Conditional law given a random element

Let $X:(\mathrm{\Omega },\mathcal{F},P)\to (E,\mathcal{S})$ and $Y:(\mathrm{\Omega },\mathcal{F},P)\to (T,\mathcal{T})$ be measurable random elements. A conditional law of X given Y is a probability kernel $K:(T,\mathcal{T})\to (E,\mathcal{S})$ such that $\omega \mapsto K(Y(\omega ),\cdot )$ is a regular conditional distribution of X given $\sigma (Y)$ in Regular conditional distribution. Thus for every $A\in \mathcal{S}$ and $H\in \sigma (Y)$, ${\int }_{H}K(Y(\omega ),A)dP=P(H\cap \{X\in A\})\mathrm{.}$

Composition with Y makes each evaluation $\sigma (Y)$-measurable, and every section remains a probability by Measure kernel and probability kernel. The notation $P(X\in A\mid Y=y)=K(y,A)$ refers to a chosen kernel; it is not a ratio involving the possibly zero probability $P(Y=y)$.

Write $P_Y(B)=P(Y^{-1}(B))$ for the law in Law or distribution of a random element. If a specified countable family $\mathcal{C}\subseteq \mathcal{S}$ determines probability measures, two conditional laws K,L agree as measures outside a single $\mathcal{T}$-measurable $P_Y$-null set. Indeed for each $A\in \mathcal{C}$ the measurable discrepancy $D_A=\{y:K(y,A)\mathrm{\ne }L(y,A)\}$ has null inverse image under Y by Conditional expectation is unique almost surely. Hence $P_Y(D_A)=0$ by the law definition; the countable union $D$ is null by Finite and countable subadditivity of measures, and off D the determining property gives equality of measures. This uses a supplied determining family and makes no AC assertion about obtaining one.

Values on a measurable $P_Y$-null subset may be replaced by a specified fixed probability on E without changing these identities. The inverse image of that subset is measurable null, and all event evaluations are bounded, so the modified event integrals agree. Empty-event evaluations remain zero and whole-target evaluations remain one. Existence on standard-Borel spaces is proved separately; the definition alone does not assert a conditional kernel exists.