From a Joint Law to a Conditional Kernel
A joint probability law records how two random elements behave together. A disintegration reorganizes that information: it expresses the joint law as a marginal law for one coordinate, followed by a conditional probability kernel for the other coordinate. In concrete problems, the central tasks are to identify the marginal, compute the conditional law at relevant conditioning values, and check that the proposed kernel reconstructs the joint law.
Let \(E\) and \(F\) be standard Borel spaces, and let \(\mu\) be a probability measure on \(E\times F\). Write \(\nu\) for the \(F\)-marginal, so that \(\nu(B)=\mu(E\times B)\) for measurable \(B\subseteq F\).
The identity is the decisive check. A formula that gives plausible conditional probabilities at typical values is not enough: it must define a probability measure at every conditioning value, be measurable in that value, and satisfy the identity. Values on a \(\nu\)-null set do not affect the integral, but the kernel itself still has to be defined there.
Worked Example: A Finite Joint Probability Table
Worked Example: A Finite Joint Probability Table
Let \(X\) and \(Y\) take values in \(\{0,1,2\}\) and \(\{0,1,2\}\), respectively, with joint probabilities given by this table. Rows correspond to \(Y\), and columns to \(X\).
| \(Y\backslash X\) | \(X=0\) | \(X=1\) | Row total |
|---|---|---|---|
| \(Y=0\) | \(1/10\) | \(1/5\) | \(3/10\) |
| \(Y=1\) | \(3/10\) | \(1/10\) | \(2/5\) |
| \(Y=2\) | \(1/10\) | \(1/5\) | \(3/10\) |
The row totals sum to \(3/10+2/5+3/10=1\), so they give the marginal law \(\nu\) of \(Y\). Dividing each entry by its row total gives the conditional distribution of \(X\) at each value of \(Y\):
Here \(\delta_x\) denotes the probability measure concentrated at \(x\). Each displayed measure has nonnegative weights summing to one. To define the kernel on all of \(\mathbb R\), set \(K(y,\cdot)=\delta_0\) for every \(y\notin\{0,1,2\}\). This choice is immaterial to the disintegration identity because \(\nu\) gives those values probability zero.
For any measurable \(A,B\subseteq\mathbb R\), the proposed kernel gives
The last expression is exactly \(\mathbb{P}(X\in A,Y\in B)\), so the disintegration identity holds. The kernel is measurable because it is defined on a finite support and is constant on its complement.
Recovering a Kernel from a Joint Density
For an absolutely continuous joint law, the usual computation is to integrate the joint density over the coordinate being conditioned out. The resulting marginal density provides the denominator. Where that marginal density is zero, division is not defined; those values must instead be assigned arbitrary probability measures.
Proof. Tonelli’s theorem implies that \(m\) is measurable and \(\int m(y)\,dy=1\). For \(m(y)>0\), the displayed formula defines a probability measure: its value at \(\mathbb{R}^d\) is \(m(y)/m(y)=1\), and countable additivity follows from that of the integral. When \(m(y)=0\), the assigned point mass is also a probability measure. For each measurable \(A\), the numerator is measurable as a function of \(y\) by Tonelli’s theorem, so the piecewise definition of \(K(y,A)\) is measurable.
The \(Y\)-marginal has density \(m\). For measurable \(A\subseteq\mathbb{R}^d\) and \(B\subseteq\mathbb{R}^k\),
For almost every \(y\) with \(m(y)=0\), the nonnegative function \(x\mapsto f(x,y)\) has integral zero and is therefore zero for almost every \(x\). Consequently, omitting those \(y\)-values does not change the joint integral. The last display thus equals \(\int_{A\times B}f(x,y)\,dx\,dy\), which is the joint probability of \(A\times B\). This proves the disintegration identity. \(\square
Worked Example: A Density on a Triangular Region
Worked Example: A Density on a Triangular Region
Let \((X,Y)\) have density \(f(x,y)=2\) on \(0<x<y<1\), and density zero elsewhere. It is a probability density because
Integrating out \(x\) gives the marginal density \(m(y)=2y\) for \(0<y<1\), and zero otherwise. For \(0<y<1\), the conditional kernel is uniform on \((0,y)\):
To make this a kernel from \(\mathbb{R}\) to \(\mathbb{R}\), define \(K(y,\cdot)=\delta_0\) for every \(y\notin(0,1)\). This includes \(y=0\), \(y=1\), and all values outside the support. For \(0<y<1\), normalization follows from \(K(y,\mathbb{R})=y^{-1}\int_0^y1\,dx=1\); at all other \(y\), normalization holds because \(\delta_0\) is a probability measure.
For measurable \(A,B\subseteq\mathbb{R}\), the marginal measure of \(Y\) has density \(2y\) on \((0,1)\), and hence
The kernel is measurable in \(y\): the integral with variable upper endpoint is measurable, and division by \(y\) is performed only on the measurable set \((0,1)\). The endpoint assignments do not alter the identity because the marginal law has no mass at either endpoint.
Worked Example: A Mixed Conditioning Law
Worked Example: A Mixed Conditioning Law
Let \(X\in\{0,1\}\), and let \(Y\in\mathbb{R}\). Define their joint law by placing mass \(1/4\) at \((0,0)\), mass \(1/12\) at \((1,0)\), and giving the two \(X\)-rows the following densities for \(0<y<1\): density \((2/3)(1-y)\) at \(X=0\), and density \((2/3)y\) at \(X=1\). The total mass is \(1/4+1/12+\int_0^1(2/3)\,dy=1\).
The marginal law of \(Y\) has an atom of mass \(1/3\) at zero and density \(2/3\) on \((0,1)\). At the atom, divide the joint masses by \(1/3\), obtaining \(K(0,\cdot)=\tfrac34\delta_0+\tfrac14\delta_1\). At a continuous conditioning value \(0<y<1\), divide the two row densities by their sum \(2/3\), obtaining \(K(y,\cdot)=(1-y)\delta_0+y\delta_1\). Define \(K(y,\cdot)=\delta_0\) for every \(y\notin[0,1)\), including \(y=1\).
Each of these measures is normalized: at zero the weights sum to \(3/4+1/4=1\); on \((0,1)\) they sum to \((1-y)+y=1\); elsewhere the kernel is a point mass. The piecewise weights are measurable, so this defines a probability kernel everywhere.
To check disintegration, let \(A,B\) be Borel sets and write \(a_0=\mathbf{1}_{\{0\in A\}}\), \(a_1=\mathbf{1}_{\{1\in A\}}\). Then
The atomic term is \(a_0\mathbf{1}_{\{0\in B\}}/4+a_1\mathbf{1}_{\{0\in B\}}/12\), exactly the two joint masses at \(Y=0\). The continuous term is the integral of the specified row densities over \(B\cap(0,1)\). Their sum is precisely \(\mathbb{P}(X\in A,Y\in B)\).
The conditional law at the atom is not required to equal the limit of the conditional laws as \(y\) decreases to zero. The atom’s conditional probabilities are determined by the joint masses at \(Y=0\); values approached through the continuous part are governed by the density there. Confusing these two sources of information can give an incorrect kernel.
What Is Unique, and What Is Not?
A disintegration is determined by the joint law only up to changes on a set of conditioning values with marginal probability zero. The next result makes this precise for standard Borel spaces.
Proof. Choose a countable collection \(\mathcal{C}\) of measurable subsets of \(E\) that contains \(E\), is closed under finite intersections, and generates \(\mathcal{B}(E)\). Such a collection exists because \(E\) is standard Borel. Fix \(A\in\mathcal{C}\). The disintegration identities for \(K\) and \(L\) give, for every measurable \(B\subseteq F\),
Thus \(K(y,A)=L(y,A)\) for \(\nu\)-almost every \(y\). Indeed, if the two bounded measurable functions differed on a set of positive measure, then for some positive integer \(n\) either \(\{K(y,A)\ge L(y,A)+1/n\}\) or its reversed counterpart would have positive measure; integrating over that set would contradict equality of the integrals. Since \(\mathcal{C}\) is countable, there is one \(\nu\)-null set outside which equality holds for every \(A\in\mathcal{C}\). For each \(y\) outside that set, \(K(y,\cdot)\) and \(L(y,\cdot)\) are probability measures agreeing on a generating collection closed under finite intersections. Uniqueness of measures on such a collection gives equality on all of \(\mathcal{B}(E)\). \(\square
In practice, this explains why formulas may assign different values at marginal-null conditioning points and still represent the same disintegration. It does not excuse leaving those values undefined: a kernel must remain a probability measure at every point of its conditioning space.
A Reliable Calculation Checklist
For a concrete joint law, the calculation can be organized into a short sequence. First identify the marginal law of the conditioning coordinate. Next compute the conditional measure wherever the marginal has positive mass or density. Then assign probability measures at all remaining conditioning values. Finally, check both that each value is normalized and that integration against the marginal recovers the joint law. The last check should be made for arbitrary measurable sets, or by a valid argument that establishes the corresponding measure identity.
Sum rows or columns in a table, or integrate the joint density over the other coordinate.
Divide joint masses by marginal masses, or divide the joint density by the marginal density where it is positive.
Specify probability measures at every conditioning value not covered by the formula, including boundary points when necessary.
Integrate the kernel against the marginal and compare with the joint probability on measurable rectangles.
Check Your Understanding
Use the examples and definitions above to answer the following questions.
- In the finite-table example, why can the kernel be assigned \(\delta_0\) at conditioning values outside \(\{0,1,2\}\)?
- For a joint density, what should be done at a value \(y\) where the marginal density \(m(y)\) is zero, and why?
- In the triangular-density example, verify directly that the conditional density \(1/y\) on \((0,y)\) integrates to one for \(0<y<1\).
- In the mixed example, why is the conditional law at the atom \(Y=0\) computed from the joint masses rather than inferred from the continuous density?
- What does almost-everywhere uniqueness say about two disintegrations, and what does it not permit one to omit from a kernel’s definition?