Conditioning on a Measurable Map and Its Fibers

Conditioning Along a Map

A measurable map can turn a random element into a coarser observation. For example, observing the sum of two coordinates does not reveal the coordinates themselves, but it restricts them to a level set of the sum map. A conditional law given that observation describes how probability is distributed along that level set.

This description needs care. If the observation has a continuous distribution, the event that it equals any one specified value may have probability zero. Conditioning on that event by dividing probabilities is then not available. On standard Borel spaces, a probability kernel provides a rigorous conditional law, defined for almost every observed value.

Disintegration Along a Measurable Map

Let $E$ and $F$ be standard Borel spaces, let $\mu$ be a probability measure on $E$, and let $T:E\to F$ be measurable. The distribution of the observed value $T(x)$ is the pushforward probability measure $\nu=\mu\circ T^{-1}$ on $F$.

Definition: A disintegration of $\mu$ along $T$ is a probability kernel $K$ from $F$ to $E$ such that, for every Borel set $A\subseteq E$ and every Borel set $B\subseteq F$, $$ \mu\bigl(A\cap T^{-1}(B)\bigr) = \int_B K(y,A)\,\nu(dy). $$ In words, $K(y,\cdot)$ is a probability measure on $E$ for each $y$, $y\mapsto K(y,A)$ is measurable for each Borel $A$, and the displayed identity holds for all such $A,B$.

The identity says that the probability of both $x\in A$ and $T(x)\in B$ can be computed by first integrating the conditional probability of $A$ at each observed value $y$, then averaging over the distribution $\nu$ of those values. The existence theorem for regular conditional probabilities on standard Borel spaces gives such a kernel. Its values are determined only $\nu$-almost surely: values at observations in a $\nu$-null set do not affect the identity.

The fiber over $y$ is the set $T^{-1}(\{y\})=\{x\in E:T(x)=y\}$. The central fact is that, for almost every observed value, the conditional law assigns probability one to its fiber.

Theorem: If $K$ is a disintegration of $\mu$ along $T$, then $$ K\bigl(y,T^{-1}(\{y\})\bigr)=1 \quad\text{for }\nu\text{-almost every }y. $$

Proof: Define a probability measure $\gamma$ on $F\times E$ as the distribution of $(T(x),x)$ when $x$ has law $\mu$. For Borel sets $B\subseteq F$ and $A\subseteq E$, the disintegration identity gives $$ \gamma(B\times A) =\mu\bigl(T^{-1}(B)\cap A\bigr) =\int_B K(y,A)\,\nu(dy). $$ Therefore $\gamma$ agrees on measurable rectangles with the measure obtained by integrating the kernel $K(y,dx)$ against $\nu(dy)$. Rectangles generate the product sigma-algebra, so the two probability measures agree on all Borel subsets of $F\times E$.

The graph $G=\{(y,x)\in F\times E:T(x)=y\}$ is measurable: it is the preimage of the diagonal in $F\times F$ under the measurable map $(y,x)\mapsto(y,T(x))$, and the diagonal is Borel because $F$ is standard Borel. By construction, $(T(x),x)\in G$ for every $x$, so $\gamma(G)=1$. Integrating the indicator of $G$ against the kernel representation of $\gamma$ yields

1 = \int_F\int_E \mathbf{1}_{\{T(x)=y\}}\,K(y,dx)\,\nu(dy) = \int_F K\bigl(y,T^{-1}(\{y\})\bigr)\,\nu(dy).

The integrand is measurable and lies between zero and one. A function with values in $[0,1]$ whose integral under a probability measure is one must equal one almost everywhere. This proves the claim.

\square

The theorem is an almost-sure statement, not a claim that every possible value $y$ has a meaningful conditional law concentrated on its fiber. The disintegration identity constrains the kernel only for $\nu$-almost every $y$.

Worked Examples

Worked Example: Conditioning on a Coordinate

Let $E=F_1\times F_2$, let $\mu=P\otimes Q$ be a product probability measure, and let $T(x,z)=x$ be the first-coordinate projection. The distribution $\nu$ of $T$ is $P$. For $x\in F_1$, define $K(x,A)=Q(\{z:(x,z)\in A\})$ for Borel $A\subseteq F_1\times F_2$.

The section $A_x=\{z:(x,z)\in A\}$ is Borel, and the section-measurability theorem for product measures ensures that $x\mapsto Q(A_x)$ is measurable. For Borel $B\subseteq F_1$, Fubini’s theorem gives

\int_B K(x,A)\,P(dx) = \int_B Q(A_x)\,P(dx) = (P\otimes Q)\bigl(A\cap(B\times F_2)\bigr).

Thus $K$ is a disintegration. The fiber at $x$ is $\{x\}\times F_2$, and $K(x,\cdot)$ assigns probability one to it. In this example, conditioning on the first coordinate leaves the second coordinate distributed according to $Q$.

Worked Example: The Sum of Two Uniform Coordinates

Let $(U,V)$ be uniformly distributed on the unit square $[0,1]^2$, and set $S=U+V$. The distribution of $S$ has density

f_S(s)= \begin{cases} s,&0<s<1,\\ 2-s,&1\leq s<2,\\ 0,&\text{otherwise}. \end{cases}

For $0<s<2$, define $a_s=\max(0,s-1)$, $b_s=\min(1,s)$, and $L_s=b_s-a_s$. Then $L_s>0$. Define a probability measure $K(s,\cdot)$ on the square by

K(s,A)=\frac{1}{L_s}\int_{a_s}^{b_s} \mathbf{1}_A(u,s-u)\,du.

At $s=0$ and $s=2$, set $K(0,\cdot)=\delta_{(0,0)}$ and $K(2,\cdot)=\delta_{(1,1)}$. These endpoint values do not affect a disintegration because $S$ has no atoms there.

To check the formula, change variables from $(u,v)$ to $(u,s)=(u,u+v)$. The inverse is $(u,s)\mapsto(u,s-u)$, whose Jacobian determinant has absolute value one. For a Borel set $A\subseteq[0,1]^2$ and $B\subseteq[0,2]$, this gives

\begin{aligned} \mathbb{P}\bigl((U,V)\in A,\ S\in B\bigr) &=\int_B\int_{a_s}^{b_s}\mathbf{1}_A(u,s-u)\,du\,ds\\ &=\int_B K(s,A)L_s\,ds\\ &=\int_B K(s,A)f_S(s)\,ds. \end{aligned}

Here $L_s=s$ for $0<s<1$, and $L_s=2-s$ for $1\leq s<2$, so $L_s=f_S(s)$ throughout the interior. This verifies the disintegration identity. At a fixed sum $s$, the conditional law is uniform in the $u$-parameter along the segment $\{(u,s-u):a_s\leq u\leq b_s\}$, which is precisely the fiber within the square.

Worked Example: Conditioning a Normal Variable on Its Square

Let $X$ have the standard normal distribution with density $\varphi(x)=(2\pi)^{-1/2}e^{-x^2/2}$, and let $Y=X^2$. For $y>0$, define

K(y,\cdot)=\tfrac12\delta_{\sqrt y}+\tfrac12\delta_{-\sqrt y}, \qquad K(0,\cdot)=\delta_0.

The pushforward law $\nu$ of $Y$ has density $f_Y(y)=\varphi(\sqrt y)/\sqrt y$ for $y>0$. For Borel sets $A\subseteq\mathbb{R}$ and $B\subseteq[0,\infty)$, changing variables $y=x^2$ separately on the positive and negative half-lines gives

\begin{aligned} \mathbb{P}(X\in A,\ Y\in B) &=\int_B \bigl(\mathbf{1}_A(\sqrt y)+\mathbf{1}_A(-\sqrt y)\bigr) \frac{\varphi(\sqrt y)}{2\sqrt y}\,dy\\ &=\int_B K(y,A)\,f_Y(y)\,dy. \end{aligned}

The second equality uses the symmetry $\varphi(-x)=\varphi(x)$. Hence $K$ is a disintegration. For $y>0$, the fiber of the squaring map is $\{-\sqrt y,\sqrt y\}$, and the conditional law gives equal weight to its two points. At $y=0$, the fiber is the singleton $\{0\}$.

Conditional Expectations from the Kernel

Disintegration also gives a direct formula for conditional expectations. For a bounded Borel function $f:E\to\mathbb{R}$, define

h(y)=\int_E f(x)\,K(y,dx).

Kernel integration ensures that $h$ is measurable. The next result identifies $h(T)$ as a version of the conditional expectation given the observation $T$.

Theorem: For every bounded Borel $f:E\to\mathbb{R}$, the random variable $h(T)$ is a version of $\mathbb{E}[f\mid\sigma(T)]$, where $h(y)=\int_E f(x)\,K(y,dx)$.

Proof: First take $f=\mathbf{1}_A$ for a Borel set $A\subseteq E$. For every Borel $B\subseteq F$, the disintegration identity gives $$ \int_{T^{-1}(B)} \mathbf{1}_A(x)\,\mu(dx) = \int_B K(y,A)\,\nu(dy) = \int_{T^{-1}(B)} K(T(x),A)\,\mu(dx). $$ The same equality holds for every bounded Borel $f$ by linearity for simple functions and bounded convergence. Since the sets $T^{-1}(B)$ form $\sigma(T)$, the defining integral property of conditional expectation holds on every set in $\sigma(T)$. Also $h(T)$ is $\sigma(T)$-measurable. These two properties establish the claimed version.

\square

Thus the kernel does more than describe how mass lies on fibers: integrating a function against the conditional law produces its conditional expectation as a function of the observed value. This is the measure-theoretic form of “average over the possibilities compatible with the observation.”

What the Fiber Statement Does and Does Not Say

The fiber theorem does not say that a conditional law is simply a normalized restriction of $\mu$ to the fiber. Fibers often have $\mu$-measure zero, as the line segments in the sum example do under two-dimensional area measure. A conditional law on such a fiber is defined through the disintegration identity, not by dividing by the probability of the fiber.

Nor does the theorem specify a unique conditional law at every $y$. Disintegrations are unique only up to changes on a $\nu$-null set. When evaluating a conditional law at an observation drawn from $\nu$, the almost-sure statement is the relevant one. For a particular value outside the support of $\nu$, the disintegration identity alone need not dictate a meaningful conditional distribution.

Key takeaway: A measurable observation $T(x)$ partitions the state space into fibers. A disintegration supplies conditional probability measures on the original space, and almost every such measure is concentrated on the fiber matching its observation.

Check Your Understanding

Use the definitions and results above to check that you can move between a measurable map, its fibers, and a conditional law.

State the disintegration identity for Borel sets $A\subseteq E$ and $B\subseteq F$.
Why is concentration on the fiber an almost-sure assertion rather than a claim about every $y\in F$?
For the sum of two uniform coordinates, what interval of $u$-values parametrizes the fiber at a value $s\in(0,2)$?
For the squaring map, why does the conditional law at $y>0$ place equal mass on the two points in its fiber?
How does integrating a bounded function against the disintegration yield a version of its conditional expectation given $T$?