MATH 7770-01
LIMIT THEOREMS
FALL 2010
Lecture 35. Improvements in limit theorems with power “tails”, continued.
The integrals in the sums over 1 < 𝛼𝑗 < 2 have been already evaluated in Lecture 14;
the summands in them are equal to
Γ(2 − 𝛼𝑗 ) [ +
𝛼𝑗 𝜋
𝛼𝑗 𝜋 ] 𝛼𝑗
−
⋅ (𝑐𝛼𝑗 + 𝑐−
− 𝑖 ⋅ (𝑐+
∣𝑡∣
𝛼𝑗 ) cos
𝛼𝑗 − 𝑐𝛼𝑗 ) sign(𝑧) ⋅ sin
𝛼𝑗 − 1
2
2
(35.1)
(see formula (14.12)); and it can be rewritten in exactly the form (34.11) if we understand
the function Γ as being continued analytically to all complex numbers except non-positive
integers (see the text after formula (14.12)). So we can extend the first sum in (34.20) to
all 𝛼𝑗 between 0 and 2.
∫
∞
∫
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥) 𝑑𝐵 + (𝑥) and
If 1 < 𝛽 < 2, the last two integrals in (34.20) are
0
0
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥) 𝑑𝐵 − (𝑥), and we handle them by integrating by parts; say, for the
−∞
second one:
[ 𝑖𝑡𝑥
]0
(𝑒 − 1 − 𝑖𝑡𝑥) ⋅ 𝐵− (𝑥) −∞ − 𝑖𝑡
∫
0
(𝑒𝑖𝑡𝑥 − 1) ⋅ 𝐵− (𝑥) 𝑑𝑥.
(35.2)
−∞
The term outside the integral is equal to 0 because lim𝑥→0− ∣(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥) ⋅ 𝐵− (𝑥)∣ ≤
lim𝑥→0− (𝑡2 𝑥2 /2)⋅𝑂(∣𝑥∣−𝛽 ) = 0, lim𝑥→−∞ ∣(𝑒𝑖𝑡𝑥 −1−𝑖𝑡𝑥)⋅𝐵− (𝑥)∣ ≤ lim𝑥→−∞ (2+∣𝑡∣∣𝑥∣)×
× 𝑂(∣𝑥∣−𝛽 ) = 0; and the absolute value of the integral does not exceed
[∫
∫
∣𝑡∣−1
const⋅
2⋅∣𝑥∣
−∞
−𝛽
0
𝑑𝑥+
∣𝑡∣∣𝑥∣⋅∣𝑥∣
−𝛽
]
𝑑𝑥 = const⋅
∣𝑡∣−1
(
2
1 ) 𝛽−1
+
⋅∣𝑡∣
. (35.3)
𝛽−1 2−𝛽
So we have:
𝑓 (𝑡) − 1
=−
∑
𝑗: 0<𝛼𝑗
[
𝛼𝑗 𝜋
𝛼𝑗 𝜋 ] 𝛼𝑗
−
−
Γ(1 − 𝛼𝑗 ) ⋅ (𝑐+
− 𝑖 ⋅ (𝑐+
∣𝑡∣
𝛼𝑗 + 𝑐𝛼𝑗 ) cos
𝛼𝑗 − 𝑐𝛼𝑗 ) sign(𝑡) ⋅ sin
2
2
<2
+ 𝑖𝑑1 𝑡 + 𝑂(∣𝑡∣𝛽 )
(𝑡 → 0)
(35.4)
(the term 𝑖𝑑1 𝑡 is infinitely small compared to all terms with 𝛼𝑗 < 1, and infinitely large
compare to those with 𝛼𝑗 > 1, so perhaps we shouldn’t put all terms with 0 < 𝛼𝑗 < 2
together, but write the terms with 0 < 𝛼𝑗 < 1 first, then 𝑖𝑑1 𝑡, then the terms with
1 < 𝛼𝑗 < 2, and finally 𝑂(∣𝑡∣𝛽 ). But of course (35.4) looks shorter as it is written).
Problem 72 . Find the first pseudo-moment for the distribution of Problem 70
with accuracy to 0.1.
So we can write an asymptotic expansion for 𝑓 (𝑡) for the distribution of Problem 70
(this way I am telling you that we can take 𝛽 > 1 in this problem).
1
If 𝛽 > 2, we subtract from the integrand in the last two integral in (34.20) and
subtract from it − 𝑡2 𝑥2 /2; the second pseudo-moment 𝑑2 is defined as
∫
∞
𝑑2 =
[
𝑥 𝑑 𝐹 (𝑥)−1+
2
0
∑
−𝛼𝑗
𝑐+
𝛼𝑗 𝑥
]
∫
0
+
−∞
𝑗: 0<𝛼𝑗 <2
∑
[
𝑥2 𝑑 𝐹 (𝑥)−
−𝛼𝑗
𝑐−
𝛼𝑗 ∣𝑥∣
]
(35.5)
𝑗: 0<𝛼𝑗 <2
(the second pseudo-moment, in contrast with the second moment, needn’t be nonnegative).
We add to and subtract from 𝑒𝑖𝑡𝑥 −1−𝑖𝑡𝑥 in (34.20) the next Maclaurin term −𝑡2 𝑥2/2:
𝑓 (𝑡) − 1
=−
∑
𝑗: 0<𝛼𝑗
[
𝛼𝑗 𝜋
𝛼𝑗 𝜋 ] 𝛼𝑗
−
−
Γ(1 − 𝛼𝑗 ) ⋅ (𝑐+
− 𝑖 ⋅ (𝑐+
∣𝑡∣
𝛼𝑗 + 𝑐𝛼𝑗 ) cos
𝛼𝑗 − 𝑐𝛼𝑗 ) sign(𝑡) ⋅ sin
2
2
<2
+ 𝑖𝑑1 𝑡 − 𝑑2 𝑡2/2
∫
∞
+
∑
[
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2/2) 𝑑 𝐹 (𝑥) − 1 +
0
−𝛼𝑗
𝑐+
𝛼𝑗 𝑥
]
𝑗: 0<𝛼𝑗 <2
∫
0
[
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2/2) 𝑑 𝐹 (𝑥) −
+
∑
]
−𝛼𝑗
𝑐−
.
𝛼𝑗 ∣𝑥∣
(35.6)
𝑗: 0<𝛼𝑗 <2
Now
we
add
to
and
subtract
from
the
expressions
in
the
brackets
the
sums
∑
∑
−
−𝛼𝑗
+ −𝛼𝑗
(the only thing we seem to be doing is adding
, 𝑗: 2<𝛼𝑗 <3 𝑐𝛼𝑗 ∣𝑥∣
𝑗: 2<𝛼𝑗 <3 𝑐𝛼𝑗 𝑥
and subtracting the same things again and again). We get instead of the last two integrals:
−∞
[
∑
∫
𝑐+
𝛼𝑗
𝑗: 2<𝛼𝑗 <3
∞
0
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2/2) 𝑑(−𝑥−𝛼𝑗 )
∫ 0
]
−
+ 𝑐𝛼𝑗
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2/2) 𝑑∣𝑥∣−𝛼𝑗
−∞
∫
∞
+
∑
[
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2/2) 𝑑 𝐹 (𝑥) − 1 +
0
∫
−𝛼𝑗
𝑐+
𝛼𝑗 𝑥
]
(35.7)
𝑗: 0<𝛼𝑗 <3
0
+
[
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2/2) 𝑑 𝐹 (𝑥) −
−∞
∑
]
−𝛼𝑗
𝑐−
.
𝛼𝑗 ∣𝑥∣
𝑗: 0<𝛼𝑗 <3
Let’s evaluate the first tow integrals. For 2 < 𝛼𝑗 < 3, integrating by parts, we get:
∫
∞
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2/2) 𝑑(−𝑥−𝛼𝑗 )
0
[
]∞
= (𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2/2) ⋅ (−𝑥−𝛼𝑗 ) 0 − 𝑖𝑡
∫
∞
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥) ⋅ (−𝑥−𝛼𝑗 ) 𝑑𝑥.
0
(35.8)
1
𝑑(− 𝑥1−𝛼𝑗 ),
The term outside the integral is equal to 0. Rewriting (−𝑥−𝛼𝑗 ) 𝑑𝑥 as
𝛼𝑗 − 1
we integrate by parts another time (again with the term outside the integral being equal
2
to 0), and get:
∫ ∞
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2/2) 𝑑(−𝑥−𝛼𝑗 ) = −
0
2
𝑡
=−
(1 − 𝛼𝑗 )(2 − 𝛼𝑗 )
∫
𝑡2
𝛼𝑗 − 1
∞
(𝑒
𝑖𝑡𝑥
∫
∞
(𝑒𝑖𝑡𝑥 − 1) ⋅ 𝑥1−𝛼𝑗 𝑑𝑥
0
(35.9)
− 1) 𝑑(−𝑥
2−𝛼𝑗
).
0
Since 0 < 𝛼𝑗 − 2 < 1, the integral here is of the kind that we have considered in Lecture
14, and it is equal
(
)
−Γ 1 − (𝛼𝑗 − 2) ⋅ 𝑒−𝑖(𝛼𝑗 −2)⋅sign(𝑡)⋅𝜋/2 ∣𝑡∣𝛼𝑗 −2 .
(35.10)
Multiplying by the factor in front of the integral in (35.9) and taking into account that
Γ(3 − 𝛼𝑗 )
−1 = 𝑒−𝑖⋅2⋅sign(𝑡)⋅𝜋/2 and
= Γ(1 − 𝛼𝑗 ) (with the analytically-continued
(2 − 𝛼𝑗 )(1 − 𝛼𝑗 )
Γ-function), we get that the 𝑗-th summand in the first sum in (35.7) is equal to
[
𝛼𝑗 𝜋
𝛼𝑗 𝜋 ] 𝛼𝑗
−
−
− 𝑖 ⋅ (𝑐+
∣𝑡∣ ,
−Γ(1 − 𝛼𝑗 ) ⋅ (𝑐+
𝛼𝑗 + 𝑐𝛼𝑗 ) cos
𝛼𝑗 − 𝑐𝛼𝑗 ) sign(𝑡) ⋅ sin
2
2
(35.11)
that is, it is expressed by exactly the same formula as for 0 < 𝛼𝑗 < 1 or 1 < 𝛼𝑗 < 2.
So we get:
𝑓 (𝑡) − 1
=−
∑
𝑗: 0<𝛼𝑗
[
𝛼𝑗 𝜋 ] 𝛼𝑗
𝛼𝑗 𝜋
−
−
− 𝑖 ⋅ (𝑐+
∣𝑡∣
Γ(1 − 𝛼𝑗 ) ⋅ (𝑐+
𝛼𝑗 − 𝑐𝛼𝑗 ) sign(𝑡) ⋅ sin
𝛼𝑗 + 𝑐𝛼𝑗 ) cos
2
2
<3
+ 𝑖𝑑1 𝑡 − 𝑑2 𝑡2 /2
∫
∞
+
∑
[
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2 /2) 𝑑 𝐹 (𝑥) − 1 +
0
∫
−𝛼𝑗
𝑐+
𝛼𝑗 𝑥
]
𝑗: 0<𝛼𝑗 <3
0
+
∑
[
(𝑒𝑖𝑡𝑥 − 1 − 𝑖𝑡𝑥 + 𝑡2 𝑥2 /2) 𝑑 𝐹 (𝑥) −
]
−𝛼𝑗
𝑐−
∣𝑥∣
.
𝛼𝑗
(35.12)
Let us formulate what we are getting this way (adding and subtracting the same
things, expressing integrals through the Γ-function, and from time to time saying “if 𝛽
is so and so” and estimating the remainder term) as a lemma (not a theorem, because it
is not a result about the limiting behavior of the distribution of the normalized sum with
improvements).
Lemma 35.1. Let the distribution function satisfy the conditions (34.3), (34.4) with
non-integer 𝛼𝑗 , 𝛽; let conditions (34.9) be satisfied if 𝛽 < 1. Then
−∞
𝑗: 0<𝛼𝑗 <3
𝑓 (𝑡) − 1
∑
[
𝛼𝑗 𝜋
𝛼𝑗 𝜋 ] 𝛼𝑗
−
−
=−
Γ(1 − 𝛼𝑗 ) ⋅ (𝑐+
− 𝑖 ⋅ (𝑐+
∣𝑡∣
𝛼𝑗 + 𝑐𝛼𝑗 ) cos
𝛼𝑗 − 𝑐𝛼𝑗 ) sign(𝑡) ⋅ sin
2
2
(35.13)
𝑗
+
⌊𝛽⌋
∑
𝑖𝑠 𝑑𝑠 𝑡𝑠/𝑠! + 𝑂(∣𝑡∣𝛽 )
𝑠=1
3
(the first sum is taken over all 𝑗).
If the conditions (34.1), (34.2) are satisfied, and condition (34.8) for 𝛽 < 1, then the
asymptotic expansion (35.13) holds with 𝑂(∣𝑡∣𝛽 ) replaced with 𝑜(∣𝑡∣𝛽 ) as 𝑡 → 0.
The proof has in fact been carried out for the remainder term 𝑂(∣𝑡∣𝛽 ); the only thing
that remained was to say “etc.” The statement with 𝑜(∣𝑡∣𝛽 ) is proved in the same way,
only ±∣𝑡∣−1 as( the limits
of integration are replaced with some functions ± 𝑢(𝑡) = 𝑜(∣𝑡∣−1 ),
)
such that 𝐵± ± 𝑢(𝑡) (or the corresponding total variations on the interval from 𝑢(𝑡) to ∞
or from − ∞ to − 𝑢(𝑡)) are 𝑜(∣𝑡∣−𝛽 ) as 𝑡 → 0.
We know what we are doing after this: we write the asymptotic expansion for ln 𝑓 (𝑡)
as 𝑡 → 0 using the Maclaurin expansion ln 𝑤 = (𝑤 − 1) − (𝑤 − 1)2 /2 + (𝑤 − 1)3 /3 − ... ;
( 𝑧 )
after this, the expansion in (negative) powers of 𝑛 of 𝑛 ln 𝑓
, and after that, one for
( 𝑧 )𝑛
{
( 𝑧 )}
𝐵𝑛
𝑓𝜁𝑛 (𝑧) = 𝑓
= exp 𝑛 ln 𝑓
.
𝐵𝑛
𝐵𝑛
4