# Proving that the joint probability of $X$ and $Y$ is the integration over $Y$ of the conditional distribution of $X$ given $Y=y$.

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.

• 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. – ad2004 Nov 30 '19 at 21:51

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.
• 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? – nomadicmathematician Dec 1 '19 at 23:30
• 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!). – iljusch Dec 1 '19 at 23:43