Full text of DOI:10.1017/S0305004100004746

Math. Proc. Camb. Phil. Soc. (2001), 130, 365
365
c
Printed in the United Kingdom ꢀ 2001 Cambridge Philosophical Society
Tail expansions for random record distributions
¨
¨
By RUDOLF GRUBEL and NIKLAS VON OHSEN
¨
Institut fu¨r Mathematische Stochastik, Universitat Hannover
Postfach 60 09, D-30060 Hannover, Germany
e-mail: rgrubel@stochastik.uni hannover.de
(Received 26 July 1999)
Abstract
The random record distribution ν associated with a probability distribution µ
P∞
can be written as a convolution series, ν = n=1 n−1(n + 1)−1µ?n. Various authors
have obtained results on the behaviour of the tails ν((x, ∞)) as x → ∞, using Laplace
transforms and the associated Abelian and Tauberian theorems. Here we use Gelfand
´
transforms and the Wiener–Levy–Gelfand Theorem to obtain expansions of the tails
under moment conditions on µ. The results differ notably from those known for other
convolution series.
1. Introduction and main result
Random record models have been introduced by Gaver [8], who gave several ap-
plications. For the variant to be discussed in the present paper let Xn, n ∈ N, be
independent random variables with distribution µ. The corresponding partial sums
P
n
S0 ÷ 0, Sn ÷ m=1 Xm for n ∈ N, constitute a renewal process with lifetime distri-
bution µ. This classical terminology refers to the case where the X-variables are non-
negative, in the general case we call (Sn)n∈N a random walk with step distribution µ.
0
Let (Yn)n∈N be another sequence of independent and identically distributed random
variables, with continuous distribution function FY and independent of (Sn)n∈N . We
0
0
regard Yn as a label attached to Sn. The index τ of the first record of the Y -sequence
is the infimum of all n ∈ N with the property that Yn > Ym for m = 0, . . . , n − 1
and Sτ is the first random record. Strictly monotone transformations of the Y -values
do not change τ, so the distribution νrr of Sτ depends on µ only and not on F. We
call νrr the random record distribution associated with µ, writing νrr(µ) if we want to
emphasize the dependence on µ.
With ‘?’ denoting convolution so that µ?n is the distribution of Sn and using the
well-known elementary fact that P(τ = n) = n−1(n + 1)−1 for all n ∈ N (see e.g.
[7, section I·5]), we obtain the basic representation
∞
X
1
νrr =
µ?n
(1·1)
n(n + 1)
n=1
of the random record distribution as a convolution series in µ. The first moment of τ
is obviously infinite, which implies that, as a rule, the expected value of Sτ does not

¨
Rudolf Gru¨bel and Niklas von Ohsen
366
exist (we exclude the trivial case where µ is concentrated at 0). This naturally raises
the question of the behaviour of the tails of Sτ , i.e. the asymptotics of νrr((x, ∞)) as
x → ∞. The first results in this direction were obtained by Westcott [21, 22], who
considered the one-sided case and gave conditions that implied νrr((x, ∞)) ∼ m1/x,
R
where m1 = m1(µ) ÷ xµ(dx) denotes the first moment associated with µ. Westcott
also considered heavy-tailed distributions, with e.g. µ((x, ∞)) of regular variation
with index −α, 0 < α < 1, and showed that in such cases the ratio of the νrr- and
µ-tail behaves like α log(x); Embrechts and Omey [4] who obtained a converse to this
statement. The proofs in these papers are based on Abel–Tauber results for Laplace
transforms (see [2] for a standard reference in this area).
Convolution series such as (1·1) appear in a variety of situations and special cases
are of considerable importance in applied probability theory. With all coefficients
P∞
?n
equal to one we obtain the renewal measure νren
÷
µ , a central object of
n=1
P∞
renewal theory. The series νharm
÷
n=1 n−1µ?n is the harmonic renewal measure,
which plays an important role in connection with the Wiener–Hopf factors of a
distribution, in turn very useful for the analysis of random walks via ladder variables
(see [10, 14, 15] and the references given therein). Generally, if the coefficients are
non-negative and sum to 1, then the series arises as the distribution of a random
sum. This includes (1·1). Cases in which the coefficients are the probabilities of a
Poisson or geometric distribution are important in insurance mathematics and risk
theory, where they are used in connection with total claim size distributions and ruin
probabilities. The Poisson case is also essential to the analysis of infinitely divisible
distributions. [7] continues to be a central reference for much of this material, but
see also [1].
In all these cases the tail behaviour of the resulting ν or of normalized versions
thereof has been investigated thoroughly. For renewal measures this is related to the
rate of convergence in the renewal theorem (see e.g. [13] and the references given
therein). A typical result is the following,
Z
∞Z
∞
νren((−∞, x]) = c1 · x + c2 + c3
µ((z, ∞)) dz dy + r(x)
(1·2)
x
y
as x → ∞ with some lower order error term r(x), where the coefficients c1, c2, c3
depend on µ via its moments. For harmonic renewal measures see [10] for the one-
sided case and [12] for the general case. Once more, we cite a typical result:
Z
∞
νharm((−∞, x]) = log x + c1 + c2
µ((y, ∞)) dy + r(x),
(1·3)
x
with c1, c2 again depending on µ via its moments. For compound distributions with
rapidly decreasing probability coefficients see again [13] and the references given
therein: if e.g. µ((x, ∞)) is regularly varying then
ν((x, ∞)) ∼ c1µ((x, ∞)),
(1·4)
i.e. the compound distribution tail is asymptotically equal to a multiple of the tail
of the input measure.
On a technique-of-proof level, a general picture that seems to have emerged from
these efforts is that the classical Abel–Tauber approach via Laplace transforms can
often be complemented (in the sense of treating the general, two-sided case and

Tail expansions for random record distributions
367
also in the sense of obtaining higher order expansions) by what has become to be
known as the Banach algebra method, essentially the use of Gelfand transforms and
´
the Wiener–Levy–Gelfand theorem. The use of this method can be traced back to
one of the first proofs of the discrete renewal theorem in [5] and has since been used
in [3, 6, 16–18] and others; a variant that produces expansions was introduced in
[11, 13].
It is one aim of the present paper to show that the Banach algebra method can
also be applied to random record distributions. This in itself might not be very
surprising, but the final result is in our view remarkable as it differs significantly from
those obtained for renewal measures, harmonic renewal measures and compound
distributions with exponentially decreasing weights. To explain this, note that in all
these cases after a finite number of normalization terms (such as c1x+c2 in (1·2) and
log x+c1 in (1·3)) we are down to an order of magnitude that depends directly on the
rate of tail decrease of µ. Indeed, it is often possible to show with elementary complex
variable arguments that an exponentially decreasing tail of µ implies an exponential
rate of convergence in some associated limit theorem for ν. A naive interpolation
of (1·2), (1·3), (1·4) and Westcott’s result would then lead us to expect for (1·1) a
major term c1/x and a remainder whose order depends on the tail of µ. However,
we will see below that random record distributions do not fit into this scheme – the
expansion will be in negative powers of x, irrespective of the rate of tail decrease of
µ. Also, it turns out that in order to handle the random record situation some new
arguments are needed. Some of these are of a rather technical nature (see e.g. Lemma
9 below) and might not yet have found their most natural form; we suspect that the
additional log-factor in our error bounds (which is not present in the results for the
other convolution series mentioned above) is due to this circumstance. In any case we
are confident that these techniques considerably enlarge the range of applicability
of the Banach algebra method.
In order to state our main result we require some more notation. We will generally
assume that µ is spread out, i.e. that µ?n has a non-vanishing absolutely continuous
component for some n ∈ N, and our assumptions on the rate of tail decrease are of
the form
µ((−∞, −x]) = O(x−γ), µ((x, ∞)) = O(x−γ
)
as x → ∞
(1·5)
for some γ > 1. If this holds then the moments
Z
mj(µ) ÷ xj µ(dx), j = 1, . . . , k = k(γ) ÷ dγe − 1
exist and we can define the inverse moments r0(µ), . . . , rk(µ) associated with µ induc-
tively by
ꢀ ꢁ
j
X
j
r0(µ) ÷ 1, rj(µ) ÷ −
ml(µ) rj−l(µ) for j = 1, . . . , k.
(1·6)
l
l=1
The reason for this name will become clear in the course of the proof of our
theorem; note that r1(µ) = −m1(µ). The following theorem gives an expansion of the
tails νrr((−∞, −x]), νrr((x, ∞)) of random record distributions in terms of negative
powers of x, with the coefficients depending on the inverse moments of µ in a simple
manner.

¨
Rudolf Gru¨bel and Niklas von Ohsen
368
Theorem 1. Let µ be a spread out probability distribution with m1(µ) > 0 and assume
that (1·5) holds for some γ > 1. Then, as x → ∞, νrr((−∞, −x]) = O(x−γ log x) and
dγe−1 rj(µ)
X
νrr((x, ∞)) = −
x−j + O(x−γ log x).
(1·7)
j
j=1
As long as some of the inverse moments of µ are not equal to zero, the right tail of
the random record distribution associated with µ will decrease at a polynomial rate
only, even after the unavoidable term m1(µ)/x has been subtracted. In this context
the exponential distribution plays a special role (as it does in the renewal case, where
the remainder term in the corresponding limit theorem vanishes).
Example 2. If µ has density fµ(x) = λe−λx, x > 0, for some λ > 0, then µ?n is the
gamma distribution with shape parameter n and scale parameter λ so that νrr has
density f given by
∞
λnxn−1
X
1
f(x) =
e−λx
(n + 1)n (n − 1)!
(1 − e−λx − λxe−λx).
n=1
1
λx2
=
This implies νrr((x, ∞)) = (λx)−1 + O(e−λx). Indeed, a simple calculation shows that
we have mk(µ) = k!/λk and therefore r1(µ) = −1/λ, rk(µ) = 0 for all k > 1.
Section 2 consists of the proof of Theorem 1, some comments on possible extensions
can be found in Section 3.
2. Proof of the theorem
We first recall some standard material from the theory of commutative Banach
algebras, which we then apply to a class of algebras suitable for our purposes. For
the general theory of these structures we refer the reader to the classic [9] or one of
the many excellent textbooks on functional analysis such as [19].
Let B be the σ-field of the Borel subsets of the real line, M denotes the linear space
of all σ-additive functions µ: B → C. Each such complex-valued finite measure µ can
be written in the form
µ = µ1 − µ2 + iµ3 − iµ4
with non-negative finite measures µ1, . . . , µ4; this decomposition is unique if we fur-
ther require that µ1 and µ2 respectively µ3 and µ4 are concentrated on disjoint Borel
sets. The corresponding total variation measure |µ| is then given by |µ| = µ1 +· · ·+µ4
and satisfies
(
)
n
X
|µ|((a, b]) = sup
|µ((ck−1, ck])|: n ∈ N, a = c0 6 c1 6 · · · 6 cn = b
k=1
R
for all a, b ∈ R with a < b. We write mk(µ) for the kth moment xk µ(dx) provided
that mk(µi) is finite for i = 1, . . . , 4. In this case the associated Fourier transform
Z
µˆ: R → C, µˆ(θ) ÷ eiθx µ(dx),

Tail expansions for random record distributions
369
ꢂ
is k times differentiable and (dkµˆ dθk)(0) = ikmk(µ). We mention in passing that
the existence of the kth derivative of µˆ at 0 does not imply the existence of the kth
moment, not even for non-negative measures, but that this formula is convenient for
the calculation of moments once their existence has been established.
With the norm kµkTV ÷ |µ|(R) the space M becomes a Banach space. Moreover,
the convolution
Z
µ ? ν(A) ÷ µ(A − x) ν(dx) for all A ∈ B,
defines a ‘multiplication’, which makes M a commutative Banach algebra with unit
element δ0, the unit measure concentrated at 0. Convolution and moments interact
as follows,
ꢀ ꢁ
k
X
k
mk(µ ? ν) =
mj(µ) mk−j(ν),
(2·1)
j
j=0
provided that the moments exist. Comparing (2·1) and (1·6) we see that the inverse
moments of µ are simply the moments of some convolution inverse of µ in M (the pre-
cise interpretation involves localization, which we carry out below). Maximal ideals
I in M are either of the form
ꢃ
I = I(θ0) ÷ µ ∈ M: µˆ(θ0) = 0
for some θ0 ∈ R, or they contain Ma, the set of all absolutely continuous measures
(see e.g. [9, section 30]). There is a one-to-one correspondence between maximal ideals
I and multiplicative functionals ψ: M → C, given by I = ψ−1({0}), and the Gelfand
transform of µ is the function µ˜ on the set of maximal ideals defined by µ˜(I) = ψ(µ).
From the above statement on the structure of the maximal ideals in M it then follows
that µ˜1 = µ˜2 implies µˆ1 = µˆ2. As Fourier transforms characterize measures, so do
Gelfand transforms: M is semisimple. It also follows from the multiplicativity of the
ψ-functionals that
|µ˜(I)| 6 kµkTV for all µ ∈ M
(2·2)
for all maximal ideals I.
We now fix a real number γ > 1. Let Mr(γ) be the set of all µ ∈ M with the
property |µ|((x, x + 1]) = O(x−γ) as x → ∞. With
kµkr,γ ÷ kµkTV + 2γ sup (1 + x)γ |µ|((x, x + 1])
x>0
this is a Banach algebra and to any maximal ideal I in this space there exists a
maximal ideal I0 in M such that I = I0 ꢀ Mr(γ) (see [17]). Reflection at 0 defines
an operator S: M → M, S(µ)(A) ÷ µ(−A) and the corresponding spaces of measures
characterized by the behaviour of |µ|([x, x + 1)) as x → −∞ are Ml(γ) ÷ {µ ∈
M: S(µ) ∈ Mr(γ)}. The space of main interest to us is M(γ) ÷ Ml(γ) ꢀ Mr(γ), with
norm kµkγ ÷ kS(µ)kr,γ + kµkr,γ . This is again a Banach algebra and a little separate
argument shows that the maximal ideals of M(γ) again arise as the intersections of
maximal ideals in M with M(γ). In particular, the range of the Gelfand transform
of some µ ∈ M(γ) with respect to M(γ) is a subset of the range of the Gelfand
transform of µ regarded as an element of M. This range is important in view of the
´
Wiener–Levy–Gelfand theorem, which we need in the following form.

¨
Rudolf Gru¨bel and Niklas von Ohsen
370
Proposition 3. Let µ ∈ M(γ) and U ⊂ C, U open, be such that the range of the
Gelfand transform of µ is contained in U. Let Ψ: U → C be an analytic function. Then
there exists a unique element ν of M(γ) with the property νˆ = Ψ ◦ µˆ.
Of course, the same statement holds if M(γ) is replaced by M throughout provided
the Gelfand transform refers to this space too. In view of the above inclusion relation
and the connection to Fourier transforms we will generally consider the range of the
Gelfand transform with respect to M. We will use Proposition 3 with the following
(U, Ψ)-pairs,
1
z
U1 ÷ {z ∈ C: z ꢀ 0}, Ψ1(z) ÷ ,
U2 ÷ {z ∈ C: Re(z) > 0 or Im (z) ꢀ 0}, Ψ2(z) ÷ log(z),
∞
X
1
U3 ÷ {z ∈ C: |z| < 1}, Ψ3(z) ÷
zk.
k(k + 1)
k=1
With the logarithm we always mean the canonical version, which is real for positive
real arguments and analytic on U2. The function Ψ3 is obviously the one that relates
directly to random record measures. Note that we can extend Ψ3 continuously to
the closure of U3 by
ꢀ
ꢁ
1
z
Ψ3(z) = 1 +
− 1 log(1 − z) if |z| 6 1, z ^ {0, 1}, Ψ3(0) = 0, Ψ3(1) = 1,
and that, with this extension,
ˆ
νrr(µ) (θ) = Ψ3(µˆ(θ)) for all θ ∈ R
(2·3)
for all probability measures µ on the real line. In particular, if µ0 denotes the expo-
nential distribution with mean 1 then
ꢀ
ꢁ
1
iθ
ˆ
νrr(µ0) (θ) = 1 + iθ log 1 −
for all θ ꢀ 0.
In order to cope with exceptional points in the range of µ˜ we need to be able to
localize. The existence of suitable partitions of unity is guaranteed by the following
auxiliary result.
Lemma 4. For any α > 0 there exists an element
∞
\
ρα ∈ Ma(∞) ÷ Ma ꢀ
M(k)
k=1
with m0(ρα) = 1, mk(ρα) = 0 for all k ∈ N and
ꢄ
1, if |θ| 6 α,
0, if |θ| > 2α.
ρˆα(R) ⊂ [0, 1], ρˆα(θ) =
Moreover, if µ ∈ M is such that µ((x, ∞)) = O(x−γ log x) for some γ > 1 as x → ∞
then also ρα ? µ((x, ∞)) = O(x−γ log x) as x → ∞ and the same statement holds with
(x, ∞) replaced by (−∞, −x].
Proof. It is well-known that there exists an infinitely often differentiable function
φ: R → C with the properties required for ρˆα. The inversion formula for character-
istic functions implies that there is a ρα ∈ Ma with Fourier transform φ and that

Tail expansions for random record distributions
371
R
ꢅ
ꢆ
|x| |ρα|(dx) < ∞ for all k ∈ N. From this |ρα| (−∞, −x] ꢁ (x, ∞) = o(x−k) as
k
x → ∞ follows with Markov’s inequality, i.e. ρα ∈ M(k). The same argument can be
applied to θ → ρˆα(θ)(iθ)l for any l ∈ N, which therefore must be the transform of
ꢅ
ꢂ
ꢆ
an element of Ma(∞) too. Finally, we obviously have dkρˆα dθk (0) = 0 and hence
mk(ρα) = 0 for all k ∈ N.
For the proof of the last statement of the lemma it is obviously enough to consider
the right tail. We split the range of integration,
Z
Z
ρα ? µ((x, ∞)) =
ρα((x − y, ∞)) µ(dy) +
ρα((x − y, ∞)) µ(dy).
(−∞,x/2]
(x/2,∞)
For the first term we use
Z
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
ρ ((x − y, ∞)) µ(dy) 6 ρ ((x/2, ∞)) kµk
,
ꢇ
α
α
TV
(−∞,x/2]
which is O(x−η) for all η > 0. For the second term we use Fubini’s theorem (integra-
tion by parts) to obtain
Z
ρα((x − y, ∞)) µ(dy)
(x/2,∞)
Z
= ρα((x/2, ∞)) µ((x/2, ∞)) +
µ((x − z, ∞)) ρα(dz)
(−∞,x/2)
and then bound the individual terms,
|ρα((x/2, ∞)) µ((x/2, ∞))| 6 kραkTV |µ((x/2, ∞))|,
Z
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
µ((x − z, ∞)) ρ (dz) 6 kρ k
sup |µ((x/2, ∞))|,
ꢇ
α
α TV
(−∞,x/2)
y>x/2
which again results in the upper bound O(x−γ log x).
The tail function Fµ associated with some µ ∈ M is given by
ꢄ
µ((x, ∞)),
if x > 0,
Fµ : R → C,
Fµ(x) ÷
−µ((−∞, x]), if x < 0.
If this function is integrable then we can define a new measure Σµ ∈ Ma, the tail
measure associated with µ, by
Z
Σµ(A) =
Fµ(x) dx for all A ∈ B,
A
and we can regard Σ as a linear operator on
Z
n
o
Dom (Σ) ÷ µ ∈ M: |Fµ(x)| dx < ∞
with values in M. It is easy to check that µ is in Dom (Σ) if µ is a probability measure
with finite first moment, or generally if µ ∈ M(γ) for some γ > 2.
The proofs of the following statements are either elementary or contained in
[12, 13] and are therefore omitted. Remember that mk(µ) denotes the kth moment
R
xk µ(dx) of µ. We write Σk for the k-fold iteration of Σ, where Σ0 is understood to
be the identity.

¨
Rudolf Gru¨bel and Niklas von Ohsen
372
Lemma 5.
(i) If µ ∈ M(γ + 1) then Σµ ∈ M(γ).
(ii) For any µ ∈ Dom (Σ),
µˆ(θ) − µˆ(0)
ˆ
ˆ
if θ ꢀ 0, (Σµ) (0) = m1(µ).
(Σµ) (θ) =
iθ
(iii) Suppose that µ1, µ2 ∈ Dom (Σ) and µ1(R) = 0. Then Σ(µ1 ? µ2) = (Σµ1) ? µ2.
(iv) If µ ∈ M(γ) is such that Σµ ∈ M(γ), then µ((−∞, −x]) = O(x−γ) and µ((x, ∞))
= O(x−γ) as x → ∞.
(v) If µ ∈ M(γ) then, for j ∈ N0, k ∈ N with j + k < γ,
1
j
m
j−1(Σkµ) = mj(Σk−1µ).
In particular, for k < γ − 1,
1
k!
(Σkµ)(R) =
mk(µ).
We next introduce a reference family of measures with known tail decrease. For
any j ∈ N let fj: R → R be defined by fj(x) ÷ 0 for x < 0 and
j
∞
ꢈ
ꢉ
xk
(k + j + 1)! xj+1
j!
x
k
X
X
fj(x) ÷ j! e−x
=
1 − e−x
k!
k=0
k=0
for x > 0, for definiteness we take fj(0) to be the right hand limit 1/(j + 1). Let
νj be the non-negative measure with Lebesgue density fj, let µ0 be the exponential
distribution with mean 1. From Example 2 we know that ν1 is the random record
distribution associated with µ0; for general j we have
∞
X
j!
νj =
µ?0k.
k(k + 1) · · · (k + j)
k=1
Lemma 6. Let j ∈ N. Then νj is a finite measure with total mass 1/j and Fourier
transform
j−1
ꢈ
ꢉ
X
1
1
iθ
νˆj(θ) =
(iθ)k + (iθ)j log 1 −
.
j − k
k=0
Moreover, νj ∈ M(j + 1) and Σlνj = νj−l for l = 1, . . . , j − 1.
Proof. We have
∞
∞
xk
kxk−1
X
X
fj0(x) = −j! e−x
= −j! e−x
+ j! e−x
(k + j + 1)!
(k + j + 1)!
k=0
k=1
∞
xk
X
ꢅ
ꢆ
k + j + 2 − (k + 1)
(k + j + 2)!
k=0
= −fj+1(x) for all x > 0,
which implies Σνj+1 = νj. This together with the known formula for νˆ1 can be used
to prove the formula for νˆj by induction, and this in turn delivers the total mass.
The next lemma shows that the localizing functions introduced above do not

Tail expansions for random record distributions
373
change the tail behaviour of the reference measures on the scale that is of inter-
est to us.
Lemma 7. With ρα and νj as above we have, for all η > 0 and as x → ∞,
ρα ? νj((−∞, −x]) = o(x−η), ρα ? νj((x, ∞)) − νj((x, ∞)) = o(x−η).
Proof. Let k ∈ N be such that j + k > η. We apply Lemma 5(iii) k times and use
Σl(ρα − δ0)(R) = 0 for l = 0, . . . , k, νj = Σkνj+k to obtain
k
X
1
(ρα − δ0) ? νj = Σkρα ? νj+k
−
Σlρα.
j + l
l=1
Both factors of the first term on the right-hand side are in M(j +k+1) and so are the
terms in the sum by Lemma 5(i) and Lemma 4, hence (ρα −δ0)?νj ∈ M(j+k+1). This
implies that the associated tails decrease at least as fast as x−j−k, which is o(x−η).
We will use operational calculus in M(γ) mainly to get rid of certain error terms
in various preliminary steps of the proof of our theorem. For the expansion itself the
rearrangement given in Lemma 8 below together with the bound in Lemma 9 will be
crucial. The proof of the latter is somewhat lengthy, but uses elementary arguments
only; it is here that the log-factors in the error bounds of Theorem 1 appear.
Lemma 8. Let µ ∈ M(γ) for some γ > 1 and put k ÷ dγe − 2. Then Σjµ is a finite
measure for j = 0, . . . , k. Let sj ÷ Σjµ(R) and let νj be as in Lemma 6. Then, for all
θ ∈ R,
ꢈ
ꢉ
ꢈ
ꢉ
1
iθ
1
k
ˆ
(µˆ(θ) − s0) iθ log 1 −
= ((Σ µ) (θ) − sk) νˆk+1(θ) −
k + 1
k−1
X
1
k
ꢈ
ꢉ
X
1
j
ˆ
−
((Σ µ) (θ) − sj) +
sj νˆj+1(θ) −
.
j + 1
j + 1
j=0
j=1
Proof. For k = 0 this is immediate from the formula for the transform of ν1. The
induction step will follow if we can show that
ꢈ
ꢉ
ꢈ
ꢉ
1
1
k
ˆ
((Σ µ) (θ) − sk) νˆk+1(θ) −
+ sk νˆk+1(θ) −
k + 1
k + 1
ꢈ
ꢉ
1
1
= ((Σk−1µ) (θ) − sk−1) νˆk(θ) −
+
((Σk−1µ) (θ) − sk−1).
ˆ
ˆ
k
k
From Lemma 5(iii) we obtain for arbitrary µ1, µ2 ∈ Dom (Σ)
ˆ
ˆ
ˆ
((Σµ1) (θ) − (Σµ1) (0))(µˆ2(θ) − µˆ2(0)) + (Σµ1) (0)(µˆ2(θ) − µˆ2(0))
ˆ
ˆ
ˆ
= (µˆ1(θ) − µˆ1(0))((Σµ2) (θ) − (Σµ2) (0)) + (Σµ2) (0)(µˆ1(θ) − µˆ1(0)),
so the equation follows on using Σνk+1 = νk and νˆk(0) = 1/k.
Lemma 9. Suppose that γ > 1 and let k ÷ dγe − 2. Let µ be a measure with bounded
density f satisfying
f(x) = O(|x|−γ
)
for x → ±∞
and let sj be the total mass of Σjµ, j = 0, . . . , k. Then, as x → ∞ and with νk+1 as in

¨
Rudolf Gru¨bel and Niklas von Ohsen
374
Lemma 6,
!
δ0) − k−1
Σjµ (I(x)) = O(x−γ log x)
X
1
1
(Σkµ − skδ0) ? (νk+1
−
k + 1
j + 1
j=0
for I(x) = (x, ∞) and for I(x) = (−∞, −x].
Proof. We first consider the case I(x) = (x, ∞). Let x > 0 and put
A(x) ÷ {(y, z) ∈ R2: 0 6 y 6 x, 0 6 z 6 x, y + z > x},
B(x) ÷ {(y, z) ∈ R2: y < 0, x < z < x − y}.
Further, µ ⊗ ν denotes the product of any two measures µ and ν in M. Using the
fact that νk+1 is concentrated on (0, ∞) we obtain
1
(Σkµ − skδ0) ? (νk+1
−
δ0)((x, ∞))
k + 1
1
= Σkµ ? νk+1((x, ∞)) − sk νk+1((x, ∞)) −
Σkµ((x, ∞))
k + 1
= Σkµ ⊗ νk+1({(y, z) ∈ R2: y + z > x})
−Σkµ ⊗ νk+1({(y, z) ∈ R2: z > x})
−Σkµ ⊗ νk+1({(y, z) ∈ R2: y > x})
= Σkµ ⊗ νk+1(A(x)) − Σkµ ⊗ νk+1(B(x))
−Σkµ ⊗ νk+1((x, ∞) × (x, ∞)).
For the last term the desired rate follows immediately from
Σkµ((x, ∞)) = O(xk+1−γ),
ν
k+1((x, ∞)) = O(x−k−1).
For the purposes of obtaining upper bounds we may replace the densities of Σkµ
and νk+1 by suitable multiples of x → (1 + |x|)−γ+k and, for x > 0, x → (1 + x)−k−2
respectively. In particular, for some suitable constant c and with the elementary
inequality
(1 + x + y)k+1 − (1 + x)k+1 6 y (k + 1) (1 + x + y)k for all x, y > 0
we obtain
Z
Z
0
x−y
1
1
|Σkµ ⊗ νk+1(B(x))| 6 c
dz
dy
(1 + z)k+2
(1 + |y|)γ−k
−∞
x
Z
∞
c
(1 + x + y)k+1 − (1 + x)k+1
=
dy
(k + 1)(1 + x)k+1
(1 + x + y)k+1(1 + y)γ−k
0
Z
∞
c
y
6
dy.
(1 + x)k+1
(1 + x + y)(1 + y)γ−k
0
We need the upper bound O(xk+1−γ log x) for the integral. For the range from 0 to
x we use
Z
Z
x
x
y
1
dy 6
(1 + y)k+1−γ dy
(1 + x + y)(1 + y)γ−k
1 + x
0
0

Tail expansions for random record distributions
375
and consider the cases k + 1 < γ < k + 2 and γ = k + 2 separately: the first leads to
Z
x
(1 + y)k+1−γ dy = O(xk+2−γ),
0
in the second case we get
Z
x
(1 + y)k+1−γ dy = O(log x),
0
i.e. in both cases the required overall rate O(xk+1−γ log x) results. For the remaining
range from x to ∞ the required rate is immediate from
Z
Z
∞
∞
y
dy 6
(1 + y)k−γ dy
(1 + x + y)(1 + y)γ−k
x
x
1
=
(1 + x)k+1−γ
,
γ − k − 1
where we used γ − k > 1.
For A(x) we use A(x) = A1(x) + A2(x) with
ꢃ
ꢃ
A1(x) ÷ (y, z) ∈ A(x): y 6 x/2 , A2(x) ÷ (y, z) ∈ A(x): y > x/2 .
Bounding the densities as for B(x) for the first of these we obtain
Z
Z
x/2
x
1
1
|Σkµ ⊗ νk+1(A1(x))| 6 c
dz
dy
(1 + z)k+2
(1 + y)γ−k
0
x−y
Z
x/2
2
k+2 c
y
6
dy.
(1 + x)k+2
(1 + y)γ−k
0
As in the above analysis of B(x) we obtain the rate O(log x) for the integral if γ = k+2
and O(xk+2−γ) if k + 1 < γ < k + 2, so this term has the desired rate O(x−γ log x)
too.
For the remaining part A2(x) we first consider the case k = 0. As for A1(x),
Z
Z
x
∞
1
1
|µ ⊗ ν1(A2(x))| 6 c
= c
dz
dy
(1 + z)2
(1 + y)γ
x/2 x−y
Z
x
1
1
dy
1 + x − y (1 + y)γ
x/2
Z
x/2
2γ c
(1 + x)γ
1
6
dy,
1 + y
0
which is O(x−γ log x).
For k > 0 a suitable integration by parts will be crucial. In extension of the
definition preceding Lemma 6 let f0: [0, ∞) → R be defined by f0(0) ÷ 1 and
1
x
f0(x) ÷ (1 − e−x
) for all x > 0.
(In contrast to the functions fk with k > 0 this is not the density of a finite measure.)
Remember that fj is the density of νj for j = 1, 2, . . ., and note that fj0 = −fj+1 also
holds for j = 0 where the derivative refers to the right derivative at x = 0. Also,

¨
Rudolf Gru¨bel and Niklas von Ohsen
376
fj(0) = 1/(j + 1) holds for all non-negative integers j. Then
Z
Z
x
x
Σkµ ⊗ νk+1(A2(x)) =
f
k+1(z) dz Σk−1µ((y, ∞)) dy
x/2 x−y
Z
x
ꢅ
ꢆ
=
=
fk(x − y) − fk(x) Σk−1µ((y, ∞)) dy
x/2
ꢀ
ꢁ
k
ꢇ
ꢇ
ꢇ
X
x
1
j!
fk−j(x − y) − yjfk(x) Σk−jµ((y, ∞))
y=x/2
j=1
ꢀ
ꢁ
Z
x
1
+
f0(x − y) − ykfk(x) µ(dy)
k!
x/2
k
k
ꢈ ꢉ
ꢈ ꢉ
ꢈꢈ
ꢉꢉ
X
X
1
x
x
=
Σ
k−jµ((x, ∞)) −
fk−j
Σ
k−jµ
, ∞
k + 1 − j
2
2
j=1
j=1
k
k
ꢈꢈ
ꢉꢉ
X
X
j
1
j!
ꢀ
1
j!
x
x
−
xjfk(x)Σk−jµ((x, ∞)) +
fk(x)Σk−jµ
, ∞
2
2
j=1
j=1
ꢁ
Z
x
1
+
f0(x − y) − ykfk(x) µ(dy).
k!
x/2
We consider these terms individually and start at the end: as for k = 0,
Z
Z
ꢇ
ꢇ
ꢇ
ꢇ
x
x/2
c
(1 + x)γ
c
1
ꢇ
f (x − y) µ(dy) 6
dy,
ꢇ
0
1 + y
x/2
0
Z
Z
ꢇ
ꢇ
ꢇ
ꢇ
x
∞
ꢇ
yk f (x) µ(dy) 6
(1 + y)k−γ dy,
ꢇ
k
(1 + x)k+1
x/2
x/2
which together yields the rate O(x−γ log x) for the integral. For the terms appearing
in the sums, with the exception of the first sum, this same rate follows from
fj(x) = O(x−j−1), Σjµ((x, ∞)) = O(x−γ+j+1
) for j = 0, . . . , k.
In total a cancellation occurs and we arrive at
Σkµ ⊗ νk+1(A2(x)) − k−1
Σjµ((x, ∞)) = O(x−γ log x).
X
1
j + 1
j=0
This means that the statement of the lemma has been proved for I(x) = (x, ∞).
For the left tail we use essentially the same arguments. With
C(x) ÷ {(y, z) ∈ R2: y 6 −x, z > −x − y}
and using the fact that νk+1 is concentrated on (0, ∞) we obtain for x > 0
ꢀ
ꢁ
1
(Σkµ − skδ0) ? νk+1
−
δ0 ((−∞, −x])
k + 1
1
= Σkµ ? νk+1((−∞, −x]) −
= Σkµ ⊗ νk+1({(y, z) ∈ R2: y + z 6 −x})
−Σkµ ⊗ νk+1({(y, z) ∈ R2: y 6 −x})
= −Σkµ ⊗ νk+1(C(x)).
Σkµ((−∞, −x])
k + 1

Tail expansions for random record distributions
377
We split C(x) = C1(x) + C2(x) with
ꢃ
ꢃ
C1(x) ÷ (y, z) ∈ C(x): y < −2x , C2(x) ÷ (y, z) ∈ C(x): y > −2x .
For the C1(x)-part we use, similar to the argument for A1(x) in the right tail situation,
Z
Z
Z
Z
−2x
∞
1
(1 + z)k+2
1
1
|Σkµ ⊗ νk+1(C1(x))| 6 c
dz
dy
(1 + |y|)γ−k
−∞
−x−y
∞
1
= c
dy
(k + 1)(1 + y − x)k+1 (1 + y)γ−k
2x
∞
1
dy,
6 c
(k + 1)(1 + y)γ+1
x
which is O(x−γ). For k = 0 we further have
Z
Z
−x
∞
1
1
|µ ⊗ ν1(C2(x))| 6 c
dz
dy
(1 + z)2
(1 + |y|)γ
−2x −x−y
Z
2x
1
1
= c
dy
1 + y − x (1 + y)γ
dy,
x
Z
x
c
1
6
(1 + x)γ
1 + y
0
which is O(x−γ log x). If k > 0 we again use a k-fold integration by parts,
Z
Z
−x
∞
Σkµ ⊗ νk+1(C2(x)) = −
f
k+1(z) dz Σk−1µ((−∞, y])) dy
−2x −x−y
Z
−x
= −
= −
fk(−y − x) Σk−1µ((−∞, y]) dy
−2x
Z
k
ꢇ−x
X
−x
X
ꢇ
ꢇ
fk−j(−y − x) Σk−jµ((−∞, y])
+
f0(−y − x) µ(dy)
y=−2x
−2x
j=1
k
k
X
1
= −
Σ
k−jµ((−∞, −x]) +
fk−j(x) Σk−jµ((−∞, −2x])
k + 1 − j
j=1
j=1
Z
−x
+
f0(−y − x) µ(dy).
−2x
For the integral we use
Z
Z
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
−x
2x
1
1
f (−y − x) µ(dy) 6 c
dy
ꢇ
0
1 + y − x (1 + y)γ
−2x
x
and the rate O(x−γ log x) follows with the now familiar arguments. The second sum
can be handled as the A2(x)-case. Again, overall a cancellation occurs and we arrive
at
−Σkµ ⊗ νk+1 C2(x) − k−1
Σjµ((−∞, −x]) = O(x−γ log x),
X
ꢅ
ꢆ
1
j + 1
j=0
which completes the proof of the second statement.

¨
Rudolf Gru¨bel and Niklas von Ohsen
378
We now relate the above constructions to the random record situation. Through-
out the following let µ be a probability distribution that satisfies the conditions of
Theorem 1. Assumption (1·5) means that Σµ ∈ M(γ) which in turn gives µ ∈ M(γ)
as µ is non-negative. Further, we may assume that m1(µ) = 1, which leads to some
simplification in the notation below. To see this it is enough to check that random
record distributions and the assertion of Theorem 1 behave in an equivariant manner
under transformations R → R, x → c · x with c > 0.
We claim that for any maximal ideal I in M we have
I ⊃ Ma
⇒
|µ˜(I)| < 1.
(2·4)
For the proof of (2·4) let l ∈ N be such that µ?l has a non-vanishing absolutely
ꢊ
ꢊ
ꢊ
ꢊ
?l
continuous part. The singular part then satisfies (µ )sing
using (2·2) and
< 1, so (2·4) follows on
TV
l
?l
?l
?l
˜
˜
|µ˜(I)| = |(µ ) (I)| = |((µ )sing) (I)| 6 k(µ )singkTV < 1.
Let µ0 be the exponential distribution with parameter 1 and let νrr(µ) and νrr(µ0)
be the random record distributions associated with µ and µ0 respectively. From (2·3)
we obtain the following formula for the Fourier transform of the difference of these
two distributions:
ˆ
(νrr(µ) − νrr(µ0)) (θ) = Ψ3(µˆ(θ)) − Ψ3(µˆ0(θ))
ꢀ
ꢁ
ꢀ
ꢁ
1
1
iθ
=
− 1 log(1 − µˆ(θ)) − iθ log 1 −
.
(2·5)
µˆ(θ)
Here the last expression has to be interpreted by continuous extension for θ = 0 or
those θ-values with µˆ(θ) = 0 or µˆ(θ) = 1 (see the discussion of Ψ3 above). Now let
α > 0 be small enough for
inf Re(µˆ(θ)) > 0
(2·6)
|θ|68α
to hold; this is possible because of µˆ(0) = 1 and the continuity of µˆ at 0. As µ is
non-lattice we then also have
ꢇ
ꢇ
ꢇ
ꢇ
sup µˆ(θ) < 1.
(2·7)
|θ|>α
We now decompose the difference of the record measures given in (2·5) on using the
localizing functions introduced in Lemma 4:
ˆ
(νrr(µ) − νrr(µ0)) (θ) = φ1(θ) − φ2(θ) + φ3(θ)
(2·8)
with
φ1(θ) ÷ (1 − ρˆ2α(θ))Ψ3((1 − ρˆα(θ))µˆ(θ)),
φ2(θ) ÷ (1 − ρˆ2α(θ))Ψ3((1 − ρˆα(θ))µˆ0(θ)),
ꢀꢀ
ꢁ
ꢀ
ꢁꢁ
ꢅ
ꢆ
1
1
iθ
φ3(θ) ÷ ρˆ2α(θ)
− 1 log 1 − µˆ(θ) − iθ log 1 −
.
µˆ(θ)
By construction,
|(1 − ρˆα(θ))µˆ(θ)| < 1 for all θ ∈ R,
which means that the range of the Fourier transform of (δ0 − ρα) ? µ is contained in

Tail expansions for random record distributions
379
U3. If I is a maximal ideal containing Ma then (2·4) together with ρα ∈ Ma implies
˜
|((δ0 − ρα) ? µ) (I)| = |µ˜(I)| < 1.
Hence the range of the Gelfand transform of the measure (δ0 − ρα) ? µ is contained
in U3, so Proposition 3 implies the existence of an element η1 of M(γ) with Fourier
transform ηˆ1(θ) = Ψ3((1 − ρˆα(θ))µˆ(θ)) (note that the total variation norm of the
measure (δ0 − ρα) ? µ might exceed the value 1, which means that a more direct
reasoning by an obvious extension of (1·1) to signed measures in the unit ball of
(M, k·kTV) does not work). Obviously, Σ(δ0 −ρ2α) ∈ M(γ), so using Lemma 5(iii) and
(iv) we see that φ1 is the Fourier transform of a measure with tail decrease O(x−γ).
By inspection,
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
ꢇ
inf µˆ0(θ) > 0, sup µˆ0(θ) < 1,
|θ|68α
|θ|>α
so we can argue similarly with µ0 in place of µ, which means that φ2 is also the
Fourier transform of a measure with tail decrease O(x−γ).
In order to analyse φ3 we first note that the factors 1 − ρˆ2α(θ) in φ1(θ) and φ2(θ)
made it possible to change µˆ(θ) in a neighbourhood of θ = 0 and that we can now
similarly alter µˆ(θ) on |θ| > 4α. We require some further auxiliary measures. First,
let µ1 ÷ Σµ + δ0 − µ; obviously, µ1 ∈ M(γ). Lemma 5(ii) shows that
ꢀ
ꢁ
ꢅ
ꢆ
1
iθ
µˆ1(θ) = 1 − µˆ(θ) 1 −
for θ ꢀ 0,
µˆ1(0) = 1.
Considering the real parts of the factors separately for θ ꢀ 0 as in [12] we see that
the range of µˆ1 is contained in U2. If a maximal ideal I contains Ma then the same
arguments as used above for φ1 yield µ˜1(I) ∈ U2, hence we can apply Proposition
3 with (U2, Ψ2) to conclude that log µˆ1(θ) is the Fourier transform of an element of
M(γ). Further, let µ2 ÷ µ + δ0 − ρ4α; obviously, µ2 and Σµ2 are in M(γ). For |θ| > 8α
we have ρˆ4α(θ) = 0 so that µˆ2(θ) = µˆ(θ) + 1 ꢀ 0 because of (2·7). For |θ| < 8α we
obtain on using (2·6)
ꢅ
ꢆ
Re µˆ2(θ) > |θi|n6f8α(Re(µˆ(θ)) + 1 − ρˆ4α(θ)) > 0,
hence µˆ2(θ) ꢀ 0. If I is a maximal ideal containing Ma then we get, as above,
µ˜2(I) = µ˜(I) + 1 ꢀ 0
ꢇ
ꢇ
ꢇ
ꢇ
because of µ˜(I) < 1. Hence a similar reasoning as used above for the logarithm of
µˆ1(θ), now with (U1, Ψ1) instead of (U2, Ψ2), shows that there exists a µ3 ∈ M(γ) with
1
µˆ3(θ) =
for all θ ∈ R.
(2·9)
µˆ(θ) + 1 − ρˆ4α(θ)
We claim that Σµ3 ∈ M(γ). To see this we note that the associated Fourier transform
can be written as
d
d
Σµ3(θ) = −µˆ3(θ) Σµ2(θ),
which shows that Σµ3 is the convolution product of two elements of M(γ).
Using µ3 and Σµ3 we can now rewrite the Fourier transform of the last term in

¨
Rudolf Gru¨bel and Niklas von Ohsen
380
the decomposition (2·8) as follows,
ꢀ
ꢁ
ꢅ
ꢆ
1
iθ
d
φ3(θ) = ρˆ2α(θ) (µˆ3(θ) − 1) log µˆ1(θ) − ρˆ2α(θ) Σµ3(θ) + 1 iθ log 1 −
.
From the above reasoning we obtain that the factors in the first product on the
right hand side are transforms of elements of M(γ). Note that the factor µ3 − δ0 has
total mass 0 and that Σ, applied to this difference, yields an element of M(γ). Hence
we obtain O(x−γ)-behaviour for the tails associated with this first product on using
Lemma 5(iii) and (iv) again.
Overall, it remains to establish a suitable expansion for the measure µ4 with Fourier
transform
ꢀ
ꢁ
1
iθ
d
µˆ4(θ) = ρˆ2α(θ) (Σµ3(θ) + 1) iθ log 1 −
.
Combining Lemmas 8 and 9 (with Σµ3 for µ) we obtain
µˆ4(θ) = ρˆ2α(θ) (µˆ5(θ) + µˆ6(θ))
with µ5((−∞, −x]) = O(x−γ log x), µ5((x, ∞)) = O(x−γ log x) as x → ∞ and
dγe−2
ꢈ
ꢉ
X
1
µ6 ÷
(Σj+1µ3)(R) νj+1
−
δ0
.
j + 1
j=1
The last statement in Lemma 4 yields ρ2α ? µ5(I(x)) = O(x−γ log x) for I(x) =
(−∞, −x] and I(x) = (x, ∞). Lemma 7 implies that the tails of ρ2α ? µ6 differ from
those of µ6 by a negligible amount.
Putting all these pieces together we see that
(νrr(µ) − νrr(µ0) + µ6)((x, ∞)) = O(x−γ log x),
and, as both νrr(µ0) and µ6 are concentrated on the right halfline,
νrr(µ)((−∞, −x]) = O(x−γ log x)
(2·10)
as x → ∞. The latter implies the statement of the theorem for the left tail, hence it
remains to check the behaviour of the (right) tail of µ6. Lemma 5(v) yields
1
(Σj+1µ3)(R) =
mj+1(µ3).
(j + 1)!
As (2·9) implies that µˆ3(θ)µˆ(θ) = 1 for all θ in some neighbourhood of 0 we have
mj(µ3) = rj(µ) by the remarks following (2·1). Further, Lemma 6 yields
ν
j+1((x, ∞)) = fj(x) = j! x−j−1 + o(e−x),
so we obtain
dγe−2 rj+1(µ)
X
µ6((x, ∞)) =
x
−j−1 + O(x−γ log x).
(2·11)
j + 1
j=1
From (2·10) and (2·11) the right tail formula (1·7) of the theorem follows.
3. Comments
Some extensions and variations of our main result are obvious, for example to
rate functions more general than x−γ, others would require some more work. In

Tail expansions for random record distributions
381
particular, we would expect that the above methods also yield tail expansions in
the case that the first moment of the input distribution vanishes (see [12] for the
corresponding situation in harmonic renewal theory). With the proper smoothness
conditions on µ it is also possible to obtain expansions for the densities rather than
the tails of random record distributions. For the lattice case, where µ is concentrated
on the multiples of some h > 0, this has been carried out in [20].
There is some similarity between our concept of inverse moments and the classical
concept of cumulants. The latter play a special role in the context of the central limit
theorem, where the distribution with vanishing remainder term is the normal dis-
tribution which has all cumulants equal to zero from order three onwards. Example
2 shows that this role is taken over by the exponential distribution in the random
record context. Are exponential distributions characterized by the requirement that
all inverse moments from order two onwards vanish? This condition implies that all
(ordinary) moments exist and do not increase too rapidly, so, as the ordinary mo-
ments can be obtained from inverse moments, such a characterization does indeed
hold.
In summary, the above shows that Gelfand theory together with some elementary,
albeit lengthy arguments leads to tail expansions for yet another type of convolution
P∞
series, namely n=1 n−1(n + 1)−1µ?n. The techniques can obviously be applied to
P∞
other series such as n=1 n−1(n + 1)−1(n + 2)−1µ?n. What we do not have at the
moment is an explanation for the qualitatively very different form of the expansions
for these series as compared to the classical renewal and harmonic renewal case.
P∞
Gelfand theory provides the connection between convolution series n=1 anµ?n and
P∞
the functions φ: U ⊂ C → C, z →
n=1 anzn, and it seems natural to look at
the nature of the singularity of φ at z = 1 for an explanation. However, the one
observation we can offer at present is the fact that 1/φ0(z) is a polynomial in the
renewal and harmonic renewal case and not in the random record case. Whether
this is an algebraic coincidence or whether such an observation might lead to a
classification of convolution series with respect to the qualitative type of their tail
expansions we do not know.
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