Positive-part moments via the Fourier-Laplace transform

0.23 MB·English

by  Iosif Pinelis

Positive-part moments
via the Fourier-Laplace transform
∗†
0 Iosif Pinelis
1
0 Abstract: Integral expressions for positive-part moments EX+p (p > 0)
2 of random variables X are presented, in terms of the Fourier-Laplace or
FouriertransformsofthedistributionofX.Anecessaryandsufficientcon-
n ditionforthevalidityofsuchanexpressionisgiven.Thisstudywasmoti-
a vatedbyextremalproblemsinprobabilityandstatistics,whereoneneeds
J toevaluatesuchpositive-partmoments.

5
AMS 2000 subject classifications: Primary 60E10, 42A38; secondary

60E07,60E15,42A55,42A61.
]
R Keywordsandphrases:positive-partmoments,characteristicfunctions,
Fouriertransforms,Fourier-Laplacetransforms,integralrepresentations.
P
.
h
t Address:
a
DepartmentofMathematicalSciences
m
MichiganTechnologicalUniversity
[ Houghton,Michigan49931,USA
E-mail:[email protected]

2 Telephone:[phone]
v
4 Runninghead:
1 Positive-partmomentsviatheFourier-Laplacetransform
2
4
.
2
0
9
0
:
v
i
X
r
a
∗Department of Mathematical Sciences, Michigan Technological University, Houghton,
Michigan,USA.
†SupportedbyNSFgrantDMS-0805946
1
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Iosif Pinelis/Positive-part moments via the Fourier-Laplace transform 2
1. Introduction
In a number of extremal problems in probability and statistics (see e.g. [11, 12,
19,20,21,22,1,2,3,23,24,25,4,5,26])oneneedstoevaluatethepositive-part
moments EXp of random variables (r.v.’s) X, where x := 0∨x = max(0,x),
+ +
the positive part of x, and xp :=(x )p, for any x∈R and p>0.
+ +
In particular, an effective procedure was needed in [26] to compute EXp
+
for p = 3 and r.v.’s of the form X = Γ +yΠ , where a,b,y are posi-
µ,a2 b2/y2
tive real numbers, µ ∈ R, Γ and Π are independent r.v.’s, Γ has
µ,a2 b2/y2 µ,a2
the normal distribution with parameters µ and a2, and Π has the Poisson
b2/y2
distributionwithparameterb2/y2.Forpurelynormalr.v.’sX (withoutthePois-
son component) such a computation is easy. However, the naive approach, by
the formula E(Γ +yΠ )p = ∞ E(Γ +yj)p P(Π = j) did not
µ,a2 b2/y2 + j=0 µ,a2 + b2/y2
work well, especially when y is smalPl. On the other hand, the series of the form
∞ ejzP(Π = j) (which is the Laplace-Fourier transform of the Poisson
j=0 b2/y2
Pdistribution) is easily computable. Therefore, a natural idea was to perform a
harmonic analysis of the function R ∋ x 7→ f (x) := xp. This can be easily
p +
done by finding its Laplace-Fourier transform. Then, once the function f is
p
decomposed into harmonics (i.e., exponential functions x7→ezx for z ∈C), one
almost immediately obtains a general expression for the positive-part moments
of a r.v. X in terms of the Laplace-Fourier transform of the distribution of X,
as provided indeed by Theorem 1 in this paper. Such expressions turn out to
be computationally effective, as well as the ones in terms of the characteristic
function(c.f.)R∋t7→EeitX thatareobtainedascorollaries.Theorem2ofthis
paper provides a necessary and sufficient condition for such a representation.
A subsequent literature search has revealed only one paper, Brown [8], that
contains formulas for EXp in terms of the c.f. of X. However, the constant-
+
sign argument used in [8] does not actually seem to work for p ∈ (0,2). The
resultsin[8]arebasedonthemethoddevelopedbyvonBahr[29]toexpressthe
absolutemomentsE|X|p intermsofthec.f.ofX.Otherpaperscontainingsuch
expressions for absolute moments include [14, 8, 9]; see also [18, §§1.8.6–1.8.8]
and [16, §11.4].
Aswasexplained,ourapproachdiffersfromthepreviousones.Webeginwith
expressions of positive-part moments in terms in the Fourier-Laplace transform
z 7→EezX with Rez 6=0, and then use the Cauchy integral theorem to express
EXp intermsofthec.f.ofX.Itshouldbeclearthatexpressionsfortheabsolute
+
moments will follow trivially from expressions for the positive-part moments.
2. Results
Let X be any r.v. Let s and s be any real numbers such that s 606s and
1 2 1 2
EesX <∞ for all s∈[s ,s ]. Let p be any positive real number, and then let
1 2
k :=k(p):=⌊p⌋ and ℓ:=ℓ(p):=⌈p−1⌉,
respectively the integer part of p and the smallest integer that is no less than
p−1, so that k 6p<k+1, ℓ<p6ℓ+1, and ℓ6k. For all complex z and all
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Iosif Pinelis/Positive-part moments via the Fourier-Laplace transform 3
m=−1,0,1,..., let
m zj m z2j
e (z):=ez − , c (z):=(−1)m+1 cosz− (−1)j ,
m 2m
X j! (cid:16) X (2j)!(cid:17)
j=0 j=0
m z2j+1
s (z):=(−1)m+1 sinz− (−1)j ,
2m+1
(cid:16) X (2j+1)!(cid:17)
j=0
with the convention −1 a := 0 for any a ’s, so that e (z) ≡ ez, c (z) ≡
j=0 j j −1 −2
cosz, and s (z)≡siPnz. For any complex number z =s+it, where s and t are
−1
real numbers and i stands for the imaginary unit, let Rez := s and Imz := t,
the real and imaginary parts of z.
In this paper, we shall present a number of identities involving certain in-
tegrals. For the sake of brevity, let us assume the following convention (unless
specifiedotherwise):whensayingthatsuchanidentitytakesplaceundercertain
conditions, we shall actually mean to say that under those conditions the cor-
responding integral exists in the Lebesgue sense (but may perhaps be infinite)
and the identity takes place.
Theorem 1. For any s∈(0,s2] and any j =−1,0,...,ℓ such that E|X|j+ <
∞,
Γ(p+1) ∞ Ee (s+it)X
EXp = j dt (1)
+ 2π Z (s(cid:0)+it)p+1 (cid:1)
−∞
Γ(p+1) ∞ Ee (s+it)X
= Re j dt. (2)
π Z (s(cid:0)+it)p+1 (cid:1)
0
Remark 1. Of course, the statement of Theorem 1 is devoid of content in the
casewhens2 =0.AsfortheconditionE|X|j+ <∞,hereweusetheconvention
00 :=1(any other realnumber inplace of 1would dohereaswell), to interpret
|X|j+ whenX =0andj ∈{−1,0}. So,thecondition E|X|j+ <∞willtrivially
hold if j ∈ {−1,0}. On the other hand, if j > 1 (and hence p > 1), then the
expressionEe (s+it)X undertheintegralsontheright-handsidesof (1)and
j
(2) would lose(cid:0)meaning w(cid:1)ithout the condition E|X|j+ <∞.
Corollary 1. If p ∈ N and E|X|p < ∞, then for any s ∈ [s ,0) and any
1
j =−1,0,...,ℓ
p! ∞ Ee (s+it)X
EXp =EXp+ j dt (3)
+ 2π Z (s(cid:0)+it)p+1 (cid:1)
−∞
p! ∞ Ee (s+it)X
=EXp+ Re j dt.
π Z (s(cid:0)+it)p+1 (cid:1)
0
Corollary 2. If E|X|p <∞ then
EXk Γ(p+1) ∞ Ee (itX)
EXp = I{p∈N}+ Re ℓ dt. (4)
+ 2 π Z (it)p+1
0
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Iosif Pinelis/Positive-part moments via the Fourier-Laplace transform 4
In particular, one has the following: for any m∈N such that EX2m <∞,
EX2m (2m)! ∞ Es (tX)
EX2m = + 2m−1 dt; (5)
+ 2 π Z t2m+1
0
for any m∈N such that E|X|2m−1 <∞,
EX2m−1 (2m−1)! ∞ Ec (tX)
EX2m−1 = + 2m−2 dt. (6)
+ 2 π Z t2m
0
Corollary 3. If Y is another r.v. then
Γ(p+1) ∞ EeitX −EeitY
EXp =EYp+ Re dt (7)
+ + π Z (it)p+1
0
provided that E|X|p+E|Y|p <∞ and EXj =EYj for all j =1,...,k.
Moreover, one has the following: for any m∈N
(2m)! ∞ EsintX −EsintY
EX2m =EY2m+(−1)m dt (8)
+ + π Z t2m+1
0
if EX2m+EY2m <∞ and EX2j+1 =EY2j+1 for all j =0,...,m−1,m− 1;
2
for any m∈N
(2m−1)! ∞ EcostX −EcostY
EX2m−1 =EY2m−1+(−1)m dt (9)
+ + π Z t2m
0
ifE|X|2m−1+E|Y|2m−1 <∞andEX2j =EY2j forallj =0,...,m−1,m−1.
2
Foranyr.v.X asinCorollary2,itisalwayspossible(andeasy)toconstruct
a r.v. Y as in Corollary 3 such that the characteristic function EeitY, and
hence its real and imaginary parts EcostY and EsintY, are easily computable.
In particular, one can always take Y to be a r.v. with finitely many values;
more specifically, k+1 values would suffice for (7) and m+2 values for (8) or
(9). An obvious advantage of the formulas given in Corollary 3 (over those in
Corollary 2) is that the corresponding integrals will converge (for p>1) faster
near ∞, and just as fast near 0. Also, Corollary 3 will be just one step closer
thanCorollary2topossibleapplicationstotheconvergenceofthepositive-part
moments in the central limit theorem; see Example 4 in Section 3 for further
details.
Remark 2. If EXp < ∞ for an even natural p = 2m, then it is clear that
+
identity(4)(or,equivalently,(5))cannotholdwithouttheconditionE|X|p <∞,
sinceoneneedsthemomentEXk =EXp ontheright-handsideof (4)toexist.
Similarly, if EXp < ∞ for an odd natural p = 2m−1, then identity (4) (or,
+
equivalently, (6)) cannot hold without the condition E|X|p <∞.
However, if p > 0 is not an integer, the term EXk I{p ∈ N} on the right-
2
hand side of (4) disappears. One may then wonder as to what, if any, moment
condition on the left tail of the distribution of X is needed in order for identity
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Iosif Pinelis/Positive-part moments via the Fourier-Laplace transform 5
(4)tohold–havinginmindthatthefinitenessofthepositivepartmomentEXp
+
depends only on the right tail of the distribution of X. Perhaps surprisingly, it
turnsoutthataleft-tailconditionisnecessaryfor(4),eveniftheintegralin(4)
is allowed to be understood as an improper one (at 0); moreover, the moment
conditionE|X|p <∞inCorollary2canbeonlyslightlyrelaxed,evenwhenpis
not an integer. Actually, one can obtain a necessary and sufficient condition for
such a representation of the positive-part moment in terms of the characteristic
function, as described in the following theorem.
Theorem 2. Take any p ∈ (0,∞)\N and any r.v. X with EXp < ∞. Then
+
the following two conditions are equivalent to each other:
Γ(p+1) ∞ Ee (itX)
(I) E|X|ℓ <∞ and EXp = Re ℓ dt;
+ π Z (it)p+1
0+
(II) P(X >x)=o(1/xp) as x→∞.
−
Here and in what follows, x :=(−x) for all x∈R; also let xp :=(x )p.
− + − −
Note that the part E|X|ℓ < ∞ of condition (I) of Theorem 2 will trivially
hold if 0 < p < 1 (and hence ℓ = 0). On the other hand, if p > 1 (and hence
ℓ > 1), then the expression Ee (itX) in condition (I) of Theorem 2 would lose
ℓ
meaning if E|X|ℓ =∞; cf. Remark 1.
Notealsothat,ifcondition(II)or,equivalently,(I)ofTheorem2holds,then
EXr <∞ for all r ∈(0,p). Indeed, condition (II) of Theorem 2 implies
−
EX−r =R0∞rxr−1 P(X− >x)dx=O R0∞rxr−1(1∧x−p)dx <∞. (10)
(cid:0) (cid:1)
On the other hand, it is easy to give examples where condition (II) of Theo-
rem2holds,whileEXp =∞,andso,Corollary2isnotapplicable;forinstance,
−
take any r.v. X such that P(X >x)=1/(xplnx) for all large enough x>0.
−
InviewofCorollary2andTheorem2,fortheidentity (4)toholdforagiven
p ∈ (0,∞)\N (with the integral understood in the Lebesgue sense) and a r.v.
X with EXp < ∞, it is sufficient that EXp < ∞ and it is necessary that
+ −
EXr <∞ for all r ∈(0,p).
−
Yet, no “simple” necessary and sufficient condition for (4) to hold – in the
Lebesguesense–foragivenp∈(0,∞)\Nappearstobeknown;itisapparently
an open question whether such a condition exists at all. However, one can see
that condition (II) of Theorem 2 is not sufficient (although, by Theorem 2, it
is necessary) for (4). Indeed, for any p ∈ (0,∞)\N, there exists a r.v. X with
EXp <∞suchthatthetwoequivalentconditions–(I)and(II)–ofTheorem2
+
hold, while the integral ∞ReEeℓ(itX)dt does not exist in the Lebesgue sense.
0 (it)p+1
For instance, one can coRnsider
Example 1. The idea is to take a r.v. X with a lacunary distribution and
then construct a matching lacunary set A ⊂ (0,∞) such that the integrand
ReEeℓ(itX) is of constant sign and large enough in absolute value for t ∈ A.
(it)p+1
Take indeed any p∈(0,∞)\N and let X be a discrete r.v. such that, for some
d∈(1,∞), one has P(X =−dn)= c for all n∈N, where c is the constant
ndnp
chosen so that P(X =−dn)=1. Then it is easy to check that condition
n∈N
P
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Iosif Pinelis/Positive-part moments via the Fourier-Laplace transform 6
(II) (and hence condition (I)) of Theorem 2 holds. Moreover, X =0 a.s., and
+
so, EXp < ∞. Further, one can show that there exist some d = d(p) > 1,
+
ε=ε(p)∈(0,1−1/d), and m=m(p)∈N such that (−1)mReEeℓ(itX) > 1
(it)p+1 tlnε
for all t ∈ A := ∞ [ε(1−ε), ε ], whence (−1)m ReEeℓ(itX)dt = ∞. Fotr
n=1 dn dn A (it)p+1
details, see [27, PrSoposition 2.6 and its proof]. R
Theorem 2 and Example 1 are similar in spirit to a number of known if-
and-only-if conditions and counterexamples, respectively; those results in the
literaturemainlyconcernsuchsimplerrelationsastheonesofthetails/moments
of the distribution with the derivatives of the characteristic function at 0 (but
alsowithotherlinearfunctionalsofthec.f.)–seee.g.[13,30,28,6,7].Ibragimov
[15] provides necessary and sufficient conditions – both in terms of (truncated)
moments/tails and the c.f. – for rates O(n−p) in the central limit theorem to
hold; cf. Example 4 in the next section.
3. Applications
Here let us give examples of known or potential applications of identities pre-
sented in Section 2.
Example 2. Let X ,...,X be independent r.v.’s such that X 6 y for some
1 n i
real y > 0 and EX 6 0, for all i. Let S := X +···+X and assume that
i 1 n
EX2 6 σ2 for some real σ > 0. Also, let ε := 1 E(X )3, so that
i i σ2y i i +
P0 < ε < 1. In the paper [26], mentioned in the IntroductiPon, an optimal in a
certain sense upper bound on the tail P(S >x) was obtained, of the form
E3(η−t )2
Pin(x):= x +,
E2(η−t )3
x +
where x ∈ R; η := Γ +yΠ˜ ; Γ and Π˜ are any indepen-
(1−ε)σ2 εσ2/y2 (1−ε)σ2 εσ2/y2
dentr.v.’swhosedistributionsare,respectively,thecenteredGaussianonewith
variance (1−ε)σ2 and the centered Poisson one with variance εσ2/y2; t ∈R is
x
theonlyrealrootoftheequationm(t )=x;andm(t):=t+E(η−t)3/E(η−t)2.
x + +
Thus,tocomputetheboundPin(x),oneneedsafastcalculationofthepositive-
part moments of the r.v. X :=η−t. While the direct approaches mentioned in
the Introduction do not work here, the formulas of Theorem 1 and Corollary 2
proveveryeffective,sincetheFourier-Laplacetransformofther.v.η−tisgiven
by a simple expression:
Eez(η−t) =exp −zt+ λ2 (1−ε)σ2+ ezy−1−zy εσ2 ∀z ∈C. (11)
2 y2
(cid:8) (cid:9)
It takes under 0.5 sec in Mathematica on a standard Core 2 Duo laptop to
produce an entire graph of the bound Pin(x) for a range of values of x using
either (2) or (4).
Example 3. Aspointedoutbyareferee,positive-partmomentsnaturallyarise
in mathematical finance. For example, (S −K) is the value of a call option
+
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Iosif Pinelis/Positive-part moments via the Fourier-Laplace transform 7
with the strike price K when the stock price is S. Exact bounds on E(S−K)
+
under various conditions were recently given in [10].
Example 4. Just as it was done in [8] for absolute moments, the identities
given in Corollary 3 could be used to prove the convergence of the positive-part
moments in the central limit theorem, or even to obtain bounds on the rate of
suchconvergence.However,weshallnotbepursuingthismatterhere,sincethe
mainmotivationforthepresentpaperwastheneedforaneffectivecomputation
of the bound Pin(x), as described in Example 2.
4. Proofs
For any two expressions a and b, let us write: a < b or, equivalently, b > a if
⌢ ⌢
|a| 6 Cb for some positive constant C depending only on p; a ≍ b if ab > 0,
a<b, and b<a.
⌢ ⌢
Letusalsowritea<<b(toberead“aismuchlessthanb”)or,equivalently,
b >> a (“b is much greater than a”) if b > 0 and |a| = o(b). Accordingly,
if A is an assertion and a > 0, then “A holds if a << b” will mean “for all
p ∈ (0,∞)\N there exists some constant K = K(p) > 0 such that A holds
whenever |a| < b/K”, that is, A will hold “eventually”. Similarly, “if a << b
and A holds, then a << b ” will mean “for all p ∈ (0,∞)\N and all K > 0
1 1 1
there exists K = K(p,K ) > 0 such that one has the implication ‘if |a| < b/K
1
and A holds, then |a |<b /K ’”.
1 1 1
Also, let us write a∼b if a/b→1.
Proof of Theorem 1. Note that (2) immediately follows from (1), since the in-
tegrandin(2)isevenint∈R.Notealsothat zj dz =0forallj =0,...,ℓ
and s 6= 0, where C := {z ∈ C: Rez = s}. SRoC,siztp+is1enough to prove (1) with
s
j = −1. The key observation here is the following: for any x ∈ R and any
complex number z =s+it with s:=Rez >0,
∞ x ∞ Γ(p+1)
ezu(x−u)p du= ezu(x−u)pdu=ezx vpe−zvdv = ezx;
Z + Z Z zp+1
−∞ −∞ 0
thisisobviousforrealz >0,andthenonecanuseanalyticcontinuation. Thus,
for each s ∈ (0,∞) the Fourier transform R ∋ t 7−→ ∞ eituesu(x−u)p du
−∞ +
of the integrable function R ∋ u 7−→ esu(x−u)p is Rthe function R ∋ t 7−→
+
Γ(p+1) e(s+it)x , which is integrable as well. So, by the inverse Fourier trans-
(s+it)p+1
form, oneobtains (1)(with j =−1)with therealconstant xinplace ofthe r.v.
X.(Equivalently,onecanuseheretheLaplaceinversion.)Now(1)followsbythe
Fubini theorem, since E e(s+it)X = EesX for all s>0 and t∈R.
(s+it)p+1 (s2+t2)(p+1)/2
(cid:12) (cid:12)
Proof of Corollary 1. H(cid:12)ere, just(cid:12)as in the proof of Theorem 1, without loss
of generality (w.l.o.g.) j = −1. Consider first the case when s > 0. W.l.o.g.,
2
s < 0. By the convexity of EesX in s, one has Ee(s+it)X 6 Ees1X∨Ees2X
1 (s+it)p+1 |t|p+1
(cid:12) (cid:12)
(cid:12) (cid:12)
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Iosif Pinelis/Positive-part moments via the Fourier-Laplace transform 8
for all s ∈ [s ,s ] and all t 6= 0, so that s2 Eej (s+it)X ds → 0 as |t| →
∞. Hence, by1th2e Cauchy integral theoremRsa1n(cid:12)d t(hs(cid:0)e+ict)opn+t1in(cid:1)u(cid:12)ity of the integral
(cid:12) (cid:12)
I(s) := ∞ Ee(s+it)X dt in s ∈ [s ,s ], the difference between the value of I(s)
−∞ (s+it)p+1 1 2
for s ∈ (R0,s ] and that of I(s) for s ∈ [s ,0) is 1 EezX dz, where C is the
circle in C o2f radius ε centered at 0 and 1traced oiuRtCcεouzpn+t1erclockwise, pεrovided
that ε∈(0,s ∧s ). So, (3) (in the case when s >0 and with j =−1) follows
1 2 2
from (1) by taking ε ↓ 0, since ezX = e(zX) +O(|zX|p+1eε|X|) over z ∈ Cε;
p
recall that here it is assumed that p∈N.
Assume now that s =0 (and, again, j =−1). Then, by what has just been
2
proved, one has (3) for any s ∈ [s ,0) with X ∧N in place of X, where N is
1
any real number. By the dominated convergence theorem and in view of the
condition E|X|p < ∞, one has E(X ∧N)p → EXp and E(X ∧N)p → EXp
+ +
as N → ∞. To complete the proof of Corollary 1, it remains to use again the
dominated convergence theorem, in view of the inequalities Ee(s+it)(X∧N) 6
(s+it)p+1
(s2E+ets2()X(p∧+N1))/2 and es(X∧N) 61+es1X for all N >0, s∈[s1,0),(cid:12)(cid:12)and t∈R. (cid:12)(cid:12)
Proof of Corollary 2. Let us first prove identity (4) in the case when the r.v.
X takes on only one value, say x ∈ R. For any ε > 0, consider the integration
contour C := {it: t 6 −ε}∪C+ ∪{it: t > ε}, where C+ := {z ∈ C: |z| =
ε ε
ε, Rez >0}. Then, by (1) (with j =−1) and the Cauchy integral theorem,
2πi ezx e (zx)
xp = dz = ℓ dz, (12)
Γ(p+1) + Z zp+1 Z zp+1
C C
because zj dz =0forallj =0,...,ℓ.Next,e (zx)= zℓ+1 xℓ+1+O(|z|ℓ+2)
C zp+1 ℓ (ℓ+1)!
as z →0R, and so,
e (zx) xℓ+1 dz
ℓ dz = +O(εℓ−p+2)
ZCε+ zp+1 (ℓ+1)!ZCε+ zp−ℓ
xℓ+1
=iπ I{p∈N}+O εℓ−p+1(I{p∈/ N}+ε)
(ℓ+1)!
(cid:0) (cid:1)
xℓ+1 xk
=iπ I{p∈N}+o(1)=iπ I{p∈N}+o(1)
(ℓ+1)! Γ(p+1)
as ε↓0. This and (12) imply
xk Γ(p+1) ∞ e (itx)
xp = I{p∈N}+ Re ℓ dt, (13)
+ 2 π Z (it)p+1
0+
since the latter integrand is even in t∈R.
Next, observe that
|e (iu)|<|u|ℓ∧|u|ℓ+1 (14)
ℓ ⌢
over all u∈R, and so, for all x∈R
∞ e (itx)
ℓ dt=c |x|p, (15)
Z (cid:12)(it)p+1(cid:12) ℓ
0 (cid:12) (cid:12)
(cid:12) (cid:12)
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Iosif Pinelis/Positive-part moments via the Fourier-Laplace transform 9
where c := ∞| eℓ(iu) |du < ∞ for p ∈ (0,∞)\N. Therefore, (4) follows from
ℓ 0 (iu)p+1
(13) by the FRubini theorem – provided that p∈(0,∞)\N.
The case p∈N is treated quite similarly. Namely, if p=2m for some m∈N,
then Re eℓ(itx) = s2m−1(tx) for all x ∈ R and t > 0, and so, (5) follows by the
(it)p+1 t2m+1
Fubini theorem from (13), the estimate |s (u)| < |u|2m−1∧|u|2m+1 for all
2m−1 ⌢
u∈R, and the condition EX2m <∞. If p=2m−1 for some m∈N, then (6)
similarly follows using the estimate |c (u)| < |u|2m−2∧|u|2m for all u ∈ R
2m−2 ⌢
and the condition E|X|2m−1 <∞.
Proof of Corollary 3. This immediately follows from Corollary 2, applied the
r.v. Y, as well as to X. (Note that k = p and ℓ = p−1 if p ∈ N, and k = ℓ if
p∈/ N.)
Proof of Theorem 2. Step 0. Take indeed any p ∈ (0,∞)\N and any r.v. X
with EXp < ∞. Recalling (10), note that either one of the conditions (I) and
+
(II) implies
E|X|ℓ <∞. (16)
Step 1. For v >0, let
∞ e (−iu)
I(v):=I (v):= Re ℓ du. (17)
p Z (iu)p+1
v
Thisdefinitioniscorrect,sincetheintegralexistsintheLebesguesense,inview
of estimate (14). Note that
I (0)=0; (18)
p
this is a special case of (4), with X = −1 almost surely (a.s.). In turn, (18)
implies
v e (−iu)
I (v)=− Re ℓ du. (19)
p Z (iu)p+1
0
Step 2. At this step, let us prove that if p∈(0,1) and v ∈(0,∞), then
I (v)>0. (20)
p
Take indeed any p∈(0,1). Then ℓ=0, and (17) and (19) can be rewritten as
∞ sinπp −sin(πp +u)
I(v)= 2 2 du (21)
Z up+1
v
v sin(πp +u)−sinπp
= 2 2 du (22)
Z up+1
0
for all v > 0. Observe next that vp+1I′(v) = sin(πp +v)−sinπp for all v >
2 2
0, whence the only local minima of I in (0,∞) are at the points 2π,4π,....
Therefore, and because of (18) and the obvious equality I(∞−) = 0, to prove
(20) it suffices to show that I(2jπ)>0 for all j ∈N.
imsart ver. 2005/05/19 file: rev2a.tex date: January 5, 2010
Iosif Pinelis/Positive-part moments via the Fourier-Laplace transform 10
Using (21) and (twice) applying the Fubini theorem, for all j ∈N and q ∈N
such that j <q one has
2qπ sin(πp +2jπ)−sin(πp +u)
I(2jπ)−I(2qπ)= 2 2 du
Z up+1
2jπ
2qπ du u πp
=− dt cos +t
Z2jπ up+1 Z2jπ (cid:16) 2 (cid:17)
2qπ πp 2qπ du
=− dt cos +t
Z2jπ (cid:16) 2 (cid:17)Zt up+1
1 2qπ πp 1 1
=− dt cos +t −
p Z2jπ (cid:16) 2 (cid:17)(cid:16)tp (2qπ)p(cid:17)
1 2qπ πp 1
=− dt cos +t
p Z2jπ (cid:16) 2 (cid:17)tp
1 2qπ πp ∞
=− dt cos +t up−1e−tudu
pΓ(p) Z2jπ (cid:16) 2 (cid:17) Z0
1 ∞ 2qπ πp
=− up−1du dt cos +t e−tu
Γ(p+1) Z0 Z2jπ (cid:16) 2 (cid:17)
1 ∞ sinπp −ucosπp
= up−1 2 2 (e−2jπu−e−2qπu)du;
Γ(p+1) Z 1+u2
0
letting now q →∞ and using again the equality I(∞−)=0, by the dominated
convergence and a comparison of the integrands one obtains
1 ∞ sinπp −ucosπp
I(2jπ)= up−1 2 2 e−2jπudu
Γ(p+1) Z 1+u2
0
1 ∞ sinπp −ucosπp
> up−1 2 2 e−2jπudu
Γ(p+1) Z 1+tan2 πp
0 2
(2jπtanπp −p) cos3 πp (jπ2−1) cos3 πp
= 2 2 > 2 >0, (23)
(2jπ)p+1p (2jπ)p+1
which completes the proof of (20).
Step 3. Still assuming that p ∈ (0,1), it follows from (23) that I(2jπ) >
(jπ2(−2j1π))cpo+s13 π2p ⌢>j−p overallj ∈N.Next,foreveryj ∈N,theonlylocalminima
ofI ontheinterval[2jπ,(2j+2)π]aretheendpoints;so,I(v)>I(2jπ)∧I (2j+
2)π >j−p >v−p overallv ∈[2jπ,(2j+2)π]andallj ∈N.Thatis,I(v)(cid:0)>v−p
⌢ ⌢ ⌢
over(cid:1)all v ∈ [2π,∞). On the other hand, by (21), |I(v)| 6 ∞ 2 du < v−p
v up+1 ⌢
over all v ∈(0,∞). Thus, I(v)≍v−p over all v ∈[2π,∞). R
Yetontheotherhand,by(22),I(v)≍ v u du≍v1−p overallv inaright
0 up+1
neighborhood of 0. R
Also, I is obviously continuous on (0,∞). So, in view of (20), for each p ∈
p
(0,1),
I (v)≍v−p(v∧1) over all v >0. (24)
p
imsart ver. 2005/05/19 file: rev2a.tex date: January 5, 2010