30 1. Preparatory material

Example 1.1.26. Let Ω = [0,

1]2

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,

1]2

: x1 + x2 = y}.

Exercise 1.1.23 (Almost uniqueness of disintegrations). Let (R , (μy)y∈R ),

(

˜

R , (˜y)y∈ μ

˜

R

) 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

˜

R

), 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) =

˜(X|Y

E = y) in such

cases.

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.