# A problem of regular distribution

This is a exercise of Shiryaev’s Probability on page 233:

Suppose that the random elements $(X, Y)$ are such that there is a regular distribution $P_x(B)=P(Y\in B\mid X=x)$. Show that if $E|g(X,Y)|<\infty$ then
$$E[g(X,Y)\mid X=x]=\int g(x,y)P_x(dy) \text{ (P_x-a.s.)}$$

PS: I think there is a typo. Should we replace “$P_x$-a.s.” by “$P_X$-a.s.”?

I firstly tried to prove it for indicator function. For $g(x,y)=I_B(x,y)$, $B\in \mathcal{B}(R^2)$, I should prove for any $A\in \mathcal{B}(R)$,
$$\int_A E[I_B(X,Y)\mid X=x]P_X(dx)=\int_A\int I_B(x,y)P_x(dy)P_X(dx)$$
Then,
$$LHS=\int_{\{X\in A\}}I_B(X,Y)dP$$
For the RHS, let $B_x=\{y\mid (x,y)\in B\}$, I deduce that
\begin{align} RHS &=\int_A\int I_{B_x}(y)Px(dy)P_X(dx)\\ &=\int_A P_x(B_x)P_X(dx)\\ &=\int_A P(Y\in B_x\mid X=x)P_X(dx)\\ &=\int_A E(I_{B_x}(Y)\mid X=x)P_X(dx)\\ &=\int_A E(I_{B}(x,Y)\mid X=x)P_X(dx) \end{align}
This problem is the origin of my another problem: “An equality about conditional expectation”

In that thread, did has pointed out that $E(I_{B}(x,Y)\mid X=x)$ is meaningless, I agree with him. But what’s wrong with my deduction above?

By the way, in the thread mentioned above, StefanHansen have recommend me the book Probability With a View Towards Statistics, Volume II by J. Hoffmann-Jørgensen to study the concept of regular conditional probabilities/distributions, but I can’t find a copy of it around me. Can anyone introduce me other materials which has an excellent treatment of regular conditional probabilities/distributions?

Thank you very much!

#### Solutions Collecting From Web of "A problem of regular distribution"

I want to answer my question by myself:

$\forall B\in \mathcal{B}(\mathbb{R}^2)$ and $\forall A\in \mathcal{B}(\mathbb{R})$, let
$$\mathcal{C}=\{B\in \mathcal{B}(\mathbb{R}^2)\mid x\mapsto\int_\mathbb{R}I_B(x,y)P_x(dy)\text{ is \mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})-measurable function and }\int_A E[I_B(X,Y)\mid X=x]P_X(dx)=\int_A\int_\mathbb{R}I_B(x,y)P_x(dy)P_X(dx)\}$$
$\forall A_1, A_2\in \mathcal{B}(\mathbb{R})$, then
$$\int_\mathbb{R}I_{A_1\times A_2}(x,y)P_x(dy)=I_{A_1}(x)\int_\mathbb{R}I_{A_2}(y)P_x(dy)=I_{A_1}P_x(A_2)$$
is $\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable function. Then
$$\int_A E[I_B(X,Y)\mid X=x]P_X(dx)=\int_{\{X\in A\}}I_{A_1\times A_2}(X,Y)P(d\omega)=P(\{Y\in A_2\}\cap\{X\in A\cap A_1\})$$
$$\int_A\int_\mathbb{R}I_{A_1\times A_2}(x,y)P_x(dy)P_X(x)=\int_A I_{A_1}(x)P(Y\in A_2\mid X=x)P_X(dx)=P(\{Y\in A_2\}\cap\{X\in A\cap A_1\})$$
So $A_1\times A_2\in\mathcal{C}.$
Let
$$\mathcal{L}=\{A_1\times A_2\mid A_1,A_2\in\mathcal{B}(\mathbb{R})\}$$
then $\mathcal{L}$ is a $\pi$-system and $\mathcal{L}\subset\mathcal{C}$.

$\forall B_1,B_2\in\mathcal{C}$ and $B_1\subset B_2$, then
$$\int_\mathcal{R}I_{B_2\backslash B_1}(x,y)P_x(dy)=\int_\mathcal{R}I_{B_2}(x,y)P_x(dy)-\int_\mathcal{R}I_{B_1}(x,y)P_x(dy)$$
is $\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable function, and
$$\int_A E[I_{B_2\backslash B_1}(X,Y)\mid X=x]P_X(dx)=\int_A E[I_{B_2}(X,Y)\mid X=x]P_X(dx)-\int_A E[I_{B_1}(X,Y)\mid X=x]P_X(dx)=\int_A\int_\mathcal{R}I_{B_2}(x,y)P_x(dy)P_X(dx)-\int_A\int_\mathcal{R}I_{B_1}(x,y)P_x(dy)P_X(dx)=\int_A\int_\mathcal{R}I_{B_2\backslash B_1}(x,y)P_x(dy)P_X(dx)$$
so $B_2\backslash B_1\in\mathcal{C}$.

Let $B_1,B_2,…\in\mathcal{B}(\mathbb{R^2})$ and $B_n\uparrow B$, then by MCT
$$\int_\mathbb{R}I_B(x,y)P_x(dy)=\int_\mathbb{R}\lim_{n\to\infty} I_{B_n}(x,y)P_x(dy)=\lim_{n\to\infty}\int_\mathbb{R}I_{B_n}(x,y)P_x(dy)$$
is $\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable function and
$$\int_A E[I_B(X,Y)\mid X=x]P_x(dx)=\lim_{n\to\infty}\int_A E[I_{B_n}(X,Y)\mid X=x]P_X(dx)=\lim_{n\to\infty}\int_A\int_\mathbb{R}I_{B_n}(x,y)P_x(dy)P_X(dx)=\int_A\int_\mathbb{R}I_{B}(x,y)P_x(dy)P_X(dx)$$
so $B\in\mathcal{C}$.

Then $\mathcal{C}$ is a $\lambda$-system. So $\mathcal{C}=\mathcal{B}(\mathbb{R}^2)$.

We have prove the problem for all indicator functions, then by the routin method, we can extense to all simple functions, then to all absolutly integrable funciton.