Proof of van Handel "Probability in High Dimension" Theorem 2.3

Question

I have a question about the proof of Theorem 2.3 in van Handel's Probability in High Dimension (page 15). At the bottom of page 15, the book writes $$E[E[f(X_1,\dots,X_n)\mid X_1,\dots,X_{k-1},X_{k+1},\dots,X_n]\mid X_1,\dots,X_{k-1}]=E[E[f(X_1,\dots,X_n)\mid X_1,\dots,X_{k-1},X_{k+1},\dots,X_n]\mid X_1,\dots,X_{k}]$$ where $X_1,\dots,X_n$ are independent. I was unable to understand why this is true. Thank you!

Answer 1

Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $X_{1},X_{2},\ldots,X_{n}$ be indpendent random variables. Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$ be a bounded Borel function (Note that boundedness can be relaxed. Here, we impose this condition so that conditional expectation makes sense). Fix an integer $k$ with $1<k<n$. Let $\mathcal{G}_{1}=\sigma(X_{1},X_{2},\ldots,X_{k-1},X_{k+1},\ldots,X_{n})$. Observe that $X_{1},X_{2},\ldots,X_{k-1},X_{k+1},\ldots,X_{n}$ are $\mathcal{G}_{1}$-measurable while $X_{k}$ and $\mathcal{G}_{1}$ are independent. Therefore, Independent Lemma is applicable. Define a Borel function $\varphi_{1}:\mathbb{R}^{n-1}\rightarrow\mathbb{R}$ by $\varphi_{1}(x_{1},\ldots,x_{k-1},x_{k+1},\ldots,x_{n})=E\left[f(x_{1},\ldots,x_{k-1},X_{k},x_{k+1},\ldots,x_{n})\right]$. By Independent Lemma, we have that $E\left[f(X_{1},\ldots,X_{n})\mid\mathcal{G}_{1}\right]=\varphi_{1}(X_{1},\ldots,X_{k-1},X_{k+1},\ldots,X_{n})$.

Let $\mathcal{G}_{2}=\sigma(X_{1},X_{2},\ldots,X_{k})$. Note that $X_{1},X_{2},\ldots,X_{k-1}$ are $\mathcal{G}_{2}$-measurable while $X_{k+1},X_{k+2},\ldots,X_{n}$ and $\mathcal{G}_{2}$ are independent. Define a Borel function $\varphi_{2}:\mathbb{R}^{k-1}\rightarrow\mathbb{R}$ by $\varphi_{2}(x_{1},x_{2},\ldots,x_{k-1})=E\left[\varphi_{1}(x_{1},\ldots,x_{k-1},X_{k+1},\ldots,X_{n})\right].$ By Independent Lemma, we have that $E\left[\varphi_{1}(X_{1},\ldots,X_{k-1},X_{k+1},\ldots,X_{n})\mid\mathcal{G}_{2}\right]=\varphi_{2}(X_{1},\ldots,X_{k-1})$.

Let $\mathcal{G}_{3}=\sigma(X_{1},X_{2},\ldots,X_{k-1})$. Clearly $\mathcal{G}_{3}\subseteq\mathcal{G}_{2}$. Denote $Y=f(X_{1},\ldots,X_{n})$. By tower property of conditional expectation, we have that \begin{eqnarray*} & & E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{3}\right]\\ & = & E\left[E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{2}\right]\mid\mathcal{G}_{3}\right]. \end{eqnarray*} However, \begin{eqnarray*} & & E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{2}\right]\\ & = & E\left[\varphi_{1}(X_{1},\ldots,X_{k-1},X_{k+1},\ldots,X_{n})\mid\mathcal{G}_{2}\right]\\ & = & \varphi_{2}(X_{1},\ldots,X_{k-1}), \end{eqnarray*} which is $\mathcal{G}_{3}$-measurable. Taking $\mathcal{G}_{3}$-conditional expectation on both sides yields $E\left[E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{2}\right]\mid\mathcal{G}_{3}\right]=\varphi_{2}(X_{1},\ldots,X_{k-1}).$

Hence, \begin{eqnarray*} & & E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{3}\right]\\ & = & E\left[E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{2}\right]\mid\mathcal{G}_{3}\right]\\ & = & \varphi_{2}(X_{1},\ldots,X_{k-1})\\ & = & E\left[E[Y\mid\mathcal{G}_{1}]\mid\mathcal{G}_{2}\right]. \end{eqnarray*}