Prove this $\phi$ mixing inequality

Question

Am trying to understand Theorem 2.2 in Serfling (1968):

Proposition. Let $(\Omega,\mathcal A,P)$ be a probability space and $\mathcal F$ a sub-$\sigma$-algebra of $\mathcal A$. Let $X$ be a real random variable on $(\Omega,\mathcal A,P)$, and assume that $\|X\|_p<\infty$, where $1<p<\infty$. Then

$$ \|E[X|\mathcal F]-E[X]\|_p\leq 2\phi(\mathcal A,\mathcal F)^{1-\frac{1}{p}} \|X\|_p $$

where $\phi(\mathcal A,\mathcal F)=\sup_{A\in \mathcal A, F\in \mathcal F, P(F)>0} \bigg|P(A|F)-P(A)\bigg|$.

Proof. (Davidson (1994))

First assume that $X$ is simple with representation

$$X=\sum_{i=1}^n a_i 1_{A_i}$$

with $a_i\in\mathbb R$, $A_i\in\mathcal A$, $\cup_{i=1}^n A_i=\Omega$, $A_i$ pairwise disjoints. Let $q=\frac{p}{p-1}$. Then

$$\Big|E[X|\mathcal F]-E[X]\Big|^p=\Big|\sum_{i=1}^n a_i P[A_i|\mathcal F]-P[A_i]\Big|^p$$$$\leq \Big[\sum_{i=1}^n |a_i| \Big|P[A_i|\mathcal F]-P[A_i]\Big|\Big]^p$$

$$=\Big[\sum_{i=1}^n |a_i| \Big|P[A_i|\mathcal F]-P[A_i]\Big|^{1/p} \Big|P[A_i|\mathcal F]-P[A_i]\Big|^{1/q}\Big]^p$$

$$\leq \Big[\sum_{i=1}^n |a_i|^p \Big|P[A_i|\mathcal F]-P[A_i]\Big| \Big] \Big[\sum_{i=1}^n\Big|P[A_i|\mathcal F]-P[A_i]\Big|\Big]^{p/q}$$

$$\leq \Big[ E[|X|^p|\mathcal F]+E[|X|^p] \Big] \Big[\sum_{i=1}^n\Big|P[A_i|\mathcal F]-P[A_i]\Big|\Big]^{p/q}$$

$P$-almost surely, where the second inequality comes from Hölder's inequality.

Now Davidson writes: Let $A^+$ denote the union of all those $A_i$ for which $P[A_i|\mathcal F]-P[A_i]\geq 0$, and let $A^-=\Omega\setminus A^+$. Then

$$\sum_{i=1}^n\Big|P[A_i|\mathcal F]-P[A_i]\Big|=\Big|P[A^+|\mathcal F]-P[A^+]\Big|+\Big|P[A^-|\mathcal F]-P[A^-]\Big|$$

Why is this last equality true?

The next step is to claim

$$|P[A^+|\mathcal F]-P[A]|\leq \phi(\mathcal A,\mathcal F)$$

$$|P[A^-|\mathcal F]-P[A]|\leq \phi(\mathcal A,\mathcal F)$$

$P$-almost surely using the inequality from here.

Thanks a lot for your help.

---

For completeness I will try to finish the proof here. From the answers we have $$\sum_{i=1}^n\Big|P[A_i|\mathcal F]-P[A_i]\Big|\leq 2\phi (\mathcal A,\mathcal F) \quad\quad P\text{-almost surely}$$

Substituing back we get

$$\Big|E[X|\mathcal F]-E[X]\Big|^p\leq \Big[ E[|X|^p|\mathcal F]+E[|X|^p] \Big] \Big[2\phi (\mathcal A,\mathcal F)\Big]^{p/q} \quad\quad P\text{-almost surely} $$

Integrating both sides gives

$$E\Big[\Big|E[X|\mathcal F]-E[X]\Big|^p\Big]\leq 2E[|X|^p] \Big[2\phi (\mathcal A,\mathcal F)\Big]^{p/q} $$

Raising both sides to the power $1/p$ we obtain

$$ \|E[X|\mathcal F]-E[X]\|_p\leq 2\phi(\mathcal A,\mathcal F)^{1-\frac{1}{p}} \|X\|_p $$

Now assume $X$ is an arbitrary $\mathcal A$ measurable real random variable with $\|X\|_p<\infty$. Then there exists a sequence $(X_n)$ of simple $\mathcal A$ measurable random variables such that $|X_n|\leq|X_{n+1}|\leq |X|$ for each $n$ and $X_n\to X$ pointwise.

From the DCT using the dominating functions $|X|$, $|X|^p$ we get

$$ E[X_n]\to E[X]$$ $$E[|X_n|^p]\to E[|X|^p]$$

From the DCT for conditional expectations using the dominating functions $|X|$, $|X|^p$ we also have

$$ E[X_n|\mathcal F]\to E[X| \mathcal F] \quad \quad P\text{-almost surely}$$

$$ E[|X_n|^p|\mathcal F]\to E[|X|^p| \mathcal F] \quad \quad P\text{-almost surely}$$

From the first part we have

$$\Big|E[X_n|\mathcal F]-E[X_n]\Big|^p\leq \Big[ E[|X_n|^p|\mathcal F]+E[|X_n|^p] \Big] \Big[2\phi (\mathcal A,\mathcal F)\Big]^{p/q} \quad\quad P\text{-almost surely} $$

for each $n$. Since a countable union of null sets is null, we can combine all these statements and pass to the limit to obtain

$$\Big|E[X|\mathcal F]-E[X]\Big|^p\leq \Big[ E[|X|^p|\mathcal F]+E[|X|^p] \Big] \Big[2\phi (\mathcal A,\mathcal F)\Big]^{p/q} \quad\quad P\text{-almost surely} $$

Taking expectations on both sides and raising to the power $1/p$ as before we get

$$\|E[X|\mathcal F]-E[X]\|_p\leq 2\phi(\mathcal A,\mathcal F)^{1-\frac{1}{p}} \|X\|_p$$

Answer 1

Regarding Davidson's proof. Let $\kappa(\cdot,\cdot)$ denote the r.c.p. w.r.t. $\mathcal{F}$, and let $$ \nu(A,\omega):=\kappa(A,\omega)-\mathsf{P}(A). $$

Then for each $\omega\in \Omega$, $$ \sum_{i=1}|\nu(A_i,\omega)|=\nu(A^+(\omega),\omega)-\nu(A^-(\omega),\omega), $$ and $$ |\nu(A^+(\omega),\omega)|\vee |\nu(A^-(\omega),\omega)| \le \max_{A\in \mathcal{A}}|\nu(A,\omega)|, $$ where $\mathcal{A}=\{\bigcup_{i\in \mathcal{I}} A_i:\mathcal{I}\subset \{1,\ldots, n\}\}$. Now you can bound the rhs by the $\phi$-mixing coefficient.

Answer 2

Choose a version of $P[\cup_{i\in \mathcal I}A_i|\mathcal F]$ for each $\mathcal I\subset {\{1,\dots,n}\}$. Put $$Z(\omega)=\max_{\mathcal I\subset {\{1,\dots,n}\}}\Bigg|P[\cup_{i\in \mathcal I}A_i|\mathcal F](\omega)-P[\cup_{i\in \mathcal I}A_i]\Bigg|$$

for each $\omega\in\Omega$. For a fixed subset $\mathcal I\subset {\{1,\dots,n}\}$, using the disjointness of the $A_i$, we have

$$ P[\cup_{i\in \mathcal I}A_i|\mathcal F]=\sum_{i\in\mathcal I}P[A_i|\mathcal F]\quad\quad P\text{-almost surely.}$$

Since there are only finitely many subsets $\mathcal I\subset {\{1,\dots,n}\}$, taking a finite union of null sets gives a set $N$ with $P(N)=0$ such that

$$ P[\cup_{i\in \mathcal I}A_i|\mathcal F](\omega)=\sum_{i\in\mathcal I}P[A_i|\mathcal F](\omega) \quad \quad (1)$$

for all $\mathcal I\subset {\{1,\dots,n}\}$ and all $\omega\in\Omega\setminus N$.

Now fix some $\omega\in\Omega\setminus N$.

Let $\mathcal I^+_\omega$ denote the set of all $i\in \{1,\dots,n\}$ such that $P[A_i|\mathcal F](\omega)-P[A_i]\geq 0$, and let $\mathcal I^-_\omega$ denote the set of all $i\in \{1,\dots,n\}$ such that $P[A_i|\mathcal F](\omega)-P[A_i]< 0$.

Then using $(1)$ we have

$$\sum_{i=1}^n\Big|P[A_i|\mathcal F](\omega)-P[A_i]\Big|=\sum_{i\in \mathcal I^+_\omega}\Big[P[A_i|\mathcal F](\omega)-P[A_i]\Big]-\sum_{i\in\mathcal I^-_\omega}\Big[P[A_i|\mathcal F](\omega)-P[A_i]\Big]$$

$$=\Big[P[\cup_{i\in \mathcal I^+_\omega}A_i|\mathcal F](\omega)-P[\cup_{i\in \mathcal I^+_\omega}A_i]\Big]-\Big[P[\cup_{i\in \mathcal I^-_\omega}A_i|\mathcal F](\omega)-P[\cup_{i\in \mathcal I^-_\omega}A_i]\Big]$$

$$\leq\Big|P[\cup_{i\in \mathcal I^+_\omega}A_i|\mathcal F](\omega)-P[\cup_{i\in \mathcal I^+_\omega}A_i]\Big|+\Big|P[\cup_{i\in \mathcal I^-_\omega}A_i|\mathcal F](\omega)-P[\cup_{i\in \mathcal I^-_\omega}A_i]\Big|$$

$$\leq 2Z(\omega)$$

As this holds for all $\omega\in\Omega\setminus N$ we have shown that

$$\sum_{i=1}^n\Big|P[A_i|\mathcal F]-P[A_i]\Big|\leq 2Z \quad\quad P\text{-almost surely} \quad\quad (2)$$

Now, the inequality from here implies that

$$\Bigg|P[\cup_{i\in \mathcal I}A_i|\mathcal F]-P[\cup_{i\in \mathcal I}A_i]\Bigg| \leq \phi (\mathcal A,\mathcal F)\quad\quad P\text{-almost surely,} \quad \text{ for each } \mathcal I\subset \{1,\dots,n\}$$

and since there are only finitely many subsets $\mathcal I\subset \{1,\dots,n\}$ we can take a finite union of null sets to obtain

$$\Bigg|P[\cup_{i\in \mathcal I}A_i|\mathcal F]-P[\cup_{i\in \mathcal I}A_i]\Bigg| \leq \phi (\mathcal A,\mathcal F)\quad\quad \text{ for each } \mathcal I\subset \{1,\dots,n\},\quad P\text{-almost surely}$$

which implies $Z\leq \phi (\mathcal A,\mathcal F)$ $P$-almost surely. Combined with $(2)$ this gives

$$\sum_{i=1}^n\Big|P[A_i|\mathcal F]-P[A_i]\Big|\leq 2\phi (\mathcal A,\mathcal F) \quad\quad P\text{-almost surely} $$