58.11 Characteristic Functions And Independence
There is a way to tell if random vectors are independent by using their characteristic
Proposition 58.11.1 If Xi is a random vector having values in ℝpi, then the random
vectors are independent if and only if
where P ≡∑
j=1ntj ⋅ Xj for tj ∈ ℝpj.
The proof of this proposition will depend on the following lemma.
Lemma 58.11.2 Let Y be a random vector with values in ℝp and let f be bounded and
measurable with respect to the Radon measure λY, and satisfy
for all t ∈ ℝp. Then f
for λYa.e. y.
Proof: You could write the following for ϕ ∈G
and now recall that the inverse Fourier transform maps G onto G. Hence
for all ψ ∈G. Thus this is also so for every ψ ∈ C0∞
by an obvious application of the
Stone Weierstrass theorem. Let
be a sequence of functions in
pointwise and in L1
2. Then for any ψ ∈ C0∞
Also, the above holds for any ψ ∈ Cc
as can be seen by taking such a
convolving with a mollifier. By the Riesz representation theorem, f
equals 0.) ■
Proof of the proposition: If the Xj are independent, the formula follows from
Lemma 58.9.6 and Lemma 58.9.4.
Now suppose the formula holds. Thus
Then from the above Lemma 58.11.2, the following equals 0 for λXn a.e. xn.
Let ti = 0 for i = 1,2,
Then this implies
By the fact that the characteristic function determines the distribution measure, Theorem
58.8.4, it follows that for these xn off a set of λXn measure zero,λXn−1 = λXn−1|xn.
Returning to 58.11.19, one can replace λXn−1|xn with λXn−1 to obtain
Next let tn = 0 and applying the above Lemma 58.11.2 again, this implies that for λXn−1
a.e. xn−1, the following equals 0.
Let ti = 0 for i = 1,2,
Then you obtain
and so λXn−2 = λXn−2|xnxn−1 for xn−1 off a set of λXn−1 measure zero. Continuing this
way, it follows that
for xn−k+1 off a set of λXn−k+1 measure zero. Thus if E is Borel in ℝpn−1 ×
One could achieve this iterated integral in any order by similar arguments to the above.
By Definition 58.9.2 and the discussion which follows, this implies that the random
variables Xi are independent. ■
Here is another proof of the Doob Dynkin lemma based on differentiation
Lemma 58.11.3 Suppose X,Y1,Y2,
,Yk are random vectors X having values in
ℝn and Yj having values in ℝpj and
Suppose X is σ
Then there exists a Borel function, g :∏
j=1kℝpj → ℝn such that
Proof: For the sake of brevity, denote by Y the vector
j=1kℝpj ≡ ℝP
. For E
a Borel set of ℝn,
Consider the function
Since dλY is a Radon measure having inner and outer regularity, it follows the above
function is equal to a Borel function for λY a.e. y. This function will be denoted by g.
Then from 58.11.20
and since Y−1
is an arbitrary element of
this shows that since X
What about the case where X is not necessarily measurable in σ
Lemma 58.11.4 There exists a unique function Z
for all F ∈ σ
such that Z is σ
measurable. It is denoted
Proof: It is like the above. Letting E be a Borel set in ℝp,
Now let g
be a Borel representative of
It follows ω → g
because by definition
ω → Y
measurable and a Borel measurable
function composed with a measurable one is still measurable. It follows that for all
Borel in ℝp,
and so Z
works because a generic set of
a Borel set in ℝp
. If both Z,Z1
work, then for all F ∈ σ
Since F is arbitrary, some routine computations show Z = Z1 a.e. ■
Observation 58.11.5 Note that a.e.
where the one on the left is the expected value of X given values of Yj
. This one
corresponds to the sort of thing we say in words. The one on the right is an abstract
concept which is usually obtained using the Radon Nikodym theorem and its description is
given in the lemma. This lemma shows that its meaning is really to take the expected
value of X given values for the Yk.