6.29. Solution 1. We will need the following Lemma: As n → ∞, ∑ni=1 (1/i) ∼ ln(n).
Proof This follows from the integral test of calculus, because
!n
!n
n
1 (dx/x) ≤ ∑i=1 (1/i) ≤ 1 + 1 (dx/x).
!
Now let us first assume that E[X1 ] = 0. Define Xi& = Xi 1{|Xi |≤i} . Also define Sn = ∑ni=1 (Xi /i) and Sn& = ∑ni=1 (Xi& /i).
By the Kolmogorov maximal inequality,
%
&
"
$
n E X 2 ; |X | ≤ i
&
2
&
# &
#
'Sn '2
1
VarSn
1
1
=
≤
.
P max #Sk − E[Sk& ]# ≥ ε ln n ≤ 2
∑
2
2
2
2
2
ε (ln n)2
i
1≤k≤n
ε ln (n) ε ln (n) i=1
Therefore,
%
&
%
&
"
' n ($
2n E X 2 ; |X | ≤ i
∞
∞ E X 2 ; |X | ≤ i
∞
# &
#
1
1
1
1
1
1
& #
2
#
≤
A
P
max
S
−
E[S
]
≥
ε
ln
2
≤
A
∑ 4n
∑ 1≤k≤22n k
∑
∑ 4n ∑
k
2
2
i
i
n=1
n=1
i=1
i=1
n≥log2 (i)
*
)
%
&
∞ E X 2 ; |X | ≤ i
1
1
1
2
≤B∑
=
BE
X
1 ∑ 2 ≤ CEX1 .
2
i
i
i=1
i≥|X |
1
By the Borel–Cantelli lemma, a.s., for all n large,
' n(
# &
#
max n #Sk − E[Sk& ]# ≤ ε ln 22 .
1≤k≤22
n
n+1
Any m is between some 22 and 22 . Therefore,
' n+1 (
#
# &
#
# &
&
&
#Sm − E[Sm ]# ≤ max #S − E[S ]# ≤ ε ln 22
= 2ε ln(2) · 2n ≤ 2ε ln(2) ln m.
k
k
n+1
1≤k≤22
& − ES& |/ ln(m) → 0. But
This proves that a.s., |Sm
∑k P{Sk (= Sk& } = ∑k P{|X1 | > k} < ∞ because 'X1 '1 < ∞. So
m
& |/ ln m → 0. But
almost surely, Sk = Sk& for all k large. It suffices to prove that |ESm
#
# #
#
#
#
#
m
m
# & # # E[X1 ; |X1 | ≤ i] # # E[X1 ; |X1 | > i] ##
#ESm # = # ∑
# = #∑
#,
#i=1
# #i=1
#
i
i
since EX1 = 0. Thus,
m
# &#
#ESm # ≤ ∑ E[|X1 |; |X1 | > i] .
i
i=1
m
For all η > 0, there exists i0 such that for al i ≥ i0 , E{|X1 |; |X1 | > i} ≤ η. Therefore, ∑m
i=i0 (· · · ) ≤ η ∑i=1 (1/i) ∼
i0
& |/ ln m ≤ η for all η, whence the result.
η ln m. But ∑i=1
(· · · ) ≤ i0 'X1 '1 . Therefore, lim supm |ESm
If µ = EX1 (= 0, then consider instead Xi◦ := Xi − µ. The preceding proves that ln−1 (n) ∑ni=1 (Xi◦ /i) → 0 a.s.
Equivalently, ln−1 (n){∑ni=1 (Xi /i) − µ ∑ni=1 (1/i)} → 0. Because ∑ni=1 (1/i) ∼ ln(n), we obtain the result in general.