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This is Exercise 44.2 from Bauer's Probability Theory.

Let $X: ( \Omega, \mathscr{A}) \to (E, \mathscr{B}(\Omega))$ and $Y:(\Omega,\mathscr{A}) \to (\Omega', \mathscr{A}')$ be random variables on a probability space $(\Omega,\mathscr{A}, P)$ with values in a Polish space $E$ or a measurable space $(\Omega', \mathscr{A}')$ , respectively. Then denote $Q$ as the Markov kernel from $(\Omega', \mathscr{A}')$ to $(E, \mathscr{B}(E))$ such that for every $B \in \mathscr{B}(E)$, $y \mapsto Q(y,B)$ is a version of $P\{X \in B | Y = y\}.$ For each $y \in \Omega'$ the probability measure $B \mapsto Q(y,B)$ is then called the conditional distribution of $X$ under the hypothesis that $Y$ equals $y$. This measure is denoted by $P_{X|Y=y}.$ Thus for every $B \in \mathscr{B}(E)$ $$P_{X|Y=y}(B) = P \{X \in B | Y = y\} \; P_Y\text{-almost surely.}$$

Now for each $y \in \Omega'$ designate by $\mu_y$ the probability measure $P_{X|Y=y}.$ Prove that the joint distribution $P_{X \otimes Y}$ of $X$ and $Y$ can be computed by the formula

$$P_{X \otimes Y} (M) = \int\left[ \int 1_M (x,y) \mu_y(dx)\right]P_Y(dy) \; (M \in \mathscr{B}(E) \otimes \mathscr{A}').$$ In doing so, observe that $y \mapsto \int 1_M (x,y) \mu_y(dx)$ is $\mathscr{A}'$-measurable.

I came across this exercise while looking at conditional distributions. However, I am lost as to how to show this obvious identity formally using only the definition of a conditional distribution, i.e., $\int_C Q(y,B) P_Y(dy) = P ( \{X \in B\} \cap \{Y \in C\})$ for all $B \in \mathscr{B}(E)$ and $C \in \mathscr{A'}$. I would greatly appreciate any help.

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  • $\begingroup$ Not sure if this will crack the puzzle, but is there a place for rewriting the conditional distribution in a form along the lines of $f(x | y) = \frac{f(x,y)}{f(y)}$ etc. I hope this helps. $\endgroup$ – ad2004 Nov 30 '19 at 21:51
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It is sufficient to prove the statement for $M = B\times C$ with $B \in \mathscr{B}(E)$ and $C \in \mathscr{A'}$ (a "rectangular" set), since these sets form a generator of $\mathscr{B}(E) \otimes \mathscr{A}'$ which is closed under finite intersections. The proof goes like this: \begin{align*} P_{X\otimes Y}(M) &= P([X\in B]\cap [Y\in C]) \\ &\stackrel{\text{your equation}}{=} \int_C Q(y,B)\, P_Y(dy) \\ &= \int_{\Omega'} 1_C(y)\, Q(y,B)\, P_Y(dy) \\ &= \int_{\Omega'} 1_C(y) \int_E 1_B(x)\, \mu_y(dx)\, P_Y(dy) \\ &= \int_{\Omega'} \int_E 1_M(x,y)\, \mu_y(dx)\, P_Y(dy). \end{align*} To be slightly more precise about the "sufficient" part above, you can thus define a pre-measure on the set of finite unions of rectangular sets and the statement follows from Caratheodory's extension theorem.

Hope this helps!

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  • $\begingroup$ Thank you. This clears it up. Can you help me understand why $y \mapsto \int 1_M(x,y) \mu_y(dx)$ is $\mathscr{A}'$-measurable? Is this simply from Fubini? $\endgroup$ – nomadicmathematician Dec 1 '19 at 23:30
  • $\begingroup$ I think it is a consequence of the Fubini-Tonelli theorem, but I didn't figure out the details. It is implicit in its statement (of e.g. the Wikipedia article en.wikipedia.org/wiki/…) that the maps $x\mapsto \int f(x,y) dy$ and $y\mapsto \int f(x,y) dx$ are measurable/integrable (which is a consequence of the theorem, not an assumption!). $\endgroup$ – iljusch Dec 1 '19 at 23:43

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