3.5. Remainder term estimate: 0< α <1 61
= 1 +ϕα,1(t) + ϕ2α,1(t)
2! + ϕ3α,1(t) 3! +. . .
!
1 +
s
X
j=2
Ajϕjα,1(t)
= 1 +ϕα,1(t) +
1 2+A2
ϕ2α,1(t) +
1
6+A2+A3
ϕ3α,1(t) +. . .
+ 1
m! +
m
X
u=2
Au (m−u)!
!
ϕmα,1(t) +· · ·+ 1 s!+
s
X
u=2
Au (s−u)!
!
ϕsα,1(t) +
∞
X
k=s+1
ϕkα,1(t) 1 k!+
s
X
u=2
Au (k−u)!
!
= 1 +ϕα,1(t) +
s
X
k=2
ϕkα,1(t) 1 k! +
k
X
u=2
Au (k−u)!
!
+
∞
X
k=s+1
ϕkα,1(t) 1 k!+
s
X
u=2
Au (k−u)!
!
= 1 +ϕα,1(t) +
∞
X
k=s+1
ϕkα,1(t) 1 k! +
s
X
u=2
Au (k−u)!
!
. Thus,
f(t)−g(t) =e
∞
X
n=1
cn/α1 α α−n
(it)n n! −
∞
X
k=s+1
ϕkα,1(t) 1 k!+
s
X
u=2
Au (k−u)!
!
.
From this expansion for fixed s we can obtain formulas for pseudomoments µ∗j. Indeed, we have
f(t)−g(t) =e
[(1+s)α]
X
n=1
cn/α1 α α−n
(it)n
n! +o|t|α(s+1) as t→0, whence from Lemma 3.25 it follows that
µ∗j =cj/α1 α
α−j for j = 1, . . . ,[(1 +s)α].
It is important to note that s can be chosen arbitrarily. This means that for the Pareto distribution we can always construct such Hf that we have as many finite pseudomomentsµ∗j :=µj(H) as we want.f
ci :=Ci(α), i= 1, ..., s, from (58) and di are suitable constants for i= 2, . . . , s.
Our goal is to build an approximation of the distribution function Fn defined by Fn(x) :=P
Sn−an
bn ≤x
,
whereSn=X1+X2+· · ·+Xnand (an), (bn) are some suitably chosen normalizing sequences such that Fn →Gα,1 as n → ∞. From the definition of DNA(Gα,1) (see Definition 3.5) it follows thatbn=a n1/α with a >0. Without loss of generality we put a= 1, and also for α ∈(0,1) we can take an= 0, n ∈N. This means, we have
Fn(x) = P X1+· · ·+Xn≤xn1/α=Fn∗(xn1/α).
It is known thatFn→Gα,1, but we want to construct a correction term forGα,1(x) in order to get a better approximation. Just as in Theorem 3.10 or Theorem 3.21 this correction function is a linear combination of derivatives of the corresponding limit distribution. The only difference is that the coefficients of this linear combination depend on new pseudomoments µ∗k =µk(F −Geα) defined by (79).
For a given distribution functionF we construct a functionGeαusing formula (69) and fix it. In terms of new pseudomoments µ∗k for some r ∈ R+, r > α and n ∈N we construct the function Wfr,n(x) for all x∈R as follows:
Wfr,n(x) =
ρ
X
k=2
ck,n
nk G(0,k)(x,1) (87)
+
p
X
k=0 mk
X
`=1
n
`
!ck,n−`
nk
m`,k
X
u=`
pu,`,k
X
v=0
G(u,k+v)(x,1)(−`/n)v(−1)u
v! n−u/α Ceu,`, where ρ = [2(R + 1)/α], p = [2R/α], m`,k = [R + 1 + α(` − 1 − k/2)], mk = 1 + [(R−αk/2)/(1−α)], pu,`,k = max{0,[(R+ 1−u)/α+`−1−k/2]}
with
R =
( [r], if [r]6=r
r−1, if [r] =r , (88)
cr, ρ= X
k0+k2+···+ks=ρ 2k2+···+sks=r
ρ!
k0!k2!· · ·ks!Ak22· · ·Akss, r, ρ∈N0, (89) s is from (85), Aj,j = 2, ..., s, are from (70),G(u,k)(x,1) are defined by (71) and
Ceu,` = X
k1+2k2+...+RkR=u k1+k2+...+kR=`
`!
k1!...kR!
µ∗1 1!
k1
...
µ∗R R!
kR
, u=`, ..., m`,k. (90)
Then we have the following analogue of Theorem 3.21.
Theorem 3.26. If for r > 1 we have 0 < γr∗ < ∞ and ν0∗ <1, then for all x ∈R and all integersn ≥2 the following inequality holds:
Fn(x)−Gα,1(x)−Wfr,n(x)≤C(1 +|x|)−rn−r−αα 1 +nαr Qn
,
3.5. Remainder term estimate: 0< α <1 63
where constant C does not depend on x and n, Wfr,n(x) is defined by (87) and Qn =ν0∗n−1+sup|t|>
eε
|f(t)|+ 2γr∗n−r/αn−1 with εe defined as follows:
εe= min
1, 1
c0, 1
(2D)1/α, 1 D1/α
cosαπ2 8D
2+ρ/2 α
,
cosαπ2 16e c0
1/(1−α)
, (91)
where c0 = (ν0∗ + 1)γr∗1/r, D = max
2≤j≤s{2|Aj|1/j}, ρ = [2(R+ 1)/α] with Aj defined by (70) and R defined by (88).
Proof. Section 4 will be devoted to the proof of this result.
Remark 3.34. If ν0∗ < 1, then for each ε > 0 we have sup|t|>ε|f(t)| < 1. Indeed, from formula (77) for egα(t) and from definition (80) ofν0∗ we have
|f(t)−geα(t)|=
Z +∞
−∞
eitxdF −Geα(x)
≤
Z +∞
0
dF −Geα(x)=ν0∗. Then, using formulas (77), (74) forgeα(t) and the fact thatν0∗ <1 we obtain:
|f(t)| ≤ν0∗+|geα(t)| ≤ν0∗+|gα,1(t)|
1 +
s
X
j=2
|Aj| |ϕjα,1(t)|
≤ν0∗+e−|t|αcos(α π/2)
1 +
s
X
j=2
|Aj| |t|α j
<1 for large |t|.
Finally, from Lemma A.16 it follows that sup|t|>ε|f(t)|<1 for each ε >0.
Remark 3.35. Note that there exists such n0 ∈ N that nr/αQn ≤ 1 for all n ≥ n0, since ν0∗ <1 and, as a result, sup|t|>
eε
|f(t)|<1 (see Remark 3.34).
Example 3.11. Let us consider a Pareto-distributed random variable X with α = 1/2 and κ = c21, where c1 = C1(1/2) = 1/√
π is defined in (58). The dis-tribution functionF of X has the form:
1−F(x) = 1
√π√
x, x≥1/π.
Comparing this representation with representation (85) we can make the following conclusions: s can be chosen equal to 3 (since 3α ≥ 1 +α), u(x) = 0 and d3 = 0.
Coefficient d2 can be chosen arbitrarily, since c2 =C2(1/2) = 0 (see formula (58)).
Let us put d2 = 3/4.
In order to apply Theorem 3.26 we have to check the condition ν0∗ < 1 and to decide for which r we have γr∗ <∞ and νr∗ < ∞. According to the definition (80) we have
νr∗ =
Z ∞ 0
xrdF −Ge1/2(x), r ≥0, where
Ge1/2(x) =G1/2,1(x) +
3
X
j=2
AjG(0,j)(x,1)
with coefficients A2 =−1/8 and A3 =−1/24 (see (69) and (70)), chosen in such a way, that
F(x)−Ge1/2(x) =Ox−5/2 as x→ ∞.
Note that we obtainO(x−5/2) and not just O(x−(s+1)α) = O(x−2), sinceC4(1/2) = 0 in (57). Therefore, γ5/2∗ <∞ and νr∗ <∞ for r <5/2. So, we apply Theorem 3.26 with r= 5/2.
Let us check whether the condition ν0∗ < 1 is satisfied. Note that the stable distribution corresponding toG1/2,1(x, λ) is a Lévy distribution which has an explicit density function
p1/2,1(x, λ) = λ 2√
πe−λ
2
4xx−32, x >0, λ >0.
Using this we get ν0∗ =
Z ∞ 0
dF −Ge1/2(x)=
Z ∞ 0
p(x)−pe1/2(x)dx, where
pe1/2(x) = 1 384√
π e−41x x−92 192x3+ 48x2−18x+ 1.
Using software Mathematica we obtain ν0∗ ≈ 0.32 < 1. Thus, Theorem 3.26 is applicable. Now, let us see now how the correction termWf5/2,n(x) looks like in this particular case.
Wf5/2,n(x) =
12
X
k=2
ck,n
nk G(0,k)(x,1) +
8
X
k=0
1+[4−k/2]
X
`=1
n
`
!ck,n−`
nk
m`,k
X
u=`
pu,`,k
X
v=0
G(u,k+v)(x,1)(−`/n)v(−1)u
v! n−u/α Ceu,`, wherem`,k =h3 + 12(`−1− k2)i, pu,`,k = maxn0,h2 (3−u) +`−1− k2ioand
cr, ρ = X
k0+k2+k3=ρ 2k2+3k3=r
ρ!
k0!k2!k3!
−1 8
k2 −1 24
k3
, r, ρ∈N0,
Ceu,` = X
k1+2k2=u k1+k2=`
`!
k1!k2!
µ∗1 1!
k1µ∗2 2!
k2
, u=`, ..., m`,k.
The first and the second pseudomoments µ∗1, µ∗2 can be found precisely using the same method as in Example 3.10. The only difference is that we do not putd2 equal to 0. We get
µ∗1 =
Z ∞ 0
x dF −Ge1/2(x) = c1/α1 α α−1 +
1 2+A2
= 3 8 − 1
π, µ∗2 =
Z ∞ 0
x2dF −Ge1/2(x) = 2!
c2/α1 α 2(α−2)−
1 24+1
2A2+A3
= 1 8− 1
3π2.
3.5. Remainder term estimate: 0< α <1 65
From Theorem 3.26 it follows that for allx∈R and n ≥2 we have
Fn(x)−G1/2,1(x)−Wf5/2,n(x)≤C(1 +|x|)−5/2n−4 1 +n5Qn, whereQn =ν0∗n−1+sup|t|>
eε|f(t)|+ 2γr∗n−5/(2α)n−1 with εedefined by (91).
4 Proof of the main result
4.1 Some auxiliary functions and plan of the proof
In order to prove Theorem 3.26 we need to introduce some auxiliary functions. The first one is a truncated distribution function. We define it for any fixed n ∈N and ξ∈[0,∞) as follows:
Fn,ξ(y) =
( F(y) for y≤n1/α(1 +ξ),
Geα(y) for y > n1/α(1 +ξ), y ∈R. (92) Thus, we have a family of truncated functions: Fn,ξ
n∈N,ξ∈[0,∞). We denote Hn,ξ(x) := Fn,ξ(x)−Geα(x), i.e.
Hn,ξ(x) =
(
H(x) =f F(x)−Geα(x) for x≤n1/α(1 +ξ),
0 for x > n1/α(1 +ξ), x∈R, (93) and consider pseudomoments µi,n,ξ = µiHn,ξ and νr,n,ξ = νrHn,ξ. They are well-defined for the same reasons as those for whichµ∗i and νr∗ are well-defined (see Remark 3.33). Moreover, for any i ∈ N0 and any r ≥ 0 we have |µi,n,ξ| < ∞ and νr,n,ξ <∞. Indeed, puttingN =n1/α(1 +ξ), taking into account the possible jump of H(x) at point x = N, and using the boundedness of ν0∗ (see Lemma 3.24 (ii)) and of Geα (see (72)), we obtain
νr,n,ξ = νrHn,ξ=
Z +∞
−∞
|x|r dHn,ξ(x)=
Z N 0
xr dH(x)f +NrH(Nf )
≤ Nr
Z N 0
dH(x)f +H(Nf )
!
≤Nrν0∗+F(N)−Geα(N)
≤ Nrν0∗ + 1 +Geα(N)<∞ ∀r≥0. (94) From (94) and from Lemma 3.17 it follows that|µi,n,ξ| ≤νi,n,ξ <∞ for any i∈N0. Note also that for alln ∈N and ξ≥0 we have
Fn,ξ(x) =Hn,ξ(x) = 0 for x <0. (95) This follows from Lemma 3.22 (i) and from the fact that we consider only non-negative random variables, i.e. we haveF(x) = 0 for x <0.
Let us consider one more function. We denote Mn,ξ(x) := Hn,ξ(x)−Hn,0(x).
Using the definition of Hn,ξ we obtain Mn,ξ(x) =
( F(x)−Geα(x), if x∈n1/α, n1/α(1 +ξ)i,
0, otherwise, x∈R. (96)
Note thatMn,ξ and Hn,ξ are functions of bounded variation. This follows from the fact thatH(x) =f F(x)−Geα(x) is a function of bounded variation (see Remark 3.33).
Thus, pseudomoments µk
Mn,ξ
, k ∈N0, and absolute pseudomoments νr
Mn,ξ
,
r ≥ 0, are well-defined. From the definition of Hn,ξ, Mn,ξ, Lemma 3.17 and in-equality (94) it follows that
µi(Mn,ξ)≤νi(Mn,ξ)≤νi(Hn,ξ) = νi,n,ξ <∞, i∈N0. (97) From Lemma 3.22 (iii) and from the definition of Mn,ξ it follows that
x→+∞lim Mn,ξ(x) = 0 for all n∈N and ξ≥0. (98) The following two lemmata give some properties of pseudomoments µi,n,ξ, µi(Mn,ξ) and νr,n,ξ as well as the connection between them and pseudomoments µ∗i and νr∗. Recall that for any r∈R+ we denote
R =
( [r], if [r]6=r,
r−1, if [r] =r. (99)
Lemma 4.1. For pseudomomentsµi,n,ξ, µiMn,ξand µ∗i the following statements hold true.
(i) µ0,n,ξ =µ0Mn,ξ= 0 for all n∈N and ξ≥0;
(ii) If for r >1 we have 0< γr∗ <∞, then
µu,n,ξ−µ∗u≤2 (n1/α(1 +ξ))u−rγr∗, u= 1, . . . , R, where R is given by (99).
Proof. (i)We put N =n1/α(1 +ξ). Then, µ0,n,ξ =
Z +∞
−∞ dHn,ξ(x) =
Z +∞
0
dHn,ξ(x) =
Z +∞
0
dFn,ξ(x)−
Z +∞
0
dGeα(x)
=
Z N 0
dF(x) +
Z +∞
N
dGeα(x) +Geα(N)−F(N)−
Z +∞
0
dGeα(x)
=
Z N 0
dF(x)−F(N) = 0.
Using the last equality and the definition of Mn,ξ(x) =Hn,ξ(x)−Hn,0(x) we obtain µ0
Mn,ξ
=
Z +∞
−∞ dMn,ξ(x) =
Z +∞
−∞ dHn,ξ(x)−
Z +∞
−∞ dHn,0(x) = µ0,n,ξ−µ0,n,0 = 0.
(ii) Using the fact that lim
x→+∞H(x) = 0, which follows from equality (73), we havef
foru= 1, . . . , R:
µu,n,ξ−µ∗u =
Z +∞
0
yudHn,ξ−Hf(y)
=
Z +∞
N
yudH(y) +f NuH(Nf )
≤
Z +∞
N
yudH(y)f +
Nu
Z +∞
N
dH(y)f
4.1. Some auxiliary functions and plan of the proof 69
≤
Z +∞
N
yudH(y)f +Nu
Z +∞
N
dH(y)f
≤ 2
Z +∞
N
yudH(y)f .
We distinguish two cases. For [r]6=r we obtain
µu,n,ξ−µ∗u≤2
Z +∞
N
yudH(y)f ≤2Nu−rNr−[r]
Z +∞
N
y[r]dH(y)f ≤2Nu−rγr∗. For [r] =r we have
µu,n,ξ−µ∗u ≤ 2
Z +∞
N
yudH(y)f ≤2Nu−rN
Z +∞
N
yr−1dH(y)f
≤ 2Nu−r
Z N
0
xrdH(x)f
+N
Z +∞
N
yr−1dH(y)f
!
≤ 2Nu−rγr∗. This completes the proof of the lemma.
Lemma 4.2. If for r >1 we have 0< γr∗ <∞, then (i)
νk,n,ξ ≤νk∗ ≤(ν0∗+ 1)γr∗k/r, k = 1, . . . , R, where R is defined by (99).
(ii)
νq,n,ξ ≤Cn1/α(1 +ξ)q−rγr∗, q > r, q ∈R, (100) where C is some constant, which depends only on q and r.
Proof. (i) As always we put N =n1/α(1 +ξ). Using the fact that lim
x→+∞H(x) = 0f
we have for k= 1, . . . , R:
νk,n,ξ =
Z +∞
−∞ xkdHn,ξ(x)=
Z +∞
0
xkdHn,ξ(x)=
Z N 0
xkdH(x)f +NkH(Nf )
=
Z N 0
xkdH(x)f +Nk
Z +∞
N
dH(x)f
≤
Z N 0
xkdH(x)f +
Z +∞
N
xkdH(x)f =νk∗.
The other inequality from(i) we obtain as follows with z =γr∗1/r: νk∗ =
Z +∞
0
xkdH(x)f =
Z z 0
xkdH(x)f +
Z +∞
z
xkdH(x)f
≤ zk
Z z 0
dH(x)f +
zk−rzr−[r]
Z +∞
z
x[r]dH(x)f , if [r]6=r, zk−rz
Z +∞
z
xr−1dH(x)f , if [r] =r
≤ zkν0∗+zk−rγr∗ ≤(ν0∗+ 1)γr∗k/r.
(ii) We denote T(x) = −Rx+∞zRdH(z)f . Using integration by parts, we obtain νq,n,ξ =
Z +∞
−∞ xqdHn,ξ(x)=
Z +∞
0
xqdHn,ξ(x)=
Z N 0
xq−RdT(x) +NqH(Nf )
= xq−RT(x)x=N
x=0 −
Z N 0
(q−R)xq−R−1T(x)dx+Nq−R
Z +∞
N
xRdH(x)f
≤ Nq−RT(N) +
Z N 0
(q−R)xq−R−1
Z +∞
x
zRdH(z)f dx−Nq−RT(N)
= (q−R)·
Z N
0
xq−r−1xr−[r]
Z +∞
x
z[r]dH(z)f dx, if r6= [r],
Z N 0
xq−r−1x
Z +∞
x
zr−1dH(z)f dx, if r = [r]
≤ (q−R)γr∗
Z N 0
xq−r−1dx≤ q−R
q−r Nq−rγr∗. The lemma is proved.
The following lemma gives us the estimation of absolute pseudomoments ofH`∗n,ξ in terms of pseudomoments ofHn,ξ.
Lemma 4.3. For all n, `∈N, ξ≥0 and all q∈N0 we have νq
H`∗n,ξ
≤`q νq,n,ξ ν`−10,n,ξ <∞. (101) In particular, the absolute pseudomoment νq
H`∗n,ξ
is finite.
Proof. Using (95), the definition of the `-fold convolution of Hn,ξ, inequality (148) from Lemma A.6 and the fact thatνr,n,ξ is finite for anyr ≥0 (see (94)), we obtain forq ∈N:
νq
H`∗n,ξ
=
Z +∞
0
xq
dH`∗n,ξ(x)
=
Z +∞
0
xq
d
Z +∞
0
H(`−1)∗n,ξ (x−y1)dHn,ξ(y1)
≤
Z +∞
0
. . .
Z +∞
0
| {z }
`
(y1+· · ·+y`)qdHn,ξ(y1). . .dHn,ξ(y`)
≤ `q
Z +∞
0
dHn,ξ(x)
`−1Z +∞
0
xqdHn,ξ(x)≤`q νq,n,ξν`−10,n,ξ <∞.
Similarly we can show that ν0
H`∗n,ξ
≤ν`0,n,ξ. The lemma is proved.
Below we give one more useful lemma about the pseudomoments of H`∗n,ξ. Lemma 4.4. The following properties hold true.
(i) For all u, `∈N we have µu
H`∗n,ξ
= X
k1+k2+···+k`=u
u!
k1!· · ·k`!µk1,n,ξ· · ·µk`,n,ξ.
4.1. Some auxiliary functions and plan of the proof 71
(ii) For all u, `∈N with u < ` we have µu
H`∗n,ξ
= 0.
(iii) For fixed R≥1, ` ∈N and u= 1, ..., R we have µu
H`∗n,ξ
=u! X
k1+k2+···+kR=`
k1+2k2+...+RkR=u
`!
k1!· · ·kR!
µ1,n,ξ 1!
!k1
· · · µR,n,ξ R!
!kR
.
Proof. (i) Using (95) and rewritingµu
H`∗n,ξ
and using formula (147) from Lemma A.5 we obtain
µu
H`∗n,ξ
=
Z +∞
0
yudH`∗n,ξ(y)
=
Z +∞
0
...
Z +∞
0
| {z }
`
(z1+· · ·+z`)udHn,ξ(z1)· · ·dHn,ξ(z`)
=
Z +∞
0
...
Z +∞
0
| {z }
`
X
k1+k2+···+k`=u
u!
k1!· · ·k`!z1k1· · ·z`k`dHn,ξ(z1)...dHn,ξ(z`)
= X
k1+k2+···+k`=u
u!
k1!· · ·k`! µk1,n,ξ· · ·µk
`,n,ξ.
(ii) If u < `, then there exists such i ∈ {1, . . . , `} that ki = 0. Therefore each productµk1,n,ξ· · ·µk`,n,ξ from above containsµ0,n,ξ. According to Lemma 4.1 (i) we haveµ0,n,ξ = 0. This impliesµu
H`∗n,ξ
= 0 for u < `.
(iii) Note that if u < `, then the right side of the equality in (iii) is equal to 0 since we have to sum up over the empty set. This fact together with statement (ii) gives us the statement of (iii) foru < `. Let us consider now u≥`. Using the properties of the inverse Fourier transformhn,ξ(t) ofHn,ξ(x) (similarly to Lemma 3.25) we find
hn,ξ(t) :=
Z +∞
−∞
eitxdHn,ξ(x) = it
1!µ1,n,ξ+· · ·+(it)R
R! µR,n,ξ+O(|t|R+1), t→0.
Using this expansion and formula (147) from Lemma A.5 we obtain fort →0:
h`n,ξ(t) = it
1!µ1,n,ξ +...+(it)R R! µR,n,ξ
!`
+O(|t|R+1)
= X
k1+...+kR=`
`!
k1!...kR!
µ1,n,ξ
1!
k1
...
µR,n,ξ
R!
kR
(it)k1+2k2+...+RkR +O(|t|R+1)
=
R
X
u=`
(it)u
u! u! X
k1+k2+···+kR=`
k1+2k2+...+RkR=u
`!
k1!· · ·kR!
µ1,n,ξ 1!
!k1
· · · µR,n,ξ R!
!kR
+O(|t|R+1).
On the other hand we can considerh`n,ξ(t) as the inverse Fourier transform ofH`∗n,ξ(x), i.e. we have
h`n,ξ(t) = it 1!µ1
H`∗n,ξ
+· · ·+(it)R R! µR
H`∗n,ξ
+O(|t|R+1), t→0.
Taking into account property (ii) and comparing coefficients in both representations of h`n,ξ we get the statement of (iii). The lemma is proved.
We need one more auxiliary function, which is similar to the function Wfr,n from (87). Forr ∈R+, r >1, n ∈N and ξ∈[0,∞) we define
Wr,n,ξ(x) =
ρ
X
k=2
ck,n
nk G(0,k)(x,1) +W∗n,ξ(x) +
p
X
k=0 mk
X
`=1
n
`
!ck,n−`
nk
m`,k
X
u=`
pu,`,k
X
v=0
G(u,k+v)(x,1)(−`/n)v(−1)u
v! n−u/α Cu,`,
(102)
where ρ = [2(R + 1)/α], p = [2R/α], m`,k = [R + 1 + α(` − 1 − k/2)], mk = 1 + [(R−αk/2)/(1−α)], pu,`,k = max{0,[(R+ 1−u)/α+`−1−k/2]}
with
R =
( [r], if [r]6=r r−1, if [r] =r , cr, ρ= X
k0+k2+···+ks=ρ 2k2+···+sks=r
ρ!
k0!k2!· · ·ks!Ak22· · ·Akss, r, ρ∈N0, (103) s and Aj,j = 2, ..., s, are from (86) and (70), respectively,
Cu,` = X
k1+2k2+...+RkR=u k1+k2+...+kR=`
`!
k1!...kR!
µ1,n,ξ 1!
!k1
... µR,n,ξ R!
!kR
, (104)
and
W∗n,ξ(x) =n G(R+1,0)(x,1)(−1)R+1
(R+ 1)! n−R+1α µR+1,n,ξ +n Gα,1(·, n)∗Mn,ξ(xn1/α)−
R+1
X
w=0
n G(w,0)(x,1)(−1)w
w! n−w/αµwMn,ξ
(105)
with Mn,ξ(x) =Hn,ξ(x)−Hn,0(x).
Lemma 4.5. For the function Wr,n,ξ defined by (102) the following holds true.
(i) Wr,n,ξ is absolutely continuous and differentiable on R with Wr,n,ξ(x) = 0 for x <0.
(ii) There exists such constant W >0 that Wr,n,ξ(x)≤W for all x∈R. (iii) lim
x→+∞Wr,n,ξ(x) = 0.
4.1. Some auxiliary functions and plan of the proof 73
Proof. (i) According to the definition, Wr,n,ξ is a linear combination of the term
Gα,1(·, n)∗Mn,ξ(xn1/α) and some derivatives of a stable distribution function Gα,1(x, λ). From Lemma 3.14 it follows that Gα,1(x, λ) is infinitely differentiable with respect tox andλ, and its derivatives are absolutely continuous. The function Gα,1(·, n)∗Mn,ξ is absolutely continuous and differentiable as a convolution of the absolutely continuous and infinitely differentiable functionGα,1 and the function of bounded variation Mn,ξ. These facts together with the property Gα,1(x, λ) = 0 for x <0 and α < 1 (see Section 3.3) and the fact that Mn,ξ(x) = 0 for x <0 give the statement (i) of the lemma.
(ii) Let us consider Gα,1(·, n) ∗Mn,ξ. Using (97) and the fact that Gα,1 is the distribution function of a stable random variable we have
Gα,1(·, n)∗Mn,ξ(xn1/α)=
Z +∞
−∞
Gα,1xn1/α−y, n dMn,ξ(y)
≤
Z +∞
−∞ 1dMn,ξ(y)=ν0(Mn,ξ)≤ν0(Hn,ξ) =ν0,n,ξ <∞.
Using the last estimate and estimate (59) from Lemma 3.14 we obtain the inequality from (ii) for all x∈R:
Wr,n,ξ(x)≤
p
X
k=2
|ck,n|
nk D0,k+
p
X
k=0 mk
X
`=1
n
`
!|ck,n−`| nk
m`,k
X
u=`
pu,`,k
X
v=0
Du,k+v(`/n)v
v! n−u/αCu,`
+n ν0,n,ξ+n DR+1,0 1
(R+ 1)!n−R+1α µR+1,n,ξ+
R+1
X
w=0
n Dw,0 n−w/α w!
µw(Mn,ξ)=:W . Note that all pseudomoments occurring inW are finite (see (94) and (97)).
(iii) Let us show that lim
x→+∞
Gα,1(·, n)∗Mn,ξ(xn1/α) = 0. Using the definition of Gα,1 and Lemma 4.1 (i) we obtain
x→+∞lim
Gα,1(·, n)∗Mn,ξ(xn1/α) = lim
x→+∞
Z +∞
−∞
Gα,1xn1/α−y, n dMn,ξ(y)
=
Z +∞
−∞ lim
x→+∞Gα,1xn1/α−y, n dMn,ξ(y) =
Z +∞
−∞ 1dMn,ξ(y) =µ0(Mn,ξ) = 0.
The convergence of all other terms to 0 asx → ∞ can be proved in the same way as in Lemma 3.22 (iii). This completes the proof of the lemma.
Now we are ready to give a plan of the proof of our main result (Theorem (3.26)), which we repeat here for the sake of readability. Recall that
Fn(x) = P X1+· · ·+Xn≤xn1/α=Fn∗(xn1/α).
Theorem. If for r >1 we have 0< γr∗ <∞ and ν0∗ <1, then for all x∈R and all integers n≥2 the following inequality holds:
Fn(x)−Gα,1(x)−Wfr,n(x)≤C(1 +|x|)−rn−r−αα 1 +nαr Qn, where Wfr,n(x) is defined by (87), Qn =ν0∗n−1+sup|t|>
eε|f(t)|+ 2γr∗n−r/αn−1 with εedefined by (91) and constant C does not depend on x and n.
Plan of the proof.
In order to estimateFn(x)−Gα,1(x)−Wfr,n(x) we add and subtract the auxiliary functions Fn∗n,ξ and Wr,n,ξ with some fixed ξ ∈ [0,∞), and come to the following inequality.
Fn(x)−Gα,1(x)−Wfr,n(x)
≤Fn(x)−Fn∗n,ξn1/αx
+Fn∗n,ξn1/αx−Gα,1(x)−Wr,n,ξ(x)+Wr,n,ξ(x)−Wfr,n(x).
We discuss and estimate each of the three summands on the right-hand side sepa-rately in the subsequent subsections. Combining the results we will then obtain the estimation from the theorem.