In the Karlin infinite occupancy scheme, balls are thrown independently into an infinite array of boxes

Let

There are deterministic and Poissonized versions of Karlin’s occupancy scheme. In a

To define the other version of the scheme we need an additional notation. Let

In a

Put

The function

In the case

In Theorems

Treatment of the situations in which

Our present proof only works provided that, for some

The following results are concerned with the cases

Theorems

Finally, we present LILs for the variables

A transfer of results available for the Poissonized version to the deterministic version is called

Following the referee’s suggestion, for the sake of comparison, we now provide a verbal description of the LILs obtained in [

Let

Under the assumption that, for each

We shall prove a LIL for

There exist independent a.s. nondecreasing stochastic processes

for each

for each

Under (A2), there exists

Before introducing our next assumption we need some preparation. In view of (A1) and

Fix any

For each

(A2a) and (A2b) entail, for

Here is an example of

for each

for each

for

Putting

The assumption imposed in [

One may expect that

A sufficient condition for (A4) is either eventual lower semi-continuity or eventual monotonicity of

A sufficient condition for (A5) is that

Assuming (A1) and (A3), fix any

The function

For sufficiently large

For sufficiently large

Now we are ready to present a LIL for infinite sums of independent indicators.

Our proof of Theorem

We start with a simple inequality which will be used in the last part of the proof of Proposition

For

For each

(a) Using the definition of

We proceed by noting that, as

For

Lemma

In view of the representation

Invoking Rosenthal’s inequality (Theorem 3 in [

Combining fragments together we conclude that (

The next result is borrowed from Lemma 2 in [

We first show that the assumption of Lemma

We start with a lemma and a proposition which are in essence Lemma 4.13 and Proposition 4.7 in [

Proposition

In view of (A4), it is enough to show that, for each

To obtain (

Fix any

By Lemma 4.1 in [

Next, in order to prove (

By Markov’s inequality and Lemma

Next, we work towards proving that

For ease of reference, we state two known results. The former is an obvious extension of Theorem 1.5.3 in [

Given next is a collection of results on the asymptotics of

For

Assume that

According to formula (6) in [

Assume that

Assume now

For

As a preparation, we start with a lemma which facilitates checking condition (B22) of Theorem

We start by proving a simple but an important inequality. Since

Therefore, it is enough to show that in the setting of the lemma, for all

We first prove Theorem

The statements of Theorem

In the remaining part of the proof we treat simultaneously Theorems

Assume now that

Assume now that

Thus, in all settings (A4) holds according to Remark

Passing to the first part of (B22), we are going to refer to the table below which contains all the necessary information. In the first line, we list the values of

1
1

The proofs of Theorems

We start with some preparatory work. It is known, see, for instance, Lemma 1 in [

The proof of Proposition

According to the last formula in the proof of Lemma 6.9 in [

If

We start by noting that, in view of (

For

Next, we intend to show that the contributions of

The argument for

Combining all the fragments together we arrive at (

With Proposition

The deterministic and Poissonized schemes discussed in Section

We prove the result in several steps.

By Lemmas

According to Lemma