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Let $(\Omega, \mathcal{A}, \mathbf{P})$ be a probability space with a measurable function $Y: (\Omega, \mathcal{A}) \rightarrow (E, \mathcal{E})$ and another measurable function $X: (\Omega, \mathcal{A}) \rightarrow (E', \mathcal{E'})$.

Let $\kappa_{Y\mid\sigma(X)}$ be a regular conditional distribution of $Y$ given $\sigma(X)$, i.e. a stochastic kernel from $(\Omega, \sigma(X))$ to $(E, \mathcal{E})$ such that $\kappa_{Y\mid\sigma(X)}(\omega, B)=\mathbf{P}[{Y \in B}\mid \sigma(X)](\omega)$ for $\mathbf{P}$-almost all $\omega\in\Omega$ and $B\in\mathcal{E}$.

Can we infer from the existence of $\kappa_{Y\mid\sigma(X)}$ the existence of a regular conditional distribution of $Y$ given $X$, i.e. a stochastic kernel $\kappa_{Y\mid X}$ from $(E', \mathcal{E}')$ to $(E, \mathcal{E})$, such that $\kappa_{Y\mid X}(x, B)=\mathbf{P}[{Y \in B}\mid X = x]$ for $P^X$-almost all $x$ and all $B\in\mathcal{E}$ ?

Is there always a measurable function $X^{-1}: (E', \mathcal{E}) \rightarrow (\Omega, \sigma(X))$ such that $\kappa_{Y\mid X}(x, B)=\kappa_{Y\mid \sigma(X)}(X^{-1}(x), B)$?

In the book Probability theory Klenke seems to suggest in definition 8.28 that a $\kappa_{Y\mid X}$ can be constructed from $\kappa_{Y\mid \sigma(X)}$ by applying the factorization lemma and defining $\kappa_{Y\mid X}(x, \centerdot)$ to an arbitrary probability measure for $x\notin X(\Omega)$. Dembo seems to argue similarly in the solution to exercise 4.4.5 in his lecture notes.

However, the factorization lemma only seems applicable for a fixed $B\in \mathcal{E}$, i.e. for each $B$ we can obtain an $f_B: E' \rightarrow E$ such that $\kappa_{Y\mid \sigma(X)}(\centerdot, B) = f_B \circ X(\centerdot)$, and I don't see how a definition like $\kappa_{Y\mid X}(x, B):= f_B(x)$ ensures that $\kappa_{Y\mid X}(x, \centerdot)$ is a probability measure for a fixed $x$. Am I missing something?

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  • $\begingroup$ Just f.y.i, you may be interested in the discussion here, where a (related) answer via an elementary argument is proposed. $\endgroup$ Commented Mar 21, 2016 at 19:30
  • $\begingroup$ @David did you mean to link to a different question? $\endgroup$ Commented Apr 13, 2022 at 11:09

2 Answers 2

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Following the notation of the Klenke's book:

Given $B\in \mathcal{E},$ let $\varphi\left(\cdot,B\right)$ be a measurable function such that $\kappa_{Y,\sigma\left(X\right)}\left(\omega,B\right)=\varphi\left(X\left(\omega\right),B\right)$ for all $\omega \in \Omega.$ Note that $\kappa_{Y,\sigma\left(X\right)}\left(\omega,B\right)$ is constant on the set $\{ \omega: X\left(\omega\right)=x\}$ if $x\in \mathcal{E'}$, hence $\varphi\left(x,\cdot\right)$ is a probability measure on $\left(E,\mathcal{E}\right)$ for every $x\in X\left(\Omega\right).$ For every $x\in \left(X\left(\Omega\right)\right)^{c},$ set $\varphi\left(x,\cdot\right)=\mu\left(\cdot\right),$ where $\mu$ is an arbitrary probability measure on $\left(E,\mathcal{E}\right).$ Therefore, $\varphi$ is a stochastic kernel from $\left(E',\mathcal{E}'\right)$ to $\left(E,\mathcal{E}\right).$

By the definition of $\kappa_{Y,\sigma\left(X\right)}$ and the transformation theorem (theorem 4.10):

$$ \mathbb{P}\left(\{Y\in B\}\cap A\right)=\int_{A} \kappa_{Y,\sigma\left(X\right)}\left(\cdot,B\right)d\mathbb{P}= \int_{A} \varphi\left(X\left(\cdot\right),B\right)d\mathbb{P}=\int_{X^{-1}\left(A\right)} \varphi\left(\cdot,B\right)d\mathbb{P}^{X},$$ for every $A\in \sigma\left(X\right).$ Also, by definition:

$$ \mathbb{P}\left(\{Y\in B\}\cap A\right)=\int_{A} \mathbb{P}\left(Y\in B| \sigma\left(X\right)\right)\left(\cdot\right)d\mathbb{P}= \int_{A} \mathbb{P}\left(Y\in B|X=x\right)_{x=X\left(\cdot\right)}d\mathbb{P}=\int_{X^{-1}\left(A\right)}\mathbb{P}\left(Y\in B|X=\cdot\right)d\mathbb{P}^{X}.$$

It is now clear that $\kappa_{Y,X}\left(x,B\right):=\varphi\left(x,B\right)=\kappa_{Y,\sigma\left(X\right)}\left(X^{-1}\left(x\right),B\right)=\mathbb{P}\left(Y\in B|X=x\right)$ for almost $\mathbb{P}^{X}\mbox{-a.a } x\in \mathbb{R}.$

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    $\begingroup$ If $A\in\sigma(X)\subseteq\mathcal A$, then $X^{-1}(A)$ is undefined, unless $E'=\Omega$. $\endgroup$ Commented Nov 9, 2018 at 21:34
  • $\begingroup$ Moreover, the OP didn't say anything about $(E,\mathcal E')$. If it is a general measurable space, $\left\{x\right\}$ might not belong to $\mathcal E'$ for $x\in E'$. So, we need to assume that the singleton sets in $E$ belong to $\mathcal E'$ (which is the case, for example, if $E'$ is a $T_1$-space and $\mathcal E'=\mathcal B(E')$). $\endgroup$ Commented Nov 9, 2018 at 21:38
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The OP didn't say anything about $(E',\mathcal E')$. If it is a general measurable space, $\left\{x\right\}$ might not belong to $\mathcal E'$. for $x\in E′$.

So, we need to assume that the singleton subsets of $E$ belong to $\mathcal E′$, which is the case, for example, if $E′$ is a $T_1$-space and $\mathcal E′=\mathcal B(E′)$.

Once the existence of a Markov kernel $\varphi$ from $(E',\mathcal E')$ to $(E,\mathcal E)$ with $$\operatorname P\left[Y\in B\mid X\right]=\varphi(X,B)\;\;\;\text{almost surely for all }B\in\mathcal E$$ is established (which can be done as shown by Fabio Andrés Gómez; but note that we assume the existence of a regular version of the conditional probability of $Y$ given $\sigma(X)$, which cannot be guaranteed in this general setting), it's obvious that $$\operatorname P\left[X=x,Y\in B\right]=\operatorname P\left[X=x\right]\varphi(x,B)$$ and hence $$\operatorname P\left[Y\in B\mid X=x\right]=\varphi(x,B)$$ for all $(x,B)\in E'\times\mathcal E$.

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