Why Condition on a Random Element?
A conditional probability given an event is a single number: it describes probabilities after restricting attention to that event. In many problems, the information being conditioned on is not one event but the value of a random element. The resulting conditional law must describe how the distribution of another random element changes across the possible values of the one observed.
A probability kernel provides the appropriate object. It assigns a probability measure to each possible conditioning value, with a measurability requirement that makes those measures usable in integration. Its defining integral identity connects the pointwise assignment to the joint law of the random elements.
Definition of a Conditional Distribution
Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space, and let \(X:\Omega\to S\) and \(Y:\Omega\to T\) be random elements. Assume \(S\) and \(T\) are standard Borel spaces. Write \(\mathbb P_X\) for the distribution of \(X\). A probability kernel from \(S\) to \(T\) assigns a probability measure \(K(x,\cdot)\) on \(T\) to each \(x\in S\), and for each measurable \(B\subseteq T\), the function \(x\mapsto K(x,B)\) is measurable.
The identity says that the joint probability of \(X\) falling in \(A\) and \(Y\) falling in \(B\) is obtained by averaging the conditional probability \(K(x,B)\) over the distribution of \(X\) on \(A\). It is an integral requirement, not a separate assertion that can be checked by conditioning on each individual value \(X=x\).
For a fixed measurable \(B\subseteq T\), the function \(K(X,B)\) is a version of the conditional expectation \(\mathbb E[\mathbf 1_{\{Y\in B\}}\mid\sigma(X)]\). Indeed, it is measurable with respect to \(\sigma(X)\), and the defining identity gives the required integrals over every event \(\{X\in A\}\). This is the connection between conditional distributions and conditional expectation.
What the Kernel Is—and Is Not
The kernel \(K\) is a family of probability measures indexed by points \(x\in S\). It is not the probability of conditioning on the event \(\{X=x\}\). When that event has positive probability, conditioning on it is possible by a ratio, and the result agrees with an average of the kernel over that event. But for a continuously distributed \(X\), individual events \(\{X=x\}\) often have probability zero, so the ratio definition cannot be used there.
The notation \(\mathbb P(Y\in B\mid X=x)\) is often used as shorthand for \(K(x,B)\). It should be read as a value of the conditional kernel, not automatically as a ratio involving the event \(\{X=x\}\). The integral identity is what gives the kernel its probabilistic meaning.
Worked Example: A Finite Conditioning Variable
Suppose \(X\) takes values \(0,1,2\) with probabilities \(0.2,0.5,0.3\), respectively. Let \(Y\) take values in \(\{0,1\}\), and suppose the conditional probabilities of \(Y=1\) at these three values are \(0.1,0.4,0.8\). Define \(K(i,\{1\})\) to be the corresponding number, and set \(K(i,\{0\})=1-K(i,\{1\})\).
For example, take \(A=\{1,2\}\) and \(B=\{1\}\). The kernel identity gives $$ \int_A K(x,B)\,\mathbb P_X(dx) =0.5(0.4)+0.3(0.8) =0.44. $$ This is exactly \(\mathbb P(X\in\{1,2\},Y=1)\), since the contributions from \(X=1\) and \(X=2\) are \(0.5(0.4)\) and \(0.3(0.8)\).
Conditioning on the event \(X\in\{1,2\}\) instead gives $$ \mathbb P(Y=1\mid X\in\{1,2\}) =\frac{0.44}{0.5+0.3} =0.55. $$ This is an average of two kernel values, not either value on its own: \(K(1,\{1\})=0.4\) and \(K(2,\{1\})=0.8\).
The Integral Identity and Conditional Expectation
The defining identity applies first to events \(Y\in B\). It also determines conditional expectations of measurable functions of \(Y\). For a nonnegative measurable function \(f:T\to[0,\infty]\), define $$ Kf(x)=\int_T f(y)\,K(x,dy). $$ The kernel integration result ensures that \(Kf\) is measurable.
First take \(f=\mathbf 1_B\) for a measurable set \(B\subseteq T\). The claimed identity is precisely the defining identity for \(K\), since \(Kf(x)=K(x,B)\).
By linearity of integrals, the identity follows for every nonnegative simple function \(f\), which is a finite sum of nonnegative multiples of indicator functions. For a general nonnegative measurable \(f\), choose an increasing sequence of nonnegative simple functions \(f_n\) with \(f_n\uparrow f\). Then \(Kf_n(x)\uparrow Kf(x)\) for each \(x\), by monotone convergence for the measure \(K(x,\cdot)\). Applying monotone convergence to both sides of the identity for \(f_n\) gives $$ \mathbb E[\mathbf 1_{\{X\in A\}}f(Y)] = \int_A Kf(x)\,\mathbb P_X(dx). $$ The random variable \(Kf(X)\) is \(\sigma(X)\)-measurable, and this equality holds for every event \(\{X\in A\}\). Such events form \(\sigma(X)\), so the defining property of conditional expectation shows that \(Kf(X)\) is a version of \(\mathbb E[f(Y)\mid\sigma(X)]\). For integrable real-valued \(f\), apply the nonnegative result to the positive and negative parts of \(f\), both of which have finite expectation. \(\square\)
This result explains why the kernel contains more than a list of conditional probabilities for events: it lets us compute conditional expectations of functions of \(Y\) by integrating those functions against the conditional measure.
Conditioning on an Event Is Averaging the Kernel
When an event determined by \(X\) has positive probability, the event-conditioned distribution is obtained by averaging the kernel over the corresponding set of values.
By the definition of conditional probability on an event of positive probability and the defining identity for \(K\), $$ \mathbb P(Y\in B\mid X\in A) =\frac{\mathbb P(X\in A,Y\in B)}{\mathbb P(X\in A)} =\frac{\displaystyle\int_A K(x,B)\,\mathbb P_X(dx)}{\mathbb P_X(A)}. $$ The denominator is positive by assumption, so the ratio is defined. \(\square\)
Worked Example: A Continuous Conditioning Variable
Let \(X\) and \(U\) be independent uniform random variables on \([0,1]\), and set \(Y=X+U\). For \(x\in[0,1]\), define $$ K(x,B)=\int_0^1 \mathbf 1_B(x+u)\,du $$ for each Borel set \(B\subseteq\mathbb R\). This is the distribution of \(x+U\) when \(x\) is fixed: it is uniform on \([x,x+1]\). The map \(x\mapsto K(x,B)\) is measurable, since it is the integral with respect to \(u\) of the jointly measurable function \((x,u)\mapsto\mathbf 1_B(x+u)\). For each \(x\), \(K(x,\cdot)\) is a probability measure, as it is the distribution of \(x+U\).
To check the integral identity, let \(A\subseteq[0,1]\) and \(B\subseteq\mathbb R\) be Borel. Independence and the uniform distributions give $$ \mathbb P(X\in A,Y\in B) =\int_A\int_0^1\mathbf 1_B(x+u)\,du\,dx =\int_A K(x,B)\,\mathbb P_X(dx). $$ Thus \(K\) is a conditional distribution of \(Y\) given \(X\). In particular, \(K(x,[x,x+1])=1\), and integrating \(y\) against this uniform distribution gives \(Kf(x)=x+\tfrac12\) for \(f(y)=y\). Accordingly, \(\mathbb E[Y\mid\sigma(X)]=X+\tfrac12\).
For each fixed \(x\), \(\mathbb P(X=x)=0\). The event-ratio formula therefore does not define \(\mathbb P(Y\in B\mid X=x)\). The kernel nevertheless gives a conditional law indexed by every \(x\in[0,1]\), with its meaning guaranteed by the integral identity.
Two Further Patterns
Worked Example: A Deterministic Function of the Conditioning Variable
Suppose \(g:S\to T\) is measurable and \(Y=g(X)\). Define \(K(x,B)=\mathbf 1_B(g(x))\). For each fixed \(x\), this is the point mass at \(g(x)\), and for each Borel \(B\), it is measurable as a function of \(x\).
For measurable \(A\subseteq S\) and \(B\subseteq T\), $$ \int_A K(x,B)\,\mathbb P_X(dx) =\int_A\mathbf 1_B(g(x))\,\mathbb P_X(dx) =\mathbb P(X\in A,\ g(X)\in B) =\mathbb P(X\in A,\ Y\in B). $$ Hence \(K\) satisfies the required identity. In this case, once \(X=x\) is specified, the conditional distribution is concentrated entirely at \(g(x)\).
These examples highlight two different ways a conditional law can behave: it can retain randomness after conditioning, as in the uniform shift, or it can collapse to a point mass when the conditioned variable determines the response.
Why the Definition Requires a Kernel
For each fixed set \(B\), the integral identity constrains the function \(x\mapsto K(x,B)\) only through integrals with respect to \(\mathbb P_X\). Values at conditioning points that have probability zero do not affect those integrals. This is one reason not to treat a kernel value at such a point as an event-ratio probability.
There is also a coherence requirement: as \(B\) varies, \(K(x,B)\) must be the values of a probability measure for each \(x\), and for each \(B\) the values must vary measurably with \(x\). Knowing conditional expectations separately for individual indicators does not, by itself, state this measure-valued structure. The kernel packages the conditional probabilities into one object that can be integrated and used consistently.
Check Your Understanding
Use the defining identity and examples to check your understanding.
- What are the two requirements that make \(K\) a probability kernel from \(S\) to \(T\)?
- How does the defining identity show that \(K(X,B)\) is a version of a conditional expectation?
- Why is conditioning on \(X=x\) by an event-probability ratio unavailable when \(\mathbb P(X=x)=0\)?
- How is \(\mathbb P(Y\in B\mid X\in A)\) computed from the kernel when \(\mathbb P(X\in A)>0\)?
- For \(Y=g(X)\), what probability measure does the kernel assign at \(x\)?