The Law of the Iterated logarithm for certain Power Series and generalized Nörlund methods^{1}

Rüdiger Kiesel, University of Ulm

1 Introduction and main result

Let ( Ω, Σ, P ) be a probability space and suppose all random variables are defined on this space. Let X, X_{0}, X_{1}, X_{2}, … be (without further comment) throughout independent, identically distributed (i.i.d.) random variables. Lai [16] discussed the rate at which convergence takes place in the strong laws of large numbers for the classical scales of summability methods Cesàro (C_{β}) – Abel (A) (the discounting method) and Euler (E_{γ}) – Borel (B) .

Theorem Lai A The following are equivalent:

(i) E(X) = 0, E(X^{2}) = 1 ;

(ii) lim sup_{n→∞} \frac{|∑_{k=0}^{n} \binom{n − k + β − 1}{n − k}X_{k}|}{(2n^{2β−1} log logn)^{\frac{1}{2}}} = \frac{1}{(2β − 1)^{\frac{1}{2}} Γ(β)^{\frac{1}{2}}} a.s. for β ≥ 1 ;

(iii) lim sup_{t→1−} \frac{|∑_{k=0}^{∞} t^{k}X_{k}|}{(\frac{1}{1−t} log |log(1 − t^{2})|)^{\frac{1}{2}}} = 1 a.s..

Theorem Lai B The following are equivalent:

(i) E(X) = 0, E(X^{2}) = 1, E(X^{4}(log^{+} |X| + 1)^{−2}) ;

11991 Mathematics subject classification: Primary 60F15; Secondary 40G05, 40G10.

Key words and phrases: Law of the Iterated logarithm, Summability.

(ii) lim sup_{n→∞} \frac{(πγn)^{\frac{1}{4}} \frac{1}{γ + 1} |∑_{k=0}^{n} \binom{n}{k}γ^{n−k}X_{k}|}{\sqrt{\frac{1}{2}(γ + 1) log n}} = 1 a.s. for γ > 0 ;

(iii) lim sup_{t→∞} \frac{(4πt)^{\frac{1}{4}} e^{−t} |∑_{k=0}^{∞} \frac{t^{k}}{k!}X_{k}|}{\sqrt{log t}} = 1 a.s..

So we observe that in the Cesàro–Abel case we have a behaviour of iterated logarithm type, whereas the Euler–Borel case exhibits a single logarithm type behaviour.

In [13, 14] the author obtained strong laws for general classes of summability methods containing the above methods as special cases. Therefore the questions arise, what the rate of convergence in these strong laws is and where the switch from iterated to single logarithm behaviour takes place. The aim of this note is to provide an answer to the first question.

Let us first define two general classes of summability methods.

We assume that (p_{n}) is a sequence of reals with

(1.1)  \{ \begin{matrix}p_{0} > 0, p_{n} ≥ 0, n ∈ N I , such that the power series \\ p(t) = ∑_{n=0}^{∞} p_{n}t^{n} has radius of convergence R = 1. \end{matrix}

For β > 0 we define the sequence (p_{n}^{β}) by the convolution

(1.2)  p_{n}^{β} := ∑_{k=0}^{n} \binom{n − k + β − 1}{n − k}p_{k}; n = 0, 1, …

We apply the following methods of summability to sequences (s_{n}) of complex numbers:

(a) the generalized Nörlund methods (N_{βp}) , defined by the transform

(1.3)  σ_{n}^{β} := \frac{1}{p_{n}^{β}} ∑_{k=0}^{n} \binom{n − k + β − 1}{n − k}p_{k}s_{k},

and we say s_{n} → s(N_{βp}) if σ_{n}^{β} → s (n → ∞) ;

(b) the power series method (J_{p}) , where we say

(1.4)  \{ \begin{matrix}s_{n} → s(J_{p}), & if p_{s}(t) = ∑_{n=0}^{∞} s_{n}p_{n}t^{n} converges for |t| < 1 \\ & and if σ_{p}(t) = \frac{p_{s}(t)}{p(t)} → s as t → 1 − . \end{matrix}

Observe that the Cesàro (C_{β}) - and the Abel (A) - method are the special cases with p_{n} = 1, n = 0, 1, . . .. For further properties of these summability methods see [11, 14] and the references therein. In order to prevent the weights p_{n} from being too irregular, we use the following regularity condition:

(1.6)  p_{n} = n^{α}L(n) (n → ∞)

with the parameter α > −\frac{1}{2} and the function L(.) slowly varying,

i.e. \frac{L(λn)}{L(n)} → 1 (n → ∞) ∀λ > 0 , compare [3, 17] for further details.

We now state our main results

Theorem 1 Let p_{n} be a sequence satisfying (1.6) and denote by B(., .) the Beta-function. The following statements are equivalent:

(i) E(X) = 0, E(X^{2}) = 1 ;

(ii) lim sup_{n→∞} \frac{|∑_{k=0}^{n} \binom{n − k + β − 1}{n − k}p_{k}X_{k}|}{(2B(2α + 1, 2β − 1)n^{2(α+β)−1}L(n) log log n)^{\frac{1}{2}}} = 1 a.s for β ≥ 1 ;

(iii) lim sup_{t→1−} \frac{|∑_{k=0}^{∞} p_{k}t^{k}X_{k}|}{(2 ∑_{k=0}^{∞} p_{k}^{2}t^{2k} log |log (∑_{k=0}^{∞} p_{k}^{2}t^{2k})|)^{\frac{1}{2}}} = 1 a.s..

We see that, for polynomial-like growth of the weight sequence (p_{n}) we have an iterated logarithm type of behaviour. Our conjecture is, that for exponentially growing weight sequences like p_{n} = exp{n^{α}}, 0 < α < 1 we have a behaviour of single logarithm type (see [12] for some results hinting in this direction) and hence the switch occurs in the intermediate grey zone.

Under the assumptions of Theorem 1 we also can identify the limit points.

Theorem 2 Under the above assumptions the cluster set in (ii) and (iii) above is [−1, 1] a.s..

Remark: That the cluster set is [−1, 1] a.s. in case of the Abel-method was quite recently shown by Bovier, Picco [4] and also by Horvàth [10], who used a different method.

In what follows we will focus on proving Theorem 1 and Theorem 2 for the power series case. We start in §2 by establishing an upper bound for the a.s. lim sup behaviour, see Proposition 1. In §3 we show by modifying an argument of Feller [8, 2, 19] that a.s. boundedness of the lim sup already implies the moment conditions of Theorem 1(i), see Proposition 2. In §4 we use de Acosta’s [1, 2] techniques to establish Theorem 2 and hence that the lim sup in Theorem 1(iii) is 1 a.s., see Proposition 3. A main tool in doing this is the Central Limit Theorem for summability methods given by Embrechts and Maejima [7]. In §5 we briefly outline how to modify the arguments to prove the Law of Iterated logarithm for the corresponding generalized Nörlund transforms. We close by supplying an appendix on the asymptotic behaviour of sums and convolutions of weight sequences satisfying (1.6), not necessarily assuming that α > −1/2 .

2 The lim sup -behavior

We start with some notation:

p_{X}(t) := ∑_{k=0}^{∞} p_{k}X_{k}e^{− \frac{k}{t}}  P_{n} := ∑_{k=0}^{n} p_{k}  P (t) := ∑_{k≤t} p_{k};

V (t) := ∑_{k=0}^{∞} p_{k}^{2}e^{− \frac{2k}{t}}  P_{n}^{(2)} := ∑_{k=0}^{n} p_{k}^{2} P^{(2)}(t) := ∑_{k≤t} p_{k}^{2};

φ(t) := (2V (t) loglogV (t))^{\frac{1}{2}} , N(t) := [t logt].

Observe that we used a parameter transformation. This transformation has been proven useful in many problems concerning power series methods, see [11, 12, 15] and the references cited there. We use c(α) as a general positive constant only depending on α. We also introduce a sequence

(2.1)  t_{n} := exp \{\frac{n}{(logn)^{α+2}} \} ,

where the parameter α depends on the weights according to (1.6).

We start by showing that the upper tail of the series is neglectible

Lemma 1 For a sequence (p_{n}) satisfying (1.6) we have

(2.2)  lim_{t→∞} φ(t)^{−1} | ∑_{k=N(t)}^{∞} p_{k}X_{k}e^{− \frac{k}{t}} | = 0 a.s..

Proof: By the Borel-Cantelli Lemma (2.2) will follow, if we prove that for every ϵ > 0

(2.3)  ∑_{n=n_{0}}^{∞} P ( sup_{t_{n}≤t≤t_{n+1}} | ∑_{k=N(t)}^{∞} p_{k}X_{k}e^{− \frac{k}{t}} | ≥ ϵφ(t_{n})) < ∞.

We have

sup_{t_{n}≤t≤t_{n+1}} | ∑_{k=N(t)}^{∞} p_{k}X_{k}e^{− \frac{k}{t}} | ≤ ∑_{k=N(t_{n})}^{∞} p_{k}e^{− \frac{k}{t_{n+1}}} |X_{k}|

= ∑_{k=N(t_{n})}^{∞} p_{k}e^{− \frac{k}{t_{n+1}}} \{|X_{k}| − E(|X_{k}|)\}

+ ∑_{k=N(t_{n})}^{∞} p_{k}e^{− \frac{k}{t_{n+1}}} E(|X_{k}|).

Set N := N(t_{n}) . Using the asymptotic results of Lemmas A1–A3, we get for n large enough and some constant c(α)

φ(t_{n})^{− \frac{1}{2}} ∑_{k=N}^{∞} p_{k}e^{− \frac{k}{t_{n+1}}} E(|X|) ≤ c(α)E(|X|)\frac{(t_{n+1} log t_{n+1})^{α+1}L(t_{n+1} log t_{n+1})}{t_{n}^{α+ \frac{1}{2}} L(t_{n})(log log t_{n})^{\frac{1}{2}} tn^{\frac{3}{4}}} = o (1) .

By Chebyshev’s inequality (and using the lemmas A1-A3 successivly) we get for the first series

P ( ∑_{k=N}^{∞} p_{k}e^{− \frac{k}{t_{n+1}}} \{|X_{k}| − E(|X_{k}|)\} ≥ ϵφ(t_{n}))

≤ \frac{∑_{k=N}^{∞} p_{k}^{2}e^{− \frac{2k}{t_{n+1}}} V ar (|X_{k}| − E(|X_{k}|))}{ϵ^{2}φ(t_{n})^{2}}

≤ c(α)\frac{(log t_{n})^{2α+1}}{(log log t_{n})tn^{\frac{3}{2}}} .

Since this is the general term of a summable sequence, the Lemma is proved.

We now compare the behavior of the series with parameter values between t_{n} and t_{n+1} .

Lemma 2 For a sequence (p_{n}) satisfying (1.6) we have

(2.4)  lim_{t→∞} sup_{t_{n}≤t≤t_{n+1}} |\frac{p_{X}(t)}{φ(t)} − \frac{p_{X}(t_{n})}{φ(t_{n})}| = 0 a.s..

Proof: Since N(t_{n+1})/N(t_{n}) → 1 (n → ∞) we can (compare the proof of Lemma 1) use N := N(t_{n}) as upper index in the range t_{n} ≤ t ≤ t_{n+1} . Put S_{n} =

∑_{k=0}^{n} p_{k}X_{k}; then we get by Abel’s partial summation

∑_{k=0}^{N} p_{k}X_{k} (\frac{e^{− \frac{k}{t}}}{φ(t)} − \frac{e^{− \frac{k}{tn}}}{φ(t_{n})})

= ∑_{k=0}^{N} S_{k} (\frac{e^{− \frac{k}{t}} − e^{− \frac{k+1}{t}}}{φ(t)} − \frac{e^{− \frac{k}{tn}} − e^{− \frac{k+1}{tn}}}{φ(t_{n})} ) + S_{N+1} (\frac{e^{− \frac{N+1}{t}}}{φ(t)} − \frac{e^{− \frac{N+1}{tn}}}{φ(t_{n})} )

=: I + II.

Consider the sup of the second part first.

II ≤ sup_{t_{n}<t≤t_{n+1}} |S_{N+1}| |\frac{e^{− \frac{N+1}{t}}}{φ(t)} − \frac{e^{− \frac{N+1}{tn}}}{φ(t_{n})} |

≤ 2 (\frac{|S_{N+1}|}{(2P_{N+1}^{(2)} loglog P_{N+1}^{(2)})^{\frac{1}{2}}} ) (\frac{(2P_{N+1}^{(2)} loglog P_{N+1}^{(2)})^{\frac{1}{2}}}{φ(t_{n})} e^{− \frac{N+1}{t_{n+1}}} ) .

The first factor is bounded by the Chow-Teicher [5] LIL for weighted means. By Lemma A1 we see that the quotient of the second factor behaves like a slowly vary-

ing function. By Lemma A3 e^{− \frac{N+1}{t_{n+1}}} ≤ 1/\sqrt{t_{n+1}}. Hence we have II = o (1) (n → ∞) . To estimate the first part we start by using the triangle inequality and get

I = |(\frac{1 − e^{− \frac{1}{t}}}{φ(t)} ) ∑_{k=0}^{N} S_{k}e^{− \frac{k}{t}} − (\frac{1 − e^{− \frac{1}{tn}}}{φ(t_{n})} ) ∑_{k=0}^{N} S_{k}e^{− \frac{k}{t}}

+ (\frac{1 − e^{− \frac{1}{tn}}}{φ(t_{n})} ) ∑_{k=0}^{N} S_{k}e^{− \frac{k}{t}} − (\frac{1 − e^{− \frac{1}{tn}}}{φ(t_{n})} ) ∑_{k=0}^{N} S_{k}e^{− \frac{k}{tn}} |

≤ |(\frac{(1 − e^{− \frac{1}{t}})}{φ(t)} − \frac{(1 − e^{− \frac{1}{tn}})}{φ(t_{n})}) ∑_{k=0}^{N} S_{k}e^{− \frac{k}{t}} |

+ \frac{(1 − e^{− \frac{1}{tn}})}{φ(t_{n})} ∑_{k=0}^{N} |S_{k}| (e^{− \frac{k}{t}} − e^{− \frac{k}{tn}} ) .

Taking the sup over the range t_{n} ≤ t ≤ t_{n+1} we get

\begin{aligned}I & ≤ & sup_{t_{n}<t≤t_{n+1}} |\frac{(1 − e^{− \frac{1}{t}})}{φ(t)} − \frac{(1 − e^{− \frac{1}{tn}})}{φ(t_{n})}| ∑_{k=0}^{N} |S_{k}|e^{− \frac{k}{t_{n+1}}} \\ & + & \frac{(1 − e^{− \frac{1}{tn}})}{φ(t_{n})} ∑_{k=0}^{N} |S_{k}| (e^{− \frac{k}{t_{n+1}}} − e^{− \frac{k}{tn}} ) := I_{1} + I_{2}. \end{aligned}

Now using the elementary inequality |1 − e^{x}| ≤ |x| for x close to 0 and Lemma A3, we get for t_{n} ≤ t ≤ t_{n+1}

|\frac{1 − e^{− \frac{1}{t}}}{φ(t)} − \frac{1 − e^{− \frac{1}{tn}}}{φ(t_{n})} | ≤ \frac{(1 − \frac{φ(t_{n})}{φ(t_{n+1})}) + (1 − \frac{t_{n}}{t_{n+1}} )}{φ(t_{n})t_{n}} ≤ \frac{α + 2}{φ(t_{n})t_{n}(log n)^{α+2}} .

To estimate |S_{n}| we use the Chow-Teicher LIL for weighted means, see [5]. We start by splitting the sum in I_{1} in two parts and set 8 Law of the Iterated logarithm N_{2} := N_{2}(t_{n}) := 2(α + 2)t_{n} loglog t_{n}. Using Lemmas A1 and A3 we get for n large enough

∑_{k=N_{2}}^{N} |S_{k}|e^{− \frac{k}{t_{n+1}}} ≤ e^{− \frac{N_{2}}{t_{n+1}}} ∑_{k=N_{2}}^{N} |S_{k}| ≤ 2e^{− \frac{N_{2}}{t_{n+1}}} N(P_{N+1}^{(2)} loglog(P_{N+1}^{(2)}))^{\frac{1}{2}}
≤ 2c(α)(log t_{n})^{−(α+2)}(t_{n} log t_{n})^{α+ \frac{3}{2}} L(t_{n} log t_{n}) log log(t_{n} log t_{n}).

Combining the last two inequalities we get

sup_{t_{n}<t≤t_{n+1}} |\frac{(1 − e^{− \frac{1}{t}})}{φ(t)} − \frac{(1 − e^{− \frac{1}{tn}})}{φ(t_{n})}| ∑_{k=N_{2}}^{N} |S_{k}|e^{− \frac{k}{t_{n+1}}}

≤ c(α)\frac{(t_{n} log t_{n})^{α+ \frac{3}{2}} L(t_{n} log t_{n}) log log(t_{n} log t_{n})}{(log t_{n})^{α+2}t_{n}(t_{n})^{α+ \frac{1}{2}} L(t_{n}) log log(t_{n})} → 0.

For the first part of the sum in I_{1} we get the similar bound

∑_{k=0}^{N_{2}} |S_{k}|e^{− \frac{k}{t_{n+1}}} ≤ ∑_{k=0}^{N_{2}} |S_{k}| ≤ 2N_{2}(2P_{N_{2}}^{(2)} loglog(N_{2}^{2}P_{N_{2}}^{(2)}))^{\frac{1}{2}}
≤ c(α)(t_{n} log log t_{n})^{α+ \frac{3}{2}} L(t_{n} log log t_{n})(log log(t_{n} log log t_{n}))^{\frac{1}{2}} .

With these estimates we get

sup_{t_{n}<t≤t_{n+1}} |\frac{(1 − e^{− \frac{1}{t}})}{φ(t)} − \frac{(1 − e^{− \frac{1}{tn}})}{φ(t_{n})}| ∑_{k=0}^{N_{2}} |S_{k}|e^{− \frac{k}{t_{n+1}}}

≤ c(α)\frac{(t_{n} log log t_{n})^{α+ \frac{3}{2}} L(t_{n} log log t_{n})(log log(t_{n} log log t_{n}))^{\frac{1}{2}}}{t_{n}^{α+ \frac{3}{2}} L(t_{n})(log log t_{n})^{\frac{1}{2}} (log n)^{(α+2)}}
≤ c(α)\frac{(log(n/(log n)^{α+2})^{α+ \frac{3}{2}}}{(log n)^{α+2}} → 0 (n → ∞).

Rüdiger Kiesel 9 So it remains to estimate I_{2} . We have

\frac{(1 − e^{− \frac{1}{tn}})}{φ(t_{n})} ∑_{k=0}^{N} |S_{k}| (e^{− \frac{k}{t_{n+1}}} − e^{− \frac{k}{tn}} )

= (e^{− \frac{1}{t_{n+1}}} − e^{− \frac{1}{tn}} ) \frac{(1 − e^{− \frac{1}{tn}})}{φ(t_{n})} ∑_{k=0}^{N} |S_{k}| ∑_{ν=0}^{k−1} e^{− \frac{ν}{t_{n+1}}} e^{− \frac{k−ν}{tn}}
≤ (e^{− \frac{1}{t_{n+1}}} − e^{− \frac{1}{tn}} ) \frac{(1 − e^{− \frac{1}{tn}})}{φ(t_{n})} ∑_{k=0}^{N} |S_{k}|ke^{− \frac{k}{t_{n+1}}} .

Repeating the above argument we get

(e^{− \frac{1}{t_{n+1}}} − e^{− \frac{1}{tn}} ) \frac{(1 − e^{− \frac{1}{tn}})}{φ(t_{n})} ≤ \frac{1}{t_{n}^{2}φ(t_{n})(log n)^{α+2}}

Using again the Chow-Teicher LIL for weighted means we get the bound

lim sup_{n→∞} ∑_{k=0}^{n} |S_{k}|k ≤ c(α)n^{2}(P_{n}^{(2)} log log P_{n}^{(2)})^{\frac{1}{2}} .

We now can repeat the previous argument splitting the sum this time at N_{2} = 2(α + 3)t_{n} log log t_{n}. We proceed using de Acosta’s (k and t - dependent) truncation technique. Set

X_{k} = X_{k}^{d}(t) + X_{k}^{u}(t)

with

X_{k}^{d}(t) = X_{k}1 (|X_{k}| ≤ τ \frac{e^{\frac{k}{t}}}{p_{k}} \sqrt{\frac{2V (t)}{log logV (t)}})
X_{k}^{u}(t) = X_{k}1 (|X_{k}| > τ \frac{e^{\frac{k}{t}}}{p_{k}} \sqrt{\frac{2V (t)}{log logV (t)}})

where τ > 0 is a real number to be specified later. Let

p_{X}^{u}(t) := ∑_{k=0}^{N(t)} p_{k}X_{k}^{u}(t)e^{− \frac{k}{t}} .

Lemma 3 Let (p_{n}) satisfy conditions (1.1) and (1.6) with α > −\frac{1}{2} , then we have:

lim_{n→∞} \frac{p_{X}^{u}(t_{n})}{φ(t_{n})} = 0 a.s..

Proof: We start by showing

(2.5)  lim_{n→∞} φ(t_{n})^{−1} ∑_{k=0}^{t_{n}} p_{k}X_{k}^{u}(t_{n})e^{− \frac{k}{tn}} = 0 a.s..

We have

|φ(t_{n})^{−1} ∑_{k=0}^{t_{n}} p_{k}X_{k}^{u}(t_{n})e^{− \frac{k}{tn}} | ≤ φ(t_{n})^{−1} ∑_{k=0}^{t_{n}} p_{k}|X_{k}|1 (|X_{k}| > τ \frac{e^{\frac{k}{t}}}{p_{k}} \sqrt{\frac{2V (t_{n})}{log logV (t_{n})}}) e^{− \frac{k}{tn}} .

Using Lemma A1 we see that

\sqrt{\frac{2V (t_{n})}{log logV (t_{n})}} ≥ (Γ(2α + 1))^{\frac{1}{2}} \sqrt{\frac{t_{n}^{2α+1}L(t_{n})}{log logt_{n}}} =: h(t_{n}).

Hence

1 (|X_{k}| > τ \frac{e^{\frac{k}{tn}}}{p_{k}} \sqrt{\frac{2V (t_{n})}{log logV (t_{n})}}) ≤ 1 (|X_{k}| > τ \frac{e^{\frac{k}{tn}}}{p_{k}} h(t_{n})) .

In the range k_{0} ≤ k ≤ t_{n} we have h(t_{n})e^{k/t_{n}} ≥ h(k)e and so (with k_{0} large enough),

(2.6) ∑_{k=k_{0}}^{t_{n}} p_{k}|X_{k}|1 (|X_{k}| > τ \frac{e^{\frac{k}{tn}}}{p_{k}} h(t_{n})) e^{− \frac{k}{tn}} ≤ ∑_{k=k_{0}}^{t_{n}} p_{k}|X_{k}|1 (|X_{k}| > τ \frac{e}{p_{k}} h(k)) .

By Kronecker’s Lemma, see [6] p.114, (2.5) will tend to 0 if

∑_{k=k_{0}}^{∞} |X_{k}|1 (|X_{k}| > τ \frac{e}{p_{k}} h(k)) \frac{p_{k}}{φ(k)} < ∞ a.s..

This will follow if the expectation of the series is finite. Using the asymptotic results for regularly varying functions we get

∑_{k=k_{0}}^{∞} \frac{p_{k}}{φ(k)}E (|X_{k}|1 (|X_{k}| > τ \frac{e}{p_{k}} h(k)))

= ∑_{k=k_{0}}^{∞} \frac{p_{k}}{φ(k)} ∑_{j=k}^{∞} E (|X_{k}|1 (eτ \frac{h(j)}{p_{j}} < |X_{k}| ≤ eτ \frac{h(j + 1)}{p_{j+1}}))

≤ ∑_{k=k_{0}}^{∞} \frac{p_{k}}{φ(k)} ∑_{j=k}^{∞} \frac{eτh(j + 1)}{p_{j+1}}P (eτ \frac{h(j)}{p_{j}} < |X| ≤ eτ \frac{h(j + 1)}{p_{j+1}})

≤ ∑_{j=k_{0}}^{∞} \frac{eτh(j + 1)}{p_{j+1}}P (eτ \frac{h(j)}{p_{j}} < |X| ≤ eτ \frac{h(j + 1)}{p_{j+1}}) ∑_{k=k_{0}}^{j} \frac{p_{k}}{φ(k)}

≤ eτ ∑_{j=k_{0}}^{∞} \frac{jL(j)}{log log j} P (eτ\sqrt{\frac{jL(j)}{log log j}} < |X| ≤ eτ\sqrt{\frac{(j + 1)L(j + 1)}{log log(j + 1)}}) < ∞,

where the convergence of the series follows from the fact that E(X^{2}) < ∞ is equivalent to the convergence of series of type ∑ kP (\sqrt{k} ≤ |X| ≤ \sqrt{k + 1}) .

We are left with showing

(2.7)  \frac{1}{φ(t_{n})} ∑_{k=t_{n}+1}^{t_{n} log t_{n}} p_{k}e^{− \frac{k}{tn}} X_{k}^{u}(t_{n}) → 0 a.s..

Set S_{j}^{u}(t_{n}) = ∑_{k=0}^{j} p_{k}X_{k}^{u}(t_{n}) . We use the notation N(t_{n}) := t_{n} log t_{n} and omit the

argument t_{n} in the following. By Abel’s partial summation we get

∑_{k=t_{n}+1}^{N} p_{k}e^{− \frac{k}{tn}} X_{k}^{u} = S_{N}^{u} exp \{− log t_{n}\}−S_{t_{n}+1}^{u} exp \{−\frac{t_{n} + 1}{t_{n}}\}+ ∑_{k=t_{n}+1}^{N}(1−e^{− \frac{1}{tn}})e^{− \frac{k}{tn}} S_{k}^{u}.

Analysing the above argument we see that we have in fact shown that for any sequence (k_{ν}) with k_{ν} → ∞ (ν → ∞) we have

(2.8)  \frac{1}{φ(k_{ν})}S_{k_{ν}}^{u}(t_{n}) → 0 (ν → ∞) a.s..

Using Lemma A1 and the above result we get

\frac{S_{N}^{u}e^{− log t_{n}}}{φ(t_{n})} ≤ (\frac{S_{N}^{u}}{φ(N)}) (\frac{φ(N)}{φ(t_{n})t_{n}} ) → 0 a.s..

12 Law of the Iterated logarithm So we are left with discussing the sum, i.e.

Z_{n} := (t_{n}φ(t_{n}))^{−1} ∑_{k=t_{n}}^{N} e^{− \frac{k}{tn}} |S_{k}^{u}|.

With Y_{k} := |S_{k}^{u}|/φ(k) we have

Z_{n} = (t_{n}φ(t_{n}))^{−1} ∑_{k=t_{n}}^{N} e^{− \frac{k}{tn}} φ(k)Y_{k} ≤ ( max_{t_{n}≤k≤t_{n} log t_{n}} Y_{k}) (t_{n}φ(t_{n}))^{−1} ∑_{k=t_{n}}^{N} e^{− \frac{k}{tn}} φ(k).

Since φ is regularly varying with index α + \frac{1}{2} , we have by Lemma A2

∑_{k=t_{n}}^{N} e^{− \frac{k}{tn}} φ(k) ≤ c(α)t_{n}^{α+ \frac{3}{2}} (log log t_{n})^{\frac{1}{2}} = O (φ(t_{n})t_{n}) .

Now by (2.8) sup_{k≥t_{n}} Y_{k} → 0 a.s and hence we obtain Z_{n} → 0 (n → ∞) a.s..

Lemma 4 We have

lim sup_{n→∞} \frac{p_{X}^{d}(t_{n})}{φ(t_{n})} ≤ 1 a.s. and lim inf_{n→∞} \frac{p_{X}^{d}(t_{n})}{φ(t_{n})} ≥ −1 a.s..

Proof: Notice first that

E (\frac{p_{X}^{d}(t_{n})}{φ(t_{n})}) = −E (\frac{p_{X}^{u}(t_{n})}{φ(t_{n})}) → 0 (n → ∞).

Let Y_{k}^{d}(t) := X_{k}^{d}(t) − E(X_{k}^{d}(t)) , then Y_{k}^{d}(t) has mean = 0 and variance

E((Y_{k}^{d}(t))^{2}) = E((X_{k}^{d}(t) − E(X_{k}^{d}(t)))^{2}) ≤ E(X_{k}^{2}) = 1.

Furthermore we have

p_{Y}^{d} (t) = ∑_{k=0}^{N(t_{n})} Y_{k}^{d}(t)p_{k}e^{− \frac{k}{tn}} = p_{X}^{d}(t) − E(p_{X}^{d}(t)).

We compute an estimate of the moment generating function by using the standard exponential inequalities, see [2] p.276 and p.278,

E (exp \{\frac{sp_{Y}^{d} (t_{n})}{φ(t_{n})}\}) = ∏_{k=0}^{N(t_{n})} E (exp \{\frac{sp_{k}Y_{k}(t_{n})e^{− \frac{k}{tn}}}{φ(t_{n})} \})

≤ ∏_{k=0}^{N(t_{n})} (1 + \frac{s^{2}p_{k}^{2}e^{− \frac{2k}{tn}}}{2φ(t_{n})^{2}} E (exp \{\frac{|s||Y_{k}(t)|p_{k}e^{− \frac{k}{tn}}}{φ(t_{n})} \})) .

Now |Y_{k}(t_{n})| ≤ τ\sqrt{2V (t_{n})/ log log V (t_{n})}e^{\frac{k}{tn}} /p_{k}, and using the inequality 1 + a ≤ e^{a} we get

E (exp \{\frac{sp_{Y}^{d} (t_{n})}{φ(t_{n})}\}) ≤ ∏_{k=0}^{N(t_{n})} (1 + \frac{s^{2}p_{k}^{2}e^{− \frac{2k}{tn}}}{2φ(t_{n})^{2}} exp \{\frac{|s|τ}{log log V (t_{n})}\})

≤ exp \{\frac{s^{2}}{2} φ(t_{n})^{−2} ∑_{k=0}^{t_{n} log t_{n}} p_{k}^{2}e^{− \frac{2k}{tn}} exp \{\frac{|s|τ}{log log V (t_{n})}\}\}

≤ exp \{\frac{s^{2}}{4 log log V (t_{n})} exp \{\frac{|s|τ}{log log V (t_{n})}\}\} .

Now we use the exponential Markov-inequality to get for all s > 0, γ > 0

\begin{aligned}P (\frac{p_{Y} (t_{n})}{φ(t_{n})} ≥ 1 + γ) & = & P (exp \{\frac{sp_{Y} (t_{n})}{φ(t_{n})}\} ≥ e^{s(1+γ)}) \\ & ≤ & e^{−s(1+γ)}E (exp \{\frac{sp_{Y} (t_{n})}{φ(t_{n})}\}) . \end{aligned}

Hence for s = 2(1 + γ) log log V (t_{n}) we obtain

P (\frac{p_{Y} (t_{n})}{φ(t_{n})} ≥ 1 + γ) ≤ exp \{−(1 + γ)^{2} log log V (t_{n}) (2 − e^{2τ(1+γ)})\} .

For any γ > 0 we can choose τ > 0 small enough such that

(1 + γ)^{2} (2 − e^{2τ(1+γ)}) = 1 + δ > 1.

Now we can bound the above probability by the general term of a summable series and using the Borel-Cantelli Lemma we obtain that the lim sup ≤ 1 a.s.. The argument for the lim inf is analogous.

We close this section by putting together all our Lemmas to the final result.

Proposition 1 Let X_{0}, X_{1}, … be an i.i.d. sequence of random variables with E(X_{0}) = 0, E(X_{0}^{2}) = 1 and let (p_{k}) be a sequence of weights satisfying (1.1) and (1.6) with α > −\frac{1}{2} . Then

lim sup_{t→∞} \frac{∑_{k=0}^{∞} p_{k}X_{k}e^{− \frac{k}{t}}}{φ(t)} ≤ 1 a.s.

and

lim inf_{t→∞} \frac{∑_{k=0}^{∞} p_{k}X_{k}e^{− \frac{k}{t}}}{φ(t)} ≥ −1 a.s..

3 A Converse to the Law of Iterated logarithm

Proposition 2 Let (X_{n}) be a sequence of i.i.d. random variables and (p_{n}) be a sequence of weights satisfying (1.1) and (1.6) with α > −\frac{1}{2} . Then

P (lim sup_{t→∞} \frac{|p_{X}(t)|}{φ(t)} < ∞) > 0

implies E(X_{0}) = 0 and E(X_{0}^{2}) < ∞.

Proof: We adapt an argument by Feller ([2, 19], pp.297–301, pp.297–298). By the 0−1 -Law we get

lim sup_{t→∞} \frac{|p_{X}(t)|}{φ(t)} < ∞ a.s..

By Lemma A1

φ(t) = \sqrt{2V (t) log log V (t)} = O (t^{α+ \frac{1}{2}} L(t)(log log t)^{\frac{1}{2}} ) = o (t^{α+1}L(t)) = O (p(t)) ,

hence using the converse of the Strong Law of Large Numbers for power series methods, c.f. [13, 14], it follows that E(|X_{0}|) < ∞ and E(X_{0}) = 0. By the usual symmetrisation argument, we can w.l.o.g. assume that the X_{n} are symmetric.

We suppose that E(X_{0}^{2}) = ∞ and show that this will lead to a contradiction.

Fix M > 0 and define Y_{n} := X_{n}1 (|X_{n}| ≤ α_{M}) , where we choose α_{M} in such a way that E(Y_{n}^{2}) > M .

By symmetry the Y_{0}, Y_{1}, … are i.i.d. with E(Y_{n}) = 0 and E(Y_{n}^{2}) < ∞. Hence they obey the LIL of Proposition 1 and using in particular the same sequence (t_{n}) as above we have

P (p_{Y} (t_{n}) > \sqrt{2MV (t_{n}) log log V (t_{n})} i.o.) = 1.

Let Z_{n} := X_{n}1 (|X_{n}| > α_{M}) and X̃_{n} := Y_{n} −Z_{n}. By symmetry X_{n} and X̃_{n} have the same distribution for each n. Therefore the two events

A_{+}^{M} := \{p_{Y} (t_{n}) > \sqrt{2MV (t_{n}) log log V (t_{n})}, p_{X}(t_{n}) − p_{Y} (t_{n}) ≥ 0 i.o.\}

A_{−}^{M} := \{p_{Y} (t_{n}) > \sqrt{2MV (t_{n}) log log V (t_{n})}, p_{X}(t_{n}) − p_{Y} (t_{n}) ≤ 0 i.o.\}

have the same probability, since they are given by {X ∈ B} resp. {X̃ ∈ B} with B a Borel-set in R I ^{N I} , compare the argument on p.299 in [2]. Now

A_{+}^{M} ∪ A_{−}^{M} = \{p_{Y} (t_{n}) > \sqrt{2MV (t_{n}) log log V (t_{n})}, i.o.\} .

Using the subadditivity of probability measures this implies

P (A_{+}^{M}) + P (A_{−}^{M}) ≥ P (A_{+}^{M} ∪ A_{−}^{M}) = 1,

and hence P (A_{+}^{M}) ≥ \frac{1}{2} . So we obtain for arbitrary M > 0 that P (A_{+}^{M}) ≥ \frac{1}{2} . Using the inclusion

\{lim sup_{n→∞} \frac{p_{X}(t_{n})}{\sqrt{2V (t_{n}) log log V (t_{n})}} = +∞\}

⊇ ⋂_{M=1}^{∞} \{p_{Y} (t_{n}) > \sqrt{2MV (t_{n}) log log V (t_{n})}, p_{X}(t_{n}) − p_{Y} (t_{n}) ≥ 0 i.o.\}

and that the sequence A_{+}^{M} is decreasing and the continuity of the probability 16 Law of the Iterated logarithm measure we get

P (lim sup_{n→∞} \frac{p_{X}(t_{n})}{\sqrt{2V (t_{n}) log log V (t_{n})}} = +∞)

≥ P ( ⋂_{M=1}^{∞} A_{+}^{M} ) = lim_{M→∞} P (A_{+}^{M}) ≥ \frac{1}{2}.

Therefore the 0−1 -Law implies that

P (lim sup_{n→∞} \frac{p_{X}(t_{n})}{\sqrt{2MV (t_{n})^{2} log log V (t_{n})^{2}}} = +∞) = 1,

which is a contradiction.

4 The cluster set

Proposition 3 To each ρ ∈ [−1, 1] there exists a sequence t_{n} → ∞ such that

lim_{n→∞} \frac{p_{X}(t_{n})}{φ(t_{n})} = ρ a.s..

Proof: Define (log n)! := ∏_{k=2}^{n} log k and with this notation

t_{n} := n!(log n)!(log log n)!,
N_{2} := N(t_{n}) := t_{n} log t_{n} ∼ n!(log n)!(log log n)!n log n,
N_{1} := N_{1}(t_{n}) := t_{n}(log n)^{−1} = n!(log n)!(log log n)!(log n)^{−1}.

Observe that N_{2}/N_{1} ∼ n(log n)^{2} . We split p_{X}(t_{n}) in three parts, namely

\begin{aligned}p_{X}(t_{n}) & = & ∑_{k=0}^{N_{1}−1} p_{k}X_{k}e^{− \frac{k}{tn}} + ∑_{k=N_{1}}^{N_{2}} p_{k}X_{k}e^{− \frac{k}{tn}} + ∑_{k=N_{2}+1}^{∞} p_{k}X_{k}e^{− \frac{k}{tn}} \\ & =: & p_{X}^{(1)}(t_{n}) + p_{X}^{(2)}(t_{n}) + p_{X}^{(3)}(t_{n}). \end{aligned}

Rüdiger Kiesel 17 By Lemma 1 we immediatly get

lim_{n→∞} \frac{p_{X}^{(3)}(t_{n})}{φ(t_{n})} = 0 a.s..

Furthermore

\frac{p_{X}^{(1)}(t_{n})}{φ(t_{n})} = \frac{| ∑_{k=0}^{N_{1}−1} p_{k}X_{k}e^{− \frac{k}{tn}} |}{\sqrt{2V (t_{n}) log log V (t_{n})}}
≤ \frac{| ∑_{k=0}^{N_{1}−1} p_{k}X_{k}e^{− \frac{k}{tn}} |}{\sqrt{2 ∑_{k=0}^{N_{1}−1} p_{k}^{2}e^{− \frac{2k}{tn}} log log V (t_{n})}} × \sqrt{\frac{∑_{k=0}^{N_{1}−1} p_{k}^{2}e^{− \frac{2k}{tn}}}{V (t_{n})}} .

By the Chow-Teicher LIL [5], we have

\frac{| ∑_{k=0}^{N_{1}−1} p_{k}X_{k}|}{\sqrt{2 ∑_{k=0}^{N_{1}−1} p_{k}^{2} log log N_{1}}} ≤ 1 a.s..

Since in our range of consideration e^{− \frac{k}{tn}} is always close to 1 , we can repeat their arguments to get

\frac{|∑_{k=0}^{N_{1}} p_{k}X_{k}e^{− \frac{k}{tn}} |}{\sqrt{2 ∑_{k=0}^{N_{1}} p_{k}^{2}e^{− \frac{2k}{tn}} log log V (t_{n})}} ≤ 1 a.s..

On the other hand by Lemma A1 we get

\frac{∑_{k=0}^{N_{1}−1} p_{k}^{2}e^{− \frac{2k}{tn}}}{V (t_{n})} ≤ c(α)\frac{(t_{n}/ log t_{n})^{2α+1}L(t_{n}/ log t_{n})}{t_{n}^{2α+1}L(t_{n})} → 0 (n → ∞).

Hence

lim_{n→∞} \frac{p_{X}^{(1)}(t_{n})}{φ(t_{n})} = 0 a.s..

Since for large n N_{1}(t_{n+1}) > N_{2}(t_{n}) we see that p_{X}^{(2)}(t_{n+1}) and p_{X}^{(2)}(t_{n}) are independent for such n. If we set

A_{n}^{ϵ} := \{|\frac{p_{X}^{(2)}(t_{n})}{φ(t_{n})} − ρ| < ϵ\}

for ϵ > 0 arbitrary, we are left with showing P (A_{n}^{ϵ} i.o.) = 1 .

So by the Borel-Cantelli Lemma our proof is finished, if we can show

Lemma 5

∑_{n=2}^{∞} P (A_{n}^{ϵ}) = ∞ for each ϵ > 0.

Proof: We represent p_{X}^{(2)}(t_{n}) as the average of q_{n} := τ^{−2}2 log log V (t_{n}) independent random variables, each of which converges in distribution to a Gaussian random variable of variance 1 by the Embrechts-Maejima Central Limit Theorem, see [7]. To do this we set

σ_{n}^{2} := ∑_{k=N_{1}}^{N_{2}} p_{k}^{2}e^{− \frac{2k}{tn}}

and define numbers p_{ln} for 0 ≤ l ≤ q_{n} and each n recursivly by

\begin{aligned}p_{0n} & := & N_{1}, \\ p_{ln} & := & inf \{m : ∑_{k=p_{l−1,n}+1}^{m} p_{k}^{2}e^{− \frac{2k}{tn}} ≥ \frac{σ_{n}^{2}}{q_{n}} \} , \\ p_{q_{n}n} & := & N_{2}. \end{aligned}

Since

\frac{p_{k}^{2}e^{− \frac{2k}{tn}}}{σ_{n}^{2}} ≤ c_{1}(α)\frac{(t_{n} log t_{n})^{2α}L^{2}(t_{n} log t_{n})}{t_{n}^{2α+1}L(t_{n})} ≤ c_{2}(α)\frac{(log t_{n})^{2α}L(t_{n} log t_{n})}{t_{n}} → 0 (n → 0)

uniformly in k the {p_{ln}; n ∈ N I , 0 ≤ l ≤ q_{n}} are well-defined. Furthermore

p_{ln} − p_{l−1,n} ≥ c(α)\frac{t_{n}}{(log t_{n})^{2α}L(t_{n} log t_{n}) log log t_{n}} → ∞ (n → ∞)

uniformly in l. We define

Z̃_{ln} := ∑_{k=p_{l−1,n}+1}^{p_{ln}} p_{k}X_{k}e^{− \frac{k}{tn}} ,

then E(Z̃_{ln}) = 0 and

E(Z̃_{ln}^{2}) = ∑_{k=p_{l−1,n}+1}^{p_{ln}} p_{k}^{2}e^{− \frac{2k}{tn}} ∼ \frac{σ_{n}^{2}}{q_{n}} .

Hence

Z_{ln} := \sqrt{\frac{q_{n}}{σ_{n}^{2}}} ∑_{k=p_{l−1,n}+1}^{p_{ln}} p_{k}X_{k}e^{− \frac{k}{tn}}

has mean 0 and variance ≈ 1 . Furthermore

∑_{l=1}^{q_{n}} Z_{ln} = \frac{\sqrt{q_{n}}}{σ_{n}} ∑_{k=N_{1}}^{N_{2}} p_{k}X_{k}e^{− \frac{k}{tn}} = \frac{\sqrt{q_{n}}}{σ_{n}} p_{X}^{(2)}(t_{n}).

Therefore

\frac{p_{X}^{(2)}(t_{n})}{φ(t_{n})} = \frac{σ_{n} ∑_{l=1}^{q_{n}} Z_{ln}}{\sqrt{q_{n}2V (t_{n}) log log V (t_{n})}} = \frac{∑_{l=1}^{q_{n}} Z_{ln}}{τq_{n}} (1 + o (1)).

Hence for any interval [c, d] ⊂ [−1, 1] with ρ ∈ [c, d] we can use the independence of the Z_{ln} to get

P (\frac{p_{X}^{(2)}(t_{n})}{φ(t_{n})} ∈ [c, d]) ≥ ∏_{l=1}^{q_{n}} P (Z_{ln} ∈ τ[c, d]) .

For the Z_{ln} the Embrechts-Maejima CLT [7], Theorem 1 (see also the Remark), applies and we have uniformly in l

P (Z_{ln} ∈ τ[c, d]) → γ(τ[c, d]) (n → ∞)

with

γ(τ[c, d]) = \frac{1}{\sqrt{2π}} ∫_{cτ}^{dτ} exp\{−x^{2}/2\}dx.

Now let c = ρ − ϵ, d = ρ + ϵ, then by ([2], pp.294,295)

γ(τ[c, d]) ≥ exp{−(τ^{2}ρ^{2}/2) + log(γ([−τϵ, τϵ]))}.

20 Law of the Iterated logarithm Therefore

P (\frac{p_{X}^{(2)}(t_{n})}{φ(t_{n})} ∈ [c, d]) ≥ exp\{−q_{n}(τ^{2}ρ^{2}/2) + q_{n} log(γ([−τϵ, τϵ]))\}
= exp{log log V (t_{n})^{2}(−ρ^{2} + τ^{−2}2 log(γ([−τϵ, τϵ])))},

and we can choose τ large enough, so that

ρ^{2} − τ^{−2}2 log(γ([−τϵ, τϵ])) = δ < 1.

Since by Lemma A1 log log V (t_{n}) ∼ log(n log n) , we get

P (\frac{p_{X}^{(2)}(t_{n})}{φ(t_{n})} ∈ [c, d]) ∼ \frac{1}{(n log n)^{δ}}

and this is the general term of a divergent series. Remark: Applicability of the Embrechts-Maejima CLT. We set

W_{ln} := \frac{∑_{k=p_{l−1,n}+1}^{p_{ln}} p_{k}X_{k}e^{− \frac{k}{tn}}}{∑_{k=p_{l−1,n}+1}^{p_{ln}} p_{k}e^{− \frac{k}{tn}}} = ∑_{k=0}^{∞} c_{k}(ln)X_{k}.

Then we have ∑_{k=0}^{∞} c_{k}(ln) = 1 and

d_{ln}^{2} := ∑_{k=0}^{∞} c_{k}(ln) = \frac{∑_{k=p_{l−1,n}+1}^{p_{ln}} p_{k}^{2}e^{− \frac{2k}{tn}}}{( ∑_{k=p_{l−1,n}+1}^{p_{ln}} p_{k}e^{− \frac{k}{tn}} )^{2}} ≤ c(α)\frac{t_{n}^{2α+1}L^{2}(t_{n})}{t_{n}^{2α+2}L^{2}(t_{n}) log log t_{n}} → 0 (n → ∞).

Since furthermore

\frac{sup_{k} c_{k}(ln)}{d_{ln}} = \frac{1}{d_{ln}} sup_{k} \{\frac{p_{k}e^{− \frac{k}{tn}}}{∑_{j=p_{l−1,n}+1}^{p_{ln}} p_{j}e^{− \frac{j}{tn}}} \} → 0 (n → ∞)

the Embrechts-Maejima [7] CLT applies to W_{ln}, which means that Z_{ln} = \frac{1}{d_{ln}} W_{ln} converges in law to a Gaussian random variable with mean 0 and variance 1 .

5 The LIL for (N_{βp}) -methods

5.1 The upper bound

We use the following notations:

\begin{aligned}S_{n}^{β} & := & ∑_{j=0}^{n} \binom{n − j + β − 1}{n − j}p_{j}X_{j}, \\ U_{n}^{β} & := & ∑_{j=0}^{n} (\binom{n − j + β − 1}{n − j})^{2} p_{j}^{2}, \\ a_{n} & := & \sqrt{2U_{n}^{β} log log U_{n}^{β}}. \end{aligned}

Observe that by Lemma A1 a_{n}^{2} ∼ 2B(2α + 1, 2β − 1)n^{2(α+β)−1} log log n. We want to show that

lim sup_{n→∞} \frac{S_{n}^{β}}{a_{n}} ≤ 1 a.s..

To do this, we use a general theorem on the Law of Iterated logarithm for triangular matrix methods by Stadtmüller [18] (we use notation of this paper in the following). Since U_{n}^{β} is regularly varying (with index 2(α + β) − 1 ), we only have to show that the conditions (1.4) and (1.5a) of Stadtmüller’s Theorem 1 hold true. In our case (1.4) reduces to

(5.1) lim_{ρ→1+} lim inf_{n→∞} min_{m:1≤n/m≤ρ} ∑_{j=0}^{m} \binom{m − j + β − 1}{m − j}\binom{n − j + β − 1}{n − j}p_{j}^{2}/U_{m}^{β} = 1.

Choosing m = n we immediately see that in (5.1) the inequality ≤ 1 holds true. Using the properties of the binomial coefficients, we get that the left-hand side of (5.1) is greater or equal to

lim_{ρ→1+} lim inf_{n→∞} (1 + \frac{(β − 1)}{n})^{(ρ−1)n} = lim_{ρ→1+} exp\{(β − 1)(ρ − 1)\} = 1.

22 Law of the Iterated logarithm Similarly, condition (1.5a) reduces in our context to

\frac{1}{n^{α+β−1}L(n)} ∑_{j=2}^{n} |\binom{n − j + β − 1}{n − j}p_{j} − \binom{n − j + β}{n − j + 1}p_{j−1}| = O (1)

and this can be seen by using the well-known identities for the binomial coefficients.

5.2 The cluster set

We want to show: To each ρ ∈ [−1, 1] there exists a sequence n_{k} → ∞ such that

lim_{n→∞} \frac{S_{n_{k+1}}^{β}}{a_{n_{k+1}}} = ρ a.s..

To prove this we set

n_{k} := k^{k}.

Using the triangle inequality we get

|\frac{1}{a_{n_{k+1}}} S_{n_{k+1}}^{β}| ≤ |\frac{1}{a_{n_{k+1}}} ∑_{j=0}^{n_{k}} \binom{n_{k+1} − j + β − 1}{n_{k+1} − j}p_{j}X_{j}|
+ |\frac{1}{a_{n_{k+1}}} ∑_{j=n_{k}+1}^{n_{k+1}} \binom{n_{k+1} − j + β − 1}{n_{k+1} − j}p_{j}X_{j}|
:= S_{k}^{(1)} + S_{k}^{(2)}.

To estimate S_{k}^{(1)} we can adjust Stadtmüller’s [18] LIL to get

lim sup_{k→∞} \frac{∑_{j=0}^{n_{k}} \binom{n_{k+1} − j + β − 1}{n_{k+1} − j}p_{j}X_{j}}{\sqrt{2s_{k}^{2} log log s_{k}^{2}}} ≤ 1 a.s.

and using the same techniques as in Lemmas A1 and A2 we have

\begin{aligned}s_{k}^{2} & := & ∑_{j=0}^{n_{k}} (\binom{n_{k+1} − j + β − 1}{n_{k+1} − j})^{2} p_{j}^{2} \\ & ≤ & cn_{k+1}^{2α+2β−1}L(n_{k+1}) ∫0^{\frac{n_{k}}{n_{k+1}}} (1 − t)^{(2α+1)−1}t^{(2β−1)−1}dt. \end{aligned}

Since

\frac{2s_{k}^{2} log log s_{k}}{a_{n_{k+1}}^{2}} ≤ ∫0^{\frac{n_{k}}{n_{k+1}}} (1 − t)^{(2α−1)−1}t^{(2β+1)−1}dt → 0 (k → ∞)

we get

S_{n_{k+1}}^{(1)} → 0 a.s..

Now for ϵ > 0 let

A_{k}^{ϵ} := \{|\frac{1}{a_{n_{k+1}}} ∑_{j=n_{k}+1}^{n_{k+1}} \binom{n_{k+1} − j + β − 1}{n_{k+1} − j}p_{j}X_{j} − ρ| < ϵ\} .

Then (A_{k}^{ϵ}) is a sequence of independent events and by the Borel-Cantelli Lemma we are done, if we can show that

∑_{k=0}^{∞} P (A_{k}^{ϵ}) = ∞.

We use the same ideas as in §4 and represent S_{n_{k+1}}^{(2)} as an average of q_{k} := τ^{−2}2 log log U_{n_{k}}^{β} random variables, each of which converges to a Gaussian random variable with mean 0 and variance 1 . Set

σ_{k}^{2} := ∑_{j=n_{k}+1}^{n_{k+1}} (\binom{n_{k+1} − j + β − 1}{n_{k+1} − j})^{2} p_{j}^{2}

and choose numbers p_{lk} for each k recursively for 0 ≤ l ≤ q_{k} by

\begin{aligned}p_{0k} & := & n_{k}; \\ p_{lk} & := & inf \{m : ∑_{j=p_{l−1,k}+1}^{m} (\binom{n_{k+1} − j + β − 1}{n_{k+1} − j})^{2} p_{j}^{2} ≥ \frac{σ_{k}^{2}}{q_{k}} \}; \\ p_{q_{k},k} & := & n_{k+1} − 1. \end{aligned}

Using the asymptotic results in much the same way as in §4 in the proof of Lemma 5 we see that the p_{lk} are well-defined and p_{l−1,k} − p_{lk} → ∞ (k → ∞) uniformly in l. We set

Z_{lk} := \frac{\sqrt{q_{k}}}{σ_{k}^{2}} ∑_{j=p_{l−1,k}+1}^{p_{lk}} \binom{n_{k+1} − j + β − 1}{n_{k+1} − j}p_{j}X_{j}

and have E(Z_{lk}) = 0 and E(Z_{lk}^{2}) ∼ 1. Furthermore

\frac{S_{k}^{(2)}}{a_{n_{k+1}}} = \frac{σ_{k}}{\sqrt{q_{k}}} \frac{∑_{l=1}^{q_{k}} Z_{lk}}{\sqrt{2U_{n_{k}}^{β} log log(U_{n_{k}}^{β})}} = \frac{∑_{l=1}^{q_{k}} Z_{lk}}{τq_{k}} (1 + o (1)).

Hence for any interval [c, d] ⊂ (−1, 1) with ρ ∈ [c, d] we have

P (\frac{S_{k}^{(2)}}{a_{n_{k+1}}} ∈ [c, d]) = ∏_{l=1}^{q_{k}} P (Z_{lk} ∈ τ[c, d]).

Now choosing c = ρ − ϵ, d = ρ + ϵ and using the Embrechts-Maejima CLT [7] for each Z_{lk} we get

P (\frac{S_{k}^{(2)}}{a_{n_{k+1}}} ∈ [c, d]) ≥ exp \{q_{k}(−τ^{2}ρ^{2}/2) + q_{k} log γ([−τϵ, τϵ])\}

= exp {log log U_{n_{k}}^{β}(−ρ^{2} + log γ([−τϵ, τϵ])}

and since log log U_{n_{k}}^{2} ∼ log(k log k) we see that this is the general term of a divergent series, if we choose τ large enough.

5.3 The converse of the LIL

Thsi follows by exactly the same arguments as in the power series case.

6 Appendix

Throughout this paragragh we assume that the sequence (p_{n}) satisfies condition (1.6), i.e.

p_{n} = n^{α}L(n) (n → ∞)

with α > −1 and L(.) a slowly varying function, see [17, 3] for details on such functions.

Lemma A1 We have the following asymptotic properties (i) p(t) ∼ Γ(α + 2)P (t) ∼ Γ(α + 2)t^{α+1}L(t) (t → ∞);

(ii) V (t) ∼ Γ(2α + 2)P^{(2)}(t) ∼ Γ(2α + 2)t^{2α+1}L^{2}(t) (t → ∞);

(iii) ∑_{k=0}^{n} \binom{n − k + β − 1}{n − k}p_{k} ∼ B(α + 1, β)n^{α+β}L(n) (n → ∞);

(iv) ∑_{k=0}^{n} (\binom{n − k + β − 1}{n − k})^{2} p_{k}^{2} ∼ B(2α + 1, 2β − 1)n^{2(α+β)−1}L^{2}(n) (n → ∞).

Proof: (i),(ii) are basically contained in [11], see also [12] for a detailed proof. (iii),(iv) are proved in [15], Lemma 1 based on classical results from [9, 20], Theorem 42 resp. Theorem 2.3.1.

Lemma A2 Let h : [0, ∞) → N I with t = O (h(t)) (t → ∞) . Then for t ≥ t_{0}(= t_{0}(L(.)))

∑_{k=h(t)}^{∞} p_{k}e^{− \frac{k}{t}} ≤ [α + 2][α + 2]!P (h(t)) exp\{−h(t)/t\}.

Proof:

\begin{aligned}∑_{k=h(t)}^{∞} p_{k}e^{− \frac{k}{t}} & = & p_{h(t)}e^{− \frac{h(t)}{t}} + ∫_{h(t)}^{∞} e^{− \frac{x}{t}} dP (x) \\ & = & p_{h(t)}e^{− \frac{h(t)}{t}} + P (x)e^{− \frac{x}{t}} |_{h(t)}^{∞} + \frac{1}{t} ∫_{h(t)}^{∞} e^{− \frac{x}{t}} P (x)dx. \end{aligned}

Using the substitution u := x/t and the fact that t^{−γ}L(t) → 0 (t → ∞) asymptotically non increasing for each γ > 0 [3], Theorem 1.5.4, Proposition 1.3.6 26 Law of the Iterated logarithm (v), we obtain

\frac{1}{t} ∫_{h(t)}^{∞} e^{− \frac{x}{t}} P (x)dx = ∫_{h(t)/t}^{∞} e^{−u}P (ut)du
= t^{[α+2]} ∫_{h(t)/t}^{∞} e^{−u}u^{[α+2]} ((ut)^{α+1−[α+2]}L(ut)) du
≤ t^{[α+2]} sup_{u≥h(t)/t} {(ut)^{α+1−[α+2]}L(ut)} ∫_{h(t)/t}^{∞} e^{−u}u^{[α+2]}du
≤ t^{[α+2]}(h(t))^{α+1−[α+2]}L(h(t)) ∑_{k=0}^{[α+2]} \frac{[α + 2]!}{k!} (\frac{h(t)}{t})^{α} e^{− \frac{h(t)}{t}}
≤ [α + 2][α + 2]!(h(t))^{α+1}L(h(t))e^{− \frac{h(t)}{t}} .

Furthermore we have for the first part

|p_{h(t)}e^{− \frac{h(t)}{t}} − P_{h(t)}e^{− \frac{h(t)}{t}} | ∼ P_{h(t)}e^{− \frac{h(t)}{t}} (t → ∞).

Therefore we obtain the proposed bound for t large enough. Lemma A3 For t_{n} := exp{−n/(log n)^{γ}}, γ > 1 , we have

exp \{\frac{−1}{(log(n + 1))^{γ}} \} ≤ \frac{t_{n}}{t_{n+1}} ≤ exp \{\frac{−1}{2(log n)^{γ}} \} .

Proof: Set f(x) := x/(log x)^{γ} then f′(x) = \frac{1}{(log x)^{γ}} (1 − \frac{γ}{log x}) ≥ 0 for x ≥ e^{γ} and f′′(x) = \frac{γ}{x(log x)^{γ+1}} (1 − \frac{γ}{log x}) ≤ 0 for x ≥ e^{γ+1} . Therefore we get by the meanvalue-theorem using the monotonicity of f′ that f′(n) ≥ f(n+1)−f(n) ≥ f′(n+1) and so

\frac{t_{n}}{t_{n+1}} = exp\{f(n + 1) − f(n)\} ≤ exp\{f′(n)\}
= exp \{\frac{1}{(log n)^{γ}} (1 − \frac{γ}{log n})\} ≤ exp \{\frac{1}{2(log n)^{γ}} \} .

\frac{t_{n}}{t_{n+1}} = exp\{f(n + 1) − f(n)\} ≥ exp\{f′(n + 1)\}
= exp \{\frac{1}{(log(n + 1))^{γ}} (1 − \frac{γ}{log(n + 1)})\} ≥ exp \{\frac{1}{(log(n + 1))^{γ}} \} .

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Universität Ulm Abteilung Stochastik 89069 Ulm Germany email: kiesel@mathematik.uni-ulm.de