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I understand how Fubini Theorem works when we have product space $\left(\Omega_1\times\Omega_2, \mathcal{A}_1\times\mathcal{A}_2,P_1\times P_2\right)$. I am looking at a slightly different question.

Suppose I have a probability triple $\left(\times_1^{2n} \Omega, \times_1^{2n}\mathcal{A},\times_1^{2n}P\right)$. I have $2n$ r.v.'s $X_1,\cdots,X_n,Y_1,\cdots,Y_{n}$ defined on this space, and for each $i$, $X_i$ only depends on the $i$-th coordinate. Let $f$ be a measurable function. In what sense, is the following true?

$$\mathbb{E}\left[f\left(X_1,\cdots,Y_n\right)\right] = \int_{\Omega_{n+1}\times\cdots\Omega_{2n}} \left\{\int_{\Omega_1\times\cdots\Omega_{n}}f\left(X_1,\cdots,Y_n\right)\,d\left(P_1\times\cdots P_n\right)\right\}\,d\left(P_{n+1}\times\cdots P_{2n}\right)$$

Does Fubini theorem apply here? Namely, is it true that

$$\int_{\Omega_1\times\cdots\Omega_n} f\left(X_1,\cdots,X_n,Y_1,\cdots,Y_n\right)\,d \left(\times_{i=1}^n P_i\right)$$

is a random variable on $\Omega_{n+1}\times\cdots\Omega_{2n}$? How would Fubini applies in this situation?

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I think at the very beginning, it should be $\sigma(A_{1}\times A_{2})$ rather than $A_{1}\times A_{2}$. Given $\sigma$-algebras $A_{1}$ and $A_{2}$, in general $A_{1}\times A_{2}$ is not a $\sigma$-algebra.

For the question, let's consider $n=2$, so we have four variables $x_{1},x_{2},y_{1},y_{2}$.

There is a canonical identification that $(x_{1},x_{2},y_{1},y_{2})=((x_{1},x_{2}),(y_{1},y_{2}))$, as a result one can actually prove that $\sigma(A_{1}\times A_{2}\times A_{3}\times A_{4})=\sigma(\sigma(A_{1}\times A_{2})\times\sigma(A_{3}\times A_{4}))$.

For the measures issue, with that identification, one can prove that $P_{1}\otimes P_{2}\otimes P_{3}\otimes P_{4}=(P_{1}\otimes P_{2})\otimes(P_{3}\otimes P_{4})$.

If $f$ is measurable with respect to $\sigma(A_{1}\times A_{2}\times A_{3}\times A_{4})$ and integrable, with that identification and the usual Fubini Theorem, one has \begin{align*} &\int_{\Omega_{1}\times\Omega_{2}\times\Omega_{3}\times\Omega_{4}}f(x_{1},x_{2},y_{1},y_{2})d(P_{1}\otimes P_{2}\otimes P_{3}\otimes P_{4})\\ &=\int_{\Omega_{1}\times\Omega_{2}}\left(\int_{\Omega_{3}\times\Omega_{4}}f((x_{1},x_{2}),(y_{1},y_{2}))d(P_{3}\otimes P_{4})\right)d(P_{1}\otimes P_{2})\\ &=\int_{\Omega_{3}\times\Omega_{4}}\left(\int_{\Omega_{1}\times\Omega_{2}}f((x_{1},x_{2}),(y_{1},y_{2}))d(P_{1}\otimes P_{2})\right)d(P_{3}\otimes P_{4}). \end{align*}

For random variables $X_{1},X_{2},X_{3},X_{4}$, if $f$ is such that $f(X_{1},X_{2},Y_{1},Y_{2})$ is measurable and integrable, then the corresponding formula holds.

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  • $\begingroup$ I probably misused notation. The book I have use $\mathcal{A}^n$ to represent "the product $\sigma$-field generated from sets of the from $A_1\times\cdots\times A_n$, where $A_i \in \mathcal{A}$". Anyway, I wonder where we use the fact that $X_i$ only depends on the $i$-th coordinate? $\endgroup$ – Yuki Kawabata Jan 27 '18 at 4:51
  • $\begingroup$ I overlooked your sentence that "depends on the $i$-th coordinate", by the way, then what is the domain of each $X_{i}$? $\endgroup$ – user284331 Jan 27 '18 at 4:52
  • $\begingroup$ If I'm understanding my textbook correctly, all $X_i$'s (as well as $Y_i$'s) are defined on the big product space, and have domain $\mathbb{R}$. In my case, all $\Omega_i$'s. $\mathcal{A}_i$'s, and $P_i$'s are the same. $\endgroup$ – Yuki Kawabata Jan 27 '18 at 5:01
  • $\begingroup$ You have several combinations. If $X_{1}=X_{1}(x_{1},x_{2},y_{1},y_{2})$ (which means its domain is the whole $\Omega_{1}\times\cdots\Omega_{4}$) and so are those $X_{2},...$, then I require that $\varphi_{1}(x_{1},x_{2},y_{1},y_{2})=f(X_{1}(x_{1},...),X_{2}(x_{1},...),Y_{1}(x_{1},...),Y_{2}(x_{1},...))$ to be measurable and integrable. $\endgroup$ – user284331 Jan 27 '18 at 5:12
  • $\begingroup$ Or you have the combination like $X_{1}=X_{1}(x_{1})$ (which means its domain is only $\Omega_{1}$) and so are those $X_{2},...$, then I require that $\varphi_{2}(x_{1},...,y_{2})=f(X_{1}(x_{1}),...,Y_{2}(y_{2}))$ to be measurable and integrable. $\endgroup$ – user284331 Jan 27 '18 at 5:14

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