27.4. CHARACTERISTIC FUNCTIONS AND INDEPENDENCE 747

Lemma 27.4.3 Suppose X,Y 1,Y 2, · · · ,Y k are random vectors X having values inRn and Y j having values in Rp j and

X,Y j ∈ L1 (Ω) .

SupposeX is σ (Y 1, · · · ,Y k) measurable. Thus

{X−1 (E) : E Borel

}⊆

{(Y 1, · · · ,Y k)

−1 (F) : F is Borel ink

∏j=1Rp j

}

Then there exists a Borel function, g : ∏kj=1Rp j → Rn such that

X = g (Y 1,Y 2, · · · ,Y k) .

Proof: For the sake of brevity, denote by Y the vector (Y 1, · · · ,Y k) and by y thevector (y1, · · · ,yk) and let ∏

kj=1Rp j ≡ RP. For E a Borel set of Rn,∫

Y −1(E)XdP =

∫Rn×RP

XRn×E (x,y)xdλ (X,Y )

=∫

E

∫RnxdλX|ydλY . (27.9)

Consider the function y→∫Rn xdλX|y. 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 27.9∫

Y −1(E)XdP =

∫Eg (y)dλY =

∫RP

XE (y)g (y)dλY

=∫

Ω

XE (Y (ω))g (Y (ω))dP =∫Y −1(E)

g (Y (ω))dP

and since Y −1 (E) is an arbitrary element of σ (Y ) , this shows that since X is σ (Y )measurable,X = g (Y ) P a.e. ■

What about the case whereX is not necessarily measurable in σ (Y 1, · · · ,Y k)?

Lemma 27.4.4 There exists a unique function Z (ω) which satisfies∫FX (ω)dP =

∫FZ (ω)dP

for allF ∈ σ (Y 1, · · · ,Y k)

such that Z is σ (Y 1, · · · ,Y k) measurable. It is denoted by

E (X|σ (Y 1, · · · ,Y k))

Proof: It is like the above. Letting E be a Borel set in Rp,∫Y −1(E)

XdP =∫Rn×RP

XRn×E (x,y)xdλ (X,Y ) =∫

E

∫RnxdλX|ydλY .