Disintegration theorem

Motivation

Consider the unit square in the Euclidean plane R², S = [0, 1] × [0, 1]. Consider the probability measure μ defined on S by the restriction of two-dimensional Lebesgue measure λ² to S. That is, the probability of an event E ⊆ S is simply the area of E. We assume E is a measurable subset of S.

Consider a one-dimensional subset of S such as the line segment L_x = {x} × [0, 1]. L_x has μ-measure zero; every subset of L_x is a μ-null set; since the Lebesgue measure space is a complete measure space,

E ⊆ L x ⟹ μ ( E ) = 0.

While true, this is somewhat unsatisfying. It would be nice to say that μ "restricted to" L_x is the one-dimensional Lebesgue measure λ¹, rather than the zero measure. The probability of a "two-dimensional" event E could then be obtained as an integral of the one-dimensional probabilities of the vertical "slices" E ∩ L_x: more formally, if μ_x denotes one-dimensional Lebesgue measure on L_x, then

μ ( E ) = ∫ [ 0 , 1 ] μ x ( E ∩ L x ) d x

for any "nice" E ⊆ S. The disintegration theorem makes this argument rigorous in the context of measures on metric spaces.

Statement of the theorem

(Hereafter, P(X) will denote the collection of Borel probability measures on a metric space (X, d).)

Let Y and X be two Radon spaces (i.e. separable metric spaces on which every probability measure is a Radon measure). Let μ ∈ P(Y), let π : Y → X be a Borel-measurable function, and let ν ∈ P(X) be the pushforward measure ν  = π_∗(μ) = μ ∘ π⁻¹. Then there exists a ν -almost everywhere uniquely determined family of probability measures {μ_x}_x∈X ⊆ P(Y) such that

the function x ↦ μ x is Borel measurable, in the sense that x ↦ μ x ( B ) is a Borel-measurable function for each Borel-measurable set B ⊆ Y;

μ_x "lives on" the fiber π⁻¹(x): for ν -almost all x ∈ X,

and so μ_x(E) = μ_x(E ∩ π⁻¹(x));

for every Borel-measurable function f : Y → [0, ∞],

In particular, for any event E ⊆ Y, taking f to be the indicator function of E,

Product spaces

The original example was a special case of the problem of product spaces, to which the disintegration theorem applies.

When Y is written as a Cartesian product Y = X₁ × X₂ and π_i : Y → X_i is the natural projection, then each fibre π₁⁻¹(x₁) can be canonically identified with X₂ and there exists a Borel family of probability measures { μ x 1 } x 1 ∈ X 1 in P(X₂) (which is (π₁)_∗(μ)-almost everywhere uniquely determined) such that

μ = ∫ X 1 μ x 1 μ ( π 1 − 1 ( d x 1 ) ) = ∫ X 1 μ x 1 d ( π 1 ) ∗ ( μ ) ( x 1 ) ,

which is in particular

∫ X 1 × X 2 f ( x 1 , x 2 ) μ ( d x 1 , d x 2 ) = ∫ X 1 ( ∫ X 2 f ( x 1 , x 2 ) μ ( d x 2 | x 1 ) ) μ ( π 1 − 1 ( d x 1 ) )

and

μ ( A × B ) = ∫ A μ ( B | x 1 ) μ ( π 1 − 1 ( d x 1 ) ) .

The relation to conditional expectation is given by the identities

E ⁡ ( f | π 1 ) ( x 1 ) = ∫ X 2 f ( x 1 , x 2 ) μ ( d x 2 | x 1 ) , μ ( A × B | π 1 ) ( x 1 ) = 1 A ( x 1 ) ⋅ μ ( B | x 1 ) .

Vector calculus

The disintegration theorem can also be seen as justifying the use of a "restricted" measure in vector calculus. For instance, in Stokes' theorem as applied to a vector field flowing through a compact surface Σ ⊂ R³, it is implicit that the "correct" measure on Σ is the disintegration of three-dimensional Lebesgue measure λ³ on Σ, and that the disintegration of this measure on ∂Σ is the same as the disintegration of λ³ on ∂Σ.

Conditional distributions

The disintegration theorem can be applied to give a rigorous treatment of conditional probability distributions in statistics, while avoiding purely abstract formulations of conditional probability.