convergence of the fractional parts of the random variables to the ...

Convergence of the fractional parts of the random variables 125
∞∏
m=1
[ii]. According to Proposition 3,
∞∑
m=1
(1 − |c m (k)|) divergent ⇐⇒
c m (k) is divergent, that is {S n } d → F, with F ≠ Exp ∗ (λ).
Corollary 2. If the sequence (c m ), of the random variables sequence (X m )
∞∑
meets the condition of (1 − |c m (k)|) to be convergent, ∀ k ∈ Z 0 , then
m=1
{S n } d → Exp ∗ (λ), where S n = n ∑
m=1
X m , n ∈ N.
Theorem 6. Let (X m ) be a sequence of independent and identically distributed
random variables with the characteristic function ϕ, X 1 ∼ Exp ∗ (λ),
and 0 < V arX 1 < ∞. Let (a m ) be a sequence of real numbers so that
lim m→∞ a m = 0. We define V n :=
(i). If
∞∑
m=1
n ∑
m=1
a m X m , n ∈ N.
a 2 m is convergent, then {V n } d → Exp ∗ (λ).
(ii). We suppose that there is k ∈ Z 0 , ϕ(2πka m ) ≠ 0. If
a 2 m is divergent,
then {V n } don’t converge to Exp ∗ (λ).
∞∑
m=1
Proof. [i]. Let k ∈ Z 0 be fixed. By Corollary 2, it is sufficient to show that
∞∑
(1 − |ϕ(2πka m )|) is convergent.
m=1
It is known that if X ∼ Exp ∗ (λ), then ϕ X (t) = λ/ (λ − it), from where
|ϕ X (t)| =
( )
λ
t 2 −
1
2
√
(1
λ 2 + t = +
.
λ) 2
If we consider the development in binomial series (1 + x) − 1 2 = 1 − x 2
(
+
3
8 x2 + ..., then we obtain 1 + ( ) )
t 2 −
1
2
λ
= 1 − 1
2λ t2 + o(t 2 ), from where the
following 1
4λ t2 < 1 − |ϕ X (t)| < 1 λ t2 .
Then
∞∑
∞∑ 1
(1 − |ϕ(2πka m )|) <
λ 4π2 k 2 a 2 m = 4π2 k 2 ∞∑
a 2
λ
m.
m=1
m=1
m=1