Enlarging a probability space to get independence

Question

Am reading Lemma 5.2 from this paper by Dvoretzky:

Let $X$ be a real random variable on a probability space $(\Omega,\mathcal A,P)$ satisfying $|X|\leq 1$, let $\mathcal F=\sigma(X)$, and let $\mathcal G$ be a sub-$\sigma$-algebra of $\mathcal A$. Then

$$E\Big[\big|E[X|\mathcal G]-E[X]\big|\Big]\leq 4\alpha(\mathcal F,\mathcal G) \quad\quad\quad (1)$$

where $\alpha(\mathcal F,\mathcal G)=\sup_{F\in\mathcal F, G\in\mathcal G} \big|P(F\cap G)-P(F)P(G)\big|$.

At some point in the proof the author writes: "Let $\tilde{X}$ be a random variable with the same distribution as $X$ and independent of $\mathcal G$ (if need be the probability space can be enlarged in order to carry such a random variable)."

I don't see how to achieve this result, and how to do it without changing the quantities that appear in $(1)$. Any ideas on how to proceed?

Thanks a lot for your help.

EDIT: I tried to write down the details below. Any comment is very appreciated.

Answer 1

Let $Y$ be a random variable on a probaility space $(\Omega',\mathcal A',P')$ having the same distribution as $X$. On $\Omega \times \Omega'$ with the product $\sigma-$ field and the measure $P\times P'$ define $\tilde{X} (\omega,\omega')=Y(\omega')$. Then $\tilde{X}$ has the same distribution as $X$ and it is independent of $\mathcal G$.

Answer 2

Consider the product space

$$(\Omega',\mathcal A',P'):=(\Omega\times\mathbb R,\mathcal A \otimes \mathcal{B}(\mathbb{R}),P\times P_X),$$

where $P_X$ denotes the distribution of $X$. Make the following definitions:

$$X':\Omega'\to \mathbb R, \quad X'(\omega,x):=X(\omega) $$

$$\tilde{X}:\Omega'\to \mathbb R, \quad \tilde{X}(\omega,x):=x$$

$$\mathcal G':=\{G\times\mathbb R: G\in \mathcal G\}$$

$$\mathcal F':=\{F\times\mathbb R: F\in \mathcal F\}$$

Note that $X',\tilde{X}$ are measurable real random variables on $(\Omega',\mathcal A',P')$ with common distribution $P_X$, that $\mathcal G',\mathcal F'$ are sub-$\sigma$-algebras of $\mathcal A'$, and that $$\mathcal F'=\sigma\{X^{-1}(B)\times \mathbb R:B\in\mathcal B(\mathbb R)\}=\sigma(X')$$ Note also that the definition of product measure implies that $\tilde{X}$ is independent from the sub-$\sigma$-algebra $\{A\times \mathbb R:A\in \mathcal A\}$ of $\mathcal A'$. In particular $\tilde{X}$ is independent from $\mathcal G'$.

Since $X$ is $P$-integrable and $X',\tilde{X}$ have the same distribution as $X$, they are $P'$-integrable and

$$\int X' \,dP'=\int \tilde{X} \,dP'=\int X \,dP$$

I claim that $$E[X'|\mathcal G'](\omega,x)= E[X|\mathcal G](\omega) \quad \quad P'-a.s.$$

Note that the RHS is $\mathcal G'$ measurable since $E[X|\mathcal G]$ is $\mathcal G$ measurable. Let $G\in \mathcal G$. Then we have

$$\int_{G\times \mathbb{R}} X' \, dP'= \int_G X \,dP= \int_G E[X|\mathcal G]\,dP =\int_{G\times \mathbb{R}} E[X|\mathcal G]\circ p_1 \, dP' $$

where $p_1:\Omega\times \mathbb R\to \Omega$ denotes the projection map. As this holds for all $G\in \mathcal G$, we conclude that indeed $E[X'|\mathcal G']= E[X|\mathcal G]\circ p_1$, $P'$-almost surely.

It follows that any version of $E[X'|\mathcal G']$ will have the same distribution as any version of $E[X|\mathcal G]$. This implies

$$E\Big[\big|E[X'|\mathcal G']-E[X']\big|\Big]=\int |x-c|\,dP_{E[X'|\mathcal G']}=\int |x-c|\,dP_{E[X|\mathcal G]}=E\Big[\big|E[X|\mathcal G]-E[X]\big|\Big]$$ where $c=E[X']=E[X]$.

We conclude that $(1)$ is equivalent to

$$E\Big[\big|E[X'|\mathcal G']-E[X']\big|\Big]\leq 4\alpha(\mathcal F',\mathcal G') \quad\quad\quad (1')$$

where $\alpha(\mathcal F',\mathcal G')=\sup_{F'\in\mathcal F', G'\in\mathcal G'} \big|P'(F'\cap G')-P'(F')P'(G')\big|$.