Page 98 - Elements of Distribution Theory

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

84 Characteristic Functions

Then

|F(x) − F(y)|≤|F(x 1 ) − F(x 0 )|≤ M|x 1 − x 0 |≤ 2M|x − y|.
That is, F satisfies a Lipschitz condition and, hence, is absolutely continuous; see
Appendix 1.
Fix x and h.By Theorem 3.3,

F(x + h) − F(x) 1 ∞
= G h (t)ϕ(t) dt
h 2π −∞
where
exp{−itx}− exp{−it(x + h)}
G h (t) = .
ith
It follows from (3.5) that, for all h, |G h (t)|≤ 1so that, by the Dominated Convergence
Theorem (see Appendix 1),
F(x + h) − F(x) 1 ∞
lim = G 0 (t)ϕ(t) dt
h→0 h 2π
−∞
where
1 − exp{−ith}
G 0 (t) = lim G h (t) = exp{−itx} lim .
h→0 h→0 ith
By Lemma A2.1,
1 − exp(−ith) = ith + R 2 (th)

2
where |R 2 (th)|≤ (th) /2. Hence,
1 − exp(−ith)
lim = 1
h→0 ith
so that G 0 (t) = exp(−itx) and
F(x + h) − F(x) 1 ∞
lim = exp(−itx)ϕ(t) dt.
h→0 h 2π −∞
Hence, F is differentiable with derivative

1 ∞

F (x) = exp(−itx)ϕ(t) dt.
2π
−∞
The result now follows from Theorem 1.9.
Hence, for those cases in which the integral of the modulus of the characteristic function
is finite, Theorem 3.8 gives an alternative to Theorem 3.3 for obtaining the distribution of a
random variable from its characteristic function. When this condition is satisfied, calculation
of the integral in Theorem 3.8 is often easier than calculation of the integral appearing in
Theorem 3.3.

Example 3.12 (Normal distribution). Let X denote a random variable with a normal
distribution with parameters µ and σ,as described in Example 3.6. The characteristic