where 0 < r < 1 and (a_{k}) ∈ ℓ_{2}, then

c max\{\sqrt{p}∥a∥_{2}∥X∥_{2}, ∥a∥_{p}∥X∥_{p}\} ≤ ∥ ∑ a_{k}X_{k}∥_{p} ≤ C max\{\sqrt{p}∥a∥_{2}∥X∥_{2}, ∥a∥_{p}∥X∥_{p}\}.

Proof: We begin with the following result.

Proposition 6.3. Let p ≥ 2, and let (X_{i}) be a sequence i.i.d. symmetric Weibull random variables with parameter r where 0 < r < 1. Then, the following is true.

a_{1}∥X∥_{p} ≤ ∥ ∑_{k≤p} a_{k}X_{k}∥_{p} ≤ Ca_{1}∥X∥_{p}.

Proof: The first inequality is trivial. To prove the second, note that

∥ ∑_{k≤p} a_{k}X_{k}∥_{p} ≤ a_{1}∥ ∑_{k≤p} X_{k}∥_{p} ≤ a_{1}∥ ∑_{k≤p} |X_{k}| ∥_{p}.

Assume first that r is bounded away from 0, say r > 1/2. Then, letting (Γ_{i}) be a sequence of i.i.d. exponential distributions with mean 1 we can write

\begin{matrix}E(∑_{k≤p} |X_{k}|)^{p} = & E ([ ∑_{k≤p}(|X_{k}|^{r})^{1/r}]^{r})^{p/r} = E ([ ∑_{k≤p} Γ_{k}^{1/r}]^{r})^{p/r} \\ ≤ & E (∑_{k≤p} Γ_{k})^{p/r} ≤ p\frac{Γ(p + p/r)}{Γ(p)} ≤ p\frac{(p(1 + 1/r))^{p}}{Γ(p)} Γ(1 + p/r) \\ ≤ & K^{p}Γ(1 + p/r), \end{matrix}

since r is bounded away from zero. This shows the upper bound for r > 1/2. To handle the case r ≤ 1/2 we will apply a method that was used in Schechtman and Zinn (1990). Let (X_{(k)}) denote the nonincreasing rearrangement of (|X_{k}|)_{k≤p}. For a number t > 0 we have

P (∑_{k≤p} |X_{k}| ≥ t) = P (∑_{k≤p} X_{(k)} ≥ t) ≤ ∑_{k≤p} P (X_{(k)} ≥ q_{k}t),

where (q_{k}) is any sequence of nonnegative numbers such that ∑_{k≤p} q_{k} ≤ 1. Now, using the inequality

P (Y_{(k)} ≥ s) ≤ \frac{(∑_{j≤p} P (Y_{j} ≥ s))^{k}}{k!} ,

valid for any sequence of independent random variables (Y_{k}) (cf. e.g. Talagrand (1989, Lemma 9)) we get that

P (X_{(k)} ≥ q_{k}t) ≤ \frac{(p exp\{−q_{k}^{r}t^{r}\})^{k}}{k!} .

Therefore,

E(∑_{k≤p} |X_{k}|)^{p} ≤ ∑_{k≤p} \frac{p^{k}}{k!} · p ∫_{0}^{∞} t^{p−1} exp\{−kq_{k}^{r}t^{r}\}dt.

Substituting u = (k^{1/r}q_{k}t)^{r} into the kth term and integrating we get

∫_{0}^{∞} t^{p−1} exp\{−(k^{1/r}q_{k}t)^{r}\}dt = \frac{Γ(p/r)}{r(k^{1/r}q_{k})^{p}} .

Therefore,

E| ∑_{k≤p} X_{k}|^{p} ≤ (p/r)Γ(p/r)\{ ∑_{k≤p} \frac{p^{k+1}}{k!k^{p/r}q_{k}^{p}} \} ≤ (p/r)Γ(p/r)\frac{1}{(inf_{k}\{k^{1/r}q_{k}\})^{p}} ∑_{k=1}^{∞} \frac{p^{k}}{k!} ,

which is bounded by C^{p}∥X∥_{p}^{p}, as long as k^{1/r}q_{k} is bounded away from zero. The choice q_{k} ≈ k^{−1/r} concludes the proof of the Proposition.

Now for the general case. By the previous proposition we can and do assume that n > p. We will establish the upper bound first.

We first observe that it suffices to prove the result under the additional assumption that the first ⌈p⌉ entries in a coefficient sequence are equal. Indeed, suppose we know the result in that case. Then, for the general sequence (a_{k}), we can write,

\begin{matrix}∥ ∑ a_{k}X_{k}∥_{p} ≤ & ∥ ∑_{k≤p} a_{1}X_{k} + ∑_{k≥1} a_{k}X_{⌈p⌉+k}∥_{p} \\ ≤ & C max \{\sqrt{p}∥X∥_{2}(\sqrt{p}a_{1} + ∥a∥_{2}), (p^{1/p}a_{1} + ∥a∥_{p})∥X∥_{p}\} \\ ≤ & C max\{\sqrt{p}∥a∥_{2}∥X∥_{2}, ∥a∥_{p}∥X∥_{p}\}, \end{matrix}

since, by Lemma 6.1, we have

pa_{1}∥X∥_{2} ≤ Ka_{1}∥X∥_{p} ≤ K∥a∥_{p}∥X∥_{p}.

Next we note that if the first ⌈p⌉ coefficients are equal, and m is defined as in Theorem 4.1, then automatically

m ≈ (∥a∥_{2}/∥a∥_{p})^{2p/(p−2)} ≥ cp,

and the inequality of that theorem takes form

∥ ∑ a_{k}X_{k}∥_{p} ≤ K \frac{∥a∥_{p}}{m^{1/p}} ∥ ∑_{k=1}^{m} X_{k}∥_{p}.

By definition m ≤ 2(∥a∥_{2} / ∥a∥_{p})^{2p/(p−2)}, so that

\frac{∥a∥_{p}}{m^{1/p}} \sqrt{pm} ≤ \sqrt{2p} ∥a∥_{2} .

Therefore, in order to complete the proof it suffices to show that

∥ ∑_{k=1}^{m} X_{k}∥_{p} ≤ K max \{\sqrt{pm}∥X∥_{2}, m^{1/p}∥X∥_{p}\}.

In other words, the problem has been reduced to the special case when all coefficients are equal.

To prove the latter inequality we will use Theorem 5.2. Fix n ∈ N. Let S = ∑_{i=1}^{n} X_{i}, and M = sup_{1≤i≤n} X_{i}, that is, M^{∗}(t) ≈ X^{∗}(t/n). Our aim is to show

∥S∥_{p} ≤ K\{\sqrt{np} ∥X∥_{2} + ∥M∥_{p} \}.

(Note that ∥M∥_{p} ≤ n^{1/p} ∥X∥_{p}.) By Lemma 5.1, it suffices to estimate ∥S∥_{p,∞} + ∥M∥_{p}. Let us recall that X^{∗}(t) = (log(1/t))^{1/r}.

¿From Theorem 5.2, we know that

S^{∗}(x) ⪯ T_{1}(x) + T_{2}(x),

where

T_{1}(x) = \frac{log(1/x)}{\sqrt{x^{−1/n} − 1}} ∥X^{∗}|_{[x/n∨(x^{−1/n}−1),1]}∥_{2} ,

and

T_{2}(x) = log(1/x) sup_{x/n≤t≤x^{−1/n}−1} \frac{X^{∗}(t)}{log (1 + \frac{x^{−1/n}−1}{t})}.

Now, ∥S∥_{p,∞} ≈ sup_{0<x<1} x^{1/p}T_{1}(x) + sup_{0<x<1} x^{1/p}T_{2}(x).

To get a handle on these quantities, we use the following approximation:

x^{−1/n} − 1 ≈ \{ \begin{matrix}(1/n) log(1/x) & if x ≥ e^{−n} \\ x^{−1/n} & if x ≤ e^{−n}. \end{matrix}

Now, we can see that if x ≤ e^{−n}, then T_{1}(x) = 0, and if x ≥ e^{−n}, then

T_{1}(x) ≤ c\sqrt{n}\sqrt{log(1/x)} ∥X∥_{2} .

Hence, sup_{0<x<1} x^{1/p}T_{1}(x) ≤ c\sqrt{np} ∥X∥_{2}.

As for T_{2}, we use similar approximations, and we arrive at the following formula. If x ≥ e^{−n}, then

T_{2}(x) ≈ sup_{x/n≤t≤(1/n) log(1/x)} \frac{log(1/x)X^{∗}(t)}{1 + log(1/t) − log(n/ log(1/x))},

and if x ≤ e^{−n}, then

T_{2}(x) ≈ sup_{x/n≤t≤x^{1/n}} \frac{log(1/x)X^{∗}(t)}{log(1/t)}.

Now let us make the substitution X^{∗}(t) = (log(1/t))^{1/r}, where 0 < r < 1. If x ≤ e^{−n}, then it is clear that the supremum that defines T_{2}(x) is attained when t is as small as possible, that is, t = x/n. But, since x ≤ e^{−n}, it follows that log(n/x) ≈ log(1/x), and hence

T_{2}(x) ≈ X^{∗}(x/n) ≈ M^{∗}(x).

Now consider the case x ≥ e^{−n}. Then

T_{2}(x) ≈ sup_{x/n≤t≤(1/n) log(1/x)} log(1/x)H(log(1/t)),

where

H(u) = \frac{u^{1/r}}{1 − log(n/ log(1/x)) + u}.

Now, looking at the graph of H(u), we see that supremum of H(u) over an interval on which H(u) is positive is attained at one of the endpoints of the interval. Hence

T_{2}(x) ≈ log(1/x) max{H(log(n/x)), H(log(n/ log(1/x)))}.

Now,

log(1/x)H(log(n/x)) = \frac{log(1/x)X^{∗}(x/n)}{1 + log(log(1/x)/x)} ≈ X^{∗}(x/n) ≈ M^{∗}(x),

because 1 + log(log(1/x)/x) ≈ log(1/x). Also,

log(1/x)H(log(n/ log(1/x))) = log(1/x)(log(n/ log(1/x)))^{1/r}.

Putting all this together, we have that

sup_{0<x<1} x^{1/p}T_{2}(x) ≈ ∥M∥_{p,∞} + sup_{x≥e^{−n}} F (log(1/x)),

where

F (u) = e^{−u/p}u(log(n/u))^{1/r}.

Thus, the proof will be complete if we can show that F (u) ≤ \sqrt{np} ∥X∥_{2} for all u ≤ n. To

this end, from Stirling’s formula, we have that ∥X∥_{2} = Γ(1 + 2/r)^{1/2} ≥ c^{−1} (\frac{2}{re})^{1/r}.

Notice that

\frac{uF′(u)}{F (u)} = −\frac{u}{p} + 1 − \frac{1}{r log(n/u)}.

This quantity is positive if u is small, it is negative if u = p or u = n, and it is decreasing. Hence, F (u) attains its supremum at u_{0}, where F′(u_{0}) = 0, and u_{0} ≤ p. But then

F (u_{0}) ≤ \sqrt{np}G(n/u_{0}),

where

G(v) = \frac{(log v)^{1/r}}{\sqrt{v}} .

Simple calculus shows us that

G(v) ≤ G(e^{2/r}) = (\frac{2}{re})^{1/r} ≤ c ∥X∥_{2} ,

which proves the right inequality in Theorem 6.2.

To prove the left inequality, notice that by the original proof of Rosenthal’s inequality (Rosenthal (1970)) we have

∥ ∑ a_{k}X_{k}∥_{p} ≥ (∑ a_{k}^{p})^{1/p}∥X∥_{p},

so it suffices to show that

∥ ∑ a_{k}X_{k}∥_{p} ≥ c\sqrt{p}(∑ a_{k}^{2})^{1/2}∥X∥_{2}.

Let us assume without loss of generality that p ≥ 3. Let δ = ^{\frac{(∑_{j≤p} a_{j}^{2})^{1/2}}{\sqrt{p}}}, and define a sequence d = (d_{k}) by the formula

d_{k} = \{ \begin{matrix}δ, & if k ≤ p; \\ a_{k}, & otherwise. \end{matrix}

Then,

∥ ∑ a_{k}X_{k}∥_{p} ≥ κ∥ ∑ d_{k}X_{k}∥_{p},

for some absolute constant κ > 0. Indeed, let C be a constant such that

∥ ∑_{k≤p} X_{k}∥_{p} ≤ C∥X∥_{p}.

Since δ ≤ a_{1} we have

\begin{matrix}∥ ∑ d_{k}X_{k}∥_{p} ≤ & δ∥ ∑_{k≤p} X_{k}∥_{p} + ∥ ∑_{k>p} a_{k}X_{k}∥_{p} ≤ Ca_{1}∥X∥_{p} + ∥ ∑_{k>p} a_{k}X_{k}∥_{p} \\ ≤ & (C + 1)∥ ∑ a_{k}X_{k}∥_{p}, \end{matrix}

so that one can take κ = 1/(C + 1). Let (ϵ_{k}) be a sequence of Rademacher random variables, independent of the sequence (X_{k}). Notice that max d_{j}/∥d∥_{2} ≤ 1/\sqrt{p}. Therefore, using the minimality property of Rademacher functions (cf. Figiel, Hitczenko, Johnson, Schechtman and Zinn (1995, Theorem 1.1), or Pinelis (1994, Corollary 2.5)) (here we use p ≥ 3) and then Hitczenko and Kwapień (1994, Theorem 1) we get

∥ ∑ a_{k}X_{k}∥_{p} ≥ κ∥ ∑ d_{k}X_{k}∥_{p} ≥ κ∥ ∑ d_{k}∥X_{k}∥_{2}ϵ_{k}∥_{p} ≥ c\sqrt{p}∥d∥_{2}∥X∥_{2} = c\sqrt{p}∥a∥_{2}∥X∥_{2}.

The proof is completed.

Remark 6.4: For r = 1 our formula gives p∥a∥_{p} + \sqrt{p}∥a∥_{2}, while Gluskin and Kwapień obtained p sup_{k≤p} a_{k} + \sqrt{p}(∑_{k>p} a_{k}^{2})^{1/2}. Although these two quantities look different, they are equivalent. Clearly p sup_{k≤p} a_{k} + \sqrt{p}(∑_{k>p} a_{k}^{2})^{1/2} ≤ p∥a∥_{p} + \sqrt{p}∥a∥_{2}. To see that opposite inequality holds with an absolute constant, notice that if a_{1} and ∑_{k>p} a_{k}^{2} are fixed, then p∥a∥_{p} + \sqrt{p}∥a∥_{2} maximized if the first several a_{k}’s are equal to a_{1}, the next one is between a_{1} and 0, and the rest are 0. In this case it is very easy to check that the required inequality holds.

Theorem 6.2 implies the following result for tail probabilities.

Corollary 6.5. Let a = (a_{k}) ∈ ℓ_{2}, a ≠ 0, and let S = ∑ a_{k}X_{k}. Then

lim_{t→∞} log_{t} ln 1/P (|S| > t) = r.

Proof: Since P (|S| > t) ≥ \frac{1}{2}P (a_{1}|X_{1}| > t) = \frac{1}{2} exp\{−(t/a_{1})^{r}\}, we immediately get

lim sup_{t→∞} log_{t} ln 1/P (|S| > t) ≤ r.

To show the opposite inequality note that if s < r then

E exp\{|S|^{s}\} = 1 + ∑_{k=1}^{∞} \frac{E|S|^{sk}}{k!} < ∞,

by Theorem 6.2. Hence P (|S| > t) ≤ exp{−t^{s}}E exp{|S|^{s}}, which implies

lim inf_{t→∞} log_{t} ln 1/P (|S| > t) ≥ s.

Since this is true for every s < r, the result follows.

Acknowledgments. We would like to thank S. Kwapień for his remarks and suggestions that led to discovery of the proof of Theorem 2.2.

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P. Hitczenko  S. J. Montgomery-Smith
Department of Mathematics,  Department of Mathematics
North Carolina State University  University of Missouri – Columbia
Raleigh, NC 27695–8205, USA  Columbia, MO 65211

e-mail: pawel@math.ncsu.edu             e-mail: stephen@math.missouri.edu

K. Oleszkiewicz Instytut Matematyki Uniwersytet Warszawski ul. Banacha 2 02-097 Warszawa, Poland

e-mail: koles@mimuw.edu.pl