Integrating Against a Probability Kernel

From a Kernel to a Joint Law

A probability kernel assigns a probability measure on one space to each point of another space. Its role is similar to a probability distribution that changes with a state: for each starting point, the kernel specifies a distribution over possible outcomes.

To use this construction in probability, two facts are essential. First, integrating a measurable function against the kernel must produce a measurable function of the starting point. Second, an initial probability law and a kernel must determine a probability measure on the product space. We establish both facts and then calculate with the resulting joint measure.

Definitions and the Kernel Integral

Let $(S,\mathcal S)$ and $(T,\mathcal T)$ be measurable spaces. A probability kernel from $S$ to $T$ is a function $K:S\times\mathcal T\to[0,1]$ such that $B\mapsto K(x,B)$ is a probability measure on $(T,\mathcal T)$ for each $x\in S$, and $x\mapsto K(x,B)$ is $\mathcal S$-measurable for each $B\in\mathcal T$.

Definition: For a nonnegative $\mathcal T$-measurable function $g:T\to[0,\infty]$, define $$ Kg(x)=\int_T g(y)\,K(x,dy),\qquad x\in S. $$ More generally, for a nonnegative $(\mathcal S\otimes\mathcal T)$-measurable function $f:S\times T\to[0,\infty]$, define $$ Kf(x)=\int_T f(x,y)\,K(x,dy). $$

For the second definition, the section $y\mapsto f(x,y)$ is measurable, so the integral is defined for each $x$. The substantive point is that $x\mapsto Kf(x)$ is also measurable. Notice that $f$ may depend on both coordinates; it is not enough to consider only functions of $y$.

Measurability of Kernel Integration

Theorem: If $K$ is a probability kernel and $f:S\times T\to[0,\infty]$ is $\mathcal S\otimes\mathcal T$-measurable, then $x\mapsto\int_T f(x,y)\,K(x,dy)$ is $\mathcal S$-measurable.

Proof. First consider an indicator function $f=1_E$, where $E\in\mathcal S\otimes\mathcal T$. For each $x$, write $E_x=\{y\in T:(x,y)\in E\}$. The section $E_x$ is in $\mathcal T$: the collection of sets $E$ with measurable sections is a sigma-algebra containing every measurable rectangle. Thus $$ K1_E(x)=\int_T 1_E(x,y)\,K(x,dy)=K(x,E_x). $$ We show that this is measurable in $x$. Let $\mathcal D$ be the collection of sets $E\in\mathcal S\otimes\mathcal T$ for which $x\mapsto K(x,E_x)$ is measurable. If $E=A\times B$, then $K(x,E_x)=1_A(x)K(x,B)$, which is measurable by the kernel assumptions. The class $\mathcal D$ contains the whole space, is closed under complements, and is closed under countable disjoint unions. For complements this follows from $K(x,T\setminus E_x)=1-K(x,E_x)$; for disjoint unions it follows from countable additivity and taking limits of partial sums. Measurable rectangles form a pi-system generating $\mathcal S\otimes\mathcal T$. The pi-lambda theorem therefore gives $\mathcal D=\mathcal S\otimes\mathcal T$. Now let $f$ be any nonnegative product-measurable function. There is a sequence of nonnegative measurable simple functions $f_n$ increasing pointwise to $f$. Each $Kf_n$ is measurable, because it is a finite linear combination of functions of the form $K1_E$. By the monotone convergence theorem, $Kf_n(x)$ increases to $Kf(x)$ for every $x$. A pointwise limit of measurable functions is measurable. $\square$

Taking $f(x,y)=g(y)$ gives measurability of $Kg$. A useful additional consequence is that if $f$ is bounded and real-valued, then $Kf$ is measurable: write $f=f^+-f^-$ and apply the theorem to the two nonnegative parts. Both integrals are finite because $K(x,\cdot)$ is a probability measure and $f$ is bounded.

Constructing the Joint Probability Measure

Let $\mu$ be a probability measure on $(S,\mathcal S)$, and let $K$ be a probability kernel from $S$ to $T$. The theorem shows that the function $x\mapsto K(x,E_x)$ is measurable for each $E\in\mathcal S\otimes\mathcal T$, so the following definition makes sense.

Theorem: The formula $$ (\mu\otimes K)(E)=\int_S K(x,E_x)\,\mu(dx), \qquad E\in\mathcal S\otimes\mathcal T, $$ defines a probability measure on $(S\times T,\mathcal S\otimes\mathcal T)$. For $A\in\mathcal S$ and $B\in\mathcal T$, $$ (\mu\otimes K)(A\times B)=\int_A K(x,B)\,\mu(dx). $$

Proof. The formula gives a nonnegative value for every measurable $E$. For the empty set, every section is empty, so $(\mu\otimes K)(\varnothing)=0$. For a measurable rectangle, the section is $B$ when $x\in A$ and is empty otherwise. Therefore $$ (\mu\otimes K)(A\times B) =\int_S 1_A(x)K(x,B)\,\mu(dx) =\int_A K(x,B)\,\mu(dx). $$ To prove countable additivity, let $E_1,E_2,\ldots$ be pairwise disjoint measurable subsets of $S\times T$. Their sections $(E_n)_x$ are pairwise disjoint for every $x$, and $(\bigcup_n E_n)_x=\bigcup_n(E_n)_x$. Countable additivity of $K(x,\cdot)$ gives $$ K\left(x,\left(\bigcup_n E_n\right)_x\right) =\sum_{n=1}^{\infty}K(x,(E_n)_x). $$ Integrating this equality and applying the monotone convergence theorem to the partial sums yields $$ (\mu\otimes K)\left(\bigcup_n E_n\right) =\sum_{n=1}^{\infty}(\mu\otimes K)(E_n). $$ Finally, the section of $S\times T$ is $T$, and $K(x,T)=1$. Hence $$ (\mu\otimes K)(S\times T)=\int_S 1\,\mu(dx)=1. $$ Thus $\mu\otimes K$ is a probability measure, with the stated rectangle formula. $\square$

The notation $\mu\otimes K$ emphasizes how the measure is built: first draw a point according to $\mu$, then use the kernel at that point to specify the distribution of the second coordinate. In general, this joint measure is not the product of two fixed marginal measures.

Integrating with Respect to the Joint Measure

The construction also gives an integration formula. For every nonnegative product-measurable $f$, one has

\int_{S\times T} f(x,y)\,(\mu\otimes K)(d(x,y)) = \int_S\left[\int_T f(x,y)\,K(x,dy)\right]\mu(dx).

For an indicator $f=1_E$, this is exactly the definition of $\mu\otimes K$. Linearity gives the formula for nonnegative simple functions. Approximating a general nonnegative $f$ from below by simple functions and applying monotone convergence on both sides proves the formula. For bounded real-valued $f$, apply it to the positive and negative parts and subtract.

Worked Examples

Worked Example: A Finite Kernel

Let $S=\{a,b\}$, $T=\{c,d\}$, and give both sets their full power-set sigma-algebras. Set $\mu(\{a\})=1/3$ and $\mu(\{b\})=2/3$. Define the kernel by $$ K(a,\{c\})=K(a,\{d\})=\frac12,\qquad K(b,\{c\})=\frac14,\qquad K(b,\{d\})=\frac34. $$ Each row sums to one, so each $K(x,\cdot)$ is a probability measure. The associated joint masses are $$ (\mu\otimes K)(\{a\}\times\{c\})=\frac16,\quad (\mu\otimes K)(\{a\}\times\{d\})=\frac16,\quad (\mu\otimes K)(\{b\}\times\{c\})=\frac16,\quad (\mu\otimes K)(\{b\}\times\{d\})=\frac12. $$ They sum to $1$. For the function with values $f(a,c)=1$, $f(a,d)=0$, $f(b,c)=2$, and $f(b,d)=3$, the joint integral is $$ \int f\,d(\mu\otimes K)=\frac16+0+\frac{2}{6}+\frac32=2. $$ Computing by kernel integration gives $Kf(a)=1/2$ and $Kf(b)=11/4$, so $$ \int_S Kf\,d\mu=\frac13\cdot\frac12+\frac23\cdot\frac{11}{4}=2, $$ as required.

Worked Example: A Continuous Uniform Kernel

Take $S=T=[0,1]$, with their Borel sigma-algebras. For $x>0$, let $K(x,\cdot)$ be the uniform probability measure on $[0,x]$; let $K(0,\cdot)=\delta_0$. To verify the kernel measurability condition, it suffices to check the generating pi-system $(-\infty,t]$ in the Borel subsets of $\mathbb R$. For $x>0$, $$ K(x,(-\infty,t])= \begin{cases} 0,&t<0,\\ t/x,&0\le t<x,\\ 1,&t\ge x, \end{cases} $$ and at $x=0$ this value is $1$ if $t\ge0$, and $0$ otherwise. For each fixed $t$, these are Borel-measurable functions of $x$. The class of Borel sets $B$ for which $x\mapsto K(x,B)$ is measurable is a lambda-system containing that pi-system, so the condition holds for all Borel $B$. Let $\mu$ be uniform on $[0,1]$, and take $f(x,y)=x+y$. For $x>0$, $$ Kf(x)=\frac1x\int_0^x(x+y)\,dy=\frac1x\left(x^2+\frac{x^2}{2}\right)=\frac{3x}{2}. $$ At $x=0$, $Kf(0)=0$, consistent with the same formula. Consequently, $$ \int f\,d(\mu\otimes K)=\int_0^1\frac{3x}{2}\,dx=\frac34. $$ The kernel integral is a measurable function of the starting point, and its integral gives the joint expectation.

Worked Example: A Deterministic Kernel

Let $S=T=[0,1]$ with Borel sigma-algebras, let $\mu$ be uniform measure, and define $K(x,\cdot)=\delta_{x^2}$. For each Borel set $B$, $K(x,B)=1_B(x^2)$, which is measurable because $x\mapsto x^2$ is continuous. The joint measure is concentrated on the curve $\{(x,x^2):0\le x\le1\}$. For any nonnegative Borel function $f$, its integral is $$ \int_{[0,1]\times[0,1]}f(x,y)\,(\mu\otimes K)(d(x,y)) =\int_0^1 f(x,x^2)\,dx. $$ For example, taking $f(x,y)=x+y$ gives $$ \int (x+y)\,d(\mu\otimes K)=\int_0^1(x+x^2)\,dx =\frac12+\frac13=\frac56. $$ This is a joint probability measure even though, given $x$, the second coordinate has no randomness.

Why the Construction Matters

The measurable-kernel integral is the basic mechanism for turning state-dependent distributions into ordinary joint probabilities. The rectangle formula records the joint probability that the first coordinate lies in $A$ and the second in $B$, while the integration formula handles functions depending on both coordinates.

A common mistake is to treat $K(x,B)$ as if it were a constant independent of $x$. The correct rectangle probability is an integral over $A$, not generally $\mu(A)$ times a fixed number. That simplification is valid only when the relevant kernel probability is constant on $A$. Another subtle point is that measurability for each fixed set $B$ alone does not immediately establish measurability for an arbitrary product-measurable event $E$; the section argument and the pi-lambda theorem supply that step.

Check Your Understanding

Use the definitions and proofs above to answer these questions.

For a nonnegative product-measurable $f$, what function of $x$ does kernel integration produce?
Why does checking $x\mapsto K(x,E_x)$ on rectangles suffice to establish its measurability for all product-measurable $E$?
Write the joint measure of a rectangle $A\times B$ in terms of $\mu$ and $K$.
Which property of the kernel and which convergence theorem establish countable additivity of $\mu\otimes K$?
For the deterministic kernel $K(x,\cdot)=\delta_{x^2}$, how can a joint integral of $f$ be written as an integral over $[0,1]$?