Disintegration theorem

In mathematics, the disintegration theorem is a result in measure theory and probability theory. It rigorously defines the idea of a non-trivial "restriction" of a measure to a measure zero subset of the measure space in question. It is related to the existence of conditional probability measures. In a sense, "disintegration" is the opposite process to the construction of a product measure.

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,

${\displaystyle E\subseteq L_{x}\implies \mu (E)=0.}$

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

${\displaystyle \mu (E)=\int _{[0,1]}\mu _{x}(E\cap L_{x})\,\mathrm {d} x}$

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).) The assumptions of the theorem are as follows:

• Let Y and X be two Radon spaces (i.e. a topological space such that every Borel probability measure on M is inner regular e.g. 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. Here one should think of π as a function to "disintegrate" Y, in the sense of partitioning Y into ${\displaystyle \{\pi ^{-1}(x)\ |\ x\in X\}}$. For example, for the motivating example above, one can define ${\displaystyle \pi ((a,b))=a,(a,b)\in [0,1]\times [0,1]}$, which gives that ${\displaystyle \pi ^{-1}(a)=a\times [0,1]}$, a slice we want to capture.
• Let ${\displaystyle \nu }$ ∈ P(X) be the pushforward measure ν = π_∗(μ) = μ ∘ π⁻¹. This measure provides the distribution of x (which corresponds to the events ${\displaystyle \pi ^{-1}(x)}$).

The conclusion of the theorem: There exists a ${\displaystyle \nu }$-almost everywhere uniquely determined family of probability measures {μ_x}_x∈X ⊆ P(Y), which provides a "disintegration" of ${\displaystyle \mu }$ into ${\displaystyle \{\mu _{x}\}_{x\in X}}$, such that:

• the function ${\displaystyle x\mapsto \mu _{x}}$ is Borel measurable, in the sense that ${\displaystyle x\mapsto \mu _{x}(B)}$ is a Borel-measurable function for each Borel-measurable set B ⊆ Y;
• μ_x "lives on" the fiber π⁻¹(x): for ${\displaystyle \nu }$-almost all x ∈ X,
  $${\displaystyle \mu _{x}\left(Y\setminus \pi ^{-1}(x)\right)=0,}$$

  
  and so μ_x(E) = μ_x(E ∩ π⁻¹(x));
• for every Borel-measurable function f : Y → [0, ∞],
  $${\displaystyle \int _{Y}f(y)\,\mathrm {d} \mu (y)=\int _{X}\int _{\pi ^{-1}(x)}f(y)\,\mathrm {d} \mu _{x}(y)\mathrm {d} \nu (x).}$$

  
  In particular, for any event E ⊆ Y, taking f to be the indicator function of E,^[1]
  $${\displaystyle \mu (E)=\int _{X}\mu _{x}\left(E\right)\,\mathrm {d} \nu (x).}$$

  

Applications[]

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 ${\displaystyle \{\mu _{x_{1}}\}_{x_{1}\in X_{1}}}$ in P(X₂) (which is (π₁)_∗(μ)-almost everywhere uniquely determined) such that

${\displaystyle \mu =\int _{X_{1}}\mu _{x_{1}}\,\mu \left(\pi _{1}^{-1}(\mathrm {d} x_{1})\right)=\int _{X_{1}}\mu _{x_{1}}\,\mathrm {d} (\pi _{1})_{*}(\mu )(x_{1}),}$

which is in particular

${\displaystyle \int _{X_{1}\times X_{2}}f(x_{1},x_{2})\,\mu (\mathrm {d} x_{1},\mathrm {d} x_{2})=\int _{X_{1}}\left(\int _{X_{2}}f(x_{1},x_{2})\mu (\mathrm {d} x_{2}|x_{1})\right)\mu \left(\pi _{1}^{-1}(\mathrm {d} x_{1})\right)}$

and

${\displaystyle \mu (A\times B)=\int _{A}\mu \left(B|x_{1}\right)\,\mu \left(\pi _{1}^{-1}(\mathrm {d} x_{1})\right).}$

The relation to conditional expectation is given by the identities

${\displaystyle \operatorname {E} (f|\pi _{1})(x_{1})=\int _{X_{2}}f(x_{1},x_{2})\mu (\mathrm {d} x_{2}|x_{1}),}$

${\displaystyle \mu (A\times B|\pi _{1})(x_{1})=1_{A}(x_{1})\cdot \mu (B|x_{1}).}$

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.^[2]

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.^[3]

See also[]

• Joint probability distribution
• Copula (statistics)
• Conditional expectation

References[]