Page 92 - Elements of Distribution Theory

P1: JZP
052184472Xc03 CUNY148/Severini May 24, 2005 2:34

78 Characteristic Functions

Theorem 3.5. Let X denote a real-valued random variable with characterstic function ϕ(·).
m
If, for some m = 1, 2,..., E(X ) exists and is finite, then
m j
(it)
j m
ϕ(t) = 1 + E(X ) + o(t ) as t → 0
j!
j=1
and, hence, ϕ is m-times differentiable at 0 with
( j)
j
j
ϕ (0) = (i) E(X ), j = 1,..., m.
If, for some m = 1, 2,..., ϕ (2m) (0) exists, then E(X 2m ) < ∞.
m
Proof. Assume that E(X )exists and is finite for some m = 1, 2,.... By Lemma A2.1
in Appendix 2, we may write
m j
(it)
exp{it}= + R m (t)
j!
j=0
where
m+1 m
|R m (t)|≤ min{|t| /(m + 1)!, 2|t| /m!}.
Hence,
m j j
(it) E(X )
ϕ(t) = E[exp{it X}] = 1 + + E[R m (tX)].
j!
j=1
It remains to be shown that

|E[R m (tX)]|
→ 0as t → 0.
m
|t|
We know that |E[R m (tX)]|≤ E[|R m (tX)|] and that
m+1
m!|R m (tX)| |t||X| m
≤ min , 2|X| .
m
|t| (m + 1)
Let
m+1 m

|t||x| m 2|x| if |x|≥ 2(m + 1)/|t|
M t (x) = min , 2|x| = m+1 .
(m + 1) |t||x| /(m + 1) if |x|≤ 2(m + 1)/|t|
Hence, it suffices to show that
E[M t (X)] → 0 as t → 0.
Note that for each x, M t (x) → 0as t → 0. Furthermore,

m
|M t (x)|≤ 2|x|
m
and,undertheconditionsofthetheorem,E(2|X| ) < ∞.ItnowfollowsfromtheDominated
Convergence Theorem (see Appendix 1) that
E[M t (X)] → 0 as t → 0.

This proves the first part of the theorem.