If $X_1, …, X_{n+1}$ are independent real random variables and $h:\mathbb{R}^{n+1} \to \mathbb{R}$ a Borel function. Now taking the conditional expectation $\mathbb{E}[h(X_1, \ldots, X_{n+1})| \sigma(X_1, \ldots, X_n)]$
Assuming $\mathbb{E}[|h(X_1, \ldots, X_{n+1})|] < \infty $ show that if
$g(x_1, \ldots, x_n) = \mathbb{E}[h(x_1, \ldots, x_n, X_{n+1})]$ then
$g(X_1, \ldots, X_n)$ is a version of $\mathbb{E}[h(X_1, \ldots, X_{n+1})| \sigma(X_1, \ldots, X_n)]$.
The claim would follow from the definition of conditional expectation, if we can show that
$$ \mathbb{E}[h(X_1, \ldots, X_{n+1}) \mathbf{1}_G] = \mathbb{E}[g(X_1, \ldots, X_{n}) \mathbf{1}_G], $$
for all $G \in \sigma(X_1, \ldots, X_n)$.
What I tried so far is trying to expand the expectations and rewrite them with Fubini's theorem.
Assuming the underlying probability space is $(\Omega, \mathcal{F}, P)$.
I thought I could rewrite with a restricted push forward $P_{|G}^{X_i}(A) = P(X_i^{-1}(A) \cap G)$ to get
$$ \mathbb{E}[h(X_1, \ldots, X_{n+1}) \mathbf{1}_G] = \int_{(x_1, \ldots x_{n+1})\in \mathbb{R}^{n+1}} h(x_1, \ldots, x_{n+1}) dP_{|G}^{X_1, \ldots, X_{n+1}} $$
All variables are $X_i$ independet so we can factor $dP_{|G}^{X_1, \ldots, X_{n+1}} = d \Pi_{i=1}^{n+1} P_{|G}^{X_i}$.
$$ = \int_{(x_1, \ldots x_{n+1})\in \mathbb{R}^{n+1}} h(x_1, \ldots, x_{n+1}) d \Pi_{i=1}^{n+1} P_{|G}^{X_i} = \int_{(x_1, \ldots x_{n})\in \mathbb{R}^{n}} \left(\int_{x_{n+1}\in \mathbb{R}} h(x_1, \ldots, x_{n+1}) dP_{|G}^{X_{n+1}} \right) d \Pi_{i=1}^{n} P_{|G}^{X_i} $$
Now $P_{|G}^{X_{n+1}}(A) = P(X_{n+1}^{-1}(A)\cap G) = P(X_{n+1}^{-1}(A))P(G)$, since $\sigma(X_{n+1})$ and $\sigma(X_1,\ldots, X_n)$ are independet. Since $\mathbb{E}[|h(X_1, \ldots, X_{n+1})|] < \infty $ Fubini applies and we get
$$ \int_{(x_1, \ldots x_{n})\in \mathbb{R}^{n}} \left(\int_{x_{n+1} \in \mathbb{R}} h(x_1, \ldots, x_{n+1}) dP_{|G}^{X_{n+1}} \right) d \Pi_{i=0}^{n} P_{|G}^{X_i} = P(G)\int_{(x_1, \ldots x_{n})\in \mathbb{R}} \left(\int_{x_{n+1} \in \mathbb{R}} h(x_1, \ldots, x_{n+1}) dP^{X_{n+1}} \right) d \Pi_{i=0}^{n} P_{|G}^{X_i} $$
$$ = P(G) \int_{(x_1, \ldots x_{n}) \in \mathbb{R}^n} E[h(x_1, \ldots, X_{n+1})] d \Pi_{i=0}^{n} P_{|G}^{X_i} = P(G) \int_{(x_1, \ldots x_{n})\in \mathbb{R}^n} g(x_1, \ldots, x_n) d \Pi_{i=0}^{n} P_{|G}^{X_i} = P(G)\mathbb{E}[g(X_1, \ldots, X_n) \mathbf{1}_G]. $$
So I the end I get
$$ \mathbb{E}[h(X_1, \ldots, X_{n+1}) \mathbf{1}_G] = P(G)\mathbb{E}[g(X_1, \ldots, X_n) \mathbf{1}_G],$$ which is wrong according to the claim!
I can't spot my mistake. So is the claim really true and if so where do I mess up?
Best Answer
Denote $X=(X_1,\dots,X_n)\sim \mu$ and $Y=X_{n+1}\sim \nu$, then $X,Y$ are independent and $(X,Y)\sim \mu \times \nu$. Also, $$ g(x)=\int_{\mathbb{R}}h(x,y)\nu(dy) \quad x\in \mathbb{R}^n.$$ Thus $$g(X)=\mathbb{E}[h(X,Y)|X].$$
In fact, $g(X)$ is $\sigma(X)$-measurable and $$ \mathbb{E}[g(X)1_{\{X\in A\}}]= \mathbb{E}[g(X)1_A(X)]= \int_{\mathbb{R}^n} g(x)1_A(x) \mu(dx)=\int_{\mathbb{R}^n} \left( \int_{\mathbb{R}} h(x,y)\nu(dy)\right)1_A(x)\mu(dx) = \int_{\mathbb{R}^n\times \mathbb{R}} h(x,y)1_A(x)\mu\times \nu(dx\times dy) = \mathbb{E}[h(X,Y)1_A(X)]= \mathbb{E}[h(X,Y)1_{\{X\in A\}}]$$ for all $A\in \mathcal{B}_{\mathbb{R}^n}$.