Do the operators $\pi_{\mathcal F_n}X:=\operatorname E\left[X\mid\mathcal F_n\right]$ converge along a family $\mathcal F_n\supseteq\mathcal F_{n+1}$?

Question

Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space and $p\ge1$. We know that if $\mathcal F\subseteq\mathcal A$ is a $\sigma$-algebra on $\Omega$, then $$\pi_{\mathcal F}X:=\operatorname E\left[X\mid\mathcal F\right]\;\;\;\text{for }X\in\mathcal L^1(\operatorname P)$$ is a linear contraction from $\mathcal L^p(\operatorname P)$ to $$\mathcal L^p(\mathcal F,\operatorname P):=\{X\in\mathcal L^p(\operatorname P):X\text{ is }\mathcal F\text{-measurable}\}.$$ If $p=2$, then $\pi_{\mathcal F}$ is an orthogonal projection from $\mathcal L^2(\operatorname P)$ to $\mathcal L^2(\mathcal F,\operatorname P)$ and hence a self-adjoint operator on $\mathcal L^2(\operatorname P)$. By the tower property of conditional expectation, we obtain $$\pi_{\mathcal G}=\pi_{\mathcal F}\pi_{\mathcal G}=\pi_{\mathcal G}\pi_{\mathcal F}\tag2$$ for all $\sigma$-algebras $\mathcal F,\mathcal G\subseteq\mathcal A$ with $\mathcal G\subseteq\mathcal F$.

Now assume $\mathcal F_n\subseteq\mathcal A$ is a $\sigma$-algebra on $\Omega$ for $n\in\mathbb N$ with $$\forall n\in\mathbb N:\mathcal F_n\supseteq\mathcal F_{n+1}\tag1$$ and let $$\mathcal F_\infty:=\bigcap_{n\in\mathbb N}\mathcal F_n.$$

Are we able to show that $$\left\|\left(\pi_{\mathcal F_n}-\pi_{\mathcal F_\infty}\right)X\right\|_{L^2(\operatorname P)}\xrightarrow{n\to\infty}0\tag3\;\;\;\text{for all }X\in\mathcal L^2(\operatorname P),$$ which is to say that $\left(\pi_{\mathcal F_n}\right)_{n\in\mathbb N}$ converges to $\pi_{\mathcal F_\infty}$ in the strong operator topology on $\mathfrak L(\mathcal L^2(\operatorname P))$?

I'd also be interested in the question whether the convergence will even hold in the uniform operator topology and whether an analogue conclusion is possible for arbitrary $p\ne2$.

Answer 1

1. Let us show that $$\left\|\left(\pi_{\mathcal F_n}-\pi_{\mathcal F_\infty}\right)X\right\|_{L^2(\operatorname P)}\xrightarrow{n\to\infty}0 \;\;\;\text{for all }X\in\mathcal L^2(\operatorname P).$$ We can assume without loss of generality that $X$ is $\mathcal F_0$-measurable, because $X=X-\pi_{\mathcal F_0}X+\pi_{\mathcal F_0}X$ and $$\left(\pi_{\mathcal F_n}-\pi_{\mathcal F_\infty}\right)(X-\pi_{\mathcal F_0}X)=0.$$ Let $X_k=\pi_{\mathcal F_k}X-\pi_{\mathcal F_{k+1}}X$. Then $$ \left\lVert X-\pi_{\mathcal F_n}X\right\rVert_2^2=\left\lVert \sum_{k=0}^{n-1}X_k\right\rVert_2^2=\sum_{k=0}^{n-1}\left\lVert X_k\right\rVert_2^2, $$ where we used the fact that $\mathbb E\left[X_kX_j\right]=0$ if $j\neq k$. As $\left\lVert X-\pi_{\mathcal F_n}X\right\rVert_2^2\leqslant 4\lVert X\rVert_2^2$, we derive that $\sum_{k=0}^\infty\left\lVert X_k\right\rVert_2^2$ converges. From this, we readily deduce that the sequence $\left(\pi_{\mathcal F_n}X\right)_{n\geqslant 1}$ is Cauchy in $\mathbb L^2$, and converges therefore to a random variable $Y$ in $\mathbb L^2$. Moreover, the random variable $Y$ is necessarily $\mathcal F_\infty$-measurable, as it can be expressed as the limit of the sequence $\left(\pi_{\mathcal F_n}X\right)_{n\geqslant n_0}$, a sequence of $\mathcal F_{n_0}$-measurable functions. In order to see that $Y$ is indeed $\pi_{\mathcal F_\infty}X$, one can see that for each $A\in\mathcal F_\infty$, the quantity $\mathbb E\left[\left(\pi_{\mathcal F_n}X-Y\right)\mathbf{1}_A\right]$ converges to $0$ and is equal to $\mathbb E\left[\left(\pi_{\mathcal F_\infty}X-Y\right)\mathbf{1}_A\right]$.

2. The convergence also holds in $\mathbb L^p$ spaces, provided of course that $X\in\mathbb L^p$. For $p>2$, we use uniform integrability of $\left\{\lvert \pi_{\mathcal F_n}X-\pi_{\mathcal F_\infty}X\rvert^p,n\in\mathbb N\right\}$ combined with the convergence in probability (which comes from the convergence in $\mathbb L^2$). For $1\leqslant p<2$, we can use a truncation argument.

3. The convergence does not hold for the uniform operator topology. To see this, take a sequence $\left(a_j\right)_{j\in\mathbb Z}$ such that $\sum_{j\in\mathbb Z}a_j^2=1$, an i.i.d. sequence $(\varepsilon_j)_{j\in\mathbb Z }$ where $\varepsilon_j$ takes the values $1$ and $-1$ with equal probability. Let $\mathcal F_n$ be the $\sigma$-algebra generated by the random variables $\varepsilon_j$, $j\leqslant -n$ and let $X^{(k)}=\sum_{j\in\mathbb Z}a_{j+k}\varepsilon_j$. By Khintchine's inequality, $$ A_p\sum_{j\leqslant -n}a_{j+k}^2\leqslant\left\lVert X^{(k)}\right\rVert_p\leqslant B_p\sum_{j\leqslant -n}a_{j+k}^2$$ and $$A_p\sum_{j\leqslant -n}a_{j+k}^2\leqslant \left\lVert \left(\pi_{\mathcal F_n}-\pi_{\mathcal F_\infty}\right)X^{(k)}\right\rVert_p\leqslant B_p\sum_{j\leqslant -n}a_{j+k}^2.$$ Therefore, the operator norm of $\pi_{\mathcal F_n}-\pi_{\mathcal F_\infty}$ is bigger than $A_p/B_p$ and does not go to zero.