1924 CHAPTER 59. BASIC PROBABILITY
Thus
eit·(
X−m√n
)= 1+ it·X−m√
n− (t·(X−m))2
2n
+
(1− f
(t·(
X−m√n
)))(t·(X−m))2
2n.
Thus
φ Zn(t) =
n
∏j=1
[1−E
((t·(X−m))2
2n
)
+E
((1− f
(t·(
X−m√n
)))(t·(X−m))2
2n
)]
=n
∏j=1
[1− 1
2nt∗Σt +
12n
E((
1− f(
t·(
X−m√n
)))(t·(X−m))2
)]. (59.18.44)
(Note (t·(X−m))2 = t∗ (X−m)(X−m)∗ t.) Now here is a simple inequality for complexnumbers whose moduli are no larger than one. I will give a proof of this at the end. Itfollows easily by induction.
|z1 · · ·zn−w1 · · ·wn| ≤n
∑k=1|zk−wk|. (59.18.45)
Also for each t, and all n large enough,∣∣∣∣ 12n
E((
1− f(
t·(
X−m√n
)))(t·(X−m))2
)∣∣∣∣< 1.
Applying 59.18.45 to 59.18.44,
φ Zn(t) =
(n
∏j=1
(1− 1
2nt∗Σt
))+ en
=
(1− 1
2nt∗Σt
)n
+ en
where
|en| ≤n
∑j=1
∣∣∣∣ 12n
E((
1− f(
t·(
X−m√n
)))(t·(X−m))2
)∣∣∣∣=
12
∣∣∣∣E((1− f(
t·(
X−m√n
)))(t·(X−m))2
)∣∣∣∣