Page 103 - Elements of Distribution Theory

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

3.3 Further Properties of Characteristic Functions 89

(ii) Let X denote a random variable with a lattice distribution and let ϕ denote its
characteristic function. A constant b is a maximal span of the distribution if and
only if ϕ(2π/b) = 1 and

|ϕ(t)| < 1 for all 0 < |t| < 2π/b.
(iii) Let X denote a random variable with range X that is a subset of

{a + bj, j = 0, ±1, ±2,...}
and let ϕ denote the characteristic function of the distribution. Then, for all −∞ <
t < ∞,
a k

ϕ(t) = exp −2πk i ϕ t + 2π , k = 0, ±1, ±2,....
b b
Thus, if a = 0, the characteristic function is periodic.

Proof. Suppose X has a lattice distribution. Then the characteristic function of X is of the
form
∞

ϕ(t) = exp{ita} exp{it jb}p j
j=−∞
where the p j are nonnegative and sum to 1. Hence,

∞

ϕ(2π/b) = exp{i2πa/b} exp{i2π j}p j = exp{i2πa/b}
j=−∞
so that
|ϕ(2π/b)|= 1.
Now suppose that |ϕ(t)|= 1 for some t = 0. Writing

ϕ(t) = y 1 + iy 2 ,
2
2
we must have y + y = 1. Let F denote the distribution function of the distribution. Then
1 2

exp{itx} dF(x) = exp{itz}
for some real number z.It follows that

[exp{itz}− exp{itx}] dF(x) = 0
and, hence, that

[cos(tz) − cos(tx)] dF(x) = [1 − cos(t(x − z))] dF(x) = 0.

Note that 1 − cos(t(x − z)) is nonnegative and continuous in x. Hence, F must be discon-
tinuous with mass points at the zeros of 1 − cos(t(x − z)). It follows that the mass points
of the distribution must be of the form z + 2π j/t, for j = 0, ±1,.... This proves part (i)
of the theorem.