Properties of Characteristic Functions
We have seen several examples on how to calculate a characteristic function when given a random variable. Equivalently we have seen examples of how to calculate the fourier transforms of probability measures. For such transforms to be useful, we need to k
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We have seen several examples on how to calculate a characteristic function when given a random variable. Equivalently we have seen examples of how to calculate the Fourier transforms of probability measures. For such transforms to be useful, we need to know that knowledge of the transform characterizes the distribution that gives rise to it. The proof of the next theorem uses the Stone-Weierstrass theorem and thus is a bit advanced for this book. Nevertheless we include the proof for the sake of completeness. Theorem 14.1 (Uniqueness Theorem). The Fourier transform μ ˆ of a probability measure μ on Rn characterizes μ: that is, if two probabilities on Rn admit the same Fourier transform, they are equal. Proof. Let f (σ, x) = and
2 2 1 e−|x| /2σ , (2πσ 2 )n/2
2 2 fˆ(σ, u) = e−|u| σ /2 .
Then f (σ, x) is the density of X = (X1 , . . . , Xn ), where the Xj ’s are independent and N (0, σ 2 ) for each j (1 ≤ j ≤ n). By Example 6 of Chapter 13 and the Tonelli-Fubini Theorem, we have n −x 2 j 1 √ f (σ, x)ei u,x dx = e( 2σ 2 +iuj xj ) dx1 . . . dxn Rn Rn j=1 σ 2π n −x 2 1 ( 2j +iuj xj ) √ = dxj e 2σ σ 2π R j=1 =
n
e−
u 2σ 2 j 2
= fˆ(σ, u).
j=1
Therefore
1 u−v ˆ f (σ, u − v) = f σ, σ2 (2πσ 2 )n/2 u −v 1 = f (σ, x)ei σ 2 ,x dx. (2πσ 2 )n/2 Rn
J. Jacod et al., Probability Essentials © Springer-Verlag Berlin Heidelberg 2004
112
14 Properties of Characteristic Functions
Next suppose that μ1 and μ2 are two probability measures on Rn with the same Fourier transforms μ ˆ1 = μ ˆ2 = μ ˆ. Then f (σ, u − v)μ1 (du) 1 i u −v 2 ,x σ = f (σ, x)e dx μ1 (du) (2πσ 2 )n/2 x v , x 1 −i 2 σ μ ˆ dx, e = f (σ, x) σ2 (2πσ 2 )n/2 (the reader will check that one can apply Fubini’s theorem here), and the exact same equalities hold for μ2 . We conclude that g(x)μ1 (dx) = g(x)μ2 (dx) for all g ∈ H, where H is the vector space generated by all functions of the form u → f (σ, u − v). We then can apply the Stone-Weierstrass theorem1 to conclude that H is dense in C0 under uniform convergence, where C0 is the set of functions “vanishing at ∞”: that is, C0 consists of all continuous functions on Rn such that limx→∞ |f (x)| = 0. We then obtain that g(x)μ1 (dx) = g(x)μ2 (dx) Rn
Rn
for all g ∈ C0 . Since the indicator function of an open set can be written as the increasing limit of functions in C0 , the Monotone Convergence Theorem (Theorem 9.1(d)) then gives μ1 (A) = μ2 (A),
all open sets A ⊂ Rn .
Finally the Monotone Class Theorem (Theorem 6.2) gives μ1 (A) = μ2 (A)
for all Borel sets A ⊂ Rn ,
which means μ1 = μ2 .
Corollary 14.1. Let X = (X1 , . . . , Xn ) be an Rn -valued random variable. Then the real-valued r.v.’s (Xj )1≤j≤n are independent if and only if ϕX (u1 , . . . , un ) =
n
ϕXj (uj )
(14.1)
j=1 1
One can find a nice treatment of the Stone–Weierstrass theorem in [20, p. 160].
14 Properties of Characteristic Functions
113
Proof. If the (Xj )1≤j≤n are independent, then n i u X ϕX (u) = E{ei u,X } = E e j =1 j j ⎧ ⎫ n n ⎨ ⎬ =E eiuj Xj = E{eiuj Xj } ⎩ ⎭ j=1
by the i
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