30 1. Preparatory material
Example 1.1.26. Let Ω = [0,
with Lebesgue measure, and let (X1,X2)
be the coordinate random variables of Ω, thus X1,X2 are iid with the uni-
form distribution on [0, 1]. Let Y be the random variable Y := X1 + X2
with range R = R. Then one can disintegrate Y by taking R = [0, 2] and
letting μy be normalised Lebesgue measure on the diagonal line segment
{(x1,x2) [0,
: x1 + x2 = y}.
Exercise 1.1.23 (Almost uniqueness of disintegrations). Let (R , (μy)y∈R ),
R , (˜y)y∈ μ
) be two disintegrations of the same random variable Y . Show
that for any event F , one has P(F |Y = y) =
P (F |Y = y) for μY -almost
every y R, where the conditional probabilities P(|Y = y) and
P (|Y = y)
are defined using the disintegrations (R , (μy)y∈R ), (
R , (˜y)y∈ μ
), respec-
tively. (Hint: Argue by contradiction, and consider the set of y for which
P(F |Y = y) exceeds
P (F |Y = y) (or vice versa) by some fixed ε 0.)
Similarly, for a scalar random variable X, show that for μY -almost every
y R, that (X|Y = y) is absolutely integrable with respect to the first
disintegration if and only if it is absolutely integrable with respect to the
second disintegration, and one has E(X|Y = y) =
E = y) in such
Remark 1.1.27. Under some mild topological assumptions on the under-
lying sample space (and on the measurable space R), one can always find at
least one disintegration for every random variable Y , by using tools such as
the Radon-Nikodym theorem; see [Ta2009, Theorem 2.9.21]. In practice,
we will not invoke these general results here (as it is not natural for us to
place topological conditions on the sample space), and instead construct dis-
integrations by hand in specific cases, for instance, by using the construction
in Example 1.1.25. See, e.g., [Ka2002] for a more comprehensive discus-
sion of these topics; fortunately for us, these subtle issues will not have any
significant impact on our discussion.
Remark 1.1.28. Strictly speaking, disintegration is not a probabilistic con-
cept; there is no canonical way to extend a disintegration when extending
the sample space. However, due to the (almost) uniqueness and existence
results alluded to earlier, this will not be a difficulty in practice. Still, we
will try to use conditioning on continuous variables sparingly, in particular,
containing their Use inside the proofs of various lemmas, rather than in their
statements, due to their slight incompatibility with the “probabilistic way
of thinking”.
Exercise 1.1.24 (Fubini-Tonelli theorem). Let (R , (μy)y∈R ) be a disinte-
gration of a random variable Y taking values in a measurable space R, and
let X be a non-negative (resp. absolutely integrable) scalar random vari-
able. Show that for μY -almost all y R, (X|Y = y) is a non-negative (resp.
Previous Page Next Page