Chapter 7
The Central Limit Theorem
7.5. [We need to know also that S and D have the same variance.] Let S = X+Y
and D = X - Y. Note that X = (S + D)/2 and Y = (S - D)/2. Thus,
h
i
h
i h
i
⇥
⇤
E eitX+isY = E 2it(S+D)/2+is(S-D)/2 = E ei(t+s)S/2 E ei(t-s)D/2 ,
by the independent of S and D. Now suppose S is N(µ,
N(⌫, 2 ). Then,
⇥
⇤
2
E eitX+isY = ei(t+s)µ/2 ei(t-s)⌫/2 e-(t+s)
1
1 2
= e 2 it(µ+⌫)- 8 t
(
2
+
2
)
1
2
2
/8 -(t-s)2
e
1
e 2 is(µ-⌫)- 8 s
2
(
2
+
) and D is
2
2
/8
)
,
which is the desired result.
7.6.
1. We first compute/recall the characteristic function of a binomial: If
X has a Bin(n , p) distribution, then
EeitX =
n
X
eitj
j=0
✓ ◆
n j
p (1 - p)n-j
j
= peit + 1 - p
n
,
thanks to the binomial theorem. Therefore, if Xk is Bin(nk , p) for
k = 1, 2, then
Eeit(X1 +X2 ) = peit + 1 - p
n1 +n2
.
This is the characteristic function of a Bin(n1 + n2 , p), and hence is
a Bin(n1 + n2 , p) thanks to the uniqueness theorem.
17
18
CHAPTER 7. THE CENTRAL LIMIT THEOREM
2. We first compute/recall the characteristic function of a Poisson: If X
has a Poisson( ) distribution, then
EeitX =
1
X
eitj
j=0
ej!
j
= exp - + eit .
This shows that E exp(it(X1 + X2 )) = exp(- + eit ) where : + 1 +
2 , and hence X1 + X2 has a Poisson ( 1 + 2 ) distribution, thanks to
the uniqueness theorem.
3. We can write Xk = µk + k Zk [k = 1, 2] where Z1 and Z2 are i.i.d.
N(0 , 1) random variables. Since E exp(itZk ) = exp(-t2 /2), it follows that
Eeit(X1 +X2 ) = eit(µ1 +µ2 ) Eeit
= eitµ-t
2
2
/2
1 Z1
Eeit
2 Z2
,
thanks to independence, where µ := µ1 + µ2 and 2 := 21 + 22 . This
shows that X1 + X2 is N(µ , 2 ).
Rb
7.11. Note that a e-itx dx = (e-ita - e-itb )/(it). Therefore,
✓
◆
Z
1 1 - 1 "2 t2 e-ita - e-itb
2
b (t) dt
e
µ
2⇡i -1
t
Z1 Zb
Z Z
1
1 b 1 - 1 "2 t2 -itx
- 12 "2 t2 -itx
b (t) dx dt =
b (t) dt dx.
=
e
e
µ
e 2
e
µ
2⇡ -1 a
2⇡ a -1
[Fubini–Tonelli is justified
R because the integrand is absolutely integrable.]
b (t) = R eity µ(dy) and use Fubini–Tonelli again. This
Now plug in µ
yields
✓
◆
Z
1 1 - 1 "2 t2 e-ita - e-itb
b (t) dt
e 2
µ
2⇡i -1
t
Z
Z Z
1 1 b 1 - 1 "2 t2 -it(x-y)
=
e 2
e
dt dx µ(dy)
2⇡ -1 a -1
Z1 Zb
=
" (x - y) dx µ(dy);
-1 a
see (7.24) on page 96. So now we compute the inner integral directly. [Do
be brave!]
Zb
Z b-y
2
" (x - y) dx =
" (z) dz = P a - y 6 N(0, " ) 6 b - y
a
a-y
=P
a-y
b-y
6 N(0, 1) 6
"
"
.
19
Firstly, this is bounded (by zero and one), uniformly for all " > 0. Secondly,
lim
Zb
"!0 a
" (x
- y) dx =
8
>
<1
1
>2
:
0
if a < y < b,
if a = y < b or a < y = b
otherwise.
The bounded convergence theorem does the rest.
7.19. Define Yi = XiP
-E[Xi ] = XiP
- 12 , so that E[Y
i ) = Var(Xi ) =
Pin] = 0 and
PVar(Y
n
n
n
1
1
1
jX
.
Evidently,
jY
=
jX
j
=
j - 4 n(n + 1).
j
j
j=1
j=1
j=1
j=1
12
2
Therefore,
n
n
X
X
4
jXj - n2 = 4
jYj + n.
j=1
j=1
P
Therefore, we need to only concentrate on 4n-3/2 n
j=1 jYj , and the Yj ’s
1
are i.i.d. with mean zero and variance 12 . Because sin t ⇠ t - 16 t3 as t ! 0,
⇥
E e
itY1
⇤
=
Z 1/2
eitx dx =
-1/2
sin(t/2)
t2
=1+ smaller terms
t/2
24
(t ! 0).
Of course, sin(0)/0 is interpreted as one here. Therefore,
"
E exp it
4
Pn
j=1 jYj
n3/2
!#
=

✓
◆
4jYj
E exp it 3/2
n
j=1
n
Y
0
1
n
2 X
2t2 j2
2t
⇡
1⇡ exp @- 3
j2 A
3n3
3n
j=1
j=1
◆
✓
◆
✓
2 Zn
2t2
2t
2
x2 dx ⇡ exp = e-t
⇡ exp - 3
3n 1
9
n 
Y
with
2
= 4/9. This proves that
4
Pn
j=1
jXj - n2
n3/2
✓
4
) N 0,
9
◆
.
7.21. Define Yj = Xj 1{|Xj |61} so that Yj ’s are i.i.d. taking the values ±1 and 0.
Evidently, P{Xj 6= Yj } = (2j2 )-1 , which is summable in j. Therefore, by
the Borel–Cantelli lemma with probability one, Xj = Yj for all but a finite
number
the same asymptotic properPof j’s. In particular, Sn /SD(Sn ) hasP
n
2 -1
ties as n
Y
/SD(S
)
a.s.
But
Var(S
)
=
) ) ⇠ (3n/2)
n
n
j=1 j
j=1 (1.5 - (2j
Pn
p
as n ! 1. Apply the CLT to the Yj ’s to find then that j=1 Yj / n )
N(0, 1). Assemble terms to find that Sn /SD(Sn ) ) N(0, 2/3).
2
/2
,
20
CHAPTER 7. THE CENTRAL LIMIT THEOREM
P
j
7.22. Convergence follows from E( 1
j=0 r |Xj |) = E(|X1 |)/(1 - r) < 1. For all
t 2 R, as r " 1,
13
2
0
1
1
X
Y
p
p
⇥
⇤
rj Xj A5 =
E exp it 1 - r rj X1
E 4exp @it 1 - r
j=0
j=0
⇡
Therefore,
1 ✓
Y
j=0
1-
t2
0
t2
⇡ exp @-
p
P
j
1-r 1
j=0 r Xj ) N(0,
2
/2).
2
2
(1 - r)r2j
2
◆
1
1
(1 - r) X 2j A
2
r
⇡ e-t
2
j=0
2
/4
.