Chapter 7
The Central Limit Theorem
k
k
7.1. Since C1
c (R ) ⇢ Cb (R ), it suffices to prove that if
Z
Z
k
lim f dµn = f dµ for all f 2 C1
c (R ),
n!1
then µn ) µ.
k
Claim. For every f 2 Cc (Rk ) and ✏ 2 (0 , 1), there exists g✏ 2 C1
c (R )
such that supx |f(x) - g✏ (x)| 6 ✏.
This Claim does the job. Indeed, the Claim implies that
Z
Z
f dµn - g✏ dµn 6 ✏µn (Rk ) = ✏, and
Z
Z
f dµ - g✏ dµ 6 ✏.
Consequently,
lim sup
n!1
Z
Z
f dµn - f dµ 6 2✏.
Since ✏ > 0 and f 2 Cc (Rk ) are arbitrary, the result follows.
Now by Fejer’s theorem, if f 2 Cc (Rk ) then for all ✏ > 0 there exists
h✏ 2 C1 (Rk ) such that supx |f(x) - h✏ (x)| 6 ✏/2. Suppose f(x) = 0 when
k
kxk > p. Then it suffices to prove that there exists g✏ 2 C1
c (R ) such that
supkxk6p |g✏ (x) - h✏ (x)| 6 ✏. But this is easy. For instance, let
✓
◆
1
(x) := exp - 2
.
(p - kxk)+
k
Then,
2 C1
(x) 6 1 for all x, and (x) > exp(-1/p(p c (R ), 0 6
1)) > 1 - ✏ when kxk < p, if p = p(✏) is chosen sufficiently large. Then
g✏ (x) := (x)h✏ (x) satisfies
(1 - ✏)h✏ (x) 6 g✏ (x) 6 h✏ (x)
17
when kxk < p.
18
CHAPTER 7. THE CENTRAL LIMIT THEOREM
Therefore,
◆
sup |g✏ (x) - h✏ (x)| 6 max sup |h✏ (x)| , 1 ✏ 6 K✏,
kxk6p
✓
x
where K := supx |f(x)| + ✏ 6 supx |f(x)| + 1.
7.2. Let X have distribution µ. Then the characteristic function of -X is
⇤
⇥
b (t).
E e-itX = E[cos(tX)] - itE[sin(tX)] = E [eitX ] = µ
b is a characteristic function.
Therefore, µ
Let Y be independent from X, and have the distribution P{Y = 1} = P{Y =
-1} = 1/2. Define Z = XY. Then, the characteristic function of Z is
⇥
⇤
⇥
⇤ 1 ⇥
⇤ 1 ⇥
⇤
b (t).
E eitZ = E eitXY = E eitX + E e-itX = E [cos(tX)] = Re µ
2
2
Therefore, Reb
µ is a characteristic function.
Let X 0 be independent from X and have also distribution µ. Then, the
characteristic function of X - X 0 is
h
i
⇥
⇤ ⇥
⇤
0
E eit(X-X ) = E eitX E e-itX = |b
µ(t)|2 .
Therefore, |b
µ|2 is a characteristic function.
If Xi has characteristic function ofPµi , and ifQthe Xi ’s are independent,
n
bi by independence.
then the characteristic function of n
i=1 Xi is
i=1 µ
Pn
Finally, suppose Y = i=1 Xi 1{Z=i} where Z is independent from the X’s,
and P
P{Z = k} = 1/n for k = 1, . . . , n. Then, the characteristic function of
bi .
Y is n
i=1 pi µ
b (t) is E[cos(tX)] - iE[sin(tX)] = µ
b (-t).
Finally, note that the conjugate of µ