Tutorials › Probability Theory (Graduate) › Probability Kernels and Measurability

Home  /  Tutorials  /  Conditional Laws and Disintegration

Conditional Laws and Disintegration · Tutorial 253 of 946

Probability Kernels and Measurability

Learn to recognize a probability kernel and verify its measurability through evaluation maps and generating classes.

Advanced 10 min read

What You'll Learn

  • State the definition of a probability kernel between measurable spaces
  • Describe the evaluation sigma-algebra on a space of probability measures
  • Prove measurability by checking evaluations on a generating pi-system
  • Verify kernels on countable spaces using probability mass functions
  • Check Euclidean kernels, including Gaussian location and deterministic kernels

Why Probability Kernels Need Two Kinds of Validity

A probability kernel assigns a probability distribution on one space to each point of another space. It is a way to describe a distribution that varies with a parameter: for each input \(x\), there is a probability measure \(K(x,\cdot)\) on the output space.

There are two requirements. For each fixed input, the assignment in the output set must be a probability measure. Across inputs, the probability assigned to each measurable output set must vary measurably. The second requirement is essential: without it, the family of probability measures need not define a measurable dependence on the input.

Definition and Evaluation Maps

Definition: Let \((E,\mathcal E)\) and \((F,\mathcal F)\) be measurable spaces. A probability kernel from \(E\) to \(F\) is a function \(K:E\times\mathcal F\to[0,1]\) such that:
  1. For each \(x\in E\), the set function \(B\mapsto K(x,B)\) is a probability measure on \((F,\mathcal F)\).
  2. For each \(B\in\mathcal F\), the function \(x\mapsto K(x,B)\) is \(\mathcal E\)-measurable.
We write \(K(x,\cdot)\) for the probability measure assigned to \(x\).

For a measurable space \(F\), let \(\mathcal P(F)\) denote the set of probability measures on \((F,\mathcal F)\). For each \(B\in\mathcal F\), its evaluation map is the function \(\mathrm{ev}_B:\mathcal P(F)\to[0,1]\) given by \(\mathrm{ev}_B(\mu)=\mu(B)\). Equip \(\mathcal P(F)\) with the evaluation sigma-algebra, namely the smallest sigma-algebra that makes every evaluation map measurable.

Given a family \(K(x,\cdot)\) of probability measures, define \(\widehat K:E\to\mathcal P(F)\) by \(\widehat K(x)=K(x,\cdot)\). Then kernel measurability is exactly the measurability of this map into the evaluation sigma-algebra.

Proposition: A family \(x\mapsto K(x,\cdot)\) of probability measures is measurable as a map into \(\mathcal P(F)\) with its evaluation sigma-algebra if and only if, for every \(B\in\mathcal F\), \(x\mapsto K(x,B)\) is measurable.
Proof. The evaluation sigma-algebra is generated by sets of the form \(\mathrm{ev}_B^{-1}(A)\), where \(B\in\mathcal F\) and \(A\) is a Borel subset of \([0,1]\). The composition \(\mathrm{ev}_B\circ\widehat K\) is exactly \(x\mapsto K(x,B)\). Thus, if \(\widehat K\) is measurable, each such composition is measurable. Conversely, if every composition is measurable, the preimage under \(\widehat K\) of every generating set \(\mathrm{ev}_B^{-1}(A)\) is measurable. Since these sets generate the evaluation sigma-algebra, \(\widehat K\) is measurable.

\(\square\)

This equivalence makes evaluation maps the natural way to check a kernel. One need not reason about measurability of a family of measures all at once; one checks the real-valued function associated with each measurable output set. Often, a generating family of output sets is enough.

Checking a Generating Pi-System

A collection \(\mathcal C\subseteq\mathcal F\) is a pi-system if it is closed under finite intersections. Suppose that \(\mathcal C\) contains \(F\) and generates \(\mathcal F\), meaning \(\sigma(\mathcal C)=\mathcal F\). The next result lets us verify evaluations on \(\mathcal C\) rather than on every set in \(\mathcal F\).

Theorem: Let \(K(x,\cdot)\) be a probability measure on \((F,\mathcal F)\) for each \(x\in E\). Suppose \(\mathcal C\subseteq\mathcal F\) is a pi-system, \(F\in\mathcal C\), and \(\sigma(\mathcal C)=\mathcal F\). If \(x\mapsto K(x,C)\) is measurable for every \(C\in\mathcal C\), then \(x\mapsto K(x,B)\) is measurable for every \(B\in\mathcal F\).
Proof. Let \(\mathcal D\) be the collection of all \(B\in\mathcal F\) for which \(x\mapsto K(x,B)\) is measurable. Since \(K(x,F)=1\), we have \(F\in\mathcal D\). If \(B\in\mathcal D\), then $$ K(x,B^c)=1-K(x,B), $$ so \(B^c\in\mathcal D\). If \(B_1,B_2,\ldots\) are pairwise disjoint members of \(\mathcal D\), countable additivity gives $$ K\left(x,\bigcup_{j=1}^{\infty}B_j\right) =\sum_{j=1}^{\infty}K(x,B_j) =\lim_{n\to\infty}\sum_{j=1}^{n}K(x,B_j). $$ Each finite sum is measurable, and a pointwise limit of measurable functions is measurable. Thus \(\bigcup_{j=1}^{\infty}B_j\in\mathcal D\). Consequently, \(\mathcal D\) is a lambda-system. By hypothesis \(\mathcal C\subseteq\mathcal D\); the pi-lambda theorem therefore gives \(\sigma(\mathcal C)\subseteq\mathcal D\). Since \(\sigma(\mathcal C)=\mathcal F\), every \(B\in\mathcal F\) belongs to \(\mathcal D\), as required.

\(\square\)

The probability-measure condition matters in this argument: it gives the complement identity \(K(x,B^c)=1-K(x,B)\). The theorem does not say that checking an arbitrary small collection of sets suffices. The collection must be a pi-system generating the output sigma-algebra.

Worked Example: A Poisson Kernel on a Countable Space

Worked Example: A Poisson Kernel on a Countable Space

Take \(E=[0,\infty)\) with its Borel sigma-algebra and \(F=\mathbb N_0\) with the power set sigma-algebra. For \(\lambda\in E\) and \(n\in\mathbb N_0\), define $$ p_n(\lambda)=e^{-\lambda}\frac{\lambda^n}{n!}, $$ using \(\lambda^0=1\), including when \(\lambda=0\). Set \(K(\lambda,B)=\sum_{n\in B}p_n(\lambda)\) for \(B\subseteq\mathbb N_0\).

For each fixed \(n\), \(p_n\) is nonnegative and continuous on \([0,\infty)\). At \(\lambda=0\), \(p_0(0)=1\), while \(p_n(0)=0\) for \(n\geq1\). For every \(\lambda\geq0\), the exponential series gives $$ \sum_{n=0}^{\infty}p_n(\lambda) =e^{-\lambda}\sum_{n=0}^{\infty}\frac{\lambda^n}{n!} =e^{-\lambda}e^\lambda=1. $$ Thus \(B\mapsto K(\lambda,B)\) is a probability measure on the countable space.

To check measurability, enumerate the elements of any \(B\subseteq\mathbb N_0\) and write \(K(\lambda,B)\) as the corresponding series of nonnegative continuous functions \(p_n(\lambda)\). Its partial sums are measurable, and their pointwise limit is measurable. Hence \(K\) is a probability kernel. In a countable discrete output space, this argument shows why checking the individual point probabilities is a convenient route to checking all evaluations.

Worked Example: A Gaussian Location Kernel

Worked Example: A Gaussian Location Kernel

Let both \(E\) and \(F\) be \(\mathbb R\) with their Borel sigma-algebras. Write $$ \varphi(u)=\frac{1}{\sqrt{2\pi}}e^{-u^2/2}, \qquad K(x,B)=\int_B\varphi(y-x)\,dy, $$ where the integral is with respect to Lebesgue measure. For each fixed \(x\), \(y\mapsto\varphi(y-x)\) is the density of a normal distribution with mean \(x\) and variance \(1\). It is nonnegative and integrates to \(1\), so \(K(x,\cdot)\) is a probability measure.

We verify measurability in \(x\) for every Borel set \(B\), in fact by proving continuity. The derivative is \(\varphi'(u)=-u\varphi(u)\), and $$ \int_{\mathbb R}|\varphi'(u)|\,du =\frac{1}{\sqrt{2\pi}}\int_{\mathbb R}|u|e^{-u^2/2}\,du =\sqrt{\frac{2}{\pi}}<\infty. $$ For any \(x,x_0\in\mathbb R\), the fundamental theorem of calculus and Tonelli's theorem yield $$ \begin{aligned} |K(x,B)-K(x_0,B)| &\leq \int_{\mathbb R}|\varphi(y-x)-\varphi(y-x_0)|\,dy\\ &\leq |x-x_0|\int_{\mathbb R}|\varphi'(u)|\,du\\ &=|x-x_0|\sqrt{\frac{2}{\pi}}. \end{aligned} $$ For the second inequality, integrate the absolute derivative along the interval between \(x\) and \(x_0\), then integrate over \(y\); translation leaves the integral of \(|\varphi'|\) unchanged. The bound tends to zero as \(x\to x_0\). Thus \(x\mapsto K(x,B)\) is continuous, and in particular Borel measurable, for every \(B\). This verifies the kernel condition.

Worked Example: A Deterministic Kernel

Worked Example: A Deterministic Kernel

Let \(f:E\to F\) be a measurable function between measurable spaces. Define \(K(x,B)=\mathbf 1_B(f(x))\), where \(\mathbf 1_B\) is \(1\) on \(B\) and \(0\) elsewhere. For fixed \(x\), this is the point-mass probability measure at \(f(x)\): it assigns probability \(1\) to sets containing \(f(x)\) and \(0\) to the other sets.

For fixed \(B\in\mathcal F\), the evaluation is $$ K(x,B)=\mathbf 1_B(f(x))=\mathbf 1_{f^{-1}(B)}(x). $$ Since \(f\) is measurable, \(f^{-1}(B)\in\mathcal E\), so its indicator is measurable. Therefore \(K\) is a probability kernel. A deterministic measurable map is thus a special case of a kernel, not an exception to the definition.

A Useful Measurable Transformation

Proposition: Let \(K\) be a probability kernel from \((E,\mathcal E)\) to \((F,\mathcal F)\), and let \(g:F\to G\) be measurable, where \((G,\mathcal G)\) is a measurable space. Define $$ K_g(x,C)=K\bigl(x,g^{-1}(C)\bigr),\qquad C\in\mathcal G. $$ Then \(K_g\) is a probability kernel from \(E\) to \(G\).
Proof. For each fixed \(x\), \(K_g(x,\cdot)\) is the image probability measure of \(K(x,\cdot)\) under \(g\), because inverse images preserve complements and countable unions, and \(g^{-1}(G)=F\). In particular, it is a probability measure. For each \(C\in\mathcal G\), measurability of \(g\) gives \(g^{-1}(C)\in\mathcal F\). The kernel property of \(K\) then implies that \(x\mapsto K(x,g^{-1}(C))=K_g(x,C)\) is measurable. Both defining conditions hold.

\(\square\)

Why Evaluation Measurability Matters

A kernel represents a family of distributions whose dependence on the input is measurable. This requirement makes it possible to use the family in further probability constructions. In particular, the evaluation perspective provides a precise check: ask whether the probability of each measurable output event is a measurable function of the input.

A common mistake is to check that \(K(x,\cdot)\) is a probability measure for each \(x\) and assume that this settles the definition. It does not. The input dependence is a separate condition. Another useful shortcut is the generating-pi-system theorem: when the output sigma-algebra has a convenient generating pi-system, checking evaluations on that system proves the full measurability condition. For Borel subsets of \(\mathbb R\), for example, intervals from a suitable countable generating family can be used.

Check Your Understanding

Use the definitions and results above to answer the following questions.

  1. What are the two independent requirements in the definition of a probability kernel?
  2. Why does measurability of all evaluation maps characterize measurability into \(\mathcal P(F)\) with its evaluation sigma-algebra?
  3. In the Poisson example, why does checking the point probabilities lead to measurability for every subset of \(\mathbb N_0\)?
  4. What role does the pi-system hypothesis play in the generating-class theorem?
  5. Why is measurability of \(f\) needed for the deterministic kernel \(K(x,B)=\mathbf 1_B(f(x))\)?

Previous

Next

Back to Tutorials
Clinical Biostats